[Study Package - Chemistry - Part- I]

CHEMISTRY
STUDY PACKAGE
PART - I
A COMPLETE TEXTBOOK FOR CLASS - XI

Brilliant
STUDY CENTRE
PALA
Mutholy Campus, Ph: 04822 - 206100, 206800
Arunapuram Campus, Ph: 04822 - 212415, 210949, 216975
Ernakulam - Ph: 0484 - 2665080, 2665090

www.brilliantpala.org., email: brilliantstudycentre@gmail.com

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CONTENTS

1. Some Basic Concepts of Chemistry--------------------------------------------------------------05

2. Structure of Atoms ----------------------------------------------------------------------------------------31

3. Classification of Elements and Periodicity in Properties---------------------------64

4. Chemical Bonding and Molecular Structure ------------------------------------------------89

5. States of Matter ------------------------------------------------------------------------------------------ 124

6. Thermodynamics ----------------------------------------------------------------------------------------- 155

7. Chemical and Ionic Equilibrium ----------------------------------------------------------------- 193

8. Redox Reactions ------------------------------------------------------------------------------------------ 243

9. Hydrogen ------------------------------------------------------------------------------------------------------ 270

10. The s-Block Elements --------------------------------------------------------------------------------- 300

11. The p-Block Elements --------------------------------------------------------------------------------- 326

12. Organic Chemistry - Some Basic Principles -Part I

(Nomenclature) -------------------------------------------------------------------------------------------- 355

13. Organic Chemistry - Some Basic Principles -Part II

( Isomerism and Reaction Mechanism)------------------------------------------------------ 379

14. Organic Chemistry - Some Basic Principles -Part III

(Purification and Characterisation of Organic Compounds) -------------------- 408

15. Hydrocarbons ----------------------------------------------------------------------------------------------- 430

16. Environmental Chemistry--------------------------------------------------------------------------- 479

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CHAPTER - 01
SOME BASIC CONCEPTS OF CHEMISTRY

INTRODUCTION
Importance of Chemistry. Chemistry is the branch of science that deals with the composition, properties,
structure and interactions of matter. Chemical principles help in designing and synthesising new materials
having specific properties, that can improve the quality of life of man.
Cooking, fermentation, glass-making, and metallurgy are all chemical processes that date from the beginning
of civilisation. Today, vinyl, teflon, liquid crystals, semiconductors, superconductors, optical fibre, ceramic
materials, fullerenes, carbon nanotubes, etc represent the fruits of chemical technology. Chemical industries
manufacturing acids, alkalies, soaps, detergents, fertilizers, drugs, insecticides, metals, alloys, polymers and
other chemicals and materials contribute in a big way to the national economy.
Chemistry plays a crucial role in meeting the demand for food materials, health care products and life saving
drugs. Drugs like taxol, cisplatin, etc., used in cancer therapy and azidothymidine (AZT) used in the treatment
of AIDS are all synthetic materials.
In recent years, chemistry has tackled some of the pressing aspects of environmental degradation. Safer
alternatives for hazardous refrigerants like chlorofluorocarbons(CFCs), responsible for ozone depletion, have
been successfully synthesised. Management of greenhouse gases, devising safer nonpolluting materials and
fuels are some of the challenges before the chemist.
1.1 NATURE OF MATTER
Matter is anything that has mass and occupies space. It exists in three physical states- solid, liquid, and gas. In
solids, the constituent particles are held very close to each other in an orderly manner with little freedom of
movement. In liquids, particles are close to each other but they can move around. In gases, particles are far
apart and their movements are easy and fast.
Solids have definite volume and definite shape. Liquids have definite volume but no definite shape. They take
the shape of the container in which they are taken. Gases have neither definite volume nor definite shape. They
completely occupy the container in which they are kept. These three states are interconvertible on changing
the conditions of temperature and pressure.
Heat Heat
Solid Liquid Gas.
Cool Cool
Phases of Matter. The six known phases of matter are solids, liquids, gases, plasma, Bose-Einstein
condensates (BEC) and fermionic condensate.

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1.1.1 Classification of Matter

(i) Mixture. A mixture contains two or more substances, present in any ratio, e.g., aqueous sugar solution,
air, tea, etc. Components of a mixture can be separated by physical methods such as mechanical separation,
hand picking, filtration, crystallisation, distillation, etc.

Homogeneous mixture has uniform composition throughout, e.g., salt solution, air, etc.

Heterogeneous mixture has non-uniform composition, e.g., salt and sugar, grains and pulses, smoke, etc.

(ii) Pure substances. Pure substances have fixed composition. Their constituents cannot be separated by
physical methods. Pure substances are further classsified into elements and compouds.

Element consists of only one type of particles (atoms or molecules). Atoms of different elements are different
in nature, e.g., H, O, He, Cu, etc. Two or more atoms combine to give molecules, e.g., H2, O2, N2, etc.

Compound is formed when atoms of different elements combine in a definite ratio, e.g., H2O, CO2, NaCl,
etc. The properties of a compound are different from those of its constituent elements.

1.2 PROPERTIES OF MATTER AND THEIR MEASUREMENT

(i) Physical properties can be observed/measured without changing the identity/composition of a substance.

(ii) Chemical properties are the characteristic reactions of different substances.

1.2.1 The International System of Units (SI system). The SI system is the common standard system of
measurement (11th General Conference of Weights and Measures,1960). The system has seven base units.

Table - 01 Seven basic physical quantities and their SI units.

Physical Quantity Symbol SI Unit Symbol

1. Length l metre m
2. Mass m kilogram kg
3. Time t second s
4. Electric current I ampere A
5. Thermodynamic temperature T kelvin K
6. Amount of substance n mole mol
7. Luminous intensity Iv candela cd

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Table - 02 Some common derived units.

Unit/Symbol in terms Unit/Symbol in terms of SI Basic
Physical Quantity Physical Quantity
of SI Basic Units Units
2 -2 -1 -2
1. Area m 9. Pressure Pascal (Pa) or N m = kg m s
3 -2
2. Volume m 10. Focre Newton (N) kg m s
-1
3. Amount of substance moles (mol) 11. Frequency Hertz (Hz) s
-3 -1 -1 -1 2 -3 -1
4. Density kg m 12. Potential difference Volt (V)JC = J A s = kg m s A
-1 2 -3
5. Velocity ms 13. Power Watt (W) kg m s
-1 2 -2
6. Speed ms 14. Work/Energy Joule (J) N m = kg m s
-2 -1
7. Acceleration ms 15. Electrical resistance ohm (W)V A
8. Electric charge Coulomb (C) = A s 16. Electrical conductance ohm-1 (W -1 ) A V-1

Temperature Scales. Temperature is the degree of hotness. The three scales of temperature are Celsius
(ºC), Farenheit (ºF), and Kelvin (K).
Conversion of Celsius (ºC) to Kelvin (K) : Kelvin (K) = ºCelsius + 273.15  C  273 .
Table - 03 Prefixes used in SI System.

Name deca- hecto- kilo- mega- giga- tera- peta exa- zeta- yotta-
Symbol da h k M G T P E Z Y
1 2 3 6 9 12 15 18 21 24
Factor 10 10 10 10 10 10 10 10 10 10
Name deci- centi- milli- micro- nano- pico- femto- atto- zepto- yocto-
Symbol d c m m n p f a z y
-1 -2 -3 -6 -9 -12 -15 -18 -21 -24
Factor 10 10 10 10 10 10 10 10 10 10

1.3 UNCERTAINTY IN MEASUREMENT

Significant figures. The number of significant figures in a measurement is the number of figures that are
known with certainity plus one that is uncertain, beginning with the first non zero digit.
Rules for determining significant figures. i. All non-zero digits are significant : 125 cm has three significant
digits, 0.125 cm has 3 significant digits, 2001 has 4 significant digits, and 2.05 has 3 significant digits.
ii. Zeros to the left of the first non-zero digit in the number are not significant : 0.005 has only 1 significant digit,
whereas 0.025 has 2 significant digits.
iii. Zeros between non-zero digits are significant : 2.05 has 3 significant digits.
iv. Zeros to the right of the decimal point are significant : 2.00, 0.020, and 0.2000 have 3, 2 and 4 significant
digits respectively.
v. If a number ends in zero that are not to the right of a decimal, the zeros may or may not be significant :
1200 has 2, 3 or 4 significant digits.
1.3.1 Scientific Notation. The number is written in the standard exponential form, N  10 n where N is a number
with a single non-zero digit to the left of the decimal point and n is an integer, e.g., 1200 g can be expressed
as : (i) 1.2×103g (2 significant digits), (ii)1.20×103g (3 significant digits), etc.
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In these expressions, all zeros to the right of the decimal point are significant.
Avogadro’s number is expressed as 6.022×1023 mol–1 and Planck’s constant as 6.62×10–34 Js.
Exact integral numbers do not have any uncertainity associated with them and as such, those numbers have
infinite number of significant figures.
Calculation involving significant figures. The accuracy of a result is limited to the least accurate
measurement or the final result cannot be more accurate than the least accurate number involved in
the calculation.
1. The result of an addition or substraction should be reported to the same number of decimal places as that
of the term with least number of decimal digits, e.g., sum of 5.245 + 4.28 should be reported as 9.52.
2. The result of multiplication or division should be reported to the same number of significant figures as is
possessed by the least precise term, e.g., 4.3 × 1.125 = 4.8
3.425×3.8 = 13.015 should be reported as 13
3. If the calculation involves a number of steps, the result should contain the same number of significant figures

as that of the least precise number involved, other than the exact numbers, e.g., 5.28  0.146  3
0.0419
= 55.19427 should be reported as 55.19.
Rounding off. i. If the digit next to the last digit to be retained is less than 5, the last digit is taken as such and
all other digits to its right are dropped, e.g. 2.214 = 2.21.
ii. If the digit next to the last digit to be retained is greater than 5, the digit to be retained is increased by 1 and
all other digits on the right are dropped, e.g. 4.2368 = 4.24.
iii. If the digit next to the last digit to be retained is equal to 5, the last significant figure is left unchanged if even
and increased by 1 if odd, e.g. 2.235 = 2.24 and 4.225 = 4.22.
1.3.2 Precision and Accuracy. Precision refers to the closeness of two or more measurements of the same
quantity. Accuracy refers to the closeness of an experimental (observed) value to the actual (true) value.
Precision is independent of accuracy.
These aspects depend on the accuracy of the measuring device and the skill of the operator. For example, let
the true value for a result be 2.00. Student A makes two mesurements and reports 1.95g and 1.94 g, student
B reports 1.94 g and 2.05 g, and student C reports 2.01 g and 1.99 g as the results.
i. Values obtained by student A are precise (close to each other) but not accurate (not close to the true value).
ii. Values reported by B are neither precise nor accurate.
iii. Results obtained by C are precise and accurate.
1.3.3 Dimensional Analysis. The method used to convert units from one system to another is called factor label

method or unit factor method or dimensional analysis. For example, inch is converted to cm as:
 1 in   2.54 cm 
  1   [Both of these are called unit factors]
 2.54cm   1 in 
(i) To convert 15 cm to inch, 15 cm is multiplied by the unit factor :
 1in 
15 cm  15 cm     5.90 in
 2.54 cm 
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In these multiplications, the numerator should have that part which is required in the desired result.
 1m   1000 mm 
(ii) 25 cm is converted to mm (1 cm = 10 mm) as :  25.0 cm   100 cm   1m   250 mm
  

 12.5 g   1kg  1000 cm

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(iii) To convert 12.5 g cm to kilograms per litre : 

–3
3     12.5 kg L-1
 cm   1000 g  L

1.4 LAWS OF CHEMICAL COMBINATION

The combination of elements to form compounds is governed by five basic laws :
i. Law of conservation of mass, ii. Law of constant composition or law of definite proportions
ii. Law of multiple proportions, iv. Law of equivalent, reciprocal, or combining proportions
v. Law of combining volumes (Gay Lussac’s law of gaseous volumes).
The first four laws are related with mass, whereas the fifth law is related with the volume of reacting gases.
1.4.1 Law of Conservation of Mass. This law states that matter can neither be created nor destroyed (French
chemist, Antonie Lavoisier, 1789).
If masses m and n of substances A and B react to form masses x and y of substances C and D, then,
m + n = x + y.
It follows that the total mass of reactants is equal to the total mass of products.
In the formation of water, g of H2 reacts with 32g of O2 to give 36g of product.

1.4.2 Law of Constant Composition or Definite Proportions. A compound, irrespective of the method by
which it is formed will always contain the same elements combined in the same proportion by mass
(French chemist, Joseph Proust - 1799).
The law is true only for compounds obtained from one isotope. For example, when C12 forms CO2, the ratio
of masses is 12 : 32 (3 : 8), but for C13 and C14 the ratios are 13:32 and 14 : 32 (7 : 16) respectively.
1.4.3 Law of Multiple Proportions. When two elements combine to form more than one compound, the
different masses of one element combining with a fixed mass of the other will bear a simple whole
number ratio (British scientist, John Dalton,1803).
For example, sulphur combines with oxygen to form SO2 and SO3. The weights of oxygen that combine with
32 parts (atomic weight) of sulphur are in the ratio of 32 : 48 (2 : 3).

Similarly carbon combines with oxygen to form CO and CO2. The weights of oxygen that combine with 12
parts by weight of carbon in the two compounds are in the respective ratio of 16 : 32 (1 : 2).
1.4.4 Law of Reciprocal, Equivalent, or Combining Proportions. The proportions in which two elements
separately combine with the same weight of a third element are also the proportions in which the first
two elements combine together (German chemist, Jeremias Richter, 1792). If two elements A and B combine

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with a fixed weight of a third element C, then A and B will combine with one another in the same ratio or a
multiple or sub-multiple of the ratio in which they combine with C.
For example, 2 parts by weight of hydrogen reacts with 16 parts of oxygen and 32 parts of sulphur to form
H2O and H2S respectively. Therefore, when sulphur and oxygen react to form products, they will do so in the
ratio 16 : 32 (1 : 2) or a multiple or a sub-multiple of this ratio.
In SO2, 32 parts of sulphur combine with 32 parts of oxygen, therefore the ratio is 32 : 32 (1 : 1). In SO3, 32
parts of S combines with 48 parts of oxygen and thus the ratio 32 : 48 (2 : 3).

The weights of elements that combine with a constant weight of a standard substance (hydrogen, oxygen, or
chlorine) are called their equivalent or combining weights. According to the law, substances react together
in the ratio of their combining or equivalent weights or their multiples or submultiples.
The reverse of the law of reciprocal proportions is thus the law of equivalent proportions.
1.4.5 Gay Lussac’s Law of Gaseous Volumes. When gases react, they do so in volumes which bear a simple
ratio to one another and to the volumes of resultant substances in the gaseous state, all volumes being
measured at the same temperature and pressure (French scientist, Joseph Louis Gay Lussac,1808).
For example, 1 volume of H2 combines with 1 volume of Cl2 to form 2 volumes of HCl gas. The ratio, by
volume, viz, 1 : 1 : 2 is simple and in accordance with the law.
H 2(g)  Cl 2(g)  2HCl(g)
1volume 1volume 2 volumes
Ratio 1 : 1 : 2
Similarly, 2 volumes of carbon monoxide combines with 1 volume of oxygen to form 2 volumes of carbon
dioxide. Their ratio by volumes, 2 : 1 : 2 is also simple multiple.

2CO(g)  O2(g)  2CO2(g)

2volumes 1volume 2volumes
Ratio 2 : 1 : 2

Gay Lussac’s law can be considered as law of definite proportions by volume.

1.4.6 Avogadro’s Law. Italian scientist, Amedeo Avogadro (1811) stated that equal volumes of all gases under
identical conditions of temperature and pressure contain identical number of molecules. Avogadro’s
law helps in determining atomicity of gases and in relating molecular weight and vapour density.
Density of gas
Vapour density = Density of H
2

 Molecular weight = 2 × Vapour density (relating molecular weight and volume)

Weight of 1 gram mole = 1 molecular weight = 22.4 litres at NTP
 Gram molecular volume = 22.4 litres at NTP (finding molecular formulae of gases)

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1.5 DALTON’S ATOMIC THEORY

In 1808, Dalton published ‘A New System of Chemical Philosophy’ in which he proposed the following :
1. Matter consists of very small indivisible particles called atoms.
2. All atoms of the same element have identical properties including size, mass, and shape. Atoms of different
elements have different masses and different chemical properties.
3. When atoms of the same or different elements combine together to form compounds, they do so in fixed
ratios. Atoms of two elements may combine in different ratios to form more than one compound.
4. In a chemical reaction, atoms neither created nor destroyed but they get reorganised. An atom is the
smallest particle which takes part in a chemical reaction.

Merits of Dalton’s Theory. Dalton’s theory could explain the laws of chemical combination.
Limitations of Dalton’s Atomic Theory. Dalton’s atomic theory could explain the following :
1. Gay Lussac’s law of combining volumes.
2. Atoms of same or different elements combine to form molecules.
3. Atoms of different elements differ from each other in mass, size, valency, etc.
4. The nature of binding force between the atoms of molecules responsible for the existence of matter in solid,
liquid and gaseous states.
5. It could not make any distinction between the ultimate particle of an element or compound.
1.6 ATOMIC AND MOLECULAR MASSES
Atom is the smallest unit of matter that has all the characteristic properties of an element.
1.6.1 Atomic Mass. The mass of atom or atomic mass is very small since atoms are extremely small. The atomic
masses of elements were determined by comparing their masses with the mass of hydrogen (1 without units).
The present system of atomic masses is based on carbon-12 as the standard.
Atomic Mass Unit. Atomic mass unit is defined as mass equal to 1/12th the mass of a carbon-12 atom.
Mass of 1 carbon-12 atom = 1.9924 × 10–23g
-23
1 1.9924  10
1 amu = 1u = th of mass of carbon-12 = g
12 12
1 amu = 1u = 1.66056 × 10–24g = 1.66056 × 10–27 kg
1.67356  10-24
Mass of hydrogen atom, 1H = = 1.0078 amu = 1.0080 amu.
1.66056  10-24
Mass of oxygen - 16, (16O) = 15.995 amu.
Presently, amu has been replaced by u, known as unified mass. amu is also called Dalton (Da).
i. Relative Atomic Mass. 12C is assigned a mass of exactly 12 atomic mass units (amu) and masses of all
other atoms are given relative to this standard.
Mass of one atom of element Mass of 1atom of element
Re lative atomic mass  
1 /12 of the mass of 12 C atom 1 amu
ii. Average Atomic Mass. Various naturally occuring elements have more than one isotope. Average atomic
mass of an element is determined by taking into account the existence of isotopes and their relative abundance.

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pi A i  (Isotopic molar mass  Percentage abundance)

Average atomic mass, A= =
100 100
Isotope of Carbon Re lative abundance Fractional abundance Atomic mass (amu)
12
C 98.892 0.98892 12
13
C 1.108 0.0118 13.00335
14 -10 -12
C 2 × 10 2 × 10 14.00317
Alternatively, the average relative atomic mass of an element may be calculated from the fractional
abundances of the isotopes of that element.
Average atomic mass, A  f i Ai  f1A1  f 2 A 2  ......
where (fi) is the fractional abundance of the isotope of mass numbers (Ai).
Average atomic mass of carbon = (0.98892)(12u) + (0.01108)(13.00335u) + (2 × 10–12)(14.00317u)
= 12.011u
Atomic masses of elements mentioned in the periodic table are the average atomic masses.
iii. Gram Atomic Mass. The atomic mass of an element expressed in grams is called gram atomic mass. It
is also called gram atom. It is the mass of 1 mole of atoms.
Gram atom = mass of atom in amu ×1.66 × 10–24 × Avogadro’s number
Gram atomic mass or one gram atom of oxygen = (16 × 1.66 × 10–24 × 6.022 × 1023g) = 16g
Atomic mass of sodium = 23 amu
Gram atomic mass or one gram atom of sodium = 23g
1.6.2 Molecular mass. Molecular mass (relative molecular mass) is the sum of atomic masses of all atoms of all
elements present in a molecule.
Molecular mass of CO2 = (12.011u) + 2(16.00u) = 44.011u
Molecular mass of propane (C3H8) = 3(12.011u) + 8(1.008u) = 44.097u
Molecular mass of sucrose (C12H22O11) = 12(12.011u) + 22(1.008u) + 11(16.00u) = 342.308u
i. Gram Molecular Mass. It is the molecular mass expressed in grams (gram-molecule of a substance).
Gram Molecular Mass = Mass of 1 gram molecule = Mass of 1 mole (NA) of molecules
Molecular mass of O2 = 32 amu = 32 × 1.67 × 10–24 g
Mass of NA molecules of O2 = 32 × 1.67 × 10 –24 g × 6.022 × 1023 = 32 g
Thus gram molecular mass of a substance is the mass in grams of 6.022 × 1023 molecules.
ii. Formula Mass and Gram Formula Mass. In ionic compounds like NaCl, KCl, etc., each ion is surrounded
by a number of oppositely charged ions. In such cases, formula mass is calculated instead of molecular mass.
It is calculated by adding up the atomic masses of the atoms present in one formula unit.
Formula mass of NaCl = Atomic mass of Na + Atomic mass of Cl = 23.0u + 35.5u = 58.5u
Formula mass of a substance expressed in grams is called gram formula mass.
1.6.3 Equivalent Mass. The equivalent mass is the number of grams of an element which will combine with or
replace 1g hydrogen, 8g of oxygen or 35.5g of chlorine (g equivalent –1 or g eq–1).
Atomic mass
(i) Equivalent mass of element =
Valency

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Ionic mass
(ii) Equivalent mass of ion =
Charge

Molecular mass
(iii) Equivalent mass of compound =
Charge on cation or anion

Molecular mass
(iv) Equivalent mass of acid =
Basicity
Basicity is the number of displaceable H+ ions present in one molecule of the acid. Basicity is 1 for HCl and
HNO3, 2 for H2SO4 and H3PO3 and 3 for H3PO4.
Molecular mass
(v) Equivalent mass of base =
Acidity
Acidity is the number of displaceable OH– ions present in one molecule of the base. Acidity is 1 for NaOH
and KOH, 2 for Ca(OH)2 and Mg(OH)2 and 3 for Al(OH)3.
Equivalent mass is not a fixed quantity; it varies from reaction to reaction.
Molecular mass
(vi) Equivalent mass of oxidising/reducing agent =
Number of electrons involved

1.7 MOLE CONCEPT

In SI system, mole (mol) is the base quantity for the amount of a substance. One mole is the amount of
substance that contains as many particles as there are atoms in exactly 12 g of the 12C isotope.
One gram atom of any element contains the same number of atoms and one gram molecule of any substance
contains the same number of molecules. This is equal to 6.022137×1023 correct upto 7 significant figures.
The value 6.022×1023 and is called Avogadro number/constant, denoted by NA.
The amount of substance containing Avogadro’s number of atoms or molecules is called a mole.
Number of molecules (or atoms) Mass of substance (in g)
mole  
NA Gram atomic / m olecular mass of subs tan ce
Volume of gas at STP(in L)
mole 
22.414L
Calculation of Avogadro’s number or constant. To determine the number precisely, gram atomic mass of
carbon-12 (12 g mol-1) is divided by the mass of a single carbon-12 atom (1.992648 × 10–23g).

12g mol-1of 12 C
Number of carbon atoms in one mole 
1.992648 10-23 g  12 C atom 
-1

 6.0221367 1023 atoms mol-1 = 6.022 × 1023 atoms mol–1 (rounding off to 4 significant figures)
1 mol (1 g atom) of C atoms = 6.022 × 1023 C atoms.
1 mol(1 g molecule) of H2O molecules = 6.022 × 1023 H2O molecules.
1 mole (1 g molecule) of NaCl = 6.022 × 1023 formula units of NaCl.
1. Molar Mass. The mass of one mole of a substance in grams is called its molar mass. Molar mass in
grams is equal to atomic or molecular formula mass in u. Units of molar mass are g mol–1 or kg mol–1.
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Molar mass of CO2 = 12.011 + 2(16.0) = 44.011g mol–1

Molar mass of NaCl = 23.0 + 35.5 = 58.5 g mol–1
2. Molar Volume. Molar volume is the volume of one mole of a gas at STP. This volume has been
experimentally found to be 22.4L at STP and is called molar volume.
1 mol of O2 gas = 32g = NA molecules = 22.4L at STP.
1 mol of CO2 gas = 44g = NA molecules = 22.4L at STP
1.8 PERCENTAGE COMPOSITION
It is the number of parts by mass of a particular element or constituent present in 100 parts by mass of the
compound.

Mass of theelement in the compound

Mass % of element = 100
Molar mass of thecompound

1.9 EMPIRICAL FORMULA AND MOLECULAR FORMULA

Empirical formula represents the simplest whole number ratio of the various atoms present in a compound.
For example, the empirical formula of glucose is CH2O, benzene is CH, and hydrogen peroxide is OH.
Molecular formula is the number of different types of atoms present in a molecule of a compound. The
molecular formula of glucose is C6H12O6, benzene is C6H6, and hydrogen peroxide is H2O2.
Relation Between Empirical Formula and Molecular Formula. The molecular formula of a compound is
a simple whole number multiple of its empirical formula,
Molecular formula = n × Empirical formula [where n is an integer]

Compound Empirical Formula n Molecular Formula

Glucose CH2O 6 C6H12O6
Benzene CH 6 C6H6
Hydrogen peroxide OH 2 H2O2

1.9.1 Calculation of Empirical Formula and Molecular Formula

If the mass percentages of various elements present in a compound are known, its empirical formula can be
calculated. Molecular formula can be calculated if the molar mass is known.
1. Calculation Empirical Formula. i. Calculation of relative atomic ratio. Divide the mass percentages
by the respective atomic masses to get relative number of atoms or the atomic ratio of the various elements
present in one molecule of the compound.
Percentage of the element
Atomic ratio = Atomic mass of the element

ii. Calculation of simplest atomic ratio. Divide the atomic ratio by the smallest quotient from amongst the
values obtained for each element. This gives the simplest atomic ratio. If the ratios are not whole numbers,
convert them to whole numbers by multiplying by suitable coefficient.
iii. Deducing the empirical formula. Empirical formula is written by mentioning the numbers (atomic ratio)
after writing the symbols of respective elements.

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2. Calculation of Molecular formula. i. Calculation of empirical formula mass. Determine the empirical
formula mass by adding together the atomic masses of all atoms in the empirical formula of the compound.
ii. Calculation of n. Divide the molecular mass by empirical formula mass
Molecular mass
 (n)
Empirical formula mass
iii. Deducing the molecular formula. Multiply empirical formula by n, to get molecular formula
Molecular formula = n × Empirical formula
Qn. An organic compound contains 40% carbon, 6.67% hydrogen and 53.33% oxygen by mass. What is
the molecular formula of the compound if its molecular mass is 180 ?

Ans. Element %of element Atomicmass Relative

percentage / Atomicmass
Atomicratio Atomic
ratio(wholenumber)
40 3.33
C 40.00 12.01  3.33  1.000 1
12.01 3.33
6.67 6.62
H 6.67 1.008  6.62  1.99 2
1.008 3.33

O 53.33 15.994 53.33 3.33 1

 3.33  1.000
15.994 3.33

Ratio of moles of atoms, C : H : O = 1 : 2 : 1

Empirical formula of compound = CH2O
Empirical formula mass = 12 + (2 × 1.008) + 16 = 30

Molar mass 180

n  6
Empirical formula mass 30
Molecular formula of compound = (CH2O) ×6 = C6H12O6

1.10 STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS

The relationship between the amounts of reactants and products is called stoichiometry of a reaction. A
balanced equation can be called a stoichiometric equation. The numbers before the formula units, used to
balance the equation are the stoichiometric coefficients. State symbols in equations show the physical state
of the reactants and products; gas (g), liquid (l), solid (s) and aqueous (aq) solution.

N2  3H 2  2NH 3
Mol ratio 1 mole 3 moles 2 moles
Molecular ratio 1 molecule 3 molecules 2 molecules
Weight ratio 28 g 6g 34 g
Volume ratio 1 volume 3 volume 2 volume (at STP)
In the above equation, 1 mol N2 = 3 mol H2 = 2 mol NH3.
Mol of N2 : Mol of H2 : Mol of NH3 : : 1 : 3 : 2

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1.11 LIMITING REAGENT

The reactant which is completely consumed, limits the amount of the product formed in a reaction, therefore
it is called the limiting reagent.

CaCO3 + 2HCl 
 CaCl2 + H 2O + CO 2
100g 73g 44g

According to the balanced chemical reaction, 100g CaCO3 reacts with 73g of HCl to give 44g of CO2. If the
mass of CaCO3 or HCl is less than these quantities, 44 g of CO2 will not be produced.
1.12 CHEMICAL EQUATIONS
A chemical equation is a brief representation of a chemical change in terms of symbols and formulae of
reactats and products. It can convey qualitative and quantitative information regarding a chemical reaction.
1.13 METHODS OF EXPRESSING CONCENTRATIONS OF SOLUTIONS
A solution is a homogeneous mixture of two or more chemically non- reacting substances, the relative amounts
of which can be varied up to a certain limit. If a solution consists of two components it is called a binary
solution. The major component is called the solvent and the minor component the solute. The concentration
of a solution (amount of solute present in a given volume) is expressed as follows :
1. Mass percent or weight percent (% w/w or mass % or % by mass). It is the mass of solute in grams
present in hundred grams of the solution.

Mass of solute
Mass percent   100
Mass of solution

x% by mass = 100 g solution contains x g solute and (100 - x) g solvent.

2. Mass - Volume percent (w/v %). The mass of solute present in 100 cm3 of solution.

w Weight of solute
% = Volume of solution 100
v
3. Volume percent (% v/v). The number of parts by volume of solute (liquid) per hundred parts by volume
of solution

v Volume of solute

%   = Volume of solution  100
V
4. Mole percent. (In the case of gases, % v/v is same as mole %)

Moles of solute
Mole = 100
Total moles

5. Molarity (M). It is defined as the number of moles of solute in 1L of the solution.

Number of moles of solute n g / mA

Molarity (M) = Volume of solution in litres  
V V

mA = molecular mass of solute; n = number of moles of solute.

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Molarity of a solution depends upon temperature since, volume is temperature dependent.

i. Molarity equation (Dilution equation). When a solution with molarity M1 and volume V1 is diluted to
volume V2, then,

M1V1
M1 × V1 = M2 × V2 or M2 
V2

M1V1
ii. For n times dilution : Final molarity, M 2 
nV1
M1V1 + M 2 V2 +.....+ M n Vn
iii. Mixing of n solutions having same solute: Molarity of mixture, M 
V1 + V2 +.....+ Vn

6. Molality (m). Molality is the number of moles of solute present in 1 kg of solvent.

Number of moles of solute
Molality (m) =
Mass of solvent in kg
Mass of solute wA
m= =  1000
Mol.mass of solute  Mass of solvent (in kg) mA  w B
where wA= mass of solute in g, mA = molecular mass of solute, and wB = mass of solvent in g.

7. Normality (N). It is the number of gram equivalents of the solute in 1 L of the solution.

Number of gram equivalents of solute

Normality (N) =
Volume of solution in litres

Mass of solute (w a ) wA wA 1000

Gram equivalent = = 
Equivalent mass of the solute (E A ) EA  V  in L EA  V  inmL
Normality equation (Dilution equation). If a solution of normality N1 and volume V1 is diluted to volume
V2, the normality becomes N2, then
N 1 × V1
N1 × V1 = N2 × V2 or N2 = V2

When solutions having normalities N1, N2, N3and volumes V1, V2, V3 respectively, are mixed, the normality
(Nn) of the resulting solution is given by,
N 1 V 1 + N2 V 2 + N 3 V 3
Nn =
V 1 + V 2 + V3
8. Mole Fraction (  ). It is the ratio of the number of moles of a particular component to the total number of
moles in the solution. If nB moles of solute B dissolves in nA moles of solvent A,
nA nB
Mole fraction of solvent A,  A  n  n Mole fraction of solute B,  B  n  n
A B A B

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9. Mass Fraction (X). It is the ratio of mass of a particular component to the total mass of the solution. If
solute B of mass wB is dissolved in solvent A of mass wA,
wA
Mass fraction of solvent A ,(XA) = w  w
A B

wB
Mass fraction of solute B, (XB) = w  w or XB = (1 – XA)
A B

10. Formality. Since, molecular weights of ionic solids are not determined accurately in solution, molecular
masses of ionic solids are often refered to as formula weight and molarity as formality.
Mass of solute
Formality =
Formula mass of solute  V (in L)
11. Parts per million (ppm). Very low concentration of solute in solution is expressed in ppm.
Mass of solute
parts per million (ppm) =  106
Mass of solution
ppm = Mass fraction × 106
1.13.1 Relation between Stoichiometric Quantities
i. Relation between mole fraction and molality. When xA is the mole fraction of a component in solution,
x A  1000
Molality, m 
1 - x A  M B
where, MB is the molar mass of solute.
ii. Relation between molality and molarity
1 d M
 - B
m M 1000
d = density of solution, m = molality, M = molarity and MB = molar mass of solute.
iii. Relation between normality and molarity
Strength of solution (g L–1) = Molarity × Molecular mass = Normality × Equivalent Mass
Normality Molecular mass
 n or Normality  n  Molarity
Molarity Equivalent mass
1.14 n-FACTOR
It is a conversion factor by which molar mass of a substance is divided to get equivalent mass.
i. n-factor for acid. It is the number of replaceable H+ ions in 1 mole of acid in a reaction. n-factor of
monobasic acids like HCl, CH3COOH, etc., is the same as its basicity (=1). n-factor of H2SO4 may be 1 or
2, depending on the number of H+ ions replaced.
ii. n-factor for bases. It is the number of replaceable OH- ions per molecule in a reaction. n-factor of
monoacidic bases like NaOH, KOH, etc., is the same as its acidity (=1). n-factor of Ca(OH)2 may be 1 or 2,
depending on the number of OH- ions replaced. n-factor of Al(OH)3 may be 1, 2, or 3.
iii. n-factor for salts. The n-factor of salts is defined as the number of moles of electrons exchanged (lost or
gained) by one mole of the salt.

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 [Study Package - Chemistry - Part- I]

QUESTIONS
LEVEL - I
1. Which among the following is/are life saving drug(s) effective in cancer therapy?
1) Azidothymidine (AZT) 2) Cisplatin and taxol
3) Chlorofluorocarbons 4) Cellulose acetate
2. Which of the following is a characteristic of both mixtures and compounds?
1) Their properties are same as those of their constituents
2) Energy is absorbed when they are formed
3) Their mass equals to the sum of the masses of their components
4) They contain components in fixed proportions
3. A mixture of sand and iodine can be separated by
1) Crystallisation 2) Sublimation 3) Distillation 4) Fractionation
4. Using thermal scanner a health worker measured the body temperature of Covid-19 suspect to be 37.5oC.
How many degrees would the thermal scanner register if the instrument was set to register the temperature in
degree Fahrenheit.
1) 99.5o F 2) 52.8o F 3) 67.5o F 4) 100.4o F
5. The true value of a result is 2.00 g. A student takes two measurements and reports the result as 1.95 g and
1.93 g. These values are
1) both precise and accurate 2) precise but not accurate
3) accurate but not precise 4) neither precise nor accurate
6. Law of multiple proportions is illustrated by one of the following pairs.
1) H2S and SO2 2) NH3 and NO2 3) Na2S and Na2O 4) N2O and NO
7. Consider the following laws of chemical combination with examples:
I) Law of multiple proportion; N2O, NO, NO2
II) Law of reciprocal proportion; H2O, SO2, H2S
Which is correct with examples?
1) I and II 2) I only 3) II only 4) I and II are incorrect
8. In compound A, 1.0 g nitrogen combines with 0.57 g of Oxygen. In compound B 2.0 g nitrogen units with
2.24 g Oxygen and in compound C. 3.0 g of nitrogen combine with 5.11 g oxygen. These results obey the law
of
1) multiple proportions 2) constant proportions
3) reciprocal properties 4) law of conservation of mass
9. Which among the following is not according to Daltons atomic theory?
1) Matter consists of indivisible atoms
2) atoms of different elements differ in mass
3) Compounds are formed when atoms of the same element combine in fixed ratio
4) Chemical reactions involve reorganisation of atoms
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10. Which one of the phrases would be incorrect to use?

1) One mole of an element 2) One mole of a compound
3) An atom of an element 4) An atom of a compound
11. Number of atoms of oxygen present in 10.6 g of Na2CO3 will be
1) 6.02 × 1022 2) 12.04 × 1022 3) 1.806 × 1023 4) 31.80 × 1028
12. How many molecules are present in 5.23 g of glucose (C6H12O6)?
1) 1.65 × 1022 2) 1.75 × 1022 3) 1.75 × 1021 4) 1.65 × 1021
13. Mass of one atom of X is 6.66 × 10–23g. Hence number of mole of atom X in 40 kg is

40  103 40  103
1) 103 mol 2) 10–3 mol 3) mol 4) mol
6.66  10-23 6.022  1023
14. The number of electrons present in 5.6 litre of H2 gas at S.T.P is

1) 6.02 × 1023 2) 3.01 × 1023 3) 12.04 × 1023 4) 3.0 × 1010

15. A certain compound has the molecular formula X4 O6. If 10g of X4O6 has 5.72 g X, the atomic mass of X is
1) 32 u 2) 37 u 3) 42 u 4) 98 u
16. Total number of atoms in 4.25 g of NH3 is:
1) 6.02 × 1023 2) 4 × 6.02 ×1023
3) 1.7 × 1024 4) 4.5 × 6.02 × 1023
17. If air contains 71% N2 by volume, the number of atoms of nitrogen per litre of air at STP is.
1) 3.8 × 1023 2) 0.38 × 1023 3) 38 × 1023 4) 0.38 × 1021
18. A metal oxide is reduced by heating it in a stream of hydrogen. It is found that after complete reduction,
3.15 g of the oxide have yielded 1.05 g of the metal. We may conclude that:
1) atomic mass of the metal is 4 2) atomic mass of the metal is 8
3) equivalent mass of the metal is 4 4) equivalent mass of the metal is 8
19. Volume occupied by one molecule of water (density = 1g/cm3) is:
1) 3 × 10–23 cm3 2) 5.5 × 10–23 cm3
3) 9 × 10–23 cm3 4) 6.02 × 10–23 cm3
20. What weight SO2 can be made by burning sulphur in 5 moles of dioxygen
1) 640 grams 2) 160 grams 3) 80 grams 4) 320 grams
21. The largest number of molecules is in
1) 54 g of nitrogen dioxide 2) 28 g of carbon dioxide
3) 36 g of water 4) 46 g of ethyl alcohol
22. In a gas S and O are 50% by mass, hence, their mol ratio is
1) 1:1 2) 1: 2 3) 2 :1 4) 3:1

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 [Study Package - Chemistry - Part- I]

23. Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The mass percentage of
carbon in cortisone is 69.98%. It’s molar mass is
1) 176.5 gmol–1 2) 252.2 gmol–1 3) 287.6 gmol–1 4) 360.1 gmol–1
24. The vapour density of a chloride of an element is 39.5. The equivalent mass of the element is 3.82. The atomic
weight of the element is
1) 15.28 2) 7.64 3) 3.82 4) 11.46
25. For a reaction, N 2 g   3H 2 g   2NH3 g  :

Identify dihydrogen (H2) as a limiting reagent in the following reaction mixtures.

1) 14 g of N2 + 4 g of H2 2) 28 g of N2 + 6 g of H2
3) 56 g of N2 + 10 g of H2 4) 35 g of N2 + 8 g of H2
26. The simplest formula of a compound containing 50% of an element X (atomic weight 10) and 50% of element
Y (atomic weight 20) is
1) XY 2) X2Y 3) XY2 4) X2Y3
27. If 30 mL of H2 and 20 mL of O2 react to form water, what is left at the end of the reaction?
1) 10 mL of H2 2) 5 mL of H2 3) 10 mL of O2 4) 5 mL of O2
28. The amount of sugar (C12H22O11) required to prepare 2 L of its 0.1 M aqueous solution is
1) 68.4 g 2) 17.1 g 3) 34.2 g 4) 136.8 g
29. An aqueous solution of glucose is 10% (w/w). The mole fraction of glucose in this solution is:
1) 0.011 2) 0.989 3) 0.101 4) 0.11
30. The correct set of co-efficients x, y, p and q for the following balanced equation is:

xFeS2  yO 2 
 pFe 2 O3  qSO 2

x y p q
1) 2 5 2 4
2) 4 5 2 8
3) 6 11 3 12
4) 4 11 2 8

LEVEL - II
1. Copper forms two oxides. For the same amount of copper, twice as much oxygen was used to form first
oxide than to form second one. What is the ratio of the valencies of copper in first and second oxides?
1) 2 : 1 2) 1 : 2 3) 3 : 1 4) 1 : 3
2. What weight of CO2 will contain the same number of oxygen atoms as are present in 3.6 g of water?
1) 8.8 g 2) 7.2 g 3) 4.4 g 4) 220 g

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3. The ratio of masses of dioxygen and dinitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of
their molecules is:
1) 1 : 4 2) 7 : 32 3) 1 : 8 4) 3 : 16
4. A compound made of two elements A and B is found to contain 25% A (atomic mass 12.5) and 75% B
(atomic mass 37.5). The simplest formula of the compound is:
1) AB 2) A2B2 3) AB3 4) A3B
5. 1 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reactant is left in excess and how much?
(At. wt. Mg = 24, O = 16)
1) Mg, 0.16 g 2) O2, 0.16 g 3) Mg, 0.44 g 4) O2, 0.28 g

6. In the reaction 4NH 3  g   5O2  g   4NO  g   6H 2 O  l  , when 1 mole of ammonia and 1 mole of O2 are
made to react to completion?
1) 1.0 mole of H2O is produced 2) 1.0 mole of NO will be produced
3) All the oxygen will be consumed 4) All the ammonia will be consumed
7. When 2.46 g of hydrated salt (MSO4.xH2O) is completely dehydrated. 1.20 g of anhydrous salt is obtained.
If the molecular weight of anhydrous salt is 120 g mol–1 what is the value of x?
1) 2 2) 4 3) 5 4) 7
8. The carbonate of a metal is isomorphous with MgCO3 and contains 6.091% of carbon. Atomic weight of the
metal is nearly:
1) 48 2) 68.5 3) 137 4) 120
9. Haemoglobin contains 0.25% iron by mass. The molar mass of Haemoglobin is 89600 gmol–1. Calculate the
no. of iron atom per molecule of Haemoglobin.
1) 4 2) 2 3) 6 4) 8
10. The molality of an H2SO4 solution is 9. The weight of the solute in 1 kg H2SO4 solution is:
1) 900.0 g 2) 469 g 3) 882.0 g 4) 9.0 g
11. 36.5% HCl has density equal to 1.20 gmL–1. The molarity (M) and molality (m) respectively, are:
1) 15.7, 15.7 2) 12, 12 3) 15.7, 12 4) 12, 15.7
12. The mass of 60% HCI required for the netralisation of 10L of 0.1 MKOH is:
1) 60.8 g 2) 21.9 g 3) 100 g 4) 219 g
13. A sample of ammonium phosphate (NH3)4 PO4 contains 3.18 moles of H atoms. The number of moles of O
atoms in the sample is:
1) 0.265 2) 0.795 3) 1.06 4) Cr3+, Fe2+, Co3+
14. The molarity of a solution obtained by mixing 750 mL of 0.5 M HCl with 250 mL of 2 M HCl will be:
1) 1.00 M 2) 1.75 M 3) 0.975 4) 0.875
15. A 6.90 M solution of KOH in water has 30% by mass of KOH. Calculate density of solution?
1) 1.33 g mL–1 2) 1.288 g mL–1 3) 1.66 g mL–1 4) 1.44 g mL–1

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 [Study Package - Chemistry - Part- I]

16. For the reaction Fe2 O3  3CO  2Fe  3CO 2 , the volume of carbon monoxide required to reduce one mole
of ferric oxide is:
1) 67.2 dm3 2) 11.2 dm3 3) 22.4 dm3 4) 44.8 dm3
17. Calculate the mass of calcium oxide required that reacts with 852 g of P4O10.
1) 2016 g 2) 1008 g 3) 504 g 4) 1800 g
18. A silver coin weighing 11.34 g was dissolved in nitric acid. When sodium chloride was added to the solution
all the silver (present as AgNO3) was precipitated as silver chloride. The weight of the precipitated silver
chloride was 14.35 g. Calculate the percentage of silver in the coin.
1) 4.8% 2) 95.2% 3) 90% 4) 80%
19. Concentrated HNO3 is 69% by mass of nitric acid. Calculate the volume of the solution which contains 23 g
of HNO3. (Density of concentrated HNO3 solution is 1.41 g ml–1)
1) 23.6 2) 32.6 3) 26.8 4) 18.5
20. 2 g of a mixture of CO and CO2 on reaction with excess I2O5 produced 2.54 g of I2. What will be the mass
percentage of CO2 in the original mixture?
1) 35% 2) 70% 3) 30% 4) 60%
21. 20 g of impure magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g
magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample?
1) 60 2) 84 3) 75 4) 96
22. 1g of a mixture of NaHCO3 and Na2CO3 is heated to 150oC. The volume of the CO2 produced at STP is
112 mL. Calculate the percentage of Na2CO3 in the mixture (Na = 23, C = 12, O = 16)
1) 20 2) 46 3) 84 4) 16
23. The mass of NaBrO3 required to prepare 150 mL of 0.75 N of a solution based on the reaction
Br3-  6H   6e -  Br -  3H 2 O is:

1) 1.42 g 2) 2.83 g 3) 3.85 g 4) 4.25 g

24. 6.0 g sample of CaCO3 reacts with 20 g solution of HCl having 20% by mass of HCl (density = 1.10 g/mL).
Calculate percentage purity of CaCO3 sample:
1) 9.13% 2) 91.33% 3) 54.8% 4) 5.48%
25. Choose the correct statements:
a) The number of atoms in 1 g of helium is 1.506 × 1022
b) The mass of 1 molecule of CO is 4.65 × 10-23g
c) The volume at STP occupied by 240g of SO2 is 22.4 litre
d) The volume at STP occupied by 240g of SO2 is 84 litre
1) a, b 2) a, c 3) b, c 4) b, d

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26. 1.8 g of Mg is burnt in a closed vessel which contains 0.8 g of oxygen. Which of the following facts are correct
for the resultant system?
a) Amount of MgO formed is 0.05 mol
b) Mass of Mg left in excess is 0.8 g
c) Amount of oxygen left is zero
d) Volume of 0.25 M H2SO4 to dissolve the formed MgO is 200 mL
1) a, d 2) b, c, d 3) a, c, d 4) a, b, c, d
27. The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 27oC. Calculate the number of moles of
NO2 in100 g mole mixture:
1) 76.6 mole 2) 3.348 mole 3) 334.8 mole 4) 33.48 mole
28. What is the empirical formula of vanadium oxide, if 2.74 g of the metal oxide contains 1.53 g of metal?
(Atomic mass of vanadium = 52 u )
1) V2O3 2) VO 3) V2O5 4) V2O7
29. The specific heat of a metal is 0.16. Its approximate atomic weight would be
1) 32 2) 16 3) 40 4) 64
xy
30. 109% labelled oleum has x mole of H2SO4 and y mole of SO3 respectively. What is the value of  
x-y
in oleum?
1) 8.81 2) 9.91 3) 10.6 4) 11.6

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 [Study Package - Chemistry - Part- I]

SOLUTION
KEY
LEVEL - I
1. 2 2. 3 3. 2 4. 1 5. 2
6. 4 7. 1 8. 1 9. 3 10. 4
11. 3 12. 2 13. 1 14. 2 15. 1
16. 1 17. 2 18. 3 19. 1 20. 4
21. 3 22. 2 23. 4 24. 2 25. 3
26. 2 27. 4 28. 1 29. 1 30. 4

LEVEL - II

1. 1 2. 3 3. 2 4. 1 5. 1
6. 3 7. 4 8. 3 9. 1 10. 2
11. 4 12. 1 13. 3 14. 4 15. 2
16. 1 17. 2 18. 2 19. 1 20. 3
21. 2 22. 4 23. 2 24. 2 25. 4
26. 3 27. 4 28. 3 29. 3 30. 2
HINTS
LEVEL - I
1. 2 Cisplatin and taxol are effective in Cancer therapy.
2. 3 Mixture and compounds, both obey Law of Conservation of Mass.
3. 2 Sublimation is used for the separation of sand and iodine.
 9
4. 1
o
F   o C    32
 5
5. 2 The values are not accurate but are precise.
6. 4
7. 1
8. 1

9. 3 Compounds are formed between atoms of different elements.

10. 4 Compounds have only molecules as smallest unit exhibiting their properties.

Number of atoms of oxygen  

10.6 
11. 3   3  6.02  10 = 1.806 × 10
23 23

 106 

5.23 5.23  6.02  1023

12. 2  5.23 g glucose  moles No. of molecules  1.75  1022 molecules
180 180

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13. 1 Atomic mass = 6.66 × 10–23 × 6.02 × 1023 = 40 g mol–1

40000g
No. of moles in 40 kg = 40 g mol -1  1000 mol

5.6
14. 2 No. of electrons in 5.6 litre of H2 at STP   6.02  1023  2  3.0110 23
22.4
Molecular weight of X 4 O 6 Weight of X 4 O6 4x  96 10
15. 1   x  32
Atomic weight of 4X Weight of X 4x 5.72
4.25
16. 1 Number of atoms  4   6.02 1023  6.02 10 23
17
17. 2 Volume of N2 per litre of air = 0.71 litre
0.71
No.of ‘N’ atoms per litre at STP   6.02  10 23  2  0.38  10 23
22.4
1.05  8
18. 3  eq.mass of metal
2.1
18 Mass
19. 1 Mass of one molecule  g Volume of one molecule
6.02  10 23 Density
20. 4 5S  5 O 2  5SO 2 ; 5 O 2  5 SO 2 ; 5  64  320 g
21. 3 36 g water contain 2 moles all other less.
22. 2
252  100
23. 4 % of carbon   69.98, molar mass  360.1g mol–1
M
24. 2 Mol. mass of metal chloride (MClx): = M + x × 35.5 = 2 × VD = 2 × 39.5 = 79.0
Let valency of metal be ‘x’ and atomic mass of element = eq.mass  x
79
MClx = (3.82 × x) + (x × 35.5) = 79 x 2
3.82  35.5
Atomic weight = eq.mass  x  3.82  2  7.64

25. 3 N 2  g   3H 2 g  
 2NH 3 g 

2 mol N 2 & 5 mol H 2 are present. H is completely consumed.

 Limiting reagent  2

26. 2 X = 50% Y = 50%

X : Y = 5 : 2.5 = 2 : 1, hence X2Y
1
27. 4 30 mL of H2 reacts with   30  15 mL O 2
2
(20 – 15) = 5 mL of O2 will left at the end of the reaction.
mass of solute 1000 W 1000
28. 1 Molarity   0.1  
molar mass V ; 342 2000

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 [Study Package - Chemistry - Part- I]

29. 1 Consider 100 g solution

Mass of glucose = 10 g, Moles of glucose = 0.056, Mass of water = 90 g
0.056
Moles of water = 5, Mole fraction of glucose   0.011
5.056
30. 4 The balanced equation is
4FeS2  11O 2 
 2Fe 2 O3  8SO 2

LEVEL - II

1. 1 Let the compounds be CuO2x and CuOx, where ‘x’ is the eq. mass of oxygen.
2x 2
Ratio of valencies of copper =  i.e., 2 :1
x 1
3.6
2. 3 No. of water molecules in 3.6 g of water   6  1023 molecules = 1.2 × 1023 molecules
18
No. of oxygen atoms = 1.2 × 1023
Now 2 × 6 × 1023 atoms of oxygen in CO2 = 44 g
44
1.2 × 1023 atoms of oxygen in CO2  1.2  1023  4.4 g
2  6  10 23

moles of O 2 mass of O 2 mass of N 2

3. 2  
moles of N 2 molar mass of O 2 molar mass of N 2
1 4 7
 
32 28 32
25 75
4. 1 Mole ratio of A to B is : = 2 : 2. MF = A2B2. The simplest formula will be AB.
12.5 37.5
1 0.56
5. 1 Moles of Mg   0.0417 mol ; Moles of O2   0.0175 mol
24 32
1
Mg  O2 
 MgO
2
1mol 0.5mol 1mol

O2 is limiting reagent ; 0.5 mol of O2 react with 1 mole Mg to form 1 mol of MgO
1 0.0175
0.0175 mol of O2 react with  = 0.035 mol Mg
0.5
Amount of Mg left unreacted = 0.16 g
6. 3 4NH3 g   5O 2 g  
 4NO g   6H 2 O l
1 mol 1 mol
O2 is the limiting reactant  0.8 mol NO is produced, 1.2 mol H2O is produced
All the O2 will be consumed and 0.2 mol NH3 will be left.

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 Brilliant STUDY CENTRE

7. 4 MSO 4 .xH 2O 
 MSO 4  xH 2 O Mass of water = 2.46 – 1.20 = 1.26 g
2.46g 1.20g

1.26 1.20
 No.of moles of H 2 O   0.07 No. of moles of MSO4   0.01
18 120
 0.07 moles of water associated with 0.01 mole of salt.
Thus, there are 7 moles of water per mole of anhydrous salt.
8. 3 Let the formula of the metal carbonate be MCO3, then percentage of carbon is
12 6.091
 where x = atomic mass of M
x  60 100
x  56
9. 1 100  0.25 x is no. of iron atoms. Atomic mass of iron = 56
89600
10. 2 9 molal H2SO4 = 9 × 98 g in 1000 g H2O
882
= 882 g H2SO4 in 1882 g H2SO4 solution =  1000 g H 2SO 4 in1kg solution  469 g
1882
% by weight  10  d 36.5  10  1.2
11. 4 M   12M
M2 36.5
36.5  1000 1000
m   15.7m
36.5  (100 - 36.5) 63.5
12. 1 M.eq.of KOH = 10000  0.1  1000
1000
 m.eq.of HCl  WHCl   36.5  36.5g pure
1000
100
60 g pure sample  100 g sample HCl 36.5 g pure sample   36.5g
60
Sample HCl  60.8 g
1 3.18 mole
13. 3 Number of moles of O atoms = moles of H   1.06 mol
3 3
14. 4 M1V1  M 2 V2  MV
M1V1  M 2 V2 0.5  750  2  250
M   0.875 M
V 1000
15. 2 KOH solution is 30% by mass.  Mass of KOH = 30 g and Mass of solution = 100 g
100 30
 Volume of solution   Molarity  6.90   d = 1.288 g mL–1
d 100
56 
1000  d
16. 1 Fe2O3 + 3CO  2Fe + 3CO2
One gram mol of any gas occupies 22.4 litre at NTP. 1 mol of Fe2O3 requires 3 mol of CO for its
reduction i.e., 1 mol of Fe2O3 requires 3 × 22.4 litre or 67.2 dm3 CO to get itself reduced.

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 [Study Package - Chemistry - Part- I]

17. 2 6CaO2  P4O10 

 2Ca 3 (PO4 )2
base acid salt

Molar mass of P4O10= 284 852

 Mole of CaO required = 6 × mole of P4O10  6 
284
6  852
 Mass of CaO required  284  56  1008g

18. 2 Ag  HNO3 
 AgNO 3
AgNO3  NaCl 
 NaNO3  AgCl
Mole of Ag in coin = mole of Ag in AgCl
14.35
a  0.1mole
143.5
10.8
Mass of Ag in coin = 0.1×108=10.8 g  % silver in coin  100%  95.2%
11.34
19. 1 69 g of HNO3 = 100 g of solution
100
23g of HNO3   23  33.3g of solution
69
33.3
Volume of solution   23.6 mL
1.41
20. 3 5CO  I 2O5  5CO 2  I2

2.54
Moles of I 2 O5  = 0.01 = 0.05 moles of CO
25.4
Weight of CO = 0.05 × 28 = 1.4 g ; Weight of CO2 = 2–1.4 = 0.6 g
0.6
Hence percentage of CO 2   100  30%
2
21. 2 MgCO3 s  
 MgO s   CO 2 g  40g of MgO are produced from 84g MgCO3
1mol 1mol
84g 40g

84 16.8
8g of MgO are produced from MgCO3   8  16.8g ; % Purity of MgCO3   100  84%
40 20
o
22. 4 2NaHCO3 
150 C
 Na 2CO3  CO 2  H 2 O
n NaHCO3 2 112
  n NaHCO 3  2n CO 2  2   0.01mole
n CO 2 1 22400

WNaHCO3  0.01 84  0.84 g ; WNa 2CO3  1.00 - 0.84  0.16 g ; % Na2CO3 = 16

23. 2 Amount of NaBrO3 in the solution = (0.75 eq dm–3) (0.150 dm–3) = 0.1125 eq
Mass of NaBrO3 required = (0.1125 eq) (0.151/6) g eq–1 = 2.83 g

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 Brilliant STUDY CENTRE

24. 2 CaCO3  2HCl  CaCl2  CO 2  H 2 O

20  20
moles of HCl present   0.1096
100  36.5
5.48
amount of CaCO3 required 0.0548 mole = 5.48 g. % purity of sample   100  91.33%
6
25. 4 b, d are correct
26. 3 a, c, d are correct
27. 4 Molecular mass of mixture (NO2 + N2O4) = 2 × V.D. = 38.3 × 2 = 76.6
Let x mole of NO2 is present in 100 mole mixture
 mass of NO2 + mass of N2O4 = mass of mixture
x × 46 + (100 – x) × 92 = 100 × 76.6
 x = 33.48 mole
1.53
28. 3 Metal oxide = 2.74 g; % of V =  100  55.83
2.74
Thus, % of O = 100 – 55.83 = 44.17
55.83 44.17
No.of moles of V   1.1 No. of moles of O =  2.76
52 16
Simplest ratio of V and O = 1:2.5 or 2:5 Hence, the empirical formula = V2O5
29. 3 Dulong-petit law states that, atomic weight × specific heat = 6.4
6.4
Approximate atomic weight   40
0.16
30. 2 109% oleum means it has 9 g H2O which reacts with free SO3 to give H2SO4
According to stoichiometry SO3  H 2 O 
 H 2SO 4
80  9
Mass of SO3 which will react with 9 g H2O=  40g
18
100 g sample of 109% labelled oleum contain 40 g SO3 and 60 g H2SO4
60 40
 Mole of H2SO4   0.6122  x moles of SO3   0.5  y
98 80

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 [Study Package - Chemistry - Part- I]

CHAPTER - 02

STRUCTURE OF ATOM

INTRODUCTION
Ancient Indian and Greek philosophers were of the view that atoms are the building blocks of matter. The
Greek philosopher, Democritus (430 BC) named them atomos, meaning ‘indivisible’. He believed that atoms
are uniform, solid, hard, incompressible and indestructible.
The atomic theory of matter was first proposed on a scientific basis by John Dalton, a British school teacher
in 1808. Dalton’s atomic theory regarded atoms as the ultimate particles of matter.
2.0 SUBATOMIC PARTICLES (ELEMENTARY PARTICLES)
It was known that substances like glass, ebonite, etc., when rubbed with silk or fur, generate electricity. In
1830, Michael Faraday showed that when electricity is passed through a solution of an electrolyte, chemical
reactions resulting in the liberation/desposition of matter occured at the electrodes. Faraday’s observations
suggested particle nature of electricity.
Experimental observations in the twentieth century established that atoms can be further divided into sub-
atomic particles such as electrons, protons, and neutrons - a concept, totally different from that of Dalton.
2.1.1 Discovery of Electron
Julius Plucker (1889) and J.J. Thomson (1898) observed that when high voltage electric current was passed
through a gas at low pressure, negatively charged particles travelled from the cathode (negative electrode) to
the anode (positive electrode). These were called cathode rays or cathode ray particles.

Figure - 01 Cathode ray discharge tube

When the pressure of the gas was about 10-4 atmospheres and potential difference was 10000 volts, current
started flowing through a stream of particles moving from the cathode towards the anode. These rays, after
passing through a perforated anode produced a bright spot on zinc sulphide coating behind it.
Properties of cathode rays. i. Cathode rays travel in straight lines from the cathode towards the anode. An
object placed in the path of cathode rays casts a shadow.
ii. Cathode rays consist of material particles. A light paddle wheel placed in its path rotates.
iii. When cathode rays are focussed on a metal foil, it gets heated up.
iv. When an electric field is applied, cathode rays get deflected towards the positive plate indicating that they
are negatively charged.

31
 Brilliant STUDY CENTRE

v. When a magnetic field is applied perpendicular to their path, cathode rays get deflected in the direction
expected for negatively charged particles.
vi. Cathode rays ionize the gas through which they pass.
vii. They produce greenish fluorescence when they strike the walls of the discharge tube.
viii. These rays produce a bright spot on zinc sulphide (phosphorescent) coating.
ix. When cathode rays fall on metals like tungsten, X-rays are produced.
x. Cathode rays penetrate through thin metal foils.
xi. The charge to mass ratio (e/m) of cathode ray particles is independent of the gas taken in the discharge
tube and the metal used as cathode.
These properties indicate that cathode rays are fast moving electrons which are constituents of all atoms.
2.1.2 Charge to mass ratio of electron
In 1897, British physicist J.J. Thomson measured the ratio of charge (e) to mass of electron (me) using
cathode ray tube by applying electrical and magnetic fields perpendicular to each other and to the path of the
electrons. Magnetic and electric fields deflected the beam in opposite directions A and C from point B.

The deflection of the beam depended on the magnitude of negative charge on the particle, mass of the particle
and strength of the electrical or magnetic field. From the amount of electric charge required to balance the
effect of the magnetic field, the e/me of the particle was calculated as 1.758820 × 1011 C kg-1.
The e/me ratio for cathode rays was found to be the same irrespective of the nature of the cathode or the gas
in the discharge tube indicating that electrons are basic constituents of all atoms.
2.1.3 Charge on the Electron
The charge, ‘e’ on the electron was determined by American scientist, Robert Millikan (1909) by his oil drop
experiment. Millikan found that the magnitude of charge ‘q’ on the droplets is always an integral multiple of the
electrical charge, e (=1.6022× 10-19C).
2.1.4 Mass of the Electron
The mass of the electron was calculated by combining the of results Millikan’s experiment with Thomson’s
e 1.6022 1019 C
e / me value, me    9.1094 1031 kg
e/me 1.758820 1011 C kg1
The mass of the electron is approximately 1/1837 of the mass of the hydrogen atom.
2.1.5 Discovery of Proton
In 1886, German physicist, Eugen Goldstein used a perforated cathode in a discharge tube and observed a
new type of rays streaming behind the cathode which he named anode rays or canal rays.
i. Anode rays consist of material particles travelling in straight lines.
ii. The behaviour of these particles in magnetic and electrical fields are opposite to that cathode rays, indicating
that they are positively charged.
iii. The e/m value of anode rays depends on the gas from which they originate (taken in the discharge tube).
iv. Some of these particles carry a multiple of the fundamental unit of electric charge.
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 [Study Package - Chemistry - Part- I]

The smallest and lightest positive ion was obtained from hydrogen and was called proton. The charge to mass
ratio of proton was found to be 9.58 × 104 C/g. The charge on the proton was found to be equal, but
opposite in magnitude to that of the electron (1.6022 × 10-19 C ). The mass of the proton worked out to be
1.6726 × 10-27 kg (about 1837 times that of an electron).
This particle was characterised in 1919. The discovery of protons can be attributed to Rutherford.
2.1.6 Discovery of Neutron
In 1932 British physicist, James Chadwick bombarded a thin sheet of beryllium with -particles and observed
electrically neutral particles having mass slightly greater than that of protons. He named these particles neutrons.
9
4 Be  42 He 12
6 C 0 n
1

The mass of the particle was found to be 1.675 × 10-27kg.

Table - 01 Characteristics of fundamental particles
Absolute Relative Approx.
Name Symbol Mass (kg) Mass (u)
charge charge Mass (u)
-19 -31
Electron e -1.6022×10 -1 9.10939×10 0.00054 0
-19 -27
Proton p +1.6022×10 +1 1.67262×10 1.00727 1
-27
Neutron n 0 0 1.67493×10 1.00867 1
2.1.7 Atomic Number and Mass Number
H.G.J. Moseley (1913) observed that the frequency of X-rays emitted when electrons in a cathode ray tube
strikes the anode depends on the charge on the nucleus of the element used as anode.
The number of unit positive charges on the nucleus of the atom is called atomic number of the element. In an
atom, the number of protons is equal to the number of electrons. Thus, atomic number of an element is equal
to the number of protons in the nucleus or the number of electrons.
Atomic number (Z) = Number of protons = Number of electrons.
The mass of the atom is mainly concentrated in the nucleus. The nucleus contains protons and neutrons which
are collectively called nucleons. The total number of nucleons in an atom is called its mass number.
Mass number (A) = Number of protons + Number of neutrons = Number of nucleons.
Thus, element ‘X’ is represented as AZ X .
2.1.8 Isotopes, Isobars, Isotones, Isodiaphers, and Isosters
i. Isotopes. Atoms having same atomic number but different mass numbers are known as isotopes, e.g.,
hydrogen has three isotopes namely, protium  11 H  , deuterium  12 D  and tritium  13 T  . The three isotopes
of carbon are 12 13 14 35 37
6 C, 6 C and 6 C . Chlorine has two isotopes, 17 Cl and 17 Cl .

Chemical properties of atoms are controlled by the number of electrons, therefore, isotopes of a given element
exhibit same chemical behaviour.
ii. Isobars. Atoms having same mass number but different atomic numbers are known as isobars, e.g.,
40
16 S, 40
17 Cl , 40
18 Ar, 40
19 K , and
40
20 Ca . The parent and the daughter nuclides in -decay are isobars.
iii. Isotones.Isotones are atoms of different elements containing the same number of neutrons,
e.g., 13 14 30
6 C and 7 N ; 14 Si,
31
15 P and 32
16 S.

33
 Brilliant STUDY CENTRE

v. Isosteres. Molecules or ions which have the same number of atoms and/or same number of valence
electrons are called isosteres, e.g., CO2 and N2O, CO and N2, CaO and KF, OF2 and HClO. The parent
and the daughter nuclides in -decay are isosters.
vi. Isoelectronics. These are species (atoms or ions) containing the same number of electrons. For example,
e.g., O2-, F-, Na+, Mg+ and Al3+ contain 10 electrons each.
2.2 ATOMIC MODELS
Several atomic models have been proposed to explain the distribution of the charged particles in the atom.
2.2.1 Thomson Model of the Atom
In 1898, J.J. Thomson proposed the plum pudding (raisin pudding or watermelon) model of the atom. The
atom is a sphere (radius 10-10 m) in which the positive charge is uniformly distributed with electrons embedded
in it in such a way as to give a stable electrostatic arrangement.

This model could explain the overall neutrality of the atom, but inconsistent with the results of later experiments.
2.2.2 Rutherford’s Nuclear Model of Atom
In 1911, New-Zealand born British chemist, Ernest Rutherford and his students (Hans Geiger and Ernest
Marsden) bombarded a very thin gold foil with  -particles. The scattered  - particles fell on a ZnS screen
placed around it and produced tiny flashes. It was observed that :
i. Most of the  -particles ( 99%) went through the gold foil undeflected
ii. A small fraction of  - particles was deflected by small angles.
iii. A very small number of particles (1 in 20,000) bounced back (deflected by about 180o).

On the basis of the  -scattering experiment, Rutherford arrived at the following conclusions:
i. Most of the mass and all the positive charge are concentrated in a very small region of the atom called
nucleus. The size of the nucleus is very small compared to the size of the atom. Calculations by Rutherford
showed that the radius of the nucleus is about 10-15 m and that of the atom is 10-10 m (radius of the atom is
about 1,00,000 times the radius of the nucleus).
ii. The positive charge of the nucleus is due to protons and the magnitude of charge is different for different
elements.
iii. The electrons revolve round the nucleus in different concentric, circular paths called orbits. The electrostatic
attraction between electrons and the nucleus is balanced by the centrifugal force of the revolving electron.

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 [Study Package - Chemistry - Part- I]

iv. Total positive charge in the nucleus is equal to the negative charge of the electrons so that the atom is
electrically neutral.
v. Most of the space in the atom is empty.
The model is similar to the solar system. Electrons are referred to as planetary electrons and Rutherford’s
nuclear model of atom is known as planetary model of the atom.
Radius of the nucleus. Radius of the nucleus is related to the mass number as :
R  A1 3  R  R o A1 3  R  1.33  1013 A1 3 cm
Since Ro = 1.33 × 10-13 (constant) and A is the mass number of the atom (p + n).
Drawbacks of Rutherford model of Atom. i. Rutherford’s model could not explain the stability of the
atom. According to the classical theory of electromagnetism (Maxwell), an accelerated charged particle must
emit radiations in the form of electromagnetic waves and lose its total energy. Since energy of electrons keep
on decreasing, the radius of the circular path should decrease and it should ultimately fall into the nucleus.
Calculations have shown that it will take only 10-8 seconds for the electron to spiral into the nucleus.
ii. It could not explain the distribution of electrons around the nucleus and their energies.
iii. Rutherford model could not explain the line spectrum of the hydrogen atom.

2.3 DEVELOPMENTS LEADING TO THE BOHR’S MODEL OF ATOM

(i) Dual character, i.e., particle and wave nature of electromagnetic radiation.
(ii) Atomic spectra which can be explained only by assuming quantised electronic energy levels.
2.3.1 Wave Nature of Electromagnetic Radiation
James Maxwell (1870) suggested that when electrically charged particles move under acceleration, alternating
electrical and magnetic fields are produced and transmitted in the form of waves. He called them
electromagnetic waves or electromagnetic radiation.
The oscillating electric and magnetic fields produced by an oscillating charged particle are perpendicular to
each other and to the direction of propagation of the wave. Electromagnetic waves, unlike sound or water
waves, do not require a medium to travel.
Electromagnetic radiations which differ from one another in wave length constitute the electromagnetic spectrum.
Visible light is a form of electromagnetic radiation, as are radio waves, microwaves, and x-rays.

Figure - 02 The electric and magnetic field components of electromagnetic wave.

The electric and magnetic components of electromagnetic wave vibrate in two mutually perpendicular planes
(the frequency, wavelength and amplitude of both components are the same).

35
 Brilliant STUDY CENTRE

Characteristics of electromagnetic radiation

i. Wave length    is the distance between two maxima of either electrical or magnetic components of the
wave (crest to crest or trough to trough). It is usually expressed in meters.
ii. Frequency ( ) is the number of waves that pass any point in unit time. It is expressed in s–1 or Hz.
iii. Velocity (c) is the distance travelled in one second. c is a constant for electromagnetic waves.
c   c/T
or
1
c = 2.997925 m s-1  3  10 ms
8

iv. Amplitude (a) of a wave is the height of the crest or the depth of the trough.
v. Wave number    is the reciprocal of wavelength. It is expressed in m–1.
  1/ 
2.3.2 Particle Nature of Electromagnetic Radiation - Planck’s Quantum Theory
The electromagnetic theory of radiation could explain experimental observations such as diffraction and
interference of light. However the following observations could not be explained on the basis of this theory.
i. Black body radiation - the nature of emission of radiation from hot bodies.
ii. Photo electric effect - the ejection of electrons from a metal surface when radiation strikes it.
iii. Variation of heat capacity of solids as a function of temperature.
iv. Line spectra of atoms (hydrogen).
These phenomena could be explained only if electromagnetic waves also possess particle nature.
Planck’s Quantum Theory. In order to explain these phenomena, German physicist, Max Plank (1901)
put forward his theory, known as Planck’s Quantum Theory, which regards electromagnetic radiations as
particles. The theory was further extended by Einstein (1905). The main postulates are:
i. Radiant energy is emitted or absorbed by atoms or molecules discontinuously, in the form of small packets
called quanta. In the case of light, these energy packets are called photons.
ii. The energy of each quantum is directly proportional to the frequency of the radiation.
hc
E  h or E 

where h is Planck’s constant (6.626×10-34 Js or 3.99×10-13 kJ sec mol-1), c is the velocity of light
(3.0  108 ms 1 ) , is the wavelength, and  is the frequency..
iii. The total energy absorbed or emitted is some whole number multiple of the quantum, E  nh , where n
is an integer.
2.3.3 Black body Radiation
When solids are heated, they emit radiations over a wide range of wavelengths. The ideal body which emits
and absorbs radiations of all frequencies is called a black body and the radiations emitted by it is called black
body radiation.
The frequency distribution of the emitted radiation from a black body depends only on its temperature. At a
given temperature, intensity of emitted radiation increases with decrease of wavelength (), reaches a maximum
value and then decreases with further decrease of wavelength.
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 [Study Package - Chemistry - Part- I]

Figure - 03 Wave length - intensity relationship.

For example, when an iron rod is heated, its colour changes from red to yellow and finally begins to glow with
white light and then blue light. In terms of frequency (), it means that the radiation emitted goes from a lower
frequency to a higher frequency (or higher to lower wavelength) as the temperature increases.
  1/ T or   T
At low temperature the spectrum is having low energy radiations in the infra-red region. As the temperature
increases, the radiations emitted becomes shorter in wavelength (higher frequency).
Failure of wave theory to explain black body radiations. Wave theory could not explain the colour
change. Radiation of only one wavelength could be possibly given out by a black body irrespective of its
temperature and one colour should be there.
Explanation of black body radiation by quantum theory. With the quantum theory, Planck was able to
explain the distribution of intensity of black body radiation as a function of frequency or wavelength at different
temperatures. When we heat a black body/iron rod its colour changes from red (longer wavelength) to yellow
(shorter wavelength) to white (still shorter wavelength) and finally becomes blue at very high temperature
(shortest wavelegth).
2.3.4 Photoelectric effect
In 1887, Heinrich Hertz observed that electrons were ejected when certain metals like potassium, caesium,
rubidium, etc., were exposed to a beam of light. This phenomenon is called photoelectric effect.The emitted
electrons are called photoelectrons.
Metals such as sodium, potassium, etc., show photoelectric effect with visible light, whereas metals like zinc,
magnesium, cadmium, etc., show photoelectric effect with ultra violet light.
(i) The electrons are ejected from the surface as soon as the beam of light (radiation) strikes the surface.
(ii)The number of electrons ejected is proportional to the intensity of light. (iii) For each metal, there is a
characteristic minimum frequency,  0 , called threshold frequency, below which photoelectric effect is not
observed. (iv) The kinetic energy of the emitted electron increases with increase in frequency of the light. E.g.,
red light [frequency (4.3-4.6)×1014 Hz] of any brightness could not eject electrons from potassium, but
yellow light [frequency (5.1-5.2)×1014 Hz] of any brightness can (threshold frequency = 5.0×1014 Hz).
Failure of wave theory to explain photoelectric effect. The above results could not be explained on
the basis of the classical theory, according to which the number of electrons and the kinetic energy associated
with them should depend on the brightness of the light (energy of a beam of light depends on its brightness).
Explanation of black body radiation by quantum theory. Einstein (1905) explained photoelectric effect
using Planck’s quatum theory. When a photon of sufficient energy strikes an electron in an atom of the metal,
it transfers its energy instantaneously to the electron and the electron is ejected without any time lag.
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 Brilliant STUDY CENTRE

The kinetic energy of the electron is proportional to the frequency of the electromagnetic radiation. The
minimum energy required to eject an electron, h 0 is called the work function, W0.
The kinetic energy of the ejected photoelectron is given by
1
h  h 0  KE or hv  h 0  m e V 2 or KE  h  W0
2
where me is the mass of the electron, V is the velocity of the ejected electron, and W0 is the work function.
This equation is known as Einstein photoelectric equation.
W0 is a constant for a particular metal and is equal to the ionisation energy of gaseous atoms of that metal.
A more intense beam of light consists of a larger number of photons, therefore, the number of electrons
ejected is also larger.
Conclusions from photoelectric effect. (i) Photoelectric effect demostrates particle nature of light. (ii) A
photon is a quantum of energy with rest mass equal to zero. (iii) There is no effect for frequency of incident
light on the number of emitted photoelectrons. (iv) There is no effect for intensity of incident light on the kinetic
energy of the emitted photoelectrons.
2.4 DUAL BEHAVIOUR OF ELECTROMAGNETIC RADIATION
Particle nature of light could explain black body radiation and photoelectric effect but failed to explain the
phenomena of interference and diffraction. Thus, the idea that light possesses both particle and wave-like
properties was accepted. When radiation interacts with matter, it displays particle-like properties in contrast
to wave -like properties which it exhibits when it propagates.
It turns out that some microscopic particles like electrons also exhibit particle-wave duality.
2.4.1 Evidence for quantised electronic energy levels - Atomic Spectra
The speed of light depends on the nature of the medium through which it passes. As a result a beam of light is
deviated (refracted) from its original path as it passes from one medium to another. When a ray of white light
is passed through a prism it spreads out into a series of coloured bands called spectrum. The study of spectra
is referred to as spectroscopy.
Types of Spectra
1. Continuous Spectrum. In a continuous spectrum, radiations corresponding to all wavelengths are present.
Ordinary white light is composed of all the seven colours merging into one another, violet (7.50  1014 Hz) to
red (4.0  1014Hz).
2. Discontinuous Spectrum. In a discontinuous spectrum, certain wavelengths may be missing because of
absorption by gases. Discontinuous spectrum are of two types, absorption spectrum and emission spectrum
i. Absorption spectrum. A continuum of radiation is passed through a sample which absorbs radiation of
certain wave lengths. The missing wavelengths corresponds to radiation absorbed, leaves dark spaces in the
continuous spectrum. Atoms produce line absorption spectra while molecules give band absorption spectra.
ii. Emission spectrum. When light is emitted by atoms or molecules, emission spectrum is obtained. To
produce an emission spectrum, energy is supplied to a sample of atoms, molecules or ions by heating or
irradiating it. The wavelength or frequency of the emitted radiation, as the sample gives up the absorbed
energy, is measured.

(a) (b)

Figure - 04 a) Atomic emission spectrum, b) Atomic absorption spectrum.

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a) Line emission spectrum. Atomic spectrum consists of distinct lines produced by an excited atom. German
chemist Robert Bunsen (1811-1899) was the first to use line spectra to identify elements.
b) Band emission spectrum. It consists of discrete bands. It is given by molecules, therefore, it is also called
molecular spectrum. The bands are composed of closely packed lines separated by dark spaces.
Elements like rubidium (Rb), Ceacium (Cs), thallium (Tl), indium (In), gallium (Ga) and scandium (Sc) were
discovered by spectroscopic methods.
2.5 LINE SPECTRUM OF HYDROGEN
When an electric discharge is passed through gaseous hydrogen, H2 molecules dissociate to produce
energetically excited H-atoms, which emit electromagnetic radiation of discrete frequencies. The hydrogen
spectrum consists of many series of lines, named after their discoverers.
Balmer (1885) showed that the visible lines of hydrogen spectrum obey the formula (Balmer formula)

 1 1 
  109, 677  2  2  cm 1
2 n 
where n is an integer greater than 2 (i.e., 3, 4, 5.....).

Figure - 05 Various spectral series in hydrogen spectrum.

Rydberg Formula. Swedish scientist, Johannes Rydberg observed that all series of lines in the hydrogen
spectrum could be described by

 1 1 
  109, 677  2  2  cm1
 n1 n 2 
where n1= 1, 2, 3.... and n2= (n1+ 1, n1+ 2), (n1+ 3).......
The value, 109, 677 cm–1 is called Rydberg constant (RH) for hydrogen.
The first series of lines for hydrogen that correspond to n =1, 2, 3, 4, 5 and 6 are known as Lyman, Balmer,
Paschen, Brackett, Pfund and Humphry series respectively.
Hydrogen has the simplest line spectrum. Line spectrum becomes complex in the case of heavier elements.

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Table - 02 Spectral series in atomic spectrum of hydrogen.

Series n1 n2 Region of spectrum Wavelength of lines (nm)
Layman 1 2,3,4, ... Ultra-violet 121.6, 102.6, 97.3, 95, 93.8, ...
Balmer 2 3,4,5, ... Visible 656.3, 486.1, 434.0, 410.2, 397.0, ...
Paschen 3 4,5,6, ... Infrared 1875.1, 1281.8, 1093.8, 1005.0, ...
Brackett 4 5,6,7, ... Infrared 4050.0, 2630.0, 2166.0, 1945.0, ...
Pfund 5 6,7,8, ... Infrared 7451.0, 4652.0, 3740.0, ...
Humphry 6 7,8,9, ... Far-infrared

 1 1 
For hydrogen-like particles, the formula is   R  2  2  Z2 , where Z is its atomic number..
 n1 n 2 
2.6 BOHR’S MODEL OF THE HYDROGEN ATOM
In order to overcome the drawbacks of the Rutherford model, Danish physicist, Niels Bohr (1913) proposed
the quantum mechanical model of the atom. The model is based on quantum theroy of radiation and classical
laws of physics. Bohr model is based on the following postulates:
1. The atom consists of a small, heavy, positively charged nucleus. The electron in the hydrogen revolves
around the nucleus in circular paths of fixed radius and energy. These are called orbits or stationary
states or allowed energy states. These orbits are arranged concentrically around the nucleus.
2. The energy of an electron in an orbit does not change with time. However, an electron in a lower orbit can
move to a higher orbit by absorbing a definite quantity of energy. Energy is emitted when an electron
moves from a higher to a lower stationary state. The energy change does not take place in a continuous
manner, but in discrete units called quanta.
3. The frequency of radiation absorbed/emitted when transition occurs between stationary states is given by
Bohr’s frequency rule
E E 2  E1
 
h h
where E1 and E2 are the energies of the lower and the higher energy states and ‘h’ is Planck’s constant.
4. The angular momentum of electron in a stationary state is given by
h
mvr  n
2
where m is the mass and v is the velocity of electron, r is radius of the orbit and n is the number of the orbit.
Thus, the electrons can move only in orbits for which the angular momentum is an integral multiple of ,
i.e., only fixed orbits are allowed.
2.6.1 Important deductions from Bohr Theory
1. The stationary states for electron, n = 1, 2, 3, .... are known as Principal quantum numbers.
n 2h 2
2. The radius of the nth stationary state, rn = 2 2 or rn = a 0n 2 or rn =52.9(n2) pm
4π me
where a0 = 52.9 pm is the radius of the first orbit of hydrogen, called Bohr radius.

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2π 2 me 4  1 
3. The energy of the electron in the nth stationary state, E n = 2 2 or E n  R H  2  J
n h n 
where RH is Rydberg constant ( 2.18  1018 J atom-1 = 1312 kJ mol-1).

1
The energy of the electron in the first orbit of H-atom (n=1), E1  2.18  10 18  2   2.18  10 18 J .
1 
18  1  18
Energy of the electron in the second orbit of H-atom (n=2), E 2  2.18  10  2   0.545  10 J .
2 
The depiction of energies of the various stationary states of H-atom is called its energy level diagram.
vo Z vo
4. Velocity of the electron in the nth orbit, v n  and v n  for hydrogen.
n n
v0 is the velocity of the electron in the first orbit of hydrogen (2.188 ×106 m s-1). Velocity increases with
increase of positive charge on the nucleus and decreases with increase of principal quantum number.
v
5. Number of revolutions of the electron in the nth orbit of hydrogen  n .
2rn
6. Bohr theory can be applied to one electron ions like He+, Li2+, Be3+, etc.

n 2h 2 a 0n 2 52.9(n 2 )
i. Radius of the nth orbit, rn = or rn = pm or rn =
4π 2 Zme 2 Z Z
where Z is the nuclear charge of the hydrogen-like ion.

2π2 Z2me4
ii. Energy of the electron in the nth orbit, En =
n 2h 2

18  Z 
2
1312Z2
or En  2.18 10  2  J   kJ mol1
n  n2
vo Z
iii. Velocity of the electron in the nth orbit, v n 
n
2.6.2 Explanation of Line Spectrum of Hydrogen
Under normal conditions, the electron in hydrogen atom resides in the first orbit. It moves to higher energy
levels depending on the amount of energy absorbed. When the electron returns to any of the lower energy
shells, energy is emitted as radiation of a definite frequency. This is observed as a line in the emission spectrum
of the hydrogen atom. The difference in energy between two orbits Ei and Ef ,

 R   R   1 1   1 1 
E  E f  Ei    2H     2H   R H  2  2   2.18 1018  2  2  J
 n f   ni   ni nf   ni n f 
The frequency of the absorbed/emitted radiation (in s-1 or Hz),

1   2.18 10 J  1  1   3.29  1015  1  1  Hz

18
E R H  1
        2 2 
h h  n i2 n f2  6.626 1034 Js  n i2 n f2   ni nf 
and in terms of wave number,
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 RH  1 1  3.29  1015 s 1  1 1   1 1 
   2 2  1  2
 2   1.09677  107  2  2  m -1
c hc  n i n f  3  10 ms  n i n f 
8
 ni nf 
In case of absorption spectrum, nf > ni, the term in parenthesis is positive (energy is absorbed), wheras in case
of emission spectrum ni > nf , E is negative (energy is released).
Each spectral line is associated with a particular transition. The brightness or intensity of spectral lines depends
on the number of photons with same wavelength/frequency.
2.6.3 Limitations of Bohr Model
1. Bohr model could not explain the spectra of multi-electron atoms.
2. Angular momentum of the revolving electron being nh/2 has not been explained.
3. On using a high resolution spectroscope, it is observed that a line in the hydrogen spectrum is a collection
of several lines lying very close to each other. This is known as fine spectrum. Bohr’s theory could not
explain the fine spectra of hydrogen.
4. Bohr theory could not explain why spectral lines split into a group of lines in a magnetic field (Zeeman
effect) and electric field (Stark effect).
5. According to Bohr, an electron follows a fixed path (circular orbit). If this is true, then position and
velocity can be determined simultaneously. This is not in agreement with Haisenberg’s uncertainty principle.
6. Bohr theory failed to explain the directional nature of the covalent bond (shape of covalent molecules).
2.7 TOWARDS QUANTUM MECHANICAL MODEL OF THE ATOM
Two important developments that led to the quantum mechanical model of the atom are (i) Dual behaviour of
matter and (ii) Heisenberg’s uncertainty principle.
2.7.1 Dual Behaviour of Matter (de Broglie Equation)
In 1924, de Broglie proposed that the electron, like light, behaves both as particle and wave. He derived an
expression for calculating the wavelength of the electron wave using Planck’s equation, and Enstein’s mass-
energy relationship.
h h
 
mv p
Momentum (p) of the moving electron is inversely proportional to its wavelength.
Let kinetic energy of the particle of mass ‘m’ be E, then
1
m v 2 or 2Em  m v or 2Em  mv  p (momentum)
2 2
E 
2
h h
 
p 2Em
de Broglie’s prediction was confirmed when it was found that an electron beam undergoes diffraction, a
phenomenon characteristic of waves. This is the principle behind electron microscope, which can achieve a
magnification of about 15 million times.
Every object in motion has wave character. The wavelengths associated with ordinary objects are so short
(due to large masses) that their wave property cannot be detected, but the wavelengths associated with
electrons and other subatomic particles, with very small mass, can can be detected.
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2.7.2 Heisenberg’s Uncertainty Principle

In 1927, Werner Heisenberg, a German physicist stated the uncertainty principle which is the consequence of
the dual nature of matter and radiation. It is impossible to determine simultaneously, the exact position
and momentum (or velocity) of a microscopic particle with absolute accuracy or certainty. If an
attempt is made to measure any one of these quantities with higher accuracy, the other becomes less accurate.
h h h
x  p  or x   (mv)  or x  v 
4 4 4m
where  x is the uncertainty in position,  p is the uncertainty in momentum and h is Planck’s constant.
 x  v is the uncertainty product. When  x = 0,  v =  and vice versa.
In terms of uncertainty in energy, E and uncertainty in time, t, the equation can be written as
h
E  t 
4
In terms of uncertainty in wavelength and uncertainty in momentum, the equation may be written as
h  h d  h 
 
p2
 p    p  dp  p 2 
 
Explanation of Heisenberg’s Uncertainty Principle. In order to measure the position and momentum of
an electron, we use light. The photon of light strikes the electron and the reflected photon is seen in the
microscope. As a result of the hit, both the position and the velocity of the electron are disturbed. Heisenberg’s
uncertainty principle does not apply to macroscopic particles.
Significance of Uncertainty Principle. The uncertainty principle rules out the concept of definite paths of
electrons as proposed by Bohr. It is only possible to predict the probable region in space where the electron
can be found. Thus, Heisenberg replaced the concept of definite orbits by the concept of probability.
Uncertainty Principle in daily life. Uncertainty principle has no significance in daily life. In case of objects
of ordinary size, the uncertainties in position and momentum are negligible. For example, the product of
uncertainty of a body of mass 1mg is negligibly small
h 6.626 1034 kg m 2s 1
x  v   6
 1028 m 2s 1
4m 4  3.14 10 kg
but for a minute particle like the electron, the product of uncertainty is not negligible
h 6.626 1034 kg m 2 s 1
xv    104 m 2s1
4m 4  3.14  (9.111031 kg)
If uncertainty in position of the electron is 10-8 m, the uncertainty in velocity is quite large.

104 m2s1
v  8
 10 m s1
10 m
Therefore, it is not possible to predict the trajectory of the electron in the atom.
Reasons for the Failure of the Bohr Model. (i) The wave character of the electron is not considered. (ii)
Bohr model regard electrons as charged particles moving in fixed circular orbits around the nucleus. This is
not possible according to Heisenberg uncertainty principle.

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2.8 QUANTUM MECHANICAL MODEL OF ATOM

Quantum mechanics (Werner Heisenberg and Erwin Schrdinger, 1926) deals with the study of the motion of
microscopic objects that have both wave-like and particle-like properties. In this model, discrete energy
levels or orbits proposed by Bohr are replaced by the mathematical function, (psi) which is related to
probability of finding electrons around the nucleus. The Schrdinger equation for a system, atom or molecule,
whose energy does not change with time,

H   E

where H is a mathematical operator called Hamiltonian and E represent the quantised energy of the system.
Solution of this equation gives the values of E and .
Hydrogen Atom and the Schrdinger Equation. When Schrdinger equation is solved for the hydrogen
atom, several wave functions called  corresponding to energy E1, E2, E3 ... are obtained. Each of
these wave functions is called an atomic orbital.
2.8.1 Important Features of Quantum Mechanical Model
1. The energy of electrons in an atom is quantized, i.e., they can have only specific values.
2. All the information about the electron in a given energy level is contained in the wave function .
3. Since it is not possible to determine the exact position and velocity of the electron in an atom, the concept
of probability of finding the electron at a point is introduced.
6. The wave function, for an electron corresponds to the atomic orbital. Each orbital has a definite energy
and can accommodate two electrons.
4. The probability of finding an electron in an atom is proportional to  and is known as probability
density. It always has a positive value and the value at different points in an atom corresponds to the
region around the nucleus where the electron is most likely to be found.
5. The wave function and its square,  have values for all locations about the nucleus, i.e., the probability
of finding the electron is maximum near the nucleus and decreases as the distance from nucleus increases.
7. In multi-electron atoms, the electrons are filled in various orbitals in increasing order of energy.
8. For a given type of atom, each orbital may be described uniquely by a set of three quantum numbers, n
(principal quantum number), l (azimuthal quantum number) and ml (magnetic quantum number).
2.8.2 Orbitals and Quantum Numbers.
The various shells and sub-shells in an atom are distinguished by their size, shape and orientation in space.
These are expressed in terms of certain numbers called quantum numbers.
1. Principal Quantum Number (n)
i) Principal quantum number n represents the major energy level to which an electron belongs.
ii) n can have values 1, 2, 3, 4..... (n = 1, K shell; n = 2, L shell; n = 3, M shell; n = 4, N shell, etc.).
iii) The number of electrons that can be accomodated in any shell is 2n2.
iv) Greater the value of n, larger the distance from the nucleus; r = (0.529 n 2 / Z)Å .
v) As the value of n increases, the energy of the electron increases.
2. Azimuthal (Subsidiary or Angular momentum) Quantum Number (l)
i) The values of l signify the shape and energy of the sub-shells or sub-levels in a major energy shell.
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ii) For a given principal quantum number n, l can have n values from 0 to (n – 1).
iii) The shapes of the sub-shells are as follows:
l = 0, s-sub-shell spherical shape
l = 1, p-sub-shell dumb-bell shape
l = 2, d-sub-shell double-dumb bell (clover-leaf shape, etc.)
l = 3, f-sub-shell complicated shape (double-clover leaf shape, etc.)
(The letters s, p, d, f designate old spectral terms : sharp , principal, diffuse, fundamental.)
iv) The energy of the sub-shells of a shell is in the order: s < p < d < f < g.
v) The number of electrons in a particular sub-shell is equal to 2(2l +1).
3. Magnetic Quantum Number (ml)
i) Magnetic quantum number describes the orientation of orbitals in space.
ii) It is denoted by ml, an integer; the values of ml range from -l through 0 to +l.
iii) Total number of values of ml for a given value of n = n2.
iv) Total values of ml for a given value of l = (2l+1)
v) The values of ml signify the possible number of orientations of a sub-shell.
vi) In the absence of magnetic field, the three p-orbitals are equivalent in energy and are said to be threefold
degenerate (orbitals having same energy are known as degenerate orbitals).
Table - 03 Quantum numbers and types of orbitals.

n (l ) (m l ) Orbitals
1 0 (1s) 0 1s
2 0 (2s) 0 2s
2 1 (2p) -1, 0, +1 2px, 2py , 2pz
3 0 (3s) 0 3s
3 1 (3p) -1, 0, +1 3px, 3py , 3pz
3 2 (3d) -2, -1, 0, +1, +2 3dxz, 3dyz, 3dz², 3dxy , 3dx²-y ²
4 0 (4s) 0 4s
4 1 (4p) -1, 0, +1 4px, 4py , 4pz
4 2 (4d) -2, -1, 0, +1, +2 4dxz, 4dyz, 4dz², 4dxy , 4dx²-y ²
4 3 (4f) -3, -2, -1, 0 1, 2, 3 4f x³, 4f y ³, 4f z³, 4fxyz, 4f x(y ²-z²), 4f y (z²-x²), 4fz(x²-y ²)

4. Spin Quantum Number (s or ms). The existence of mutiple structure (doublet, i.e., two lines closely
spaced) led to the introduction of the spin quantum number (G..Uhlenback, S. Gouldsmit).
i) For every value of ml, s has two values, (+½) and (-½) indicating opposite spins. These are called two
spin states of the electron represented by arrows, (spin up) and (spin down).
ii) The values of s signify the direction of rotation or spin of an electron in its axis during its motion.

11  3
iii) Angular momentum of a spinning electron  s  s  1     1   
22  2

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2.8.3 Shapes of Atomic Orbitals

For different orbitals, the plots of wave function as a function of distance from the nucleus (r) are different. For
1s radial wave function,  is positive everywhere (the probability density is maximum at the nucleus and
decreases sharply with distance), but for 2s orbital, it is positive in some regions and negative in others. The
point where the wave function changes sign is called nodal surface or node. In general, the ns-orbital has (n
-1) nodes, i.e., number of nodes increases with increase in principal quantum number n.

(a)

Figure - 06 Plots of orbital wave function  (r) for 1s 2s and 2p orbitals against (r)

(b)

Figure - 07 Plots of radial probability density ψ 2 (r) for 1s, 2s and 2p orbitals against (r)
The variation of probability density as a function of (r) for various orbitals is also different (German physicist,
Max Born). In these graphs,  2 is positive throughout. In case of 1s and 2s orbitals, the probability of finding
the electron in the nucleus is not known, but there is a probability of finding electron just outside the nucleus as
s-orbital lies very close to the nucleus. For 2p orbital, probability is zero at r = 0.
Charge Cloud Picture of Orbitals. In this representation, the probability of finding the electron in a particular
region of space is directly proportional to the density of dots in that region.

(a) (b)

Figure - 08 (a) Probability density plot (b) Boundary surface diagram

Boundary Surface Diagrams. In this representation, the boundary surface is drawn encompassing regions
of maximum probability (about 90%) of finding the electron ( ψ 2). A boundary surface diagram enclosing
100% probability is not possible since probability is has some value even at infinite distance.
i. Shape of s-orbitals. s-orbitals do not have directional dependence; they are spherically symmetrical. Their
size and energy increases with increase in the value of n. The probability of finding the 1s electron is maximum
near the nucleus and decreases as the distance from the nucleus increases.
1s- orbital has no node. 2s- orbital has one node, 3s- orbital has two nodes and so on. In general, ns orbital
has (n-1) nodes, i.e., number of nodes increases with increase in principal quantum number.
All orbitals with l  0 have angular dependence. Thus, p, d, f and other higher angular momentum orbitals
are not spherically symmetrical.

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a) (b)

Figure - 09 Boundary surface diagrams of (a) s-orbitals (b) p-orbitals.

ii. Shape of p-orbitals: A p-orbital consists of two lobes that form a dumb-bell shaped structure.
The three p-orbitals along x, y, z-axes named px, py, pz orbitals are perpendicular to each other.
All the three p-orbitals of a sub-shell have the same size and shape but differ from each other only in orientation.
Such orbitals of a sub-shell having same energy are referred to as degenerate orbitals. The subscripts x, y,
and z indicate the axes along which the orbitals are oriented and possess maximum electron density.
The lobes of a p-orbital are separated from each other by a point of zero electron density called nodal plane.
2p-orbital has no nodal surface. p-orbitals of higher energy levels are also dumb-bell shaped and have nodal
surfaces. The number of nodal surfaces in any np orbital is (n-2).
iii. Shape of d-orbitals: The five d- orbitals are designated as dxy, dyz, dxz, dx²-y² and dz2. The dxy, dxz and dyz
orbitals lie in the x-y, x-z, and y-z planes, respectively. The dx²-y² orbital has its lobes pointing along the x and
y axes. These four orbitals have double dumb-bell or clover leaf structure (four lobes each). The dz2 orbital
lie along the z-axis. It is dumb-bell shaped with a doughnut around its middle.

Figure - 10 Boundary surface diagrams of d-orbitals.

2.8.4 Nodes. A node is a plane (nodal plane) or point (nodal point) where the electron density is zero ( ψ 2  0 ).
There are two types of nodes; radial node (spherical node) and angular node.
(a) Radial Node (Spherical Node) : For a radial node, radial probability function is zero.

Figure - 11 Nodes in s-orbitals; zero node for 1s, one node for 2s and three nodes for 3s.
(b) Angular Node: For an angular node, the probability density function becomes zero at certain planes
passing through the nucleus (origin). Number of angular nodes is given by ’l’(azimuthal quantum number.).
i. s-orbital (l = 0) do not have any nodal plane
ii. p-orbital (l = 1) has one nodal plane,e.g., nodal plane for pz is the XY plane.

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iii. d-orbital (l = 2) has one nodal planes, e.g., dxy orbital, has two nodal planes passing through the origin and
bisecting the xy plane containing z-axis. he total number of nodes are given by (n1), i.e., sum of l angular
nodes and (n-l- 1) radial nodes.
2.8.5 Energy Level Diagrams
A diagram representing relative energies of orbitals in an atom is called energy level diagram.
(a) Energy Level Diagram of Hydrogen Atom. In hydrogen and hydrogen-like atoms, all the orbitals of a
particular energy level have the same energy (degenerate). The principal quantum number solely determines
the energy of the electron. Angular momentum quantum number l merely determines the shape of the orbital.
The energy of orbitals in hydrogen and hydrogen like species increases as follows: 1s < 2s = 2p < 3s = 3p =
3d < 4s = 4p = 4d =4f < etc.

(a) (b)

Figure - 12 Energy level diagram of (a) hydrogen atom (b) multielectron atom.
(b) Energy Level Diagram of Multielectron Atoms. In a multi-electron atom, the energies of the various
orbitals depend not only on the nuclear charge but also on the other electrons present. The attraction between
the electron and the nucleus as well as repulsive interactions between the electrons, i.e., the effective nuclear
charge experienced by the electron, Zeff should be considered.
Zeff = Z - σ
where Z is the atomic number and σ is the screening constant.
Thus, different sub-shells of a particular shell have different energies; the sub-shell with higher value of l has
higher energy. The increasing order of energies is s < p < d < f.
As the value of n increases, some sub-shells of a lower energy level may have higher energy than some sub-
shells of a higher energy level, e.g., energy of 3d sub-shell is higher than 4s.
Bohr-Bury’s (n + l) rule: The increasing order of energies of the various sub-shells in a multielectron atom
can be obtained by Bohr-Bury’s (n + l) rule. According to this rule:
i) In neutral atoms, a sub-shell with lower value of (n + l) has lower energy. For example, 4s orbital has
lower energy than 3d orbital.
For 4s orbital n = 4 and l = 0. Hence, n + l = 4 + 0 = 4.
For 3d orbital, n = 3 and l = 2. Hence, n + l = 3 + 2 = 5.
ii) If two sub-shells have the same (n + l) value, the sub-shell with lower value of n has lower energy.
For 4s orbital, n = 4 and l = 0. Hence, n + l = 4 + 0 = 4
For 3p orbital, n = 3 and l = 1. Hence, n + l = 3 + 1 = 4

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Thus, 3p orbital has lower energy than 4s orbital since it has lower value of n.
The increasing order of energies of the various sub-shells of a multielectron atom is
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p .....
2.9 RULES FOR FILLING OF ORBITALS IN AN ATOM
The filling of orbitals in the ground state of an atom is governed by the aufbau principle which is based on
Pauli’s exclusion principle, the Hund’s Rule of maximum multiplicity and the relative energies of the orbitals.
2.9.1 Aufbau Principle
Aufbau (aufbau in German means building up) principle gives a sequence in which various sub-shells are
filled up depending on the relative order of the energy of the sub-shells.
In the ground state of the atom, orbitals are filled in the order of increasing energies.
The sub-shell of lowest energy is filled up first, then the next sub-shell of higher energy and so on. The
sequence in which various sub-shells of an atom are filled is as follows:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f, 5d, 6p, 7s, ...

Figure - 13 Sequence of filling various energy levels.

Exceptions to aufbau Principle. In the case of some elements, the actual electronic arrangement is slightly
different from the arrangement expected on the basis of Aufbau principle. The reason behind this is that half-
filled and fully filled sub-shells have extra stability.
24
Cr 1s2, 2s2, 2p6, 3s2, 3p6, 3d4, 4s2 (expected) ; 1s2, 2s2, 2p6, 3s2, 3p6, 3d5, 4s1 (actual)
29
Cu 1s2, 2s2, 2p6, 3s2, 3p6, 3d9, 4s2 (expected) ; 1s2, 2s2, 2p6, 3s2 3p6, 3d10, 4s1 (actual)
2.9.2 Pauli’s exclusion principle
Pauli’s exclusion principle (Wolfgang Pauli, 1925) states that no two electrons in an atom can have the
same values for all the four quantum numbers. An orbital cannot have more than two electrons and if two
electrons are present, their spins should be opposite (+1/2 and -1/2). E.g., in the case of helium, quantum
numbers n, l and m of the two electrons are the same but their spin quantum numbers are different.
2.9.3 Hund’s rule of Maximum Multiciplicity
Hund’s rule (Friedrich Hund, 1925) states that electrons are distributed among the orbitals of a sub-
shell in such a way as to give the maximum number of unpaired electrons with parallel spins. This
means that the degenerate orbitals in a sub-shell are first filled singly before they begin to pair up.
Pairing of electrons occurs with the fourth electron in the p-orbitals, the sixth electron in the d-orbitals and the
eighth electron in the f-orbitals. The electrons thus occupy different orbitals of the sub-shell so as to minimize

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inter-electronic repulsion and increase the stability of the atom.

Half-filled and compeletly filled sub-shells are more stable on account of symmetry. E.g., the electronic
configuration of nitrogen is 1s2, 2s2, 2px1, 2py1,2pz1 and not 1s2, 2s2, 2px2, 2py1,2pz0.
2.9.4 Electronic Configuration of Elements
The distribution of electrons in the orbitals of an atom is called its electronic configuration. The electronic
configuration of different atoms can be represented in two ways:
(i) sa pb dc ...... notation. The subshell is represented by the respective letter symbol and the number of
electrons present in the subshell by the superscript ( a, b, c, ..., etc). Similar subshells in different shells is
differentiated by giving the principal quantum number before the respective subshell. For example, electronic
configuration of carbon is 1s2, 2s2, 2px1, 2py1, 2pz0.
(ii) Orbital diagram. Each orbital of the subshell is represented by a box and the electron is represented by
a upward arrow (  ) for positive spin or a downward arrow (  ) for negative spin. The advantage of of this
type of notation is that it represents all the four quantum numbers.

Electronic configuration and orbital diagram of elements (also illustrating Hund’s rule).

Hydrogen 1H 1s1

Helium 2He 1s2

Lithium 3Li 1s2 2s2 2px0 2py0 2pz0

Beryllium 4Be 1s2 2s2 2px1 2py0 2pz0

Boron 5B 1s2 2s2 2px1 2py0 2pz0

Carbon 6 C 1s2 2s2 2px1 2py1 2pz0

Nitrogen 7 N 1s2 2s2 2px1 2py1 2pz1

Oxygen 8 O 1s2 2s2 2px2 2py1 2pz1

Fluorine 9 F 1s2 2s2 2px2 2py2 2pz1

Neon 10Ne 1s2 2s2 2px2 2py2 2pz2

2.9.5 Stability of Completely Filled and Half-filled Sub-shells
The ground state electronic configuration of an element corresponds to the state of lowest electronic energy.
However, in elements like Cu and Cr, where the two subshells (4s and 3d) differ only slightly in their energies,

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an electron shifts from a subshell of lower energy (4s) to a subshell of higher energy (3d), provided such a shift
results in all orbitals in the subshell of higher energy becoming either half-filled or completely filled.
For example, the valence electronic configurations of Cr and Cu are 3d5 4s1 and 3d10 4s1 respectively (not
3d4 4s2 and 3d9 4s2) because there is extra stability associated with these electronic configurations.
Causes of stability of completely filled and half-filled sub-shells
1. Symmetrical Distribution of Electrons. Symmetry leads to stability. Completely filled or half filled
subshells have symmetrical distribution of electrons in them and, are therefore, more stable.

(less symmetrical); (more symmetrical)

(less symmetrical); (more symmetrical)

Electrons in the same subshell (3d) have equal energy but different spatial distribution; their shielding of one
another is relatively small and the electrons are more strongly attracted by the nucleus.
2. Exchange Energy. When two or more electrons with same spin are present in degenerate orbitals, they
exchange their positions. The energy released due to this exchange is called exchange energy.
The maximum number of exchanges occur when the subshell is either half-filled or completely filled. As a
result, exchange energy and stability are maximum. Exchange energy is the basis of Hund’s rule of maximum
multiplicity.

Figure - 14 The number of possible exchanges for d3 configuration is 6 and for d5 configuration is 10
The extra stability of half-filled and competely filled subshells is due to (i) relatively small shielding, (ii) smaller
coulombic repulsion energy, and (iii) large exchange energy.

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QUESTIONS
LEVEL - I
1. The absolute value of charge on the electron was determined by:
1) J.J. Thomson 2) R.A. Millikan 3) Rutherford 4) Chadwick
2. Rutherford’s experiment on scattering of   particles showed for the first time that the atom has:
1) electrons 2) protons 3) neutrons 4) nucleus
3. The ratio of the volume of the atom and the volume of the nucleus is:
1) 1010 2) 1012 3) 1015 4) 1020
4. The triad of nuclei that is isotonic is:
12 14
1) 14
6 C, 14
7 N, 19
9 F
2) 6 C, 7 N, 19
9 F 3) 14
6 C, 14
7 N, 17
9 F
4) 14
6 C, 15
7 N, 17
9 F

5. Which of the following does not characterise X-rays?

1) The radiations can ionise gases 2) It causes ZnS to fluorescence
3) Deflected by electric and magnetic fields 4) Have wavelengths shorter than ultraviolet rays
6. A 500W mercury lamp emits monochromatic radiation of wavelength 331.3 nm. How many photons are
emitted from the lamp per second? (h = 6.626 × 10–34 J-s, velocity of light = 3 × 108 ms–1)
1) 8.35 × 1019 2) 8.35 × 1021 3) 8.35 × 1020 4) 8.35 × 1023
7. Which one of the following is not the characteristic of Planck’s quantum theory of radiation?
1) The energy is not absorbed or emitted in whole number multiple of quantum.
2) Radiation is associated with energy.
3) Radiation energy is not emitted or absorbed continuously but in the form of small packets called quanta.
4) The magnitude of energy associated with a quantum is proportional to the frequency.
8. Light of frequency 6 × 1014 Hz is incident on a metal whose work function is 2eV (h = 6.63 × 10–34 Js,
leV = 1.6 × 10–19 J). The maximum energy of electrons emitted will be:
1) 2.49 eV 2) 4.49 eV 3) 0.49 eV 4) 5.49 eV
9. Which of the following statements is incorrect?
1) The frequency of radiation is inversely proportional to its wavelength.
2) Energy of radiation increases with increase in frequency.
3) Energy of radiation decreases with increase in wavelength.
4) The frequency of radiation is directly proportional to its wavelength.
10. The frequency of one of the lines in Paschen series of hydrogen atom is 2.340 × 1014Hz. The quantum number
n2 which produces this transition is
1) 6 2) 5 3) 4 4) 3
11. In the hydrogen atomic spectrum, the emission of the least energetic photon takes place during the transition
from n = 6 energy level to n = ......... energy level.
1) 1 2) 3 3) 5 4) 4
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12. In Bohr’s stationary orbits:

1) Electrons do not move
2) Electrons move emitting radiations
3) Energy of the electron remains constant
4) Angular momentum of the electron is h/2
13. The distance between 4th and 3rd Bohr orbits of He+ is :
1) 2.645 × 10–10 m 2) 1.322 × 10–10 m 3) 1.851 × 10–10 m 4) 2.33 × 10–10 m
14. Line spectra is characteristic of :
1) molecules 2) atoms 3) radicals 4) both A and B
15. The number of possible spectral lines emitted when electron in n = 4 Bohr orbit reaches to n = 1 Bohr orbit is:
1) 1 2) 2 3) 4 4) 6
16. The highest energy in Balmer series, in the emission spectra of hydrogen is represented by:
(RH = 109737 cm–1)
1) 4389.48 cm–1 2) 2194.74 cm–1 3) 5486.85 cm–1 4) 27419.25 cm–1
17. Calculate the wavelength of electron in an orbit of Be3+ ion having radius equal to the Bohr’s radius (a0).
1) 1.34 Ao 2) 1.53 Ao 3) 1.66 Ao 4) 1.46 Ao
18. The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.
1) 857.6 nm 2) 896.7 nm 3) 845.5 nm 4) 860 nm
19. The de Broglie wavelength of a tennis ball of mass 60g moving with a velocity of 10 m s–1 is approximately:
[Planck’s constant = 6.63 × 10–34 J s]
1) 10–33 m 2) 10–31 m 3) 10–16 m 4) 10–25 m
20. In an atom, an electron is moving with a speed of 600 ms–1 with an accuracy of 0.005%. Certainty with which
the position of the electron can be located is: (h = 6.6 × 10–34 kg m2 s–1, mass of electron me = 9.1 ×10–31 kg)
1) 1.52 × 10–4 m 2) 5.10 × 10–3 m 3) 1.92 × 10–3 m 4) 3.84 × 10–3 m
2
21. The graph between  and r(radial distance) is shown below. This represents :

1) 3s orbital 2) 1s orbital 3) 2p orbital 4) 2s orbital

22. Maximum number of electrons in a subshell with l = 3 and n = 4 is:
1) 14 2) 16 3) 10 4) 12
23. Two electrons occupying the same orbital are distinguished by:
1) Principal quantum number 2) Magnetic quantum number
3) Azimuthal quantum number 4) Spin quantum number

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24. Which one of the following sets of quantum numbers is possible?

1 1
1) n = 3, l = 3, ml = –3. ms = + 2) n = 2, l = 1, ml = 2, ms = –
2 2
1
3) n = 2, l = 0, ml = 0, ms = + 4) n = 1, l = 0, ml = 0, ms = 0
2
25. The number of possible orientations of d orbitals in space is:
1) 2 2) 3 3) 4 4) 5
26. The orbital angular momentum of an electron in a d-orbital is:
h h h 2h
1) 6 2) 2 3) 4)
2 2 2 2
27. The 71st electron of an element X with an atomic number of 71 enters into the orbital:
1) 4f 2) 6p 3) 6s 4) 5d
28. The subshells which are filled just before and just after the filling of 5p subshell are respectively:
1) 5s, 5d 2) 4d, 6s 3) 4d, 4f 4) 6s, 4f
29. The orbital diagram in which both the Pauli’s exclusion priciple and Hund’s rule are violated, is :

1)     2)    

3)     4)   

30. Which of the following rules could explain the presence of three unpaired electrons in N-atom?
1) Hund’s rule 2) Aufbau’s principle
3) Heisenberg’s uncertainty principle 4) Pauli’s exclusion principle
31. Increasing order (lowest first) for the values of e/m (charge/mass) for electron (e), proton (p), neutron (n) and
 - particle (  ) is:
1) e, p, n,  2) n, p, e,  3) n, p,  , e 4) n,  , p, e

LEVEL - II
1. Which of the following pairs is isodiaphers?
14 23 24 23 234 238 12
1) 6 C and 11 Na 2) 12 Mg and 11 Na 3) 90 Th and 92 U 4) 6 C and 15
7 N

2. The energy required to break one mole of Cl–Cl bonds in Cl2 is 242 kJ mol–1. The longest wavelength of light
capable of breaking Cl–Cl bonds is: (c = 3 × 108 m s–1 and NA = 6.02 × 1023 mol–1)
1) 494 nm 2) 594 nm 3) 640 nm 4) 700 nm
3. A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the
other is at:
1) 325 nm 2) 743 nm 3) 518 nm 4) 1035 nm
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4. Which of the graphs shown below does not represent the relationship between incident light and the electron
ejected from metal surface?
number of K.E. of
electrons electrons

1)  2) 
0 0
 Frequency of light  Intensity of light
K.E. of K.E. of
electrons electrons

3)  4) 
0 0
 Frequency of light  Energy of light
5. A metal is irradiated with light of wavelength 600 nm. Given that the work function of the metal is 1.0 eV, the
de Broglie wavelength of the ejected electron is close to:
1) 6.6 × 10–7 m 2) 8.9 × 10–11 m 3) 1.19 × 10–9 m 4) 6.6 × 10–13 m
6. The maximum kinetic energy of the photoelectrons is formed to be 6.63 × 10–19 J, when the metal is irradiated
with a radiation of frequency 2 ×1015 Hz. The threshold frequency, of the metal is about?
1) 1 × 1015 s–1 2) 2 × 1015 s–1 3) 3 × 1015 s–1 4) 1.5 × 1015 s–1
7. A photon with initial frequency 1011 Hz scatters off electron at rest. Its final frequency is 0.9 × 1011 Hz. The
speed of scattered electron is close to: (h = 6.63 × 10–34 Js. me = 9.1×10–31 kg)
1) 4 × 103 ms–1 2) 3 × 102 ms–1 3) 2 × 106 ms–1 4) 30 ms–1

8. Photons of frequency, v, fall on metal surface for which the threshold frequency is  0 . Then:

1) All ejected electrons have the same kinetic energy, h (v –v0).

2) The ejected electrons have a distribution of kinetic energy from zero to h (v –v0).
3) The most energetic electron has kinetic energy hv.
4) The average kinetic energy of ejected electrons is h (v –v0).
9. A stream of electrons from a heated filament was passed between two charged plates kept at a potential
difference V esu. If ‘e’ and ‘m’ are charge and mass of an electron respectively, then the value of h /  (where
 is wave length associated with electron wave) is given by:
1) 2meV 2) meV 3) 2meV 4) meV

10. The first emission line in the atomic spectrum of hydrogen in the Balmer series appear at.

9R 7R 3R 5R
1) cm 1 2) cm 1 3) cm 1 4) cm 1
400 144 4 36

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11. Which point does not pertain to electron in Bohr’s model of atom?

1) Angular momentum is an integral multiple of h/  2 

2) The path of the electron within an atom is circular

3) Force of attraction of electron towards nucleus is balanced by centrifugal force
4) For a moving electron, energy changes continuously
12. Wavelength of high energy transition of hydrogen atoms is 91.2 nm. Calculate the corresponding wavelength
of He atoms.
1) 22.8 nm 2) 20.5 nm3) 14.6 nm 4) 17.9 nm
13. An excited He+ ion emits photon of wavelength ‘  ’ in returning to ground state from nth orbit. If ‘R’ is Rydberg’ss
constant, then:

4R 4R 4R  1 4R  1

1) n  2) n  3) n  4) n 
4R  1 4R  1 4R 4R
14. An electron in a hydrogen atom in its ground state absorbs 1.5 times as much energy as the minimum required
for it to escape from the atom. What is the velocity of the emitted electron?
1) 1.54 ×106 m/s 2) 1.54 ×108 m/s 3) 1.54 ×103 m/s 4) 1.54 ×104 m/s
15. Calculate the wavelength (in nanometer) associated with a proton moving at 1.0 × 103 ms–1 s:
(mass of proton = 1.67 × 10–27 kg and h = 6.63 × 10–34 J s).
1) 0.032 nm 2) 0.40 nm 3) 2.5 nm 4) 14.0 nm
16. Calculate the uncertainty in velocity of an oxygen molecule (5.3 × 10–26 kg), trapped within a sac of radius
5 × 10–10 m.
1) 3.11 m/sec 2) 2.65 m/sec 3) 4.31 m/sec 4) 1.99 m/sec
17. If the uncertainty in velocity and position is same, then the uncertainity in momentum will be:

hm h h 1 h
1) 2) m 3) 4)
4 4 4m m 4
18. Select the correct statements among the following:
a) Outside any orbital, the probability of finding electron is zero.
b) For single electronic atom or ion, the most probable distance of electron in an orbital having no radial node
n 2a 0
is from the nucleus, where a0 is the first Bohr radius.
Z
c) The average distance of electron (belonging from the same orbit) form the nucleus decreases with the
increase in the value of angular momentum quantum number for the orbital.
d) The angular wave function of any s-orbital is independent from  and  .
1) a, b, c 2) b, c, d 3) a, c, d 4) a, b, c, d
19. An electron has a spin quantum number +1/2 and a magnetic quantum number –1. It cannot be present in:
1) d-orbital 2) f-orbital 3) s-orbital 4) p-orbital
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20. With increasing principal quantum number, the energy difference between adjacent energy levels in H-atom:
1) decreases 2) increases 3) remains constant
4) decreases for low value of Z and increases for higher value of Z.
21. Which of the following statements is correct?
1) The 3d orbitals remain degenerate in the presence of magnetic field.
2) The electron densities in the xy and yz planes are zero in 3dxz orbital.
3) The electron density in the xy plane in 3d z2 orbital is zero.
4) The electron density in the xy plane in 3dxy orbital zero.
22. Choose the correct statement among the following:
1) Number of orbitals in nth shell are n2.
2) Number of orbitals in a subshell are (2l–1).
3) Number of subshell in nth shell are (n – 1).
4) Number of electrons in an orbital of a subshell are 2 (2l + 1).
23. Which of the following statements is correct?
1) Total number of electrons in a subshell is 2l + 1.

2) p z , d x 2  y2 and d z 2 orbitals are non-axial.

3) Only s orbital has directional orientation while pz, d and f orbitals have non-directional properties.
4) Spin multiplicity of nitrogen atom is 4.
24. Which of the following subshells can accommodate as many as 10 electrons?
1) 2d 2) 3d 3) 4f 4) 5g
25. Which of the following statements on quantum numbers is not correct?
1) Quantum numbers n, l, m and ms are needed to describe an electron in an atom completely.
2) Quantum numbers n, l, m and s are obtained by solving the Schrodinger wave equation.
3) A subshell in an atom can be designated with two quantum numbers n and l.
4) The maximum value of l is equal to n – 1 and that of m is  l.50. The quantum numbers n and l of four
electrons are given:
i) n = 4, l = 1 ii) n = 4, l = 0 iii) n = 3, l = 2 iv) n = 3, l = 1
The correct order of increasing energy for the orbitals is
1) (i) < (iii) < (ii) < (iv) 2) (iii) < (iv) < (ii) < (i) 3) (iv) < (ii) < (iii) < (i) 4) (ii) < (iv) < (i) < (iii)
26. Which of the following graph represents the radial probability function of 3d electron?

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27. For the energy levels in an atom which one of the following statements is correct?
1) The 4s sub-energy level is at a higher energy than the 3d sub-energy level
2) The M-energy level can have maximum of 32 electrons
3) The second principal energy level can have four orbitals and contain a maximum of 8 electrons
4) The 5th main energy level can have maximum of 72 electrons
28. Select the correct statement for Ne.
1) It is not isoelectronic with H2O 2) Its last electron enters in s orbital
3) The value of m must be zero for the last electron 4) The value of l must be 1 for the last electron
29. Consider the ground state of Cu atom (Z = 29). The number of electrons with the azimuthal quantum numbers
l = 1 and 2 are, respectively,
1) 12, 5 2) 12, 10 3) 16, 6 4) 14,8
30. Select the correct statements:
a) Heisenberg’s principle is applicable to stationary electron.
b) Pauli’s exclusion principle is not applicable to photons.
c) For an electron in hydrogen atom, the product of velocity and principal quantum number will be indepen-
dent of principal quantum number.
d) Quantum numbers l and m determine the value of angular wave function.
1) a, b, c, d 2) a, b, c 3) b, c, d 4) a, c, d

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SOLUTIONS
LEVEL - I
1. 2
2. 4
3. 3
4. 4
5. 3
6. 3 Power  500W  500Js 1 ;   331.3nm  3.313  107 m

hc 6.62  1034 Js  3.0  108 ms 1

E  7
 5.99  1019 J
 3.313 10 m

500 J s 1
Number of photons emitted per second   8.35  1020
5.99 1019 J
7. 1
8. 3 Absorbed energy = threshold energy + K.E. photo electron

3.9756 1019
Energy of light  h = 6.626 × 10–34 ×6 ×1014 = 3.9756 × 10–19 J   2.49 eV
1.6  1019
Kinetic energy of photoelectron = (2.49 – 2.0) eV = 0.49 eV
9. 4

1 1 1 1 1 1 
10. 2 v  R H  2  2    R H  2  2   n 2  5 for Paschen series.
  n1 n 2   3 n2 
11. 3
12. 3
n2 o  16 9  o 10
13. 3 rn  0.529 A ; r4  r3  0.529    A  1.851 10 m
Z  2 2
14. 2
n(n  1)
15. 4 No. of lines produced when an electron jumps from nth level to ground level  . The possible
2
spectral emissions correspond to the transfer n = 4 to n =3, n = 4 to n = 2, n = 4 to n =1, n = 3 to n = 2,
n = 3 to n = 1 and n = 2 to n = 1. There are 6 emissions.
1 1 1 2  1 1  1
16. 4 v  R H z 2  2  2   109677 1  2  2   27419.25cm
  n1 n 2  2  
(series limit will be of highest energy)

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n2 n2
17. 3 In the expression, rn  a 0 , substituting Z = 4 and r = a0, we get a0  a0  n  2
Z2 42

Z2  42
Now, the ionization enthalpy is En    13.6 eV  13.6  54.4e V.
n2 22
So, the kinetic energy = –En= 54.4 eV. Thus, the wavelength can be calculated as
h 6.626  1034 o
   1.66  1010  1.66 A
2mE 2  9.111031  54.4  1.6  1019
1
18. 2 Since K.E. = mv 2
2
1 1

 2 K.E.  2  2  3.0 10 kg.m s  2

25 2 2
1
 v   31   812 ms
 m   9.110 kg 
h 6.626  1034 Js
  31 1
 8967  1010 m  896.7 nm
mv (9.110 kg) (812 ms )

h 6.63  1034
19. 1 According to de Broglie relationship,     1.105  1033 m
mv 60 103  10
h h
20. 3 Using Heisenberg’s uncertainty principle, x.mv = or x =
4 4mv
0.005
Given that v = 0.005% of 600 m s– 1, so v = 600   0.03 m s 1
100
6.6  1034
Hence, x  31
 1.92  103 m
4  3.14  9.1 10  0.03
21. 4
22. 1 n = 4, l = 3  4f subshells
Since maximum no. of electrons in a subshell = 2 (2 l +1)
So, total no.of electrons in 4 f subshell = 2 (2 × 3 + 1) = 14 electrons.
23. 4
24. 3 A) For n = 3, value of l and n cannot be the same  l  3
B) For l = 1, m l  2 as ml  l is not possible
1
C) n = 2, l = 0, ml = 0, ms   is a possible set
2
D) ms = 0 is incorrect.
25. 4

h  2 (2  1) h  6 h
26. 1 Orbital angular momentum  l (l  1) (Here, l = 2, for d-orbitals)
2 2 2

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27. 4 For 71st electron:

Z = 70 : [Xe]54 4f14 6s2 ; Z = 71 : [Xe] 4f14 5d1 6s2
28. 2
29. 4
30. 1
31. 4
LEVEL - II
1. 3 Isodiaphers have same neutron excess.
2. 1 Energy required to break Cl2 molecule = (242 × 103)/NA J

c 6.626 1034  3108  6.022 1023

From E  hv  E  h We have    494.7 nm
 242 103

1 1 1 1 1 1
3. 2 From the relation      Solving, we get  2 = 743 nm.
 absorbed 1  2 355 680  2

1 1
4. 3 hv  hv 0  mv 2 ; mv 2  K.E.  hv  hv 0
2 2
Plot of K.E. vs v cannot have intercept equal to zero.

5. 3 hc 6.626  1034 Js  3  108 ms 1

E absorbed   9
 0.0331 1017 J
 600  10 m
E absorbed  E 0  KE , KE  E absorbed  E 0 = 3.31×10–19 J – 1.6 × 10–19 J = 1.71 × 10–19 J

h 6.626  1034 Js
de Broglie wavelength     1.188  109 m
31 19
2Em 2  9.1 10 kg  1.71 10 J
6. 1 KE = h (v-v0)

KE 6.63  1019
v0 = v   2  1015   1 1015 s 1
h 6.63  1034
7. 1 Initial energy of photon = Energy of photon after scattering + K.E. of scattered electron

1 1
h1  h 2  mv 2  mv 2  h (1   2 )
2 2

34
2h (1   2 ) 2  6.63  10 (10  0.9 10 )
11 11

v 
2
 31 ; v  14615384  3.8  10  4 10 ms
3 3 1

m 9.110

8. 2 Due to distribution of kinetic energy of ejected photoelectrons inside the crystal.

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9. 3 As electron of charge ‘e’ is passed through ‘V’ volt, kinetic energy of electron becomes = ‘eV’
h
As wavelength of e– wave    
2m.K.E.
h h
   2meV
2meV 
10. 4 For Balmer series n1 = 2 and n2 = 3 for the first line.
11. 4
12. 1 For maximum energy, n1 = 1 and n2 =  . Hence, we have
1  1 1  1  Z2 1
 Z2  2  2  Since R is a constant and transition remains the same,  Z2  He  2H 
  n1 n 2  H   H ZHe 4
1
Hence,  He   91.2  22.8 nm
4
1 2 1 1 1  n 2  1
13. 1  R  2   2    4R  2 
 1 n    n 
1 4 R  1 4R
 n
n 2
4R 4 R  1
14. 1 Energy absorbed = 13.6 ×1.5 = 20.4 eV of this 6.8 eV is converted to K.E.
1
6.8 eV  6.8 1.6  1019 J; 6.8 1.6 1019  K.E.    mv 2
2

2KE 2  1.088  1018

v  = 1.54 ×106 m/s
m 9.11031
h 6.63  1034
15. 2 Using de Broglie relationship     3.97  1010 m  0.40 nm
mv 1.67  1027  103
16. 4 Uncertainity in position x  Diameter of sac  5  1010 m

h 6.626  1034
x    ,    1.99 m / sec
4m 4  3.14  5.3  1026  5  10 10

h h mh
17. 4 mv.v   v  ; p  mv 
4 4m 4
18. 2 b, c, d
19. 3
20. 1
21. 2 3dxz involves xy and yz nodal planes.
22. 1
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1 3
23. 4 Nitrogen has 3 unpaired electrons, therefore, S  3   .
2 2
 3
Spin multiplicity of nitrogen atom : 2S  1   2    1  4
 2
24. 2
25. 2 The quantum number s is not the outcome of the solution of Schrodinger equation.
26. 1
27. 3
28. 4

h  1  (6.626 10 34 J s) 3 1

29. 2    m x   (4)  (3.14)(1.67 1027 kg)(5 106 m)  6.3 10 m s
4  p 
30. 3 b, c, d

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Chapter 03
CLASSIFICATION OF ELEMENTS AND
PERIODICITY IN PROPERTIES

3.0 INTRODUCTION
The Periodic Table provides an organising principle that coordinates and rationalises the diverse physical and
chemical properties of elements. The periodic classification is a consequence of the electronic configuration of
atoms. Trends in physical and chemical properties across periods and down groups is the basis of periodicity.
Need for Classification. The rapid expansion of chemical knowledge necessitated classification of the elements
based on certain criteria to facilitate the systematic study of the properties of the elements and their compounds.
3.1 GENESIS OF PERIODIC CLASSIFICATION
The earliest attempt to classify the elements was in 1789, when Antoine Lavoisier grouped the elements into
gases, non-metals, metals and earths.
1. Dobereiner’s Triads (Law of Triads). German chemist, Johann Dobereiner (1827) arranged similar
elements in groups of three called triads. The atomic mass of the central element was the arithmetic mean
of the atomic masses of the other two elements. E.g., the atomic mass of Na, the middle element of the Li,
Na, K triad is the average of the atomic masses of Li and K.
Element Atomic Weight Element Atomic Weight Element Atomic Weight
Li 7 Ca 40 Cl 35.5
Na 23 Sr 88 Br 80
K 39 Ba 137 I 127
Dobereiner’s relationship worked only for a few elements and therefore, dismissed as mere coincidence.
2. A.E.B de Chancourtois. French geologist, A.E.B de Chancourtois (1862) was the first to arrange the
chemical elements in order of atomic weights. He made a cylindrical table of elements displaying periodic
recurrence in properties (Teeluric helix) but did not attract much attention.
3. Newland’s Law of Octaves. English chemist, Alexander John Newlands (1864) was the first to correlate
chemical properties of elements with their atomic masses. When elements were arranged in the order of
increasing atomic masses, the eighth element had properties similar to the first one (like the eighth note
of a musical scale). The law appeard to be true only for elements up to calcium.

Li (7) Be (9) B (11) C (12) N (14) O (16) F (19)

Element/Atomic weight Na (23) Mg (24) Al (27) Si (29) P (31) S (32) Cl (35.5)
K (40) Ca (40)

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4. Lothar Meyer’s Classification (Atomic volume curve). German chemist, J Lother Meyer (1868)
classified elements by plotting atomic masses against atomic volumes. In the plot, elements with similar properties
occupied similar positions.
3.2 MENDELEEV’S PERIODIC TABLE
Mendeleev’s Periodic Law (Russian chemist, Dimitri Ivanovich Mendeleev,1869) states that the physical
and chemical properties of elements are a periodic function of their atomic weights.

He arranged the elements in rows and columns in the order of increasing atomic masses so that elements with
similar properties occupied the same vertical columns and leaving out blank spaces wherever necessary.
3.2.1 Characteristics of Mendeleev’s Periodic Table
(i) It consists of 9 vertical columns called groups and 7 horizontal rows or periods (Mendeleev called them
Series). The groups are numbered I to VIII and zero.

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(ii) Each group, with the exception of zero and VIII, is divided into two sub-groups, A and B. Elements on the
left hand side of each group constitute sub-group A while those at the right hand side constitute sub-
group B.
(iii) Group VIII consists of three sets each of three elements.
(iv) Group zero has no sub groups.
(v) Mendeleev proposed that some elements were still undiscovered and left several gaps in the table. E.g.,
he left gaps for Ga and Ge under Al and Si and called them Eka-Aluminium and Eka-silicon. Later on it
was found that the properties predicted for these elements were almost similar.
3.2.2 Merits of Mendeleev’s Periodic Table
(i) It simplified and systematised the study of elements and their compounds.
(ii) It helped in predicting new elements on the basis of the blank spaces given in the periodic table.
(iii) Atomic weights of elements like Be, U, In, Au and Pt were corrected.
(iv) Left several gaps in the table for undiscovered elements. e.g., for gallium and germanium are called these
elements as Eka Aluminium and Eka Silicon.
3.2.3 Demerits of Mendeleev’s Periodic Table
(i) Position of hydrogen in group IA is controversial since it also resembles halogens.
(ii) No separate positions for isotopes.
(iii) Lanthanides and actinides are not given proper positions.
(iv) Order of increasing atomic weights is not strictly followed in the case of certain elements, e.g., Ar (39.94)
is placed before K (39.08) and Te (127.6) is placed before I (126.9).
(v) Similar elements are placed in different groups (Cu in IB and Hg in IIB) and dissimiliar elements are placed
in the same group (alkali metals in IA and coinage metals in IB).
3.3 MODERN PERIODIC LAW AND THE PRESENT FORM OF PERIODIC TABLE
English Physicist, Henry Moseley (1913) observed that the frequency () of X-rays produced by the
bombardment of electrons on metal targets is directly proportional to the atomic number (Z) of the metal. A
plot of v against atomic number (not atomic mass) gave a straight line. He suggested that atomic number is
a more fundamental property than atomic weight and should be the basis of classification of the elements.
The Modern Periodic Law or Moseley’s Periodic Law states that the physical and chemical properties
of elements are a periodic function of their atomic number.
Cause of Periodicity. The periodic repetition of properties of elements is due to recurrence of similar valence
shell electronic configuration after regular intervals.
3.3.1 The Long Form (Present Form) of the Periodic Table
The modern periodic table is also referred to as Bohr’s Table since it follows Bohr’s scheme of arrangement
of electrons in atoms. It consists of 7 rows, called periods and 18 columns called groups or families.
Periods are numbered 1, 2, 3, 4, 5, 6 and 7 and can accommodate 2, 8, 8, 18, 18, 32, and 32 elements
respectively. The groups are numbered, 1, 2 ,3,........,18.

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3.4 NOMENCLATURE OF ELEMENTS WITH ATOMIC NUMBER ABOVE 100

In order to avoid confusion regarding naming of newly discoverd elements, IUPAC has adopted a systematic
nomenclature derived from the atomic number of the element using the numerical roots for 0 and numbers 1 to
9 and adding the suffix ium.

Digit Name Abbreviation Digit Name Abbreviation

0 nil n 5 pent p
1 un u 6 hex h
2 bi b 7 sept s
3 tri t 8 oct o
4 quad q 9 enn e
The roots are put together in the order of digits in the atomic number to get the name and the abbreviations to
get a three-letter symbol for the element. Later, a permanent name and symbol are given.

Atomic IUPAC Recommended IUPAC Official

number Name Symbol Name Symbol
101 Unnilunium Unu Mendelevium Md
105 Unnilpentium Unp Dubnium Db
110 Ununnillium Uun Darmstadtium Ds
111 Unununnium Uuu Roentgenium Rg
118 Ununoctium Uuo Oganesson Og
3.5 ELECTRONIC CONFIGURATION OF ELEMENTS AND THE PERIODIC TABLE
In the long fom of the periodic table, classification of elements is based on the electronic configuration. The
location of an element in the periodic table indicates the quantum numbers of the orbital last filled.
3.5.1 Electronic Configurations in Periods
(i) Each period corresponds to a new principal quantum number (n) and can accommodate electrons obeying
Aufbau Principle. Thus, the successive periods can accommodate 2, 8, 8, 18, 18, 32, and 32 elements in
K, L, M, N, O, P, Q shells.
(ii) Each period starts with an alkali metal (ns1) and ends with a noble gas (ns2np6) except helium (1s2).
(iii) The number of elements in each period is twice the number of atomic orbitals in that energy level.
Period 1 is the shortest with two elements. Periods 2 and 3 contain 8 elements each and are called short
periods. Periods 4 and 5 contain 18 elements each and are called long periods. Period 6 is the longest with
32 elements. Period 7 also has 32 elements, most of which are artificially synthesised.
In the first period, filling of electrons take place in the first shell (n = 1, l = 0) in the only orbital (1s1-2).
The second period can accomodate 8 elements; electrons in (n = 2, l = 0, 1) 4 orbitals (2s1-2 and 2p1-6).
The third period also has only 8 elements. Even though, the third period (n = 3, l = 0, 1, 2) with 9 orbitals (3s,
3p, 3d) can accommodate 18 elements, energy of the 3d orbitals is greater than 4s orbital and hence, they are
not filled immediately after 3p.
3d orbitals are filled after 4s orbitals. 21Sc to 30Zn are included in the fourth period (first transition series).
Similarly, the fourth period beginning with 4s accommodates 18 elements. The fourth period (n = 4, l = 0, 1,
2, 3) with 16 orbitals (4s, 4p, 4d and 4f) can accommodate 32 electrons, but the energy of 4f orbitals is
greater than 4d, 5p and 6s orbitals.
4d orbitals are filled after 5s (second transition series). 39Y to 48Cd are included in the fifth period.
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4f orbitals are filled after 5p and 6s, in fact, after 5d1 (57La -5d1, 6s2) and hence called inner transition
elements (transition within transition - 4f within 5d) and included in the sixth period.
The fifth period (n = 5, l = 0, 1, 2, 3, 4) has 25 orbitals (5s, 5p, 5d, 5f, 5g) and starts with 5s followed by
4d1-10 and 5p1-6. 5d and 5f are included in fifth and sixth periods respectively.
The sixth period (n = 6) starts with filling of 6s orbital followed by 5d1, 4f1-14, 5d2-10 and 6p1-6. The 4f- filling
elements (58Ce to 71Lu) are placed separately and called lanthanoid series (since they come after 57La).
The seventh period (n = 7) starts with filling of 7s orbital followed by 6d1, 5f1-14, 6d2-10 and 7p1-4. The 5f- filling
elements (90Th to 103Lr), called actinoid series is placed separately along with lanthanoids.
3.5.2 Groupwise Electronic Configurations
All elements belonging to a particular group have similar valence shell electronic configuration, e.g., ns1 for
alkali metals, ns2 for alkaline earth metals, ns2 np5 for halogens, ns2 np6 for noble gases, etc.
3.6 ELECTRONIC CONFIGURATIONS AND TYPES OF ELEMENTS
In the long form of the periodic table, elements are classified into four blocks (s, p, d and f ) depending on the
type of orbital that receives the last electron. Hydrogen and helium are exceptions to this categorisation.
Hydrogen (1s1) can be placed in group 1. It can accept an electron to attain noble gas configuration (1s2) and
exhibit similarities with halogens. Thus, it is placed separately at the top of the Periodic Table.
Helium belongs to s-block, but its position in the p-block (group 18) is justified by a completely filled valence
shell (1s2) and properties characteristic of noble gases.
3.6.1 s-block elements
Elements of groups 1 and 2 (alkali and alkaline earth metals and helium) constitute the s-block.
s-block elements (except helium) lie on the extreme left of the periodic table. In s-block elements, shells upto
(n-1) are completely filled and the electron enters the outermost (n) shell.
General electronic configuration is [inert gas] ns1-2.
s - block consists of soft, highly reactive metals with low melting and boiling points (except hydrogen).
They are good conductors of heat and electricity.
s- block elements except Be impart colour to the flame (in flame-test).
s- block elements, especially alkali metals have low ionisation energy since the valence electrons are in a new
principal quantum number (away from the nucleus).
They are the most electropositive elements, therefore, they form ionic compounds.
They are strong reducing agents because they can lose their valence electrons easily.
The hydroxides are strong bases due to the large difference in electronegativities of alkali metals and oxygen
than between oxygen and hydrogen in M-O-H.
Common oxidaion states of groups 1 and 2 elements are +1 and +2 respectively.
3.6.2 p-block elements
Groups 13 to 18 constitute the p-block.
p-block elements lie on the extreme right of the periodic table. They have shells upto (n -1) completely filled
and the differentiating electron enters the p-orbital of the nth orbit.
General electronic configuration is [inert gas] ns2 np1-6.
This block inculdes some metals, all nonmetals and metalloids.
s-block and p-block elements are collectively called normal or representative elements.
Group 15 elements are called Pnicogens (due to chocking or suffocating property of nitrogen)

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group 16 elements are called Chalcogens (ore - forming), group 17 elements Halogens, and group 18
elements Noble Gases.
Many elements show more than one oxidation state. They mostly form covalent compounds. Their ionic
character increases from top to bottom in a group.
Elements of group 13 and lower elements of group 14 are metallic solids while those of 14, 15 and 16 form
molecular solids.
In a period, reducing character decreases and oxidising character increases.
These elements usually does not impart colour to the flame (in flame test).
Group 18 (Zero group or Noble gases) elements have completely filled s and p sub-shells. Their outer electronic
configuration is ns2 np6, except for helium (1s2). Because of stable configuration, these elements do not undergo
normal chemical reactions. Xe reacts with highly electronegative elements such as fluorine and oxygen to form
compounds like XeF2, XeF4, XeF6, XeOF2, XeOF4 and XeO3.
3.6.3 d-block elements (Transition elements)
Ten elements each of groups 3 to 12 constitute the d-block of the periodic table. When the outermost (n) and
the penultimate (n-1) shells are incompletely filled and the differentiating electron enters the (n-1)d orbitals, the
elements are called d-block or transition elements. Transition elements form a bridge between chemically
active metals of s-block and the less active elements of p-block.
General electronic cofiguration is [inert gas] (n-1) d1-10 ns1-2.
These are hard, malleable and ductile metals. They show lustre and high thermal and electrical conductivities.
They have relatively low ionisation energy. They are less electropositive than s-block elements and form ionic
or covalent bonds. Atoms in lower oxidation states are ionic and in higher oxidation states are covalent.
Atomic sizes are much lower than those of groups 1 and 2. As the inner (n-1)d orbitals are filled, the increased
nuclear charge pulls the electrons inwards. Density increases with reduction of atomic size.
Most of them form coloured complexes or ions due to electron transitions among vacant d-orbitals (d-d
transitions). They form both ionic and covalent compounds.
They form co-ordination compounds (complex ions).
They exhibit variable valency and variable oxidation states because electrons from d-orbitals also participate
in bond forming.
They have high melting and boiling points due to covalent bonding between half-filled d-orbitals and metallic
bonding by valence electrons.
They exhibit a tendency to remain unreactive or noble as the ionisation energy increases (e.g. Au, Pt).
They are paramagnetic due to the presence of unpaired electrons in d-orbitals.
They form non-stoichiometric compounds (e.g. Fe0.94O to Fe0.84O).
Many d-block elements and their compounds are used as catalysts (due to variable oxidation states).
Most transition elements form alloys (due to comparable atomic sizes).
However, Zn, Cd and Hg with electronic configuration, (n-1) d10ns2 do not show most of the properties of
transition elements.
3.6.4 f-block Elements (Inner-transition elements)
In f-block elements, the (n-1) and (n-2) shells are incompletely filled and the last electron enters into the f-
orbital of the pre-penultimate or antepenultimate (n-2) shell.
General electronic configuration is (n-2) f1-14 (n-1) d0-1 ns2.

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They are called inner-transition elements as they have three incomplete outer shells and form a transition
series within the transition elements (d-block).
The elements of f-block have been classified into two series, Lanthanoids and Actinoids.
Lanthanoids. First inner-transition series or 4 f-series, has 14 elements 58Ce to 71Lu. Filling of electrons takes
place in 4f subshell. They are called Lanthanoids because they come after 57 La (group-3) belonging to 5d
series. They are also called rare earths.
Actinoids. Second inner-transition series or 5 f-series, has 14 elements from 90 Th to103 Lr. Filling of electrons
take place in 5f subshell. They are called Actinides since they come after 89 Ac in the 6d- transition series
(group-3). All actinoids are radioactive. Th, Pa and U are naturally occuring and the remaining 11 (Np to Og)
elements are artificially synthesised. Elements after uranium are called transuranium elements.
Within each series, the properties of the elements are quite similar. Lanthanoides and actinoids are all metals.
They have high melting and boiling points. The most common oxidation state for lanthanids is +3.
They are paramagnetic due to the presence of unpaired electrons in their orbitals. They form coloured compounds
and complexes.
The lanthanoids and actinoids have been placed at the bottom to avoid expansion of the periodic table.
3.6.5 Classification of Elements into metals, non-metals and metalloids
(a) Metals. Metals comprises more than 78% of all known elements and appear on the left hand side of the
periodic table.
Metals are usually solids at room temperature (except mercury and gallium).
They are malleable and ductile.
They have high melting and boiling points and are good conductors of heat and electricity.
The metals are characterised by their nature of readily giving up elecron(s) and produce lustre.
Oxides of these metals are generally basic (some are acidic in their higher oxidation states, e.g., CrO3).
(b) Non-metals. Nonmetals are located at the top right hand side of the periodic table.
They are usually solids, liquids (bromine) or gases at room temperature with low melting and boiling points.
They are poor conductors of heat and electricity.
Non-metals do not lose electrons but take up electrons to form corresponding anions.
Oxides of nonmetals are generally acidic in nature.
(c) Metalloids (Semi-metals or Metalloids). In the periodic table, nonmetallic character increases from left
to right across a row. Elements which lie at the border of metallic and nonmetallic behavior, possess properties
that are characteristic of both metals and nonmetals. These elements are called semi-metals or metalloids.The
metalloids comprise of the elements B, Si, Ge, As, Sb and Te.
Oxides of metalloids are generally amphoteric in nature.
(d) Typical elements. Elements of the third period (Na, Mg, Al, Si, P, S, Cl.) are called typical elements.
Properties of all elements belonging to a particular group resemble those of the corresponding typical element
of that group, e.g., the general properties of alkali metals can be predicted from the properties of Na (not Li,
the first member of the group).
The properties of elements of second period differ in many respects from other members belonging to the
same group due to smaller atomic size and absence of d-orbitals.

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3.7 PERIODIC TRENDS IN PHYSICAL PROPERTIES OF ELEMENTS

There are periodic variations in the physical properties of elements down any group or across any period.
3.7.1 Atomic Radius. It is the distance from the centre of the nucleus to the outermost shell containing
electrons or the point upto which the density of the electron cloud is maximum. Atomic radius refers to
both covalent and metallic radius, depending on whether the element is a non-metal or metal. Atomic radius is
measured by X-ray or other spectroscopic methods and expressed in nm.
(i) Covalent radius. It is one-half of the distance between the centres of two atoms bonded by a single
covalent bond. For a homodiatomic molecule,
d A  A  rA  rA or 2rA or rA = (d A-A )/2
E.g., bond distance in Cl2 is 198 pm, therefore, the atomic radius of chlorine is 99 pm.
Table - 01 Atomic radii of s-block and p-block elements (in pm)
Period Group-1 Group-2 Group-13 Group-14 Group-15 Group-16 Group-17 Group-18
1 H (37) He (140)
2 Li (152) Be (111) B (88) C (77) N (74) O (66) F (64) Ne (154)
3 Na (186) Mg (160) Al (143) Si (117) P (110) S (104) Cl ( 99) Ar (188)
4 K (231) Ca (197) Ga (153) Ge (122) As (121) Se (117) Br (114) Kr (202)
5 Rb (244) Sr (215) In (167) Sn (158) Sb (141) Te (137) I (133) Xe (216)
6 Cs (262) Ba (224) Tl (171) Pb (175) Bi (182) Po At (140) Rn
(ii) Metallic radius (Crystal radius). It is one-half of the distance between the nuclei of two adjacent
metal cores in the metallic crystal lattice. E.g., distance between Cu atoms in solid copper is 256 pm;
hence metallic radius of copper is 128 pm.
Metallic radius of an element is always greater than the covalent radius. This is because metallic bond (electrical
attraction between nucleus and mobile electrons) is weaker than covalent bond.
Factors Influencing Atomic Radii. Effective nuclear charge, size of valence shell, multiplicity of bond,
percentage ionic character, and ionic charge influence atomic radii.
i. Atomic radius decreases with increase in effective nuclear charge, i.e., atomic radius (1/Zeff).
ii. Atomic radii increase with increase in principal quantum number, i.e., atomic radii increase down the group.
iii. Covalent radius decreases with increase in the number of bonds between atoms : C-C > C=C > CC.
iv. Increase in percentage ionic character results in shortening of bond and lowering of atomic radii.
v. Greater the charge on the cation, smaller the ionic radii (due to higher Zeff) : Fe3  Fe 2  Fe   Fe
vi. Greater the charge on the anion, larger the ionic radii (due to lower of Zeff.) : O  O   O 2
Periodicity in Atomic Radii. The atomic size generally decreases across a period. As electrons are added
to the same valence shell, Zeff increases with atomic number resulting in increased attraction of electrons
towards the nucleus.
Within a group, atomic radius increases regularly with atomic number. For alkali metals and halogens, the
principal quantum number (n) increases down the groups, and the valence electrons are farther from the
nucleus. The filled inner energy levels shield the outer electrons from the pull of the nucleus. Consequently the
size of the atom increases.

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Atomic radii of noble gases are very high. Atomic radius of inert gases is the highest in a period because it
is expressed as van der Waals radius (which is larger than the covalent radius).

(a) (b)

Figure - 01 Variation of atomic radius in (a) second period (b) alkali metals and halogens.
Among transition elements (d-block), covalent radii decrease from left to right until near the end and then
increases slightly. This is because d-electrons do not shield nuclear attraction for the outer electrons completely.
3.7.2 Ionic radius
It is the effective distance from the centre of the nucleus of the ion up to which it has influence in the
ionic bond. Ionic radius is determined by X-ray analysis. For an ionic molecule AB, the bond length
dAB  rc  ra
where rc = radius of cation and ra = radius of anion.
(i) Periodicity in Ionic Radii. Among representative elements, ionic radii decrease from left to right in a
period. In any group of representative elements, ionic radii increase with atomic number :
Li+ (76) < Na+(102) < K+ (138) < Rb + (150) ; Be2+ < Mg2+ < Ca2+ < Sr2+ ; F- < Cl- < Br- < I-
Ionic radius of Na+ is much smaller than that of Na (metallic radius = 186 pm and ionic radius =102 pm)
because there is complete removal of one shell leading to decrease in size.
Ionic radius of Cl- is larger than that of Cl (covalent radius = 99 pm, ionic radius = 184 pm) because there is
reduction in effective nuclear charge (due to additional electron) and hence, expansion of the electron cloud.
Ionic radius decreases with increase in charge on the cation :
Cr2+ (80) > Cr3+ (61.5) > Cr4+ (55) > Cr5+ (49)> Cr6+ (44).
(ii) Lanthanoid and Actinoid Contraction: Among lanthanoids, the f-orbials do not effectively shield nuclear
charge, therefore, there is reduction in size of ions with increasing atomic number. This is called lanthanoid
contraction. A similar phenomenon among actinoids (90Th to 103 Lr) is called actiniod contraction.
(iii) Isoelectronic species. Atoms, ions, etc., containing the same number of electrons, but differ in the
magnitude of nuclear charge are called isoelectronic species, e.g., N3-, O2-, F-, Ne, Na+, Mg2+ and Al3+ are
isoelectronic (10 e-); S2-, Cl-, K+, Ca2+, Sc3+ (19 e-) ; SO2, NO3-, CO32- (32 e-); N2, CO, CN- (14 e-).
Ionic radii of isoelectronic species decrease with increase in nuclear charge :
Al3+ < Mg2+ < Na+ < F- < O2- < N3-.
3.7.3 Ionisation Enthalpy
Ionisation enthalpy is defined as the energy required to remove the most loosely bound electron from an
isolated gaseous atom in the ground state. It is measured in kJ mol-1, k cal mol-1 or eV.
M(g) 
(IE1 )
 M  (g)  e ; M  (g) 
(IE 2 )
 M 2  (g)  e ; M 2 (g) 
(IE3 )
 M 3 (g)  e 
IE1, IE2 and IE3 are the first, second and third ionization enthalpies. Ionization enthalpies are always positive.
In general, (IE)1 < (IE)2 < (IE)3 because attraction between the nucleus and the remaining electrons increases
considerably when the number of electrons decreases.

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(a) (b) (c)

Figure - 02 Ionisation enthalpies of (a) second period elements (b) alkali metals (c) elements up to Z= 60.
Factors Influencing lonisation enthalpy. i. Size of the atom : Ionisation enthalpy decreases with
increase in atomic size. As the distance between the outermost electrons and the nucleus increases, the force
of attraction decreases. As a result, outermost electrons are held less firmly and lesser amount of energy is
needed for removing them. Thus, ionisation energy generally decreases in any group from top to bottom.
ii. Nuclear Charge. Ionisation enthalpy increases with increase in the nuclear charge. With increase in
the nuclear charge, electrons of the outermost shell are held more firmly by the nucleus, therefore, greater
amount of energy is required to pull out an electron. Ionisation energy increases as we move from left to right,
along a period, due to increase in nuclear charge.
iii. Shielding or screening effect. The electrons in the inner shells act as a shield between the nucleus and the
electrons in the outermost shell. This is called shielding effect. As the number of electrons increases shielding
effect increases and ionisation enthalpy decreases.
iv. Penetration effect of the electron. Ionization enthalpy also depends on the type of electron that is
removed. An s electron penetrates closer to the nucleus and therefore, more tightly held than a p-electron.
Similarly, p-electron is held more firmly than a d-electron and so on. The extent of penetration of the orbitals
and the ionisation energies of electrons in them are in the order, s > p > d > f.
Ionisation enthalpy of Al is less than that of Mg because the outer most electron has to be removed from 3p-
orbital (lesser penetration) in Al, but from 3s orbital in Mg (greater penetration).
v. Electronic configuration. If an atom has exactly half-filled or completely filled orbitals, then such an
arrangement has extra stability. The removal of an electron from such an atom requires more energy. E.g. first
ionisation energy of beryllium is greater than boron because beryllium has filled outermost orbital (2s) while
boron has partially filled outermost orbital (2p).
Be (Z = 4) 1s2, 2s2 ;
B (Z = 5) 1s2, 2s2, 2p1
Similarly, noble gases have completely filled outermost shells and hence they have the highest ionisation energies
in their respective periods.
Table - 02 First ionisation enthalpies of representative elements and noble gases (kJ mol-1)
1 2 13 14 15 16 17 18
1 H (1312) He (2372)
2 Li (520) Be (899) B (801) C (1086) N (1402) O (1314) F (1681) Ne (2081)
3 Na (496) Mg (737.6) Al (577) Si (786) P (1011) S (999) Cl (1255) Ar (1520)
4 K (419) Ca (590) Ga (579) Ge (760) As (946) Se (941) Br (1142) Kr (1350)
5 Rb (403) Sr (549) In (558) Sn (708) Sb (834) Te (869) I (1009) Xe (1170)
6 Cs (374) Ba (502) Tl (589) Pb (715) Bi (703) Po (813) At (917) Rn (1037)

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Periodic Trends in Ionisation Energy. In any period, ionisation energy generally increases because effective
nuclear charge increases and atomic size decreases.
Exceptions: Ionisation energy of Be (2s2 2p0) is greater than that of Al (3s2 3p1) due to higher stability of fully
filled s-orbital. Ionisation energy of N ( 2s2 2px1 2py1 2pz1) is greater than O (2s2 2px2 2py1 2pz1) and P (3s2
3px1 3py1 3pz1) is greater than that of S (3s2 3px2 3py1 3pz1) due to stability of half-filled orbitals.
Group Trends in Ionisation Energy. Generally, ionisation energy decreases down any group since atomic
size and shielding effect increases.
3.7.4 Electron Gain Enthalpy
The electron gain enthalpy, egH, is the change in standard molar enthalpy when a neutral, gaseous
atom gains an electron to form an anion.
X(g)  e (g) 
 X  (g)
It is a measure of the ease with which an atom adds an electron to form anion.
O(g)  e  (g) 
Exothermic
 O  (g) ; E a  140.9kJ mol 1 ;  eg H  140.9 kJ mol1
The second electron gain enthalpy is invariably positive because electron repulsion outweighs nuclear attraction.
O (g)  e (g) 
Endothermic
O2 (g) ; Ea  744.9kJ mol1; eg H  744.9 kJ mol1
Factors Influencing Electron Gain enthalpy. i) Atomic size. Electron gain enthalpy  (1 / Atomic size)
ii) Effective nuclear charge. Electron gain enthalpy  Effective nuclear charge (Zeff )
iii) Screening effect. Electron gain enthalpy  (1 / Screening effect)
Periodic trends in electron gain enthalpy among s and p-block elements.In a period, electron gain
enthalpies becomes more negative (except for inert gases) because effective nuclear charge, Zeff, increases (it
will be easier to add an electron to a small atom).
Table - 03 First electron gain enthalpies of representative elements and noble gases (kJ mol-1)
1 2 13 14 15 16 17 18
1 H (-73) He (+48)
2 Li (-60) Be (+66) B (-83) C (-122) N (+31) O (-141) F (-328) Ne (+116)
3 Na (-53) Mg (+67) Al (-50) Si (-199) P (-74) S (-200) CI (-349) Ar (+96)
4 K (-48) Ca Ga (-36) Ge (-116) As (-77) Se (-195) Br (-325) Kr (+96)
5 Rb (-47) -
Sr In (-29) Sn (-120) Sb (-101) Te (-190) I (-295) Xe (+77)
6 Cs (-46) -
Ba Tl (-30) Pb (-101) Bi (-110) Po (-174) At (-270) Rn (+68)
-
In a group, electron gain enthalpy becomes less negative because the size of the atom increases and the added
electron is at a larger distance from the nucleus.
Electron gain enthalpies of alkali metals are high and negative because ns1 electronic configuration is unstable.
Electron gain enthalpies of group 2 metals are low or positive because of ns2 configuration is fairly stable.
Negative electron gain enthalpy of O and F are less than those of S and Cl because the electron is added to the
smaller, n = 2 energy level where it experiences significant repulsion from other electrons present in this level.
In S and Cl, the electron goes to the larger n = 3 energy level and consequently occupies a larger region of
space leading to much less electron-electron repulsion.
Group 17 elements (halogens) have very high negative electron gain enthalpies because they can attain stable,
noble gas configuration by accepting an electron.

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Noble gases have large positive electron gain enthalpies because the electron has to enter the next higher
energy level, leading to a very unstable electronic configuration.
Electron gain enthalpy for addition of electron to the same orbit of an atom : s > p > d > f.
Half-filled and completely filled orbitals of a subshell are comparatively more stable. The addition of an extra
electron to such a system is difficult and hence electron gain enthalpy decreases. For example, nitrogen has
very low electron gain enthalpy because there is greater electron repulsion when the incoming electron enters
an orbital that is already half-filled (2s2 2Px1 2Py1 2Py1).
3.7.5 Electronegativity
Electronegativity,  (kii), is the tendency of an atom to attract the shared pair of electrons towards
itself in a covalently bonded molecule. It is not a measurable quantity.
Electronegativity scales : Pauling scale, Mulliken-Jaffe scale, and Allred-Rochow scale.
American scientist, Linus Pauling (1922) arbitrarily assigned a value of 4.0 to fluorine, the element with the
greatest ability to attract electrons. Based on this, approximate values were assigned to other elements.
The electronegativity of an element is not constant; it varies depending on the element to which it is bound.
When two atoms have similar electronegativities, the bond between them is largely covalent, but when the
electronegativity difference is large, the bond becomes polar.
Table - 04 Electronegativity values of representative elements (Pauling scale)

1 2 13 14 15 16 17
1 H (2.1)
2 Li (1.0) Be (1.5) B (2.0) C (2.5) N (3.0) O (3.5) F (4.0)
3 Na (0.9) Mg (1.2) Al (1.5) Si (1.8) P (2.1) S (2.5) Cl (3.0)
4 K (0.8) Ca (1.0) Ga (1.6) Ge (1.8) As (2.0) Se (2.4) Br (2.8)
5 Rb (0.8) Sr (1.0) In (1.7) Sn (1.8) Sb (1.9) Te (2.01) I (2.5)
6 Cs (0.7) Ba (0.9) TI (1.8) Pb (1.9) Bi (1.9) Po (1.76) At (2.2)
Factors affecting electronegativity. i) Atomic size. Electronegativity  (1/atomic size). With increase in
atomic size, force of attraction between the nucleus and the valence electrons decreases.
ii) Effective nuclear charge. Electronegativity  Zefff .With increase in nuclear charge, force of attraction
between the nucleus and the valence shell electrons increases.
iii) Oxidation state. Electronegativity  oxidation state (charge on cation). In higher oxidation states, elements
have higher magnitude of positive charge and higher polarising power, hence electronegativity increases.
iv) s-character of the hybrid orbital. Electronegativity increases with s-character of hybrid orbitals ;
sp (50%) > sp2 (33%) > sp3 (25% ).
v) Bond length. Electronegativity is inversely proportional to bond length ; C  C < C = C < C  C .
(vi) Non-metallic character. Non-metallic character increases with increase in electronegativity (metallic
character decreases, i.e., electropositive character increases).
Variation of electronegativity in a period. On moving across a period, Z and Zeff increases; force of
attraction between the nucleus and the valence shell electrons increases, therefore, electronegativity increases.
Variation of electronegativity in a group. In a group, Z increases, but Zeff almost remains constant; number
of shells and atomic radius increases, therefore, electronegativity decreases.
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Exceptions: Electronegativities of zero group elements is zero because they do not form compounds (the
values for Kr and Xe may not be zero since they form compounds).

Figure - 03 Periodic trends in physical properties of elements.

3.8 PERIODIC TRENDS IN CHEMICAL PROPERTIES
3.8.1 Periodicity of Valence or Oxidation States. Generally, the valence of representative elements is equal to
the number of electrons in the outermost orbitals or eight minus the number of outermost electrons. Nowadays
the term oxidation state is also used for valence. The oxidation state of an element in a compound is the charge
acquired by its atom by virtue of its electronegativity difference with the other atom(s) in the molecule.

Group 1 2 13 14 15 16 17 18
No. of valence electrons 1 2 3 4 5 6 7 8
Valence 1 2 3 4 3, 5 2, 6 1, 7 0, 8
Among representative elements, valency with respect to oxygen increases from 1 to 7 and valency with respect
to hydrogen increases from 1 to 4 and then decreases to 1.
Table - 05 Periodic trends in valence of elements as shown by the formulas of their compounds

Group 1 2 13 14 15 16 17
Formula of oxide Li2O MgO B 2O 3 CO2 Na2O3, Na 2O5
Na2O CaO Al2O3 SiO2 P 2O6, P 4O10 SO3 Cl2O7
K 2O SrO Ga2O3 GeO2 As2O3, As2O5 SeO3
BaO In2O3 SnO2 Sb2O3, Sb2O5 TeO3
PbO2 Bi2O3
Formula of hydride LiH B2H6 CH4 NH3 H 2O HF
NaH CaH2 AlH3 SiH4 PH3 H 2S HCl
KH GeH4 AsH3 H2Se HBr
SnH4 SbH3 H2Te HI
3.8.2 Anomalous Properties of Second Period Elements. The first elements of groups 1 and 2 and groups 13 to
17 differ in many respects from the other members of their respective groups. E.g., Li, unlike other alkali
metals and Be unlike other alkaline earth metals form compounds with considerable covalent character, while
other members form ionic compounds. In fact, Li and Be exhibit greater similarities with diagonal elements,
Mg and Al. The similarity in properties between two diagonally placed elements in the periodic table is called
diagonal relationship.

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Element Li Be B
Metallic radius, M (pm)/Ionic radius, M (pm) 152/76 111/31 88
Element Na Mg Al
+
Metallic radius, M (pm)/Ionic radius, M (pm) 186/102 160/72 143

The anomalous behaviour of the first member of a group in the s- and p-blocks may be attributed to (i) small
size, (ii) large charge/radius ratio, (iii) high electronegativity and (iv) absence of d-orbitals in the valence shell.
The first member has only 4 valence orbitals (2s, 2p) available for bonding, whereas the second member has
9 valence orbitals (3s, 3p, 3d). The maximum covalency of the first member is 4, e. g., B only forms BF4-.
Other members can expand their valence shell to accomodate more than 4 pairs of electrons due to the
presence of d-orbitals, e.g., Al forms [AlF6] 3-.
Among p-block elements, the first member of any group displays greater ability to form pπ - pπ multiple
bonds to itself (C=C, CC, N=N, NN) and to other second period elements (C=O, C=N, CN, N=O),
due to their small size.
3.8.3 Periodic Trends in Chemical Reactivity. In a period, atomic and ionic radii decreases from left to right.
Ionisation enthalpies increase and electron gain enthalpies become more negative (except for noble gas). This
results in enhanced chemical reactivity at the two extremes with the lowest at the centre.
The maximum chemical reactivity at the extreme left (alkali metals) is by the loss of electron (oxidation) forming
cation and at the extreme right (halogens) is by the gain of electron (reduction) forming anion. Thus, metallic
character (reducing property) decreases and non-metallic character (oxidising property) increases from left to
right in the period.
The normal oxide formed by the element at the extreme left is the most basic (e.g., Na2O) and the element at
the extreme right is the most acidic (e.g., Cl2O7). Oxides formed by elements in the centre are amphoteric
(e.g., Al2O3, As2O3) or neutral (e.g., CO, NO, N2O).
In the case of transition metals (3d series), change in atomic radii is much smaller compared to those of
representative elements in the period. The change in atomic radii is still smaller among inner-transition metals
(4f series). The ionization enthalpies are intermediate between those of s- and p-blocks, therefore, they are
less electropositive than alkali and alkaline earth metals.
In the case of main group elements, increase in atomic and ionic radii with increase in atomic number results in
decrease of ionisation enthalpies and electron gain enthalpies. Metallic character increases and non-metallic
character decreases down the group.

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QUESTIONS
LEVEL - I
1. The law of triads is not applicable to:
1) Cl, Br, I 2) Na, K, Rb 3) S, Se, Te 4) Ca, Sr, Ba
2. Eka-aluminium and Eka-silicon are known as:
1) Gallium and germanium 2) Aluminium and silicon
3) Iron and sulphur 4) Carbon and silicon
3. Considering the elements from left to right in the third period of the periodic table, the atomic volume of the
elements :
1) first decreases then increases 2) decreases
3) increases at constant rate 4) remains unchanged
4. According to Moseley, a straight line graph is obtained on plotting.
1
1)  vs. Z 2)  2 vs. Z 3)  vs.Z 4) vs. Z

5. The statement that is not correct for periodic classification of elements is:
1) The properties of elements are a periodic function of their atomic numbers
2) Non-metallic elements are less in number than metallic elements
3) The first ionization energies of elements along a period do not vary in a regular manner with increase in
atomic number
4) For transition elements, the d-subshells are filled with electrons monotonically with increase in atomic
number
6. The IUPAC name for the element with atomic number. Z = 119 is:
1) Unp 2) Uns 3) Uno 4) Uue
7. The atom having the valence shell electronic configuration 4s 4p would be in:
2 2

1) Group 2 and period 3 2) Group 2 and period 4

3) Group 14 and period 4 4) Group 14 and period 3
8. The element with the electronic configurations as (Ar) 3d104s24p3 represents:
1) Metal 2) Non-metal 3) Metalloid 4) Transition metal
9. Which pair of atomic numbers represents s-block elements?
1) 7, 15 2) 6, 12 3) 9, 17 4) 3, 12
10. Which of the following statements is correct with respect to the property of elements with an increase in
atomic number in the carbon family (group 14)?
1) Atomic size decreases 2) Ionization energy increases
3) Metallic character decreases 4) Stability of +2 oxidation state increases
11. In a period, the elements having least melting point are:-
1) Noble gas 2) Alkali metals 3) Chalcogens 4) Pnicogens
12. The atomic radius increases as we move down a group because:
1) effective nuclear charge increases
2) atomic mass increases
3) added electrons are accommodated in new electron level
4) atomic number increases
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13. The best reason to account for the general tendency of atomic diameters to decrease as the atomic numbers
increase within a period of the periodic table is the fact that
1) Outer electrons repel inner electrons
2) Closer packing among the nuclear particles is achieved
3) The number of neutrons increases
4) The increasing nuclear charge exerts a greater attractive force on the electrons
14. The covalent radius of an element depends on:
a) Number of bonds formed between the atoms of that element
b) Type of hybridization involved by its atoms in the covalent bond formation
c) Ionic character of the covalent bond formed by its atom with the atoms of other element
d) Oxidation state of the atom in its covalent compounds
1) a and b 2) a, b and d 3) a, b, c 4) a, b, c, d
15. Consider the isoelectronic series : K , S , Cl and Ca ; the radii of the ions decrease as
+ 2– – 2+

1) Ca2+ > K+ > Cl– > S2– 2) Cl– > S2– > K+ > Ca2+ 3) S2– > Cl– > K+ > Ca2+ 4) K+ > Ca2+ > S2– > Cl–
16. In which of the following pairs, the ionisation energy of the first species is less than that of the second
1) O–, O2– 2) S, P 3) N, P 4) Be+, Be
17. The set representing the correct order of first ionization potential is
1) K > Na > Li 2) Be > Mg > Ca 3) B > C > N 4) Ge > Si > C
18. Ionisation energy of F is 320 kJ mol–1. The electron gain enthalpy of fluorine would be
1) –320 kJ mol–1 2) –160 kJ mol–1 3) +320 kJ mol–1 4) +160 kJ mol–1
19. Amongst the following elements (whose electronic configurations are given below), the one having the highest
ionization energy is
1) [Ne] 3s2 3p1 2) [Ne] 3s2 3p3 3) [Ne] 3s2 3p2 4) [Ar] 3d10 4s2 4p3
20. Ionisation potential and electron affinity of fluorine are 17.42 and 3.45 eV respectively. Calculate the
electronegativity of fluorine.
1) 3.726 2) 2.726 3) 1.726 4) 5.726
21. The graph of IE1 or  i H1 versus atomic number (Z) is given below:

Which of the following statement is correct?

1) Alkali metals are at the maxima and noble gases at the minima.
2) Noble gases are at the maxima and alkali metals at the minima.
3) Transition elements are the maxima.
4) Minima and maxima do not show any regular behaviour.
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22. The correct order of decreasing first ionization energy is:

1) Si > Al > Mg > Na 2) Si > Mg > Al > Na
3) Al > Si > Mg > Na 4) Mg > Li > Al > Si
23. When the first electron gain enthalpy (  eg H) of oxygen is –141 kJ/mol, its second electron gain enthalpy is:
1) almost the same as that of the first
2) negative, but less negative than the first
3) a positive value
4) a more negative value than the first
24. Second electron affinity of an element
1) is always negative 2) is always positive
3) can be positive or negative 4) is always zero
25. Which of the following are the incorrect statements?
a) The second ionisation potential of an atom is always less than the first ionisation potential
b)The addition of an electron to a neutral atom is an endothermic process
c) Fluorine has the maximum electronegativity
d) The size of the cation is always less than the size of the neutral atom
1) a, b 2) a, b, c
3) a, b, d 4) b, c, d
26. Which of the following represents a correct sequence of electronegativity values:
1) F > N > O > C 2) F > N < O > C
3) F > N > C > O 4) F < N < O < C
27. If the atomic number of an inert gas element is Z, then an element with which of the following electronic
configurations will have the highest electronegativity (according to Pauling scale)?
1) Z – 1 2) Z – 2
3) Z + 1 4) Z + 2
28. Diagonal relationship is due to:
1) Identical valency
2) Identical oxidation state
3) Identical maximum covalency
4) Compensation of change in electronegativity across a period by that in a group.
29. In comparison to boron, beryllium has:
1) lesser nuclear charge and greater first ionisation enthalpy
2) lesser nuclear charge and lesser first ionisation enthalpy
3) greater nuclear charge and greater first ionisation enthalpy
4) greater nuclear charge and lesser first ionisation enthalpy
30. Which is/are amphoteric oxides?
1) BeO 2) SnO
3) ZnO 4) All of these

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LEVEL - II
1. Which of the following statement is incorrect?
1) Mendeleev’s periodic law was based on atomic numbers of the element.
2) Zero group was not present in the periodic table when Mendeleev presented it.
3) The effective nuclear charge (Zeff) is the atomic number minus shielding effect.
4) There are four transition series in the periodic table each one consists of 10 elements.
2. The element with Z = 120 (not yet discovered) will be an/a:
1) Inner-transition metal 2) Transition metal 3) Alkaline earth metal 4) Alkali metal
3. In which block, 106th element belongs:
1) s-block 2) p-block 3) d-block 4) f-block
4. An element X belongs to fourth period and fifteenth group of the periodic table. Which one of the following is
true regarding the outer electronic configuration of X? It has :
1) partially filled d-orbitals and completely filled s-orbital
2) completely filled s-orbital and completely filled p-orbitals
3) completely filled s-orbital and half-filled p-orbitals
4) half-filled d-orbitals and completely filled s-orbital
5. In the long form of the periodic table, the valence shell electronic configuration of 5s2 5p4 corresponds to the
element present in:
1) Group 16 and period 6 2) Group 17 and period 5
3) Group 16 and period 5 4) Group 17 and period 6
6. In which of the following sets of atomic numbers, all the elements do not belong to the same group?
1) 3, 11, 37 2) 12, 38, 56 3) 10, 16, 24 4) 10, 18, 54
7. Which of the following element is just below in the E.C. of [Ne]3s2 3p6 4s2 3d10 4p3
1) As 2) Ge 3) Sb 4) Sn
8. In general, the properties that decrease and increase down a group in the periodic table, respectively, are:
1) electronegativity and electron gain enthalpy 2) electronegativity and atomic radius
3) atomic radius and electronegativity 4) electron gain enthalpy and electronegativity
9. Elements a, b, c, d and e have the following electronic configuration
a) 1s2 2s2 2p1 b) 1s2 2s2 2p6 3s2 3p1 c) 1s2 2s2 2p6 3s2 3p3
d) 1s2 2s2 2p6 3s2 3p5 e) 1s2 2s2 2p6 3s2 3p6 4s2
Which among these will belong to the same group in the periodic table?
1) a & d 2) b & c 3) a & b 4) a & e
10. Which of the following is correct regarding the variation of properties in a group of the periodic table?

1) 3 Li 19 K : Ionization enthalpy increases.

2) 9 F 35 Br : Electron gain enthalpy with negative sign increases.
3) 6 C 32 Ge : Atomic radii increases.
4) 18 Ar 54 Xe : Noble character increases.
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11. In which of the following sets of elements first element is more metallic than second?
a) Ba, Ca b) Sb, Sn c) Ge, S d) Na, F
1) a only 2) a and b 3) a, b and d 4) a, c, d
12. The ionic radii of N , O and F are respectively given by
3– 2– –

1) 1.36, 1.40, 1.71 2) 1.36, 1.71, 1.40 3) 1.71, 1.40, 1.36 4) 1.71, 1.36, 1.40
13. In which of the following arrangements, the order is not correct according to the property indicated against it?
1) Increasing size : Al3+ < Mg2+ < Na+ < F– 2) Increasing size IE1: B < C < N < O
3) Increasing size EA1: I < Br < F < Cl 4) Increasing metallic radius : Li < Na < K < Rb
14. The first ionisation potential of Na, Mg, Al and Si are in the order
1) Na < Mg > Al < Si 2) Na > Mg > Al > Si
3) Na < Mg < Al > Si 4) Na > Mg > Al < Si
15. Ionization potential of phosphorus is greater than that of sulphur because
1) of its smaller size. 2) of more penetrating power of p-orbitals.
3) its nuclear force of attraction on electrons. 4) half-filled orbitals are more stable.
16. The incorrect statement among the following is :
1) the first ionization potential of Al is less than the first ionization potential of Mg
2) the second ionization potential of Mg is greater than the second ionization potential of Na.
3) the first ionization potential of Na is less than the first ionization potential of Mg.
4) the third ionization potential of Mg is greater than the third ionization potential of Al.
17. The sucessive ionisation energy values for an element X are given below :
a) 1st ionisation energy = 410 kJ mol –1
b) 2nd ionisation energy = 820 kJ mol –1
c) 3rd ionisation energy = 1100 kJ mol –1
d) 4th ionisation energy = 1500 kJ mol –1
e) 5th ionisation energy = 3200 kJ mol–1
Find out the number of valence electron for the atom, X :
1) 4 2) 3 3) 5 4) 2
18. First and second ionisation energies of magnesium are 7.646 and 15.035 eV respectively. The amount of
energy in kJ needed to convert all the atoms of magnesium into Mg2+ ions present in 12 mg of magnesium
vapours is: (Given 1eV = 96.5 kJ mol-1)
1) 1.1 2) 1.5 3) 2.0 4) 0.5
19. The transformation, Na(s)  Na+(g) involves:
1) ionization energy 2) sublimation energy
3) both ionization energy and sublimation energy 4) vapourisation energy

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20. The first eight ionisation energies for a particular neutral atom is as given below. Which oxidation states are not
possible of the atom?
IE1 IE2 IE3 IE4 IE5 IE6 IE7 IE8
1000 2252 3363 4556 7004 8496 27108 31724
-1 -1 -1 -1 -1 -1 -1 -1
kJmol kJmol kJmol kJmol kJmol kJmol kJmol kJmol
a) –2 b) –3 c) –6 d) +6
1) c 2) c, d 3) b, c 4) a, b, c
21. The formation of the oxide ion O2- (g) requires first an exothermic and then an endothermic step as shown below,
O  g   e  O   g  ; H o  142 kJ mol1

O  g   e  O2  g  ; Ho  844 kJ mol1

This is because
1) oxygen is more electronegative
2) oxygen has high electron affinity
3) O– ion will tend to resist the addition of another electron
4) O– ion has comparatively larger size than oxygen atom
22. Which of the following transitions involves maximum amount of energy?

1) M (g)  M g  2) M (g)  M   g 

2 3

3) M (g)  M2g  4) M (g)  M  g 

23. Which of the following characteristics regarding halogens are correct?

a) Ionization energy decreases with increase in atomic number
b) Electronegativity decreases with increase in atomic number
c) Electron affinity decreases with increase in atomic number
d) Enthalpy of fusion increases with increase in atomic number
1) a, b, c, d 2) a, b, d
3) a, c, d 4) a, b
24. Which of the following element has the highest EN?
1) As 2) Sb 3) P 4) S
25. The electronegativity of H, O and X are 2.1, 3.5 and 0.7 respectively. The correct nature of compound
X–O–H is
1) acidic 2) basic
3) amphoteric 4) cannot be predicated

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26. Which of the following affects the electronegativity of an atom?

a) s-character in hybridization b) Multiplicity of bond between atoms
c) Oxidation number d) The number of neutrons in the nucleus
1) a, b, c 2) a, c, d 3) a, c 4) a, b
27. Which of the following is not concerned to effective nuclear charge?
1) Higher ionization potential of carbon than boron.
2) Higher ionization potential of magnesium than aluminium.
3) Higher values of successive ionization energy.
4) Higher electronegativity of higher oxidation state.
28. Beryllium and aluminium exhibit many properties which are similar. But the two elements differ in:
1) exhibiting amphoteric nature in their oxides 2) forming polymeric hydrides
3) forming covalent halides 4) exhibiting maximum covalence in compounds
29. Consider the electronic configurations, (i) 1s22s22p63s1 (ii) 1s22s22p63s0
Which of the following statements is not true?
1) Energy is required to change (i) to (ii)
2) (i) represents electronic configuration of Na-atom
3) (i) and (ii) may represent ground state electronic configuration of different or same elements
4) Less energy is required to remove one electron from (ii) than from (i)
30. The electronic configuration of four elements are :
1) [Xe] 6s1 2) [Xe] 4f14, 5d1, 6s2 3)[ Ar] 4s2, 3d10, 4p5 4) [Ar] 3d7, 4s2
Which one of the following statements about these elements is not correct?
1) 1 is a strong reducing agent 2) 2 is a d-block element
3) 3 has high electron affinity 4) 4 shows variable oxidation state

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SOLUTIONS
LEVEL - I
1. 2 The law of triads is not applicable to Na, K, Rb. Atomic mass of K is not average of Na and Rb.
2. 1
3. 1
4. 3
5. 4
6. 4 Ununennium (or Eka-francium)
7. 3
8. 3 Z = 33 is Arsenic, a metalloid.
9. 4 The elements with atomic number 3 and 12 belongs to s-block.
Z = 3 : 1s2 2s1
Z = 12 : 1s2 2s2 2p6 3s2
10. 4 The stability of +2 oxidation state increases due to inert pair effect.
11. 1
12. 3 Down the group, atomic size increases as extra shell of electron being added.
13. 4
14. 4 a, b, c, d
Co-valent radius depends on all these factors.
15. 3 The correct order is S2- > Cl- > K+ > Ca2+
16. 2
17. 2 IE decreases down the group and increases along the period. The correct order of IE1 is Be > Mg > Ca
18. 1 IE and EA of an element are equal in magnitude but opposite in sign.
 IE   EA
19. 2 The highest IE is of (Ne) 3s2 3p3 due to half filled p-subshell which has extra stability.
20. 1 According to Mulliken’s equation.
IP  EA
 when both IP and EN are taken in eV;;
5.6
17.42  3.45
F   3.726
5.6
21. 2 Noble gases are at the maxima which have closed electron shells and very stable electronic configuration.
On the other hand, minima occurs at the alkali metals and their low IE1 can be correlated with their high
reactivity.
22. 2
23. 3 Second electron gain enthalpy is always positive for an element because energy is required to add the
second electron to negatively charged ion.
24. 2 O  g   e   
 EA1
O g  e    
EA 2
 O 2 g 
Energy is required to add an electron to the negatively charge species due to electron–electron repulsion.

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25. 3 a, b, d
26. 2
27. 1 The element having atomic number (Z – 1) is a halogen and has the highest electronegativity in that period.
28. 4
29. 1 In case of Be electron is to be removed from filled 2s subshell while in case of B, the electron is to be
removed from outer 2p subshell (2s22p1). Hence, Be has higher ionization enthalpy than B though its
nuclear charge is less.
30. 4 BeO, SnO, ZnO, all are soluble in alkalies as well as in acids, thus are amphoteric oxides.
MO  2HCl 
 MCl2  H 2 O

MO  2NaOH 
 Na 2 MO 2  H 2 O
LEVEL - II
1. 1 This law was based on atomic masses of the elements.
2. 3 The element with Z = 120 will have the electronic configuration; [Noble gas] ns2. It will belong to alkaline
earth metals.
3. 3 Element belongs to d-block is unnilhexium (Unh) 106
4. 3
5. 3 The highest principal quantum number available in the electronic configuration is the period number of that
element, therefore, the period for given electronic configuration is 5. 5p4 corresponds to group 16.
6. 3 All the elements in the same group will have similar valence shell electronic configurations.
10
Ne, 16S and 24Cr are elements of different groups.
7. 3 The electronic configuration given is of As. The element just below As33 is Sb51.
8. 2 Electronegativity decreases as we go down the group and atomic radius increases as we go down the
group.
9. 3 Similar valence e–, 2s2 2p1 and 3s2 3p1.
10. 3 Second and higher electron gain enthalpy values are positive.

6 C 32 Ge : Moving top to bottom atomic radii increases in a group.

11. 4 a, c, d
Sn is more metallic than Sb.
12. 3
13. 2
14. 1 IE1 of Mg is higher than that of Na because of increased nuclear charge and also than that of Al because
in Mg an s-electron has to be removed. While in Al it is 3p electron that has to be removed. Mg also has
stable fully filled configuration. The IE1 of Si is, however, higher than those of Mg and Al because of
increased nuclear charge. Thus, the overall order is Na < Mg > Al < Si.
15. 4 Phosphorus contains half-filled 3p orbitals, have extra stability. So, the removal of an electron from
phosphorus atom requires more energy than sulphur.
16. 2 Na+ has noble gas configuration.
17. 1

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18. 1 Total energy needed to convert one Mg atom into

Mg 2g ion = IE1 + IEn = 22.681 eV = 2188.6 kJ mol-1

12 mg of Mg = 0.5 × 10-3 mole


The process is Na  s   Na  g   Na  g  . Therefore, total energy = SE + IE
SE IE
19. 3
20. 3 b, c
There is a larger jump in the IE6 to IE7, so removal of 7th electron is very difficult.
Hence, the oxidation state of the neutral atom can be +6 or (6 – 8) = –2.

21. 3 O  g   e  
  O2  g  ; H o  844 kJ mol1
This process is unfavourable in the gas phase because the resulting increase in electron-electron repulsion
overweighs the stability gained by achieving the noble gas configuration.
22. 4
23. 2 a, b, d electron affinity of Cl is more than that of F.

24. 4 EN down the group    decreases and increases along the period  
  P, As, Sb elements belong
to group 15 and EN decreases from P  As  Sb . But S belongs to group 16 and EN increases from
P  S . Hence, S has the higest EN among the above elements.
25. 2 The polarity of X–O bond is greater than O–H bond because the electronegativity difference of X–O
bond (3.5 – 0.7 = 2.8) is greater than that of O–H bond (3.5 – 2.1 = 0.4). Hence, X–O bond will be
broken first, thus X–O–H is basic in nature.
26. 1 a, b, c
Increases in ‘s’character increases electronegativity. Higher the oxidation state, more is the electronegativity.
27. 2
28. 4 Maximum co-valence of Be is 4 where as that of Al is 6.
29. 3
30. 2 It is 71Lu, the last member of Lanthanide series.

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CHAPTER - 04
CHEMICAL BONDING

INTRODUCTION
A chemical bond is any of the interactions that account for the association of atoms into molecules, ions,
crystals and other stable species that make up the familiar substances of the world. It may be stated that every
system tends to attain greater stability and bonding is nature’s way of lowering energy and attaining stability.
A molecule is formed only if it is more stable (lower energy) than the individual atoms. Generally, electrons in
the outermost shell (valence shell) of atoms are involved in bond formation.
Different theories have been put forward to explain chemical bonding. These are Kossel-Lewis approach,
Valence Shell Electron Pair Repulsion (VSEPR) Theory, Valence Bond (VB) Theory and Molecular Orbital
(MO) Theory. These are related to the developments in understanding the structure of the atom, electronic
configuration of elements and the periodic table.
4.1 KOSSEL - LEWIS APPROACH TO CHEMICAL BONDING (OCTET RULE)
In 1916, Kossel and Lewis gave a satisfactory explanation for the formation of the chemical bond in terms of
electrons.
Lewis developed the electronic theory of chemical bonding in which he pictured the atom in terms of a positively
charged kernel made up of the nucleus and the inner electrons and an outer shell that could accomodate a
maximum of eight electrons. He further assumed that eight electrons occupy the corners of a cube which
surrounds the ‘kernel’ and that atoms achieve stable octet when they are linked by chemical bonds.
Octet Rule. Atoms combine either by transfer of valence electrons from one atom to another or by
sharing of valence electrons in order to have an octet of electrons in their valence shells.
In NaCl, this happens by the transfer of an electron from Na to Cl, thereby forming Na+ and Cl– ions which are
stabilized by electrostatic attraction. In case of molecules like H2, F2, Cl2, etc., bond formation is by the sharing
of a pair of electrons between two atoms by which each atom attains stable outer octet of electrons.
4.1.1 Lewis Symbols
American Chemist G. N. Lewis introduced simple notations called Lewis symbols to represent valence electrons
in an atom. The number of dots around the symbol represents the number of valence electrons.

The group valence is generally either the number of dots in Lewis symbols or 8 minus the number of dots.
Lewis Representation of Simple Molecules. The Lewis dot structure provides an overall picture of bonding
in molecules and ions in terms of shared pairs of electrons and the octect rule.The dots represent electrons.
.. .. . .. . ... .. ... .. .. .. ..
.. .. ..F.. N
2-

[ [
2-
O F .  ...
[ [
...O. ... ; .. ...... .. F N F ... ; ..
O C ..
.. O  .. = C O
O ..
.. ..
..
..
.... ..

..

..

O .O

..

.
=

.. O F
..
.. .

O .. O
..

O
.

..F..
..
..
..

..
..

..
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In relation to chemical bonding, Kossel considered the following facts :

i. In the periodic table, the highly electronegative halogens and the highly electropositive alkali metals are
separated by the noble gases.
ii. The formation of a negative ion from a halogen atom and a positive ion from an alkali metal atom is by the
gain and loss of an electron by the respective atom.
iii. The negative and positive ions thus formed have noble gas electronic configurations. Noble gases (except
helium with a duplet of electrons) have stable outer shell configuration of eight (octet) electrons, ns2np6.
iv. The negative and positive ions are stabilized by electrostatic attraction. For example, in NaCl, electropositive
sodium is in combination with electronegative chlorine such that both atoms attain stable, inert gas
configuration; sodium attains the configuration of neon and chlorine attains the configuration of argon.

 Na   e  ; Cl  e  
Na   Cl  overall Na   Cl  
 NaCl

The bond formed, as a result of the electrostatic attraction between the positive and negative ions is called
electrovalent bond. The electrovalence is equal to the number of unit charge(s) on the ion. Thus, sodium is
assigned a positive electrovalence of one and chlorine is assigned a negative electrovalence of one.
Similarly, formation of CaCl2 occurs by the transfer of two electrons from Ca to two Cl atoms.

Both calcium and chlorine attain noble gas (argon) configuration.

4.2 THE COVALENT BOND
Langmuir (1919) refined Lewis theory by abandoning the idea of stationary cubical arrangement of the octet
and introducing the term covalent bond. According to Lewis- Langmuir theory, formation of a molecule is by
the sharing of a pair of electrons between two atoms, each contributing one electron to the shared pair. In the
process, both the atoms attain octet.
When two atoms having similar electronegativities combine, they share a pair of electrons and attain noble gas
configuration (covalent bonds are shown as straight lines joining two atoms). E.g., chlorine molecule is formed
by the sharing of a pair of electrons by two chlorine atoms.

..
..Cl .. . .. . .. .
..Cl
.. . .Cl
.. . .. . or Cl
.. .Cl Cl

Ammonia is formed by the sharing of electrons between one nitrogen atom and 3 hydrogen atoms.

.. ..
. .. H or H N H ..
.N. . 3[ H.
[
H.N
..
H H
Hydrogens attain duplet while nitrogen attains octet of electrons. One pair of electrons on the N atom does
not involve in bond formation and it is called a lone pair.

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In a similar way water, methane, ethane etc. are formed as follows.

.. .
..O .. . ..
. +2 [ H.] ..O
... H or ..O – H
H
H
H H
...
.C.. . + 4[H]. H..C
.. . H or H C H
H H
... H
H ... H H
H ..C
...C...H or H C C H
HH H H
4.2.1 Multiple Covalent Bond
Covalent bond can be single bond, double bond or triple bond according to the number of electron pairs
shared between two atoms. When two atoms share two pairs of electrons, the bond is called a double bond,
e.g., CO2, C2H4, O2, etc.
H. .. ..H H H
.. .. .. .. .
H. C C. H or H
C = C
. .. O.. (O = O)
.O
H
Oxygen molecule (double bond) Ethene (double bond)
When two atoms share 3 pairs of electrons, it is said to be a triple bond, e.g., ethylene, N2 etc.
.. ..
H . C ..
.... C .. H H C CH N ...
... N or (N N)
Ethyne (triple bond) Nitrogen molecule (triple bond)
4.2.2 Formal Charge
In a polyatomic ion, a formal charge can be assigned to each atom, based on the difference between the
number of electrons assigned to it in the Lewis structure.
1
Formal charge on an atom in Lewis structure  V  N  B
2
where V is the number of valence electrons on the atom, N is the number of non-bonding electrons on the
atom and B is the number of bonding (shared) electrons on the atom.
Counting of electrons is based on the assumption that the atom in the molecule owns one electron of
each shared pair and both the electrons of the lone pair.
1. Formal charge in ozone molecule
1..
O
O
(a) 2 .. (b)
O.. O 3 O O
..
..
..

In structure (a): Formal charge on oxygen atom marked as 1  6  2  [(1 2)  6]  1

Formal charge on oxygen atom marked as 2  6  4  [(1 2)  4]  0 and
Formal charge on oxygen marked as 3  6  6  [(1 / 2)  2]  1
Hence, the structure of ozone molecule with formal charges is (b).
Formal charges do not indicate real charge separation within the molecule.
Generally, the structure having lowest energy is the one with the smallest formal charges on the atoms.

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2. Formal charge on sulphur atom in sulphuric acid

..
O O

..
..
..
..
.. 2+ .. .. ..
H–O
.. – S – O
.. – H H–O
.. – S – O
.. – H
O.. O

..
..

..
..
(I) (II)
In structure (I): FC sulpur  6  0  (8 / 2)  2 ; FC oxygen  6  6  (2 / 2)  1 (O bonded to S only)
FC oxygen  6  4  (4 / 2)  0 (O bonded to S and H) ; FC hydrogen  1  0  (2 / 2)  0
In structure (II): FC sulpur  6  0  (12 / 2)  0 ; FC oxygen  6  4  (4 / 2)  0 (O bonded to S only)
The H and O atoms bonded to S atom are similar to those in structure (I).
Structure (II) is more stable since structure (I) has greater potential energy due to separation of charges.
4.2.3 Limitations of Octet Rule
The octet rule applies mainly to the second period elements. There are three types of exceptions.
1. Incomplete octet of the central atom: In some compounds, especially those with valency less than four,
the number of electrons surrounding the central atom may be less than eight, e.g., LiCl, BeH2, BCl3, BF3,
AlCl3, etc. (electron- deficient molecules).
Cl
..
Li .. Cl H .. Be .. H Cl .. B .. Cl
2. Odd-electron molecules: In case of odd-electron molecules like NO, NO2, etc., octet rule is not satisfied
for all atoms.
.. .. .. . + .. .
N. O.. O.. N ...
O
3. The expanded octet: In elements having 3d orbitals available for bonding, there can be more than 8
electrons around the central atom. This is termed expanded octet. e.g., SF6, PF5, H2SO4, and many coordination
compounds.

(10 e– around P atom) (12 e– around S atom) (12 e– around S atom)

Other drawbacks of octet theory. i. Octet theory is based on the chemical inertness of noble gases. However,
some noble gases like Kr and Xe form compounds like XeF2, XeF4, XeF6, XeF3, XeOF2, KrF2, etc.
ii. The theory does not account for the shapes and stability (energy) of molecules.
4.3 IONIC OR ELECTROVALENT BOND
According to Kossel and Lewis, the formation of ionic compounds depends on (i) the ease of formation of
ions from the respective atoms and (ii) the arrangement of positive and negative ions in the ionic solid or lattice
structure of the crystalline compound.
The formation of positive ion involves ionisation, i.e., removal of electron(s) from the neutral atom and negative
ion involves the addition of electron(s) to the neutral atom.
4.3.1 Factors Favouring the Formation of Ionic Bond. Ionic bonds are formed between elements with low
ionisation enthalpies and high negative electron gain enthalpies.
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Ionisation enthalpy is the enthalpy change accompanying the removal of electron (s) from the outermost
shell to form a positive ion in gaseous phase.
 M + (g) + e 
M(g) 
Lower the ionisation enthalpy (usually endothermic), greater will be the ease of formation of ionic bond.
Electron gain enthalpy (  egH) is the enthalpy change accompanying the addition of an electron to the
gaseous atom in the ground state to form an anion.
 X  (g) ; Electron gain enthalpy,  eg H
X(g) + e  
Greater the electron gain enthalpy, greater will be the energy released during the formation of the anion and
greater will be the ease of formation of the ionic compond.
Metals form cations easily because they have low ionisation and electron gain enthalpies.Non-metals have
large ionisation enthalpiess and negative electron gain enthalpies, therefore, they can easily accept an electron
and get converted to the anion. Thus, ionic componds in crystalline state consists of orderly, three-dimensional
arrangement of cations and anions held together by coulombic interaction energies.
For any stable compond to form from its elements, there must be a net lowering of the potential energy, i.e., the
reaction must be exothermic. In the formation of NaCl, ionisation enthalpies of sodium and chlorine are:
 
Na  Na   e   ie H  495.8 kJ mol 1 ; Cl  e  Cl  eg H  348.7 kJ mol1
The sum of the energy changes associated with the formation of both the ions is +147.1 kJ mol–1. It is more
than compensated by the enthalpy of lattice formation of NaCl (-788 kJ mol-1). Thus lattice enthalpy is a
qualitative measure of the stability of ionic compounds
Lattice enthalpy: Lattice enthalpy of an ionic solid is the energy required to separate one mole of a solid ionic
compond into gaseous constituent ions.
 M  (g)  X (g)
MX(s) 
Greater the lattice enthalpy, greater will be ease of formation and stability of the ionic compond.
The lattice enthalpy of NaCl has been found to be + 787.0 kJ mol–1.
Na (g)  Na  (g)  e   495.4 kJ (Ionisation enthalpy of Na)

Cl(g)  e   Cl(g)

348.8 kJ (Electron gain enthalpy of Cl)
 
Na (g)  Cl(g)  NaCl (s) 787.0 kJ (-Lattice enthalpy)

Na (g )  Cl (g )  NaCl (s)  640.4 kJ

Thus, the release of lattice energy results in lowering of potential energy which favours the formation of NaCl.
4.4 BOND PARAMETERS
4.4.1 Bond Length.
Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule
(R=rA+rB). rA and rB are the covalent radii of the atoms in the diatomic molecule. It is measured by
spectroscopic, X-ray diffraction, and electron-diffraction techniques. The covalent radius is the approximate
radius of an atom’s core which is in contact with the core of an adjacent atom in a bonded situation.

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(a) (b)

Figure - 01 (a) Bond length in covalent molecule (b) covalent radius (rc) and van der Waals radius (rvdw).
van der Waals radius. It represents the overall size of the atom which includes its valence shell in a non-
bonded situation. It is half of the distance between two similar atoms in separate molecules in a solid.
4.4.2 Bond Angle
It is the angle between the orbitals containing the bonding electron pairs around the central atom in a molecule
or complex ion. It is expressed in degree. Bond angle is determined by spectroscopic methods. It helps in
determining the shape of molecules. E.g., the HOH bond angle in water is represented as under:

4.4.3 Bond Enthalpy

The amount of energy required to break one mole of bonds of a particular type between two atoms in the
gaseous state is called bond enthalpy. The unit of bond enthalpy is kJ mol–1.
H 2(g) 
 H (g) + H (g) ;  a H  = 435.8 kJ mol-1

 Cl(g) + Cl(g) ; a H  = 242.7 kJ mol-1

Cl2(g) 
The bond enthalpy for molecules containing multiple bonds like O2 and N2 will be
O2 (O = O) (g)  O(g) + O(g) ;  a H  = 498 kJ mol-1


 N (g) + N (g) ;  a H  = 946.0 kJ mol-1

N 2 (N  N) (g) 
Larger the bond dissociation enthalpy, stronger will be the bond in the molecule.
For a heteronuclear diatomic molecules like HCl,
HCl (g) 
 H(g) + Cl (g);  a H  = 431.0 kJ mol-1
In case of polyatomic molecules, the mean or average bond enthalpy is calculated.
H 2 O(g) 
 H (g) + OH (g) ;  a H1 = 502 kJ mol-1

OH (g) 
 H (g) + O(g) ;  a H 2 = 427 kJ mol-1
Total bond dissociation enthalpy
Average bond enthalpy =
Number of bonds broken
Average bond enthalpy of water = (502+ 427)/ 2 = 464.5 kJ mol-1
4.4.4 Bond Order
The bond order is the number of bonds between two atoms in a molecule. E.g., bond order of H2 (with a single
shared electron pair) is 1, O2 (with two shared electron pairs) is 2 and N2 (with three shared electron pairs) is
3. Similarly in CO (three shared electron pairs between C and O) the bond order is 3.

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For N2, bond order is 3 and its  a H  is 946 kJ mol-1; one of the highest for a diatomic molecule. Isoelectronic
molecules like F2 and O22- have identical bond orders (1). Bond order of N2, CO and NO+ is 3.
With increase in bond order, bond enthalpy increases and bond length decreases (stability increases).
4.4.5 Polarity of Bonds - Dipole Moment
A covalent bond between two atoms acquires partial polar character when electronegativity values of the two
bonded atoms differ. The charged ends of the bond behave as an electrical dipole and the degree of polarity is
measured in terms of dipole moment. Dipole moment is defined as the product of the magnitude of charge
on any one of the atoms and the distance between them.
Dipole moment,  = Q  r
where, ‘Q’ is the charge on any one of the atoms and ‘r’ is the distance between the atoms.
The unit of dipole moment is called Debye (1×10–18 e.s.u. cm) represented by D.
In SI units, dipole moment is the Coulomb-meter abbreviated as C m.
1 D = 1×10–18 e.s.u. cm. = 3.33564 × 10–30 C m
Thus, dipole moment of HCl, 1.03 ×10–18 e.s.u. cm is expressed as 1.03 D.
It is a vector quantity as it has direction as well as magnitude. The direction of dipole moment is usually
represented by an arrow pointing from the positive end towards the negative end, e.g.,
Dipole moment gives an idea about the polar character of a molecule.
I. Dipole Moment and Molecular Structure
a) Diatomic molecules. In the case of diatomic molecule, the dipole moment of the bond is also the dipole
moment of the molecule, e.g., dipole moment of HCl molecule is the same as that of H–Cl bond (  = 1.03 D).
Larger the electronegativity difference between the atoms, larger will be the dipole moment of the molecule.
E.g., dipole moment of HF is 1.91 D while that of HCl is 1.03 D (F is more electronegative than Cl).
b) Polyatomic molecules. The dipole moment of individual bonds in a polyatomic molecule is referred to as
bond dipole. The overall dipole moment of the molecule depends on the orientations of the various bond
dipoles. E.g., CO2 and H2O are both triatomic molecules, but dipole moment of carbon dioxide is zero while
that of water is 1.83 D. This can be explained on the basis of their structures.

(a) (b)

CO2 is linear with the two C=O bonds oriented in opposite directions (180o). The bond dipole of C=O bond
is 2.3 D. Due to linear structure, the bond dipoles of two C=O bonds cancel each other and the net dipole
moment of the molecule is zero. Hence, CO2 is nonpolar. Water molecule has a bent structure in which two O–
H bonds are oriented at an angle 104.5o. Therefore, the bond dipoles of the O–H bonds do not cancel each
other. Hence, the resulting dipole moment is 1.83 D.
In BF3, the bond dipoles of the three B–F bonds give a net sum of zero because the resultant of any two is
equal and opposite to the third (parallelogram law of forces).

(c) (d) (e)

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c) Molecules with one lone-pair. The orbital dipole of the lone pair also has to be considered in explaining
the observed values of dipole moment. Although N–F bond is more polar than N–H bond, dipole moment of
NF3 is 0.25 D while that of NH3 is 1.49 D. This is because in NH3, the orbital dipole due to lone pair is in the
same direction as the resultant dipole moment of the three N–H bonds and therefore, it adds on the resultant
of three N–H bonds.
In case of NF3, the orbital dipole due to lone pair is in the opposite direction to the resultant of N–F bonds
and consequently decreases the net dipole moment.
Table - 01 Dipole moments of some common compounds.
Type of Dipole Dipole
Molecule Geometry Molecule Geometry
Molecule Moment Moment
AB H2, N2, O2, F2 0 linear Cl2 0 linear
HF 1.78D linear HBr 0.79D linear
HCl 1.07D linear HI 0.38D linear
AB2 H 2O 1.85D bent SO2 1.60D Linear
H 2S 0.95D bent CO2 0 Linear
AB3 NH3 1.47D pyramidal BF3 0 planar
NF3 0.23D pyramidal SO3 0 planar
AB4 CH4 0 tetrahedral CCl4 0 tetrahedral
CHCl3 1.04D tetrahedral CH2Cl2 1.57D tetrahedral

4.4.6 Resonance
When a molecule can be represented by more than one Lewis formula, none of which is able to describe the
molecule accurately, then the actual structure is a resonance hybrid of the various Lewis formulae (canonical
forms). The resonance between canonical forms can be represented by a double headed arrow (  ). E.g.,
ozone can be represented by two Lewis structures:

In structures I and II, the O = O bond length (121 pm) should be smaller than O – O (148 pm) bond length.
But experiments have shown that both bonds are of equal length, 126 pm, which is intermediate between that
of single and double bonds. Thus, the actual structure of ozone is intermediate between structures I and II and
is called a resonance hybrid. Structures I and II are called resonating structures, contributing structures
or canonical forms.

i) Carbon dioxide molecule (CO2):

ii) Carbonate ion (CO32–):

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Characteristics of Resonance. i. Resonating structures are hypothetical and do not have real existence. The
actual molecule has a single structure which is a resonance hybrid of the various canonical forms.
ii. Bond length in hybrid structure is intermediate between the bond lengths of the various resonating forms,
e.g., C– C bond length in benzene (139 pm) is intermediate between C– C (154 pm) and C = C (134 pm).
iii. Resonance hybrid is more stable, i.e., having lower energy than any of the resonating forms. Canonical
structures of similar energy contribute equally to the resonance hybrid.
iv. The difference in energy between the resonance hybrid and the most stable resonating form is called
resonance energy. Larger the value of resonance energy, greater will be the stability of the resonance
hybrid.
4.4.7 Covalent Character of Ionic Bond
When two oppositely charged ions come together, the positive ion tends to distort the electron cloud of the
negative ion towards itself. Thus, the ionic bond does not remain 100% ionic, but develops some covalent
character. Covalent character of ionic bond depends on: (i) polarising power of cation and (ii) polarisability
of anion.The power of cation to cause distortion in the electron cloud of the negative ion is referred to as its
polarising power and the ability of anion to undergo distortion is called its polarisability.
The extent of covalent character can be decided on the basis of a set of rules called Fajan’s rules.
(i) Smaller the size of cation, larger will be its polarising power.
Li+ is smaller than K+ ion, therefore, LiCl has more covalent character than KCl.
Polarising power of cation is usually called ionic potential or charge density.
Polarisation of the anion increases as the size of the cation decreases. In a group, ionic character increases
from top to bottom. For example, Be2+ shows maximum covalent character:
BeCl2< MgCl2< CaCl2< SrCl2< BaCl2
In a period, covalent character increases as the size of cation decreases: Na+<Mg2+<Al3+<Si4+
(ii) For two cations of similar size, the one with pseudo noble gas configuration (ns2np6nd10) has larger
polarising power than one with noble gas configuration (ns2np6).
CuCl is more covalent than NaCl because polarising power of Cu+ ion, which has pseudo noble gas
configuration (3s2 3p6 3d10), is greater than Na+ ion (2s2 2p6). Zeff of ns2np6(inert) < Zeff of ns2 np6 nd10
(pseudo-inert).
(iii) Larger the size of anion, higher will be its polarisability. LiI is more covalent than LiF. AlF3 is ionic,
but AlCl3 is covalent. For the same cation, covalent character increases with size of the anion:
CaF2<CaCl2<CaBr2<CaI2
(iv) Charge on cation and anion. Polarisation increases with increase in charge on cation and anion (ionic
character decreases or covalent character increases): Na+<Mg2+<Al3+<Si4+
4.5 VALENCE SHELL ELECTRON REPULSION (VSEPR) THEORY
The theory was proposed by Sidgwick and Powell (1940) and modified by Nyholm and Gillespie (1957). It
is based on the repulsive interactions of electron pairs in the valence shell of atoms. The shapes of covalent
molecules can be predicted using this theory. The main postulates are:

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1. The shape of a molecule depends on the number of valence shell electron pairs, bonded or non-bonded,
around the central atom.
2. The electron pairs in the valence shell repel one another due to electrostatic repulsion between their
electron clouds.
3. The electron pairs occupy positions in space in such a way that repulsion is minimised and distance
between them is maximum. The valence shell is considered as a sphere with the electron pairs localised on
the surface at maximum distance from one another.
4. A multiple bond is treated like a single electron pair and the two or three electron pairs of a multiple bond
are considered as a single super pair.
5. When a molecule has two or more resonance structures, VSEPR model can be applied to any one
structure. Formal charges are not included in the representation.
6. The magnitude of repulsions between bonding pairs of electrons depends on the electronegativity difference
between the central atom and the other atoms in the molecule.
7. The lone pair (lp) of electrons takes up more space around the central atom than a bond pair (bp) since
lone pair is attached to one nucleus while bond pair is shared by two nuclei. Thus, the presence of lone
pairs on the central atom causes repulsions which result in distortion of bond angles. The order of repulsion is:
Lone pair – lone pair > lone pair – bond pair > bond pair – bond pair.
8. Sometimes, the lone pair may be transferred from the filled shell of an atom to the unfilled shell of the
adjacent bonded atom. This phenomenon is called back bonding.
4.5.1 Molecular Geometry
For the prediction of geometrical shapes of molecules, molecules are divided into two categories; (i) molecules
in which the central atom has no lone-pair and (ii) molecules in which the central atom has ione-pair(s).
In compounds of the type, AB2, AB3, AB4, AB5 and AB6, the arrangement of electron pairs and the B atoms
around the central atom A are : linear, trigonal planar, tetrahedral, trigonalbipyramidal and octahedral, respectively.
Such arrangement can be seen in the molecules like BF3 (AB3), CH4 (AB4) PCl5 (AB5) and SF6(AB6) .

1. Based on number of electron pairs: Molecules formed from elements A and B.

2. In presence of lone pairs on the central atom. Presence of lone -pairs affect the geometry of molecules.
There are three types of repulsive forces between lone-pairs (lp) and bond-pairs (bp):
lp-lp repulsion > lp-bp repulsion> bp-bp repulsion.

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Molecule No. of No. of Arrangement of

Shape of molecule
Type Bond pairs Lone pairs Electron pairs

AB2E 2 bp 1 lp (Bent)

The O=S=O bond angle is 119.5o (<120o) due to lp-bp repulsions. Other examples are O3, SnCl2, etc.

AB3E 3 bp 1 lp (Trigonal)

H-N-H bond angle is 107o(<109o 28’) due to lp-bp repulsions. Other examples are NCl3, PH3, PCl3, etc.

AB2E2 2 bp 2 lp (Bent)

Larger lp-bp interactions push O-H bonds closer, thereby decreasing H-O-H bond angle to 104.5o
(<109o 28’). Other examples are H2S, NH2-, etc.

AB4E 4 bp 1 lp (See-saw)

The lone-pair occupies equitorial position, since repulsive interactions are minimum (120o) compared to
axial positions (90o). Therefore, the molecule has irregular tetrahedral geometry or see-saw shape.

AB3E2 3 bp 2 lp (T-shape)

The lone-pairs occupy equitorial positions to minimise repulsive interactions (120o). Therefore, the molecule
has T-shape. Other examples are BrF3, IF3, etc.

AB2E3 2 bp 3 lp (Linear)

All lone-pairs occupy axial positions (120o). Other examples are I3-, Br3-, etc.

AB5E 5 bp 1 lp (Square pyramidal)

Other examples are BrF5 and ClF5.

AB4E2 4 bp 2 lp (Square planar)

The lone-pairs occupy axial positions, therefore, lp-lp repulsion is minimum.

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Applications of VSEPR theory. (i) The theory is sucessful in determining the geometry even when the
energy difference between possible structures is very small. (ii) The VSEPR theory could accurately predict
the geometry of molecules, especially of componds of p-block elements.
Limitations of VSEPR Theory. (i) The theory could not predict the shapes of many molecules, and therefore,
has limited application. (ii) Effect of electron pair repulsion on shape of molecules is not very clear.
QUANTUM MECHANICAL THEORY OF CHEMICAL BONDING
In order to overcome these limitations, two theories based on quantum mechanical principles were introduced;
(i) Valence bond theory (VBT) and (ii) Molecular orbital theory (MOT).
4.6 VALENCE BOND THEORY (VBT)
This theory was proposed by Heitler and London (1927) and developed further by Pauling and Stater. According
to this theory, atoms with unpaired electrons combine with each other to form molecules. In this way, the
electrons are paired up and the participating atoms attain stable electronic arrangement.
Formation of H2 molecule. When two hydrogen atoms HA and HB approach each other, the following
interactions occur : (i) Attractive forces between electron eA and nucleus HB and electron eB and nucleus HA
tend to bring the atoms close together. (ii) Repulsive forces between electrons eA and eB and nuclei HA and HB
tend to push them apart.

(a) (b)

Figure - 02 (a) Attractive and repulsive forces during formation of hydrogen molecule, (b) Variation of potential
energy during bond formation.
As the magnitude of attractive forces is greater than the repulsive interactions, the atoms approach each other
and the potential energy of the system decreases. At a certain distance r0, the attractive forces exactly balance
the repulsive forces and the energy of the system is minimum. The distance r0 is called bond length and the
energy released corresponding to the minimum in the curve is called bond enthalpy or bond dissociation
enthalpy. The bond length of the H2 molecule is 74 pm and the bond dissociation enthalpy is 435.8 kJ mol-1.
4.6.1 Essential conditions for the formation of a chemical bond
i. There should be maximum overlap of the atomic orbitals of the participating atoms.
ii. Each of the two atomic orbitals must have one unpaired electron with opposite spin.
By the overlap of atomic orbitals, the electron pair becomes concentrated in the region of overlap and helps to
bind the nuclei together. The lowering of potential energy due to bond formation is determined by the extent of
overlap of orbitals.
The number of bonds formed by an atom in the ground state is usually the number of unpaired electrons, but
the atom may form more bonds than the number of unpaired electrons when it is excited.
By providing energy, electrons which are paired up in the ground state can be unpaired and promoted to
suitable empty orbitals. This increases the number of unpaired electrons and hence the number of bonds that
can be formed.
The shape of the molecule is determined by the directions in which the orbitals point.
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4.6.2 Bonding as Overlap of Atomic Orbitals

A covalent bond is formed when the half-filled valence atomic orbitals of two atoms overlap with one another
and the electrons pair up in the overlap region (undergo partial interpenetration). Partial merging of atomic
orbitals is called overlap and the extent of overlap decides the strength of the bond.
Positive or in phase overlap : Overlap of lobes of the same sign leads to attractive interactions and bond
formation.

Negative (or out of phase) overlap : Overlap of lobes of opposite sign leads to repulsive interacions. No
bond is formed.

Zero (out of phase due to different orientation) overlap : Inability for interactions. According to latest
convention z-axis is taken as the internuclear axis. Thus, s-atomic orbital can combine with pz-atomic orbital
but not with px or py atomic orbitals (a). p-orbitals combine only if their orientations in space are the same (b).

(a) (b)

4.6.3 Types of Overlap and Nature of Covalent Bonds.

Covalent bonds are classified into sigma (  ) bonds, pi (  ) bonds, delta () bonds and phi () bonds.
1. Sigma bond: This type of covalent bond is formed by the end-to-end (head-on or axial) overlap of bonding
orbitals along the internuclear axis.
a) s-s overlap. Overlapof two half-filled s-orbitals, e.g., H2 molecule.
b) s–p overlap. Overlap of a half-filled s-orbital of one atom with a half-filled p-orbital of another, e.g.. HX
molecule (X = F, Cl, Br, I).

(a) (b)
c) p-p overlap. Overlap of half-filled p-orbitals of the participating atoms, e.g., X2 molecule (X = F, Cl, Br, I).
Bond strength is proportional to area of overlap. The relative strength is in the order: s-s < s-p < p-p .
The relative strength is maximum for p-p overlap since area of overlap is maximum in p-p overlap.

2. Pi () bond. A - bond is formed by the lateral (sideways) overlap of p-orbitals (and d-orbitals). In a
-covalent bond, the region of overlap of atomic orbitals lie parallel to the line joining the two nuclei, both
above and below the inter nuclear axis. bond merely shortens the bond length.

The relative strengths of -bonds increases with internuclear distance in the order: 2p– 2p > 2p – 3d >
2p – 3p > 3p – 3p

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4.6.4 Strength of Bond

The strength of a bond depends on the extent of overlap between orbitals of the participating atoms.  -bonds
in s-s, s-p and p-p overlap are cylindrical about the bond axis (internuclear axis).
- bond is always formed in addition to a  - bond (in componds having multiple bonds). In - bonds, lobes
of unhybridised orbitals overlap above and below the bond axis. This overlap is weaker than the  -bond.
4.6.5 Directional Propertiess
Strength of the bond depends on the extent of overlap, and therefore, orbitals overlap only in certain directions
to attain maximum overlap. VB theory explains the directional properties of bonds and the geometry of molecules
in terms of overlap and hybridisation.
4.7 HYBRIDISATION
In order to explain the valencies of atoms like Be, B, C, etc., and the characteristic geometrical shapes of
polyatomic molecules like CH4, NH3, H2O, etc., Linus Pauling (1931) introduced the concept of hybridisation.
It is defined as the process of intermixing of orbitals of slightly different energies resulting in the formation of a
new set of orbitals of equivalent energies and shape. The driving force for hybridisation is the formation
of a bonding geometry with the lowest net potential energy.
4.7.1 Salient Features of Hybridisation
1. The number of hybridised orbitals formed is equal to the number of atomic orbitals hybridised.
2. The hybridised orbitals are always equivalent in energy and shape.
3. The atomic orbitals involved in hybridisation can be empty, half-filled or completely filled.
4. The hybridised orbitals have two lobes, one larger and one smaller; larger lobe is more effective in overlapping
with other atomic orbitals and forming bonds. Smaller lobes are generally not shown since they are close to
the nucleus and do not participate in bond formation.
5. The hybridised orbitals are oriented in certain directions in space. The type of hybridisation determines the
geometry of the molecule.
6. Hybridised orbitals are usually involved only in the formation of  - bonds (  - bonds result from the
overlap of unhybridised orbitals).
7. Hybridisation occurs only in covalent bond formation with other atoms.
4.7.2 Important Conditions for Hybridisation
1. Usually, only the orbitals of the valence shell undergo hybridisation. The inner atomic orbitals which are not
used remain unaffected by hybridisation.
2. The energy of the mixing orbitals must be the same or nearly the same.
3. Promotion of electrons to empty orbitals is not a necessary step in hybridisation.
4. It is not essential that only half-filled orbitals take part in hybridisation. In some cases, filled orbitals of the
valence shell participate in hybridisation.
4.7.3 Types of Hybridisation
There are various types of hybridisation involving s, p, and d orbitals.
1. sp-Hybridisation (Diagonal or Linear Hybridisation). sp-hybridisation involves the mixing of one
s-orbital and one p-orbital to form two sp-hybrid orbitals of equivalent shape and energy. Generally, the pz-
orbital is involved in hybridisation. The two sp-hybrid orbitals are oriented at an angle of 180o, hence it is
called diagonal or linear hybridisation. Each sp-hybrid orbital has equal s and p-character of 50% each.
Molecules in which the central atom is sp-hybridised and linked to two other atoms have linear shape.
i) Formation of beryllium chloride. Beryllium has a ground state configuration of 1s2, 2s2, 2p0. In the excited
state, one of the 2s-electrons is promoted a vacant 2p-orbital. The 2s-orbital and one 2p-orbital undergo
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hybridisation to form two sp-hybridised orbitals. Each of the sp-hybridised orbitals overlaps axially with the
half-filled p-orbital of chlorine atoms to form two Be–Cl sigma bonds. Thus, BeCl2 has linear shape.

(1) (2)

2. sp2-Hybridisation (Trigonal planar hybridisation). This type of hybridisation involves one s-orbital and
two p-orbitals to form three sp2 hybrid orbitals. The sp2 hybrid orbitals lie in the same plane and are directed
to the three corners of an equilateral triangle. Each sp2 hybrid orbital has 33.33% s-character and 66.66%
p-character. Molecules in which the central atom is sp2 hybridised and linked to three other atoms have
triangular planar shape with bond angle 120o.
i) Formation of boron trichloride (BCl3). Ground state configuration of boron atom is 1s2 , 2s2, 2p1. In the
excited state, the configuration is 1s2 , 2s1, 2px1, 2py1. The 2s-orbital of boron intermixes with two 2p-orbitals
to form three sp2 hybrid orbitals. The sp2 hybrid orbitals are directed to the corners of an equilateral triangle
and lie in the same plane. sp2 hybrid orbitals of boron overlap axially with half-filled orbitals of chlorine atoms
to form three B-Cl sigma bonds. BCl3 has triangular planar shape with Cl–B–Cl bond angle 120o.
3. sp3 -Hybridisation (Tetrahedral hybridisation). It involves mixing of one s-orbital and three p-orbitals of
the valence shell to form four sp3-hybrid orbitals of equivalent energy and shape. In order to minimise repulsive
interactions among them, the four sp3 hybrid orbitals are directed towards the four corners (vertices) of a
regular tetrahedron. The angle between the sp3 hybrid orbitals is 109o 28'. Each sp3 hybrid orbital has 25% s-
character and 75% p-character. Molecules in which the central atom is sp3 hybridised and linked to four
other atoms have tetrahedral shape.
i) Formation of methane (CH4). In methane, the 2s orbital and three 2p orbitals of excited carbon atom
undergo hybridization to form four sp3 hybrid orbitals.
Each of the sp3 hybrid orbitals overlaps axially with half-filled 1s-orbitals of four hydrogen atoms forming
sp3-s  -bonds. The H–C–H bond angles are 109.5o. CH4 molecule has tetrahedral shape.

(i) (ii) (iii)

iii) Formation of ammonia (NH3) molecule. In NH3, nitrogen atom is in sp3 hybridised state. Three sp3
hybrid orbitals of the N atom are used for forming sp3-s sigma bonds with H atoms.
The fourth sp3- hybrid orbital has a lone pair of electrons. The relatively large lp-bp (lone pair-bond pair)
repulsion causes H-N-H bond angle to decrease from 109o 28' to 107o.
iv) Formation of water (H2O) molecule. In H2O, oxygen atom is in the sp3 hybridised state. Two of the sp3-
hybrid orbitals contain lone-pairs of electrons. The other two hybrid orbitals form sp3-s sigma bonds with H
atoms. lp-lp and lp-bp repulsions result in H–O–H bond angle to decrease from 109o 28' to 104.5o.

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Table - 02 Types of hybridisation of orbitals and their geometries (E= lone pair).

Type of Bond Lone Electron Geometry of

Hybridisation Examples
molecule pairs pairs pairs molecule
AB 2 2 0 2 sp Linear BeF 2
2 - 2-
AB 3 3 0 3 sp Trigonal planar BCl3, BF 3 ,NO 3 , CO 3
2
AB 2E 2 1 3 sp V-shaped O 3, SO 2, SnCl2, P bCl2
3 +
AB 4 4 0 4 sp Tetrahedral CH 4, NH 4 ,SiF 4
3 -
AB 3E 3 1 4 sp Trigonal pyramidal NH 3, P X3, CH 3
3
AB 2E 2 2 2 4 sp V-shaped H 2O, NH 2, SF2
4.7.4 Other Examples of sp3, sp2, and sp hybridisation
i) Formation of ethane (CH3 – CH3). In ethane, carbon atoms are sp3 hybridised. One hybrid orbital on each
carbon overlaps axially to form an sp3-sp3  -bond. The other three hybrid orbitals on each carbon forms
sp3-s  -bonds with hydrogen atoms. C–H bond length is 109 pm and C–C bond length is 154 pm. All
C–C–H bond angles are 109o28'.

(i) (ii)

ii) Formation of ethene (CH2=CH2). Both carbon atoms are sp2-hybridised. sp2 hybrid orbitals of carbon
atoms overlap axially to from a C–C  -bond. The other two sp2 hybrid orbitals on each carbon atom form
sp2-s  -bonds with two hydrogen atoms each. Thus, in ethylene, all the six atoms lie in one plane. The
unhybridised 2p-orbital on the carbon atoms overlap sidewise to form two  -electron clouds (one
 -bond) distributed above and below the plane of the carbon and hydrogen atoms.
The C=C bond consists of one sp2-sp2  -bond and one p  -bond; bond length of 134 pm is less than C–C
single bond length of 154 pm. C–H bonds are sp2-s sigma bonds with bond length 108 pm. The H–C–H bond
angles are 117.6o while H–C–C angles are 121o.

iii) Formation of acetylene ( CH  CH ). Carbon atoms in acetylene are sp-hybridsed. The two sp-hybrid
orbitals are linear (angle180o). The unhybridised p-orbitals are perpendicular, both mutally and also to the sp-
hybrid orbitals. sp-hybrid orbitals of carbon atoms overlap axially to form a C–C  -bond. The second sp-
hybrid orbital on each carbon overlaps with 1s-orbitals of H atoms to form C–H  -bonds.

The unhybridised px-orbitals on the carbon atoms overlap laterally to form the first  -bond and py-orbitals
overlap to form the second  -bond. The electron clouds of the  -bonds lie along the internuclear axis and are
mutally perpendicular. The four  -electron clouds (two  -bonds) so formed further merge into one another to
form a single cylindrical electron cloud around the inter-nuclear axis (surrounding the C–C sigma bond.)

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Thus, in acetylene molecule, the C  C bond consists of one sp-sp  -bond and two  -bonds. The C  C
bond length is 120pm. The C–H bond length is 108 pm and bonds are sp-s  -bonds. The H–C–C bond
angle is 180o, therefore, the molecule is linear.
Table - 03 Average bond lengths and bond energies of carbon–carbon bonds.
Bond Hybridisation s  Character Bond length (pm) Bond energy (kJ mol 1 )
CC sp3 25% 154 348
CC sp 2 33% 134 615
CC sp 50% 120 812
4.7.5 Hybridisation of Elements Involving d-Orbitals
The elements beyond the second period contain d -orbitals which can also take part in hybridisation and
account for geometries such as trigonal bipyramidal, octahedral, etc. In case of elements of the third period,
energy of 3d orbitals is similar to those of both 3s and 3p orbitals and 4s and 4p orbitals and can undergo
hybridisation with both. Hybridisation involving outer shell d-orbitals and inner-level d-orbitals are possible.
1. sp3d-Hybridisation
It is the mixing of one s-orbital, three p-orbitals one d-orbital to form five sp3d-hybridised orbitals which adopt
trigonal bipyramidal geometry. Three of the hybrid orbitals lie in a horizontal plane at an angle of 120o to one
another. These are called equatorial orbitals.The other two hybrid orbitals, lie along the vertical axis at right
angles to the plane of the equatorial orbitals and are called axial orbitals.
Formation of PCl5. The ground state electronic configuration of phosphorus is 1s2 2s2 2p6 3s2 3p3. On
excitation, the 3s-electrons get unpaired and one of them is promoted to the vacant 3dz2 orbital.

The five half-filled orbitals (one 3s, three 3p and one 3dz2) hybridise to form five sp3d hybrid orbitals, which
point towards the five corners of a trigonal bipyramid.
In PCl5, the five sp3d hybrid orbitals overlap axially with half filled 3p orbitals of Cl atoms to form five P–Cl
sigma bonds. Three of them lie in one plane at an angle of 120o to one another. These are called equatorial
bonds. The remaining two are at right angles to the plane of the equatorial bonds, one above and the other
below. These are called axial bonds.
As the axial bond-pairs suffer greater repulsion from the equatorial bond-pairs, axial P–Cl bonds become
slightly longer (219 pm) than the equatorial bonds (204 pm). The axial bonds are slightly weaker than the
equatorial bonds which makes PCl5 reactive.
2. sp3d2- Hybridisation. One s, three p and two d orbitals undergo intermixing to form six identical
sp3d2- hybrid orbitals. The six hybridised orbitals are directed towards the corners of a regular octahedron at
an angle of 90o to one another.
Formation of SF6. The ground state outer configuration of sulphur is 3s2 3p4. In the excited state, the electron
pairs in 3s and 3px orbitals unpair and gets promoted to vacant 3dz2 and 3d x  y orbitals. The 3s, three 3p and
2 2

two 3d orbitals hybridise to form six sp3d2 hybrid orbitals.

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In SF6, the six sp3d2 hybrid orbitals overlap with half-filled orbitals of fluorine to form six
S–F sigma bonds. SF6 has regular octahedral geometry. All the six S–F bonds have the same bond length and
all F–S–F bond angles are 90o.
Table - 04 Types of hybridisation of orbitals and their geometries.
Type of Bond Lone Electron Hybridisation
Geometry of molecule Examples
molecule pairs (B) pairs (L) pairs of atom A
3 +
AB5 5 0 5 sp d Trigonal bipyramidal PF5, SF5 , Fe(CO)5
3
AB4E 4 1 5 sp d See-saw shape SF4
3
AB3E2 3 2 5 sp d T-shaped ClF3
3
AB2E3 2 3 4 sp d Linear XeF2
3 2
AB6 6 0 6 sp d Octahedral SF6
3 2
AB5E 5 1 6 sp d Square pyramidal IF5, BrF5, ClF5
3 2 -
AB4E2 4 2 6 sp d Square planar XeF4, ICl4, IF4
3 3
AB7 7 0 7 sp d Pentagonal bipyramidal IF7
3 3 -
AB5E2 5 2 7 sp d Pentagonal planar XeF5
4.8 MOLECULAR ORBITAL THEORY (MOT)
The molecular orbital theory was put forward by F. Hund and R.S. Mulliken (1932) and further developed by
John Slater, Lennard-Jones and Charles Coulson. According to this theory, the nuclei of the two atoms lie at
appropriate distance and all the atomic orbitals of one atom overlap with all the atomic orbitals of the other
atom, provided the overlapping orbitals are of the same symmetry and similar energy. The resulting polynuclear
molecular orbitals contain all the electrons of the molecule.
The basic principles of molecular orbital theory
1. When nuclei of two atoms come close to each other, their atomic orbitals interact leading to the formation
of molecular orbitals. The atomic orbitals completely lose their identity after the formation of molecular
orbitals.
2. Each molecular orbital is described by a wave function,  known as molecular orbital wave
function.
3. The molecular orbital wave function,  is such that 2 represents the probability density or electron charge
density.
4. Each molecular orbital wave function,  is associated with a set of quantum numbers which determine the
energy and shape of the molecular orbital.
5. Each  is associated with a definite energy and the total energy of the molecule is the sum of the energies
of the occupied molecular orbitals.
6. Electrons fill the molecular orbitals in the same way as in atomic orbitals, obeying aufbau principle, Pauli’s
exclusion principle and Hund’s rule of maximum multiplicity.
7. Each electron in a molecular orbital belongs to all the nuclei present in the molecule.
8. Each electron moving in a molecular orbital has a spin of + ½ or - ½ .

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The basic difference between an atomic orbital and a molecular orbital is that while an electron in an atomic
orbitals is influenced by one nucleus only (monocentric), an electron in molecular orbital is influenced by all the
nuclei contained in a molecule (polycentric).
4.8.1 Linear Combination of Atomic Orbitals (LCAO) Method
Molecular orbitals are formed by the combination of atomic orbitals of bonded atoms. In wave mechanics
each electron is described by a wave fuction . Just as in case of atomic orbitals, Schrodinger wave equation
can be written to describe the behaviour of electrons in molecules also. The complete solution of the wave
equation for an atom with many electrons is difficult and it is much more difficult for a molecule with two or
more nuclei. A simplified approach is the LCAO approximation.

Figure - 03 Energies of bonding and antibonding molecular orbitals.

1. A linear combination of two atomic orbitals A and B leads to the formation of two molecular orbitals b
(bonding molecular orbital, BMO) and a (antibonding molecular orbital, ABMO). Mathematically,
 MO   A   B or
b   A   B (bonding molecular orbital, BMO)
a   A   B (antibonding molecular orbital, ABMO)
2. The energy Eb of molecular orbital b is lower than either EA or B, the energies of the atomic orbitals.
3. The energy Ea of molecular orbital a is higher than either of EA or B.
4. The extent of lowering of energy of the bonding molecular orbital is slightly lower than the extent of increase
of energy of the antibonding molecular orbital, as shown in figure.
5. The greater the overlap of the two combining atomic orbitals Aand B , the lower would be the energy of
the bonding molecular orbital b )formed.
6. The combining atomic orbitals A and B are completely used up during the formation of the molecular
orbitals band a . The electrons originally present in Aand B now occupy the molecular orbitals b ) and
(a) obeying aufbau principle, Pauli’s exclusion principle and Hund’s rule of maximum multiplicity.
Table - 05Comparison of bonding and antibonding molecular orbitals.
Bonding Molecular Orbitals Antibonding Molecular Orbitals
1 Formed by linear combination of atomic orbitals when Formed by linear combination of atomic orbitals when
their wave functions are added ;  bYAYB their wave functions are subtracted ;  aYAYB
2 Formed when the combining orbitals have the same Formed when the combining orbitals have the same
sign. signs.
3 These are designated as....etc. These are designated as *** ,.... etc.
4 Energy is less than the energy of the combining Energy is more than the energy of the combining
orbitals. orbitals.
5 Electron density between the nuclei increases - Electron density between the nuclei decreases -
stabilises the molecule. destabilises the molecule.
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4.8.2 Conditions for combination of Atomic Orbitals

i) The combining atomic orbitals should have comparable energies. For example, in case of homonuclear
diatomic molecules of the type A2 (e.g., H2, N2 , O2, etc.), 1s atomic orbital of one atom can combine with
1s atomic orbital of another atom. Similarly, 2s atomic orbital can combine with 2s atomic orbitals because
they have same energy. In case of heteronuclear diatomic molecules of the type AB (e.g., HCl) 1s atomic
orbital of H can combine with 3pz atomic orbital of Cl as they have comparable energies.
ii) The extent of overlap between the atomic orbitals should be large.
iii) The combining atomic orbitals must have proper orientations. Only those atomic orbitals can overlap to
form molecular orbitals which have same symmetry about the internuclear axis.
4.8.3 Designation of Molecular Orbitals.
The molecular orbitals are designated as  (sigma),  (pi),  (delta), (phi), etc.
1. Sigma molecular orbitals. The molecular orbitals which are cylindrically symmetrical around the internuclear
axis are called sigma (  ) molecular orbitals. The molecular orbital formed by addition of wave functions of
two 1s-atomic orbitals is designated as  1s and the molecular orbital formed by subtraction of wave functions
of two 1s-atomic orbitals is designated as  *1s. Similarly, combination of two 2s-atomic orbitals results in the
formation of two sigma molecular orbitals designated as  2s and  *2s.

Among 2p-atomic orbitals, molecular orbital formed by the addition overlap of two 2pz atomic orbitals is
designated as  2pz and the molecular orbital formed by substraction overlap of two 2pz atomic orbitals, is
designated as  *2pz .

2. Pi molecular orbitals: The molecular orbitals formed by the lateral (sidewise) overlap of p-orbitals are
called -molecular orbitals. For example, the addition and substraction overlap of two 2px atomic orbitals
result in the formation of 2px and *2px molecular orbitals respectively.

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Similarly, addition and subtraction overlap of two 2py atomic orbitals result in the formation of 2py and
*2py molecular orbitals.
In these cases the bonding molecular orbital has one nodal plane while the antibonding molecular orbital has
two nodal planes which are mutually perpendicular.
4.8.4 Energy Level Diagram for Molecular Orbitals
The combination of two 1s atomic orbitals results in two molecular orbitals designated as  1s and *1s.
Similarly, the 2s and 2p atomic orbitals, i.e., eight atomic orbitals of two atoms give rise to eight new molecular
orbitals, viz., 2s * 2s2p x  * 2p x 2p y  * 2p y 2p z  * 2p z .

The increasing order of energies of the various molecular orbitals is

1s   * 1s  2s   * 2s  σ2p z  2p x  2p y   * 2p x   * 2p y   * 2p z .

However, it has been observed that in diatomic molecules such as Li2, Be2, C2, and N2, the molecular orbital
energy level diagram is slightly different; the energy of σ 2pz molecular orbital is higher than that of 2px and
2py molecular orbitals. The increasing order of energies of molecular orbitals in these molecules is
1s   * 1s   2s   * 2s  2p x  2p y  σ2p z   * 2p x   * 2p y   * 2p z
The difference in the molecular orbital configurations of B2, C2, and N2 molecules is due to orbital-orbital
interactions which come into play during combination.
O2 and F2 do not exhibit these mixing interactions, presumably because of relatively larger difference of energy
between their 2s and 2pz orbitals. The fact is also supported by spectroscopic measurements.

Figure - 04 MO occupancy and molecular properties of homonuclear diatomic molecules (B2 to Ne2)

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4.8.5 Electronic Configuration and Molecular Behaviour

The molecular orbitals are filled, with the total number of electrons, in the order of increasing energy. Orbitals
of equal energy are half-filled with parallel spins before they begin to pair up (Hund’s rule).
Stability of Molecules. The stability of molecules can be determined by the relative number of electrons in
the bonding (Nb) and antibonding (Na) molecular orbitals.
1.Bond Order. Bond order (b.o.) is defined as one half of the difference between the number of electrons
present in the bonding and antibonding orbitals.
1
Bond order  (N b  N a )
2
i. A positive bond order (Nb > Na) implies that the molecule is more stable.
ii. A negative (Nb < Na) or zero (Nb = Na) bond order implies that the molecule is unstable.
Integral values 1, 2 or 3 correspond to single, double or triple bond.
iii. A bond order of 0.5 indicates a bond character equivalent to about half a bond. The bond order may
not always be a whole number.
2. Bond length. The bond order in a molecule is an approximate measure of its bond length. The bond length
decreases with increase in bond order.
3.Bond Energy. Higher the bond energy, stronger the bond. For double and triple bonds, the bond energy is
higher.
4. Magnetic behaviour. The magnetic behaviour of molecules can be predicted, from the electronic
configuration. If all the electrons in a molecule are paired, the substance is diamagnetic and if there are unpaired
electrons, the substance is paramagnetic. If the bond order is a whole number, the molecule may or may not be
paramagnetic, e.g., O2.
4.8.6 Bonding in Some Homonuclear Diatomic Species
1. Hydrogen Molecule (H2). The two hydrogen atoms with one electron each in 1s atomic orbitals combine
to form 1s and  *1s molecular orbitals. The two electrons fill the bonding 1s molecular orbital.

The molecular orbital electronic configuration of hydrogen molecule, H 2 :  1s  ,   *1s 

2 0

Bond order of H2 molecule  ( N b  N a ) / 2  ( 2  0) / 2  1

The two hydrogen atoms are bonded together by a single covalent bond. The bond dissociation energy of
hydrogen molecule has been found to be 435.8 kJ mol–1 and bond length equal to 74 pm. Since no unpaired
electron is present in the molecule, it is diamagnetic.
2. Helium Molecule (He2). The electronic configuration of helium atom is 1s2. Therefore, in He2 molecule
there are 4 electrons. These electrons will be accommodated in 1s and 1s * molcular orbitals.
M.O. configuration of He2 :(1s)2 (1s*)2

Bond order  (N b  N a ) / 2  (2  2) / 2  0 .
Since bond order for He2 is zero, the molecule does not exist.

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3. Lithium molecule (Li2). Electronic configuration of Li atom : 1s2 2s1.

M.O. Configuration of Li2 : (1s) 2 ( *1s) 2 (2s)2 or KK  2s 

2

The notations, KK represent (1s) 2 and ( *1s)2 molecular orbitals formed from 1s orbitals. Since they
contain the same number of electrons, the bonding and antibonding effect of these orbitals may be neglected.
Bond order  (N b  N a ) / 2  (4 - 2) / 2  (2  1) / 2  1 2
The bond order (1/2) implies that Li2 is stable; Li2 molecules are found to exist in the vapor state. However, in
the solid state, it is energetically more favorable for lithium to form metallic structure.
Since there are no unpaired electrons, the molecule is diamagnetic. Bond energy of Li2 is 110kJ mol–1.
4. Carbon molecule (C2). Electronic configuration of carbon atom : 1s 2 2s 2 2p1x 2p1y 2p z0 .

M.O. configuration of C2 : KK(2s) 2 ( * 2s) 2 (2p x ) 2  (2p y )2 . Bond order  (4  0) / 2  2 .

Since bond order is 2, the molecule is stable.
Bond length =131 pm. Bond energy of B2 is 620 kJ mol–1.
Since there are no unpaired electrons, the molecule is diamagnetic.
5. Nitrogen Molecule (N2). Electronic configuration of nitrogen atom : 1s 2 2s 2 2p1x 2p1y 2p1z .

Molecular orbital configuration of N2 : KK (2s)2 (  * 2s)2 ( 2p x ) 2  ( 2p y )2 (2p z ) 2

Bond order  (10  4) / 2  3 . Bond length =110 pm.
The bond order indicates triple covalent bond and very high bond energy (945 kJ/mole). Since all the electrons
are paired, it is diamagnetic.
6. Oxygen Molecule (O2)
M.O. configuration of O2 : KK(2s) 2 ( * 2s)2 ( 2p z ) 2 (2p x )2  (2p y )2 ( * 2p x )1  ( * 2p y )1
Bond order  (10  6) / 2  2 . Bond energy is 495 kJ/mole and bond length is 121 pm.
O2 contains two unpaired electrons in 2p x * and 2p y * molecular orbitals, therefore, it is paramagnetic;
 mm  2(2  2)  8 BM .
Thus M.O. theory sucessfully explains the paramagnetic nature of oxygen (the VB theory failed to
explain the paramagnetic nature of oxygen).
7. Fluorine molecule (F2). Electronic configuration of fluorine is 1s2, 2s2, 2p5.
F2 : KK (2s) 2 ( *2s) 2 ( 2p z )2 (2p x ) 2  (2p y ) 2 ( *2p x ) 2  ( * 2p y ) 2 (  * 2p z )0
Bond order = (10-8)/2 =1. Bond energy = 159 kJ/mole. Bond length =143 pm.
F2 is diamagnetic since it has no unpaired electrons.
8. Neon (Ne2). Electronic configuration of neon is 1s2, 2s2, 2p6. The 20 electrons in Ne2 molecule are
distributed into five bonding and five anti-bonding molecular orbitals.
Ne2: KK (2s) 2 ( *2s)2 ( 2p z ) 2 (2p x ) 2  (2p y ) 2 (  * 2p x ) 2  ( *2p y ) 2 (  *2p z ) 2
Bond order =(10-10)/2 = 0. Since bond order for Ne2 is zero, it is highly unstable and does not exist.
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4.9 HYDROGEN BONDING

In any chemical species where hydrogen is bonded to a highly electronegative atom like F, O or N, the bond
becomes highly polar. The hydrogen atom acquires a partial positive charge, while the electronegative atom
acquires a partial negative charge. When a number of such species come together, the positive end of one
molecule attracts the negative end of the other developing a weak electrostatic attraction called H-bonding.
   
   
X H X H X H
The extent of H-bonding depends on the physical state of the compound. It is maximum in solids and minimum
in gases. H-bonding can strongly influence the structure and properties of compoundse.g., the high boiling
point of water compared to H2S. The hydrogen bond length (197 pm) is much larger than covalent bond length
(96 pm). Thus, H-bond is much weaker than covalent and ionic bonds.
1. Intermolecular hydrogen bond : Intermolecular H-bonding occurs between two different molecules of
the same or different substances,e.g., HF, alcohol, water, ammonia, etc. It results in association of many
molecules as a group. Its effect is reflected in many properties of the compounds such as increase in melting
point, boiling point, viscosity, surface tension, solubility, etc.
H H H

O H O H O H
    
2. Intramolecular hydrogen bond: This type of H-bond is formed between hydrogen and an electronegative
atom within the molecule. It does not affect the physical properties of the componds to a large extent.
Intramolecular H-bonds are present in o-chlorophenol, salicylaldehyde, o-nitrobenzoic acid, etc.

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QUESTIONS
LEVEL - I
1. All of the following lists include at least one ionic compound except:
1) NO2, NaNO2, KNO3 2) CF4, CaF2, HF 3) NaCl, MgCl2, OCl2 4) H2S, SO2, SF6
1 2
2. What are the formal charges of the atoms in the nitrite ion O N O in the order of nitrogen, oxygen 1 and
oxygen 2?
1) 0, 0, –1 2) 0, –1, –1 3) –1, 0, –1 4) –1, 0, 0
3. Which pair is not correct order of lattice energy?
1) KCl > MgO 2) AlN > MgO 3) BeCO3 > MgCO3 4) BeCO3 = MgCO3
4. All the following species have all their bond lengths identical except:
1) AsF3 2) AsF4 3) AsF4 4) AsF6
5. Which of the following is the correct order of dipole moment of molecules BF3, NF3, NH3?
1) BF3 < NH3 < NF3 2) NH3 > BF3 > NF3 3) BF3 > NF3 > NH3 4) NH3 > NF3 > BF3
6. Which of the following facts regarding Fajan’s rules is not correct?
1) A small positive ion favours covalency
2) A large negative ions favours covalency
3) Large charge on either ion, or on both ions, favours covalency
4) Covalency is favoured if the polarizing ion has a noble gas configuration in preference to pseudo inert gas
configuration
7. The dipole moment of hydrogen chloride (bond distance: 127 pm) is 1.03 D. The per cent ionic character of
its bond is about:
1) 15 2) 17 3) 19 4) 21
8. Which of the following is arranged in order of increasing dipole moment?
1) BCl3  NH3  H 2 O  SO 2 2) BCl3  NH3  SO 2  H 2 O

3) NH 3  SO2  H 2 O  BCl3 4) H 2 O  SO 2  NH 3  BCl3

9. Which of the following statements regarding I3 is not correct?

1) The central I atom has five lone pairs of electrons
2) The lone pairs occupy equatorial positions
3) The bond angle I — I — I is 180o
4) There two I atoms around central I atom occupy equatorial positions
10. H2O molecule is dipolar, whereas BeF2 is non polar. It is because:
1) the electronegativity of F is greater than that of O
2) H2O involves hydrogen bonding whereas BeF2 is a discrete molecule
3) H2O is linear and BeF2 is angular
4) H2O is angular and BeF2 is linear

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11. Shapes of certain interhalogen compounds are stated below. Which one of them is not correctly stated?
1) IF7 : pentagonal bipyramid 2) BrF5 : trigonal bipyramid
3) BrF3 : planar T-shaped 4) ICl3 : planar dimeric
12. Select incorrect statement:
1) Double bond is shorter than a single bond 2)  -bond is weaker than a  -bond
3) Double bond is stronger than a single bond 4) Covalent bond is stronger than a hydrogen bond
13. Which of the following statements regarding carbon monoxide is correct?
1) A triple bond exists between C and O
2) O atom carries negative charge and C carries positive charge
3) Dipole moment of CO is low as compared to its calculated value from the charge and bond distance
4) Bond order of CO is 3
14. A covalent bond is most likely to be formed between two elements which:
1) have similar electronegativities 2) have low ionization energies
3) have low melting points 4) form ions with small charges
15. Which of the following statements regarding valence bond method is not true?
1) In the valence-bond method, the molecule is considered to be the collection of atoms and then interactions
between different atoms are considered
2) For a molecule to be stable, the electrostatic attractions must predominate over the electrostatic repulsions
3) The potential energy of a diatomic molecule is less than the sum of corresponding energies of free atoms
4) The net force of attraction acting on the atoms in a molecule is greater than that of repulsion
16. The number of lone pairs of electrons on the central atoms of H2O, SnCl2, PCl3 and XeF2 respectively, are
1) 2, 1, 1, 3 2) 2, 2, 1, 3 3) 3, 1, 1, 2 4) 2, 1, 2, 3
17. The sp3d2 hybridization of central atom of a molecule would lead to
1) square planar geometry 2) tetrahedral geometry
3) trigonal bi pyramidal geometry 4) octahedral geometry
18. The hybridization of chlorine orbitals in the compound ClF3 is
1) sp3 2) sp2d 3) sp3d 4) sp3d2
19. Which of the following statements is not correct?
1) Hybridization involves the mixing of atomic orbitals of the same atom
2) Hybridization of orbitals results into the formation of equivalent orbitals in all respect
3) Hybridization may involve combination of s, p and d orbitals of same energy level
4) Hybridization of orbitals predicts the molecular geometry of the molecule and not vice-versa
20. SF2, SF4 and SF6 have the hybridisation at sulphur atom respectively as:
3 2 2 3
1) sp2 , sp3 , sp2 d 2 2) sp3 , sp3 , sp3d 2 3) sp3 , sp3d, sp3d 2 4) sp , spd , d sp
21. In which of the following pairs the two species are not isostructural?
1) CO32 and NO3 2) PCl4 and SiCl4 3) PF5 and BrF5 4) AlF63 and SF6
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22. In which of the following pairs of molecules/ions, both the species are not likely to exist?
1) H 2 , He 22  2) H 2 , He 22 3) H 22  , He 2 4) H 2 , He 22

23. During the change of O2 to O 2 , the incoming electron goes to the orbital:
1) * 2p z 2) 2p y 3) * 2p x 4) 2p x
24. The energy of 2s is greater than *1s orbital because:
1) 2s is bigger than *1s MO
2) 2s is bonding whereas *1s is an ABMO
3) 2s orbital has a greater value of n than *1s MO
4) 2s orbital is formed only after *1s
25. The bond order of CO and NO is
1) 3 and 2 2) 3 and 2.5 3) 3 and 1.3 4) 3 and 3.5
26. Which of the following molecules is paramagnetic?
1) C2 2) N2 3) O2 4) F2
27. Which of the following species has the shortest bond length?
1) N 2 2) N2 3) N2 4) N 22

28. The number and type of bonds in C 22  ion in CaC2 are:

1) One  bond and one  - bond 2) One  bond and two  - bonds
3) Two  bonds and two  - bonds 4) Two  bonds and one  - bond
29. H2O is a liquid while H2S a gas. This is due to:
1) difference in the state of hybridisations of O and S in their compounds
2) high molecular mass of H2S as compared to H2O
3) strong hydrogen bonding in H2O molecule as compared to H2S molecule
4) H–O–H bond angle in H2O is 104.5o while H–S–H bond angle in H2S is 92o
30. Which of the following molecules form no intermolecular hydrogen bonds?
1) CH3CH2OH 2) HF 3) CH3COCH3 4) CH3COOH
LEVEL - II
1. MgSO4 is soluble while BaSO4 is insoluble in H2O. This is because:
1) lattice energy of BaSO4 is greater than MgSO4 2) BaSO4 is more covalent than MgSO4
3) hydration energy of Mg is greater than Ba
2+ 2+
4) lattice energy of MgSO4 is greater than BaSO4
2. Maximum bond angle at nitrogen is present in which of the following?
1) NO 2 2) NO 3 3) NO2 4) NO 2

3. The correct order of Cl — O bond lengths in ClO , ClO2 , ClO3 and ClO4 is:
1) ClO  ClO2  ClO3  ClO4 2) ClO4  ClO3  ClO2  ClO
3) ClO3  ClO4  ClO2  ClO 4) ClO4  ClO3  ClO2  ClO
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4. Which of the following hydrocarbons has the lowest dipole moment?

H3 C H
1) C C 2) H 3C  C  C  CH 3
H CH2 – CH3

3) H 3C  CH  C  CH 2 4) H 3C  CH 2  C  CH
5. Arrange the following compounds in the order of increasing dipole moment : toluene (I),
m-dichlorobenzene (II), o-dichlorobenzene (III), p-dichlorobenzene (IV).
1) I < IV < II < III 2) IV < I < II < III 3) IV < I < III < II 4) IV < II < I < III

6. The cyanide ions CN and N2 are isoelectronic, but in contrast to CN , N2 is chemically inert because of
1) Low bond energy
2) Absence of bond polarity
3) Unsymmetrical electron distribution
4) Presence of more number of electrons in bonding orbitals
7. For which of the following molecules,   0 ?

a) b) c) d)

1) a and b 2) only c 3) c and d 4) only a

8. The dipole moment of KCl is 3.336 × 10–29 Cm. The interatomic distance K+ and Cl– ion in KCl is 260 pm.
Calculate the dipole moments of KCl, if there were opposite charges of the fundamental unit located to each
nucleus.
1) 4.165 × 10–29 Cm 2) 4.325 × 10–27 Cm 3) 5.321× 10–29 Cm 4) 5.012× 10–27 Cm
9. The dipole moment of NF3 is very less than that of NH3 because:
1) Number of lone pairs in NF3 is much greater than in NH3.
2) Unshared electron pair is not present in NF3 as in NH3.
3) Both have different shapes.
4) Of different directions of moments of N–H and N–F bonds.
10. Which of the following statements is correct?
1) CO2 has linear whereas SO2 has nonlinear structures
2) CO2 has nonlinear whereas SO2 has linear strucutres
3) Both SO2 and CO2 has linear structures
4) Both SO2 and CO2 has non-linear structures
11. According to VSEPR theory which of the following molecules is (are) bent (non-linear)?
I. CO2 II. C2H2 III. O3 IV. BeCl2 V. KrF2
1) only III 2) only III and IV 3) only III, IV and V 4) All of these

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12. In PO34 , P  O bond order is:

1) 1.25 2) 2 3) – 0.75 4) –3
13. The bond dissociation energy of B–F in BF3 is 646 kJ mol , whereas that of C–F in CF4 is 515 kJ mol–1. The
–1

correct reason for higher B–F bond dissociation energy as compared to that of C–F is the:
1) smaller size of B atom as compared to that of C atom.
2) stronger  bond between B and F in BF3 as compared to that between C and F in CF4.
3) significant p  p interactions between B and F in BF3, whereas there is no possibility of such interaction
between C and F in CF4.
4) lower degree of p  p interactions between B and F in BF3 than that between C and F in CF4.
14. The geometry of CN groups around Ni in [Ni(CN)4]2– is:
1) tetrahedron 2) square planar 3) trigonal bipyramid 4) octahedron
15. The bond length of C = O bond in CO is 1.20 Å and in CO2 it is 1.34Å. Then C = O bond length in CO32 will
be:
1) 1.50 Å 2) 1.34 Å 3) 1.29 Å 4) 0.95 Å
16. The correct order of hybridisation of the central atom in the following species NH3, [PtCl4]2–, PCl5, and
BCl3 is:
1) dsp2, dsp3, sp2 and sp3 2) sp3, dsp2, sp3d and sp2
3) dsp2, sp2, sp3 and dsp3 4) dsp2, sp3, sp2 and dsp3
17. The planar shape of N(SiH3)3 is explained by the:
a) type of hybrid orbitals of nitrogen b) additional d  p overlap along the N - Si bond
c) higher electronegativity of nitrogen d) higher electronegativity of silicon
1) a 2) b and c 3) a and b 4) c and d
18. Among the following, the molecules expected to be stabilized by anion formation is: C2, O2, NO, F2
1) NO 2) C2 3) F2 4) O2
19. The geometries of the molecules BF3 and NF3 are trigonal planar and trigonal pyramidal, respectively. Which
statement best accounts for the difference?
1) N is more electronegative than B
2) BF3 is ionic, NF3 is covalent
3) B utilise sp2 hybridization, while N does not
4) N is sp3 hybridised and has non-bonding pair of valence electrons, while B does not
20. Which of the following order of energies of molecular orbitals of N2 is correct?

1) E  2p x   E  2p y   E  2p z   E   * 2p x   E   * 2p y 

2) E  2p x   E  2p y   E  2p z   E   * 2p x   E   * 2p y 

3) E  2p x   E  2p y   E  2p z   E   * 2p x   E   * 2p y 

4) E  2p x   E  2p y   E  2p z   E   * 2p x   E   * 2p y 

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21. Which of the following orders regarding the bond order is correct?
1) O2  O2  O2 2) O2  O2  O2 3) O2  O2  O2 4) O2  O2  O2
22. Which of the following is a positive overlap which leads to bonding?

1A) 2) 3) 4)

23. The nodal plane in the -bond of ethene is located in:

1) The molecular plane
2) A plane parallel to the molecular plane
3) A plane perpendicular to the molecular plane which bisects the (C – C) -bond at a right angle
4) A plane perpendicular to the molecular plane which contains the (C – C) -bond
24. Consider the statements.
I. Bond length in N 2 is 0.02Å greater than N2

II. Bond length in NO  is 0.09Å less than in NO

III. O 22  has shorter bond length than O2

Which of the above statements are true?
1) I and II 2) I and III 3) I, II and III 4) II and III
25. Stability of the species Li2, Li 2 and Li 2 increases in the order of:

1) Li 2  Li 2  Li 2 2) Li 2  Li 2  Li 2 3) Li 2  Li 2  Li 2 4) Li 2  Li 2  Li 2
26. According MO theory:
1) O 2 is paramagnetic and bond order greater than O2.

2) O 2 is paramagnetic and bond order less than O2.

3) O 2 is diamagnetic and bond order less than O2.

4) O 2 is diamagnetic and bond order more than O2.

27. Which of the following statements are not correct?
a) Hydrogen bonds are not present amongst gaseous hydrogen fluoride.
b) Both SO24 and SO32 have tetrahedral geometry..
c) In ozone both the bond lengths between oxygen atoms are not identical because it contains one double
bond and one coordinate - covalent bond.
d) In CO32 , all the three carbon-oxygen bond length are identical.
1) a and b 2) a, b and c 3) a, c and d 4) b, c and d

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28. Which species contain sigma covalent, pi    - covalent, coordinate covalent as well as ionic bonds?
1) H2SO4 2) NH4NO3 3) NaOCl 4) K2CrO4
29. CO2 is a gas at room temperature while SiO2 is a high melting solid. The best explanation of this difference is
that
1) Si has more electrons than C
2) The atoms in CO2 are bonded by covalent bonds while SO2 is an ionic compound
3) van der Waals’ forces are stronger in SiO2
4) CO2 consists of discrete molecules while SiO2 has network structure
30. Which of the following statements are not correct?
a) beryllium, like nitrogen, forms diatomic molecule Be2
b) He2 molecule does not exist but He2 does exist
c) The dipole moment of CH3F is greater than that of CH3Cl
d) HBr is a stronger acid than HI because of hydrogen bonding
1) a, b and c 2) b, c and d 3) a and d 4) a, c and d

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SOLUTIONS
LEVEL - I
1. 4

2. 1 The nitrite ion can be represented as

Formal charge on any atom = Number of valence electrons – Number of non-bonding electrons – 1/2
(number of bonding electrons)
1
Formal charge on nitrogen = 5  2   6   0 Formal charge on oxygen atom 1= 6  4  1  4   0
2 2
1
Formal charge on oxygen atom 2= 6  6   2   1
2
3. 1
4. 2 AsF4 has unsymmetrical shape, while all others are symmetrical.
5. 4 Both NF3 and NH3 have pyramidal shapes and a lone pair of electrons on N atom. The dipole moment
of NH3 is higher than NF3 because of same direction of N – H bonds and lone pair of electrons. BF3
has zero dipole moment.
6. 4 A noble gas configuration is the most effective at shielding the nuclear charge and has a poor polarizing
effect.

7. 2 percent ionic character 

1.03  3.3356  10 C m   100  16.9%
30

1.602  10 kg 127 10 m 

19 12

8. 2 Dipole moment is defined as the product of the magnitude of charge on any one of the atoms and the
distance between them.

9. 4 The lone pairs occupy equatorial positions and the three I atoms occupy top, middle and bottom
positions in the trigonal bipyramid giving linear structure with bond angle 180o.
10. 4

11. 2 BrF5 is square pyramidal with sp3d2 hybridization.

12. 2

13. 2 The CO molecule is represented as O C
Dipole moment is low due to pull of bonding electrons towards more electronegative oxygen atom.
14. 1 Atoms having similar electronegativity form covalent bond.
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15. 4
16. 1 H 2 O : l.p.  2,SnCl 2 : l.p.  1, PCl3 : l.p.  1, XeF2 : l.p.  3
17. 4
18. 3 sp3d hybridization is involved in ClF3. There will be two lone pairs and three unpaired electrons which
bond with three fluorine atoms.
19. 4 Molecular geometry must be known to predict the type of hybridization in the molecule.
20. 3
21. 3 CO32  and NO3 both are trigonal planar,, PCl4 andSiCl4 both are tetrahedral, PF5 is sp3d hybridized
and the shape is trigonal bipyramidal, whereas BrF5 is sp3d2 hybridized and its shape is square
pyramidal. AlF63 and SF6 both are octahedral.

22. 3 Bond order of H 22  and He2 are zero and therefore, these do not exist.

23. 3 The electron goes to * 2p x MO .

24. 3
25. 2
26. 3 O2 has two unpaired electrons and hence paramgnetic.
27. 2 N2 has the shortest bond length as its bond order is largest.
28. 2
29. 3
30. 3 CH3COCH3 : For H-bonding there must be a H-bonded to highly electronegative atoms like F, O
or N.
LEVEL - II
1. 3
2. 1 NO 2 is linear and bond angle of 180o.
3. 2 Bond length is inversely proportional to bond order. Greater the delocalization of -electrons
(–ve charge) shorter is the bond length.
4. 2

5. 2

p-Dichlorobenezene < Toluene < m-Dichlorobenzene, o-Dichlorobenzene.

6. 2

7. 3

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8. 1 From the given data, charge of electron, q = 1.602 × 10–19 C

r = 260 pm = 260×10–12 m = 2.6×10–10 m
Magnitude of dipole moment for 100% ionic character.
 = qr = (1.602 × 10–19) (2.6 × 10–10) = 4.165 × 10–29 Cm
9. 4 In NF3 N–F bond moments and lone pair moments are in opposite direction but in NH3 they are in
same direction.
10. 1
11. 1
5
12. 1 P – O bond order = Total number of bonds/Total number of resonating structures   1.25
4
13. 3 Bond strength of B–F bond is higher in BF3 as compared to C–F bond in CF4 due to formation of extra
p  p bond between B and F. p  p bonding is not possible in CF4.

14. 2

15. 3 In CO32 due to resonance, C = O bond length is in between triple and double bond, ie, in between 1.2
and 1.34. Thus, answer is 1.29 Å.

 Pt Cl4 
2
16. 2 NH3 :sp3 ;  C.N = 4, dsp2 hybridised ; PCl5 : sp 3d ; BCl3 : sp 2
17. 3
18. 2 In case of C2, the incoming electron will enter in bonding MO and therefore, bond order will increase.
Hence, stability will increase.
19. 4
20. 1
21. 2
22. 3
23. 1 The molecular plane does not have any  electron density.
24. 1
1
25. 1 Bond order  (bonding electrons -antibonding electrons)
2
Li2 = 1s 2 *1s 2 2s 2 Bond order = (4 – 2)/2 = 1
Li 2 = 1s  1s 2s
2 * 2 1
Bond order = (3 – 2)/2 = 0.5
Li 2 = 1s  1s 2s  2s
2 * 2 2 * 1
Bond order = (4 – 3)/2 = 0.5
Stability  Bond order. So, although bond order of both Li 2 and Li 2 are same but Li 2 will be more
stable as compared to Li 2 as the number of antibonding electrons are more in Li 2 than that of Li 2 .

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26. 1 Since it contains one unpaired electron, O 2 is paramagnetic; bond order  10  5 2  2.5

Hence, bond order of O 2 is greater than O2.

27. 2 A) Fluorine is the most electronegative element. A strong polarization of bonding electrons occurs in
HF leading to hydrogen bonding even in the gaseous state.
B) Due to the resonance, both bonds are identical

28. 2 NH4NO3 : Its shape is :

29. 4
30. 4 Be2 and He2 molecules do not exist. CH3F involves lesser C–F distance but more charge separation as
compared to those in CH3Cl. Here, bond distance has more dominating effect causing dipole moment
of CH3Cl greater than that of CH3F.

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CHAPTER - 05

STATES OF MATTER

INTRODUCTION
Under ordinary conditions, matter is known to exist in three states - solid, liquid and gas. A substance is
considered to be a solid if its melting point is above room temperature, a liquid if its freezing point is below
room temperature, and a gas of its boiling point is above room temperature, all under one atmosphere pressure.
The atoms, ions or molecules of a solid are fixed rigidly in definite positions in a crystal lattice. There is little
space in between them. They have no translatory or rotatory motion, but only vibrate about their fixed positions.
Hence, a solid has definite volume and definite shape. Pressure has no significant effect on its volume.
The forces of attraction in liquids are not so strong as in solids. Hence, a liquid has no definite shape but has
definite volume and the voids in between molecules are larger. The molecules have rotatory, vibratory and
small translatory motion. The volume changes only slightly with change in temperature or pressure.
Solids and liquids are sometimes referred to as condensed phases to emphasise the much closer spacing of
molecules in these states than in gaseous state.
Intermolecular forces of attraction in gases are very weak. Therefore, there are large voids and the molecules
can have all three types of motion, viz., translatory, rotatory and vibratory motion. Due to random motion of
molecules, a gas has no bounding surface. Hence, it has neither definite shape nor definite volume. Pressure
has very significant effect on its volume.
5.1 INTERMOLECULAR FORCES
Forces of attraction exist between polar as well as non-polar molecules. These intermolecular forces are
known as cohesive forces. Attractive intermolecular forces including dipole-dipole, dipole-induced dipole,
and dispersion forces are collectively called van der Waals forces (Dutch scientist, Johannes van der Waals,
1837-1923). These forces originate from the following interactions:
1. Dipole-Dipole Interaction. Polar molecules such as NH3, SO2, HF, HCl, etc., have permanent dipoles
and appreciable dipole-dipole interaction between molecules. Because of this attractive interaction, the above
gases can be easily liquefied. The magnitude of this interaction increases with increase in the dipole moment of
the molecule. The magnitude of this interaction increases with increase in the dipole moment of the molecule.
The existence of these forces was studied by Keesom (1912), therefore, these forces are also called Keesom
forces and the effect is called orientation effect.

Dipole-dipole interaction energy between stationary polar molecules (in solids) is proportional to 1/r3 and that
between rotating polar molecules is proportional to 1/r6, where r is the distance between polar molecules.
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2. Dipole-Induced dipole Interaction. A polar molecule may polarise a neutral molecule in its vicinity,
thereby inducing dipolarity in the molecule (Debye, 1920). The induced dipole then interacts with the polar
molecule and are attracted to each other. The magnitude of this interaction depends on the magnitude of the
dipole moment of the polar molecule and the polarisability of the neutral molecule.

The interaction energy is proportional to 1/r6, where r is the distance between the molecules.
3. Dispersion Forces or London Forces. The electrons of a neutral molecule keep on oscillating with
respect to the nuclei of the atoms. As a result, at a given instant, positive charge may be concentrated in one
region and negative charge in another region of the molecule. Thus, a non-polar molecule may become
momentarily self-polarised. This polarised molecule may induce a dipole moment in a neighbouring molecule
with antiparallel orientation, as shown in the figure.

The electrostatic forces of attraction between induced momentary dipoles are known as dispersion forces or
London forces. The van der Waals attraction in nonpolar molecules is exclusively due to London forces. These
forces are always attractive and interaction energy is inversely proportional to 1 r 6 of the distance ‘r’ between
the interacting particles. These forces operate only at short distances (~500 pm) and their magnitude depends
on the polarisability of the particles.
4. Ion-Dipole Interactions. It is attraction between ions and polar molecules, e.g., when NaCl is dissolved
in water, the polar water molecule is attracted towards N+ ion as well as Cl- ion (hydration of ions). The
strength of the interaction depends on the charge and size of the ion and the dipole moment and size of the
polar molecule. This interaction is usually stronger with the cation due to greater charge density on it.
5. Ion-induced Dipole Interactions. A non-polar molecule may become polarised by the presence of a
nearby ion, i.e., it forms an induced dipole, e.g., I2 molecule is polarised by nitrate ion (NO3-) to (I   I ) .
6. Hydrogen Bond. This is a strong dipole-dipole interaction between molecules in which hydrogen atom is
covalently bonded to a highly electronegative atom such as N, O or F. Species such as Cl– may also involve in
H-bonding. The strength of H-bond is determined by the coulombic attraction between the lone pair of electrons
of the electronegative atom of one molecule and the hydrogen atom of the other. The energy of hydrogen
bonds is between 10 and 100 kJ mol–1.
5.2 THERMAL ENERGY
Thermal energy of a substance is the energy due to motion of its atoms, ions or molecules. It is directly
proportional to the temperature of the substance. It is the measure of average kinetic energy of the particles
and is thus responsible for the thermal motion of particles.
Intermolecular Forces versus Thermal Interactions. Intermolecular forces tend to keep molecules together
but thermal energy tends to keep them apart. When molecular forces are very weak, molecules do not come
together to form liquid or solid unless thermal energy is reduced by lowering the temperature. Most gases do
not liquify on compression alone. They can be liquified by lowering the temperature so that thermal energy of
molecules is reduced. Predominance of thermal energy and molecular interaction energy of a substance in the
three states is as follows :
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Predominance of intermolecular forces : Gas  Liquid  Solid

Predominance of thermal energy : Solid  Liquid  Gas
5.3 THE GASEOUS STATE
Only 11 elements in the Periodic Table are gases. The gaseous state is characterised by high compressibility,
diffusability, very low density, and absence of definite volume and shape. Gases assume the shape of the
container and exert pressure equally in all directions. They mix evenly and completely in all proportions without
mechanical aid. All the three types of motion, vibratory, rotatory and translatory, are significant in gases.
Variables of State. The physical state of a substance is defined by its physical properties. Two samples of the
same substance having the same physical properties are in the same state. The variables required to specify the
state of a system are the amount of substance, n, the volume, V, the pressure P, and the temperature, T.
5.4 THE GAS LAWS
The gas laws are the result of research carried out on the physical properties of gases for several centuries.
The first reliable study on the properties of gases was made by Anglo-irish scientist Robert Boyle in 1662.
Attempts to fly with the help of hot air balloons motivated French scientists Jaccques Charles and Joseph
Lewis Gay Lussac to discover other gas laws. Contribution from Italian scientist Amedeo Avogadro, British
scientist John Dalton and others provided lots of information on the gaseous state.
5.4.1 Boyle’s Law (Pressure - Volume Relationship)
Robert Boyle studied the effect of change of pressure on the volume of a given mass of gas, found that at
constant temperature, the volume is inversely proportional to the applied pressure. This statement is
known as Boyle’s law and may be mathematically expressed as:
1
V (at constant temperature and mass) or P V = constant
P
where P is the pressure and V is the volume of the given mass of gas at constant temperature.
If V1 is the volume occupied by a given mass of gas at pressure P1 and V2 is its volume when pressure changes
to P2 (temperature remaining constant), it follows that
P1 V2
P1V1 = P2V2 (T and n are constants) or 
P2 V1

(a) (b) (c)

Figure - 01 Graphical representations of Boyle’s law; (a) P vs V (isotherms); (b) P vs 1/V, (c) log P vs log V
Fig.01a is the graph of Boyle’s law at different temperatures. Each curve corresponds to a different temperature
and is known as an isotherm (constant temperature graph). The value of k1 for each curve is different because
for a given mass of gas, it varies with temperature. The volume of the gas doubles when pressure is halved.
Fig.01b represents the graph between P and 1/V. It is a straight line passing through origin. However, at high
pressures, gases deviate from Boyles law; under such conditions a straight line graph is not obtained.
Practical importance of Boyle’s law

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i. Air at sea level is denser because it is compressed by the mass of air above it.
ii. Density and pressure of air decreases with altitude, e.g., atmospheric pressure above Mount Everest
is  0.5 atm. Decrease in pressure at high altitudes causes altitude sickness (sluggish feeling, headache,
etc.) due to decrease in O2 levels (anoxia).
iii. Jet planes, which fly at a height of about 10000 m is specially maintained at normal pressure.
Table - 01 Effect of pressure on the volume of 0.09 mole of CO2 gas at 300 K.
4 –3 3 –3 2 3
Pressure (10 Pa) Volume (10 m ) 1/V (m ) PV (10 Pa m )
2.0 112 8.9 22.4
2.5 89.2 11.2 22.3
3.5 64.2 15.6 22.47
4.0 56.3 17.7 22.5
6.0 37.4 26.7 22.44
8.0 28.1 35.6 22.48
10.0 22.4 44.6 22.4
Relationship between density and pressure of a gas. Being highly compressible, gases become denser at
high temperature. Therefore, density of a gas
m m
d or V 
V d
Substituting the value of V from Boyle’s law equation,
m
d P  k 'P
k1
Thus, for a fixed mass of gas at constant temperature, pressure is directly proportional to the density.
5.4.2 Charles’ Law (Temperature - Volume Relationship)
This law gives the relationship between volume and temperature of a gas at constant pressure. Charles’ law
states that the volume of a given mass of gas increases or decreases by 1/273 of its volume at 0oC for
each degree rise or fall of temperature, provided pressure is kept constant.
If the volumes of a gas at 0oC and toC are V0 and Vt respectively, then,

t  t   273.15  t 
Vt = V0  V0 = V0 1    Vt  V0  
273.15  273.15   273.15 
Absolute temperature scale or Kelvin scale: If the volumes of a given mass of gas are plotted against
temperatures at constant pressure, the graph obtained will be a straight line. If the graph is extrapolated to read
zero volume, the corresponding temperature is found to be –273oC (–273.15oC). This temperature, is termed
absolute zero. A scale of temperature based on -273.15oC as zero is called absolute scale of temperature or
Kelvin scale.
The relationship between the Celsius (toC) and the Kelvin (K) scales is: K = (-273.15 + toC)
Thus, -273.15 oC = 0 K, 0oC = 273(273.15) K, 100oC = 375(375.15) K
The Kelvin scale is significant in scientific study and is justified by thermodynamic arguments. Therefore it is
also called thermodynamic scale of temperature.
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(a) (b) (c)

Figure - 02 Graphical representation of Charles’ law; (a) V vs T, (b) V vs 1/T , (c) log V vs log T.
The relationship between volumes of a gas at different temperatures can be obtained by putting Tt = 273.15 +
t and T0 = 273.15 in the Charles’ law equation
Tt Vt Tt
Vt  V0  
T0 V0 T0
The general equation can be written as

V2 T2 V1 V2 V
    = constant = k2 (at constant p and n)
V1 T1 T1 T2 T

or V  T (at constant p and n)  V  k 2T

k2 is a constant whose value depends on the pressure of the gas, quantity of the gas and the units of volume.
Thus, Charles’ law may also be stated as ‘the volume of a given mass of gas at constant pressure is directly
proportional to its absolute temperature (Kelvin scale).”
If V1 and V2 be the volumes of a certain mass of gas at temperatures T1 and T2 at constant pressure, then
V1 V2
 (pressure and mass are constant)
T1 T2
The law has been verified experimentally by measuring the volumes of a given mass of gas at different temperatures
keeping the pressure constant. The V/T ratio was found to be constant.
Charles found that for all gases at a given pressure, graph of volume vs temperature (isobar) is a straight line
(Fig.02a) and on extending to zero volume, each line intercepts the temperature axis at -273.15oC (Fig.02b).
Thus, according to Charles’ law, the volume of a gas at -273.15oC will be zero, which means that gas will not
exist (V = 0). In fact all gases liquify before this temperature. The lowest hypothetical temperature at which
gases are supposed to occupy zero volume is called Absolute zero.
Gases obey Charles’ law only at very low pressures and high temperatures.
Table - 02 Volume-temperature ratio for 1 mole of N2 gas at 1 bar pressure.
3
°C (Celsius) K (Kelvin) Volume (dm ) of N2 V/T
-50 223.15 18.52 0.083
0 273.15 22.67 0.083
50 323.15 26.82 0.083
100 373.15 30.97 0.083
150 423.15 35.12 0.083

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5.4.3 Gay-Lussac’s Law (Pressure - Temperature Relationship)

This law gives the relationship between pressure and temperature of a gas at constant volume. It is also called
Amonton’ s law. Gay-Lussac’s Law states that for a given mass of gas at constant volume, the pressure
varies directly as the absolute temperature.
P
P  T (at constant n and V) or = k3 (constant)
T
The vale of k3 depends on the volume of the gas, the quantity of gas, and the units of pressure.
If P1 be the pressure of a certain mass of gas at T1 and P2 be its pressure at T2, then
P1 P
= 2 (at constant n and V)
T1 T2
This relationship can be derived from Boyle’s law and Charles’ law.
The plot of pressure versus temperature (Kelvin) for a definite mass of gas at constant volume (isochore) is a
straight line. Any isochore on extrapolation meets the temperature axis at 0K.

(a) (b) (c)

Figure - 03 Gay Lussac’s law; (a) P vs T graph (isochore), (b) Plot of P vs 1/T, (c) Plot of log P vs log T.
5.4.4 Avogadro’s Law (Volume - Amount of Substance Relationship)
Avogadro, in 1811, put forward his famous hypothesis which is now termed a law. It states that equal volumes
of all gases, measured at the same temperature and pressure, contain equal number of molecules.
It follows that, at constant temperature and pressure, volume occupied by a gas varies directly as the number
of molecules.
V  n or V = k4n (when T and P are constant)
where n is the number of moles of the gas and k4 is a constant.
According to this law, molar volume (the volume occupied by one mole) of any substance in the gaseous state
(22.414 dm3 at STP, when pressure is in atm, and 22.71 dm3 when pressure is in bar) would contain the same
number of molecules. This is called Avogadro’s number/constant, NA. Its numerical value is 6.02 × 1023
If m is the mass of a gas having molar mass M, then the number of moles, n = m/M.
m m
Thus, V  k4 or M  k 4  k.d
M V
where d is the density of the gas. Thus, density of a gas is directly proportional to its molar mass.
5.4.5 THE IDEAL GAS EQUATION (EQUATION OF STATE)
The combination of Boyle’s law, Charle’s law, and Avogadro’s law gives the ideal gas equation. It is a
relation between four variables and describes the state of any gas. It is also called equation of state.
Boyle’s law, V  1 P (when T and n are constant)
Charle’s law, V  T (when P and n are constant)

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Avogadro’s law, V  n (when T and P are constant)

Combining Boyle’s law,Charle’s law, and Avogadro’s law,
nT nT PV
V or V  R or R  or PV = nRT
T
P P nT
where R is the proportionality constant known as the universal gas constant.
The word ideal is used because there is no real gas which completely obeys this equation.
Real gases behave almost ideally under conditions of high temperature and low pressure.
From the equation, it can be seen that at constant temperature and pressure, n moles of any gas will have the
same volume since n, R, T, and P are constants. Volume of one mole of ideal gas under STP (273.15K and 1
bar pressure) will be 22.710981 L mol-1.
Standard temperature and pressure (STP). Standard temperature = 0oC or 273.15 K, and Standard
pressure = 1 bar = 105 Pa.
Nature of Universal Gas Constant, R. For one mole of an ideal gas,
Pressure×Volume Force×Area×Length Force  Length Work
R = = 
Temperature Area×Temperature Temperature Temperature
Thus, R represents work done per degree per mole.
Numerical Value of R. Since one mole of an ideal gas occupies a volume of 22.414 dm3 at STP,

PV 1 atm   22.414 dm  3

R 
1 mol  273.15 K  = 0.08206 dm atm K mol or L atm K mol .
3 –1 –1 –1 –1
nT
When pressure is expressed in bar and volume in L, then

1 bar  22.7 L
R
1 mol  273.15 K = 0.0831 L bar K mol .
–1 –1

When pressure is expressed in Nm-2, and volume in m3, then

R
101325 Nm  22.414 10
2 3
m3 
= 8.314 N m K–1 mol–1 = 8.314 J K–1 mol–1
1 mol  273.15 K 
In CGS Units (pressure in dynes per cm2 and volume in cm3:
1 atm = 76 cm × 13.596 g cm–3 × 980.6 cm s–2 = 1013250 g cm–1 s–2 = 1013250 dyn cm–2

1013250 dyn cm 2  22400 cm 3

R = 8.314 × 107 erg K–1 mol–1
1 mol  273.15 K
Further, we know that 4.184 × 107 ergs =1 cal, therefore,

8.314 107
R = 1.987 cal K–1 mol–1 or R = 2 cal K–1 mol–1.
4.184  107

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Table - 03 Values of R in different units.

Value of R Unit of P Unit of V Value of R Unit of P Unit of V
–1 –1 –1 –1
0.0831 bar L K mol bar L 0.0821 L atm K mol atm L
7 –1 –1 –2 3 –1 –1 –2 3
8.314 × 10 erg K mol dynes cm cm 8.314 J K mol Pa or Nm cm
-1 -1 –1 -1 –2 3
1.987 cal K mol 1.967 cal K mol dyn cm cm
3 –1 –1 3 -1 -1
8.314 k Pa dm K mol k Pa dm 62.3 L mm K mol mm L
1. Combined gas law. The ideal gas equation is the relationship for the simultaneous variation of the variables.
If temperature, volume and pressure of a fixed amount of gas vary from T1, V1 and P1 to T2, V2 and P2 then
P1V1 P2 V2
 nR and nR
T1 T2

P1V1 P2 V2
Combining these equations 
T1 T2
This equation is also known as combined gas law. If five out of six variables are known, the value of the
unknown variable can be calculated using this equation.
2. Density and Molar mass of a Gaseous Substance. Ideal gas equation can be rearranged as:
n P m P d P dRT
 or  or  or M=
V RT MV RT M RT P
where n = m/M (m = mass and M = molar mass of the substance), and density, d = m/V.
5.4.6 Dalton’s Law of Partial Pressures
Dalton’s law of partial pressures states that the total pressure exerted a mixture of nonreactive gases is equal
to the sum of the partial pressures of the individual gases, i.e., the pressure which each gas would exert if it
were present alone, occupying the same total volume.
If p1, p2, p3, ..... are the partial pressures of various gases present in a mixture, the total pressure, P, of the
gaseous mixture is given by
P = p1 + p2 + p3 + ..... (at constant T and V)
Utility of Dalton’s Law. Gas collected over water contains water vapour. The observed pressure of moist
gas is the sum of the pressures of the dry gas and the pressure due to water vapour (aqueous tension).
Therefore, at a particular temperature
Pobserved = Pgas + Aqueous tension or Pgas = Pobserved – Aqueous tension
Partial pressure in terms of mole fraction. Suppose at constant temperature T, three gases enclosed in
volume V exert partial pressures P1, P2, and P3 respectively, then
n1RT n RT n RT
P1  , P2  2 and P3  3
V V V
where n1, n2 and n3 are number of moles of the gases.
RT n 2 RT n 3 RT RT
Ptotal  P1  P2  P3  n1   =  n1  n 2  n 3 
V V V V

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P1 n1 n1
Dividing P1 by Ptotal , P  n  n  n  n 1
total  1 2 3
where n1 + n2 + n3 = n and 1 is the mole fraction of the first gas. Thus,
P1 = 1 Ptotal, P2 = 2 Ptotal, and P3 = 3 Ptotal
The general equation can be written as : Pi = i Ptotal
where Pi and i are partial pressure and mole fraction of the gas respectively. The partial pressure of a gas
in a mixture is the product of its mole fraction and the total pressure of the mixture.
Table - 04 Aqueous tension of water at different temperatures.
Temperature (K) Pressure (bar) Temperature (K) Pressure (bar)
273.15 0.006 293.15 0.023
283.15 0.0121 299.15 0.0331
288.15 0.0168 303.15 0.0418
5.5 THE KINETIC MOLECULAR THEORY OF GASES
At the time of their enunciation, the gas laws were only empirical generalisations based on experimental
observations. There was no theoretical background to justify them. However, in the nineteenth century Kronig,
Clausius, Maxwell and Boltzmann developed the kinetic molecular theory of gases which provided theoretical
basis for the gas laws. The postulates of this theory are:
1. A gas consists of a large number of minute particles, called molecules. The molecules are so small that their
actual volume is a negligible fraction of the total volume occupied by the gas.
2. The molecules are in a state of constant rapid motion in all possible directions, colliding in a random manner
with one another and with the walls of the container.
3. The molecular collisions are perfectly elastic so that there is no net loss of energy when gas molecules
collide with one another or against the walls of the vessel. The kinetic energy may be transferred from one
molecule to another but it is not converted into any other form of energy such as heat.
4. There are no attractive forces between molecules or between molecules and the walls of the vessel in which
the gas is contained. The molecules move independently of one another. In fact, there are no intermolecular
forces of attraction or repulsion in a perfect (ideal) gas.
5. The pressure of a gas is due to the bombardment of the molecules on the walls of the container.
6. Since molecules are moving with different velocities, they possess different kinetic energies. However, the
average kinetic energy of the molecules of a gas is directly proportional to the absolute temperature.
The postulates of the kinetic theory are valid only for ideal gases.
The pressure of a gas can be evaluated by calculating the momentum transferred on unit area of the wall of the
container by the molecules of the gas.
5.5.1The Kinetic Gas Equation
It is a mathemetical equation based on the postulates of the kinetic theory of gases. All the gas laws can be
deduced from this equation. According to kinetic gas equation,
1
PV  mnc 2
3
where P = pressure of the gas, m = mass of each molecule of the gas, n = total number of molecules of the gas
present in volume V and c = root mean square velocity of the gas molecules.
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Root mean square velocity, denoted by c, u, crms, urms, etc., is the mean of the squares of the different velocities
possessed by molecules of a gas at a given temperature. It is given by the relation,

  n i vi 2 
12
n1v12  n 2 v 22  n 3 v32  .....
c    
n1  n 2  n 3  .....   n i 
where v1, v2, v3,.....,vn are velocities possessed by groups of n1, n2, n3,....., nn molecules, respectively.
5.6 BEHAVIOUR OF REAL GASES (DEVIATIONS FROM IDEAL BEHAVIOUR)
The equation of state, PV = nRT is valid only for ideal gases. Real gases obey this equation only at low
pressures and high temperatures. However, gases which are soluble in water or easily liquefiable like CO2,
NH3, SO2, etc., show larger deviations than gases like H2, N2, O2, etc.
A plot of PV against P at constant temperature for an ideal gas should be a straight line parallel to the pressure
axis. However, real gases do not give such graphs indicating significant deviation from ideal behaviour.
The curves for H2 and He exhibit increase in PV with P, whereas those for CO and CH4 first decreases and
then increases.

(a) (b)

Figure - 04 (a) PV vs P for ideal and real gases (b) Pressure vs volume for ideal and real gases.
Deviation from ideal behaviour is apparent also when pressure is plotted against volume. The P vs V plot of
real gas does not coincide with that for ideal gas. At very high pressure, the measured volume is more than
ideal volume. At low pressures, measured and calculated volumes approach each other.
Compressibility factor or Compression factor. For an ideal gas, compressibility factor
PV
Z 1
nRT
Any deviation of Z from unity is a measure of the imperfection of the gas.
PVreal
For a real gas, Z  ...(i)
nRT
nRT
For an ideal gas, Videal 
P
nRT V
Putting the value of in Eqn. (i) we have Z  real ...(ii)
P Videal
Thus, compressibility factor is the ratio of the actual molar volume of a gas to the molar volume if it
were an ideal gas at that temperature and pressure.
The graphs of Z for a number of gases over a range of pressures at constant temperature (0oC) are shown in
Fig. 04. At low pressures, all gases have Z close to unity which means that the gases behave almost ideally.
At very high pressures, all gases have Z more than unity indicating that they are less compressible than ideal
gas. This is because molecular repulsive forces are dominant at high pressures.

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At moderately low pressures, CO, CH4 and NH3 are more compressible than ideal gases; PV < (PV)ideal so
that Z < 1. This is due to the fact that at low pressures attractive forces are dominant and favour compression.
With increase in pressure, compressibility factor decreases, passes through a minimum and then increases;
PV > (PV)ideal so that Z >1.

(a) (b)

Figure - 05 (a) Variation of compressibility factor for some gases, (b) Effect of Temperature on compressibility
H2 and He at 0oC are less compressible than ideal gases at all pressures ( Z >1). However, if the temperature
is sufficiently low (below –165oC for H2 and below –240oC for He), these gases also give the same type of
Z-P plots as NH3, CO and CH4 at 0oC. If the temperature is sufficiently high, the Z – P plots of NH3, CO and
CH4 will be similar to those of H2 and He at 0oC (Z will increase continuously with increase in pressure).
5.6.1 Effect of Temperature on Deviations form Ideal Behaviour.
The dip in the curve (Fig 05), becomes smaller as the temperature is raised. At 50oC, the curve remains almost
horizontal for an appreciable range of pressures indicating that Z becomes almost equal to unity. This temperature
is called the Boyle point or Boyle temperature. Below this temperature, the value of Z at first decreases,
reaches a minimum and then increases as the pressure is increased. Above 50oC, the value of Z shows a
continuous rise with increase in pressure.
The temperature at which a real gas obey ideal gas law over an appreciable range of pressure is called Boyle
temperature or Boyle point. Boyle temperature is different for different gases.
5.6.2 Explanation for the Deviations. In order to explain the deviations from ideal behaviour, it is necessary to
modify the first two postulates of kinetic theory of gases.
(i) The volume occupied by the molecules is negligible compared to the total volume of the gas.
(ii) The forces of attraction/repulsion between gas molecules are negligible.
These two postutates hold good only at low pressures or high temperatures because the molecules are far
apart. However, at high pressures and low temperatures, the gas molecules come close together. Hence,
under conditions of high pressure/low temperature
(i) The volume occupied by the molecules is not negligible compared to the total volume of the gas.
(ii) The forces of attraction/repulsion between gas molecules is not negligible.
5.6.3 Equations of State for Real Gases (van der Waals Equation)
1. Volume correction : The ideal gas equation PV = nRT is derived on the assumption that gas molecules are
point masses, i.e., they do not have finite volume. van der Walls suggested that a correction term nb for volume
occupied by the molecules should be subtracted from the total volume V in order to get the ideal volume.
Thus, the compressible volume per mole of the gas is (V – b). If volume, V of the gas contains n moles, then
the excluded volume is nb. Hence, the ideal volume is (V – nb). The volume b per mole is also known as
co-volume (incompressible volume).
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(a) (b)

Figure - 06 (a) Excluded volume, (b) Molecular attraction

2. Pressure correction (Correction for intermolecular forces of attraction). In the derivation of the ideal
gas equation, it was assumed that there are no intermolecular forces of attraction. Actually, it is not so.
A molecule ‘C’ somewhere in the middle of the container (Fig.06b) is being attracted uniformly on all sides by
the neighbouring molecules. These forces cancel each other and there is no resultant attractive force on the
molecule. However, as the molecule approaches the wall of the container ‘W’, it experiences attractive forces
from the bulk of the molecules behind it. Thus, it will strike the wall with a lower velocity, and therefore, exert
a lower pressure than it would have if there was no force of attraction from behind. Thus, it is necessary to add
a correction factor ‘p’ to the pressure of the gas in order to get the ideal pressure. Therefore, the correct
pressure should be (P + p). The value of p for one mole of gas will be inversely proportional to the square of
the volume V. Hence, pressure correction
1 a
p 2
 2
V V
where a is a constant depending on the nature of the gas.
The pressure correction factor p is also called the internal pressure of the gas.
The kinetic gas equation for one mole of a real gas, therefore, becomes
 a 
 P  2   V  b   RT
 V 
This is known as the van der Waals equation. The constants a and b are known as the van der Waals
constants. These are characteristic of each gas.
For n moles of gas occupying volume V, the excluded volume is given by nb and the compressible volume will
be (V – nb). Hence, van der Waals equation for n moles of gas becomes
 n 2a 
 P    V  nb   nRT
 V2 
This equation is more accurate than the ideal gas equation for expressing the P–V–T behaviour of real gases.
5.7 LIQUEFACTION OF GASES
A gas can be liquefied by cooling or by the application of pressure or the combined effect of both.
5.7.1 Critical Constants of a Gas
1. Critical Temperature (Tc). Critical temperature of a gas is the temperature above which it cannot be
liquefied howsoever high the pressure may be. For example, the critical temperature of CO2 is 31.1oC. This
means that it is not possible to liquefy carbon dioxide above 31.1oC.
2. Critical Pressure (Pc). It is the pressure required to liquify a gas at its critical temperature. For example,
at 31.1oC, CO2 can be liquefied at 73 atm (72.9) pressure. Thus, the critical pressure of CO2 gas is 72.9 atm.
The critical temperature of O2 is –118oC and that of H2 is –240oC. Their critical pressures are 50.1atm and
12.8 atm, respectively.

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3. Critical Volume(Vc). The volume occupied by one mole of a gas at its critical temperature and critical
pressure is known as the critical volume (Vc). For example, critical volumes of CO2, O2, H2 are 94.0, 78.2
and 65.5 ml per mole, respectively.
5.7.2 The P-V Isotherms (Andrews Isotherms) of carbon dioxide. The importance of critical temperature of a
gas was discovered by T. Andrews (1861) in his experiments on pressure-volume relationship of CO2 at
various temperatures. In the first isotherm (Fig.08) at 13.1oC (point A) represents CO2 in the gaseous state
having a certain volume at a certain pressure. On increasing the pressure, volume decreases along curve AB.
At B (49.8 atm), liquefaction of CO2 begins and thereafter a rapid decrease in volume takes place at the same
pressure, as more and more of the gas is converted into liquid. At C, the gas has been completely liquefied. As
the liquid is only slightly compressible, further increase of pressure produces only a very small decrease in
volume. This is shown by a steep line CD which is almost vertical.
Thus, along curve AB, CO2 exists as gas; along BC, it exists partly as gas and partly as liquid while along CD,
it exists as liquid. A considerable decrease of volume (along BC) takes place when the gas changes into liquid
at constant pressure.
The isotherm EFGH at 21.5oC shows similar behaviour except that liquefaction starts at a higher pressure and
the horizontal portion FG, representing decrease in volume, becomes smaller. At still higher temperatures, the
horizontal portion of the curves become shorter and shorter until at 31.1oC it reduces to a point (X). Above
31.1oC, the isotherm is continuous and there is no liquefaction at all.

Figure - 07 P-V isotherms of carbon dioxide.

The experimental study on CO2 and other gases revealed that for every gas there is a certain temperature
above which it cannot be liquefied howsoever high the pressure may be. Above critical temperature, the
kinetic energy of the gas molecules is sufficient to overcome the attractive forces.
The difference between ‘permanent’ gases (H2, O2, N2, etc.) and ‘temporary’ gases (CO2, Cl, HCl, NH3,
etc.) is the fact that while permanent gases have very low critical temperatures, temporary gases have critical
temperatures within the range of ordinary temperatures.
5.7.3 Continuity of State. Isotherms in Fig.07 shows that it is possible to convert liquid CO2 into gas and vice-
versa, without any ‘discontinuity’, that is without having, more than one phase present at any time. On joining
the ends of the horizontal portions of the various isotherms, a boundary curve CGXFB (dotted line) is obtained.
At the top lies the critical point X (31.1oC). Within the area of the boundary curve, both liquid and gaseous
states coexist, but outside this area, either liquid or gaseous state alone exist.

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Suppose, a certain volume of CO2, represented by point x is heated at constant volume to a temperature at
which the pressure increases to a point y, lying above the critical pressure of the gas. Let the gas be now
cooled at the same pressure. The temperature and volume both decrease along yz. At z, the temperature is
21.5oC and CO2 exists as liquid. During this transition from gas to liquid, there is only one phase present at any
time. Thus, the transition from gaseous state to liquid state or vice-versa, can be regarded as continuous.
Table - 05 Critical Constants of Gases.
3 –1
Gas Pc (atm) Vm,c (dm mol ) Tc (K) Zc
He 2.26 0.0579 5.3 0.307
Ne 26.9 0.0417 44.4 0.308
H2 12.8 0.0655 33.2 0.309
O2 50.1 0.0782 154.3 0.308
N2 33.5 0.0901 126.2 0.291
CO2 72.8 0.094 304.1 0.274
NH3 111.5 0.0725 405 0.243
CH4 54.6 0.0987 190.6 0.288
Difference between vapour and gas. A gas can be liquefied below its critical temperature by applying
pressure, therefore, below critical temperature, it is vapour and above critical temperature, it is gas.
Importance of critical temperature. Critical temperature, Tc and van der Waals constant ‘a’ are measures
of intermolecular forces of attraction. The value of ‘a’ increases in the same order as Tc, therefore the ease of
liquefaction of gases is in the order: SO 2 > Cl 2 > NH 3 > HCl > CO 2 > CH 4 > CO > N 2 >
Ne > H2 > He.
Liquefaction of Gases. It is necessary to cool a gas below its critical temperature before it can be liquefied.
In the case of gases like NH3, Cl2, SO2 or CO2, which have fairly high critical temperature, the application of
pressure alone is sufficient to cause liquefaction. Gases such as H2, O2, N2 and He which have very low critical
temperatures could not be liquefied by this simple technique. These gases were regarded as ‘permanent
gases.’ But these gases can also be liquefied if they are first cooled below their critical temperatures and then
compressed.
5.8 THE LIQUID STATE
A liquid may be regarded as a condensed gas or a molten solid. In a liquid, the molecules are not rigidly fixed
as in solids. They have some freedom of motion which is much less than that of gases. Thus, a liquid has a
definite volume though not a definite shape. It is much less compressible and far denser than a gas.Since the
molecules of a liquid are not far apart from one another, intermolecular forces are fairly strong. The characteristic
properties of liquids arise from the nature and magnitude of intermolecular forces.
Structure of Liquids: In 1961, Ering and Ree proposed that intermolecular space in a liquid is not randomly
distributed but contains molecular-sized ‘holes’ or vacancies. Molecules of the liquid continually move into
these holes and the vacancies continually move around. This process permits flow of liquids.
Physical Properties of Liquids. Molecules of a liquid are held closer to one another than in a gas due to
much stronger van der Waals forces of attraction. Consequently, liquids, unlike gases, have definite volume.
There is little or no space between molecules, therefore, liquids are almost incompressible.
Density. Liquids have much higher density than gases because molecules are more closely packed. For
example, the density of water at 373 K and 1 atm pressure is 0.958 g cm–3, while that of water vapour at the
same temperature and pressure is 0.000588 g cm–3.
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Diffusion. Liquids diffuse, but the rate of diffusion is much lower than gases due to smaller intermolecular
spaces and restricted motion of molecules.
5.8.1 Vapour Pressure
When an evacuated container is partially filled with liquid, it evaporates filling the container with vapour. The
pressure exerted by the vapours on the walls of the container is called vapour pressure. After some time, an
equilibrium is established between liquid phase and vapour phase. The vapour pressure at this stage is known
as equilibrium vapour pressure or saturated vapour pressure at that temperature (vapour pressure is
temperature dependent).
When molecules with higher kinetic energies escape from the liquid into the vapour phase, the average kinetic
energy and hence temperature of the liquid decreases. Thus, a liquid on evaporation cools itself.
1. Factors affecting rate of evaporation. i) Temperature. Evaporation increases with temperature because
of increase in kinetic energy of molecules.
ii) Surface area of liquid. Evaporation is a surface phenomenon. Greater the surface area of the liquid greater
is the evaporation.
iii) Nature of liquid. Liquids having weaker intermolecular forces tend to evaporate faster. Thus, dimethyl
ether evaporates faster than ethyl alcohol. A liquid that evaporates more rapidly is said to be more volatile.

Figure - 08 Vapour pressure curves of various liquids

2. Vapour pressure - temperature curves and boiling points of liquids. A liquid, heated in an open
vessel, boils at the temperature at which its vapour pressure becomes equal to the atmospheric pressure. Free
vapourisation occuring throughout the liquid is called boiling. The boiling temperature at 1 atm pressure is
called normal boiling point. When pressure is 1 bar, the boiling point is called standard boiling point of the
liquid. Standard boiling point is slightly lower than the normal boiling point because 1 bar pressure is slightly
less than 1 atm. The normal boiling point of water is 100oC (373 K), the standard boiling point is 99.6oC
(372.6 K).
3. Effect of External Pressure on boiling point. i) At high altitudes, atmospheric pressure is low, and
therefore, liquids boil at lower temperatures than at sea level. Since water boils at lower temperature, pressure
cooker is used for cooking food.
ii) In hospitals, surgical instruments are sterilized in autoclaves in which boiling point of water is increased by
increasing the pressure above the atmospheric pressure.
iii) If the pressure above a liquid is lowered by connecting the vessel to a vacuum pump, the liquid boils at a
much lower or any desired temperature. Many organic liquids, which decompose before their normal boiling
points are distilled under reduced pressure.
4. Critical temperature. A liquid does not boil when heated in a closed vessel, but its vapour pressure
increases steadily. When the density of the liquid and its vapours becomes the same, the boundary between the
liquid and the vapours (meniscus) disappears. This temperature is called critical temperature.
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5. Heat of Vaporisation. The quantity of heat required to convert one mole of a liquid into vapour at its
boiling point is known as molar heat of vaporisation, HV.
Table - 07 Molar heats of vaporisation (HV) and normal boiling points (Tb) of liquids.
-1 -1
Liquid  HV (kJ mol ) Tb (K) Liquid  HV (kJ mol ) Tb (K)
Water 40.67 373 Benzene 31.38 353
Ethyl alcohol 39.54 351 CCl4 29.96 350
Acetic acid 24.31 391 Ether 25.64 307.6

The molar heat of vaporisation of water at 100oC is 40.67 kJ mol–1. The same amount of heat is evolved when
one mole of steam condenses into liquid water at 100oC.
The molar heat of vaporisation of water is much higher than that of ether, benzene or carbon tetrachloride. This
indicates that intermolecular forces of attraction in water are much stronger than those in most organic liquids.
5.8.2 Surface Tension
Liquids have a tendency to minimise their surface area. The force that tends to contract the surface of a liquid
is known as surface tension. Thus, surface tension may be defined as the force acting at right angles to
the surface of a liquid along unit length of the surface. It is generally represented by the symbol  and is
expressed in dynes cm–1. In SI units, surface tension is defined as the force in newtons (1 newton = 105 dynes)
acting at right angles to the surface of a liquid along 1 metre length of the surface.
The existence of strong intermolecular forces of attraction in liquids gives rise to surface tension. Consider a
molecule P somewhere in the bulk of the liquid. It is attracted equally in all directions by other molecules which
surround it and, therefore, cancel the effect of one another.

Figure - 09 Forces acting on a molecule on liquid surface (R) and inside bulk (P).
For a molecule near the surface (at R), the downward attractive forces are greater than the upward forces
because there are more molecules of the liquid below than in the air above the surface. These unbalanced
attractive forces, acting downward, tend to draw the surface molecules into the bulk of the liquid and therefore,
tend to reduce the surface area to a minimum.
Table - 08 Surface tension of some common liquids (in Nm-1) at 20oC.
Liquid Surface Tension Liquid Surface Tension Liquid Surface Tension
Water 0.0728 CCl4 0.0268 CS2 0.0335
Acetic acid 0.0276 Benzene 0.0289 Acetone 0.0237
Ethyl alcohol 0.0223 Toluene 0.0284 Ethyl ether 0.017
For most liquids, surface tension at room temperature varies between 0.027 and 0.042 Nm-1. For water,
however,  is 0.0728 Nm–1 at 20oC. This high value is due to extensive hydrogen bonding in water.
Surface Energy. The work required to increase the surface area of a liquid by one cm2 is called surface
energy of the liquid.

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While surface tension has units of force per unit length, i.e, dynes cm–1, surface energy has units of work
(energy) per unit area, i.e., ergs cm–2. The S.I. units of surface energy is joules per sq. metre or J m-2.
1 dyne cm–1 = 1 erg cm–2 = 10–3 N m–1
The dimensions of surface energy, viz., ergs cm–2 may also be put as (dynes × cm)/cm–2, i.e., dynes cm–1,
which are also the dimensions of surface tension. Thus, surface tension is equal to surface energy numerically
as well as dimensionally.
Effect of Temperature on Surface Tension. Since intermolecular forces decrease with increase in energy
of molecules, surface tension of a liquid decreases with rise in temperature.
At critical temperature, since the surface of separation between a liquid and its vapour disappears, the surface
tension becomes zero.
Effects of Surface Tension. i) The effect of surface tension is to reduce the area of the surface to a minimum.
Hence, drops of a liquid or bubbles of a gas are spherical in shape (a sphere has minimum surface area for a
given volume). ii) Liquids wet things because they spread across their surfaces as thin film. iii) The rise of a
liquid in a capillary tube (e.g., rise of oil in the wick of a lamp, rise of underground water on to the surface of
earth, rise of water or sap in plants) is a well known phenomenon.
iv) Surface Active Agents. Substances which, when added to water, lower its surface tension are called
surface active agents or surfactants. Liquids like methanol, ethanol, acetone and acetic acid when added to
water, lower its surface tension. Soaps, salts of higher sulphonic acids and higher amines are surface active
materials which act as detergents.
5.8.3 Viscosity
Viscosity may be defined as the force of friction between two layers of a liquid moving past one another with
different velocities. It is intermolecular forces that resist the movement of layers past one another.
When a liquid flows through a tube, the layer immediately in contact with the wall of the tube is stationary. The
velocity of the successive layers increases as we move away from the surface. This type of flow where there is
regular change of velocity between layers is called laminar flow.

Figure - 10 The velocity profile in laminar flow.

When the velocity of the layer at a distance dz is changed by du, the velocity gradient is given by du/dz. In
order to maintain the flow of the layers, an external force must be applied. This force is proportional to the
area of contact of layers, A and the velocity gradient, du/dz.

du du du f.dz
f  A and f  or f A or f  A or 
dz dz dz A.du
where the constant of proportionality  is called the coefficient of viscosity.
Thus, viscosity coefficient is the force when velocity gradient and area of contact are unity. In cgs system, the
unit of is poise (named after Jean Louise Poiseuille). SI unit of viscosity coefficient is newton second per
square metre (N s m-2) = pascal second (Pa s = 1kg m-1 s-1).

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1 poise = 1 g cm-1 s-1 = 10-1 kg m-1 s-1

The above equation called Newton’s law of viscosity applies only to laminar or streamlined flow and not to
turbulent flow.
5.8.4 Fluidity. The reciprocal of viscosity is called fluidity, denoted by . Thus, fluidity,  = 1/.
Effect of nature of liquid on viscosity. Greater the viscosity, the more slowly the liquid flows. H-bonding
and van der Waals forces cause high viscosity. Polyhydric alcohols like glycerol have very high viscositiy due
to extensive H-bonding between molecules. Glass is an extremely viscous liquid; so viscous that many of its
properties resemble solids.
Table - 08 Coefficient of viscosity of some liquids.

Liquid  Liquid  Liquid 

Water 1.008 Ethyl ether 0.233 Acetic acid 1.229
Chloroform 0.563 Benzene 0.647 Nitrobenzene 2.01
Ethyl alcohol 1.216 Acetone 0.329 CCl4 0.968
Effect of temperature on viscosity. The viscosity of a liquid decreases with rise in temperature. The decrease
is about 2 per cent per degree rise of temperature in many cases.
Effect of Pressure. The viscosity of liquids increases with increase in pressure.

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QUESTIONS
LEVEL - I
1. Which among the following statements is not true?
1) Individual molecules of a liquid doesnot boil but the bulk boils
2) Water has wetting property but individual water molecules do not wet
3) Water can exist as liquid water, ice and water vapour. The physical and chemical properties of ice water
and water vapour are different
4) Characteristics of solid, liquid and gaseous state of water depends on energies of molecules and manner in
which water molecules aggregate
2. Which among the following is not a type of vander wall’s force?
1) dipole-dipole interaction 2) dipole-induced dipole interaction
3) ion-dipole interaction 4) dispersion forces
3. Which among the following is not true regarding dispersion forces?
1) Dispersion force is another name for London force
2) These forces are attractive in nature and is inversely proportional to sixth power of distance between two
interacting particles.
3) Their magnitude depends on the polarisability of particle
4) These forces are important only at long distance i.e., more than ~500 pm
4. Hydrogen bonded to which all elements among the following may participate in hydrogen bonding?
(i) N (ii) O (iii) F (iv) Cl
1) (i) and (ii) 2) (ii) and (iii) 3) (i), (ii) and (iii) 4) (i), (ii), (iii) and (iv)
5. Of the officially accepted 118 elements, how many are gases at 1 atmosphere pressure and 298 K?
1) 9 2) 11 3) 13 4) 21
6. Which among the following is not a property of gaseous state?
1) They mix evenly and completely in all proportions without any mechanical aid.
2) Volume and shape are not fixed.
3) They are highly compressible.
4) They exert pressure unequally in all directions.
7. Which of the following graphs is not according to Boyle’s law?

1) 2) 3) 4)

8. If a gas is expanded at constant temperature:

1) The pressure decreases 2) The kinetic energy of the molecules remains the same
3) The kinetic energy of the molecules decreases 4) The number of molecules of the gas increases

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9. At 25oC and 730 mm pressure, 730 mL of dry oxygen was collected. If the temperature is kept constant what
volume will oxygen gas occupy at 760 mm pressure?
1) 701 mL 2) 449 mL 3) 569 mL 4) 621 mL
10. V vs T curves at different pressures P1 and P2 for an ideal gas are shown below:

Which one of the following is correct?

1) P1 > P2 2) P1 < P2 3) P1 = P2 4) P2/P1 = 1/2
11. A balloon indoors where the temperature is 27oC has a volume of 2.00 L. What will be the volume of
outdoors where the temperature is-23oC? Assuming pressure remains constant.
1) 1.67 L 2) 2.23 L 3) 0.53 L 4) 1.26 L
12. A gas cylinder containing cooking gas can withstand a pressure of 14.9 atm. The pressure of the cylinder
indicates 12 atm, at 27oC. Due to sudden fire in the building, its temperature starts rising. At what temperature
will the cylinder explode?
1) 90.5o C 2) 99.5oC 3) 87.3o C 4) 34oC
13. The molecular weight of O2 and SO2 are 32 and 64 respectively. If one litre of O2 at 15oC and 750 mm
pressure contains N molecules, the number of molecules in two litres of SO2 under the same conditions of
temperature and pressure will be:
1) N/2 2) N 3) 2N 4) 4N
14. Which of the given sets of temperature and pressure will cause a gas to exhibit the greatest deviation from
ideal gas behaviour?
1) 100oC and 4 atm 2) 100oC and 2 atm 3) –100o C and 4 atm 4) 0oC and 2 atm
15. A gas was compressed to half of its volume at 30oC. To what temperature it should be heated so that its
volume increases to double of its original volume? (At constant pressure)
1) 60oC 2) 303 K 3) 1212 K 4) 606 K
16. 0.2 g of a gas X occupies a volume of 440 mL. If 0.1 g of carbon dioxide gas occupies a volume of 320 mL
at the same temperature and pressure, gas X could be:
1) O2 2) SO2 3) NO 4) C4H10
17. Equal weights of ethane and hydrogen are mixed in an empty container at 25oC. The fraction of the total
pressure exerted by hydrogen is:
1 2 1 15
1) 2) 3) 4)
2 15 16 16
18. 1.22 g of a gas measured over water at 15 C and a pressure of 775 mm of mercury occupied 900 mL.
o

Calculate the volume of dry gas at NTP (vapour pressure of water at 15oC is 14 mm).
1) 372.21 mL 2) 854.24 mL 3) 869.96 mL 4) 917.76 mL
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19. 3.7 g of a gas at 25oC occupied the same volume as 0.184 g of hydrogen at 17oC and at the same pressure.
The molecular mass of the gas is:
1) 0.024 2) 39.14 3) 41.33 4) 59.14
20. Equal masses of methane and oxygen are mixed in an empty container at 25oC. The fraction of the total
pressure exerted by oxygen is:
1 273
1) 1/2 2) 2/3 3)  4) 1/3
3 298
21. According to kinetic theory of gases for a diatomic molecule:
1) the pressure exerted by the gas is proportional to the mean square speed of the molecules
2) the pressure exerted by the gas is proportional to the root mean square speed of the molecules
3) the root mean square speed is inversely proportional to the temperature
4) the mean translational KE of the molecules is directly proportional to the absolute temperature
22. The molecular velocity of any gas is:
1) inversely proportional to the square root of temperature
2) inversely proportional to absolute temperature
3) directly proportional to square of temperature
4) directly proportional to square root of temperature
23. At STP, the order of mean square velocity of molecules of H2, N2, O2 and HBr is:
1) H 2  N 2  O 2  HBr 2) HBr  O 2  N 2  H 2
3) HBr  H 2  O 2  N 2 4) N 2  O2  H 2  HBr
24. A gas bulb of 1mL capacity contains 2.0 × 1021 molecules of nitrogen exerting a pressure of
7.57 × 103Nm-2. The root mean square speed of the gas molecules is:
1) 274 ms-1 2) 494 ms-1 3) 690 ms-1 4) 988 ms-1
25. Which one of the following statements is not true about the effect of an increase in temperature on the distribution
of molecular speeds in a gas?
1) The area under the distribution curve remains the same as under the lower termperature
2) The distribution becomes broader
3) The fraction of the molecules with the most probable speed increases
4) The most probable speed increases
26. A 3 : 2 molar mixture of N2 and CO is present in a vessel at 500 bar pressure. Due to hole in the vessel, the
gas mixture leaks out. The composition of mixture effusing out initially is:
1) n N2 : n CO ::1: 2 2) n N2 : n CO :: 6 :1 3) n CO : n N 2 ::1: 2 4) n CO : n N 2 :: 2 : 3
27. 1 mol of N2 gas at 0.8 atm takes 38 s to diffuse through a pinhole, whereas 1 mol of an unknown gas at 1.6
atm takes 57 s to diffuse through the same pinhole. The molecular weight of unknown gas is
1) 126 2) 64 3) 80 4) 252
28. What is the compressibility factor of water vapour at 10oC and 1atm pressure? Its molar volume is 33.18 dm3.
1) 0.896 2) 1.42 3) 1.986 4) 1.896
29. Which among the following is the highest temperature at which liquid carbon dioxide can be observed to be
existing is:
1) Inversion temperature 2) Critical temperature 3) Boyle temperature 4) Kraft temperature
30. Which among the following is not a unit of viscosity co-efficient?
1) Ns m–2 2) 1 Pa s 3) 1 Nm–1 4) 1 poise
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LEVEL - II
1. Strength of hydrogen bond is determined by:
1) Coulombic interaction between the lone pair electrons of electronegative atom of one molecule and hydrogen
atom of other molecules.
2) Coulombic interaction between hydrogen atom bonded to electronegative atom of one molecule and
another similar hydrogen of other molecules.
3) Number of hydrogen atoms bonded to electronegative atom in one molecule.
4) The electron gain enthalpy value of the electronegative atom.
2. A 15.0 L cylinder of Ar gas is connected to an evacuated 235.0 L tank. If the final pressure is 750 mm Hg,
what have been the original gas pressure in the cylinder?
1) 76 atm 2) 12.56 atm 3) 16.45 atm 4) 23 atm
3. A spherical air bubble is rising from the depth of a lake when pressure is P atm and temperature is T K. The
percentage increase in the radius when it comes to the surface of a lake will be: (Assume temperature and
pressure at the surface to be, respectively, 2 TK and P/4).
1) 100% 2) 50% 3) 40% 4) 200%
4. I, II and III are three isotherms, respectively, at T1, T2, and T3. Temperature will be in order:

1) T1 = T2 = T3 2) T1 < T2 < T3 3) T1 > T2 > T3 4) T1 > T2 = T3

5. An open flask containing air is heated from 300 K to 500 K. What percentage of air will be escaped to the
atmosphere, if pressure is keeping constant?
1) 80 2) 40 3) 60 4) 20
6. A rigid container containing 5 mole H2 gas at same pressure and temperature. The gas has been allowed to
escape by simple process from the container due to which pressure of the gas becomes half of its initial
pressure and temperature become (2/3)rd of its initial. The mass of gas remaining is:
1) 7.5 g 2) 1.5 g 3) 2.5 g 4) 3.5 g
7. Two closed vessels of equal volume containing air at pressure p1 and temperature T1 are connected to each
other through a narrow tube. If the temperature in one of the vessels is now maintained at T1 and that in the
other at T2, what will be the pressure in the vessels?
2p1T1 T1 2p1T2 2p1
1) 2) 3) 4)
T1  T2 2p1T2 T1  T2 T1  T2
8. 15 L of gas at STP is subjected to four different conditions of temperature and pressure as shown below. In
which case the volume will remain unaffected?
1) 273 K, 2 bar pressure 2) 273oC, 0.5 atm pressure
3) 546oC, 1.5 atm pressure 4) 273oC and 2 atm pressure

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9. How many millilitres of H2O vapour measured at 327oC and 760 torr are formed when 50 mL of ammonia at
950 torr and 127oC reacts with oxygen according to the following reaction?

NH3  g   O2  g   N 2  g   H 2O  g 
1) 75 mL 2) 125 mL 3) 140.625 mL 4) 241.4 mL
10. Starting out on a trip into the mountains, you inflate the tires on your automobile to a recommended pressure
of 3.21 × 105 Pa on a day when the temperature is –5.0oC. You drive to the beach, where the temperature is
28.0oC. Assume that the volume of the tire has increased by 3%. What is the final pressure in the tyres?
1) 350 Pa 2) 3500 Pa 3) 3.5 × 105 Pa 4) 3.5 Pa
11. A gas of volume 100 cc is kept in a vessel at pressure 104Pa maintained at temperature 24oC. If now the
pressure is increased to 105Pa, keeping the temperature constant, the volume of gas becomes:
1) 10 cc 2) 100 cc 3) 1 cc 4) 1000 cc
12. A balloon weighing 50 kg is filled with 685 kg of helium at 1 atm pressure and 25oC. What will be its pay load
if it displaced 5108 kg of air?
1) 4373 kg 2) 4423 kg 3) 5793 kg 4) 5192 kg
13. A flask containing 12 g of a gas of relative molecular mass 120 at a pressure of 100 atm was evacuated by
means of a pump until the pressure was 0.01 atm. Which of the following is the best estimate of the number of
molecules left in the flask (N0 = 6 × 1023 mol–1)?
1) 6 × 1019 2) 6 × 1018 3) 6 × 1017 4) 6 × 1013
14. A chemist isolated a gas in a glass bulb with a volume of 255 mL at a temperature of 25oC and a pressure (in
the bulb) of 10.0 torr. The gas weighed 12.1 mg. What is the molecular mass of the gas?
1) 78.9 g mol–1 2) 35.2 g mol–1 3) 88.2 g mol–1 4) 96.3 g mol–1
15. Dry ice (solid CO2) has occasionally been used as an explosive in mining. A hole is drilled, dry ice and a small
amount of gun powder are placed in the hole, a fuse is added, and the hole is plugged. When lit, it explodes up
with an immense pressure. Assume that 500.0 g of dry ice is placed in a cavity with a volume of 0.800 L and
the ignited gunpowder heats the CO2 to 700 K. What is the final pressure inside the hole?
1) 416 atm 2) 816 atm 3) 616 atm 4) 1216 atm
16. The compressibility factor for nitrogen at 330 K and 800 atm is 1.90 and at 570 K and 200 atm is 1.10. A
certain mass of N2 occupies a volume of 1 dm3 at 330 K and 800 atm. Calculate volume occupied by same
quantity of N2 gas at 570 K and 200 atm.
1) 1 L 2) 2 L 3) 3 L 4) 4 L
17. At STP, a container has 1 mole of He, 2 mole Ne, 3 mole O2 and 4 mole N2. Without changing total pressure
if 2 mole of O2 is removed, the partial pressure of O2 will be decreased by:
1) 26% 2) 40% 3) 58.33% 4) 66.66%
18. A vessel is filled with a mixture of oxygen and nitrogen. At what ratio of partial pressures will the mass of
gases be identical?
1) PO2  0.785PN 2 2) PO2  8.75PN 2 3) PO2  11.4PN 2 4) PO2  0.875PN 2

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19. A manometer attached to a flask contains ammonia gas have no difference in mercury level initially as shown
in diagram. After sparking into the flask, ammonia is partially dissociated as 2NH 3  g  
 N 2  g   3H 2  g 
now it have difference of 6 cm in mercury level in two columns, what is partial pressure of H2(g) at equilibrium?

1) 9 cm Hg 2) 18 cm Hg 3) 27 cm Hg 4) 15 cm Hg
20. The ratio among most probable velocity, mean velocity and root mean square velocity is given by:
1) 1 : 2 : 3 2) 1: 2 : 3 3) 2 : 3 : 8 /  4) 2 : 8 /  : 3
21. The ratio of average speed of an oxygen molecule to the rms speed of a nitrogen molecule at the same
temperature is:
1/2 1/2 1/ 2 1/ 2
 3   7   3   7 
1)   2)   3)   4)  
 7   3   7   3 
22. Distribution of molecules with velocity is represented by the curve

Velocity corresponding to point A is:

3RT 2RT 8RT RT
1) 2) 3) 4)
M M M M
23. A small quantity of gaseous NH3 and HBr are introduced simultaneously into the opposite ends of an open
tube that is 1 m long. Calculate the distance of the white solid NH4Br formed from the end that was used to
introduce NH3:
1) 68.55 cm 2) 63.5 cm 3) 60.09 cm 4) 65.24 cm
24. Rate of effusion of LPG (a mixture of n-butane and propane) is 1.25 times that of SO3. The mole fraction of
n-butane in LPG is:
1) 0.752 2) 0.256 3) 0.514 4) 0.667
25. The Van der Waals parameters for gases, W, X, Y and Z are:
Gas a (litre2-atm/mole2) b(litre/mole)
W 4.0 0.027
X 8.0 0.030
Y 12.0 0.027
Z 6.0 0.024
Which one of the gases has the highest Boyle temperature?
1) W 2) X 3) Y 4) Z
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26. The compressibility factor for definite amount of van der Waal’s gas at 0oC and 100 atm is found to be 0.5.
Assuming the volume of gas molecules negligible, the van der Waal’s constant a for a gas is:
1) 1.256 atm L2 mol–2 2) 0.256 atm L2 mol–2 3) 2.256 atm L2 mol–2 4) 0.0256 atm L2 mol–2
27. Which one is not correct for gaseous state obeying van der Waals’ equation?
1) Compressibility factor at critical temperature  0.375.
2) For a gas if van der Waals’ constant a = 0, TC = 0
3) Ideal gases do not have critical temperature
4) Gaseous molecules showing H-bonding show minimum deviations from Z  0.375.
28. On heating a liquid in a closed container, the temperature at which the density of liquid and vapour becomes
the same is called
1) Boyle temperature 2) Curie point 3) Critical temperature 4) triple point
29. A drop of liquid acquires spherical shape because
1) of its viscous nature 2) of capillary action
3) surface tension tends to minimise the surface area 4) all of these
30. Fire polishing of glass rod is an application of which among the following properties of liquids?
1) viscosity 2) surface tension 3) vapourisation 4) thermal expansion

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SOLUTIONS
LEVEL - I
1. 3 Physical properties of ice, water and vapour are different, but chemical composition of water remain
the same.
2. 3 ion-dipole forces are not-van der Waals force.
3. 4 dispersion forces are important only at short distance ~ 500 pm
4. 4 Species such as Cl can also participate in hydrogen bonding like N, O and F.
5. 2 H, He, N, O, F, Ne, Cl, Ar, Kr, Xe and Rn are gases.
6. 4 Gases exert pressure equally in all directions.
1 1
7. 3 According to Boyle’s law, p  therefore, graph between p and cannot be a straight line parallel
V V
to x-axis.
1
8. 1 According to Boyle’s law, V  at constant T..
P
P1V1 730  730
9. 1 P1V1  P2 V2  V2    701 mL
P2 760
10. 2

V1 V2 V  V1T2  2  250  1.67 L

11. 1 From Charle’s law,  , 2
T1 T2 T1 300

P1 P2 12 14.9
12. 2     T2 = 372.5 K = 99.5o C
T1 T2 300 T2

13. 3 Vn
14. 3 Greatest deviation from ideal behaviour is exhibited by real gases at low temperature and high pressure.
V
15. 3 T1 = 303 K; V1  ; V2 = 2 V
2
V / 2 303
T2 = ?   T2 = 1212 K
2V T2

16. 2 Vn; 440 0.2 / M


320 0.1 / 44
W W 15
17. 4 n C2 H 6  , n H2  , X2 
30 2 16
18. 2 Pressure of dry gas = Pressure of moist gas – Vapour pressure of water = 775 – 14 = 761 mm
p1V1 p 2 V2 p1V1T2 761 900  273
From,   V2   = 854.24 mL
T1 T2 T1p 2 288  760
initial NTP

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19. 3 For hydrogen, w = 0.184 g, T = 17 + 273 = 290 K, M = 2

w 0.184
From pV  RT   R  290 .....(i)
M 2
For unknown gas, w = 3.7 g, T = 25 + 273 = 298 K, M = ?
3.7
pV   R  298 .....(ii)
M
3.7 0.184 3.7  298  2
From Eqs. (i) and (ii)  R  298   R  290 or M   41.33
M 2 0.184  290
x
20. 4 PO2  O2  P   O2  32  1  P  1 P
O2
x x 3 3

32 16
21. 4
8RT 3RT
22. 4 Molecular velocity can be average velocity  , root mean square velocity 
M M
2RT
most probable velocity  . In all cases molecular velocity  T
M
1
23. 1 rms 
Mw

2.0  1021
24. 2 Number of moles of the gas  mol = 3.32 × 10-3 mol
6.02  1023
pV 7.57  103  103
From, pV = nRT , T    274.25K
nR 3.32  103  8.314
3RT 3  8.314  274.25
 Root mean square speed, v rms   3
 494.26 ms  1
M 28  10
25. 3 Distribution of molecules (N) with velocity (u) at two temperatures T1 and T2 (T2 > T1) is shown below:
At both temperatures, distribution velocity first increases, reaches a maximum value and then decreases.
Most probable speed increases with temperature increases, but fraction possessing most probable
speed decreases.

26. 4 Molar ratio of N2 and CO is 3 : 2, ie, 300 bar and 200 bar, respectively.
n N2 m CO PN2 300 3
   
n CO m N2 PCO 200 2

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n1 t 2 p1 M2 1 57 0.8 M 2
27. 4        M 2  252
t1 n 2 p 2 M1 38 1 1.6 28
PVm
28. 2  1.42
nRT
29. 2 It is crtical temperature.
30. 3 1 Nm–1 is the SI unit of surface tension.
LEVEL - II
1. 1 Strength of hydrogen bond is determined by strength of interaction between electronegative atom of one
molecule and hydrogen atom of other molecules.
2. 3 From Boyle’s law p1V1 = p2V2

p 2 V2 750  250
p1    V  235  15  250 L = 12500 mm Hg = 16.45 atm
V1 15

P1V1 P2 V2  V1  initial 
3. 1  ;
T1 T2  
 V2  final volume 
3
PV1 P V2 V V2 r 
   V1  2  V2  8V1 ;  8   2   r2  2r1
T 4 2T 8 V1  r1 
4. 3 Draw a line at constant P parallel to volume axis. Take volume corresponding to each temperature.
From volume axis, V1  V2  V3 Hence, T1  T2  T3

5. 2 V1  V, T1  300 K, T2  500 K, V2  ?

V1T2 V  500 5V
At constant pressure V1T2 = V2T1  V2   
T1 300 3

5V 2V
Volume of air escaped = final volume – initial volume  V 
3 3
2V / 3
 % of air escaped =  100  40%
5V / 3
10
6. 1 PV  RT .........(1)
M
P 2T
Let x g of the gas remain in the flask when final pressure ,
2 3
P  x   2T 
  V   R   .........(2)
2 M  3 
10 3 3  10
Dividing (1) by (2)  2    x  7.5 g
x 2 4

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7. 3 Volume of the system and amount of gas remains constant. Applying Charle’s law,
p1 p1 p p 2p1 T T 
     p  1 2  or p  2p1T2
T1 T1 T1 T2 T1  T1T2  T1  T2
8. 4 V1 = 15L, P1 = 1 atm, T1 = 0 C or 273.15 K,
o
P2 = 2 atm, T2 = 273oC or 546.15 K
P1V1 T2 1 15 546.15
V2 = ?  V2      15 L
T1 P2 273.15 2
9. 3 We first require volume of ammonia at the same condition as it is for water vapour. Hence, we use
p V pV  p  T 
combined gas law equation 2 2  1 1  V2   1   2  V1  950  600  50 mL  93.75 mL
T2 T1  p 2   T1  760 400
Now considering the following balanced chemical equation, we get
4H3  g   3O2  g   2N 2  g   6H 2 O  g 
4 vol of NH3 = 6 vol of H2O  93.75 mL of NH3 will produce
6
  93.75 mL  140.625 mL H 2 O vapour
4
Pi Vi Pf Vf PVT Pi Vi Tf
10. 3 Because the number of moles is constant,  ; Pf  i i f ; Pf 
Ti Tf Vf Ti Vf Ti

 3.21 105 Pa 
Vi

 273  28.0  = 3.50 × 105 Pa
1.03Vi  273  5.00 

11. 1 P1V1  P2 V2 ; 104 . 102 = 105V  V = 10 cc

12. 1 Mass of the filled balloon = 50 + 685 = 735 kg
Pay load = Mass of displaced air – Mass of balloon = 5108 – 735 = 4373 kg
13. 2 P1V1 = n1RT1 ; P2V2 = n2RT2
n 2 P2 0.01
  n 2  0.1   10 5 mol  6  1018 molecules
n1 P1 100

 1 atm   1L 
10.0 torr     255 mL  
PV  760 torr   1000 mL 
14. 3 n  = 1.37 × 10–4 mol
RT  L atm 
 0.0821   298.2 K 
 mol K 
 1g 
12.1 mg  
Mass  1000 mg 
Molecular mass =   88.2g mol1
Number of moles 1.37 104 mol
15. 2 We know, that 44 g of CO2 = 1 mol
500
Therefore, 500 g of CO2 =  11.36 mol
44
Also given that T = 700 K, V = 0.8 L. Now, using the ideal gas equation, we get
nRT 11.36  0.082  700
p   816 atm
V 0.8
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PV 1 800 1 800 V  200

16. 4 Z ; 1.90  ;n ; Z  1.10 
nRT n  R  330 1.90  R  330 n  R  570
V  200  1.90  R  330
1.10  V=4L
800  R  570
3 1
17. 3 PO2   PT ; After removing 2 mole of O2  PO 2   PT
10 8
3PT PT

Decreasing in p.pr. of O2 = 10 8  100 = 58.33
3PT
10
w w w PO2 28
PV  RT , PO2 V  RT ; PN2 V  RT , 
18. 4
m 32 28 PN2 32 PO2  0.875 PN2

19. 1 2NH3  g  
 N 2  g   3H 2  g 
76 0 0 Before sparking
76 – 2x x 3x After sparking at eqm
Partial pressure of H2 = 3 × 3 = 9 cm Hg
8
20. 4 : v:u  2 : : 3

8RT 8RT 3RT 3RT
21. 2 u av  So, u av O2   ; u rms  so u rms N2  
M  32 M 28

u av O2  8  28 7  7 
1/ 2

   
v rms N 2   32  3 3  3 

22. 2 Point A represents the most probable distribution of molecules. Hence, the most probable velocity is
2RT / M .
rNH3 M HBr 81
23. 1 By Graham’s law of diffusion    2.18
rHBr M NH3 17
Thus, NH3 travels 2.18 times faster than HBr. In other words, NH3 will travel 2.18 cm in the same time
in which HBr travels 1 cm. Given that the length of the tube = 100 cm.
2.18
Therefore, distance travelled in the tube by NH3   100  68.55 cm
2.18  1

rSO3 M mix 1 M mix

24. 3 Since,     M mix  51.2
rmix M SO3 1.25 80

Since M mix =x butane M butane +x propane M propane ; Therefore, 51.2 = 58x butane  44 1  x butane 

 7.2  14x butane ,  x butane  0.514

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a
25. 3 TB 
Rb
zRT  a  a
26. 1 Vm   0.11207 ;  P  2  V  RT, PV   RT
P  V  V
a
z  1 a  1.256 atm L2 mol2
RTVm

PC VC 3R
27. 4  ; Also, gaseous molecules showing H-bonding show maximum deviations in Z due to
TC 8
increase in molecular attractions (e.g., Z for NH3, H2O, CH3OH  0.22 to 0.24)

28. 3 It is critical temperature

29. 3
30. 2 Fire polishing of glass rod in an application of surface tension

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CHAPTER - 06

CHEMICAL THERMODYNAMICS

INTRODUCTION
The study of energy transformations in chemical processes is referred to as chemical thermodynamics. It is
primarily based on three generalisations which have been arrived at purely on the basis of human experience.
These are known as first, second and third laws of thermodynamics. The laws apply only to matter in bulk and
not to individual atoms or molecules and only when the system is in equilibrium or moves from one equilibrium
state to another. Most of the generalisations in physical chemistry, can be deduced from the laws of
thermodynamics. It helps to lay down the criteria for predicting feasibility or spontaneity of a process under a
set of conditions.
6.1 THERMODYNAMIC TERMS
6.1.1 The System and the Surroundings
From a thermodynamic point of view, the universe can be divided into two parts; the system and the surroundings.
The Universe = The System + The Surroundings
System. It is a specified part of the universe with real or imaginary boundaries under investigation. A system
may consist of one or more substances.
Surroundings. The part of the universe other than the system is known as surroundings. For practical purposes,
the region in the vicinity of the system under investigation constitute the surroundings.
Homogeneous and Heterogeneous Systems. A system is said to be homogeneous if it consists of one
phase only and heterogeneous if it consists of more than one phase (a phase is a homogeneous, physically
distinct, and mechanically separable part of a system which is bounded by a surface).
6.1.2 Types of Systems
There are three types of systems; open system, closed system and isolated system.
(a) Open System. A system which can exchange matter as well as energy with the surroundings is
called an open system. For example, hot water in a beaker. In this case, heat can flow from hot water to the
surroundings and water vapours can escape into the surroundings.
(b) Closed system. A system which can exchange energy with the surroundings but not matter is called
a closed system. For example, hot water in a closed bottle. In this case heat can flow out through the walls of
the bottle but water vapour cannot escape into the surroundings.
(c) Isolated system. A system which can exchange neither matter nor energy with the surroundings is
called an isolated system. Hot water in a thermos flask is an example. In this case neither heat nor water
vapour can escape from the system.

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Adiabatic system. An adiabatic system is one in which no heat enters or leaves the system. Energy is transferred
only as work.
Macroscopic system and Macroscopic properties. A system containing a large number of chemical species
(atoms, ions, or molecules) is called a macroscopic system. The properties of the system which arise from
the collective behaviour of large number of species are called macroscopic properties, e.g., temperature,
pressure, volume, composition, refractive index, etc.
6.1.3 Extensive and Intensive Properties
Physical properties of a system can be classified into extensive properties and intensive properties.
(a) Extensive properties. The properties of a system which depend on the quantity of matter contained in
it are called extensive properties. Mass, volume, energy, heat capacity, etc., are examples.
(b) Intensive properties. These are properties of a system which are independent of the quantity of matter
present in it, e.g., temperature, pressure, density, viscosity, specific heat, refractive index, etc.
Sometimes an extensive property becomes an intensive property by specifiying unit amount of the substance,
e.g., mass and volume are extensive properties but mass per unit volume, i.e., density, is an intensive property.
A molar property m is the value of an extensive property  for 1 mol of a substance. If n is the amount of
matter,  m   / n is independent of the amount of matter. Other examples are molar volume, Vm and molar
heat capacity, Cm.
6.1.4 State of the System
The state of a system is described by its measurable or macroscopic (bulk) properties.
a) State Functions or State Variables. The macroscopic properties which change with change in the state
of the system are called state variables or state functions. E.g., the state of a gas can be described by stating its
pressure (P), volume (V), temperature (T), amount (n), etc. Their values depend only on the state of the
system and not on how it is reached. State functions have two important properties.
Internal energy (U) and enthalpy (H) are other commonly used state variables.
b) Path Functions or Path Variables. The value of path function depends on path connection of two states.
There can be infinite values of path function between two states depending on the path of the process.
Heat and work are two important path functions.
6.1.5 Thermodynamic process
A thermodynamic process is one which brings about changes in the thermodynamic state of a system.
(a) Isothermal process: A process carried out at constant temperature is called isothermal process. Isothermal
process occurs when the system (called diathermic) has conducting walls (diathermic wall).
(b) Adiabatic process: A process in which no heat is exchanged between the system and the surroundings is
called adiabatic process. Adiabatic process occurs when the system (called adiabatic) has non-conducting
walls (adiabatic wall).
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(c) Isobaric process: A process carried out at constant pressure is called isobaric process.
(d) Isochoric process: A process carried out at constant volume is called isochoric process.
(e) Cyclic process: A process which brings back a system to its original state after a series of changes is
called a cyclic process.

Figure - 01 Pressure - volume changes in various thermodynamic processes.

6.1.6 Thermodynamic equilibrium
A system is said to be in thermodynamic equilibrium if its fundamental properties do not change with time. For
a system to be in thermodynamic equilibrium, it must simultaneously exhibit the following equilibria:
(a) Thermal equilibrium. System attains thermal equilibrium if the temperature throughout the system is the
same. As such no heat flows from any part of the system to any other part.
(b) Mechanical equilibrium. System attains mechanical equilibrium if the process throughout the system is
constant.
(c) Chemical equilibrium. System attains chemcial equilibrium if the composition of the system becomes
constant and do not change with time.
6.1.7 Reversible and Irreversible Processess
1. Reversible Processess. A process or change is said to be reversible if
(i) The various stages of the process can be traversed back in the opposite direction and in the opposite order.
(ii) In every part of the process, the amount of energy transferred in the form of heat or work is the same,
whichever direction the change takes place.
(iii) At every stage, an infinitesimal change of the factors controlling it will make it proceed in the opposite
direction. Thus, a reversible process is one in which the opposing forces are always infinitely close to
equilibrium.
A process carried out infinitesimally slowly can be considered to be a reversible process. Similarly, an isothermal
change (expansion or contraction) carried out slowly is also reversible.
2. Irreversible Processess. An irreversible process is one which is not carried out infinitesimally slowly so
that the sucessive steps of the direct process cannot be retraced. For example, rapid expansion or contraction
(isothermal or adiabatic), heat produced by friction, heat produced by the passage of electric current through
a resistance, diffusion of gases or liquids, etc. In all these processes, the change cannot be traversed back by
an infinitesimal change in the factors controlling it.
All natural processes are irreversible.
6.2 INTERNAL ENERGY, U
The energy possessed by a substance by virtue of its existence is called internal energy or intrinsic energy.
It is generally represented by U (or E). It depends on factors such as quantity of substance, its chemical nature,
temperature, pressure, etc.
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Internal energy is a state function. It is the sum of the chemical, electrical, mechanical or any other type of
energy possessed by a substance. Energies which contribute towards internal energy are: (i) Translational
energy of molecules (Ut), (ii) Rotational energy of molecules (Ur), (iii) Vibrational energy of molecules (Uv),
(iv) Electronic energy (Ue), (v) Nuclear energy (Un), and (vi) Interactional energy of molecules (Ui).
Internal energy, U = Ut + Ur + Uv + Ue + Un + Ui
Internal energy of a system may change when (i) heat passes into or out of the system, (ii) work is done on or
by the system (iii) or matter enters or leaves the system.
6.2.1 Change in Internal Energy. The absolute value of internal energy cannot be determined because it is not
possible to determine the exact values for the constituent energies. However, change in internal energy, U
accompaniying chemical reactions can be determined.If a system at temperature T1 undergoes a change of
state to temperature T2, the change in internal energy, U is given by,
U  U 2  U1 (change in temperature,  T  T2  T1 )
where U2 and U1 are the internal energies of the final and initial states respectively. If UR and UP are the internal
energy of the reactants and products respectively, then change in internal energy,
U  U P  U R
The value of U depends only on the initial and final states of the system and not on the manner in which the
change has been brought about. Therefore, it is a state function.
(i) The internal energy of a system depends on the quantity of substance contained in the system. Thus, it is
an extensive property.
(ii) The internal energy of a definite mass of ideal gas is a function of temperature only, therefore, in an
isothermal process there is no change in internal energy, i.e., U = 0.
Internal energy as a state function can be further explained by considering a certain amount of water taken in
an insulated container. Let the internal energy of the system change by doing some adiabatic work (wad). In
such a case, temperature rises from T1 to T2. Thus, the same amount of work, irrespective of how it is done,
produces the same change in state as measured in terms of temperature change, therefore,
U  U 2  U1  w ad
Sign of U. If U2 > U1 (or UP > UR), i.e., the system absorbs energy, U = U2 - U1 is positive.
If U2 < U1 (or UP < UR), i.e., the system gives out energy, U = U2 - U1 is negative. Thus, U is negative if
energy is evolved and U is positive if energy is absorbed.
Units of U. In S.I. system, internal energy is measured in joules. In C.G.S. system, unit of energy is erg.
1 J = 107 erg.
6.2.2 Transference of Energy
The important modes in which a system exchanges energy with the surroundings are heat and work.
1. Heat. Transference of energy takes place as heat, if the system and surroundings are at different temperatures.
Transfer of energy goes on until the system and the surroundings attain the same temperature.
If a system absorbs heat q from the surroundings, q is positive (+q) and if the system gives out heat q to the
surroundings, q is negative (–q).
2. Work. Work is expressed as the product of two factors, intensity factor and capacity factor,
Work = Intensity factor × Capacity factor

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Intensity factor is a measure of force against which work is done and capacity factor is a measure of the extent
of work done.
Mechanial work = Force × Displacement, i.e., w = F  dl
where F is the magnitude and dl is the displacement. This type of work is known as pressure-volume work.
Work ‘w’ is negative if work is done by the system (expansion) and positive if work is done on the system by
the surroundings (contraction).
Electrical work is important in reactions taking place in electrochemical or electrolytic cells.
Electrical work = Charge × Potential difference
Sign convention. When ‘w’ or ‘q’ is positive, internal energy of the system increases.
Units of Heat and Work: As both heat and work are interconvertible, they are measured in the same units.
The S.I. unit of heat and work is joule or kilojoule. One joule (J) is equal to one newton metre (1 J = 1 Nm).
In C.G.S. system, the unit of heat or work is erg.
6.3 FIRST LAW OF THERMODYNAMICS
German chemist, Herman Helmhotlz (1821-1894) proposed the first law of thermodynamics which is also
known as law of conservation of energy. It can be stated in a number of ways:
(i) Energy can neither be created nor destroyed, although, it may be converted from one form to another.
(ii) Total energy of the universe, i.e., system and the surroundings taken together is always constant during
any physical or chemical process.
(iii) The mass and energy of an isolated system remain constant.
This law is based on human experience that energy cannot be generated without consuming energy of
some other form.
6.3.1 Mathematical expression for the first law of thermodynamics
When a system having internal energy U1 absorbs heat q, its internal energy becomes U1 + q. Let work w be
done on the system so that its internal energy changes to U2, then
U2 = U1 + q + w or U2 – U1 = q + w
U = q + w ...(i)
Change in internal energy = Heat absorbed + Work done on the system.
If there is only pressure - volume work, then,
w  PV
where V is the change in volume at constant external pressure, P. Therefore, Eqn.(i) can be written as
U = q - PV
For an isolated system, q and w are zero. The first law equation becomes
U  0
which means that the energy of an isolated system is constant. Since the universe, by definition, is an
isolated system, the first law can also be stated as: The energy of the universe is conserved.
Inside the universe, energy can be transferred from one part to another or it can be converted from one form
to another, but it can neither be created nor destroyed. This statement identifies the first law as the law of
conservation of energy.
Unit conversions : 1 cal = 4184 J; 1 J = 107 ergs; 1 Nm = 1 J; 1 atm = 1.013 × 105 Nm–2 ; 1 L atm = 101.325 J
or 1.013 × 102 J.
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6.3.2 Pressure-volume work

Consider a cylinder fitted with a weightless and frictionless piston of area of cross-section A. Let it contain an
ideal gas of volume Vi and pressure P. If the external pressure, Pex acting on the piston is greater than P, the
piston moves inward till Pex becomes equal to P. Let the change be achieved in a single step and the final
volume be Vf.

Figure - 02 Work done on an ideal gas when compressd in a cylinder (represented by ‘area’)
If l is the distance through which the piston moves, then the change in volume,
l ×A = V = (Vf - Vi)
Since Pressure = Force/Area and Force = Pressure × Area
Force acting on the piston = Pex × A
Work done on the system, w = force × displacement = Pex× A× l
w  Pex (  V)   Pex  V   Pex (V f  Vi ) ... (i)
Since work is done on the system, w will be positive (the negative sign is to obtain conventional sign for w). If
the pressure is not constant at every stage of compression, but changes in a number of finite steps, work done
on the gas will be the sum of all the steps involved and will be equal to PV .
If the pressure changes in infinitesimally small steps, then, the volume decreases by an infinitesimal amount, dV
at each stage of compression. The work done on the gas is given by the relation
Vf

w    Pex dV ... (ii)

Vi

In this case ( compression), Pex at each stage is equal to (Pin + dp) [Fig. 02]. In case of expansion under similar
conditions, the external pressure will be less than the pressure of the system (Pex = Pin- dP). In general we may
write, Pex = (Pin  dP). Such a process is called reversible process.

(a) (b)

Figure - 03 (a) PV - plot for (a) reversible compression of gas (b) irreversible compression of gas (work done
on the gas is represented by the shaded area).

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6.3.3 Work done in isothermal reversible expansion of an ideal gas

Work can be related to internal pressure of the system under reversible conditions by writing Eqn.(ii) as:
Vf Vf

w rev    Pex dV    (Pin  dP)dV

Vi Vi

Since dPdV is very small, we may write

Vf

w rev    Pin dV ... (iii)

Vi

For n mols of an ideal gas, pV = nRT, or P = nRT/V, therefore,

Vf
dV V V
w rev    nRT  nRT ln f  2.303 nRT log f ... (iv)
Vi
V Vi Vi
Pi  V2 P1 
or w rev  2.303 nRT log Since at constant temperature,  
Pf  V1 P2 
The negative sign indicates work on expansion.
6.3.4 Maximum work
The work done by a gas during expansion is given by PexV where V is the increase in volume. If the external
pressure, Pex is only infinitesimally smaller than the perssure, P of the gas, expansion takes place reversibly.
However, if the external pressure is much smaller than the gas pressure, expansion takes place rapidly, i.e.,
irreversibly and the work done is much smaller.
Thus, the condition for maximum work coincides with that for thermodynamic reversibility.
6.3.5 Free expansion
Expansion of a gas in vacuum (Pex = 0) is known as free expansion. Whether reversible or irreversible, no
work is done during free expansion of an ideal gas.
Substituting w   Pex V in the First law equation, ( U  q  w ),
U  q  Pex V
If the process is carried out at constant volume ( V  0) , then
U  q v
where qv is the heat supplied at constant volume.
6.3.6 Isothermal and free expansion of an ideal gas
For isothermal expansion of an ideal gas into vacuum, w = 0 since Pex= 0. Joule determined experimentally that
q = 0, therefore, U = 0 . Thus, the first law equation, U = q + w changes as
(i) For isothermal irreversible change
q   w  Pex (V f  Vi )
(ii) For isothermal irreversible change
Vf Vf
q   w  nRT ln  2.303 nRT log
Vi Vi
(iii) For adiabatic change, no heat is allowed to enter or leave the system, q = 0, therefore,
U  0  w or w  U or U  w ad

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6.4 ENTHALPY OR HEAT CONTENT, H

Change in internal energy is the heat change accompanying a process when it is carried out at constant volume,
U = qV. But most chemical reactions are carried out under constant atmospheric pressure.
At constant pressure, the first law equation can be written as
U = qP - PV
where qp is heat absorbed and PV is the work done by the system. The above equation can be written as
U2 - U1 = qp - P(V2 - V1) or
qp = (U2 + PV2) - (U1 + PV1) ... (i)
A thermodynamic function called enthalpy is used to describe the total energy associated with a system which
includes its internal energy and energy due to environmental factors such as pressure-volume conditions.
Enthalpy of a system may be defined as the sum of its internal energy and pressure-volume (PV)
energy. Enthalpy is also known as heat content of the system. It is denoted by H.
H = U + PV ... (ii)
Eqn. (i) becomes
q P  H 2  H1 or q P  H
Although q is a path function, H is a state function because it depends on U, P and V, all of which are state
functions. It is also an extensive property like U and V. LikeH, qp is also independent of path.
For finite changes at constant pressure, Eqn.(ii) can be written as :
H  U  PV
Since P is constant, we can write
H  U  PV ... (iii)
Heat absorbed by a system at constant pressure changes its enthalpy, i.e., H  q p .Thus, change in enthalpy
of a system at constant pressure is equal to the increase in internal energy plus pressure volume work done.
H is negative for exothermic reactions and positive for endothermic reactions.
At constant volume, U = qV, therefore, Eqn.(iii) becomes
H = U = qV
The difference between H and U is not significant for systems consisting of only solids and/or liquids
because solids and liquids do not undergo significant volume change on heating.
For a reaction between ideal gases, taking place at constant temperature, PV = nRT,
For gaseous reactants, PVA = nA RT ... (iv)
For gaseous products, PVB = nB RT ... (v)
where VA and VB are the volumes and nA and nB are the number of moles of reactants and products.
Subtracting Eqn. (iv) form Eqn. (v),
P(VB – VA) = (nB – nA) RT  PV  n g RT ... (vi)
ng is the change in number of moles of gaseous products and gaseous reactants. Substituting in Eqn. (iii),
H  U  n g RT ... (vii)
Physical significance of enthalpy. Enthalpy or heat content of a system is the energy stored within the
substance of a system that is available for conversion into heat.

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Conditions under which H = U. (i) Reactions which do not involve any gaseous component. (ii) Reactions
carried out in closed vessels so that volume does not change, i.e., V = 0. (iii) Reactions which involve
gaseous components, but without change in the number of moles.
Conditions under which H  U. For reactions having ng  0H and U are different, i.e., reactions in
which there is increase or decrease in the number of moles.
6.4.1 Exothermic and Endothermic Reactions in terms of H
For exothermic reactions, H is negative: C(s)  O 2 (g) 
 CO 2 (g) ; H = -393.5 kJ mol-1

For endothermic reaction, H is positive : N2 (s)  O2 (g) 

 2NO(g) ; H = +180.7 kJ mol-1
6.4.2 Heat Capacity
Heat absorbed by a system appears as rise in temperature of the system. Heat capacity is the quantity of heat
required to raise the temperature of a system/substance through 1ºC.
q
q  C T or C 
T
where C is the heat capacity and T is the rise in temperature.
Heat capacity depends on the size, composition, and nature of the system. The heat supplied can be measured
by monitoring the temperature rise, provided we know the heat capacity. Since heat capacity varies with
temperature, its value should be considered for a very small range of temperature,
q
C
dT

where q is is a small quantity of heat absorbed and dT is the rise in temperature.

i. Specific heat (c) or specific heat capacity. It is the quantity of heat required to raise the temperature of
unit mass of a substance through one kelvin. The heat, q, required to raise the temperature of a sample is
q  c  m  T = C ×ΔT
where c is the specific heat capacity, and m is the mass of the substance.
Units of specific heat capacity : J g–1 K–1 or J g–1 °C–1.
ii. Molar heat capacity. It is the quantity of heat required to raise the temperature of one mole of the
substance by one degree celsius (or one kelvin).
Cm  C / n
iii. Molar heat capacity at constant pressure (Cp) and at constant volume (C)
Heat, q absorbed at constant volume, q v  C v T  U or C v  U / T

and at constant pressure, q p  C p  T   H or C p   H /  T .

The difference between Cp and CV for 1 mole of an ideal gas is given by
H  U  (PV)  U   (RT) or H  U  RT
On putting the values of andU in the above equation
C p T  C v T  R  T

C p  C v  R or C p  C v  R
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For n moles of gas, Cp  Cv  nR

This relationship, known as Meyer’s relationship, indicates that Cp is greater than Cv by the gas constant R
(8.314 JK–1 mol–1).
iv. Ratio between molar heat capacities. The ratio between molar heat capacities, Cp/Cv provides useful
information about the atomicity of the gas.
Cp/Cv = 
Table - 01 - values for monoatomic, diatomic and triatomic gases

Atomicity of gas Cv Cp (=Cv+R)  = Cp /Cv

Monoatomic (He, Ne, Ar) (3/2)R (5/2)R 1.66
Diatomic( H2, O2, CO) (5/2)R (7/2)R 1.4
Triatomic (CO2, H2S) (7/2)R (9/2)R 1.3

6.4.3 Measurement of UandH

Calorimetry. Energy changes associated with chemical/physical processes can be measured using calorimetry.
The process is carried out in a vessel called calorimeter. The calorimeter is kept immersed in a known volume
of a liquid. If the heat capacity of the liquid in which the calorimeter is immersed and the heat capacity of the
calorimeter are known, it is possible to determine the heat evolved in the process by measuring temperature
change. Measurements can be made at (i) constant volume, qV or (ii) at constant pressure, qP.
(a) Determination of Internal Energy Change U). The internal energy change for a reaction at constant
volume can be determined using Bomb calorimeter (Fig. 04a).

(a) (b)

Figure - 04 (a) Bomb calorimeter (b) Calorimeter for measuring heat changes at constant pressure.
A bomb calorimeter consists of a very strong vessel made of heavy steel which could withstand high pressure.
It is kept immersed in an insulated water bath provided with a stirrer and a sensitive thermometer. A small
quantity of the substance is taken in the platinum cup. The vessel is filled with excess of oxygen at a pressure
of 20 to 25 atmospheres and sealed. Combustion is initiated by passing electric current through the firing leads.
The heat evolved during the chemical reaction raises the temperature of water. From the heat capacity (C) of
the calorimeter system (including water), and the rise in temperature (T), change in internal energy per mole
(U) can be calculated.
Heat produced (q) by m g of substance at constant volume = Heat gained by the calorimeter system.
As the volume remains constant, work done is zero, therefore, heat produced by the combustion of one mole
of the substance at constant volume is equal to the change in internal energy per mole.
U  (C  T  M) / m
where M is the molar mass of the substance.

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(b) Measurement of Enthalpy. Measurement of heat change at constant /atmospheric pressure is done in a
calorimeter (coffee-cup calorimeter) as shown in Fig.4(b). The heat absorbed or evolved, qp at constant
pressure is called heat of reaction or enthalpy of reaction.
H = qp
In exothermic reaction, heat is evolved (system loses heat to the surroundings), therefore, qp as well as rH will
be negative. In endothermic reaction, heat is absorbed, therefore, qp and H will be positive.
6.4.4 ENTHALPY CHANGE, rH OF A REACTION - REACTION ENTHALPY
The enthalpy change accompanying a chemical reaction is called heat of reaction or reaction enthalpy,
rH. If the reaction takes place at constant temperature and pressure, H gives the heat of reaction.
rH = (sum of enthalpies of products) - (sum of enthalpies of reactants)
 r H   a i H products   bi H reac tan ts
i i

where stands for summation, ai and bi are the stoichiometric coefficients of products and reactants in the
balanced chemical equation. For the reaction
CH 4 (g) + 2O 2 (g) 
 CO 2 (g) + 2H 2 O (l )
rH = [Hm (CO2 ,g) + 2Hm (H2O, l)]  [Hm (CH4 , g) + 2Hm (O2, g)]
where Hm is the molar enthalpy.
Enthalpy change is useful for planning heating/cooling needed to maintain industrial chemical reactions at constant
temperature. It is also required to calculate temperature dependence of equilibrium constant.
6.4.5 Standard enthalpy of reactions ( ΔHo )
Standard enthalpy of reaction is the enthalpy change for a reaction when all participating substances are in their
standard states. The standard state of a substance at a particular temperature is its pure form at 1 bar pressure.
For example, the standard state of liquid water at 298 K is pure liquid water at 1 bar.
Standard enthalpy conditions are denoted by adding the superscript ‘°’ to H, i.e., H°.
6.4.6 Enthalpy changes during phase transformations
1. Standard Enthalpy of Fusion (fusHo). The enthalpy change accompanying the melting of one mole of a
solid substance in its standard state is called standard enthalpy of fusion or molar enthalpy of fusion,fusHo.
H 2 O  s  
 H 2 O  l  ;  fus H   6.0 kJ mol1
Melting of a solid is endothermic, so all enthalpies of fusion are positive.
Freezing is accompanied by decrease in enthalpy. The value for the freezing of one mole of water vapour at
298K is -6.0 kJ.
2. Standard Enthalpy of Vaporisation (vapH°). The heat required to vaporize one mole of a liquid at
constant temperature and under standard pressure (1bar) is called its standard enthalpy of vaporization or
molar enthalpy of vaporization. Evaporation of a liquid is accompanied by increase in enthalpy.
H2O  l  
 H 2O  g  ; vap H   40.79 kJ

H2O  g 
 H2O  l  ; cond H   40.79 kJ
The value for the condensation of one mole of water vapour at 25oC is also 40.79 kJ.

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Table - 02 Standard enthalpy changes of fusion and vaporisation

-1 -1
Substance Tf /K  fus H°/kJ mol Tb /K  vap H°/kJ mol
N2 63.15 0.72 77.35 5.59
NH3 195.40 5.65 239.73 23.35
HCl 159.00 1.99 188.00 16.15
CO 68.00 6.84 82.00 6.04
CH3 COCH3 177.80 5.72 329.40 29.10
CCl4 250.16 2.50 349.69 30.00
H2 O 273.15 6.01 373.15 40.79
NaCl 1081.00 28.80 1665.00 170.00
C6 H6 278.65 9.83 353.25 30.80
3. Standard Enthalpy of Sublimation (subH°). It is the change in enthalpy when one mole of a solid
substance sublimes at constant temperature under standard pressure (1bar).
I 2 (s)   I 2 (g); H  62.3 kJ
The enthalpy of sublimation of solid CO2 (dry ice) at 195K is 25.2 kJ and of naphthalene at 353K is 73.0 kJ.
Enthalpy of sublimation is equal to the sum of the enthalpy of fusion and enthalpy of vaporisation.
 sub H   fus H   vap H
The magnitude of enthalpy change depends on the strength of intermolecular forces.
4. Standard Enthalpy of Formation, fH°. The standard enthalpy change for the formation of one mole of
a compound from its elements in their most stable states of aggregation (reference states) is called standard
molar enthalpy of formation. The reference state of an element is its most stable state of aggregation at
298K and 1 bar pressure. For example, the reference states of dihydrogen, dioxygen, carbon and sulphur are
H2gas, O2gas, Cgraphite and Srhombic respectively.
H 2 g   1 2 O 2 g   
 H 2 O  l  ;  f H o  285.8 kJ mol 1
C  graphite,s   2H 2  g  
 CH 4 (g) ;  f H o  74.81 kJ mol 1
2C  graphite,s   3H2  g  +1 2O2 (g) 
 C2 H5OH(l ) ;  f H o  277.7 kJ mol 1
Enthalpy of formation and enthalpy of reaction.(i) Standard molar enthalpy of formation is only a special
case of r H, where one mole of a compound is formed from its constituent elements.
(ii) The enthalpy change for the formation of a compound from other compound(s) cannot be termed enthalpy
of formation. Thus, the enthalpy change for the reaction
CaO  s   CO2  g  
 CaCO3  s  ;  r H o  178.3 kJ mol 1
is not enthalpy of formation of CaCO3, since it has been formed from other compounds (CaO and CO2), and
not from its constituent elements.
(iii) The enthalpy change for the following reaction is not standard enthalpy of formation, since two moles,
instead of one mole, of product is formed from the elements
H 2  g   Cl 2  l   
 2HCl  g  ;  r H o  184.62 kJ mol 1
Here,  r H° = 2f H° .
The expression for enthalpy of formation of HCl(g) is obtained by dividing all coefficients in the balanced
equation by 2. Thus, the expression for enthalpy of formation of HCl(g) can be written as
1 2 H 2  g   1 2Cl2  l  
 HCl  g  ;  r H o  92.31kJ mol 1
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By convention, standard enthalpy of formation,  f H, of an element in reference state (most stable
state of aggregation) is taken as zero.
Table - 03 Standard molar enthalpies of formation, H of , of some substances

The general equation for calculating the enthalpy change of a reaction can be written as
 r H   a i  f H(products)   bi  f H(reac tan ts)
i i

where ‘a’ and ‘b’ are the coefficients of the products and the reactants in the balanced equation.
The heat required to decompose CaCO3 to CaO and CO2,
 r H    f H [CaO(s)]   f H [CO 2 (g)]   f H [CaCO 3 (s)]
=1(-635.1 kJ mol-1) + 1(-393.5 kJ mol-1) -1(-1206.9 kJ mol-1) = 178.31 kJ mol-1
Thus, decomposition of CaCO3(s) is an endothermic process, therefore, it must be heated to get the products.
6.5 THERMOCHEMICAL EQUATIONS
A thermochemical equation is a chemical equation which also indicates the quantity of heat evolved or absorbed
during the chemical reaction. Thermochemical equations can be written in two ways:
i) Heat effect can be written as one of the terms along with the products.
2SO2  g   O 2  g   2SO3  g   694.0 kJ
ii) Heat evolved or absorbed can be expressed in terms of H.
2SO2  g   O2  g   2SO3  g  ; H  694.0 kJ
Physical state of reactants and products must be mentioned in thermochemical equations because change of
state is also accompained by enthalpy changes.
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Reactants and products must have the same temperature and pressure. E.g.,H for the formation of liquid
water and water vapour from the elements are different.
H 2  g   1 2 O2  g   H2O  l  ;  r H  286.0 kJ

H 2  g   1 2O2  g   H 2O  g  ;  r H  249.0 kJ
The difference in H for these reactions is the amount of energy released when 1 mole of water at 25C
changes to 1 mole of liquid water.
6.5.1 Important Features of Thermochemical Equations
i) The coefficients of various substances in the equation represents the number of their respective moles.
ii) The value of H corresponds to enthalpy change with a specified number of values of various reactants
and products involved in the reaction.
iii) Thermochemical equations, like algebraic equations, can be added, subtracted, multiplied or divided by
any numerical factor.
Heat of reaction can be calculated from standard enthalpy of formation.
The reaction of 1 mole of Fe2O3 with H2 can be writen as
Fe2O3 (s)  3H 2 (g) 
 2Fe(s)  3H 2O(l )

From Table (03),  f H(H 2O, l )  285.83 kJ mol 1 and  f H (H 2 , g)  0 .

Also  f H(Fe2O3 ,s)  824.2 kJ mol 1 and  f H (Fe,s)  0 .

Thus,  r H   (sum of a i  f H  of products  b i  f H  of reactants)

rH°1= 3(-285.83 kJ mol-1) - (-824.2 kJ mol-1) = -33.3 kJ mol-1
The equation may be written for 1 mole of Fe as
1 2 Fe 2O3 (s)  3 / 2H 2 (g) 
 Fe(s)  3 / 2H 2O(l )

rH°2= 3/2(–285.83 kJ mol-1) - 1/2(-824.2 kJ mol-1) = -16.6 kJ mol-1 = 1 2 Δ r Hf

This shows that enthalpy is an extensive property.
iv) When a chemical equation is reversed, the magnitude of H remains the same, but its sign is reversed.
 CO 2 (g);  r H   393.5 kJ mol1
C(s)  O 2 (g) 

 C(s)  O 2 (g);  r H  393.5 kJ mol1

CO 2 (g) 
6.5.2 HESS’S LAW OF CONSTANT HEAT SUMMATION
Hess’s law states that the quantity of heat evolved or absorbed in a process is the same, whether the process
takes place in one step or in several steps (G. H. Hess, Russian chemist, 1840).
Suppose in a process, the system changes from state A to state B in one step and the enthalpy change q = rH.
Now suppose the system changes from state A to state B in three steps involving a change from A to C, C to
D and finally from D to B.
If q1= rH1, q2=rH2 and q3=rH3 are the heats exchanged in the first, second and third steps respectively,
then according to Hess’s law
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q  q1  q 2  q3 or  r H   r H1   r H 2   r H 3

Figure - 05 Hess’s law of constant heat summation

This means that the enthalpy of a reaction depends only on the reactants and products and not on intermediate
products that may be formed. It is also independent of the path/manner in which the change is brought about.
According to Hess’s Law, thermochemical equations can be treated as algebraic equations which can be
added, subtracted, multiplied or divided.
6.5.3 Applications of Hess’s Law
1. Calculation of enthalpies of reactions. Hess’s law makes it possible to calculate enthalpies of reactions
which cannot be determined experimentally. For example, heat change when carbon burns to form CO cannot
be determined accurately since some CO2 is always formed.
Enthalpy of formation of 1 mole of CO2 is the same (-393.5 kJ mol-1), whether the reaction takes place in a
single step (i) or in two steps (ii, iii)
C  graphite,s   O2  g  
 CO2  g  ;  r H  393.5 kJ mol 1 ... (i)
C  graphite,s   1 2O2  g  
 CO  g  ;  r H  x kJ mol1 ... (ii)
CO  g   1 2 O 2  g  
 CO 2  g  ; H  y kJ mol 1 ... (iii)
Although x, the enthalpy of combustion of carbon to CO cannot be determined easily, y, the enthalpy of
combustion of CO to CO2 has been found to be –283.0 kJ mol-1.
According to Hess’s law, x + y = –393.5 kJ mol-1
or x = –393.5 – y = –393.5 – (–283.0) = –110.5 kJ mol-1
Thus, H for the combustion of carbon to carbon monoxide is – 110.5 kJ mol-1.
2. Calculation of enthalpies of formation. Enthalpy of formation of benzene from the elements can be
calculated from the enthalpy of combustion of benzene and the enthalpies of formation of H2O and CO2.
Step - 1. The thermochemical equations for the known values are written as
15
C6 H 6 (l )  O 2  g  
 6CO 2  g   3H 2 O  l  ; H  3267.7 kJ ... (i)
2
C  s   O 2  g  
 CO 2  g  ; H  393.5kJ ... (ii)
1
H 2  g   O2  g    H 2O  l  ; H  285.9 kJ ... (iii)
2
Step - 2. Multiplying Eqn. (ii) by 6 and Eqn. (iii) by 3
6C  s   6O 2  g  
 6CO 2  g  ; H  2361.0 kJ ... (iv)
3
3H 2  g   O 2  g  
 3H 2O  l  ; H  857.7 kJ ... (v)
2
Adding (iv) and (v) and substracting (i), we get
6C  s   3H 2  g  
 C6 H 6  l  ; H  49.0 kJ
Thus, the enthalpy of formation of benzene is + 49.0 kJ.
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6.5.4 ENTHALPIES FOR DIFFERENT TYPES OF REACTIONS

1. Enthalpy of Combustion (cH ). The enthalpy of combustion of a substance at a given temperature
under atomospheric pressure is defined as the enthalpy change accompying the complete combustion of one
mole of the substance at that temperature.
If the temperature is 25oC, the enthalpy change is called the standard enthalpy of combustion, (cH ).
CH 4  g   2O2  g  
 CO2  g   2H 2O  g  ;  c H  890.3 kJ
C6H12O6  s  6O2  g 
6CO2  g  6H2O l  ;  c H  2802.0 kJ
Enthalpy of combustion depends on the physical state of the substances involved. For example, enthalpy of
combustion of hydrogen to form liquid water at 25oC is – 285.8 kJ but if combustion takes place at 100oC or
above so that water vapour cannot condense, the enthalpy of combustion will be –245.5 kJ.
2. Enthalpy of Atomisation (aH ). It is defined as the enthalpy change when one mole of bonds are
broken to obtain atoms in the gaseous phase.
CH 4 (g)   C(g)  4H(g); aH = 1665.0 kJ mol–1
In case of diatomic molecules like H2, enthalpy of atomisation is also the bond dissociation enthalpy.
H 2 (g)  2H(g); aH = 435.0 kJ mol–1
Enthalpy of atomisation is the same as enthalpy of sublimation.
Na(s) 
 Na(g); aH = 108.4 kJ mol–1
3. Bond Enthalpy (bond H ). Bond enthalpy is defined as the average energy required to dissociate one
mole of bonds of that type present in the compound. Two different terms used with reference to enthalpy
changes associated with chemical bonds are (i) Bond dissociation enthalpy and (ii) Mean bond enthalpy.
i) Diatomic Molecules: The process in which the bonds in one mole of dihydrogen gas (H2) are broken is the
bond dissociation enthalpy of H  H bond. It is the same as the enthalpy of atomisation of dihydrogen.
H 2 (g)   2H(g); H–H H = 435.0 kJ mol–1
This is true in case of all diatomic molecules :
Cl2 (g) 
 2Cl(g); Cl–Cl H = 242.0 kJ mol–1
O 2 (g) 
 2O(g); O = O H = 428.0 kJ mol–1
 2N(g);  N  N H = 937.4 kJ mol–1
N 2 (g) 
ii) Polyatomic Molecules: In case of a polyatomic molecules like methane, all the four C  H bonds are
identical in bond length and energy. The overall thermochemical equation for its atomisation is:
CH 4  g  
 C(g)  4H  g  ;  a H = 1665.0 kJ mol–1
However, the energy required to break the individual C  H bonds in the successive steps differ:
CH 4  g  
 CH3 (g)  H  g  ;  bond H = 427.0 kJ mol–1

CH3  g  
 CH 2 (g)  H  g  ;  bond H = 439.0 kJ mol–1

CH 2  g  
 CH(g)  H  g  ;  bond H = 452.0 kJ mol–1

CH  g  
 C(g)  H  g  ;  bond H = 347.0 kJ mol–1

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In such cases, the mean bond enthalpy is considered. Thus, C  H bond enthalpy in CH4,
 C – H H =   a H  / 4  (1665.0 kJ mol1 ) / 4 = 416.0 kJ mol-1
Mean bond enthalpies differ slightly from compound to compound.
The standard enthalpy of reaction,  r H  is related to bond enthalpies of the reactants and products in gas
phase reactions as:
 r H   bond enthalpiesreac tan ts   bond enthalpies products
Thus, the net enthalpy change of a reaction is the amount of energy required to break all the bonds in the
reactant molecules minus the amount of energy required to break all the bonds in the product molecules when
all reactants and products are in gaseous state.
Isomeric compounds have the same enthalpy of formation. In any homologous series, increase in enthalpy of
formation for each CH2 group is almost constant. This shows that enthalpy of formation of a bond of a
particular type is an additive property.
Table - 04 Some mean single bond enthalpies in kJ mol-1 at 298 K

Table - 05 Some mean multiple bond enthalpies in kJ mol-1 at 298 K

4. Enthalpy of Solution (  sol H ). It is the enthalpy change when 1 mole of solute is dissolved in a large
excess of solvent so that further addition of solvent does not produce any heat change.
The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving a substance in
infinite amount of solvent so that interactions between the ions/solute molecules are negligible.

The enthalpy of solution of an ionic substance AB(s), in water is the sum of the lattice enthalpy, and enthalpy of
hydration of ions.
 sol H   lattice H   hyd H
For most ionic compounds, enthalpy of solution is positive and the dissociation is endothermic. Therefore, the
solubility of those salts in water increases with temperature. If the lattice enthalpy is very high, the compound
may not dissolve at all. Thus, fluorides are less soluble than the corresponding chlorides.
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6.5.5. Lattice Enthalpy (  lattice H ). The lattice enthalpy of an ionic compound is the enthalpy change when one
mole of ionic compound dissociates into ions in gaseous state.
 Na  (g)  Cl  (g);  lattice H = + 788 kJ mol–1
Na  Cl (s) 
Born-Haber Cycle. It is impossible to determine lattice enthalpies directly, therefore, an indirect method
where we construct an enthalpy diagram called a Born-Haber Cycle (Max Born and Fritz Haber, 1919) is
employed. The sequence of steps shown in Fig. 07 is known as Born-Haber cycle. The sum of the enthalpy
changes round the cycle is zero.

Figure - 06 Enthalpy diagram for lattice enthalpy of NaCl

The lattice enthalpy of NaCl(s) can be calculated as follows:
 NaCl(s);  f H = – 411.2 kJ mol–1
Heat of formation of NaCl(s), Na  (g)  Cl (g) 
i) Sublimation of sodium metal
Na(s)   Na(g);  sub H = 108.4 kJ mol–1
ii) Ionisation of sodium atoms (ionisation enthalpy)
Na(g)   Na  (g)  e 1 (g);  i H  496 kJ mol 1
iii) Dissociation of chlorine; reaction enthalpy is half the bond dissociation enthalpy
 Cl(g);    bond H
1/ 2Cl 2 (g)   121 kJ mol 1
iv) Electron gained by chlorine atoms (electron gain enthalpy or electron affinity)
Cl(g)  e1 (g)  Cl  (g);  eg H  348.6 kJ mol1
v) Condensation of gaseous ions to form solid NaCl
 Na  Cl  (s) ;  lattice H
Na  (g)  Cl  (g) 
The summation of the above five reactions and their H values allows the calculation of (  lattice H ). According
to Hess’s law,  f H =  sub H +    bond H +  i H +  eg H +  lattice H
 –411.2 = 108.4 + 121 + 496 – 348.6 +  lattice H
Lattice enthalpy of NaCl, (  lattice H ) = 411.2 + 108.4 + 121 + 496 – 348.6 = + 788 kJ
 
For Na Cl (s)  Na  (g)  Cl  (g), internal energy is smaller by 2RT (since ng = 2) which is equal to
+ 783 kJ mol–1.
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Calculation of enthalpy of solution using Born-Haber Cycle. Enthalpy of solution of NaCl may be calculated
using lattice enthalpy of NaCl (+ 788 kJ mol–1) and enthalpy of hydrationfor NaCl (–784 kJ mol–1).
 sol H   lattice H   hyd H
 sol H = +788 kJ mol–1 + (–784 kJ mol-1) = + 4 kJ mol–1
Thus, dissolution of NaCl(s) is accompanied by very little heat change.
6.5.6 Limitations of the First Law of Thermodynamics
1. The First law establishes the relationship between heat absorbed and work done by a system in a process,
but it puts no restriction on the direction of flow of heat. It is known that transfer of heat from a lower to
a higher temperature is not possible without expense of energy (doing some external work). On the other
hand, heat flows spontaneously from a higher to a lower temperature.
2. According to the First law, the energy of an isolated system remains constant during a change of state, but
it does not tell whether a change/ process can occur spontaneously, i.e., whether it is feasible.
3. The First law states that energy of one form can be converted into an equivalent amount of energy of
another form, but it does not tell that heat energy cannot be completely converted into work.
6.6 SPONTANEITY
Spontaneous or Feasible or Probable Process. A process which has a natural tendency to occur with or
without initiation under a given set of conditions is known as spontaneous process. A spontaneous process is
an irreversible process and may only be reversed by some external agency.
i) Vaporisation of water : H 2O  l  
 H2O  g 
ii) Dissolution of sugar in water : Sugar  Water 
 Sugar solution
iii) Hydrogen reacts with oxygen to form to water. The reaction is initiated by passing an electric discharge.
2H 2  g   O 2  g   2H 2O  l 
Electric discharge

Spontaneous processes do not reverse themselves. For example, water does not run uphill.
A process which has no natural tendency to occur is said to be non-spontaneous.
6.6.1 Criteria of Spontaneity
1. Decrease in Enthalpy as the Criterion for Spontaneity. In order to attain stability, every system tries to
have minimum energy, e.g., water flows down hill, stone at a height falls down, a wound spring unwinds
(decrease potential energy), heat flows from a hot body to a cold body, etc.
C  g   O2  g  
 CO2  g  ; H  394 kJ

H 2  g   1 / 2 O 2  g  
 H 2 O  g  ; H  286 kJ
Thus, processes which are accompanied by a decrease in energy content are spontaneous.
A negative value of ΔH is not the only criterion for spontaneity. Although H values are negative for
most spontaneous processes, it is not the only criterion of spontaneity.
a) Spontaneous occurrence of endothermic reactions.
E.g., decomposition of CaCO3 on heating : CaCO3  g  

 CaO  s   CO 2  g  ; H  177.8 kJ

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(a) (b)

Figure - 07 Enthalpy diagram for (a) exothermic reactions (b) endothermic reactions
b) Spontaneous occurrence of physical processes. Many physical processes are spontaneous despite the
fact that they are accompanied by increase in energy.
i) Evaporation of water : H 2 O  l  
 H 2 O  g  ; H  40.6 kJ
ii) Dissolution of salts like NH4Cl in water : NH 4 Cl  s   aq 
 NH 4  aq   Cl (aq); H  15 kJ
c) Reactions do not go to completion. Even those reactions for which H is negative rarely go to completion.
d) Reversible reactions also occur. In reversible reactions, if the forward reaction is exothermic, then the
backward reaction will be endothermic. If negative value of H were the only critierion of spontaneity, then
the backward reaction for which H is positive should not have occurred.

2HI  g    H 2  g   I2  g 
Endothermic
Exothermic

The above facts indicate that a negative value of H is not the only critertion of spontaneity..
2. Tendency to have Maximum Randomness – Another Criterion of Spontaneity
The spontaneity of a process is also related to increase in disorder or randomness of a system.
Mixing of gases which do not react chemically. When bromine vapours (reddish brown) and air (colourless)
enclosed in two separate bulbs are connected to each other through a tube, the two gases mix completely
within a very short time.
During mixing, there is negligible energy change, therefore, the criterion of decrease of energy for
feasibility of the process is not valid.
After mixing, each molecule has a larger volume for movement which means that there is more disorder or
randomness on mixing. It can be inferred that gases mix to acquire greater randomness or disorder. As diffusion
is a spontaneous process, it is irreversible. Thus, the process proceeds spontaneously in a direction in which
the randomness of the system increases.
The arrangement of constituent particles in a solid becomes more disorderly and more random as the solid
changes into liquid. Randomness increases further when the liquid changes into gas. Thus, entropy is maximum
in the gaseous state and minimum in the solid state.
i) Dissolution of ammonium chloride : NH 4Cl  s   aq 
 NH 4  aq   Cl   aq  ; H  15 kJ
Ions which were held together in the crystal lattice become free and moving in all possible directions.
ii) Decomposition of N2O4 to yield 2 molecules also creates a more random or disordered situation.
N 2 O4  g  
 2NO 2  g  ; H  positive
iii) Dissociation of hydroiodic acid : 2HI  g  
 H 2  g   I 2  g  ; H  9.00 kJ
Dissociation of HI gives two different kinds of molecules, viz., H2 and I2. The mixing of two different kinds of
gaseous molecules is a more random situation as compared to same type of molecules in gaseous HI.
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iv) Melting of ice into water is a more random situation : H 2O  s  

 H 2 O  l  ; H  6 kJ
v) Evaporation of water to water vapour : H 2O  l  
 H 2 O  g  ; H  40.64 kJ
It is clear that processes where H is zero or positive are also spontaneous, provided there is a tendency for
maximum randomness or more disorder, i.e., increase of entropy. Thus, spontaneity of a process depends
on as well as tendency for maximum randomness.
6.6.2 Overall Tendency as the Driving Force for a Process. The overall tendency for a process to occur is
the resultant of (i) tendency for minimum energy and (ii) tendency for maximum randomness which
may be called the driving force for a process.
6.6.3 ENTROPY, S
Increase in disorder or randomness is expressed in terms of a quantity called entropy. It is denoted by S.
Greater the disorder of the system, greater is its entropy. It is denoted by the symbol S.
Characteristics of entropy. (i) It is the thermodynamic measure of randomness.
(ii) Greater the randomness of a system, higher is its entropy.
(iii) Every substance has entropy as one of its characteristic properties.
(iv) The entropy of a system is equal to the sum of the entropies of its components.
(v) Unlike enthalpy and free energy, absolute entropy can be calculated.
(vi) For solids, entropy is minimum and for gases, it is maximum.
(vii) It is a state function and an extensive property.
A system at higher temperature has greater randomness than one at lower temperature. Heat added to a
system at lower temperature causes greater randomness than at higher temperature. Thus, entropy change is
inversely proportional to the temperature.
Entropy Change. The entropy change of a system is given by
S  Sfinalstate  Sinitialstate or S   Sproducts   Sreactants
Entropy change is inversely proportional to the temperature. For a reversible process, change in entropy,
q rev.
S 
T
where qrev. is the quantity of heat supplied reversibly and isothermally to the system at T K.
The total entropy change of the system and surroundings (universe) for a spontaneous process
Suniverse or Stotal  Ssystem  Ssurr  0
Entropy for a spontaneous process increases till it becomes maximum and at equilibrium the change in entropy
becomes zero (S . Change in entropy for a reversible process is given by
q sys,rev
Ssys 
T
For a small differential change in entropy
dq
dS 
T
where dq is the infinitesimal quantity of heat absorbed under reversible conditions at temperature, T.
Both for reversible and irreversible expansion of ideal gas under isothermal conditions,U = 0, but Stotal is
not zero for irreversible process. Thus, U will be same for reversible and irreversible process, but not S.

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Units of Entropy : Since entropy is expressed as S = qrev/T = Joules/K = Joules per Kelvin (J K–1). This is
known as entropy unit (e.u.). Since entropy is an extensive property, its value depends on the amount of
substance. Therefore, the unit of entropy is JK–1 mol–1.
6.6.4 Factors affecting Entropy
(i) Volume changes. For gases, entropy increases with increase in volume. A gas expands spontaneously to
achieve a more probable (higher entropy) particle distribution.
(ii) Temperature changes. The higher the temperature, the larger is the entropy. Due to increase in kinetic
energy, particles move more randomly.
(iii) Physical state. Formation of gases from liquids and solids are almost always accompanied by increase
in entropy.
(iv) Number of particles. Entropy increases with increase in the number of particles, e.g., decomposition of
substances.
6.6.5 Entropy Change Accompanying Change of Phase
a) From solid phase to liquid phase. Entropy of fusion is defined as the change in entropy when 1 mole of
a substance changes from solid state to liquid state at its melting point at constant pressure.
 fus H
Entropy of fusion,  fusS 
Tf
b) From liquid phase to vapour phase. Entropy of vaporisation is the change in entropy when 1 mole of a
substance changes from liquid state to vapour state at its boiling point. When one mole of a substance changes
from liquid state to vapour state, reversibly at its boiling point Tb, under constant pressure.
 vap H
Entropy of vaporisation,  vapS 
Tb
c) From solid phase to vapour phase. Entropy of sublimation is defined as the change in entropy when 1
mole of a substance changes from solid state to vapour state at a particular temperature.
 sub H
Entropy of sublimation,  subS 
T
d) From one crystalline form to another. The change in entropy when one mole of a solid substance
undergoes change of state from one crystalline form (e.g., rhombic form) to another (e.g., monoclinic form) at
the transition temperature, Ttrans.
 trans H
Entropy of transition,  transS  T
trans

6.6.6 Entropy Changes in Chemical Reactions. From the values of standard absolute entropies of reactants and
products in a chemical reaction, the standard entropy change ( S ) for reaction can be calculated.
Sum of the standard absolute  Sum of the standard absolute 
S     entropies of the reactants. 
entropies of products.   
 S products  S reactants
For example, for the hypothetical reaction,
aA  bB 
 lL  mM
S  l S  L   m S  M     a S  A   b S  B  
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6.6.7 Entropy Change in Reversible and Irreversible Processes. In a reversible process, the entropy of the
system and the surroundings taken together remains constant while in an irreversible process, the entropy of
the system and the surroundings taken together increases.
For a reversible isothermal process :  Ssys  Ssur   0

For an irreversible isothermal process :  Ssys  Ssur   0

Combining the two, we have,  Ssys  Ssur   0

6.6.8 Physical Significance of Entropy
1. Entropy as a measure of the disorder of the system. Increase in entropy implies increase in disorder.
2. Entropy as unavailable energy. Entropy is unavailable energy of the system.
Entropy = Unavailable energy/Temperature in Kelvin
6.7 THE SECOND LAW OF THERMODYNAMICS
The first law of thermodynamics rules out all changes in which the energy of the universe is not conserved.
However, it does not require that all transformations which conserve energy must take place. Several processes
which are consistent with the first law of thermodynamics do not occur.
Statement of the Second Law of Thermodynamics. The second law can be stated in many ways:
All spontaneous processes are thermodynamically irreversible.
Without the help of an external agency, a spontaneous process cannot be reversed, e.g., heat cannot by
itself flow from a colder to a hotter body.
The complete conversion of heat to work is impossible without leaving some effect elsewhere.
Nature tends to pass from a less probable to a more probable condition (Ludwig Boltzmann,1866).
The entropy of the universe always increases in the course of every spontaneous or natural change.
Thus, for a spontaneous change in an isolated system, the entropy must increase ( S > 0).
6.7.1 GIBBS ENERGY AND SPONTANEITY
The spontaneity of a process is determined by the tendency to have minimum energy and maximum randomness.
The resultant of the two tendencies is expressed in terms of a thermodynamic function called Gibbs energy or
Gibbs function, G. It is defined as the amount of energy available from a system that can be put to useful work.
G = H – TS
where H is the enthalpy of the system, S is the entropy of the system, and T is the temperature on kelvin scale.
For a process carried out at constant pressure and constant temperature,
G1 = H1 – TS (initial state) ...(i)
G2 = H2 – TS2 (final state) ...(ii)
Subtracting Eqn.(i) from Eqn.(ii),
G2 – G1 = (H2 –H1) – T (S2 – S1)
or G  H  TS
This equation is known as Gibb’s - Helmholtz equation or Gibbs energy equation.
G  G 2  G1 (Change in free energy of a system)
Since H and S are extensive properties, G also is an extensive property and a state function.
Units of G: Units of G is J.
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6.7.2 Relation Between Gibbs Energy and Reaction Spontaneity

Stotal  Ssystem  Ssurr
If the system is in thermal equilibrium with the surroundings, then the temperature of the surroundings is same
as that of the system. Increase in enthalpy of the surroundings is equal to decrease in the enthalpy of the
system. Therefore, entropy change of surroundings,
Hsurr H
Ssurr    system
T T
 H 
Stotal  Ssystem    system 
 T 
Rearranging, TStotal  TSsystem  H system
For a spontaneous process, Stotal  0 , therefore,
TSsystem  H system  0 or  (H system  TSsystem )  0
Since G  H  TS , the above equation can be written as
G  0 or G  H  TS  0
where Hsystem is the enthalpy change and TSsystem is the energy that is not available to do useful work.
G, the net energy available to do useful work is a measure of the free energy. Therefore, it is also called free
energy of the reaction.
6.7.3 Criteria for spontaneity
At constant pressure and temperature, a process is spontaneous if G < 0, and non-spontaneous if G > 0.
If enthalpy and entropy are positive, then the process may be spontaneous when TS is larger than H. This
can happen in two ways:
(i) The positive entropy change of the system is large and T is small.
(ii) The positive entropy change of the system is small and T is large. This is one of the reasons why reactions
are often carried out at high temperature.
Table - 06 Effect of temperature on spontaneity of reactions

r H  rS  rG Type of Reaction
– + – Spontaneous at all temperatures
– – – (at low T) Spontaneous at low temperatures
– – + (at high T) Nonspontaneous at high temperatures
+ + + (at low T) Nonspontaneous at low temperatures
+ + – (at high T) Spontaneous at high temperatures
+ – + (at all T) Nonspontaneous at all temperatures
Thus, an endothermic reaction which may be spontaneous at low temperature may become spontaneous at
high temperature, whereas an exothermic reaction which may be non-spontaneous at high temperature may
become spontaneous at low temperature.

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6.7.4 Gibbs Energy Change and Equilibrium

Free energy changes in reversible reactions. A reversible reaction is one which is carried out in such a
way that the system is in perfect equilibrium with its surroundings at all times. Such a reaction can proceed in
either direction, so that a dynamic equilibrium is set up. The criterion for such an equilibrium is  r G = 0
Gibbs energy for a reaction in which all reactants and products are in standard state,  r G = 0 is related to
the equilibrium constant of the reaction as
0 =  r G + RT ln K or  r G = – RT ln K
or  r G  2.303RT log K ...(i)
Also, Δ r G   r H  T r S   RT ln K ...(ii)
For strongly endothermic reactions,  r H is large and positive, therefore, value of K will be much smaller than
1 and the reaction is unlikely to give much product.
For exothermic reactions,  r H is large and negative, and  r G also is likely to be large and negative. The
value of K will be much larger than 1.
Strongly exothermic reactions have large values for K, and hence go to near completion.
Since  r G depends on  r S , the value of K will also depend on whether  r S is positive or negative.
6.8 THE THIRD LAW OF THERMODYNAMICS AND ABSOLUTE ENTROPY
The decrease in temperature of a substance at constant pressure decreases its disorder. As temperature
approaches 0 K, the disorder approaches zero and the substance tends to become perfectly ordered. Based
on these facts, Nernst (1906) formulated the third law of thermodynamics which states that,
‘At absolute zero, the entropy of a perfectly crystalline substance is zero.’
6.9 ZEROTH LAW OF THERMODYNAMICS
The zeroth law of thermodynamics is also called law of thermal equilibrium. If a body A is in thermal
equilibrium with body C and body B is also in thermal equilibrium with body C, then bodies A and B are in
thermal equilibrium with each other. In other words, two objects in thermal equilibrium with a third one are in
thermal equilibrium with each other.
The use of thermometer for comparing temperatures of two or more systems is based on this principle. The
thermometer may be likened to the body C in the above statement. While comparing temperatures of two
bodies, A and B, the thermometer is allowed to come into equilibrium, first with body A and then with body B.
The temperatures as read from the thermometer give a comparative idea of the degrees of hotness of the two
bodies A and B. Since thermometer is a very small body, there is no significant exchange or loss of heat during
the measurement of temperature.

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QUESTIONS
LEVEL - I
1. Laws of thermodynamics deals with:
1) microscopic system involving few molecules
2) macroscopic system involving large number of molecules
3) individual molecules in a system
4) fundamental subatomic particles forming molecules in a system
2. Which among the following, statement is false?
1) work is a state function
2) temperature is a state function
3) change in state is completely defined when the initial and final states are fixed
4) Work appears at the boundary of the system
3. Among the following, the state functions are:
(i) internal energy (ii) irriversible expansion work (iii) reversible expansion work (iv) molar enthalpy
1) (i) and (ii) 2) (ii) and (iv) 3) (i) and (iv) 4) (i), (ii) and (iv)
4. Which of the following statements is correct?
1) The presence of reacting species in a covered beaker is an example of open system
2) There is an exchange of energy as well as matter between the system and the surroundings in a closed
system
3) The presence of reactants in a closed vessel made up of copper is an example of a closed system
4) The presence of reactants in a thermoflask or any other closed insulated vessel is an example of a closed
system
5. Calculate the work done when 1 mol of an ideal gas is compressed reversibly from 1 bar to 4 bar at a constant
temperature of 300 K :
1) 3.45 kJ 2) –8.02 kJ 3) 18.02 kJ 4) –14.01 kJ
6. Temperature of 1 mole of a gas is increased by 1 at constant pressure. Work done is:
o

1) 2 cal 2) 3 cal 3) 4 cal 4) 5 cal

7. The net work done through a series of changes reported in figure for an ideal gas is :

1) –600 kJ 2) –700 kJ 3) –1200 kJ 4) +1200 kJ

8. If at 298 K the bond energies of C – H, C – C, C = C and H – H are respectively 414, 347, 615, and
435 kJ mol–1, the value of enthalpy change for the reaction,
H 2 C  CH 2  g   H 2  g   H3C  CH3  g  at 298 K will be,
1) +250 kJ 2) –250 kJ 3) +125 kJ 4) –125 kJ
9. The gas absorbs 100 J heat and is simultaneously compressed by a constant external pressure of 1.50 atm
from 8 L to 2 L in volume. Hence, U will be :
1) –812 J 2) 812 J 3) 1012 J 4) 912 J

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10. The heat liberated from the combustion of 0.5 g of carbon raised the temperature of 2000 g of water from
24oC to 26oC. The heat of the combustion of carbon is:
1) – 4 kcal 2) – 8 kcal 3) – 62 kcal 4) – 96 kcal
11. The specific heats of I2 in vapour and solid states are 0.031 and 0.055 cal/g respectively. The heat of sublimation
of iodine at 200oC is 6.096 kcal mol-1. The heat of sublimation of iodine at 250oC will be:
1) 3.8 kcal mol-1 2) 4.8 kcal mol-1 3) 2.28 kcal mol-1 4) 5.8 kcal mol-1
12. The enthalpy change for a reaction does not depend upon:
1) the physical states of reactants and products
2) use of different reactants for the same product
3) the nature of intermediate reaction steps
4) the differences in initial or final temperatures of involved substances
13. For the reaction of one mole zinc dust with one mole sulphuric acid in a bomb calorimeter, U and w
corresponds to:
1) U  0, w  0 2) U  0, w  0 3) U  0, w  0 4) U  0, w  0
14. Assuming that water vapour is an ideal gas, the internal energy  U  when 1 mol of water is vapourized at
1 bar pressure and 100oC. (Molar enthalpy of vapourization of water at 1 bar and 373 K = 41 kJ mol–1 and
R = 8.3 J mol–1 K–1) will be:
1) 4.1 kJ mol–1 2) 3.7904 kJ mol–1 3) 37.904 kJ mol–1 4) 41.0 kJ mol–1
15. The enthalpy change  H  for the reation, N 2  g   3H 2  g   2 N H 3  g  is -92.38 kJ at 298 K. The internal
energy change U at 298 K is:
1) -92.38 kJ 2) -87.42 kJ 3) -97.34 kJ 4) -89.9 kJ
16. Given the following thermo chemical equations:
2NO g   O 2 g   2NO 2 g  ; H o  113.2 kJ mol1 , 2N 2 O g   3O 2 g   4NO 2  g  ; H o  28.0 kJ mol 1
1
calculate H o for this reaction: N 2 O g   O 2 g   2NO g 
2
1) –99.2 kJ mol –1
2) +99.2 kJ mol–1 3) +101.2 kJ mol–1 4) –101.2 kJ mol–1

17. Which of the following equations correctly represents the standard heat of formation  H f  of methane?

1) C (diamond)  2H 2 
 CH 4  g  2) C (graphite)
 2H 2 g  
 CH 4l 

3) C (graphite)
 2H 2 g  
 CH 4 g  4) C (graphite)
 4H 
 CH 4 g 

18. Consider the reaction, N 2  g   3H 2  g  

 2NH3  g  ; carried out at constant temperature and pressure.
If H and U are enthalpy change and internal energy change respectively, which of the following expressions
is true?
1) H  0 2) H  U 3) H  U 4) H  U
19. Standard molar enthalpy of formation of CO2 is equal to :
1) Zero.
2) The standard molar enthalpy of combustion of gaseous carbon.
3) The sum of standard molar enthalpies of formation of CO and O2.
4) The standard molar enthalpy of combustion of carbon (graphite).
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20. The enthalpy of reaction for the reaction : 2H 2  g   O 2 g  

 2H 2 Ol  is  r H  572 kJ mol1
What will be the standard enthalpy for the formation of H2O(l)?
1) –572 kJ mol–1 2) +57 kJ mol–1 3) +286 kJ mol–1 4) –286 kJ mol–1
21. Diborane is a potential rocket fuel which undergoes combustion according to the reaction,
B2 H 6 g   3O2 g   B2O3s   3H 2 O g 
From the following data, calculate the enthalpy change for the combustion of diborane:
3 1
2B s   O 2 g   B2 O3 s  ; H  1273kJ mol1 ; H 2 g   O 2 g   H 2 O l  ; H  286 kJ mol 1
2 2
1
H 2 O l   H 2O g  ; H  44 kJ mol ;
2B s   3H 2 g   B2 H 6 g  ; H  36 kJ mol1
1) 1273 kJ mol–1 2) –1273 kJ mol–1 3) 2035 kJ mol–1 4) –2035 kJ mol–1
22. What is the value of internal energy change  U  at 27 o C of a gaseous reaction
2A 2  g   5B2  g  
 2A 2 B5  g  (whose heat change at constant pressure is –50700 J)?

 R  8.314 JK 1
mol1 
1) –50700 J 2) –63171 J 3) –38229 J 4) +38229 J
23. What would be the amount of heat released when an aqueous solution containing 0.5 mole of HNO3 is mixed
with 0.3 mole of OH– (enthalpy of neutralisation is –57.1 kJ)?
1) –28.5 kJ 2) –17.1 kJ 3) –45.7 kJ 4) –1.7 kJ
24. In a calorimeter, the temperature increases by 6.12 K, the heat capacity of the system is 1.23 kJK–1. What is
the molar heat of decomposition for NH4NO3?
1) –7.53 kJ/mol 2) –398.1 kJ/mol 3) –16.1 kJ/mol 4) –602 kJ/mol
25. In which of the following conditions a chemical reaction cannot occur?
1) H and S increases and TS  H 2) H and S decrease and H  TS
3) H increases and S decreases 4) H decreases and S increases
26. Standard entropies of X2, Y2 and XY3 are 60, 40 and 50 JK–1 mol–1 respectively. For the reaction:
1 3
X 2  Y2  XY3
2 2
H  30 kJ to be at equilibrium, the temperature will be :
1) 1000 K 2) 1250 K 3) 500 K 4) 750 K
27. For a particular reaction, H = – 38.3 kJ and S  113J K 1 mol 1. This reaction is :
1) spontaneous at all temperatures 2) non-spontaneous at all temperatures
3) spontaneous at temperatures below 66oC 4) spontaneous at temperatures above 66oC
28. The heat liberated when 1.89 g of benzoic acid is burnt in a bomb calorimeter at 25oC increases the temperature
of 18.94 kg of water by 0.632oC. If the specific heat of water at 25oC is 0.998 cal g –1 deg –1, the value of heat
of combustion of benzoic acid is:
1) 88.1 kcal 2) 771.1 kcal 3) 981.1 kcal 4) 871.2 kcal

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29. The standard Gibbs energy change for a reaction is G o  41.8 kJ mol 1 at 700 K and 1 atm. Calculate the
equilibrium constant of the reaction at 700 K.
1) 1.314 ×104 2) 3.431 ×104 3) 3.431 ×103 4) 1.314 ×103
30. Molar heat capacity of water in equilibrium with ice at constant pressure is:
1) Zero 2) Infinity    3) 40.45 kJ K–1 mol–1 4) 75.48 J K–1 mol–1
LEVEL - II
1. In thermodynamics, process is called reversible when:
1) surroundings and system change into each other
2) there is no boundary between system and surroundings
3) surroundings are always in equilibrium with system
4) system changes into surroundings spontaneously
2. How many joules of heat are absorbed when 70.0 g of water is completely vaporised at its boiling point?
(LV= 2260 kJ/kg)
1) 23.4 kJ 2) 7.0 kJ 3) 15.81 kJ 4) 158.7 kJ
3. An ideal gas is taken around the cycle ABCA as :

The work done in the cyclic process is :

1) 12P1V1 2) 6P1V1 3) 3P1V1 4) P1V1
4. A cylinder filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant
temperature of 37.0o C. As it does so, it absorbs 208 J of heat. The value of q and w for the process will be:
(R = 8.314 J mol–1 K–1) (ln 7.5 = 2.01)
1) q = –208 J, w = –208 J 2) q = –208 J, w = +208 J
3) q = +208 J, w = +208 J 4) q = +208 J, w = –208 J
5. The standard heat of combustion of Al is –837.8 kJ mol–1 at 25oC. If Al reacts with O2 at 25oC, which of the
following release 250 kJ of heat?
1) The reaction of 0.624 mol of Al 2) The formation of 0.624 mol of Al2O3
3) The reaction of 0.312 mol of Al 4) The formation of 0.150 mol of Al2O3
6. Which of the following statement is not correct?
1) Final temperature in reversible adiabatic expansion is greater than that in irreversible adiabatic expansion.
2) When heat is supplied to an ideal gas in isothermal process, kinetic energy of gas remains constant.
3) When an ideal gas is subjected to adiabatic expansion, it gets cooled.
4) Entropy increases when an ideal gas expands isothermally.

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7. 4 kg of ice at–20oC is mixed with 10 kg of water at 20oC in an insulated vessel having a negligible heat
capacity. Calculate the final mass of water remaining in the container. Given : Specific heat capacities of
water and ice are 4.184 kJ K–1 kg–1 and 2.092 kJ K–1 kg–1, respectively. Molar enthalpy of fusion of ice is
334.7 kJ kg–1.
1) 8 kg 2) 10 kg 3) 12 kg 4) 14 kg
8. 4.48 L of an ideal gas at S.T.P requires 12 calories to raise its temperature by 15 C at constant volume. The
o

CP of the gas is :
1) 3 cal mol–1 2) 4 cal mol–1 3) 7 cal mol–1 4) 6cal mol–1
9. The enthalpy of neutralization of oxalic acid by a strong base is –25.4 k cal mol–1. The enthalpy of neutralization
of strong acid and strong base is –13.7 kcal eq–1. The enthalpy of dissociation of oxalic acid is:
1) 1 kcal mol–1 2) 2 kcal mol–1 3) 18.55 kcal mol–1 4) 11.7 kcal mol–1
10. Combustion of sucrose is used by aerobic organisms for providing energy for the life sustaining processes. If
all the capturing of energy from the reaction is done through electrical process (non P–V work) then calculate
maximum available energy which can be captured by combustion of 34.2 gm of sucrose
Given: Hcombustion  sucrose   6000 kJ mol1 ; Scombustion  180 J / K / mol and body temperature is 300 K
1) 600 kJ 2) 594.6 kJ 3) 5.4 kJ 4) 605.4 kJ
11. If  f H  CO, g   110.5 kJ mol and  f H  CO2 , g   393.5kJ mol , the mass of oxygen consumed in
1 1

the reaction C (graphite)  O2  g   mixture of CO  g   CO2  g  for which  r H  313.8 kJ mol 1 will
be about
1) 24.0 g 2) 25.5 g 3) 27.5 g 4) 29.0 g
15
12. Consider the reaction at 300 K, C6 H 6  l   O 2  g    6CO 2  g   3H 2 O  l  ; H  3271kJ
2
What is U for the combustion of 1.5 mole of benzene at 27oC?
1) – 3667.25 kJ 2) – 4900.88 kJ 3) – 4806.5 kJ 4) –3274.75 kJ
13. The enthalpy of combustion of H2(g) at 298 K to give H2O(g) is –249 kJ mol and bond enthalpies of H–H
–1

and O O are 433 kJ mol–1and 492 kJ mol–1 respectively. The bond enthalpy of O – H is
1) 464 kJ mol–1 2) –464 kJ mol–1 3) 232 kJ mol–1 4) –232 kJ mol–1
14. If bond enthalpies of  C  H   413.4 kJ mol ,  C  C   347.7 kJ mol , and  C  C   615.1kJ mol ,
1 1 1

 sub H(C, graphite)  718.4 kJ mol1 and  f H(H, g)  218 kJ mol1 , the enthalpy of formation of gaseous
isoprene will be about  CH 2  C  CH3  CH  CH 2  :
1) 206.4 kJ mol–1 2) –206.4 kJ mol–1 3) 103.2 kJ mol–1 4) 144 kJ mol–1
15. The species which by definition has zero standard molar enthalpy of formation at 298 K are

(i) Br2(g) (ii) Cl2(g) (iii) CH4(g) (iv) H aq 
1) (i) and (ii) 2) (ii) and (iv) 3) (ii) and (iii) 4) (i), (ii), (iii) and (iv)
 2NH 3  g 
16. The Haber’s process for production of ammonia involves the equilibrium : N 2  g   3H 2  g  
Assuming H o and So for the reaction do not change with temperature, which of the statements is true?
 H o
 95 kJ and So  190 JK 1 
1) Ammonia dissociates spontaneously below 500 K.
2) Ammonia dissociates spontaneously above 500 K.
3) Ammonia dissociates at all temperatures.
4) Ammonia does not dissociate at any temperature.
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17. Oxygen gas weighing 64 g is expanded from 1 atm to 0.25 atm at 30oC. Calculate entropy change, assuming
the gas to be ideal?
1) 23.1 JK–1 2) 25.0 JK–1 3) 26.1 JK–1 4) 22.4 JK–1
18. For the process H 2O(l)  H 2O  g  at T = 100oC and 1 atmosphere pressure, the correct choice is:
1) Ssystem  0 and Ssurroundings  0 2) Ssystem  0 and Ssurroundings  0
3) Ssystem  0 and Ssurroundings  0 4) Ssystem  0 and Ssurroundings  0
19. The lattice energy of NaCl is +788 kJ mol–1. The enthalpies of hydration of Na+(g) and Cl– (g) ions are
–406 kJ mol–1 and –378 kJ mol–1 respectively. The enthalpy of solution of NaCl(s) in water is
1) 786 kJ mol–1 2) 4 kJ mol–1 3) –4 kJ mol–1 4) –792 kJ mol–1
1
20. H 2  g   O 2(g)  H 2O l  ; H 298K  68.32 kcal . Enthalpy of vapourization of water at 1 atm and 25oC is
2
10.52 kcal. The standard enthalpy of formation (in kcal) of water vapour at 25oC is:
1) 10.52 2) –78.84 3) +57.80 4) –57.80
21. Given that, NH3 g   3Cl2 g   NCl3 g   3HCl g  ; H1
N 2 g   3H 2 g   2NH3 g  ; H 2
H 2 g   Cl2 g   2HCl g  ; H3
The enthalpy of formation of NCl3(g) in terms of H1 , H 2 and H3 is:
H 2 3 H 2 3
1)  f H  H1   H 3 2)  f H  H1   H 3
2 2 2 2
H 2 3 H 2 3
3)  f H  H1   H 3 4)  f H  H1   H 3
2 2 2 2
22. The enthalpy of neutralization of a weak monoprotic acid (HA) in 1 M solution with a strong base is
–55.95 kJ/mol. If the unionized acid is required 1.4 kJ/mol heat for it’s complete ionization and enthalpy of
neutralization of the strong monobasic acid with a strong monoacidic base is –57.3 kJ/mol. What is the %
ionization of the weak acid in molar solution?
1) 1% 2) 3.57% 3) 35.7% 4) 10%
23. Nitrogen dioxide, NO2 , an air pollutant, dissolves in rain-water to form a dilute solution of nitric acid. The
equation for the reaction is : 3NO 2  g   H 2O  l   2HNO3  l   NO  g 
Calculate So for the reaction in J K–1. The standard entropy of HNO3(l), NO(g), NO2 (g) and H2O (l) are
155.6, 210.6, 240.5 and 69.96 J mol–1 K–1 respectively.
1) 155.6 J K–1 2) –155.6 J K–1 3) 269.7 J K–1 4) –269.7 J K–1
24. What is the melting point of benzene if H fusion  9.95 kJ / mol and Sfusion  35.7 JK 1mol 1 ?
1) 278.7oC 2) 278.7 K 3) 300 K 4) 298 K
25. Standard entropies of X2, Y2 and XY3 are 60, 40 and 50 JK–1 mol–1 respectively. For the reaction:
1 3 
X 2  Y2  XY3 ; H  30 kJ to be at equilibrium, the temperature should be :
2 2
1) 750 K 2) 1000 K 3) 1250 K 4) 500 K

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26. Assuming H and S do not change with temperature. Calculate, the boiling point of liquid A using the
thermodynamic data given below :
Thermodynamic data A l  A  g 
H o  kJ / mol  130 100
So  J K 1mol1  100 200

1) 300 K 2) 130 K 3) 150 K 4) 50 K

27. Which among the following substance has highest entropy?
1) CH4(g) 2) C3H8(g) 3) CH3OH(l) 4) PbBr2(s)
28. The densities of graphite and diamond at 298 K are 2.25 g cm and 3.31 g cm , respectively. If the standard
–3 –3

Gibbs energy difference  G o  is equal to 1895 J mol–1, the pressure at which graphite will be transformed
into diamond at 298 K is:
1) 11.1×108 Pa 2) 11.1×107 Pa 3) 11.1×106 Pa 4) 11.1×1010 Pa
29. Calculate the equilibrium constant for the the reaction given below at 400K, if H o  77.2 kJ mol 1 and
So  122 J K 1 mol 1 : PCl5 g  
 PCl3 g   Cl2  g 
1) 1.958×10–3 2) 1.958×10–4 3) 1.958×10–2 4) 1.958×102
30. When two ideal gases are mixed
1) S mix   0 H  mix   0 2) S mix   0 H  mix   0
3) S mix   0 H  mix   0 4) S mix   0 H  mix   0

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SOLUTIONS
LEVEL - I
1. 2 Laws of thermodynamics deals with macroscopic system.
2. 1 Work is not a state function. It depends on path of the process .
3. 3 Internal energy and molar enthalpy are state functions.
4. 3 Presence of reaction species in a covered beaker - Closed system. Exchange of matter as well as
energy-Open system. Presence of reaction in a closed vessel - Closed system. Presence of reactants in
thermoflask - Isolated system.
P  4
5. 1 w  2.303nRT log  2   2.303 1 8.314  300 log  3458.3 J  3.458 kJ
 P1  1
6. 1 At temp. T, PV = RT
When temp. is increased by 1o, at constant pressure, suppose volume increases by V
P  V  V   R  T  1 or PV  PV  RT  R
But PV  RT  PV  R , i.e., work done = R = 2 cal.
7. 2 Net work done = Area covered by PV curve
ab  5m3  2m3 
Area of trapezium   h     2 10 Pa  7  10 J = –700 kJ
5 5

 2   2 
8. 4 H   Σ B.E.of the reactants  Σ B.E.of the products 

 615   4  414   435  347   414  6    2706  2831  125 kJ

9. 3 U  q  W, q  100 J , W  1.5  6  9 L atm  911.9 J  U  1011.9 J
10. 4 The amount of heat liberated by 0.5 g of C = mst = 2000 × 1 × (26 – 24) = 4000 cal = 4 kcal
4 kcal  1
Calorific value   8 kcal per g  Heat of combustion = –8 × 12 = –96 kcal per mole
0.5
H 2  H1
11. 4 I 2 s   I2 g  CP   0.031  0.055  103  254 ;  C P
T2  T1
H 2  6.096
  0.031  0.055  103  254  H 2  5.79 kcal mol 1
50
12. 3 This is because the enthalpy change is a state function.
13. 1 In bomb calorimeter, w = 0, U= q(  ve)
14. 3 The reaction involved is H 2 O  l   H 2O  g  , where n g  1mol,  vap H  41kJ mol1. Now,,
U  H  n g RT
 41kJ mol1  1 8.3  103 kJ K 1 mol1  373K  41  3.096 kJ mol 1  37.904 kJ mol1
15. 2  n g  2  4  2 ; H  U  n g RT

U  H  n g RT = -92.38 × 1000 – (–2) × 8.314 × 298 = -87424 J = –87.424 kJ

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16. 2 First, we divide the second reaction by 2:

3
N 2 O  g   O 2  g   2NO 2  g  H  14.0 kJ mol 1
2
Now, reversing the first reaction, we get
2NO2  g   2NO  g   O2  g  H  113.2 kJ mol1
Adding the above reactions, we get
3
N 2 O  g   O 2  g   2NO 2  g  H 0  14.0 kJ mol 1
2
2NO 2  g   2NO  g   O 2  g  H 0  113.2 kJ mol 1
1
N 2 O  g   O 2  g   2NO  g  H o  99.2 kJ mol 1
2
17. 3 Standard heat of formation of methane is represented by C(graphite) + 2H 2 g  
 CH 4 g  because the
elements taken in their standard state.
18. 3 H  U  nRT, n  2  4  2  H  U
19. 4 Standard molar enthalpy of formation of CO2 and the standard molar enthalpy of combustion of carbon
(graphite) refer to the same chemical equation: C(graphite)  O 2  g  
 CO2  g 
20. 4 According to the definition of standard enthalpy of formation, the enthalpy change for the following
1
reaction will be standard enthalpy of formation of H2O(l). H 2 g   O 2 g  
 H 2Ol  .
2
or the standard enthalpy of formation of H2O(l) will be half of the enthalpy of the given equation i.e.,  r H 
1 572 kJ mol1
is also halved.  f H H2O (l )    r H   = –286 kJ / mol.
2 2
21. 4 From the given data, the heat of combustion reaction of diborane can be calculated as
B2 H 6  g   3O 2  g   B2 O3  s   3H 2O  g 
 r H o  H fo  B2 O3   3   H fo  H 2 O   H o vap  H 2 O    H fo  B2 H 6 

Now,  vap H o  H 2O   H fo  H 2O   44  242 kJ mol1  c H o  B2 H 6   1273  3  242   36

Therefore, = –2035 kJ mol–1
22. 3 H  U  n g RT ; 50700  U   5   8.314  300 ; U  38229 J
23. 2 0.3 mole OH– ion will be completely neutralised,  H  57.1 0.3  17.13 kJ
24. 4 Molecular weight of NH4NO3 = 80 g mol –1

 Molar heat of decomposition, H  mst  80  1.23  6.12  602 kJ / mol

25. 3 If H   ve and S   ve then the reaction is non-spontaneous.
1 3
26. 4 X 2  Y2  XY3
2 2
1 3 1 3
So Reaction  So XY3  So X 2  So Y2  50   60   40  40 JK 1 mol1
2 2 2 2
 H 30  1000
G  H  T S  0 at equilibrium.  T    750 K
S 40
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27. 4 G  H  T S ; For spontaneous process, G  0.

H
 H  T S  0 ,  T ; 38.3  1000  T i.e., T  338.93 K,i.e., 66o C
S 113
28. 2 Heat liberated by 1.89 g of benzoic acid, q = msT = 18940×0.998×0.632 = 11946.14 cal
Heat liberated by the combustion of 1 mol benzoic acid, i.e., 122 g benzoic acid
11946.14
  122  771126.5 cal = 771.12 kcal mol–1
1.89
29. 4 The standard Gibbs energy and equilibrium constant are related by the equation:
G o   RT ln K  2.303RT log K

Given that G o  41.8 kJ  41800 J, T  700 K and R  8.314 J K 1 mol1. Substituting these
values in the equation, we get 41800  2.303  8.314  700 log K
41800
41800  13403 log K or log K   3.1187  K  1.314  103
13403
dq P
30. 2 By definition CP,m   H 2 O  s  , temperature does not change if some heat is
. For H 2 O  l  
dT
 ve
given to the system. Hence CP,m  
Zero
LEVEL - II
1. 3 In thermodynamics a process is said to be reversible when system is always in equilibrium with
surroundings.
2. 4 Since the process involves phase change, heat absorbed, Q = mass × latent heat of vaporisation
Given, mass = 70.0 g = 0.07 kg , Lv = 2260 kJ/kg  Q = 0.07 × 2260 kJ = 158.2 kJ
3. 3 The work done in the cyclic process
1 1
= Area bounded (ABCA)   AC  AB   2V1  3P1  3P1V1
2 2
4. 4 Process is isothermal reversible expansion, hence U  0. Therefore, according to first law of
thermodynamics q = –w. As q = +208 J, hence w = –208 J.
3 1
5. 4 Al  O 2 
 Al2O3 .
2 2
27 g 51g

H  837.8 kJ mol1 .
A) 0.624 mol of Al = 837.8 × 0.624 on combustion gave = 523 kJ. Hence, false.
B) Formation of 0.624 mol of Al2O3 gave = 837.8 × 2 × 0.624 = 1045 kJ. Hence, false.
C) 0.312 mol of Al on combustion gave = 261 kJ. Hence, false.
D) Formation of 0.150 mol of Al2O3 gave = 251.3 kJ. Hence, true.

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w
6. 1 For adiabatic process, U  w  w  C V T  T 
CV

7. 3 The conservation of heat gives mi Ci  20 K   m1 fus H  mw Cw  20 K   0

 4 kg   2.092 kJK 1kg 1   20 K   m1  334.7 kJ kg 1   10 kg   4.184 kJ K 1 kg 1   20 K   0

This gives
 10  4.184  20    4  2.092  20    836.8  167.36   669.44 
m    kg    kg    kg  2.0 kg
 334.7   334.7   334.7 

Hence, mass of water in the container is m  10 kg    2 kg   12 kg.

U 4.48 12
8. 4 CV  , n  0.2 ; Cv   4 cal
n T 22.4 0.2  15
CP  CV  R  4  2  6 cal
9. 2 Oxalic acid is dibasic acid hence expected heat of neutralization will be (2 × –13.7 kcal).
The dissociation energy of oxalic acid = 2 × 13.7 – 25.4 = 2 kcal mol–1
34.2
10. 4 No. of moles of sucrose   0.1 ;   G T,P  useful work done by the system
342
180  0.1 300
G  H  TS    6000  0.1  = 605.4 kJ
1000
11. 3 Given : The data are
1
C  graphite   O 2  g   CO  g  :  f H o  110.5 kJ mol1
2
C  graphite   O2  g   CO2  g  :  f H o  393.5 kJ mol1

C  graphite   O2  g   a mixture of CO  g  and CO2  g  :  r Ho  313.8kJ mol1

If x is the amount fraction of CO in the mixture, then the amount of O2 consumed will be
x x
 1  x   1  ; where x may be calculated from the expression x  110.5 kJ mol1 
2 2
 1  x   393.5 kJ mol 1   313.8 kJ mol 1 which gives x = 0.282. Mass of O2 consumed
 0.282  
 mol   32 g mol   27.5 g
1
 1 
 2  
12. 2 For 1 mole of combustion of benzene; n g  1.5
1.5  8.314  300
H  U  n g RT  3271  U   U  3267.25 kJ
1000
For 1.5 mole of combustion of benzene;
U  3267.25 1.5 = – 4900.88 kJ

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13. 1 We have
1
H2  g   O2  g  H
  H 2O  g 
2

 1
( H  H)
o o 2 O H
2
2H  g   O g
1  1 
Hence  H  H   O O  2 O  H   433   492  kJ mol 1  2O  H  249 kJ mol1
2  2 
1  1  
 O  H   433   492   249  kJ mol1  464 kJ mol1
2  2  
14. 3 The formation reaction is
5C  graphite   4H 2  g  
f H
 CH 2 C  CH 3  CH CH 2

5  718.4 kJ 8  218kJ 2  C C

mol1 mol1 2  C C
8  C H
5C  g   8H  g 
Hence,  f H  5  718.4  8  218  2  615.1  2  347.7  8  413.4) kJ mol1  103.2 kJ mol1
15. 2 Cl2 g  and Haq  have zero standard enthalpy of formation.
16. 1 G o  H o  TSo  95 1000  500   190   95000  95000  0
Thus, ammonia will dissociate below 500 K because G o will be negative.
p 
S  2.303nR log  1   2.303  2  8.314 log 
1 
17. 1   23.053J K 1
 p2   0.25 
18. 2 Stotal  Ssystem  Ssurroundings
H system
Ssurroundings  ; Hsystem is positive
T
19. 3 NaCl s   Na   g   Cl  g  H  788 kJ  Lattice energy 
Na   g   H 2 O l  Na   aq  H  406 kJ  Hydration 
Cl  g   H 2 O l  Cl   aq  H  378 kJ  Hydration 
The net reaction of dissoultion of NaCl(s) is NaCl s   2H 2O l  Na   aq   Cl  aq 
 sol H  788   406  378   4 kJ mol 1
1
20. 4 The required equation is H 2 g   O 2 g   H 2 O g 
2
1
The equation can be obtained as H 2  g   O 2 g   H 2 O l  ; H 298K  68.32 kcal (1)
2
H 2 O l   H 2 O  g  ; H  10.52 kcal (2)
1
Adding Eqs. (1) and (2), H 2  g   O 2 g   H 2 Og  ; H  57.80 kcal
2
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21. 1 The given equations are

NH3 g   3Cl2 g   NCl3 g   3HCl g  ; H1 (1)
N 2 g   3H 2 g   2NH3 g  ; H 2 (2)
H 2  g   C l 2  g   2H C l  g  ;  H 3 (3)
We aim at: N 2 g   3Cl2 g   2NCl3 g  ; f H  ?
Adding 2  Eq. 1  Eq.  2   3  Eq.  3 , we get
H 2 3
2 f H  2H1   H 2  3H 3    f H  H1   H 3
2 2
22. 2  H   A  ;  r H  1.4 kJ / mol ;
HA  H neutralization  H ionization   r H
H  OH   H 2 O  55.95  H ionization  57.3 ; Hionization for1M HA  1.35 kJ / mol


1.35
% heat utilized by 1 M acid for ionization   100  96.43% so acid is 100 – 96.43 = 3.57%
1.4
ionized
23. 4 The entropy change is given by So  So  products   So  reactants 
  2So  HNO 3  l    So  NO  g     3So  NO 2  g    So  H 2 O  l   
Substituting values, we get So   2mol  155.6 J mol 1K 1   1mol   210.6 J mol 1K 1  
 3mol   240.5 J mol 1K 1   1mol   69.96 J mol1K 1   ; Solving, we get So  269.7 J K 1.
H fusion  H 9.95  1000
24. 2 at M.P. Sfusion  T   278.7 K
T S 35.7
1 3 
25. 1 S  SProducts  SReactants  50    60   40   40J K 1mol 1
2 2 
H 30  1000
At equilibrium, G  H  TS  0 i.e., H  TS ; T   750 K
S 40
26. 1  A  g 
A  l  
H  H oA  g   H oA l   100   130   30 kJ ; S  SoA g   SoA l   200  100  100 J K 1 mol 1
H 30  1000
At equilibrium G  H  TS  0  T    300 K
S 100

27. 2 Gases have highest entropy.

28. 1 G  pV
 12 12  6 3 1895
1895  p     10 m   p  6
 1.1147  109 Pa  11.1  108 Pa
 2.25 3.31   1.7  10
1
29. 2 H  77.2 kJ mol ; S  122 J K 1 mol 1
o o

T = 400 K  G  H  TS ; G o  77200  400  122  28400 J

o o o

Also, we have G o  2.303 RT log10 K C where, KC is equilibrium constant

 28400  2.303  8.314  400 log10 K C  K C  1.958  10 4
30. 4 For mixing of two ideal gases S mix  is positive and H mix  zero.

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CHAPTER - 07
CHEMICAL AND IONIC EQUILIBRIUM

INTRODUCTION
It has been observed that many chemical reactions, carried out in closed vessels, do not proceed to completion.
The reactants may still be present but they do not appear to change into products any more. What happens in
such a case is that the products react at the same rate to form the reactants, i.e., the rate of the backward
reaction becomes equal to rate of the forward reaction. The reaction in such conditions is said to be in
dynamic equilibrium. The mixture of reactants and products in equilibrium is called equilibrium mixture.
The composition of the equilibrium mixture at a given temperature is the same irrespective of the initial state of
the system - whether we start with the reactants or the products.
Equilibrium attained in physical processes is called physical equilibrium and in chemical processes is called
chemical equilibrium. Equilibrium involving ions in aqueous solutions is called ionic equilibrium.
Reactions may be fast or slow depending on the experimental conditions and the nature of the reactants.
Based on the extent of reaction, chemical equilibria may be classified into three groups :
(i) Reactions that proceed nearly to completion and only negligible amounts of reactants are left unreacted.
(ii) Reactions in which only small amounts of products are formed and most of the reactants remain unchanged.
(iii) Reactions in which the concentration of reactants and products are comparable at equilibrium.
The study of chemical equilibrium helps in the elucidation of optimum conditions for maximum yield of products
from reversible reactions.
7.1 EQUILIBRIUM IN PHYSICAL PROCESSES
The most familiar examples of physical processes are phase transformations such as
solid  liquid ; liquid  gas ; solid  gas
1. Solid-Liquid Equilibrium. Ice and water inside a perfectly insulated flask at 273K and atmospheric pressure
are in equilibrium. There is no change in the masses of ice and water as the rates of transfer of molecules from
ice to water and water to ice are equal.
Rate of melting = Rate of freezing or Ice  Water
For a pure substance at atmospheric pressure, the temperature at which the solid and liquid phases are in
equilibrium is called the normal melting point or normal freezing point.
2. Liquid-Vapour Equilibrium. Evaporation of a liquid in a closed container gives rise to a constant vapour
pressure at a given temperature.
Rate of evaporation = Rate of condensation
The pressure exerted by the vapours of a liquid at a given temperature is a constant. This is called the equilibrium
vapour pressure or vapour pressure of the liquid. Vapour pressure increases with temperature.
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For a pure liquid at one atmosphere pressure, the temperature at which the liquid and its vapour are in equilibrium
is called the normal boiling point of the liquid. Water and water vapour are in equilibrium at 1 atmospheric
(1.013 bar) pressure and 100°C (373.15K).
H2O(l)  H2O(g)
Boiling point of a liquid depends on the altitude of the place. At high altitude, the boiling point decreases due to
decrease in atmospheric pressure.
3. Solid - Vapour Equilibrium. When a solid sublimes in a closed vessel, the vessel gets filled with its vapours
until a dynamic equilibrium is attained. For example, iodine sublimes to give iodine vapours and vapours
condense to give solid iodine until equilibrium is attained (the intensity of colour becomes constant).
I2 (solid)  I2 (vapour)
Camphor, naphthalene, NH4Cl, etc., exhibit similar equilibria.
7.1.1 Equilibrium Involving Dissolution of Solid or Gases in Liquids
a. Dissolution of Solids in liquids. The concentration of the solute in a saturated solution depends on the
temperature. A dynamic equilibrium exists between the solute molecules in solid state and in solution.
Solute (solution)  Solute (solid)
Rate of dissolution of solute = rate of crystallisation of solute
Equality of rates and dynamic nature of equilibrium have been established using radioactive solutes.
When radioactive sugar is added to a saturated solution of non-radioactive sugar, radioactivity is observed
both in solution and in solid sugar after some time. This is due to dynamic nature of equilibrium where exchange
occurs between radioactive and non-radioactive solute molecules in the two phases. The ratio of radioactive
to non-radioactive molecules in solution increases till it attains a constant value.
b. Dissolution of Gases in liquids. When a soda water bottle is opened, carbon dioxide gas dissolved in it
fizzes out. This is due to the difference in solubility of carbon dioxide at different pressures. The equilibrium
between CO2 in the gaseous phase and CO2 dissolved in the liquid under pressure is given by,
CO2 (gas)  CO2 (in solution)
The quantity of gas that dissolves in a liquid increases with pressure and decreases with temperature. The
equilibrium is governed by Henry’s law, which states that the mass of gas that dissolves in a given mass of
solvent at a given temperature is proportional to the pressure of the gas above the solvent.
m  P or m = kP
where k is the proportionality constant.
A soda water bottle is sealed at high pressure. As soon as the bottle is opened, some of the dissolved gas
escapes so that a new equilibrium is attained at the lower pressure, namely its partial pressure at atmospheric
pressure. Thus, soda water left open for some time turns ‘flat’.
7.1.2 Features of Physical Equilbrium
(i) Solid - liquid equilibrium. There is only one temperature at 1 atm at which the two phases coexist
(melting point). When there is no exchange of heat with the surroundings, the masses of the two phases
remain constant.
(ii) Liquid - vapour equilibrium. The vapour pressure is constant at a given temperature.
(iii) Dissolution of solids in liquids. The solubility is constant at a given temperature.
(iv) Dissolution of gases in liquids. The concentration of the gas in the liquid is proportional to the pressure
(concentration) of the gas above the liquid.

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Process Example Property that becomes constant at equilibrium

Liquid  Vapour H2O(l)  H2O(g) Vapour pressure, PH2 O at a given temperature
Solid  Liquid H2O(s)  H2O(l) Melting point at constant pressure
Solute(s)  Solute(soln) Sugar(s)  Sugar(soln) Concentration of solute in solution at given temperature
Gas (g)  Gas (aq) CO2(l)  CO2(aq) Gas (aq)/Gas (gaseous phase) at given temperature
7.1.3 General Characteristics of Equilibria Involving Physical Processes
(i) Equilibrium is possible only in a closed system at a given temperature.
(ii) The opposing processes occur at the same rate and there is a dynamic but stable condition.
(iii) All measurable properties of the system remain constant.
(iv) Equilibrium is characterised by constant value of one of its parameters at a given temperature.
(v) The magnitude of such parameters at any stage indicates the extent to which the process has proceeded
before reaching equilibrium.
7.2 EQUILIBRIUM IN CHEMICAL PROCESSES
A reaction which takes place not only in the forward direction but also in the backward direction under the
same conditions is called reversible reaction. At equilibrium, the rates of the forward and the reverse
reactions become equal and the concentrations of the reactants and products remain constant. Thus, the
equilibrium is dynamic in nature. For example

 C  D
A  B 
The equilibrium can be attained even if we start with only C and D, as the equilibrium can be reached from
either direction.

Figure - 01 (a) Variation of concentration with time

Dynamic Nature of Equilibrium. The dynamic nature of chemical equilibrium has been illustrated in the
synthesis of ammonia by Haber’s process. Haber conducted a series of experiments with known quantities
of N2 and H2 at high temperature and pressure and at regular intervals determined the quantity of ammonia
formed and the quantities of unreacted H2 and N2. He found that after a certain time the composition of the
mixture remained the same even though some of the reactants were still present. The constancy in composition
indicated attainment of equilibrium.
Dynamic Nature of Reaction. In order to study the dynamic nature of the reaction, synthesis of ammonia
was carried out under exactly same conditions of partial pressure and temperature, but separately using H2
and D2 along with N2.
 2NH (g)
N2(g) + 3H2(g)  ... (i)
3

 2ND (g)

N2(g) + 3D2(g)  ... (ii)
3

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The reaction mixtures reached equilibrium with the same composition, except that D2 and ND3 were present
(instead of H2 and NH3) in the second case.
After equilibrium was attained, the mixtures (H2, N2, NH3 and D2, N2, ND3) were mixed together. Later, when
the mixture was analysed, it was found that the concentration of ammonia was the same as before. However,
when the mixture was analysed using mass spectrometer, it was found that NH3, all deuterium containing forms
of ammonia (NH2D, NHD2 and ND3), H2 and its deutrated form (HD) and D2 were present.
It was concluded that scrambling of H and D atoms in the molecules must be the result of continuation of the
forward and reverse reactions even after the attainment of equilibrium. If the reactions had stopped when they
reached equilibrium, there would not have been any mixing of isotopes.
The use of isotopes indicate that chemical reactions reach a state of dynamic equilibrium in which the rates of
the forward and the reverse reactions are equal and there is no net change in composition.
Equilibrium can be attained from either side. The reaction can be conducted by starting with H2(g) and
N2(g) to obtain NH3(g) or by taking NH3(g) and decomposing it into N2(g) and H2(g).
 N (g) + 3H (g)
2NH3(g)  2 2

(a) (b)

Figure - 02 Variation of concentration of H2, N2 and NH3 with time (b) Equilibrium in a reaction can be
attained from either direction.
Similarly, when the reaction of H2 and I2 to form HI proceeds in the forward direction, the concentration of H2
and I2 decreases while that of HI increases, until all of these become constant at equilibrium.
 2HI(g)
H2(g) + I2(g) 
It is also possible to start with HI alone and make the reaction proceed in the reverse direction.
7.2.1 Characteristics of Chemical Equilibrium
i. Equilibrium can be attained only in closed systems.
ii. Equilibrium can be attained from either direction.
iii.Properties of the system such as pressure, concentration, etc., become constant at equilibrium.
iv.Catalyst does not alter the equilibrium.
7.3 LAW OF CHEMICAL EQUILIBRIUM AND EQUILIBRIUM CONSTANT
Law of Mass Action. The law of mass action (Norwegian chemists Cato Maximillian Guldberg and Peter
Waage, 1864) states that at constant temperature, the rate at which a substance reacts is proportional to
its active mass and the rate of a chemical reaction is directly proportional to the product of the active
masses (concentrations) of the reacting substances.
For a general reaction, A + B   Products
Rate of reaction, r  [A][B] or r = k [A][B]
where [A] and [B] are the molar concentrations of the reactants A and B respectively and k is a proportionality
constant called rate constant.
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For a general reaction, aA + bB   Products

Rate of reaction, r = k [A]a [B]b
7.3.1 Law of Chemical Equilibrium
Law of mass action can be applied to reversible reactions in equilibrium to derive a simple mathematical
expression known as law of chemical equilibrium.
 C + D,
For a general reversible reaction, A + B 
Rate of forward reaction  [A][B] = kf [A] [B]
where kf is the velocity constant of the forward reaction and [A] and [B] are the molar concentrations of
reactants A and B respectively.
Rate of backward reaction  [C] [D] = kb [C] [D]
where kb is the velocity constant of the reverse reaction and [C] and [D] are the molar concentrations of
products C and D respectively.
At equilibrium, Rate of forward reaction = Rate of backward reaction,
kf [C][D] kf [C][D]
kf [A] [B] = kb [C] [D]  =  K = =
kb [A][B] c kb [A][B]
This equation is called equilibrium equation and the constant Kc is called equilibrium constant. The subscript
‘c’ indicates that Kc is expressed in terms of equilibrium concentrations (mol L-1).
 cC + dD,
For a general reaction, aA + bB 

[C]c [D]d
Kc =
[A]a [B]b
[C]c [D]d
The concentration ratio, is called concentration quotient, Qc.
[A]a [B]b
According to law of chemical equilibrium, Qc = Kc, at equilibrium.
In general, for a homogeneous reversible reaction at a particular temperature,
`  n B + n B + n B + ...
m1A1 + m2A2 + m3A3 + ...  1 1 2 2 3 3

At equilibrium, K f [A1 ]m1 [A 2 ]m2 [A3 ]m3 ... = K b [B1 ]n1 [B2 ]n 2 [B3 ]n3 ...

Kf [B1 ]n1 [B2 ]n 2 [B3 ]n 3 ....

or   Kc
K b [A1 ]m1 [A 2 ]m2 [A3 ]m3 ...
Equilibrium constant is independent of (i) initial concentration of reactants (ii) presence of catalyst (iii) direction
from which equilibrium is attained and (iv) presence of inert materials.
7.3.2 Characteriatics of Equilibrium Constant
i) The equilibrium constant for a particular reaction is always constant, depending only on the temperature
and independent of the concentrations of the reactants and the direction from which the equilibrium is
approached.

 2HI(g) ; K c =
[HI]2
H 2 (g) + I 2 (g) 
[H 2 ] [I 2 ]
For reactions having zero heat of reaction, temperature has no effect on the value of equilibrium constant.

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ii) The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward
reaction.
1
 H 2 (g) + I 2 (g) ; K c 
'
2HI(g) 
Kc
The equilibrium constants Kc and K'c have different numerical values, therefore, it is important to specify
the form of the balanced chemical equation when quoting the value of equilibrium constant.
iii) If the stoichiometric coefficients of a chemical equation are changed by multiplying throughout by a factor,
the expression for equilibrium constant should also reflect that change. For example,
12
[HI]  [HI]2 
2
1H 2 (g) + I
1
2 2 (g) 
 HI(g) ; K "
=
c
=    Kc
[H 2 ]1 2 [I 2 ]1 2  [H 2 ] [I2 ] 
On multiplying an equation by ‘n’, the equilibrium constant for the reaction becomes equal to (Kc) n.
2
 2

 2HI(g)]; K"c =  [HI]  = (K )2
2×[H2(g) + I2(g)  c
 [H 2 ] [I 2 ] 
iv) If the equation is written/split into two steps having equilibrium constants K1 and K2, then Kc = K1  K2.

 2NOCl(g) ; K c = [NOCl]2

For example, N2(g) + O2(g) + Cl2(g) 
[N 2 ][O 2 ][Cl 2 ]
The above equation can be split into two as :
[NO]2
 2NO(g) ; K1 =
N2(g) + O2(g)  .....(i)
[N 2 ] [O 2 ]
2
2NO(g) + Cl2(g)  2NOCl(g) ; K = [NOCl] .....(ii)
2
[NO]2 [Cl 2 ]
Combining equations (i) and (ii),

[NO]2 [NOCl]2 [NOCl]2

K c  K1  K 2   =
[N 2 ] [O2 ] [NO]2 [Cl2 ] [N 2 ] [O 2 ][Cl2 ]
Table - 01 Relation between equilibrium constant for a general reaction and its multiples.
Reaction Equilibrium constant
Forward reaction Kc
Reverse reaction 1/Kc
n
Reaction multiplied by a number or fraction Kc
Reaction divided into two steps Kc= K1K2
vi) The value of equilibrium constant is not affected by the addition of a catalyst. This is because a catalyst
increases the rate of the forward and the reverse reactions to the same extent.
Effect of Temperature on Equilibria. The equilibrium constant for a particular reaction changes with
temperature; the value increases for endothermic reaction and decreases for exothermic reaction. This is
because the forward and the backward reactions have different activation energies.
Temperature has no effect on reactions having zero heat of reaction.

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7.3.5 Units of equilibrium constant

The value of equilibrium constant Kc can be calculated by substituting the concentration terms in mol L-1 and for
Kp partial pressure in Pa, kPa, bar or atm.
n general, the unit of Kc = [M]n and the unit of Kp = [P]n
where n = (number of moles of products) - (number of moles of reactants).
 PCl3 (g)  Cl2 (g) ; K has units (mol L-1) and K has units bar or atm  n = 1].
PCl5 (g)  c p

 2NH 3 (g) K has units (mol L-1)-2 and K has units bar-2 or atm-2  n = -2]
N 2 (g)  3H 2 (g)  c p

Equilibrium constants can also be expressed as dimensionless quantities by specifying standard states
of reactants and products.
For a pure gas, the standard state is 1bar, therefore, a pressure of 3bar in standard state can be expressed as
3bar/1bar = 3, which is dimensionless.
Standard state (c0) for a solute is 1 molar solution and all concentrations are measured with respect to it.
The numerical value of equilibrium constant depends on the standard state chosen. In this system, both Kc
and Kp are dimensionless but have different numerical values.
7.4 HOMOGENEOUS EQUILIBRIA
Equilibria in which all the substances are in the same phase are known as homogeneous equilibria.
i) Equilibria in which the number of product molecules is equal to the number of reactant molecules.
N2(g) + O2(g)  2NO(g) ; H2(g) + CO2(g)  H2O(g) +CO(g)
CH3COOC2H5 (aq) + H2O (l)  CH3COOH (aq) + C2H5OH (aq)
ii) Equilibria in which the number of product molecules does not equal the number of reactant molecles.
N2O4(g)  2NO2 ; PCl5(g)  PCl3(g) + Cl2(g) ; N2(g) + 3H2(g)  2NH3(g)
2SO2(g) + O2(g)  2SO3(g) ; Fe3+ (aq) + SCN–(aq)  Fe(SCN)2+ (aq)
7.4.1 Equilibrium Constant in Gaseous Systems
Use of partial pressures instead of concentrations. For reactions involving gases, it is more convenient to
express the equilibrium constant in terms of partial pressures. The ideal gas equation can be written as,
n
PV = nRT or P  RT
V
where, n/V is concentration expressed in mol L-1. If concentration ‘c’ is in mol L-1, and P is in bar then,
P = cRT or P = [gas]RT
At constant temperature, the pressure of a gas is proportional to its concentration, P  [gas] .
 cC(g)  dD(g) in equilibrium,
For the reaction, aA(g)  bB(g) 

c d
[C]c [D]d (P ) (P )
Kc = a b and
KP  C a D b
[A] [B] (PA ) (PB )

Since PA = [A(g)]RT , PB = [B(g)]RT , PC = [C(g)]RT , and PD = [D(g)]RT

(PC )c (PD )d [C]c [D]d [RT](c+d) [C]c [D]d [C]c [D]d

KP  a b  a b (a+b)
 a b
[RT](c+d)-(a+b)
 a b
[RT]n
(PA ) (PB ) [A] [B] [RT] [A] [B] [A] [B]
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K p  K c [RT]n
where,n = (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced
chemical equation.
Kp is not the same as Kc. Kp will be equal to Kc only when (c+d)-(a+b) = 0, i.e., n = 0.
 2HI(g) in equilibrium,
(i) Case I (n = 0) : For the reaction, H2(g) + I2(g) 

[HI(g)]2 (PHI ) 2
Kc  and Kp 
[H 2 (g)][I 2 (g)] (PH 2 )(PI2 )

Since PHI = [HI(g)]RT , PI2 = [I2 (g)]RT , and PH2 = [H2 (g)]RT

[PHI ]2 [HI(g)]2 [RT]2

KP   
[PH 2 ][PI2 ] [H 2 (g)]RT [I 2 (g)]RT  K c [RT] = K c
Thus, Kp = Kc, i.e., both equilibrium constants are equal.
 N 2 (g)  3H 2 (g) at equilibrium,
(ii) Case II (n > 0) : For the reaction, 2NH 3 (g) 

(PN 2 )(PH 2 )3 [N 2 (g)]RT [H 2 (g)]3 (RT)3 [N2 (g)][H2 (g)]3[RT]2

KP     Kc (RT)2
(PNH3 )2 2
[NH3 (g)] [RT] 2
[NH3 (g)]2

Here, Kp  Kc , i.e., Kp > Kc

 2NH (g) at equilibrium,
(iii) Case III (n <0) : For the reaction, N2(g) + 3H2(g)  3

(PNH3 )2 [NH3 (g)]2 [RT]2

KP   [NH 3 (g)]2 [RT]-2
(PN2 )(PH2 )3   K c (RT) -2
[N 2 (g)]RT [H 2 (g)]3 (RT)3 [N 2 (g)][H 2 (g)]3

Here, K p  K c , i.e., Kp < Kc

Thus, when n = 0, K p  K c ; when n > 0 then Kp > Kc ; and when n < 0 then Kp < Kc.
While calculating the value of Kp, pressure is expressed in bar because the standard state for pressure is 1 bar
(1pascal, Pa = 1Nm–2, and 1bar = 105 Pa) .
Table - 02 Equilibrium constant, Kp for a few reactions.
Reaction Temperature/Kp Temperature/Kp Temperature/Kp
H2(g) + I2(g)  2HI(g) 298 795 500 160 700 54
5 -2
N2(g) + 3H2(g)  2NH3(g) 298 6.810 400 41 500 3.610
24 10 4
2SO2(g) + O2(g)  2SO3(g) 298 4.010 500 2.510 700 3.010
N2O4(g)  2NO2(g) 298 0.98 400 47.9 500 1700

7.5 HETEROGENEOUS EQUILIBRIA

Equilibrium in a system having more than one phase is called heterogeneous equilibrium.
i) Equilibrium between liquid and vapour/gas in a closed container.
 H O(g)
H2O(l)  2

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ii) Equilibrium between a solid and its saturated solution.

 Ca 2+ (aq) + 2OH - (aq)
Ca(OH)2 (s) + (aq) 
iii) Equilibrium between solid and gas.
 NH (g) + HCl(g)
NH4Cl(s)  3

For the existence of heterogeneous equilibrium, pure solids or liquids must also be present at equilibrium, but
their concentrations or partial pressures do not appear in the expression for the equilibrium constant, i.e., if a
substance X is involved, then X(s) or X(l) is constant, irrespective of the amount of X taken. Contrary to this,
X(g) and X(aq) will vary as the amount of X in a given volume varies. For example,
[CaO(s)][CO 2 (g)]


CaCO3 (s)  CaO(s)  CO 2 (g) ; Kc 
[CaCO3 (s)]
Since [CaCO3(s)] and [CaO(s)] are constant, the equilibrium can be modified as
K'c = [CO2(g)] or K p  PCO2
Thus, at a particular temperature, there is a constant concentration/pressure of CO2 in equilibrium with CaO(s)
and CaCO3(s). At 1100 K, the pressure of CO2 in equilibrium with CaCO3(s) and CaO(s), is 2.0 ×105 Pa and
the equilibrium constant
K p  PCO 2  2  105 Pa / 105 Pa  2.0
Consider the reaction of steam on heated iron in a closed vessel. The following equilibrium is set up :

P 4 H 2  PFe3O4
 Fe O (s) + 4H (g) ; K 
3Fe(s) + 4H2O(g)  3 4 2 p
P3Fe  P 4 H2O

But since active mass of solids is taken to be unity,

P4H2 PH2
Kp  4 or
4 Kp 
P H2O PH2O

Thus, according to law of chemical equilibrium, the ratio of partial pressure of hydrogen to that of steam at
equilibrium should be constant.
7.6 APPLICATIONS OF EQUILIBRIUM CONSTANTS
Important features of equilibrium constants:
(i) It is applicable only when concentrations of the reactants and products have attained equilibrium state.
(ii) It is independent of the initial concentrations of the reactants and products.
(iii) It is temperature dependent, having a unique value for a particular reaction at a given temperature.
(iv) Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the forward
reaction.
(v) The equilibrium constant for a reaction is related to the equilibrium constant of the corresponding reaction,
whose equation is obtained by multiplying or dividing the equation for the original reaction by a small
integer.
Equilibrium constant can be used to predict the extent of a reaction, direction of reaction and to calculate
equilibrium concentrations.

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7.6.1 Predicting the Extent of a Reaction

The magnitude of Kc or Kp is directly proportional to the concentrations of products and inversely proportional
to the concentrations of the reactants. Thus, a high value of K indicates high concentration of products.
(a) Kc > 103. If Kc > 103, products predominate over reactants, i.e., if Kc is very large, the reaction proceeds
nearly to completion.
 2H O(g) at 500 K;
i. 2H2 (g) + O2(g)  Kc = 2.4 × 1047.
2

 2HCl(g) at 300K;

ii. H2(g) + Cl2(g)  Kc = 4.0 × 1031.
(b) Kc < 10–3. If Kc < 10–3, reactants predominate over products, i.e., if Kc is very small, the reaction
proceeds rarely.
 2H (g) + O (g) at 500 K;
i. 2H2O (g)  Kc = 4.1 × 10–48.
2 2

 2NO(g), at 298 K;

ii. N2(g) + O2(g)  Kc = 4.8 ×10–31.
(c) Kc = 10–3 to 103. If Kc is in the range of 10–3 to 103, appreciable concentrations of both reactants and
products are present.
 2HI(g) at 700K;
i. H2 (g) + I2 (g)  Kc = 57.0 .
 2NO (g) at 298K;
ii. N2O4 (g)  Kc = 4.64 × 10–3
2

The above values of Kc are neither too small nor too large. Hence, equilibrium mixtures contain appreciable
concentrations of both products and reactants.

Figure - 03 Dependence of extent of reaction on Kc.

7.6.2 Predicting the Direction of Reaction
Equilibrium constant helps to predict the direction in which a reaction will proceed at any stage. For this, the
reaction quotient Q (Qc with molar concentrations and QP with partial pressures) is calculated. Reaction
quotient is similar to equilibrium constant Kc except that concentrations are not necessarily equilibrium values.
[C]c [D]d
 c C + d D ; Qc =
a A + b B 
[A]a [B]b
i) If Qc < Kc, the reaction will proceed in the direction of the products (forward reaction).
ii) If Qc > Kc, the reaction will proceed in the direction of reactants (reverse reaction).
iii) If Qc = Kc, the reaction mixture is at equilibrium.

Figure - 04 Predicting the direction of reaction.

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For example, consider the gaseous reaction, H2(g) + I2(g)   2HI(g) ; K = 57.0 at 700 K.
c
Suppose the molar concentrations at time ‘t’ be [H2]t = 0.10M, [I2]t = 0.20 M and [HI]t = 0.40M. The
reaction quotient, Qc at this stage of the reaction is given by,
[HI]2t (0.40)2
Qc =  = 8.0
[H2 ]t [I2 ]t (0.10)(0.20)
As Qc (8.0) is less than Kc (57.0), H2(g) and I2(g) will react to form HI(g) untill Qc = Kc.
7.6.3 Calculating Equilibrium Concentrations
Knowing the initial concentrations, the equilibrium concentrations can be calculated as follows:
Step 1. Write the balanced equation for the reaction.
Step 2. Under the balanced equation, make a table that lists the following details about the substances involved
in the reaction : (i) initial concentration, (ii) change in concentration on going to equilibrium, and (iii) equilibrium
concentration.
Assume ‘x’ as the amount of one of the substances that reacts to attain equilibrium and then use the stoichiometry
of the reaction to determine the concentrations of the other substances in terms of ‘x’.
Step 3. Substitute the equilibrium concentrations in the equilibrium equation and solve for ‘x’. On solving a
quadratic equation, choose the mathematical solution that makes chemical sense.
Step 4. Calculate the equilibrium concentrations from the value of x.
Step 5. Check the results by substituting them in the equilibrium equation.
7.7 EQUILIBRIUM CONSTANT, REACTION QUOTIENT, AND GIBBS ENERGY
The value of Kc for a reaction does not depend on the rate of the reaction. However, it is directly related to the
change in Gibbs energy, G.
i) If, G is negative, the reaction is spontaneous and proceeds in the forward direction.
ii) If G is positive, the reaction is non-spontaneous. As the reverse reaction will have a negative value for G,
products of the forward reaction shall be converted back to reactants.
iii) If G is 0, the reaction is in equilibrium. At this point, there is no free energy left to drive the reaction.
The thermodynamic view of equilibrium can be described by the following equation:
G = G + RT lnQ
where, G is the standard Gibbs energy and Q is the reaction quotient.
At equilibrium, when G = 0 and Q = Kc, the above equation becomes,
G = G + RT ln K = 0 or G = – RT ln K or ln K = – G / RT
Taking antilog on both sides,
G / RT G / 2.303 RT
K = e– or K = 10–
Using this equation, the reaction spontaneity can be interpreted in terms of G .

If G , then –G /RT is positive, and e–G / RT >1, giving K >1. This implies a spontaneous reaction or
a reaction which proceeds in the forward direction to such an extent that products are formed predominantly.

If G > 0, then –G /RT is negative, and e–G / RT < 1, that is , K < 1. This implies a non-spontaneous
reaction or a reaction which proceeds in the forward direction to a small degree that only a very minute
quantity of product is formed.

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7.8 FACTORS AFFECTING EQUILIBRIA

French chemist, Le Chatelier studied the effect of concentration, temperature and pressure on chemical equilibria.
Le Chatelier’s principle states that when a system at equilibrium is subjected to a change of concentration,
pressure or temperature, the equilibrium shifts in the direction that tends to undo the effect of the change.
7.8.1 Effect of Change of Concentration
In a chemical equilibrium, increasing the concentration of reactants shifts the equilibrium in favour of the products
while increasing the concentration of products shifts the equilibrium in favour of the reactants. For example,
 2HI(g)
H2(g) + I2(g) 
If H2 is added to the reaction mixture at equilibrium, the equilibrium shifts in the forward direction wherein H2
is consumed, i.e., more of H2 and I2 react to form HI.

Figure - 05 Effect of addition of H2 on concentration of reactants and products in H2(g) + I2 (g)  2HI(g).
This can be explained in terms of the reaction quotient,
[HI]2
Qc =
[H 2 ] [I2 ]
Addition of H2 at equilibrium results in the value of Qc becoming less than Kc. In order to attain equilibrium
again, reaction occurs in the forward direction. Removal of a product also enhances the forward reaction.
In the manufacture of ammonia, ammonia gas formed is liquified and removed from the reaction mixture so that
the reaction proceeds in the forward direction.
N2(g) +3H2(g)  2NH (g)
3
Continuous removal of products keeps Qc below Kc so that the reaction continues in the forward direction.
In the production of CaO from CaCO3, removal of CO2 from the kiln drives the reaction to completion.
CaCO3(s)   CaO(s) + CO (g)
2
The equilibrium can be shifted in either direction by adding or removing a reactant or product.
7.8.2 Effect of Change of Pressure
Low pressure favours reactions which are accompanied by increase in the total number of moles and high
pressure favours reactions which are accompanied by decrease in the total number of moles. Pressure has no
effect on reactions which proceed with no change in the number of moles.
In the following reaction, 4 mols of gaseous reactants give 2 mols of gaseous products.
 CH (g) + H O(g)
CO(g) + 3H2(g)  4 2

If the system is compressed to half its volume, at constant temperature, the pressure is doubled.
According to Le Chatelier’s principle, increase in pressure shifts the equilibrium to the right, i.e., increases the
equilibrium concentration of the products.
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The reaction quotient is calculated by replacing each equilibrium concentration by double its value.
[CH4 (g)] [H2O(g)]
Qc =
[CO(g)] [H2 (g)]3
As Qc < Kc , the reaction proceeds in the forward direction.
In the synthesis of ammonia, the forward reaction is accompanied by decrease in the total number of moles.
 2NH (g)
N2(g) + 3H2(g)  3

Increase in pressure shifts the equilibrium to the right, i.e., increases the equilibrium concentration of ammonia.
In the synthesis of hydrogen iodide, there is no net change in the number of moles.
 2NH (g)
N2(g) + 3H2(g)  3

Therefore, pressure has no effect on the equilibrium.

In the following reaction, there is a net increase in the number of moles (n =1)
 2CO(g)
C(s) + CO2(g) 
Increase in pressure shifts the equilibrium to the left, i.e., decreases the equilibrium concentration of CO2.
7.8.3 Effect of Addition of Inert Gas
If the volume is kept constant, the addition of an inert gas (which does not take part in the reaction) does not
alter the equilibrium. This is because the addition of inert gas at constant volume does not change the partial
pressures or molar concentrations of the substances involved in the reaction.
The reaction quotient changes only if the added gas is a reactant or product involved in the reaction.
7.8.4 Effect of Change of Temperature
Change in temperature affects the value of the equilibrium constant, Kc and the rates of reaction. The temperature
dependence of equilibrium constant depends on the sign of H for the reaction.
The equilibrium constant for an exothermic reaction (H -ve) decreases with increase in temperature. However,
the equilibrium constant for an endothermic reaction (H +ve) increases with increase in temperature.
i) The synthesis of ammonia is an exothermic process.
 2NH (g) ; H= – 92.38 kJ mol–1
N2(g) + 3H2(g)  3

According to Le Chatelier’s principle, increase in temperature shifts the equilibrium to the left, i.e., decreases
the equilibrium concentration of ammonia. Thus, low temperature favours high yield of ammonia, but slows
down the reaction. Therefore, a catalyst is required.
ii) The effect of temperature on equilibrium can be demonstrated in the laboratory by the dimerisation of NO2
gas (brown) to N2O4 gas (colourless).

 N 2O 4 (g); H = -57.2 kJ mol-1

2NO2 (g) 
Re ddish brown Colourless

As the reaction is exothermic, low temperature favours the forward reaction, i.e., formation of colourless
N2O4. High temperature favours the reverse reaction, i.e., formation of brown NO2 (brown colour intensifies).
iii) Effect of temperature on endothermic reaction. Formation of [CoCl4]2- from [Co(H2O)6]3+ is endothermic.

 [CoCl 4 ]2- (aq) +6H 2O(l )

[Co(H 2 O)6 ]3+ (aq) + 4Cl- (aq) 
Pink Colourless Blue

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At room temperature, the equilibrium mixture is blue due to the formation of [CoCl4]2- by the forward reaction.
When cooled in a freezing mixture, the colour turns pink due to the formation of [Co(H2O)6]3+.
7.8.5 Effect of Catalyst
A catalyst increases the rate of a chemical reaction by providing a low energy pathway for the conversion of
reactants into products. It increases the rate of the forward and the reverse reactions that passes through the
same transition state and does not affect the equilibrium.
Manufacture of ammonia. The formation of NH3 from N2 and H2, is highly exothermic and proceeds with
decrease in the total number of moles of products. According to Le Chatelier’s principle, low temperature and
high pressure should favour the formation of NH3.
It has been observed that at low temperatures, the rate is very low, whereas at high temperatures the reaction
is fast but the yield is poor.
A catalyst consisting of iron catalyse the reaction to occur at satisfactory rates at moderate temperatures with
satisfactory yields (German chemist, Fritz Haber). Thus, optimum conditions for the synthesis of NH3 using
catalyst are around 500°C and 200 atm.
Manufacture of sulphuric acid. Platinum or divanadium pentoxide (V2O5) is used as catalyst for the oxidation
of SO2 to SO3.
 2SO (g);
2SO2(g) + O2(g)  Kc = 1.7 × 1026
3

Though the value of Kc is suggestive of the reaction going to completion, the reaction is very slow. If the rate
constant for a reaction is exceedingly small, a catalyst will be of little help.
7.9 IONIC EQUILIBRIUM IN SOLUTION
Michael Faraday (1824) classified substances into electrolytes and nonelectrolytes on the basis of conductivity
of their aqueous solutions. He further classified electrolytes into strong and weak electrolytes. Svante Arrhenius
(1880) found that electrolytes when dissolved in water, split into charged particles called ions. This process is
called ionisation or dissociation.
Strong electrolytes like NaCl, KCl, HCl, HNO3, KOH, NaOH, NH4Cl etc., on dissolution in water are
almost completely ionised. They are excellent conductors of electricity.
Weak electrolytes are only partially ionised in solution, e.g., CH3COOH, H3PO4, NH4OH, H3BO3 etc. They
are poor conductors of electricity. For example, ionisation of acetic acid in aqueous solution is less than 5%.
 H 3O  (aq)  CH 3COO  (aq)
CH 3COOH(aq)  H 2O(l ) 
In the case of weak electrolytes, an equilibrium is established between ions and the unionised molecules. This
type of equilibrium involving ions in aqueous solution is called ionic equilibrium.
Degree of ionisation. Degree of ionisation,  may be defined as the fraction of the total number of molecules
on an electrolyte which dissociate into ions.
% ionisation Number of molecules dissociated int o ions
 
100 Tota l number of molecules in solution
7.10 ACIDS, BASES AND SALTS
Acids, bases and salts come under the category of electrolytes and may act as either strong or weak electrolytes.
The word acid has been derived from a latin word acidus meaning sour. Acids turn blue litmus paper into red
and liberate dihydrogen on reacting with some metals. Bases turn red litmus paper blue, taste bitter and feel
soapy. Acids and bases react with each other to give salts.

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7.10.1 Arrhenius Concept of Acids and Bases

According to Arrhenius, acids are substances that dissociate in water to give hydrogen ions H+(aq) and bases
are substances that produce hydroxyl ions OH-(aq).
The ionisation of an acid HX in aqueous solution may be represented as
 H+(aq) + X– (aq)
HX (aq) 
A proton, H+ is very reactive and cannot exist freely in aqueous solution. Therefore, it bonds to the oxygen
atom of a water molecule to give hydronium ion, H3O+.
 H O+(aq) + X–(aq)
HX(aq) + H2O(l)  3

Acids such as HCl and HNO3, which are almost completely ionised in aqueous solution are termed strong
acids, whereas acids such as CH3COOH which are weakly ionised are called weak acids.
 H  (aq)  Cl  (aq) ; HNO3 (aq) 
HCl(aq)    
 H (aq)  NO3 (aq)

 H  (aq)  CH 3COO  (aq)

CH 3COOH(aq) 
Similarly, the ionisation of a base, MOH in aqueous solution may be represented as
 M+(aq) + OH–(aq)
MOH(aq) 
The hydroxyl ion also exists in the hydrated form in aqueous solution.
Bases which are almost completely ionised in aqueous solution are called strong bases, e.g., NaOH and
KOH. Bases such as NH4OH are only weakly ionised are called weak bases.
 Na+(aq) + OH–(aq) ; NH OH(aq) 
NaOH(aq)   NH +(aq) + OH–(aq)
4 4

According to this theory, neutralisation of acids and bases is the reaction between H+ and OH- ions in solution.
 H O(l)
H+(aq) + OH-(aq)  2

Hydronium and Hydroxyl Ions: The charge density on H+ ion is very high, therefore, it is present in the
hydrated form in combination with a water molecule as H3O+. In aqueous solution, the hydronium ion is further
hydrated to give species like H5O2+, H7O3+ and H9O4+.
Similarly the hydroxyl ion is hydrated to give several ionic species like H3O2–, H5O3– and H7O4–, etc.
Limitations of Arrhenius concept: Arrhenius concept is applicable only to aqueous solutions. It does not
account for the basicity of substances like ammonia which do not possess hydroxyl group.
7.10.2 Proton Transfer Theory - Concept of Lowry and Brönsted
According to Danish chemist, Johannes Brönsted and English chemist, Thomas M. Lowry (1923), acid is a
substance capable of donating a hydrogen ion, H+ and base is a substance capable of accepting a hydrogen
ion. Thus, an acid is a proton-donor (protogenic) and a base is a proton-acceptor (protophilic).
In the ionisation of HCl in water, HCl molecule acts as proton donor and H2O molecule acts as proton
acceptor and are thus, Lowry-Brönsted acid and base, respectively.

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In the reverse reaction, H+ is transferred from H3O+ to Cl–. In this case, H3O+ acts as the Bronsted acid while
Cl– acts as the Brönsted base.
Thus, Cl– is the conjugate base of the acid HCl and HCl is the conjugate acid of the base Cl–. Similarly, H2O
is the conjugate base of the acid H3O+ and H3O+ is the conjugate acid of the base H2O.
Thus, any acid-base reaction involves two conjugate pairs, i.e., when an acid reacts with a base, another acid
and base are formed.
Acid1  Base 2   Base1  Acid 2
An acid-base pair that differs only by one proton is called a conjugate acid-base pair.
In the dissolution of NH3 in H2O, basic solution is formed due to the presence of hydroxyl ions. Water acts as
proton donor and ammonia acts as proton acceptor, i.e., Lowry-Brönsted acid and base, respectively.

In the reverse reaction, H+ is shifts from NH4+ to OH–. NH4+ acts as Bronsted acid while OH– acts as Brönsted
base. Thus, OH– is the conjugate base of the acid, H2O and NH4+ is the conjugate acid of the base, NH3.
If the Brönsted acid is a strong acid, its conjugate base is a weak base and vice versa.

HCl(aq)  H 2O(l ) 

 H 3O  (aq)  Cl  (aq)
Acid (strong) Base (weak) Conjugate acid (strong) Conjugate base (weak)

CH3COOH(aq)  OH  (aq) 

 H 2O(l )  CH3COO  (aq)
Acid (weak) Base (strong) Conjugate acid (weak) Conjugate base (strong)

The strength of an acid depends on its tendency to lose a proton and the strength of a base depends on its
tendency to gain a proton.
The dual role of water as acid and base. In the reaction with HCl, water acts as a base by accepting a
proton while in case of ammonia it acts as an acid by donating a proton. Such a solvent which has dual
character of acid and base is called amphiprotic or amphoteric solvent.
H H

H 2 O  OH 
  ; 
H 2 O  H3O
 
H Base H
Acid Base Acid
Influence of solvent on acid strength. The capacity of the acid to dissociate also depends on the basic
strength of the solvent which acts as base. The acids, HClO4, H2SO4, HCl and HNO3 which have nearly the
same strength in water are in the order HClO4 > H2SO4 > HCl > HNO3 in acetic acid, since the proton
accepting tendency of acetic acid is much less than water. On the basis of proton interaction, solvents can be
classified into four types:
i. Protophilic solvents: Solvents which have great tendency to accept protons, e.g., water, alcohol, liquid
ammonia, etc.
ii. Protogenic solvents: Solvents which have the tendency to produce protons, e.g., water, liquid hydrogen
chloride, glacial acetic acid, etc.
iii. Amphiprotic solvents: Solvents which act both as protophilic or protogenic, e.g., water, ammonia, ethyl
alcohol, etc.
iv. Solvents which neither donate nor accept protons, e.g., benzene, CCl4, CS2, etc.
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HCl acts as strong acid in H2O, stronger acid in NH3, weak acid in CH3COOH, neutral in C6H6, and weak
base in HF.

HCl  HF 
 H 2 Cl  F
Base Acid Acid Base

Influence of solvent on base strength. The nature of solvent plays an equally important role in the dissociation
and relative strengths of bases as well. For example, aniline, a very weak base and sodium hydroxide, a very
strong base in water appear to be equally strong in acetic acid.
7.10.3 The Lewis Concept
G. N. Lewis (1923) defined an acid as a species which accepts a pair of electrons and base as a species which
donates a pair of electrons. According to Lewis concept, many acids do not furnish protons, e.g., the reaction
of electron deficient BF3 with NH3. BF3 has no protons, but still it acts as an acid and reacts with NH3 by
accepting a lone pair of electrons.
BF3  : NH 3 
 BF3 : NH 3
Thus, Lewis definition does not attribute acidity to any particular element but rather to unique electronic
arrangement.
Lewis acids : (a) Compounds in which the central atom has incomplete octet, e.g., BF3, BCl3, AlCl3, FeCl3,
GaCl3, etc.
(b) Compounds in which the central atom has vacant d-orbitals, e.g., SnF4, SnCl2, SnCl4, PF, PF4, SF4, TiCl4,
etc.
(c) Simple cations such as Ag+, Cu2+, Al3+, Co3+, Mg2+, etc.
(d) Compounds which have multiple covalent bonds, e.g., CO2, CS2, SO2, etc.
Lewis bases: (a) All simple anions, e.g., NO3-, Cl-, OH-, etc.
(b) Molecules containing unshared pairs of electrons (lone pairs), e.g., H2O, NH3, ROH, NX3, R2S, etc.
(c) Multiple bonded compounds which can donate the lone pair of electrons, e.g., CO, NO, C2H4, C2H2, etc.
All Bronsted bases are also Lewis bases but all Bronsted acids are not Lewis acids. Lewis bases generally
contain one or more lone pairs of electrons and, therefore, they can also accept a proton (Bronsted base).
Thus, all Lewis bases are also Bronsted bases. On the other hand Bronsted acids are those which can give a
proton, e.g., HCl, H2SO4, HNO3, etc. They may not be capable of accepting a pair of electrons, hence, all
Bronsted acids are not Lewis acids.
7.11 IONISATION OF ACIDS AND BASES
Acids like HClO4, HCl, HBr, HI, HNO3, H2SO4 and bases like LiOH, NaOH, KOH, CsOH, Ba(OH)2 are
almost completely ionised in aqueous medium. According to Arrhenius concept, they are strong acids and
bases as they dissociate completely and produce H3O+ and OH– ions respectively in the medium.
The strength of an acid or base may also be gauged in terms of Brönsted Lowry concept wherein a strong acid
is a good proton donor and a strong base is a good proton acceptor.
The conjugate base of a strong acid is a weak base and the conjugate base of a weak acid is a strong base.
Similarly, the conjugate acid of a strong base is a weak acid and the conjugate acid of a weak base is a strong
acid.
Strong acids like perchloric acid (HClO4), hydrochloric acid (HCl), hydrobromic acid (HBr), hydroiodic acid
(HI), nitric acid (HNO3) and sulphuric acid (H2SO4) will give conjugate base ions ClO4–, Cl–, Br–, I–, NO3–
and HSO4– , which are much weaker bases than H2O.
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Weak acids like nitrous acid (HNO2), hydrofluoric acid (HF) and acetic acid (CH3COOH) have very strong
conjugate bases. For example, NH2–, O2– and H– are very good proton acceptors and thus, much stronger
bases than H2O.
A very strong base would give a very weak conjugate acid.
Certain water soluble organic compounds like phenolphthalein and bromothymol blue behave as weak acids
and exhibit different colours in their acid (HIn) and conjugate base (In –) forms.

HIn(aq)  H 2O (l ) 
 H3O  (aq)  In  (aq)
Acid Base Conjugate acid Conjugate base
Indicator (colour A) (colour B)

Such compounds are useful as indicators in acid-base titrations and finding out H+ ion concentration.
7.11.1 The Ionisation Constant of Water and its Ionic Product
Water can act both as acid and base. In presence of acid, it accepts a proton and acts as the base while in
presence of a base, it acts as an acid by donating a proton. In pure water, one molecule can donate proton and
act as acid while another molecule can accept the proton and act as base.

H 2 O(aq)  H 2 O (l ) 
 H3O  (aq)  OH  (aq)
Acid Base Conjugate acid Conjugate base

[H 3O  ][OH  ]
The dissociation constant, K 
[H 2 O]
As water is a pure liquid and its concentration remains constant, it is omitted in the denominator. [H2O] is
incorporated in the equilibrium constant to get a new constant, Kw, called the ionic product of water.
Kw = [H+][OH–]
The concentration of H+ has been found out experimentally as 1.0 × 10–7 M at 298 K.
As dissociation of water produces equal number of H+ and OH– ions, the concentration of hydroxyl ions,
[OH–] = [H+] = 1.0 × 10–7 M.
Thus, the value of Kw at 298K, Kw = [H3O+][OH–] = (1 × 10–7)2 = 1 × 10–14 M2
As it is an equilibrium constant, the value of Kw is temperature dependent.
Molarity of pure water, [H2O]. Density of pure water = 1000 g / L
Molar mass of water = 18.0 g /mol.

1000 g L1
Therefore, molarity of pure water, [H2O] = = 55.55 M.
18 g mol 1

107 mol L1

Degree of dissociation of water  = 1.8 × 10–9
55.55 mol L1
Thus, the equilibrium lies mainly towards undissociated water.
Acidic, neutral and basic aqueous solutions can be distinguished by the relative values of the H3O+ and OH- ion
concentrations: [H3O+] > [OH–] for acidic solution, [H3O+] = [OH–] for neutral solution and [H3O+] < [OH–]
basic solution.

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7.11.2 The pH Scale

Hydronium ion concentration in molarity can be expressed on a logarithmic scale known as the pH scale
(potenz de hydrogen in Danish means power of hydrogen -P L Sorensen, 1909). The pH of a solution is
defined as the negative logarithm to base 10 of the activity (aH+ ) of hydrogen ion. In dilute solutions (<
0.01 M), activity of H+ is equal in magnitude to molarity. Activity has no units and is defined as:
a H   [H  ] / mol L1
pH can be related to activity as:
[H  ] 1
pH   log a H    log 1 or pH  log  or pH   log[H  ]
mol L [H ]
The concentration of [H3O ] can be written as:
+

[H3O  ]  110 pH
Thus, pH is the negative power to which 10 must be raised in order to express the hydronium ion
concentration of a solution.
Just as pH indicates the hydrogen ion concentration, pOH indicates the hydroxyl ion concentration.
pOH   log[OH  ]
But [H  ][OH  ]  K w  1 1014
Taking logarithm on both sides,
log[H  ]  log[OH  ]  log K w  log (1 1014 )
or  log[H  ]  log[OH  ]   log K w   log (11014 )
or pH  pOH  pK w  14
Thus, the sum of pH and pOH is equal to 14 in any aqueous solution at 298K.
pKw controls the relative concentrations of hydrogen and hydroxyl ions as their product is a constant.
i. For pure water or neutral solutions at 298K, the concentration of hydrogen ions, [H+] = 10–7 M. Hence, pH
of pure water or neutral solutions is given as: pH = –log(10–7) = 7
ii. For acidic solutions, [H3O+] concentration is greater than 10–7 M. Therefore acidic solutions have pH < 7.
The pH of an acidic solution of 10–2 M HCl will be
pH=-log [H+]=-log [10-2] =2.
+ - pH pOH
[H ] [OH ]
-7 -7
Acidic solution > 10 < 10 <7 >7
-7 -7
Neutral solution 10 10 7 7
-7 -7
Basic solution < 10 > 10 >7 <7
iii.For basic solutions, pH > 7.
As the pH scale is logarithmic, a change in pH by one unit means change in [H+] by a factor of 10. That is why
change in pH with temperature is often ignored.

Figure - 06 Range of pH and [H3O+].

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Measurement of pH. The pH of a solution can be roughly estimated using pH paper that has different colour
in solutions of different pH. Now-a-days pH papers are available with four strips on it. The different strips
have different colours at the same pH. The pH in the range of 1-14 can be determined with an accuracy of
~0.5 using pH paper.
For greater accuracy, pH meters are used. pH meter is a device that measures the pH-dependent electrical
potential of the test solution within 0.001 precision.
Table - 05 The pH of some common substances.
Substance pH range Substance pH range Substance pH range
Gastric juice 1.0 - 3.0 Urine (Human) 4.8 - 8.4 NaOH (sat. sol) ~15
Soft drinks 2.0 - 4.0 Saliva (Human) 6.5 - 7.5 0.1 M NaOH 13
Lemon 2.2 - 2.4 Blood plasma 7.3 - 7.5 Lime water 10.5
Vinegar 2.4 - 3.4 Milk (Cow) 6.3 - 6.6 Egg white 7.8
Apple 2.9 - 3.3 Urine (Cow) 7.4 - 8.4 Black coffee 5
1 M HCl ~0 Milk of magnesia 10.5 Tomato juice ~4.2
Conc. HCl ~-1 Sea water 8.5 Ground water 6.0-8.5
pH of Strong acids or Strong Bases
i. In an aqueous solution of a strong acid, the acid is the only source of H+ (H3O+) ions unless it is extremely
dilute (e.g., < 10-6 M).
ii. In an aqueous solution of a strong base, the base is the only source of OH- ions unless the solution is
extremely dilute (e.g., < 10-6 M).
iii. Total H3O+ or OH- in a mixture of strong acids or strong bases is given by
[H3O ]  NV / V and [OH  ]  NV / V
where N is the normality and V is the volume of the solutions. The pH can be calculated accordingly.
Limitations of pH Scale
i) The pH values do not give an exact idea about the relative strengths of solutions.
A solution of pH =1 has hydrogen ion concentration 100 times that of a pH=3 solution. A 4.0× 10-5 N HCl
solution is twice concentrated than a 2.0 × 10-5 N solution, but the pH values are 4.4 and 4.7 respectively.
ii) pH=0 is obtained for 1N solution of strong acid. In case the concentrations are 2N, 3N, 10N, etc, the
respective pH values will be negative. Similarly, for a concentrated solution of a strong base (>1M), pH can be
greater than 14. For example, for 10 N NaOH solution, [OH-] = 10 M or [H+] = 10-15 M so that pH = 15.
iii) An acid solution of very low concentration, say 10-8 N cannot have pH = 8, as indicated by the formula, but
the actual value will be slightly less than 7.
Similarly, a 10-8 M NaOH solution cannot have pH = 6. The value will be slightly greater than 7.
7.11.3 Ionisation Constants of Weak Acids
The equilibrium in case a weak monobasic acid, HX in aqueous solution can be represented as:
 H3O  (aq)  X  (aq)
HX(aq)  H 2 O(aq) 
Initial concentration [M] c 0 0
Let extent of ionisation be  [M] -c +c +c
Equilibrium concentration (M) c- c c c
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where c is the concentration of undissociated HX at time t = 0 and is the extent of ionisation of HX.
The equilibrium constant for the acid dissociation equilibrium is

c 2 2 c 2
Ka  
c(1   ) 1  
where Ka is the dissociation or ionisation constant of the acid.
Since for weak acids, is very small, 1- in the denominator may be taken as 1. Thus,

Ka = c2 or   K a / c ...(i)
It can also be represented in terms of molar concentration as Ka = [H ][X ] / [HX]
+ -

At temperature T, Ka is a measure of the strength of the acid HX. The larger the value of Ka, the stronger is the
acid. Ka is dimensionless since standard state concentration of all species is 1M.
Table - 06 The ionisation constants of some weak acids at 298K.
Acid Ka Acid Ka Acid Ka
-4 –5 –5
Hydrofluoric Acid 3.5 × 10 Acetic Acid 1.74 × 10 Benzoic Acid 6.5 × 10
-4 –8 –10
N itrous Acid 4.5 × 10 Hypochlorous Acid 3.0 × 10 P henol 1.3 × 10
-4 -10
Formic Acid 1.8 × 10 Hydrocyanic Acid 4.9 × 10 Niacin (C 5H 4 NCOOH) 1.5 × 10–5
The pH scale has been extended to other species and quantities. Thus, we have: pKa = –log (Ka)
Knowing the ionisation constant, Ka of an acid and its initial concentration, c, it is possible to calculate the
equilibrium concentrations of all species, the degree of ionisation of the acid, and the pH of the solution.

7.11.4 Ionisation of Weak Bases

The ionisation of a base MOH can be represented as:
 M+(aq) + OH–(aq)
MOH(aq) 
In a weak base, there is only partial ionization of MOH into M+ and OH–. The equilibrium constant for base
ionisation is called base ionisation constant and is represented by Kb. It can be expressed in terms of concentration
in molarity of various species in equilibrium.

[M  ][OH  ]
Kb 
[MOH]
If c is the initial concentration of base and  is its degree of ionisation, the equilibrium constant

(c) 2 c 2
Kb  
c(1  ) (1  )
Since for a weak base, is very small (1-)  1
K b  c 2 or   Kb / c

[OH  ]  c  c K b / c  cK b
The pH scale for the hydrogen ion concentration has been extended to get: pKb = –log (Kb).

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Table - 07 The Ionisation Constants of Some Weak Bases at 298K.

Base s Kb Base s Kb Base s Kb
-5 -5 –9
Ammonia 1.77× 10 Triethylamine 6.45 × 10 P yridine 1.77 × 10
–4 -10 –8
Methyl amine 3.7 × 10 Aniline 4.27 × 10 Hydroxylamine 1.1 × 10
-4 -6 –14
Dimethyl amine 5.4 × 10 Hydrazine 1.7× 10 Urea 1.3 × 10
7.11.5 Relation between Ka and Kb
Ka and Kb represent the strength of an acid and a base, respectively. A conjugate acid-base pair is related in
a simple manner so that if one is known, the other can be deduced. For example, in case of NH4+ and NH3,

 H O+(aq) + NH (aq) ; K  [H3O ][NH3 ]  5.6  1010
NH4+(aq) + H2O(l)  a
[NH 4 ]
3 3

 
 NH +(aq) + OH–(aq) ; K  [NH 4 ][OH ]  1.8  105
NH3(aq) + H2O(l)  4 b
[NH3 ]

Net:  H O+(aq) + OH–(aq) ; K = [H O+][ OH–] = 1.0 × 10–14 M

2H2O(l)  3 w 3

where, Ka is the strength of NH4+ as acid and Kb is the strength of NH3 as base.
From the net reaction, it can be seen that the equilibrium constant is equal to the product of the equilibrium
constants Ka and Kb for the reactions added.

[H3O  ][NH 3 ] [NH 4 ][OH  ]

Ka  Kb   = [H3O+][ OH–] = Kw
[NH 4  ] [NH3 ]

Kw = (5.6×10–10) × (1.8 × 10–5) = 1.0 × 10–14 M

Thus, the equilibrium constant for the net reaction obtained by the addition of two (or more) reactions is equal
to the product of the equilibrium constants for the individual reactions.
KNET = K1 × K2 × .....
Similarly, in case of a conjugate acid-base pair,
Ka × Kb = Kw
The above expression, Kw = Ka × Kb, can also be obtained from the base-dissociation equilibrium,
 
 BH+(aq) + OH–(aq) ; K  [BH ][OH ]
B(aq) + H2O(l)  b
[B]
As the concentration of water remains constant, it can be omitted from the denominator and incorporated
within the dissociation constant. Multiplying and dividing the above expression by [H+],
[BH  ][OH  ][H  ]   [BH ]

Kb  or K b  [OH ][H ] or Kb = Kw / Ka
[B][H  ] [B][H  ]
Ka × Kb = Kw
Taking negative logarithm of the above equation, pK values of conjugate acid-base pair are related as :
pKa + pKb = pKw = 14 (at 298K)
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7.11.6 Dissociation constants of Di- and Polybasic acids

Polybasic acids contain more than one replacable hydrogen. They always dissociate in stages.
[H  ][HCO3 ]
 H3O   HCO3 ; K a1 
H 2 CO3  H 2 O   K a1 = 4.53  10-7
[H 2 CO3 ]

[H  ][CO32 ]

 H3O  CO3 ; K a 2
HCO3  H 2O   2   K a 2 = 5.6  10-11
[HCO3 ]

The second dissociation constant is found to be 1/10000th of the first dissociation constant.
K a 2 is always smaller than K a which indicates that the second dissociation takes place to a much smaller
1

extent than the first dissociation.

Phosphoric acid (tribasic acid) dissociates in three stages:
 H 3O   H 2 PO 4  ;
H 3 PO 4  H 2O  K a1  7.52  103

 H 3 O   HPO 4 2  ; K a  6.23  10 8

H 2 PO 4   H 2 O  2

 H 3O  PO 43 ;

HPO 4 2   H 2O  K a3  4.20 1013
The reason for the decrease in dissociation constant of the successive stages is the increase in negative charge
on the species from which H+ ion has to detach itself from.
7.11.7 Factors Affecting Acid Strength
The extent of dissociation of an acid HA depends on the strength and polarity of the H-A bond. In general, HA
becomes a stronger acid when strength of H-A bond decreases.
In any group of the periodic table, H-A, bond strength is the more important factor in determining acidity than
its polar nature. As the size of A increases down the group, H-A bond strength decreases and acid strength
increases. For example, size as well as acid strength increases in the order:
HF << HCl << HBr << HI)
Similarly, H2S is a stronger acid than H2O
H2O < H2S < H2Se < H2Te
In any period, H-A bond polarity is the deciding factor for determining acid strength. As the electronegativity
of A increases, the strength of the acid also increases.
CH4 < NH3 < H2O < HF
Among oxoacids, the strength of acid increases as the electronegativity of the halogen increases.
HClO > HBrO > HIO ; HClO2 > HBrO2 > HIO2
HClO3 > HBrO3 > HIO3 ; HClO4 > HBrO4 > HIO4
Among oxoacids of the same halogen, acidic strength is in the following order:
HXO4 > HXO3 > HXO2 > HXO ; HClO4 > HClO3 > HClO2 > HClO
This is because the stability of conjugate base due to resonating structures is in the order:
ClO4- > ClO3- > ClO2- > ClO-
7.11.8 Common Ion Effect in the Ionisation of Acids and Bases
The suppression of the dissociation of a weak acid or a weak base on the addition of its own ions is called
common ion effect. This phenomenon, based on Le Chatelier’s principle.
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Consider the dissociation equilibrium of acetic acid,

CH COOH(aq)   H+(aq) + CH COO– (aq)
3 3

  K [Salt]
 H+ (aq) + Ac– (aq) ; K  [H ][Ac ] or [H ]  a

or HAc(aq)  a
[HAc] [Acid]
Addition of acetate ions to acetic acid solution results in decrease of concentration of H+ ions. If H+ ions are
added insted, the equilibrium shifts in the direction of undissociated acetic acid. As a result of common ion
effect, the concentration of the ion not in common in the two electrolytes is decreasd.
7.11.9 Hydrolysis of Salts and the pH of their Solutions
A reaction in which the cation or anion or both of a salt reacts with water to produce acidity or alkalinity is
called salt hydrolysis. The pH of the solution is affected by this interaction.
 Acid  Base
Salt  Water 
In cationic hydrolysis, cation reacts with water to give acidic solution
 MOH  H +
M   H 2 O 
weak base
In anionic hydrolysis, anion reacts with water to give basic solution

A   H 2 O  HA  OH 
weak acid
In terms of hydrolysis, salts can be divided into four categories:
(i) Salts of strong acid and strong base, e.g., KCl, NaCl, NaNO3, etc.
(ii) Salts of weak acid and strong base, e.g., CH3COONa, KCN, Na2CO3, etc.
(iii) Salts of strong acid and weak base, e.g., NH4Cl, FeCl3, CuSO4, etc.
(iv) Salts of weak acid and weak base, e.g., CH3COONH4, (NH4)2CO3, NH4HCO3, etc.
Hydrolysis Constant. The general equation for the hydrolysis of a salt is given by
BA  H 2O  HA  BOH
Aplying law of chemical equilibrium and considering the large excess of water present
[HA][BOH]
Kh 
[BA]
where Kh is called hydrolysis constant.
Degree of hydrolysis. Degree of hydrolysis of a salt is the fraction of the total salt which undergoes hydrolysis.
Number of moles of salt hydrolysed
Degree of hydrolysis, h 
Tota l number of moles taken
Calculation of h, Kh and pH of salt solutions
(i) Salts of strong acid and strong base. Salts of strong acids and strong bases do not hydrolyse. They
simply get hydrated and therefore, their solutions are neutral (pH = 7).
(ii) Salts of weak acid and strong base. CH3COONa being the salt of a weak acid, CH3COOH and a
strong base, NaOH gets completely ionised in aqueous solution.
CH 3COONa(aq) 
 CH 3COO- (aq) + Na + (aq)
Acetate ion thus formed undergoes hydrolysis in water to give acetic acid and OH– ions
 CH 3COOH(aq) + OH - (aq)
CH 3COO- + H 2O(l ) 
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Acetic acid being a weak acid (Ka = 1.8 × 10–5) remains mostly in the unionised form in solution. This results
in increase of OH– ion concentration in solution making it alkaline (pH >7).
pH = -½ [logKw + logKa - log c] = ½ [pKw + pKa + log c] = 7 + ½[pKa + log c]
where c moles per litre is the initial concentration of the salt in aqueous solution and Ka and Kw are the
ionisation constants of acid and water at equilibrium.
Kw K Kw
Hydrolysis constant, K h = and degree of hydrolysis, h = h or h =
Ka c K a .c
(iii) Salts of weak base and strong acid. NH4Cl, being the salt of a weak base, NH4OH and a strong acid,
HCl dissociates completely in aqueous solution.
NH4Cl(aq)   NH4+ (aq) + Cl-(aq)
Ammonium ions undergo hydrolysis to NH4OH and H+ ions
 NH OH(aq) + H+(aq)
NH4+(aq) + H2O(l)  4

NH4OH, being a weak base (Kb = 1.77 × 10–5) remains mostly unionised in solution. This results in increase
of H+ ion concentration in solution making it acidic (pH< 7).
pH = -½ [log Kw - log Kb + log c] = ½ [pKw - pKb - log c] = 7 - ½[pKb + log c]
where c moles per litre is the initial concentration of the salt in aqueous solution and Kb and Kw are the
ionisation constants of base and water at equilibrium.
Kw K Kw
Hydrolysis constant, K h = and degree of hydrolysis, h = h or h =
Kb c K b .c
(iv) Salts of weak acid and weak base. Consider the hydrolysis of CH3COONH4 formed from weak acid
and weak base. The ions undergo hydrolysis as :
 CH 3COOH  NH 4OH
CH 3COO  + NH 4   H 2O 
The NH4+ and CH3COO- ions furnished by the salt combine with OH- and H+ ions of water to form feebly
ionised NH4OH and CH3COOH.
 CH 3COO- + H + ; NH 4OH 
CH 3COOH   NH 4 + + OH - ; H 2O 
 H + + OH -
Here, both H+ and OH- ions are removed in equivalent amounts and therefore, the solution remains neutral
although hydrolysis of the salt takes place. The degree of hydrolysis is independent of concentration of solution,
and pH is determined from the pK values as:
pH = -½ [ log Ka + log Kw - log Kb] = ½ [pKw + pKa - pKb] = 7 + ½ [pKa - pKb]
Thus, it is evident that the weaker the acid and the base, the greater is the degree of hydrolysis of the salt. In
this case the degree of hydrolysis is independent of the concentration of the solution. Since Kw increases with
temperature much more rapidly than either Ka or Kb, the degree of hydrolysis increases with rise in temperature.

Kw Kw
Hydrolysis constant, K h = and degree of hydrolysis, h  K h or h =
Ka Kb Ka K b
7.12 BUFFER SOLUTIONS
A solution which resists change in pH on dilution or with the addition of small amounts of acid or alkali is called
buffer solution. For example, a mixture of acetic acid and sodium acetate acts as buffer solution around pH
4.75. A mixture of ammonium chloride and ammonium hydroxide acts as a buffer around pH 9.25.
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7.12.1 Types of Buffer Solutions

a) Solution of single substance. Solution of the salt of a weak acid and a weak base, e.g., CH3COONH4.
b) Solutions of mixtures. These are of two types:
i) Acidic Buffer. Solution of weak acid with its salt with a srong base. E.g., CH3COOH + CH3COONa,
Orthoboric acid + Sodium tetraborate (H3BO3 + Na2B4O7.10H2O), Phthalic acid + Potassium hydrogen
phthalate [C6H4(COOH)2 + HOOCC6H4COOK], etc.
ii) Basic Buffer. Solution of weak base with its salt with a strong acid. E.g., NH4OH + NH4Cl, Glycine
+ Glycine hydrochloride (NH2CH2COOH + NH2CH2COOH.HCl), etc.
c) Mixture of an acid salt and a normal salt of a polybasic acid, e.g., Na2HPO4 + Na3PO4.
d) Solution of ampholyte (amphoteric electrolyte). Proteins and amino acids.
7.12.2 Buffer action
i. Acidic buffer. In case of CH3COOH + CH3COONa buffer, acetic acid dissociates to a small extent while
sodium acetate dissociates completely in aqueous solution. The dissociation of acetic acid is suppressed by
common ion effect. Thus, there are excess of CH3COO– ions and some H+ ions in solution.
 H   CH 3COO  ; CH 3COONa 
CH 3COOH   Na   CH 3COO 
On adding a few drops of acid, H+ ions combine with acetate ions of buffer to form feebly ionised acetic acid/
on adding a few drops of base, OH– ions combine with H+ ions of buffer to form feebly ionised water.
CH 3COO   H  
 CH 3COOH ; OH   H  
 H 2O
Thus, the pH of the buffer solution remains unaltered on adding a small quantity of acid or base..
ii. Basic buffer. In NH4OH + NH4Cl buffer, NH4OH dissociates to a small extent, while NH4Cl dissociates
completely in aqueous solution. The dissociation of acetic acid is suppressed by common ion effect. Thus,
there are excess of NH4+ ions and some OH– ions in solution.
 NH 4 + + OH  ; NH 4 Cl 
NH 4 OH   NH 4   Cl
On adding a few drops of acid, H+ ions combine with OH– ions from the buffer to form water/on adding few
drops of base, OH– ions combine with NH4+ ions from the buffer to form feebly ionised NH4OH.

 NH 4 OH
 H 2 O ; NH 4 + OH 
OH   H  
+

Thus, pH of the solution remains unchanged.

7.12.3 pH of Buffer Solution (Henderson-Hasselbalch Equation)
i. Acidic Buffer. The pH of an acidic buffer is is given by Henderson-Hasselbalch equation.
[Salt] [Conjugate base]
pH  pK a  log or pH  pK a  log
[Acid] [Acid]
If [Salt] = [Acid], then pH = pKa.
When [Salt]/[Acid]  10 , then pH = pKa+1 and when [Acid]/[Salt]  10 , then pH = pKa-1
A weak acid may be used for preparing buffer solutions having pH values lying within the range pKa+1 and
pKa-1. Therefore, acetic acid having pKa value of about 4.8 may be used for making buffer solutions with pH
values lying within the range 3.8 to 5.8.

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i. Basic Buffer. The pH of a basic buffer is is given by,

[Salt] [Base] [Base]
pOH  pK b  log or pH  pK a  log or pH  pK a  log
[Base] [Salt] [Conjugate acid]
A mixture of NH4OH and NH4Cl acts as buffer around pH 9.25.
7.12.4 Importance and Uses of Buffer Solutions
i) In Biological Systems. Maintenance of pH is essential for sustaining life because enzyme catalysis is pH
sensitive. Many medical and cosmetic formulations are kept and administered at definite pH.
ii) The normal pH of blood plasma is 7.4. The pH of blood is maintained with the help of (i) Buffer of
carbonic acid (H2CO3 and NaHCO3) and (ii) Buffer of phosphoric acid (H2PO4-, HPO42-). Body fluids
such as blood, urine, etc., have definite pH and any deviation in pH indicates malfunctioning of the body.
iii) In Agriculture. In farming, dairy products and in the preservation of foods and fruits.
iv) In industrial Processes. For the manufacture of paper, dyes, inks, paints, drugs, etc.
v) In Analytical Chemistry. (i) To determine pH with the help of indicators. (ii) In qualitative inorganic analysis,
CH3COOH + CH3COONa buffer is used for the removal of phosphate ions after Group II. NH4Cl +
NH4OH buffer is used for the precipitation of Group III metals. (iii) A buffer solution of NH4Cl, NH4OH
and (NH4)2CO3 is used for the precipitation of carbonates of Group V. (iv) CH3COOH + CH3COONa
buffer is used for the precipitation of lead chromate quantitatively in gravimetric analysis.
7.13 SOLUBILITY EQUILIBRIA OF SPARINGLY SOLUBLE SALTS
For a salt to dissolve in a solvent, its solvation enthalpy must be greater than its lattice enthalpy. In case of non-
polar solvents, solvation enthalpy is low and hence, salts do not dissolve in them. Each salt has its characteristic
solubility which depends on temperature. On the basis of solubility, salts can be classified as :
Category I Soluble Solubility > 0.1M
Category II Slightly soluble 0.01M< Solubility<0.1M
Category III Sparingly soluble Solubility < 0.01M
7.13.1 Solubility Product Constant
The equilibrium between undissolved solid and the ions in saturated solution of AgCl may be represented as:
[Ag  ][Cl ]

water
AgCl(s)  
 Ag (aq)  Cl (aq) ; K 
 
[AgCl]
For pure solid substance, the concentration remains constant, therefore,

K sp  K[AgCl]  [Ag  ][Cl  ]

where Ksp is the solubility product constant or solubility product.
Thus,solubility product of an electrolyte may be defined as the product of the molar concentrations of its ions
in saturated solution at a given temperature raised to the power equal to the the number of ions produced on
dissolution of one molecule of the electrolyte.
The solubility products of Ba(SO)4, Al(OH)3 and Ca3(PO4)2 may be represented as:
 Ba 2 (aq)  SO 4 2  (aq) ; K sp  [Ba 2  ][SO 4 2 ]
BaSO 4 (s) 
 Al3 (aq)  3OH (aq) ; K sp  [Al3 ][OH  ]3
Al(OH)3 (s) 
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 3Ca 2  (aq)  2PO 4 3 (aq) ; K sp  [Ca 2 ]3 [PO 43 ]2

Ca 3 (P O) 4 (s) 
Solubility product principle can also be applied to saturated solutions of freely soluble salts.
7.13.2 Relation between solubility product and molar solubility of a sparingly soluble salt
A solid salt, Mxp+ Xyq-, with molar solubility S in equilibrium with its saturated solution may be represented as:
 xM p  (aq)  yX q 
M x X y (s) 
The solubility product constant is given by :
1 ( x y )
p x q y K sp  K 
K sp  [M ] [X ]  (xS) (yS)  x  y  S
x y x y (x+y)
or S (x+y)
 or S   x sp y 
x x  yy  x y 
i. Binary electrolyte AB : K sp = [A + ][B- ] = S2 ; S  K sp mol L-1

ii. Ternary electrolyte, AB2 : Ksp  [A2 ][B ]2  S×(2S)2 = 4S3 ; S = 3 Ksp / 4 mol L-1

iv. Quaternary electrolyte, AB3 : Ksp = [A3+ ][B- ]3 = S×(3S)3 = 27S4 ; S = 4 Ksp / 27 mol L-1

v. For zirconium phosphate, Zr3(PO4)4 , K sp  [Zr 4 ]3[PO34 ]4 = (3S)3 (4S) 4 = 6912(S)7 or S = 7 Ksp / 6912 .
Table - 08 The Solubility Product Constants of Some Ionic Salts at 298K.

S alt Ks p S alt Ksp S alt Ksp S alt Ksp S alt Ks p

-10 -6 -3 8 -3 6 -1 8
A gCl 1.8×10 BaF2 1.0×10 Fe(OH)3 1.0×10 CuS 6.3×10 Hg 2 Cl2 1.3×10
-13 -6 -1 1 -5 3 -5
A gBr 5.0×10 CaSO4 9.1×10 M g(OH)2 1.8×10 HgS 4.0×10 PbCl2 1.6×10
-17 -9 -1 5 -1 3 -8
A gI 8.3×10 CaC2 O 4 4.0×10 Pb(OH)2 1.2×10 M nS 2.5×10 PbF2 7.7×10
-12 -9 -1 5 -5 -5
A g 2 CrO4 1.1×10 CaF2 5.3×10 Ni(OH)2 2.0×10 NiS 4.7×10 PbBr2 4.0×10
-5 -9 -7 -2 8 -9
A g 2 SO 4 1.4×10 CaCO3 2.8×10 SrSO 4 3.2×10 PbS 8.0×10 PbI2 7.1×10
-12 -6 -9 -2 5 -8
A g 2 CO 3 8.1×10 Ca(OH)2 5.5×10 SrF2 2.5×10 SnS 1.0×10 PbSO 4 1.6×10
-10 -20 -1 0 -1 8 -1 1
BaSO4 1.1×10 Cu(OH)2 2.2×10 SrCO 3 1.1×10 FeS 6.3×10 ZnCO 3 1.4×10
-10 -33 -2 7 -6 -2 4
BaCrO 4 1.2×10 A l(OH)3 1.3×10 CdS 8.0×10 CuCl 1.7×10 ZnS 1.6×10
When the concentration of one or more species is not the equilibrium concentration, Ksp is given by Qsp. At
equilibrium, Ksp = Qsp, but otherwise it gives the direction of the process of precipitation or dissolution.
7.13.3 Difference between solubility product and ionic product
The term ionic product applies to all types of solutions, saturated or unsaturated, but solubility product applies
only to saturated solutions in which a dynamic equilibrium exists between undissociated salt and its ions.
The solubility product of a salt is a constant at constant temperature while ionic product depends on
concentrations of ions in solution.
The solubility product represents the upper limit for ionic product of a saturated solution of a salt at a given
temperature. Thus, a substance gets precipitated when the ionic product exceeds the solubility product.
7.13.4 Common Ion Effect on Solubility of Ionic Salts
If the concentration of one of the ions in a solution is increased, it should combine with ions of opposite charge
to form the salt which gets precipitated till once again Ksp = Qsp (Le Chatelier’s principle). Similarly, if the
concentration of one of the ions is decreased, more salt will dissolve till Ksp= Qsponce again. This is applicable
even to saturated solutions of highly soluble salts like NaCl except that activities are considered instead of
molarities in the expression for Qsp (due to higher concentrations of ions).
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1. Purification of NaCl (Salting out). When HCl gas is passed through a saturated solution of impure
sodium chloride, pure sodium chloride is precipitated and the soluble impurities remain in solution. This
phenomenon is known as salting out.
 Na   Cl  ;
NaCl  K sp  [Na  ][Cl  ]
Due to increased concentration of chloride ions available from the dissociation of HCl, the ionic product
exceeds the concentration solubility product of NaCl, and therefore, it precipitates.
2. Salting out of soap. Soap is sodium (or potassiun) salt of higher fatty acids. In the manufacture of soap, in
order to separate out soap completely, NaCl is added to the solution.
 C17 H 35 COO   Na  ;
C17 H 35 COONa  K sp  [C17 H35COO ][Na  ]
On adding NaCl, concentration of Na+ increases, ionic product exceeds solubility product of sodium stearate
which therefore, precipitates out.
4. In quantitative analysis. In gravimetric estimation, common ion effect is employed for the complete
precipitation of an ion as its sparingly soluble salt, e.g., silver ion as silver chloride, barium ion as barium
sulphate, ferric ion as hydrated ferric oxide, etc.

 AgCl + NaNO3 ; BaCl 2 + H 2SO4 

AgNO3 + NaCl   BaSO4 + 2HCl

Fe 2 (SO4 )3 + 6NH 4OH- 

 2Fe(OH)3 + 3(NH 4 ) 2SO 4
5. Solubility of salts of weak acids with pH. The solubility of salts of weak acids like phosphates, carbonates
and sulphides increases at lower pH due to decrease in concentration of the anion by protonation. Two
equilibria exist simultaneously,
Ksp = [M+] [X–]
+ -
 H+(aq) + X-(aq) ; K = [H (aq)][X (aq)] [X  ] Ka
HX (aq)  a
or =
[HX(aq)] [HX] [H + ]
Taking inverse of both sides and adding 1 we get

[HX] [H + ] [HX] + [X  ] [H + ] + K a
+1 = +1 or =
[X - ] Ka [X  ] Ka
Now, again taking inverse, we get

[X  ] Ka

f 
[X ]  [HX] K a  [H  ]
It can be seen that ‘f’’ decreases as pH decreases. If S is the solubility of the salt at a given pH, then
12
Ka  [H  ]  K a 
K sp  [S][fS]  S2
and S   K sp 
K a  [H  ]  Ka 

Thus, solubility S increases with increase in [H+] or decrease in pH.

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7.14 HYDROGEN ION INDICATORS OR ACID-BASE INDICATORS

Hydrogen ion or acid-base indicator is a substance which changes colour according to H+ ion concentration
(or pH) of the solution to which it is added. They are usually weak acids or weak bases.
The Range of an Indicator. The range of an indicator is the pH range over which it changes colour.
Methyl orange gives full acid colour (red) when added to a solution of pH  3 and full basic colour (yellow) in
a solution of pH  4.4. In solutions having pH between 3 and 4.4, methyl orange gives a colour intermediate
between red and yellow. Thus, the pH range over which methyl orange can be used as indicator is 4 to 4.4.
Phenolphthalein, another common indicator, is colourless in a solution of pH 8.3 or less and pink in a solution
of pH 10 or above. Thus, phenolphthalein can be used to determine pH over the range 8.3 to 10.
Table - 09 Indicators and their pH range.
Indicator pH range Colour in acid Colour in alkali
Alizarin yellow 10.1 - 12.0 Yellow Violet
Bromophenol blue 3.1 - 4.6 Yellow Purple
Litmus 4.5 - 8.3 Red Blue
Methyl orange 3.1 - 4.5 Pink Yellow
Methyl red 4.2 - 6.2 Red Blue
Phenol red 6.4 - 8.2 Yellow Red
Phenolphthalein 8.3 - 10.0 Colourless Pink
Thymol blue 8.1 - 9.6 Yellow Blue
A universal indicator is a mixture of indicators which gives a gradual change from one colour to another over
a wide range of pH. The approximate pH of a solution can be determined from the colour obtained when a few
drops of the indicator are added to the solution.

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QUESTIONS
LEVEL - I
1. A cylinder fitted with a movable piston contains liquid water in equilibrium with water vapour at 25oC. Which
operation result in a decrease in the equilibrium vapour pressure?
1) Moving the piston downward a short distance 2) Removing a small amount of vapour
3) Removing a small amount of the liquid water 4) Dissolving salt in the water
2.  3C  g   D  g 
For the reaction : 2A  g   B  g  
Two moles each of A and B were taken into a flask. The following must always be true when the system
attained equilibrium:
1) [A] = [B] 2) [A] < [B] 3) [B] = [C] 4) [A] > [B]
3. Consider the following reaction: 2NO 2  g  
 2NO  g   O 2  g 
In the figure below, identify the curves X, Y, and Z associated with the three species in the reaction:

1) X = NO, Y = O2, Z = NO2 2) X = O2, Y = NO, Z = NO2

3) X = NO2, Y = NO, Z = O2 4) X = O2, Y = NO2, Z = NO

4.  4P g   6Q g  . The equilibrium constant K has units:

For a hypothetical reaction : 4A  g   5B g   c

1) mol L–1 2) mol–1 L 3) (mol L–1)–2 4) unitless

5. Given :  2NH 3  g  ; K1
N 2  g   3H 2  g  
 2NO  g  ; K 2 ,
N 2  g   O 2  g  
1  H 2 O  g  ; K 3
H 2  g   O 2 
2
5  2NO  g   3H 2 O  g  will be:
The equilibrium constant for 2NH 3  g   O 2  g  
2
K1K 2 K 1K 3 2 KK3
1) K1K 2 K 3 2) 3) 4) 2 3
K3 K2 K1
6.  C , the equilibrium concentrations are [A] = 0.06,[B] = 0.12 and [C] = 0.216.
For a system, A  2B 
The value of Kc for the reaction is
1) 120 2) 400 3) 4 × 10–3 4) 250
7.  C  D, the initial concentration of A and B are equal, but the equilibrium
For the reaction, A  B 
concentration of C is twice that of equilibrium concentration of A. The equilibrium constant is:
1) 4 2) 9 3) 1/4 4) 1/9

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8.  2NO  g   O 2  g  ; Kc = 1.8 × 10–6 at 184oC

For the reaction : 2NO2  g  
when Kp and Kc are compared at 184oC, it is found that:
1) Kp is greater than Kc 2) Kp is less than Kc
3) Kp = Kc 4) None of the above

9.  2SO3 g  , the partial pressure of SO2, O2 and SO3 are 0.662, 0.101
In the equilibrium, 2SO 2 g   O 2 g  
and 0.331 atm respectively. What should be the partial pressure of oxygen so that the equilibrium concentration
of SO2 and SO3 are equal:
1) 0.404 atm 2) 1.01 atm 3) 0.808 atm 4) 0.2475 atm
10. Pure ammonia is placed in a vessel at a temperature where its dissociation constant    is appreciable at
equilibrium:
1) Kp does not change with pressure
2)  does not change with pressure
3) concentration of NH3 does not change with pressure
4) concentration of H2 is less than that of N2
11. A mixture containing 8.07 mol of hydrogen and 9.08 mol of iodine was heated at 448oC till equilibrium was
attained when 13.38 mol of hydrogen iodide was obtained. Calculate the percentage dissociation of hydrogen
iodide at 448oC.
1) 13.2% 2) 19.8% 3) 18.9% 4) 21.4%
12. A vessel at 1000 K contains carbon dioxide with a pressure of 0.5 atm. Some of the carbon dioxide is
converted to carbon monoxide on addition of graphite. Calculate the value of Kp if total pressure at equilibrium
is 0.8 atm.
1) 0.36 atm 2) 6.8 atm 3) 1.8 atm 4) 3.2 atm
13. If the value of an equilibrium constant for a particular reaction is 1.6 × 10 , then at equilibrium the system will
12

contain:
1) mostly reactants 2) mostly products
3) similar amounts of reactants and products 4) all reactants
14. Equilibrium constants for four different reactions are given as K1 = 106, K2 = 10–4, K3 = 10, and
K4 = 1. Which reaction will produce least amount of products at equilibrium?
1) K1 = 106 2) K2 = 10–4 3) K3 = 10 4) K4 = 1

15.  A  g   B  g  at a given temperature, one-third of AB is dissociated. The

For the equilibrium AB  g  
equilibrium pressure of the system is:
1) 8 times Kp 2) 16 times Kp 3) 4 times Kp 4) 9 times Kp
16. At constant temperature, the equilibrium constant (KP) for the decomposition reaction. N 2 O 4  2NO2 is
expressed by Kp= 4x2 P/(1 – x2) where, P is pressure, x is extent of decomposition. Which of the following
statement is true?
1) Kp increases with increase of P 2) Kp increases with increase of x
3) Kp increases with decrease of x 4) Kp remains constant with change in P or x
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17.  H 2  g   X 2  g  the equilibrium constant is 1.0×10–5. What is the

For the equilibrium system 2HX  g  
concentration of HX if the equilibrium concentration of H2 and X2 are 1.2×10–3 M, and 1.2×10–4 M
respectively?
1) 12×10–4 M 2) 12×10–3 M 3) 12×10–2 M 4) 12×10–1 M
18.  CS2  g   4H 2  g 
At 1400 K, Kc = 2.5 × 10–3 for the reaction CH 4  g   2H 2S  g  
A 10 L reaction vessel at 1400 K contains 2.0 mol of CH4, 3.0 mol of CS2, 3.0 mol of H2 and 4.0 mol of H2S.
In which direction does the reaction proceed to reach equilibrium?
1) Forward 2) Backward
3) May be forward or backward 4) Reaction is in equilibrium
19. The volume of the reaction vessel containing an equilibrium mixture in the reaction,
SO 2Cl 2 g   SO 2 g   Cl2 g  is increased. When equilibrium is re-established:
1) the amount of SO2(g) will decrease 2) the amount of SO2Cl2(g) will increase
3) the amount of Cl2(g) will increase 4) the amount of Cl2(g) will remain unchanged
20. Consider the following reversible reaction at equilibrium:
 2H 2 g   O 2 g  ; H   24.7kJ
2H 2 O g  
which one of the following changes in conditions will lead to maximum decomposition of H2O(g)?
1) Increasing both temperature and pressure 2) Decreasing temperature and increasing pressure
3) Increasing temperature and decreasing pressure 4) Increasing temperature at constant pressure
21. Which among the following reactions will be favoured at low pressure?
 2NO g 
1) N 2 g   O 2  g    2HI g 
2) H 2 g   I2 g  
 PCl3 g   Cl2  g 
3) PCl5 g    2NH 3 g 
4) N 2 g   3H 2 g  

22. Consider the reactions, i) PCl5 g    2NO 2 g 

 PCl3 g   Cl 2  g  ii) N 2 O 4 g  
The addition of an inert gas at constant volume:
1) will increase the dissociation of PCl5 as well as N2O4
2) will reduce the dissociation of PCl5 as well as N2O4
3) will increase the dissociation of PCl5 and step up the formation of NO2
4) will not disturb the equilibrium of the reactions
23. When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind. At equilibrium:
1) Addition of NaNO2 favours reverse reaction 2) Addition of NaNO2 favours forward reaction
3) Increasing temperature favours forward reaction 4) Decreasing pressure favours reverse reaction

24.  PCl3 g   Cl2 g  . The forward reaction at constant temperature is favoured by:
For the reaction, PCl5 g  
1) Introducing an inert gas at constant volume 2) Introducing chlorine gas at constant volume
3) Introducing an inert gas at constant pressure 4) None of these

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25. Which is correct statement if N2 is added at equilibrium condition? N 2  3H 2  2NH 3 ?

1) The equilibrium will shift to forward direction because according to IInd law of thermodynamics the
entropy must increases in the direction of spontaneous reaction.
2) The condition for equilibrium is G N2  3G H 2  2G NH3 where G is Gibbs free energy per mole of the
gaseous species measured at that partial pressure. The condition of equilibrium is unaffected by the use of
catalyst, which increases the rate of both the forward and reverse reactions to the same extent.
3) The catalyst will increase the rate of forward reaction by  and that of reverse reaction by  .
4) Catalyst will not alter the rate of either of the reaction.
26. In the reaction, HC 2 O 4  PO34  HPO 42  C 2 O 42 the Bronsted base are:

1) PO34 , C 2 O 42 2) PO34 , HPO 42 3) HC 2 O 4 , HPO 24  4) HC 2 O 4 , C 2 O 42 

27. At 25oC the dissociation constant of HCN is 4.9 × 10–10 M. Calculate the degree of dissociation of HCN if
the concentration is 0.1 M.
1) 7 × 10–5 2) 5 × 10–5 3) 6 × 10–5 4) 8 × 10–5
28. For pure water,
1) Both pH and pOH decrease with increase in temperature.
2) Both pH and pOH increase with increase in temperature.
3) pH decreases and pOH increases with increase in temperature.
4) pH increases and pOH decreases with increase in temperature.
29. The equilibrium constants for the reaction are :

K1
H3 PO4    
 H  H 2 PO 4 ; K1


H 2 PO 4 
2K
  2
 H  HPO 4 ; K2


HPO 24 
3
K  3
 H  PO 4 ; K3
 3H   PO34 will be:
The equilibrium constant for H 3PO 4 
1) K1 / K2 K3 2) K1 × K2 × K3 3) K2 / K1 K3 4) K1 + K2 + K3
30. What is the pH value of N/1000 KOH solution?
1) 10–11 2) 3 3) 2 4) 11
31. The pH of solution A is 3. It is mixed with an equal volume of another solution B having pH 2. What is the
resultant pH of the solution?
1) 3.2 2) 2.26 3) 2.5 4) 3.5
32. 25.0 mL of 0.1 M NaOH is titrated with 0.1 M HCl. Calculate pH when:
i) 20 mL ii) 24 mL of acid is added
1) 12.0, 11.30 2) 11.30, 12 3) 2.0, 2.70 4) 2.70, 2.0
33. The number of H ions present in 1 mL of solution having, pH = 13 is:
1) 6.02 × 1010 2) 6.02× 107 3) 6.02 × 1013 4) 1013
34. The degree of hydrolysis of which of the following salts is independent of the concentration of salt solution?
1) CH3COONa 2) NH4Cl 3) CH3COONH4 4) NaCl

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35. The following equilibrium exists in aqueous solution CH 3COOH  CH 3COO   H  . If dil. HCl is added
without change in temperature, the:
1) concentration of CH 3COO  will increase. 2) concentration of CH 3COO  will decrease.
3) the equilibrium constant will increase. 4) the equilibrium constant will decrease.
36. The pH of a solution containing 0.1 mol of CH3COOH, 0.05 mol of NaOH, and 0.2 mol of CH3COONa, in
1 L. (pKa of CH3COOH = 4.74) is :
1) 5.44 2) 5.20 3) 5.04 4) 4.74
37. The expression to calculate pH of sodium acetate solution at 25oC is:
1 1 1 1
1) pH  7  pK b  CH 3COOH   log salt  2) pH  7  pK a  CH 3COOH   log salt 
2 2 2 2
1 1 1 1
3) pH  7  pK b  CH 3COOH   log salt  4) pH  7  pK a  CH 3COOH   log salt 
2 2 2 2
38. In which of the following acid-base titrations, pH is greater than 8 at the equivalence point?
1) Acetic acid vs ammonia 2) Acetic acid vs sodium hydroxide
3) Hydrochloric acid vs ammonia 4) Hydrochloric acid vs sodium hydroxide
39. The solubility of PbSO4 at 25oC is 1.1 × 10-4 mol/L. Then its solubiliity product (Ksp) is:
1) 1.21 × 10-8 2) 12.1 × 10-6 3) 121 × 10-11 4) 1.21 × 10-10
40. The solubility product of a sparingly soluable salt AB at room temperature is 1.21 × 10–6. Its molar solubility
is:
1) 1.21 × 10–6 2) 1.21 × 10–3 3) 1.1 × 10–4 4) 1.1 × 10–3
41. Let the solubilities of AgCl in H2O, 0.01 M CaCl2; 0,01 M NaCl and 0.05 M AgNO3be S1, S2, S3, S4,
respectively. What is the correct relationship between these quantities?
1) S1 > S2 > S3 > S4 2) S1 > S2 = S3 > S4 3) S1 > S3 > S2 > S4 4) S4 > S2 > S3 > S1
42. For a fairly concentrated solution of a weak electrolyte AxBy, the degree of dissociation is given by:
1/  x  y 
K eq xy K eq C K eq C  K eq 
1)   2)   3)   4)    x  y1 x y 
C xy  x  y C x y 
43. The addition of NaH2PO4 to 0.1 M H3PO4 will cause:
1) No change in pH value 2) Increase in its pH value
3) Decrease in its pH value 4) Change in pH but cannot be predicted
44. The pH of blood is maintained by the balance between H2CO3 and NaHCO3. If the amount of CO2 in blood
is increased, how will it affect the pH of blood?
1) pH will remain same 2) pH will be 7
3) pH will increase 4) pH will decrease
45. For which of the following sparingly soluble salt, the solubility (S) and solubility product (Ksp) are related by
1/3
K 
the expression: S   sp 
 4 
1) BaSO4 2) Ca3(PO4)2 3) Hg2Cl2 4) Ag3PO4

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LEVEL - II
1.  COCl 2  g 
The equilibrium of formation of phosgene is represented as : CO  g   Cl 2  g  
The reaction is carried out in a 500 mL flask. At equilibrium, 0.3 mol of phosgene, 0.1 mol of CO, and 0.1 mol
of Cl2 are present. The equilibrium constant of the reaction is:
1) 30 2) 15 3) 5 4) 25
2. In a 10 L vessel, HI was heated to attain equilibrium. At equilibrium 3276.8 g HI, 406.4 g I2, 15.6 g H2
were present in the mixture. Calculate Kc, if the mixture is transferred to 5L vessel.
1) 0.029 2) 0.059 3) 0.019 4) 0.91
3. The equilibrium constant for the reaction; N 2 g   O2 g   2NO g  at temperature T is 4  104 . The value of

1 1
Kc for the reaction. NO g   N 2 g   O 2 g  at the same temperature is :
2 2
1) 0.02 2) 50 3) 4 × 10–4 4) 2.5 × 10–2
4. For the following three reactions (i), (ii) and (iii), equilibrium constants are given:
 CO 2 g   H 2 g  ; K1
i) CO g   H 2O g  
 CO g   3H 2 g  ; K 2
ii) CH 4 g   H 2 O g  
 CO 2 g   4H 2 g  ; K 3
iii) CH 4  g   2H 2 O g  
Which of the following relation is correct?
1) K 3 K 32  K12 2) K1 K 2  K 3 3) K2K3= K1 4) K3= K1K2

5.  2HI g  , K  47.6, if the initial number of moles of each reactant and
For the reaction, H 2  g   I 2 g  
product is 1 mole, then at equilibrium:
1)  I 2    H 2  ;  I2    HI  2)  I 2    H 2  ;  I 2    HI
3)  I 2    H 2  ;  I2    HI 4)  I 2    H 2  ;  I2    HI 
6. For a given exothermic reaction, KP and K P are the equilibrium constants at temperature T1 and higher
temperature T2 respectively. Assuming that heat of reaction is constant in temperature range between T1 and
T2, it is readily observed that:
1
1) K P  K P 2) K P  KP 3) K P  K P 4) K P  K 
P

7.  CO  g   H 2 O  g  , if the initial concentration of [H2] = [CO2]

For the reaction, H 2  g   CO 2  g  
and x mol L–1 of H2 is consumed at equilibrium, the correct expression of Kp is :
1  x 
2
x2 1 x2 x2
1) 1  x 2 2) 3) 2  x 2 4)
  1  x 
2
  1 x2

8.  SO 2  g   Cl 2  g  is 50. The

The vapour density of the equilibrium mixture of the reaction: SO2 Cl2  g  
percent dissociation of SO2Cl2 is:
1) 33.00 2) 35.0 3) 30.0 4) 66.00
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9. The heat of reaction at constant volume for an endothermic reaction in equilibrium is 1200 cal more than at
constant pressure at 300 K. Calculate the ratio of equilibrium constants Kp (atm) and Kc (mol L–1).
1) 2.846 × 10–3 2) 6.481 × 10–3 3) 1.856 × 10–3 4) 1.648 × 10–3
10.  H 2 O g   CO g  is 1.80 at 1000oC. If 1.0 mole of H2 and
The equilibrium constant for, H 2 g   CO 2 g  
1.0 mole of CO2 are placed in one litre flask, the final equilibrium concentration of CO at 1000oC will be:
1) 0.573 M 2) 0.385 M 3) 5.73 M 4) 0.295 M
11. The partial pressures of CH3OH, CO and H2 in the equilibrium mixture for the reaction
CO g   2H 2 g   CH3OH g 
at 427oC are 2.0, 1.0 and 0.1 atm, respectively. The value of Kp for the decomposition of CH3OH to CO and
H2 is:
1) 1 × 102 atm 2) 4 × 102 atm 3) 5 × 10–3 atm 4) 5 × 103 atm
12.  2NO 2 g  , the observed molecular weight of N O is 80 g mol–1 at 350 K. The
At equilibrium: N 2 O 4 g   2 4

percentage dissociation of N2O4(g) at 350 K is:

1) 10% 2) 15% 3) 20% 4) 18%
13.  XY g   Y g  . Initial pressure of XY is 600 mm Hg. The total pressure at
XY2 dissociates as: XY2  g   2
equilibrium is 800 mm Hg. Assuming volume of system to remain constant, the value of Kp is:
1) 50 2) 100 3) 200 4) 400
14.  2NH 3 g   CO 2 g  . If equilibrium pressure is 3 atm for the above reaction: K will be:
NH 4 COONH 2 s   p
1) 4 2) 27 3) 4/27 4) 1/27
15. At 540 K, 0.10 mole of PCl5 are heated in a 8 litre flask. The pressure of the equilibrium mixture is found to
be 1.0 atm. Calculate Kc and Kp for the reaction.
1) 2 102 mol litre1 ,1.69 atm 2) 4 102 mol litre1 ,1.72 atm
3) 2.5  102 mol litre 1 ,1.69 atm 4) 4  102 mol litre1 , 2.63atm
16. A sample of HI was found to be 22% dissociated when equilibrium was reached. What will be the degreee of
dissociation if hydrogen is added in the proportion of 1 mol for every mole of Hl present originally? Assume
temperature and pressure to be constant.
1) 0.065 2) 0.085 3) 0.037 4) 0.052
17. Vapour density of the equilibrium mixture of NO2 and N2O4 is found to be 40 for the equilibrium:
N 2O 4  g   2NO2  g  . Calculate percentage of NO2in the mixture.
1) 26.08% 2) 21.52% 3) 19.24% 4) 24.62%
18.  
The equilibrium constant of the reaction A2(g) + B2(g)  2AB(g) at 100 C is 50. If a one-litre flask
o

containing one mole of A2 is connected to a two litre flask conatining two moles of B2, how many moles of AB
will be formed at 373 K?
1) 1.86 2) 0.93 3) 2.32 4) 0.46
19. At temperature T, a compound AB2(g) dissociates according to the reaction:
 2AB  g   B2  g  with a degree of dissociation ‘x’ which is small compared to unity. Deduce
2AB2  g  
the expression for ‘x’ in terms of the equilibrium constant Kp and the total pressure P.
2K p 3K p 2K p 3K p
1) 2) 2 3) 3
4)
P P P P2

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 Brilliant STUDY CENTRE

20.  PCl3  Cl2 .

The degree of dissociation is 0.4 at 400 K and 1.0 atm for the gaseous reaction. PCl5 
Assuming ideal behaviour of all the gases, calculate the density of equilibrium mixture at 400 K and 1.0
atmosphere. (Atomic mass of P = 31.0 and Cl = 35.5)
1) 7.39 g/L 2) 3.54 g/L 3) 6.92 g/L 4) 4.53 g/L
21. A certain amount of N2O4(g) is enclosed in a closed container at 127 C when following equilibrium got setup
o

 2NO 2  g 
at a total pressure of 10 atm. N 2 O4  g  
If the concentration (moles) of NO2(g) in the equilibrium mixture be 8 × 105 ppm, the Kc
(in mol L–1) for the above reaction at 127oC is equal to
1) 3.189 2) 2.051 3) 0.974 4) 1.842
22.  PCl3  g   Cl2  g  . The equilibrium, Kc for the dissociation of PCl5 is 4.0 × 10–2 at 250oC in
PCl5  g  
a 3.0 L flask when equilibrium concentration of Cl2 is 0.15 mol/L. What was the pressure of PCl5 before any
dissociation? (R = 0.082 L-atm K–1 mol–1)
1) 37.0 atm 2) 30.59 atm 3) 24.05 atm 4) 6.745 atm
23. Assume that the decomposition of HNO3 can be represented by the following equation

4HNO3 (g)   4NO 2 (g)+2H 2 O  g   O 2  g 
and at the given temperature 400 K and pressure 30 atm the reaction approaches equilibrium.
At equilibrium partial pressure of HNO3 is 2 atm. Find Kc in (mol/lit)3:
1) 30 2) 24 3) 18 4) 16
24.  Water , which phenomenon will happen?
On applying pressure to the equilibrium : Ice 
1) More ice will be formed 2) More water will be formed
3) Equilibrium will not be disturbed 4) Water will evaporate
25. Acetic acid undergoes dimerization in benzene solution. When the solution is diluted to twice the original
volume, the position of equilibrium in the reaction 2CH 3COOH   CH 3COOH  2 is shifted:
1) to the right 2) to the left 3) neither to left nor to right4) none of these
26. Ammonia under a pressure of 15 atm at 27oC is heated to 347oC in a closed vessel in the presence of catalyst.
Under the conditions, NH3 is partially decomposed according to the equation 2NH3  N 2  3H 2 . The
vessel is such that the volume remains effectively constant, whereas pressure increases at 50 atm. Calculate
the percentage of NH3 actually decomposed.
1) 61.3% 2) 63.5% 3) 65.3% 4) 66.6%
27.  H3O  OH, K w  1014 at 25oC, hence Ka is
2H 2 O 
1) 10–7 2) 5.55 × 10–13 3) 10–14 4) 18 × 10–17
28. The first and second dissociation constants of an acid, H2A are 1.0 × 10-5 and 5.0 × 10-10 respectively. The
overall dissociation constant of the acid will be:
1) 0.2 × 105 2) 5.0 × 10-5 3) 5.0 × 10-15 4) 5.0 × 1015
29. The pH of a solution obtained by mixing 50 mL of 0.4 N HCl and 50 mL of 0.2 N NaOH is:
1) -log 2 2) -log 0.2 3) 1 4) 2
30. Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H+ ion
concentration in the mixture?
1) 3.7 × 10-3M 2) 1.11 × 10-3 M 3) 1.11 × 10-4M 4) 3.7 × 10-4M

230
 [Study Package - Chemistry - Part- I]

31. An acid-base indicator (pKa = 4.5271) has the acid form red and basic form blue. If we need 75% red to be
converted into 75% blue form in solution, the change in pH of solution should be :
1) 4.05 2) 5.0 3) 0.95 4) 0.80
32. What volume must 1L of 0.5 M CH3COOH solution should be diluted with water in order to double pH?
Ka = 1.8×10–5.
1) 3.7×104 L 2) 2.76×103 L 3) 3.10×103 L 4) 1.05×104 L
33. A definite volume of a N/20 CH3COOH (pKa=4.7447) is titrated with a strong base (NaOH). It is found that
80 equal sized drops of NaOH, added from a burette effects the complete neutralisation. Find the pH, when
the acid solution is neutralised to the extent of 20%.
1) 4.14 2) 9.86 3) 5.34 4) 8.68
34. The ionization constant of NH4 in water is 5.6×10 at 25 C. The rate constant for the reaction of NH4+ and
+ –10 o

OH– to form NH3 and H2O at 25oC is 3.4×1010 L mol–1 s–1 . Calculate the rate constant for proton transfer
from water to NH3.
1) 8.23 × 105 2) 6.07 × 105 3) 12.14 × 104 4) 10.3 × 104
35. The pH of blood stream is maintained by a proper balance of H2CO3 concentrations. What volume of 5M
NaHCO3 solution be mixed with 10 mL sample of blood which is 2 M in H2CO3 in order to maintain a pH of
7.4 Ka for H2CO3 in blood is 7.8×10–7?
1) 41.86 mL 2) 83.2 mL 3) 78.36 mL 4) 52.43 mL
36. The pH of a buffer solution prepared by adding 10 mL of 0.1 M CH3COOH and 20 mL of 0.1 M sodium
acetate will be: (given: pKa for CH3COOH = 4.74)
1) 4.05 2) 3.04 3) 5.04 4) 3.05
37. 0.2M KCN and 0.06M AgNO3 solutions are mixed in equal volumes. At 25oC Kc for the reaction is
 Ag   2CN  is 1.6 × 10–19. The conc. of Ag+ present in solution is:
Ag  CN  

2
1) 1.5 × 10–19 M 2) 1.5 × 10–18 M 3) 3 × 10–19 M
4) 3 × 10 M
–18

38. In a saturated solution of AgCl, addition of NaCl is made drop by drop in excess. Which graph correctly
represents the change?

1) 2) 3) 4)

39. When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8 × 10–10) will occur
only with :
1) 10 4 M  Ag   and104 M  Cl  2) 10 5 M  Ag   and10 5 M  Cl  

3) 10 6 M  Ag   and106 M  Cl  4) 1010 M  Ag   and10 10 M  Cl 

40. The solubility of Pb(OH)2 in water is 6.7×10–6 M Calculate solubility of Pb(OH)2 in a buffer of
pH = 8.
1) 12.03×10–2 mol/L 2) 1.203×10–3 mol/L
3) 3.102 ×10–3 mol/L 4) 3.102 ×10–2 mol/L
41. The Ksp of Ca (OH)2 is 4.42×10–5 at 25oC. A 500 mL of saturated solution of Ca (OH)2 is mixed with equal
volume of 0.4 M NaOH. How much Ca(OH)2 in mg is precipitated?
1) 527.3 mg 2) 638.4 mg 3) 218.3 mg 4) 758.2 mg
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 Brilliant STUDY CENTRE

42. An aqueous solution of a metal bromide MBr2 (0.05 M) is saturated with H2S. What is the minimum pH at
which MS will precipitate? Ksp for MS = 6.0×10–21 concentration of saturated H2S = 0.1 M; K1=10–7 and
K2= 1.3×10–13 for H2S.
1) 0.9826 2) 1.3213 3) 2.6931 4) 2.1897
2 
43. H2S is bubbled into 0.2 M NaCN solution which is 0.02 M in each Cd  CN 4  and  Ag  CN 2   .
H2S produces 1×10–9 M sulphide ion in the solution Given, K sp Ag 2S  11050 M3 ; K sp Cds  7.11028 M 2
1 2
K inst  Ag  CN 2   110 20 M 2 .K inst Cd  CN 4   7.8  1018 M1 . Identify the correct statement.
1) Ag2S precipitates first from the solution
2) Ag2S precipitates at a sulphide concentration 1×1015 M
3) CdS precipitates first from the solution
4) None of them precipitates under the given conditions
44. The solubility products of MA, MB, MC, and MD are 1.8×10–10, 4×10–3, 4×10–8 and 6×10–5 respectively.
If a 0.01 M solution of MX is added dropwise to a mixture containing A , B , C , and D ions, then the one
to be precipitated first will be:
1) MA 2) MB 3) MC 4) MD
45. CaCO3 and BaCO3 have solubility product values 1×10 and 5×10 , respectively. If water is shaken up
–8 –9

with both solids till equilibrium is reached, the concentration of CO 3 2 ion is:
1) 1.5×10–8 2) 1.225×10–4 3) 2.25×10–9 4) 2.5×10–8

232
 [Study Package - Chemistry - Part- I]

SOLUTIONS
LEVEL - I
1. 4 Vapour pressure of a liquid depends only on temperature.
2. 2
3. 1
4. 1
K 2  K 33
5. 4 K
K1
6. 4
7. 1
As K p  K c  RT  , n  3  2  1 ; Kp > Kc
n
8. 1
1 1
9. 1 PSO3  PSO2 then, PO    0.404atm
2
K p 2.475
10. 1 The equilibrium constant does not change with pressure.

11. 4 The reaction of formation of HI is as follows : H 2  I 2  2HI

8.07 9.08 0
8.07-x 9.08x 2x

1 1
K C  54.279 ; But K C    0.01842
K C 54.279
2
 x  1 1
‘x’ can be calculated as    ; x  106.75 103 mol
 1  2x  54.279 9.3674
  2x  2 106.75  10 3  213.5 10 3 ; Percentage dissociation = 21.35%
p   2  0.3
2 2
0.36
12. 3 Kp  CO    1.8atm
pCO 2 0.2 0.2
13. 2 The value of K (K = 1.6 × 1012) is very high so the system will proceed more towards forward reaction
and thus will contain mostly products at equilibrium.
14. 2 Because the minimum value of equilibrium constant, minimum is the extent of reaction.
15. 1
16. 4 KP is a characteristic constant for a given reaction and changes only with temperature.
17. 3
18. 2 K > 2.5 ×10–3 (given). Hence, the reaction will proceed in backward direction.
19. 3
20. 3 Reaction is endothermic and n  0 , hence, the formation of product will be favoured by increasing the
temperature and decreasing the pressure.
21. 3 On lowering the pressure, equilibrium favours the direction of higher volume.
22. 4 At constant volume, inert gas will not affect any of the equilibrium.
23. 3 Addition of NaNO3 or NaNO2 has no effect since both are solids.
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 Brilliant STUDY CENTRE

24. 3 On addition of inert gas at constant P, volume increases. To have Kc constant, x must increase.
25. 2 Because change in Gibbs free energy is given by : G (reaction) = G (products) – G (reactants)
Also, catalyst only speeds up the reaction without affecting the equilibrium conditions.
26. 1 Acid - H+ = Conjugate base
Ka 4.9  1010
27. 1  ;    7  105
c 0.1
28. 1 With increase in temperature the ionisation of H2O increases and hence the  H   and  OH  increases.
29. 2 K1  K1  K 2  K 3
30. 4
31. 2 For solution A, pH  3   H    103 M and for solution B, pH  2   H    102 M.

103 M  102 M
For the resultant solution,  H     5.5  103 ; pH = 2.26
2
32. 1 mmole of NaOH = 25 × 0.1 = 2.5
i) 20 mL of 0.1 M HCl is added.
mmol of HCl = 20 × 0.1 = 2
mmol of NaOH left = 2.5 – 2.0 = 0.5 mmol
0.5 m mol
Volume = 25 + 20 = 45 mL,  OH    0.01  102 M , pOH = 2, pH = 12
45 mL
ii) 24 mL of 0.1 M HCl is added.
mmol of HCl = 24 × 0.1 = 2.4, mmol of NaOH left = 2.5 –2.4 = 0.1
0.1
 OH    0.002  2 103 M , pOH   log  2  103   0.3  3  2.7
 25  24   
pH = 14 – 2.7 = 11.3
33. 2 pH = 13 
1013
 H    1013 M  1013 mol 1000 mL  
1
 1016 mol mL1
1000
Number of  H   10  6.023  10
 16 23
= 6.023 × 107 ions
34. 3 It is the salt of weak acid and weak base, hence its degree of hydrolysis will be independent of
Kw
concentration. h 
Ka  Kb
35. 2 CH 3COOH  CH 3COO   H 
If dil. HCl is added to the above equilibrium, due to the common ion effect the concentration of CH3COO–
decreases.
36. 1 Total moles of CH3COONa = 0.2 + 0.05 = 0.25
Moles of CH3COOH left = (0.1–0.05) = 0.05
Thus, acidic buffer is formed

pH  pK a  log
Salt   0.25 
 4.74  log   = 5.44
 Acid   0.05 
234
 [Study Package - Chemistry - Part- I]

1
37. 4 pH of salt of WA/SB is  pK w  pK a  log C 
pH 
2
38. 2 Salt formed at equivalence point is sodium acetate. It is the salt of weak acid and strong base.
2
39. 1 PbSO4 is a binary electrolyte, hence its Ksp can be calculated as : K sp  S 2  1.1  10  4   1.21  10  8
6 3
40. 4 Given that Ksp = 1.21×10–6, so the molar solubility is, S  K sp  1.21 10  1.1 10
41. 3 S1 > S3 > S2 > S4 is the correct order of solubility.
42. 4 The reaction can be expressed as A x By  xA y  yBx 
1/  x  y 
x x y y Cx  y x  y  K eq 
 Taking1    1   x x y y C x  y1 x  y ;    x y x  y1 
C x y C 
43. 3
44. 1  H  HCO3 . If CO is increased, according to Le Chatelier’s principle,  H  
H 2 O  CO 2  2  
will increase, thereby pH will decrease.
1/3
 Hg 22  2Cl ; K sp   Hg 22   Cl   S 2S2  4S3 ; S   sp 
2 K
45. 3 Hg 2 Cl2 
2s    
s s
 4 
LEVEL - II
0.6
1. 2 Kc   15
0.2  0.2
0.78  0.16
2. 3 Kc   0.019 . Kc is constant at a given temperature.
 2.56 
2

1 1
3. 2 K c2    50
K c1 4  104
4. 4 Reaction (iii) is obtained by adding (i) and (ii) hence K3 = K1 × K2
5. 3 Kc = 47.6. i.e, concentration of products more than that of reactants.
6. 1 K p  K p if T2 > T1.
7. 1
Dd
8. 2  ; % dissociation = 35.0%
 n  1 d
9. 4 Given that U  H  1200 cal where U  q v and H  q p

1200
nRT  1200 or n   2 (R = 2 cal)
2  300
Kp
K p  K c  RT  ,   0.0821 300  = 1.648 × 10–3
n 2
Now using
Kc

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 Brilliant STUDY CENTRE

10. 1  H 2O  CO
CO2  H 2 

x2 x
 Kc  or  K c  1.8 ;
1  x 
2
1 x
x
  1.34,  x  0.573
1 x

11. 3 CO  2H 2  CH 3OH; K p 
p CH 3OH  
2
 200
 pCO   p H  1 0.1
2 2

1 1
For CH 3OH  CO  2H 2 ; Kp    5 103 atm
K p 200
12. 2 Degree of dissociation may be calculated as,
Mm
x
 n  1 m ; Percentage dissociation = 0.15 × 100 = 15

13. 2 XY2 g   XY g   Y g 

600 – x x x
600 + x = 800 or x = 200
200  200
Applying law of mass action, K p   100
400
14. 1 PNH3 : PCO2  2 :1  p NH3  2 atm; p CO2  1atm 
2
K p   p NH3   p CO2   2 2  1  4

15. 2  PCl3  Cl2

PCl5 
0.1 0 0 mole before dissociation
0.1–x x x mole after dissociation
Given, volume of container = 8 litre
x x

Now K c 
 PCl3 Cl2 
 8 8  x2
......(1)
 PCl5   0.1  x  8  0.1  x 
8
Also PV = nRT for the equilibrium mixture at 540 K, 1 × 8 = (0.1 + x) × 0.082 × 540
 x  0.08 ......(2)
0.08  0.08
Thus from Eq (1) and (2) Kc   4 102 mol litre1
8  0.1  0.08

Also K p  K c  RT   4 102  0.08  540  = 1.72 atm

n
 n  1

236
 [Study Package - Chemistry - Part- I]

16. 3 Given that the percentage dissociation = 22, so degree of dissociation = 0.22. The reaction involved is
2HI  H2  I2
Initial moles 1 0 0
Moles at equilibrium 1    0.78  / 2  0.11  / 2  0.11
 H 2  I2   0.11 0.11  0.0199
Therefore, K c 
 HI  0.78 
2 2

When 1 mol of hydrogen is added starting with 1 mol of HI, we get

Kc 
 H 2  I2     / 2     / 2  1  0.0199
 HI 1   
2 2

Solving, we get   0.037. So, the addition of 1 mol of H2 suppresses the dissociation of HI.
17. 1 N 2 O 4  2NO2
Initial moles 1 0
Moles at equilibrium (1–x) 2x
Hence, the total moles at equilibrium = (1+x) = 1 + 0.15 = 1.15
moles of NO 2 2  0.15
  100 , Percentage of NO2 = 26.08%
total moles 1.15
18. 1  2AB(g)
A2(g) + B2(g) 
1 2 0 initial moles
(1–x) (2–x) 2x moles at equilibrium
(1–x) (2–x) 2x molar concentration at equi.
3 3 3
Where x is moles of A2 converted to AB at equilibrium and the total volume of the container
2
 2x 
 AB 
2  
 3  4x 2
Kc  ,  50  given K p  50 at 373 K  , x = 0.93
 A 2   B 2   1  x  2  x  1  x  2  x 
 ][ 
 3 3 
(The other value of x obtained by solving may be neglected as it is greater than 1)
Moles of AB = 2 × 0.93 = 1.86
19. 3  2AB  g   B2  g 
2AB2  g  
1 0 0 mole before dissociation
x
(1–x) x mole after dissociation
2
x x n  n AB  p n
Total mole at equilibrium  1  x  x   1  . Now K p  B2  
2 2 N AB2 
 n 
1
x  
x 
2
P 
Kp  2   
x 3P
 
x 2K p
1  x   1  
2
x or K p x is small 1  x  1 and 1   1 or x  3
2 2 P
 2
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20. 4  PCl3  Cl2

PCl5 
1 0 0 initial mole
1–0.4 0.4 0.4 mole at equilibrium
Total mole at equilibrium  1  0.4  0.4  0.4  1.4

Normal m PCl5 208.5

 1    1.4 Observed m PCl5 
Observed m PCl5 1.4

w w Pm 1 208.5
From PV  RT at equilibrium ;    4.54 g / litre
m v RT 1.4  0.082  400

21. 3  2NO 2

N 2O 4 
a – initial mole
a–x 2x equi mole

0.8  105 0.8 2

   , K p  4 D2  32 , K  32 atm
2x 2
 , x=
1 x 106 1.2 3 1  p

K p  K c  RT  , 32 = K c  0.082  400 ,
n
K c  0.975 mol / lit

22. 2  PCl3  g   Cl 2  g 

PCl5  g  
a – – initial conc
a–x x x equi conc.

 0.15
2
x2
Kc  , 0.04 = , a = 0.7125
ax a  0.15
PV = nRT, P = (0.7125) (0.0821) 523 = 30.59 atm
23. 1 
4HNO3 (g)  4NO2 (g) + 2H 2  O 2
P0 – – – initial press
P0 – 4x 4x 2x x equi press
P0 + 3x = 30
P0 – 4x = 2
x=4
 4 n
Kp = (16)4 (8)2  4  , K p  K c (RT) ; 1048576 = Kc[0.082 × 400]3, Kc = 30 (mol/litre)3
2 
24. 2 Volume of ice > volume of water and thus increase in pressure favours forward reaction showing decrease
in volume.

25. 2 2CH 3COOH   CH 3COOH  2 . As volume increases concentration decreases, reaction quotient
increases for the equilibrium.
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26. 1 Let  be the degree of dissociation of ammonia under these conditions. According to the balanced
equation  moles of NH3 decompose to produce  /2 mol of N2 and 3  /2 mol of H2.
2NH 3  N 2  3H 2
Initial moles 1 0 0
Moles at equilibrium 1  /2 3 / 2
Total number of moles after decomposition  1   . Using Gay Lussac’s law
P1 P2 T 273  347 620
 , P2  P1  2  15   15   31atm
T1 T2 T1 273  27 300
Number of moles without decomposition at temperature T P
 a
Number of moles after decomposition at the same temperature T Pb
1 31atm
    0.613 , % decomposition of ammonia  0.613  100  61.3%
1   50 atm
 H    OH  Kw 10 14
27. 4 Ka       18  1017 (M of H2O = 55.6)
 H 2O  H 2O  55.6
28. 3 K a  K a1  K a 2 = 1 × 10-5 × 5 × 10-10 = 5 × 10-15
29. 3 N1V1acid  N 2 V2base  N R  V1  V2 
0.4  50  0.2  50  N R  100 ; NR = 0.1   H    0.1M
pH   log  H     log 0.1  1
30. 4  H    103 M,  H    104 M,  H    105 M for the given acids.
M mix Vmix  M1V1  M 2 V2  M 3 V3 ; M mix  3  10 3  1  104  1  105 1
105 100  10  1 111 10 5
M mix    37  10 5 M = 3.7 × 10-4 M
3 3
HI  H  In

31. 3 For 75% red form  Redn 25  Blue 

75 25

H  I 
 
25
K a      , pH = pKa + log
n
 4.5271  0.4771  4.05
 HI n  75
75
For 75% blue form pH = 4.5271 + log
25  4.5271  0.4771  5.0
pH  5.0  4.05  0.95
32. 1 pH of 0.5 M CH3COOH is 2.523
After dilution pH = 5.046, H+ = 9×10–6 M
 CH 3CO2  H 
CH3COOH 
x2
 Ka  ,  C  1.35  105 M
Cx
M1V1  M 2 V2 , 1  0.5   1.35  10 5  V V  3.7  104 L

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33. 1 WA (CH3COOH) is titrated with SB (NaOH) and 20% of the WA is neutralised, so 20% of the salt of
WA/SB (CH3COONa) is formed and 80% of the WA (CH3COOH) is left. So an acidic buffer is formed.
 Acid   80; Salt   20
 Salt   Salt  1
pH = pKa + log    4.7447  log    4.7447  log
 Acid   Acid  4
 4.7447  2 log 2  4.7447  2  0.3010  4.14
34. 2 NH 4  NH 3  H  K1  5.6  10 10
H 2 O  H   OH  K 2  110 14
From the above reactions
  K K1 5.6  1010
NH 4  OH  1
  NH 3  H 2 O K  K  5.6 104
K 2 K2 , 110 14

K1 K 3.4  1010
Further, at equilibrium, K  ; K2  1   6.07  105
K2 K 5.6  10 4

35. 3 [H2CO3] in blood = 2 M, Volume of blood = 10 mL

[NaHCO3] = 5 M. Let volume of NaHCO3 used = V mL
2  10 5 V
[H2CO3] in mixture  , [NaHCO3] in mixture 
V  10 V  10
Salt   5  V  V  10  ,
pH = pKa+log , 7.4   log 7.8  107  log V = 78.36 mL
 Acid  2 10  V  10 
MV 0.110
36. 3 Number of moles of CH 3COOH    0.001
1000 1000
Number of moles of CH 3 COONa  MV  0.1  20  0.002
1000 1000

pH  pK a  log
Salt   0.002 
 4.74  log   5.04
 Acid   0.001 

37. 2  Ag  CN 2   Ag   2CN 
– 0.06 0.2 initial conc.
0.06 x 0.08 equi conc.
 Ag 2   CN 
2
 x  0.08 
2
Kc  , 1.6  10 
 19 , x  1.5 1018 M.
 Ag  CN    0.06
 2

K sp
38. 2  Ag   Cl    K sp or  Ag   
Cl 

if  Cl   , increases, then too have Ksp constant and thus [Ag+] decreases. Also after sufficient addition

of Cl   ,  Ag   will become almost constant.

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104  V  10 4  V 
39. 1 Q for AgCl   Ag  Cl   
 
  2V   2.5  109  Q  K sp , precipitation will occur..
 2V  

K sp of Pb (OH) 2  4S3  4   6.7  106   1.203  1015 . The buffer contains pH = 8

3
40. 2

 pOH  6 or OH    106 , Now let solubility of Pb(OH)2 be S mol/L in it.Thus

2 1.203 1015
 Pb 2    OH    K sp ,  Pb2  10 6   1.203 1015 ;  Pb 2  
 1.203 10 3 mol / L
   1012
41. 4 500 mL of 0.4 M NaOH are mixed with 500 mL of Ca(OH)2 a saturated solution having Ca(OH)2
solubility as S M

4.42  105
 Ca 2  2OH 
For Ca(OH)2  K sp  S   2S   4S
2 3
S  3  0.0223M
4
Now Ca  OH 2  NaOH are mixed  Solution has Ca2+ and OH– out of which some Ca2+ will
0.0223  500
precipitate. On mixing  Ca 2     0.01115  111.5  104 M
1000
0.0223  2  500 500  0.4
 OH     [from Ca (OH)2] [from NaOH] = 0.2223 M
1000 1000
Ca 2   OH    K sp , Ca 2  left  0.2223  4.42 105
2 2

4.42  105
Ca 2    8.94 10 4 mol / L
 mole of Ca (OH)2 precipitated = Mole of  Ca 
2
 0.2223
left 2

precipitated  111.5  10 4  8.94  104  102.46  104

 wt of Ca(OH2) precipitated  102.46  104  74  7582.04  104 g  758.2 mg

42. 1  MBr2  aq  

MBr2  g    M 2   2Br 

MBr2  H 2S  MS  2HBr

K sp of MS   M 2  S2   , 6 10 21   0.05 S2  

 S2    1.2  1019 M

Thus MS will be precipitated if H2S provides 1.2×10–19 M ions of S2–
2
 H   S2 
 2H   S2  ;
Now for H2S, H 2S  K1  K 2 
 H 2S
2
 H   1.2  1019 
107  1.3  1013  ,  H    1.04  10 1 and pH  0.9826
 
0.1
43. 3 Ionic product > Ksp then ppt forms

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44. 1 Since all compounds (MA, MB, MC, and MD) are univalent type. All will have the same formula of
solubility in H2O (i.e., K sp ). Same concentration of common ion  M   is added, so solubilities of all
will be suppressed. Hence, the compound with least Ksp value will be precipitated, i.e., MA with
Ksp = 1.8×10–10.
45. 2 Since Ksp of CaCO3 and BaCO3 are very close. So concentration of any species cannot be neglected.
Let the solubility of CaCO3 and BaCO3 are x and y M.
 Ba 2  CO32 
 Ca 2  CO32 , BaCO3 
 CaCO3 
x y y
x

Total  CO 32     x  y 

 K sp of CaCO 3   Ca 2   CO32    x  x  y  , K sp of BaCO3   Ba 2   CO3 2   y  x  y 

 x  x  y   108 , y  x  y   5 109

5  109 y  x  y  , 
y
 5  10 1
  y = 0.5x
108 x  x  y x

Substitute the value of y, x  x  0.5x   108 , x = 0.8×10–4, y = 0.5×0.8×10–4

  Ca 2    x  0.8 104 ,   Ba 2    y  0.4 104 ,  CO32     x  y   1.2  10 4

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CHAPTER - 08
REDOX REACTIONS

INTRODUCTION
Several chemical reactions involve transfer of electrons from one chemical species to another. Any such reaction
involves both reduction and a complimentary oxidation process. Redox reactions include all chemical reactions
in which atoms have their oxidation states changed. All these reactions are accompanied by energy changes in
the form of heat, light or electricity.
Processes like rusting of iron articles, fading of colour of fabrics, burning of fuel, etc., fall in the category of
redox reactions. Industrial processes like electroplating, extraction of metals like aluminium, and sodium,
bleaching of wood pulp, manufacture of caustic soda, etc., are also redox processes. Redox reactions also
form the basis for electrochemical and electrolytic cells.
8.1 OXIDATION AND REDUCTION - CLASSICAL CONCEPT
1. Oxidation. Oxidation is the process of addition of oxygen/electronegative element to a substance or removal
of hydrogen/ electropositive element from a substance.
i) Addition of oxygen : C(s)  O 2 (g) 
 CO 2 (g) ; S(s)  O 2 (g) 
 SO 2 (g)
2Mg(s)  O 2 (g) 
 2MgO(s) ; CH 4 (g)  2O 2 (g) 
 CO 2 (g)  2H 2 O(l )
ii) Removal of hydrogen : 4HI  O 2 
 2H 2 O  2I 2 ; 4HCl  MnO 2 
 MnCl 2 +2H 2 O  Cl2
 SnCl 4 ; Mg(s)  Cl2 (g) 
iii) Addition of electronegative element : SnCl 2  Cl 2   MgCl2 (s)
iv) Removal of electropositive element : 2KI(aq)+ H 2 O(l ) + O 3 (g) 
 2KOH(aq) + I 2 (s) + O 2 (g)
Oxidising agent or oxidant. An oxidant is a substance which supplies oxygen or electronegative element or
removes hydrogen or electropositive element. During oxidation, the oxidising agent undergoes reduction.
2. Reduction. Reduction is the process removal of oxygen/electronegative element from a substance or
addition of hydrogen/ electropositive element to a substance.

i) Removal of oxygen : 2HgO(s)   2Hg(l )  O 2 (g)
ii) Removal of electronegative element : 2FeCl3 (aq) H 2 (g) 
 2FeCl 2 (aq) 2HCl(aq)
ferric chloride ferrous chloride

iii) Addition of hydrogen : CH 2  CH 2 (g)  H 2 (g) 

 CH 3  CH 3 (g)
iv) Addition of electropositive element : SnCl 4  Sn   2SnCl 2 ; HgCl2  Hg   Hg 2 Cl2
Reducing agent or Reductant. A reductant is a substance which supplies hydrogen or any electropositive
element or removes oxygen or any electronegative element. Reductant undergoes oxidation during reduction.

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8.1.1 Simultaneous occurance of Oxidation and Reduction

In a chemical reaction, oxidation occurs only if reduction is also taking place simultaneously, i.e., oxidation
and reduction are complementary. E.g., in the reaction of Mg with F2, Mg is oxidised while F is reduced.

Since oxidation and reduction take place simultaneously, the word redox is used for this class of reactions.
8.2 OXIDATION AND REDUCTION- ELECTRON TRANSFER (MODERN) CONCEPT
Oxidation is the process in which an atom, ion or group of atoms loses one or more electrons , i.e., removal of
electrons or de-electronation.
Mg 0  Oxidation
 Mg 2+ + 2e ; Fe 2 
Oxidation
 Fe3  e  ; MnO 4 2  
Oxidation
 MnO 4   e 
The species which undergoes loss of electron(s) is called reducing agent or reductant.
Reduction is the process in which an atom, ion or group of atoms gains one or more electrons, i.e., addition of
electrons or electronation.
Cl2  2e  
Reduction
 2Cl  ; Cu 2   e  
Reduction 3 
 Cu + ; [Fe(CN)6 ]  e 
Reduction
 [Fe(CN)6 ]4
The species which gains electron(s) in a reaction is called oxidising agent or oxidant.
Redox Reaction. The process of electron transfer is described as redox reaction. As a result of transference
of electron(s) from the reducing agent to the oxidising agent, the oxidising agent gets reduced while the reducing
agent gets oxidised.

The above reaction can be considered as the sum of two half-reactions, one involving oxidation called oxidation
half-reaction and the other involving reduction called reduction half-reaction.

A  e 
 A  B  e
; B 
Reduction half  reaction Oxidation half  reaction

The sum of the half-reactions gives the overall reaction.

Thus, Zn(s)  Cu 2  (aq) 
 Zn 2 (aq)  Cu(s) can be split into two half reactions as
 Zn 2   2e  [Oxidation]
Zn  ; Cu 2   2e  
 Cu [Reduction]
The reaction: 2Al(s)  3Cu 2  (aq) 
 2Al3 (aq)  3Cu(s) can be split into the following half reactions
 2Al3  6e  [Oxidation]
2Al  ; 3Cu 2  (aq)  6e  
 3Cu(s) [Reduction ]
8.2.1 Competitive Electron Transfer Reactions
There is competition among elements for electrons. E.g., when a zinc rod is dipped in cupric nitrate solution, it
gets coated with reddish metallic copper and the blue colour of the solution disappears.
Zn(s)  Cu 2  (aq)   Zn 2 (aq)  Cu(s)
The overall reaction is the simultaneous oxidation of zinc to Zn2+ and reduction of Cu2+ to copper.

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Similarly, when a copper rod is dipped in silver nitrate solution, copper dissolves and displaces silver completely
from solution (silver gets deposited on the copper rod). The equilibrium greatly favours the products.
Cu(s)  2Ag (aq)   Cu 2 (aq)  2Ag(s)
In the case of metallic cobalt placed in nickel sulphate solution, neither the reactants nor the products are
greatly favoured (cobalt does not displace nickel completely from solution).
Co(s)  Ni 2  (aq)   Co 2  (aq)  Ni(s)
It is clear that zinc releases electrons to copper and copper releases electrons to silver, therefore, the electron
releasing tendency of the metals is in the order : Zn > Cu > Ag.
The arrangement of elements in the order of increasing electron releasing tendency is called metal activity
series or electrochemical series.
The competition between metals for electrons is utilised in designing a class of cells, named Galvanic cells in
which chemical reactions are the source of electrical energy.
8.3 OXIDATION NUMBER
Oxidation number of an element may be defined as the residual charge which its atom has when all other atoms
from the molecule are assumed to be removed as ions and counting the shared electrons with the more
electronegative atom(s). Thus, oxidation number denotes the oxidation state of an element in a compound.
(i) In case of electrovalent compounds, oxidation number of an element or radical is the same as the charge
on the ion, e.g., in NaCl, the charge and oxidation number of sodium and chlorine are +1 and -1 respectively.
0 0 1 1
2 Na  Cl2   2Na  2Cl
(ii) Oxidation number of atoms in covalent compounds is derived by assigning the electrons of each bond to
the more electronegative atom. E.g., in HCl molecule, chlorine is more electronegative than hydrogen,
therefore, the oxidation number of chlorine is -1 and hydrogen is +1.
0 0 1 1
H 2 (g)  Cl2 (g) 
 2HCl(g)
(iii) Electrons shared between two similar atoms are divided equally between the bonded atoms. For example,
in Cl2 molecule, the oxidation number of each chlorine atom is zero.
Atoms can have positive, zero or negative oxidation numbers depending on the state of combination.
If two or more atoms of a particular element are present in a molecule or ion, the oxidation number of that
element will be the average of the oxidation numbers of all the atoms of that element.
8.3.1Rules for assigning oxidation numbers
(i) The oxidation number of an element in the free or elementary state is always zero irrespective of the
allotropic form. E.g., He (noble gases), hydrogen in H2, chlorine in Cl2, oxygen in O2 or O3, phosphorus
in P4, sulphur in S8, are all equal to zero.
(ii) The oxidation number of fluorine is always -1 in all its compounds. Other halogens also have -1 oxidation
number when they occur as halide ions in their compounds. However in oxoacids, and oxoanions they
have positive oxidation number.
(iii) The oxidation number of an element in monoatomic ion is equal to the charge on the ion. E.g., in KCl, the
oxidation number of K is +1 and that of Cl is -1. Oxidation number of alkali metals is +1 and alkaline earth
metals is +2 in their compounds.
(iv) Hydrogen is assigned oxidation number +1in all its compounds except metal hydrides. In metal hydrides
like LiH, NaH, MgH2, CaH2, etc., the oxidation number of hydrogen is -1.

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(v) Oxygen is assigned oxidation number -2 in most of its compounds. However in peroxides like H2O2,
Na2O2, BaO2, etc., its oxidation number is -1. In compounds of fluorine and oxygen, like OF2 and O2F2,
the oxidation number of oxygen is +2 and +1 respectively.
(vi) The algebraic sum of the oxidation numbers of all the atoms in a molecule is zero. In case of a polyatomic
ion, the sum of oxidation numbers of all its atoms is equal to the charge on the ion.
(vii) In a binary compound of metal and non-metal, the metal atom has positive O.N., while the non-metal
atom has negative O.N. E.g., in KI, oxidation number of K is +1 and of I is –1.
(viii) In binary compounds of non-metals, the more electronegative atom has negative O.N., while the less
electronegative atom has positive oxidation number. E.g.,O.N. of Cl in ClF3 is +3 and in ICl is -1.
Transition elements exhibit several positive oxidation states. The highest oxidation number of a representative
element is the group number for the first two groups and the group number minus 10 for the other groups.
In a period, the highest oxidation number exhibited by an element generally increases, e.g., in the third period,
the highest oxidation number increases from 1 to 7 in the compounds of the elements.
Group-1 Group-2 Group-13 Group-14 Group-15 Group-16 Group-17
Element Na Mg Al Si P S Cl
Compound NaCl MgSO4 AlF3 SiCl4 P 4O10 SF6 HClO4
Highest oxidation +1 +2 +3 +4 +5 +6 +7
number/state
8.3.2 Calculation of oxidation state
(i) Oxidation number of an element in a compound is calculated as follows:
(i) Mn in KMnO4 (+1) + x + (–8)  x = +7
(ii) Cr in Cr2O 7
2-
2x + 7(-2) = –2  x = +6
(iii) Fe in [Fe(CN)6]4- x + 6(–1) = –4  x = +2
(ii) Using Lewis formula. In the Lewis dot structure for the molecule or ion, assign all the electrons in each
bond to the more electronegative atom in the bond. If two atoms have identical electronegativities, divide
the bonding electrons equallly between them.

 Number of electrons in the   Number of valence electrons assigned 

Oxidation state =   
 free non-bonded atom   to the atom in Lewis dot structure 
8.3.3 The paradox of fractional oxidation number
In certain compounds, oxidation number of a particular element may be a fraction. Examples are:
(i) I in KI3 (Potassium tri-iodide), 1+ 3x = 0 x = –1/3
(ii) Pb in Pb3O4 (Lead tetroxide) 3x + 4 (–2) = 0 x = 8/3
(iii) C in C3O2 (Carbon sub-oxide) 3x + 2 (–2) = 0 x = 4/3
(iv) Br in Br3O8 (Tribromooctaoxide) 3x + 8 (–2) = 0 x = 16/3
(v) S in Na2S4O6 (Sodium tetrathionate) 2 + 4x + 6 (–2) = 0 x = 10/4
In certain cases, fractional oxidation state is only the average oxidation number of the element in different species,
e.g., KI3 is a mixture of KI (O.N. of iodine =-1) and I2 (O.N. of iodine = 0). Similarly, red lead, Pb3O4 is a
stoichiometric mixture of 2 moles of PbO (O.N. of Pb = +2) and 1 mole of PbO2 (O.N. of Pb = +4).

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In certain compounds, fractional oxidation states is due to different oxidation states of the same element in a
compound. E.g., in case of C3O2, Br3O8 and S4O62-, the oxidation state (oxidation number) of the element
marked with asterisk is different from the rest of the atoms of the same element in each species.

(a) (b) (c)

Figure -01 (a) Carbon suboxide (b) Tribromooctaoxide, Br3O8, (c) Tetrathionate anion, S4O62-.
However, oxidation states may be in fraction as in O2+ and O2–, where it is +½ and –½ respectively.
8.3.4 Other examples of oxidation numbers
i) Oxidation numbers of two N atoms in NH4NO3 are different. NH4NO3 is made up of NH4+ and NO3-
ions. In NH4+ ion, oxidation number of N is -3 while in NO3- ions, oxidation number of N is +5.
ii) Oxidation number of sulphur in (CH3)2SO (dimethyl sulphoxide). O.N. of oxygen =-2; O.N. of each
methyl group = +1; Let O.N. of sulphur = x. Therefore, 2(+1) + x + (-2) = 0 or x = 0
ii) Permonosulphuric acid (Caro’s acid), H2SO5, two O atoms are involved in peroxide linkage (O.N. of
oxygen in peroxo linkage = –1). O.N. x of sulphur : 2 (+1) + x + 3 (-2) + 2 (-1) = 0 or x = +6

(a) (b) s

Figure - 02 (a) Permonosulphuric acid (b) Perdisulphuric acid

iv) Perdisulphuric acid, H2S2O8. In H2S2O8, 2 oxygen atoms are linked through peroxo linkage (O.N. = –1).
O. N. of sulphur : +2+(–12)+2x–2 = 0 or x = +6.
v) In chromium peroxide, CrO5, 4 out of 5 oxygen atoms are involved in peroxide linkages. The oxidation
number of Cr in the compound is calculated as : x + 4 (-1) + 1 (-2) = 0 or x = +6

vi) In the coordination complex, K4[Fe(CN)6], the sum of oxidation numbers of C and N in (CN-) are taken
together as -1, oxidation number of Fe is calculated as: 4 (+1) + x + 6 (-1) = 0 or x = +2
8.3.5 Oxidation and reduction in terms of oxidation number
Oxidation is a process in which oxidation number of the element increases, whereas reduction is a process in
which oxidation number of the element decreases. In the following reaction, oxidation number of bromine
decreases from 0 to -1 (reduction) and the oxidation number of sulphur increases from -2 to 0 (oxidation).

Based on the concept of oxidation number, oxidising agent is a substance which undergoes decrease in
oxidation number and reducing agent is a substance which undergoes increase in oxidation number.

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8.3.6 Valency and Oxidation Number

Vale ncy Oxidation numbe r
1 Valency is the combining capacity of the It is the residual charge on an atom or appears to
element. It is the number of hydrogen atoms or have when other atoms from the molecules are
double the number of oxygem atoms which removed as ions by counting the shared electrons
combine with one atom of the element. with more electronegative atoms.
2 It is a number which does not carry any plus or It refers to the charge which can be positive,
minus sign. negative or zero.
3 Valency is always an whole number Oxidation number may have fractional value.
4 Valency of the element is never zero except for
Oxidation number of an element may be zero.
noble gases.
5 Many elements like C, N and O exhibit constant Even elements like C, N, O which exhibit constant
valency valency exhibit variable O. N.

8.3.7 Oxidation number and nomenclature (Stock notation)

The oxidation number of a metal in a compound can be written in Roman numerals within parenthesis after the
symbol of the metal in the molecular formula (German chemist, Alfred Stock). This is known as Stock notation.
Formula Name O.N. Stock notation
Cu2O Cuprous oxide Cu(+1) Copper(I)oxide Cu2(I)O
CuO Cupric oxide Cu(+2) Copper(II)oxide Cu(II)O
FeSO4 Ferrous sulphate Fe(+2) Iron(II)sulphate Fe(II)SO4
Fe2(SO4)3 Ferric sulphate Fe(+3) Iron(III)sulphate Fe 2(III)(SO4)3
SnCl2 Stannous chloride Sn(+2) Tin(II)chloride Sn(II)Cl2
SnCl4 Stannic chloride Sn(+4) Tin(IV)chloride Sn(IV)Cl4
KMnO4 Potassium permanganate Mn(+7) Potassium manganate(VII) KMn(VII)O4
K2Cr2O7 Potassium dichromate Cr(+6) Potassium dichromate(VI) K2Cr2(VI)O7

8.4 TYPES OF REDOX REACTIONS

8.4.1 Combination Reactions. A compound is formed by chemical combination of two or more elements. All
combustion reactions, which make use of elemental oxygen and other reactions involving elements other than
oxygen are redox reactions.
0 0 4  2 4 1 0 4  2 1  2
 CO 2 (g) ;
C (s)  O 2 (g)  CH 4 (g)  2 O 2 (g) 
 CO 2 (g)  2 H 2 O(l )
8.4.2 Decomposition Reactions. Decomposition reactions are the reverse of combination reactions. A
decomposition reaction results in the breakdown of a compound into two or more components at least one
of which must be in the elemental state.
+1  2 0 0 1 5 2 1 1 0

2 H 2 O (l )   2 H 2 (g) + O 2 (g) ; 2KClO3 (s) 
Δ
 2KCl(s) + 3O2 (g)
There is no change in the oxidation number of potassium in potassium chlorate in the above reaction. All
decomposition reactions are not redox , e.g., decomposition of calcium carbonate is not a redox reaction.
+2 +4 -2 2  2 4  2
CaCO3 (s) 
Δ
 CaO (s) + CO 2 (g)

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8.4.3 Displacement Reactions. An ion or atom in a compound is replaced by another element.

(a) Metal Displacement Reactions. A metal in a compound is displaced by another metal.
2  6  2 0 0 2  6  2 3  2 0 0 3  2
 Cu (s) + ZnSO 4 (aq) ; Fe 2 O3 (s) + 2 Al (s) 
CuSO 4 (aq) + Zn (s)   2 Fe (s) + Al2 O3 (s)
4 1 0 0 2 1 3  2 0 0 3  2
TiCl 4 (l ) + 2 Mg (s) 
Δ
 Ti (s) + 2MgCl 2 (s) ; Cr2 O3 (s) + 2 Al (s) 
 2 Cr (s) + Al 2O3 (s)
In each case, the reducing metal is a better reducing agent than the one that is being reduced.
(b) Non-metal Displacement Reactions. Non-metal displacement redox reactions include hydrogen
displacement and a very rare reaction involving oxygen displacement.
(i) All alkali metals and some alkaline earth metals (Ca, Sr, and Ba) which are very good reductants, displace
hydrogen from water in the cold.
0 1  2 1  2 1 0 0 1  2 2  2 1 0
 2 NaOH (aq) + H 2 (g) ; Ca (s) + 2 H 2 O (l ) 
2 Na (s) + 2 H 2 O (l )   Ca(OH ) 2 (aq) + H 2 (g)
(ii) Less reactive metals such as Mg and Fe react with steam to produce dihydrogen gas.
0 1  2 2  2 1 0 0 1  2 3 2 0
 Mg(OH ) 2 (s) + H 2 (g) ; 2 Fe (s) + 3 H 2 O (l ) 
Mg (s) + 2 H 2 O (l )   Fe 2 O 3 (s) + 3H 2 (g)
(iii) Many metals, including those which do not react with cold water/steam, react with acids displacing hydrogen.
0 1 1 2 1 0 0 1 1 2 1 0
 ZnCl2 (aq) + H 2 (g) ; Fe (s) + 2 HCl (aq) 
Zn (s) + 2 HCl (aq)   FeCl 2 (aq) + H 2 (g)
The reactions of Zn and Mg with HCl are used in the preparation of H2 in the laboratory.
(iv) Metals such as cadmium and tin which do not react with steam, reacts with acids to produce H2 gas.
0 1 1 2 1 0 0 1 1 2 1 0

Cd (s) + 2 HCl (aq)   CdCl 2 (aq) + H 2 (g) ; Sn (s) + 2 HCl (aq) 

 SnCl 2 (aq) + H 2 (g)
(v) Very less reactive metals such as silver and gold do not react with HCl.
Reactivity of metals. Reactivity of metals is reflected in the rate of evolution hydrogen gas,
Fe < Zn < Mg. Zinc, copper and silver exhibit reducing activity in the order Zn > Cu > Ag.
Reactivity of non-metals. Activity series exists for the halogens also. The power of these elements as
oxidising agents decreases from fluorine to iodine F > Cl > Br > I.
Fluorine is so reactive that it replaces chloride, bromide and iodide ions in solution.
F2  2Cl  
 Cl 2  2F ; F2  2Br   
 I 2  2F
 Br2  2F ; F2  2I 
Fluorine also displaces oxygen from water, therefore, displacement reactions of chlorine, bromine and iodine
using fluorine are not carried out in aqueous solution.
1  2 0 1 1 0
2 H 2O (l ) + 2 F2 (g) 
 4 HF (aq) + O2 (g)
Chlorine can displace bromide and iodide ions in aqueous solution (layer test for halogens).
0 1 1 0 0 1 1 0
Cl2 (g) + 2 Br  (aq) 
 2 Cl (aq) + Br2 (g) ; Cl2 (g) + 2 I (aq) 
 2Cl (aq) + I 2 (g)
Bromine displaces iodide ions in aqueous solution.
0 1 1 0
Br2 (g) + 2 I (aq) 
 2 Br (aq) + I 2 (g)

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Halogen replacement reactions are of great industrial importance. Halogens except fluorine can be recovered
from halides by oxidation.
2X   X 2  2e 
Since fluorine is the strongest oxidising agent, there exists no chemical method to convert F– to F2 ; the only
method being electrolysis.
8.4.4 Disproportionation (dismutation) reactions. It is a reaction in which the same species undergoes simultaneous
oxidation and reduction, e.g., cuprous (Cu+) ions in solution disproportionates into Cu2+ ions and metallic
copper (Cu0).

Cu 2 
oxidation
 2Cu  
reduction
 Cu 0
One of the reacting substances always contains an element that can exist in at least three oxidation states. The
element in the reacting substance is in the intermediate oxidation state and both higher and lower oxidation
states of that element are formed in the reaction.
(a) In the decomposition of hydrogen peroxide, oxygen undergoes disproportionation.
1 1 1 2 0
2 H 2 O 2 (aq) 
 2 H 2 O(aq)  O 2 (g)
(b) Phosphorous, sulphur and chlorine undergo disproportionation in alkaline medium:
0 -3 1
P4 (s) + 3OH  ( aq )  3H 2 O(l ) 
 PH 3 (g) + 3H 2 P O 2  (aq)
Phosphine Dihydrohypophosphite ion

0 -2 +2
S8 (s) + 12OH  (aq) 
 4S2- (aq) + 2 S2 O32- (aq) + 6H 2 O(l )
Sulphide Thiosulphate ion

The reaction of Cl2 with Ca(OH)2 describes the formation of bleaching powder.
0 +1 1
Cl2 (g) + 2OH  (aq) 
cold
Cl O(aq) + Cl  (aq) + H 2 O(l )
Hypochlorite ion

The hypochlorite ion (ClO–) formed in the reaction oxidises coloured stains to colourless compounds.
(c) Whereas bromine and iodine follow the same trend as exhibited by chlorine in the above reaction, fluorine
shows deviation from this behaviour when it reacts with alkali.
0 1 1
Br2 (g) + 2OH  (aq) 
cold
Br O  (aq) + Br  (aq) + H 2 O(l )
Hypobromite ion

0 1 1
I 2 (s) + 2OH  (aq) 
cold
I O (aq) + I  (aq) + H 2 O(l )
Hypoiodite ion
1 2 2
2 F2 (g) + 2OH  (aq)   2 F (aq) + O F 2 (g) + H 2 O(l )
Fluorine attacks water to produce oxygen. This is due to the fact that, being the most electronegative element,
it cannot exhibit any positive oxidation state. Therefore, fluorine does not undergo disproportionation.
8.5 BALANCING OF REDOX EQUATIONS
During redox reactions, there is change in oxidation number of the elements due to transference of electrons.
The number of electrons lost during oxidation is equal to the number of electrons gained during reduction. This
is the basic principle of balancing redox equations. The important methods used are the oxidation number
method and the ion electron method.
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8.5.1 The oxidation number method

Step 1: Write the skeletal ionic equation representing the chemical reaction.
Step 2: Assign oxidation number to all the elements and identify the atoms which undergo change in oxidation
number.
Step 3: Find out the increase or decrease in oxidation number per atom and the molecule or ion in which it
occurs. If these are not equal, multiply by suitable coefficients so that these become equal.
Step 4: If the reaction is taking place in water, add H+ or OH– ions to the expression on the appropriate side
so that the total ionic charges of reactants and products are equal. If the reaction is carried out in acidic
solution, use H+ ions in the equation; if in basic solution, use OH– ions.
Step 5 : Make the number of hydrogen atoms on both sides of the expression equal by adding H2O molecules
to the reactants or products as required. If the number of oxygen atoms in the reactants and products are the
same, the equation represents the balanced redox reaction.
E.g. 1. The net ionic equation for the reaction of potassium dichromate (VI), K2Cr2O7 with ferrous sulphate
(FeSO4) in sulphuric acid to ferric (Fe3+) ion and chromium (III) ion is obtained by :
Step 1: The skeletal ionic equation : Fe 2   Cr2O 7 2  
 Fe3  Cr 3
All atoms other than oxygen and hydrogen are equal on both sides.
Step 2: Assigning oxidation numbers to all elements and identifying atoms which undergo change in oxidation
number:
2 6  2 3 3
2 2
Fe  Cr2 O 7   Fe  Cr 3
3

Dichromate ion is the oxidant and ferrous ion is the reductant.

Step 3: Calculating the increase and decrease of oxidation number and making them equal:
2 6  2 3 3
2 2
6Fe  Cr2O7   6 Fe  2Cr 3
3

Step 4: As the reaction occurs in acidic medium and the ionic charges are not equal on both sides, 14H+ are
added on the left to make the ionic charges equal.
6Fe2  Cr2O7 2  14H  
 6Fe3  2Cr 3
Step 5: Making the number of hydrogen atoms equal on both sides by adding 7H2O on the right hand side to
compensate for 14H+ on the left:
6Fe2+ + Cr2O7 2- +14H +   6Fe3+ + 2Cr 3+ + 7H2O
The number of oxygen atoms in the reactants and products are the same, therefore, the equation represents
the balanced redox reaction.
8.5.2 Ion Electron Method or Half Reaction Method
In this method, the two half equations are balanced separately and then added together to give the balanced
ionic equation.
Step 1: Write the unbalanced equation for the reaction in ionic form. Identify the elements which undergo
change in oxidation number.
Step 2: Separate the equation into two half-reactions; the oxidation half and the reduction half and balance the
atoms which undergo change in oxidation number.
Step 3: Balance the atoms other than oxygen and hydrogen in each half reaction individually.
Step 4: For reactions taking place in acidic medium, add H2O to balance O atoms and H+ to balance H atoms.
For reactions occuring in basic medium, add proper number of H2O molecules to the side short of H atoms
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and an equal number of OH– ions to the other side of the equation. Wherever H+ and OH– appear on the
same side of the equation, combine these to form H2O.
Step 5: Add electrons to the side deficient in electrons so as to balance the charge on both sides. Make the
number of electrons in the two half reactions equal, by multiplying one or both half reactions by appropriate
coefficients, if needed.
Step 6: Add the two half reactions to get the overall reaction and cancel the electrons on each side.
Step 7: Verify whether the equation contains the same type and number of atoms and the same charges
on both sides.
Eg. 1. The steps to arrive at the balanced equation for the oxidation of Fe2+ ions to Fe3+ ions by dichromate
ions (Cr2O7)2– in acidic medium, wherein, Cr2O72– ions are reduced to Cr3+ ions are :
Step 1: Unbalanced equation for the reaction in ionic form: Fe 2   Cr2 O 7  Fe3  Cr 3
Step 2: Separating the equation into two half-reactions; the oxidation half and the reduction half.
2 3 6  2 3
Fe2 
 Fe3  Cr 3
; Cr2 O 7 
Oxidation half  reaction
Re duction half  reaction

Step 3: Balancing the atoms other than oxygen and hydrogen in each half reaction individually.
2
Fe 2   Fe3 ; Cr2 O 7   2Cr 3
Step 4: As the reaction is taking place in acidic medium, adding H2O to balance O atoms and H+ to balance H
atoms.
Fe 2   Fe3 ; Cr2 O 7 2  14H    2Cr 3  7H 2 O
Step 5: Adding electrons to one side of the half reactions to balance charges:
2  
 2Cr 3  7H 2 O
 Fe3  e  ; Cr2 O 7  14H  6e 
Fe 2  
Multiplying oxidation half-reaction by 6 to make the number of electrons in both the half-reactions equal:
2  
6Fe 2  6Fe3  6e ; Cr2 O 7  14H  6e   2Cr 3  7H 2 O
Step 6: Adding the two half reactions and cancelling electrons on both sides.
6Fe2+ + Cr2O7 2- +14H+ 
 6Fe3+ + 2Cr 3+ + 7H 2O
8.6 REDOX REACTIONS AS THE BASIS FOR TITRATIONS
The titration method can be adopted to find out the strength of a reductant/oxidant using a redox sensitive
indicator. A redox indicator (oxidation-reduction indicator) is a compound which undergoes a definite colour
change at a specific electrode potential, e.g., diphenyl amine, methylene blue, etc.
1. Self-indicator. In a redox titration, one of the reagents is intensely coloured and acts as oxidant as well as
indicator (e.g., permanganate ion, MnO4– has deep pink colour). The visible end point is the appearance
(retention) of light pink colour immediately after the complete oxidation of the reductant (at MnO4– concentration
as low as 10–6 mol L–1).
Permanganometry : KMnO4 in acidic medium is a powerful oxidising agent.
2KMnO 4  3H 2SO 4 
 K 2SO 4  2MnSO 4  3H 2 O  5[O]
It oxidises Fe2+ to Fe3+, C2O42– to CO2, and H2O2 to O2
2FeSO 4  H 2SO 4  [O] 
 Fe2 (SO 4 )3  H 2O

H 2 C2 O4  [O] 
 H 2 O  2CO 2 ; H 2 O2  [O] 
 H 2O  O2

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2. Redox indicator. A redox indicator has a reduced and oxidised form with different colours and the redox
process is reversible. The indicator is oxidised immediately after all of the reactant is consumed, producing
colour change.
Electrode potential Colour
Indicator
(EV) Reduced form Oxidised form
Diphenyl amine 0.76 Colourless Violet
N-Phenyl anthranilic acid 1.08 Green Violet-red
Methylene blue 0.01 Colourless Blue
Ferroin 1.06 Red Cyan
Dichrometry : Dichromate ion, Cr2O72–, which is not a self-indicator, oxidises the indicator substance (e.g.,
diphenylamine) just after the equivalence point to produce another colour (blue), signalling the end point.
K2Cr2O7 in acidic medium is a powerful oxidising agent.
K 2Cr2O 7  4H 2SO4 
 K 2SO4  Cr2 (SO 4 )3  4H 2 O  3[O]
It oxidises Fe2+ to Fe3+ : 2FeSO 4  H 2SO 4  [O] 
 Fe2 (SO 4 )3  H 2O
3. Starch as indicator. This method is restricted to reagents which oxidise I– ions (Cu2+, MnO4-, Cr2O72-) or
reduce I2 (thiosulphate, S2O32-; arsenite, AsO43-; sulphite, SO32-). Iodine gives intense blue colour with starch
due to the formation of a complex. When I- ions are oxidised to I2, blue colour appears and when I2 is reduced
to I- ions, blue colour disappears (at the end point).
Iodine, though insoluble in water, remains in solution containing KI as KI3.
Iodimetry and Iodometry. Volumetric analysis based on the use of iodine as oxidising agent.
Iodimetry - estimation of reducing substances using standard iodine solution as the oxidising agent and starch
as indicator. At the end-point, blue colour disappears as soon as iodine is consumed by thiosulphate ions.

2Cu 2 (aq)  4I  (aq) 

 Cu 2 I 2 (s)  I 2 (aq)
Iodometry - estimation of oxidising substances involving the liberation and subsequent estimation of I2.On
addition of starch, after the liberation of iodine by the reaction of oxidising agent with iodide ions, an intense
blue colour appears. The colour disappears as soon as the whole of iodine is consumed by thiosulphate ions.

2Cu 2 (aq)  4I  (aq) 

 Cu 2 I 2 (s)  I 2 (aq) ; I 2 (aq)  2S2 O32 (aq) 
 2I (aq)  S4 O6 2 (aq)
8.7 REDOX REACTIONS AND ELECTRODE PROCESSES
a. Direct redox reaction. A redox reaction in which oxidation and reduction take place in the same vessel is
called direct redox reaction, e.g., the reaction when a zinc rod is dipped in copper sulphate solution.
Zn 0  Cu 2   Zn 2  Cu 0
Zn is oxidised to Zn2+ ions and Cu2+ ions are reduced to metallic copper by the direct transfer of electrons.
Heat is evolved during the reaction (in a direct redox reaction, chemical energy appears as heat).
b. Indirect redox reaction. A redox reaction in which oxidation and reduction take place in separate vessels
is called indirect redox reaction. E.g., the above reaction can be modified by placing the zinc rod in zinc
sulphate solution and the copper rod in copper sulphate solution.
No reaction takes place in the beakers or at the metal/salt solution interfaces, where the reduced and oxidised
forms of the same species are in contact. These represent the species in the reduction and oxidation half
reactions. Each beaker constitutes a half-cell, also called electrode or redox couple.
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Redox couple. A redox couple is the combination of the oxidised and reduced forms of a substance taking
part in an oxidation or reduction half-reaction. A redox couple is represented by separating the oxidised and
the reduced forms by a vertical line or a slash representing an interface with the oxidised form written before
the reduced form. Thus, the two redox couples mentioned above may be represented as Zn2+|Zn and Cu2+|Cu.
8.7.2 Electrochemical cell (Daniell cell). The solutions in the two half cells can be connected using a salt bridge
(a U-tube filled with an inert electrolyte like potassium chloride or ammonium nitrate in agar agar jelly) which
provides electric contact between the two solutions without allowing them to mix with each other. The zinc and
copper rods are connected by a metallic wire with provision for an ammeter and a switch. This set-up is
known as Daniell cell.

 Zn 2  2e
Left beaker : Zn(s)  (Oxidation half-cell or anode)

Right beaker : Cu 2 (aq)  2e 

 Cu(s) (Reduction half-cell or cathode)

Zn(s)  Cu 2 (aq) 
 Zn 2 (aq)  Cu(s) (Overall reaction)
The Daniell cell may be represented as
Zn Zn 2 (c1 ) Cu 2 (c 2 ) Cu or Zn ; Zn 2 (c1 ) Cu 2 (c 2 ) ; Cu
The double line represents the salt-bridge.
In the switched off position, no reaction occurs in either half cell and no current flows through the wire. When
switched on, transfer of electrons takes place from Zn to Cu2+ through the metallic wire. The flow of electricity
from one solution to the other is by the migration of ions through the salt bridge.
The flow of current is possible only if there is a potential difference between the copper and zinc electrodes,
called electromotive force (emf) or cell potential.

(a) (b)

Figure - 01 (a) Daniel Cell (b) Standard Hydrogen Electrode (SHE)

8.7.3 Electrode Potential. The potential difference set up between the metal and its ions in solution is called
electrode potential or half-cell potential. It is the tendency of an electrode to lose or gain electrons.
Electrode potential is termed oxidation potential if the electrode loses electron(s) and reduction potential
if it gains electrons. If the oxidation potential of an electrode is x-volts, its reduction potential is -x volts.
Electromotive Force (emf) or Cell Potential. It is defined as the difference between the electrode potentials
(reduction potentials) of the cathode and anode when no current is drawn through the cell.
Cell Potential, E cell  E cathode  E anode  E right  E left

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8.7.4 Standard Electrode Potential. Electrode potential depends on the concentration of the ions in solution and
temperature. If the concentration of every species taking part in the electrode reaction is unity and the reaction
is carried out at 298K and 1 bar pressure, the electrode potential called standard electrode potential (E0).
Standard hydrogen electrode (SHE). The standard hydrogen electrode consists of a platinised platinum
electrode dipping in 1M solution of H+ ions (1M HCl) at 298K through which pure H2 gas at 1 bar is bubbled.
Pt, H2(g, 1bar), H+ (aq, conc = c) or Pt(s)|H2(g, 1bar)| H+ (aq, 1M)
By convention, the standard electrode potential (E0) of hydrogen electrode is taken as 0.00 volts.
Measurement of standard electrode potential. For measuring the electrode potential of an electrode, it is
connected to SHE (anode) to form a cell.
Standard hydrogen electrode || Second half-cell
Since electrode potential of SHE is zero, the measured EMF is the electrode potential of ‘Second half-cell’.
The emf of the cell, E ocell  E oright  Eleft
o
or E ocell  E oright  0 = E oright
Standard electrode potential is a measure of the tendency of the half cell reaction to occur in the forward
direction (reduction). The higher its positive value, the greater is the tendency of the oxidised form to get
reduced by accepting electrons.
8.7.5 Electrochemical Series
The values of standard electrode potential of the elements arranged in the decreasing order is called
electrochemical series or activity series.
Applications of electrochemical series
i) To compare relative oxidising and reducing powers of elements: Elements having higher reduction
potentials are strong oxidising agents while those having lower reduction potentials are strong reducing
agents.
ii) For predicting spontaneity of redox reaction: If emf of a redox reaction is positive, the reaction is
spontaneous in the given direction.
iii) Reaction of metal with dilute acids to liberate hydrogen : Metals with negative reduction potentials
displace hydrogen from acids, whereas those with positive reduction potentials do not.
iv) Comparison of reactivity of metals (Displacement reactions) : A metal lower down the series can
displace metals lying above it from their solutions.
v) Determination of EMF of a cell : It is useful in determining the standard EMF of a galvanic cell.
vi) Thermal stability of metallic oxides : As the electropositivity increases from top to bottom, the thermal
stability of the oxides also increases from top to bottom.
8.8 APPLICATIONS OF REDOX REACTIONS
(i) Extraction of metals. Metal oxides are reduced to metals using suitable reducing agents, e.g., Fe2O3 to iron
using coke. Metals such as Al, Li, Na, K, Mg, Ca, etc., are obtained by electrolysis. (ii) Electrochemical cells,
batteries and fuel cells. (iii) Photosynthesis. CO2 is reduced to carbohydrates while water is oxidised to O2.
(iv) Production of energy: burning of fuels, oxidation of glucose in the body, etc. (v) Production of chemicals:
production of caustic soda, chlorine, fluorine, hydrogen, etc.

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Table - 03 Electrochemical series

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QUESTIONS
LEVEL - I
1. Which of the following processes does not involve oxidation of iron?
1) Decolourization of blue CuSO4 solution by iron
2) Formation of Fe(CO)5 from Fe
3) Liberation of H2 from steam by iron at high temperature
4) Rusting of iron sheets
2. In which SO2 acts as oxidant, while reacting with:
1) acidified KMnO4 2) acidified K2Cr2O7 3) H2S 4) acidified C2H5OH
3. Which one is not correct about the change given below? K 4 Fe  CN 6 
oxi
 Fe3  CO2  NO3

1) Fe is oxidised Fe 2  to Fe3 2) Carbon is oxidised from C2  to C 4

3) N is oxidised from N3– to N5+ 4) Carbon is not oxidised
4. Which of the following reactions do not involve oxidation reduction?
I) 2Cs  2H 2O  2CsOH  H 2 II) 2CuI 2  2CuI  I 2

III) NH 4 Br  KOH  KBr  NH3  H 2 O IV) 4KCN  Fe  CN  2  K 4  Fe  CN 6 

1) I, II 2) I, III 3) I, III, IV 4) III, IV
5. Consider the metals: Mn, Mg, Zn, Ag, Cu. Based on their reactivity order, pick the correct statement.
1) All five metals produce hydrogen gas on reacting with acids
2) Ag will substitute Cu from its aqueous solution
3) Mn will substitute Zn from its aqueous solution
4) Cu will substitute Mg from its aqueous solution
6. In a reaction, 4 mole of electrons are transferred to 1 mole of HNO3, the possible product obtained due to
reduction is:
1) 0.5 mole of N2 2) 0.5 mole of N2O 3) 1 mole of NO2 4) 1 mole of NH3
7. In which of the following compounds, the oxidation number of carbon is not zero?
1) C12H22O11 2) HCHO 3) CH3CHO 4) CH3COOH
8. The correct order of the oxidation states of nitrogen in NO, N2O, NO2 and N2O3 is :
1) NO2 < N2O3 < NO < N2O 2) NO2 < NO < N2O3 < N2O
3) N2O < N2O3 < NO < NO2 4) N2O < NO < N2O3 < NO2
9. Which of the following species contain an element in an oxidation state that is not a whole number?
1) VO 34 2) Mn2O3 3) S4 O 26  4) Cl2O7
10. In which pair does the named substance have the same oxidation number?
1) Mercury in HgCl2 and Hg2Cl2
2) Sulphur in H2S2O7 and H2SO4
3) Oxygen in Na2O2 and H2O
3
4) Cobalt in Co (NH 3 )5 and [Co(NH 3 )5 (NO3 )](NO3 )
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11. Average oxidation number of carbon in C3O2, Mg2C3 are respectively:

1) – 4/3, + 4/3 2) + 4/3, – 4/3 3) – 2/3, + 2/3 4) – 2/3, + 4/3
12. The oxidation number of sulphur in S8, S2F2 and H2S are:
1) 0, + 1 and –2 2) +2, +1 and –2 3) 0, +1 and +2 4) –2, +1 and –2
13. Arrange the following in the increasing order of oxidation state of Mn:
(i) Mn 2  (ii) MnO 2 (iii) KMnO 4 (iv) K 2 MnO 4
1) (i) > (ii) > (iii) > (iv) 2) (i) < (ii) < (iv) < (iii)
3) (ii) < (iii) < (i) < (iv) 4) (iii) < (i) < (iv) < (ii)
14. The oxidation number of nitrogen atoms in NH 4 NO3 are:
1) +3, +3 2) +3, –3 3) –3, + 5 4) –5, +3
15. When methane is burnt in oxygen to produce CO2 and H2O, the oxidation number of carbon changes by:
1) – 8 2) zero 3) +8 4) +4
16. Which of the following can act as an oxidising agent as well as a reducing agent?
1. H2O2 2. H2S 3. SO2 4. HNO2
1) 1, 2, 3 2) 2, 3, 4 3) 1, 3, 4 4) All
17. In which of the following pairs of compounds, nitrogen has maximum positive and maximum negative values of
oxidation number?
1) NH3 and HN3 2) HNO3 and HCN 3) N2O and NH2OH 4) N2O4 and NO2
18. Which of the following chemical reactions depicts the oxidizing behaviour of H2SO4?
1) 2HI  H 2SO 4  I 2  SO 2  2H 2 O 2) Ca  OH 2  H 2SO 4  CaSO4  2H 2O

3) NaCl  H 2SO 4  NaHSO 4  HCl 4) 2PCl5  H 2SO 4  2POCl 3  2HCl  SO 2 Cl 2

19. What products are expected from the disproportionation reaction of hypochlorous acid?
1) HClO3 and Cl 2 O 2) HClO 2 and HClO 4 3) HCl and Cl2 O 4) HCl and HClO3
20. Which of the following is not a disproportionation reaction?

1) P4 + 5 OH H2PO2 + PH3 2) Cl2 + OH Cl + ClO

3) 2H 2 O 2  2H 2 O  O 2 4) Na 2 O 2  2H 2 O  2NaOH  H 2 O 2
21. i) H 2 O 2  O3 
 H 2 O  2O 2 ii) H 2 O 2  Ag 2 O 
 2Ag  H 2 O  O 2
Role of hydrogen peroxide in the above reactions is respectively:
1) oxidizing in (i) and reducing in (ii) 2) reducing in (i) and oxidizing in (ii)
3) reducing in (i) and (ii) 4) oxidizing in (i) and (ii)
22. Which of the following is not a disproportionation reaction?
 
I) NH 4 NO3 
A
 N 2O  H 2O II) P4 
A
 PH3 + HPO2

III) PCl5 
A
 PCl3  Cl2 IV) IO3 + I I2
1) I, II 2) I, III, IV 3) II, IV 4) I, III

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23. Consider the redox reaction : Cu 2S  KMnO 4  H 2SO 4 

 CuSO 4  MnSO 4  K 2SO 4  H 2 O. Moles
of KMnO4 reduced by one mole of Cu2S are:
1) 0.5 2) 1.5 3) 2.0 4) 3.0
24. When SO2 is passed in a solution of potassium iodate, the oxidation state of iodine changes from:
1) +5 to 0 2) +5 to –1 3) –5 to 0 4) –7 to –1
25. A sulphur containing species that cannot be an oxidising agent is :
1) H2SO4 2) H2S 3) SO2 4) H2SO3
26. The oxidation states of the most electronegative element in the products of the reaction between BaO2 and
H2SO4 are :
1) 0 and –1 2) –1 and –2 3) –2 and 0 4) –2 and +1
27. The following reaction describes the production of metallic iron: 2Fe 2 O3  3C  4Fe  3CO 2
How many electrons are lost by each carbon atom in this reaction?
1) 1 2) 2 3) 3 4) 4
28. Complete the balancing of the following half reaction, taking place in basic media, Br   aq   BrO3  aq 
How many electrons are needed to balance it?
1) 2 electrons, left side 2) 2 electrons, right side 3) 4 electrons, right side 4) 6 electrons, right side
29. Number of moles of K2Cr2O7 reduced by 1 mol of Sn2+ is:
1) 1/3 2) 1/6 3) 2/3 4) 1
30. In alkaline medium ClO2 oxidises H2O2 to O2 and itself gets reduced to Cl–. How many moles of H2O2 are
oxidised by 1 mole of ClO2?
1) 1 2) 1.5 3) 2.5 4) 3.5
31. The value of ‘n’ in, MnO 4  8H   ne 
 Mn 2  4H 2 O is :
1) 5 2) 4 3) 3 4) 2
32. For the redox reaction, Zn  NO 3  Zn 2   NH 4 in basic medium, the coefficients of Zn, NO 3 and
OH  in the balanced equation are respectively:
1) 4, 1, 7 2) 7, 4, 1 3) 4, 1, 10 4) 1, 4, 10
33. 2 mole of FeSO4 are oxidised by ‘X’ mole of KMnO4 whereas 2 mole of FeC2O4 are oxidised by ‘Y’ mole
of KMnO4. The ratio of ‘X’ and ‘Y’ is:
1) 1 : 3 2) 1 : 2 3) 1 : 4 4) 1 : 5
34. In the reaction, 3Br2 + 6CO32–+ 3H2O 5Br + BrO3 + 6HCO3
1) Bromine is oxidised and carbonate is reduced 2) Bromine is reduced and water is oxidised
3) Bromine is neither reduced nor oxidised 4) Bromine is both reduced and oxidised
35. The stoichiometric numbers appearing from left to right in the reaction
MnO 24  H   MnO 4  MnO 2  H 2 O are:
1) 3, 2, 2, 1, 2 2) 3, 4, 2, 1, 2 3) 2, 4, 1, 2, 2 4) 2, 3, 1, 1, 2

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36. When acidified K2Cr2O7 solution is added to Na2S solution, green Cr3+ ions and free S are formed. When
acidified K2Cr2O7 solution is added to NaCl, no change occurs. Of the substances involved in these reactions,
which one is the best reducing agent?
1) K2Cr2O7 2) Na2S 3) Cr3+ 4) S
37. Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine
as indicator. The number of moles of Mohr’s salt required per mole of dichromate is:
1) 3 2) 4 3) 5 4) 6
38. When KMnO4 acts as an oxidising agent and ultimately form MnO 24 , MnO 2 , Mn 2O 3 and Mn 2  , then the
numbers of electrons transferrred in each case, respectively, are:
1) 4, 3, 1, 5 2) 1, 5, 3, 7 3) 1, 3, 4, 5 4) 3, 5, 7, 1
39. In the electrolytic cell, flow of electrons is from:
1) cathode to anode in solution 2) cathode to anode through external supply
3) cathode to anode through internal supply 4) anode to cathode through internal supply
 2
40. In the galvanic cell with cell reaction Zn s   2Ag  aq   Zn  aq   2Ag  s 

The correct statement given below is:

1) Zinc act as anode
2) Silver act as anode
3) Zinc ions are reduced during the cell reaction
4) Silver gets oxidised as cell operates
LEVEL - II
1. Which of the following is intermolecular redox reaction?
CHO  CH2OH
1) 2 
OH

CHO COOH
Al OC2 H5 3
2) 2C6 H 5CHO  C6 H5COOH  C6 H5 CH 2 OH
 2Cr2  SO 4 3  6H 2 O  7O2
3) 4CrO5  6H 2SO4 
4) As 2S3  HNO3 
 H 3 AsO 4  H 2SO 4  NO

2. Cr2O 72  X 
H
Cr 3  H 2 O  oxidized product of X, X in the above reaction cannot be:
1) C 2 O 24  2) Fe 2  3) SO 24  4) S2
3. Which of the following statements is wrong?
1) Acidified KMnO4 solution decolourises on the addition of sodium oxalate
2) In the reaction between Br2 and CsI, Br2 is an oxidising agent and CsI is a reducing agent
3) In the reaction 2K 2S2 O3  I 2 
 2KI  K 2S4 O 6 , the change in the oxidation number of S is 0.5.
4) C has the same oxidation number in both CH4 and CO2.
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4. Which of the following can be oxidized further with a strong oxidising agent?
1) SO 2 b) MnO 2 c) Al 2 O3 d) CrO3
1) a and b 2) a and c 3) b and d 4) c and d

5. In the reaction, 8Al  3Fe3O 4  4Al 2 O3  9Fe , the number of electrons transferred from reductant to oxidant is:
1) 8 2) 4 3) 16 4) 24
6. Equivalent weight of H3PO2 (molecular weight = M) when it disproportionates into PH3 and H3PO3 is:
1) M 2) M/2 3) M/4 4) 3M/4
7. In the reaction, C4 H10 l  Cr2 O 27 aq  H  aq  H 6 C4O 4 s  Cr 3 aq  H 2 O l the change in the oxidation
        
number of the chromium atom is:
1) a decrease by six units 2) a decrease by three units
3) an increase by three units 4) an increase by five units
8. A compound contains atoms of three elements A, B, and C. If the oxidation number of A is +2, B is +5 and
that of C is –2, the possible formula of the compound is:

1) A 3  BC 4  2 2) A3  B4C  2 3) ABC 2 4) A 3  BC3 2

9. A compound of Xe and F is found to have 53.3% Xe. Oxidation number of Xe in this compound is:
1) –4 2) zero 3) +4 4) +6
10. Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen?
1) HNO3 , NO, NH 4 Cl, N 2 2) HNO3 , NO, N 2 , NH 4 Cl
3) HNO3 , NH 4 Cl, NO, N 2 4) NO, HNO3 , NH 4 Cl, N 2
11. In the compound YBa2Cu3O7 which shows super conductivity, what is the oxidation state of Cu?
Assume that the rare earth element yttrium is in its usual +3 oxidation state.
7 7 5 5
1)  2)  3) 4) 
3 3 3 3
12. If 2.68  10 3 mol of a solution containing an ion An+ requires 1.61 × 10–3 mol of MnO 4 for the oxidation of
An+ to AO3 in acid medium, what is the value of n?
1) 3 b) 5 C) 2 D) 4
13. Which are not correct about CH2= CCl2?
a) Both carbons are in +2 oxidation state b) Both carbons are in –2 oxidation state
c) One carbon has +2 and other has –2 oxidation state d) The average oxidation number of carbon is zero
1) a only 2) a and b 3) a and b 4) c and d
14. MnO 24  undergoes disproportionation reaction in acidic medium but MnO 4 does not because:

1) In MnO 24  , Mn is in intermediate oxidation state 2) In MnO 4 , Mn is in lowest oxidation state

3) In MnO 4 , Mn is in intermediate oxidation state 4) None of the above

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15. The reaction of white phosphorous with aqueous NaOH gives phsophine along with another phosphorous
containing compound. The reaction type; the oxidation states of phosphorous in phosphine and the other
product are respectively:
1) redox reaction; –3 and –5 2) redox reaction; + 3 and + 5
3) disproportionation reaction; –3 and +1 4) disproportionation reaction; –3 and +3
16. Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the statement is
incorrect in this reaction?
1) Evolved I2 is reduced 2) CuI2 is formed 3) Na2S2O3 is oxidised 4) Cu2I2 is formed
17. Which of the following are disproportionation reaction:

a) F2  H 2 O 
 HOF  HF b) 2HCHO  NaOH 
 HCOONa  CH 3OH

c) P4s   3NaOH  3H 2 O 
 PH 3  3NaH 2 PO2 d) 2NO 2  2KOH 
 KNO 2  KNO 3  H 2 O

1) a and b 2) b and c 3) c and d 4) b, c and d

18. Which of the following show intramolecular redox change?

(I) 2KClO3  2KCl  3O 2 (II)  NH 4 2 Cr2 O7  N 2  Cr2O3  4H 2O

(III) NH 4 NO 2  N 2  2H 2 O (IV) 2Cu 2  Cu 2  Cu

1) II, III 2) I, IV 3) I, II, III 4) II, IV
19. In the reaction :
As 2S3  HNO3  H 3 AsO 4  H 2SO 4  NO the element oxidised is:
1) As only 2) S only 3) N only 4) As and S both
20. In which of the following compounds, the oxidation number of iodine is fractional?
1) IF7 2) I3 3) IF5 4) IF3
21. The correct order of compounds in the decreasing order based on the oxidation state of oxygen in them is
1) OF2  HOF  KO 2  Sr  IO3 2 2) OF2  KO 2  HOF  Sr  IO3 2

3) HOF  KO2  Sr  IO3 2  OF2 4) KO2  OF2  HOF  Sr  IO3 2

22. For the redox reaction, MnO 4  C 2 O 42   H   Mn 2   CO 2  H 2 O

The correct coefficients of the reactants for the balanced reaction are:
MnO 4 C 2 O 24  H
1) 2 5 16
2) 16 5 2
3) 5 16 2
4) 2 16 5

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23. Number of moles of MnO 4 required to oxidise one mole of ferrous oxalate completely in acid medium will be:
1) 7.5 moles 2) 0.2 moles 3) 0.6 moles 4) 0.4 moles
24. When the ion Cr2 O 72  acts as an oxidant in acidic aqueous solution the ion Cr3+ is formed. How many mole of
Sn2+ would be oxidised to Sn4+ by one mole of Cr2 O 72  ions?
1) 2/3 2) 3/2 3) 2 4) 3
25. The values of ‘x’ and ‘y’ in the following redox reaction are : xCl 2  6OH  
 ClO 3  yCl  3H 2 O
1) x = 2, y = 4 2) x = 5, y = 3 3) x = 3, y = 5 4) x = 4, y = 2

26. C2 H 2  MnO4 
H
 Mn 2  CO 2
The number of H  ions in the balanced equation of the above redox reaction is:
1) 8 2) 6 3) 4 4) 2
27. Consider the chemical change which is occuring in basic medium: ClO 3  N 2 H 4  NO  Cl   H 2 O .
The correct statement is:
1) N 2 H 4 is acting as an oxidizing agent in the reaction
2) 3 mol of N2H4 are required per mole of ClO3
3) The electrons transferred per mol of N 2 H 4 reduce 1.33 mol of ClO3
4) 5 mol of electrons are gained by per mole of ClO3
28. For the reaction: I   ClO3  H 2SO 4 
 Cl   HSO 4  I 2
The correct statements in the balanced equation are:
a) stoichiometric coefficient of HSO 4 is 6 b) iodine is oxidised
c) sulphur is reduced d) H2O is one of the products
1) a only 2) a and b 3) c and d 4) a, b and d
29. Which among the following statements is incorrect?
1) In permanganometric titration MnO 4 act as self indicator..

2) In dichrometry Cr2 O72  is not a self indicator, diphenyl amine is used as indicator..

3) Cu 2  in faintly acidic medium oxidise iodide to iodine

4) I2 is soluble in water and remains in aqueous iodine solution as I2(aq)
30. Amount of oxalic acid present in a solution can be determined by its titration with KMnO 4 solution in the
presence of H 2SO 4 . The titration gives unsatisfactory result when carried out in the presence of HCl, because
HCl
1) gets oxidized by oxalic acid to chlorine
2) furnishes H+ ions in addition to those from oxalic acid
3) reduces permanganate to Mn2+
4) oxidizes oxalic acid to carbon dioxide and water
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31. A 15 mL sample of a solution containing oxalic acid, H2C2O4, was titrated with 0.02 M KMnO4. The titration,
required 18.30 mL of the KMnO4 solution. What was the molarity of the H2C2O4 solution? In the reaction,
oxalate ion  C2 O 42  is oxidized to CO2 and MnO 4 is reduced to Mn2+.
1) 0.082 2) 0.061 3) 0.61 4) 0.82
32. Select the correct statements.
M
a) Equivalent weight of Ca(HC 2 O 4 ) 2 is when it is a reducing agent.
2
b) Equivalent weight of Ca(HC2 O 4 ) 2 is M when it behaves as an acid.
c) Ca(HC2 O 4 ) 2 can be estimated by MnO 4 / H  .
d) Ca(HC2 O 4 ) 2 can be estimated by an acid.
1) a and c 2) b and c 3) a, b and c 4) a, c and d
33. H2C2O4 and NaHC2O4 behave as acids as well as reducing agents. Which are correct statements?
a) Equivalent weight of H2C2O4 and NaHC2O4 are equal to their moleculear weights when acting as reducing
agents.
b) Equivalent weight of H2C2O4 and NaHC2O4 are equal to half their molecular weights when acting as
reducing agents
c) 100 mL of 1 M solution of each is neutralized by equal volumes of 1 N Ba(OH)2
d) 100 mL of 1 M solution of each is oxidized by equal volumes of 1 M KMnO4
1) a and c 2) b and c 3) b and d 4) c and d
 
34. When the following half-reaction is balanced: CN  CNO
Which of the following statement is (are) true regarding the balanced half-reaction?
a) Carbon is losing two electrons per atom
b) Oxidation number of carbon increases from +1 to +3
c) Oxidation number of nitrogen remains constant
d) Oxidation number of nitrogen is decreases from +3 to –3
1) a and c 2) a and d 3) a, b and c 4) a, b and d
35. Which of the following statements is correct?
1) An element in the lowest oxidation state acts only as a reducing agent.
2) An element in the highest oxidation state acts only as a reducing agent
3) The oxidation number of V in Rb4 K(HV10O28) is +4.
4) The oxidation number and valency of Hg in calomel is +1.
36. The standard reduction potential values of three metallic cations X, Y and Z are 0.52 V, – 3.03 V and – 1.18V
respectively. The order of reducing power of the correponding metals is
1) Y > Z > X 2) X > Y > Z 3) Z > Y > X 4) Z > X > Y
37. In aqueous medium H2 gas can reduce which among the following metal cations?
2
1) Zn 2 2) Cu 2  3) Mg 4) Fe 2 
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38. The statement “the compound is unstable. However it formed the compound acts as a strong oxidising agent.”
as best applicable to
1) CuF2 2) AgF2 3) PbCl2 4) SnCl2

39. For the electrochemical cell M | M  ||X  | X E oM / M  0.44 and E oX/ X  0.33V .

From this data one can deduce that

1) M  X  M   X  is spontaneous reaction.

2) M   X   M  X is the spontaneous reaction.

3) Ecell = 0.77 V
4) Ecell = –0.7 V
40. The standard electrode potential of hydrogen electrode is
1) 1.0 V 2) 0.00 V 3) 4.00 V 4) 2.00 V

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SOLUTIONS
LEVEL - I
1. 2
2. 3
3. 4 Carbon is also oxidised.
4. 4 NH4 NH3 (No change in oxidation number)
x = –3 x = –3
5. 3 The order of reactivity of these metals is Mg > Mn > Zn > Cu > Ag.
6. 2 2N5   N    8e , 2 mole HNO3 gives one mole of N2O
2

7. 3 In CH3CHO, 2x + 4 – 2 = 0  x  1
1 2 3 4
8. 4 N 2 O  N O  N 2 O3  N O 2
9. 3
10. 2 It is +6 in both cases.
11. 2 Carbon has negative oxidation no. in Mg2C3 and positive oxidation number in C3O2;
12. 1
13. 2 Mn 2   MnO 2  K 2 MnO 4  KMnO 4
 2   4   6  7

14. 3  NH 4  NO3

NH 4 NO3 
15. 3 Oxidation number change = +4 – (–4) = +8
16. 3
17. 2 Nitrogen in HNO3 and HCN has +5 and –3 oxidation number, respectively.
18. 1 2HI  H 2SO 4  I 2  SO 2  2H 2 O
19. 4
20. 4 Disproportionation involves simultaneous oxidation and reduction of the same atom in a molecule.
21. 3 i) H2O2 is oxidised and thus reductant.
ii) Ag2O is reduced and thus H2O2 is reductant.
22. 2 Reaction II is disproportionation, while I, III, and IV are not.
23. 3 Cu 2S  2KMnO 4  4H 2SO 4 
 2CuSO 4  2MnSO 4  K 2SO 4  4H 2 O
5 0
24. 1 2 KI O3  5SO 2  4H 2 O  2KHSO 4  3H 2SO 4  I 2
25. 2 S2– has minimum oxidation number and thus, can act only as reducing agent.
26. 2 Oxygen in H2O2 and BaSO4 (formed) has –1 and –2 oxidation number respectively.
27. 4 In the given reaction, there is a change involved in the number of electrons of C atom from 0 to 4 which
indicates the oxidation of C atom.
28. 4 The balanced equation in basic media is, Br   6OH   BrO3  3H 2 O  6e 

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29. 1 Cr2 O 27   3Sn 2   14H   3Sn 4   2Cr 3  7H 2 O

3 mol of Sn2+ react with 1 mol of Cr2 O 72  . So, 1 mol of Sn2+ will react with (1/3) mol of Cr2 O 72 
30. 3 The balanced chemical equation is : 2ClO 2  5H 2 O 2  2OH   2Cl  5O 2  6H 2 O
2 mol ClO2  5 mol H2O2; 1 mole ClO2  2.5 mol H2O2
31. 1
32. 3
2
33. 1 5 mole FeSO4  1 mole KMnO4 'X '  mole
5
3 2
5 mole FeC2O4  3 mole KMnO4  'Y ' 
5
34. 4 It is disproportionation reaction, and hence, Br2 is both reduced and oxidised.
35. 2 The balanced equation is 3MnO 24   4H   2MnO 4  MnO 2  2H 2 O
36. 2 Na2S reducing K2Cr2O7 but NaCl does not.

37. 4 The ionic equation is; Cr2 O 72  14H   6Fe 2   2Cr 3  6Fe3  7H 2 O

38. 3 e– + MnO4 MnO42– 3e– + MnO4 MnO2

–
4e– + 2MnO4 Mn2O3 5e + MnO4 Mn2+
39. 3
40. 1 Zn is the anode and it gets oxidised to Zn2+.
Ag acts as Cathode Ag+ are reduced to Ag.
LEVEL - II
1. 4 Intermolecular redox change involves oxidation of one molecule and reduction of other molecule.
2. 3 SO 24  cannot be oxidized since the oxidation state (+6) of S is highest.
3. 4
4. 1
5. 4
1 3 3
6. 4  P H 3  2H 3 P O3
The reaction involved is, 3H3 P O 2 
The number of electrons transferred in oxidation half reaction is 4 and in the reduction half reaction is 2.
Hence, the effective number of electrons transferred = 4, and total number of moles involved = 3.
Molecular weight M 3M
Therefore, Equivalent Weight   
n factor 4/3 4
7. 2 The oxidation state of Cr changes from +6 to +3.
8. 1 The formation of the compound is A3(BC4)2
9. 4 In XeF6 the O.N of Xe is +6.
5 2 0 3
10. 2 H N O3 , NO, N 2 , NH 4 Cl

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7
11. 1  YBa 2Cu 3O7  : 3 + (2 × 2) + 3x – 14 = 0;  x  
3
12. 3 Since in acidic medium, An+, is oxidized to AO3 , the change in oxidation state from (+n) to (+5) = 5 – n.
Total number of electrons that have been given out during oxidation of 2.68 × 10–3 mol of
A n   2.68 103   5  n 

Thus, the number of electrons added to reduce 1.61 10 3 mol of MnO 4 to Mn 2  , that is (+7) to (+2)
= 1.61 × 10–3 × 5; Therefore, 1.61 × 10–3 × 5 = 2.68 × 10–3 × (5 – n); Solving, we get n = 2
13. 2

14. 1 In MnO 24  , the oxidation number of Mn is +6. It can increase its oxidation number (to +7) or decrease
its oxidation number to +4, +3, +2, 0).
In MnO 4 , Mn is in its highest oxidation state, i.e., +7.
15. 3 The balanced disproportionation reaction is
3 1
P40  3NaOH  3H 2 O 
 PH 3  3NaH 2 PO 2
16. 2 2CuSO 4  4KI 
 Cu 2 I 2  2K 2SO 4  I 2
I 2  2Na 2S2 O3 
 Na 2S4 O 6  2NaI
This liberated I2 is reduced and NaS2O3 is oxidised.
0 2 0 1 1
17. 4 F2  H 2 O 
 HOF H F
(Since Fis the most electronegative element,it can not exist in  1oxidation state )
18. 3 It involves redox change of two species in a molecule.

19. 4  As 
3
2
 2As5  4e ;  S2   3S6  24e
3

20. 2 –1/3
21. 1 Oxidation state of oxygen in OF2 = +2 Oxidation state of oxygen in HOF = 0
Oxidation state of oxygen in KO2 = –1/2 Oxidation state of oxygen in Sr(IO3)2 = –2
22. 1 2MnO 4  5C 2 O 42   16H   2Mn 2   10CO 2  8H 2 O
23. 4 2MnO 4  16H   5C 2 O 42  2Mn 2   10CO 2

24. 4  Cr26   6e 
 2Cr 3   1; Sn 2  
 Sn 4  2e    3

25. 3 3Cl 2  6OH  

 ClO3  5Cl  3H 2 O
26. 2 6H   2MnO 4  C 2 H 2  2Mn 2  2CO 2  4H 2 O

27. 3 The balanced chemical equation is : 4ClO 3  3N 2 H 4  4Cl   6NO  6H 2 O

28. 4 The balanced equation is: ClO 3  6I   6H 2SO 4 

 3I 2  Cl  6HSO 4  3H 2 O

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29. 4 I2 is insoluble in water but dissolves in presence of KI forming KI3.

2
30. 3 MnO 4 gets easily reduced to Mn in the presence of HCl, hence is not used.

31. 2 First we need a balanced equation : C 2 O 42  CO 2  CO 2 ; MnO 4  Mn 2   Mn 2 

2 1 0.02000 5
Moles of C2 O 4  18.30     0.0009150 mol of C 2 O 24 
1000 1 2
 H 2C2O4   0.06100 M
32. 3 As an acid : HC 2 O 4 
 H  (n  1)

As reducing agent C 2 O 24  
 2CO 2  2e  (n  2)

Ca(HC2O4)2 can be estimated by MnO 4 / H 

33. 3 only b and d are correct.
34. 1 In the given redox reaction :  C  N   O  C  N
Oxidation no. of N is remaining unchanged at –3. Oxidation no. of carbon is increasing from + 2 to +4.
35. 1 An element in the lowest oxidation state can only attain higher oxidation state. So it is a reductant and
undergoes oxidation.
Calomel is Hg 2 Cl2 . The oxidation number of Hg is +1 and valency is 2.
36. 1 Since the std electrode potential is the most negative Y is the strongest reducing agent.
37. 2 E oCu  /Cu is 0.34 V while that of Hydrogen zero.
38. 2 AgF2 is unstable. It is a strong oxidising agent if it is formed.
39. 2 E ocell is positive for X  | X || M  | M and the cell reaction is X   M   X  M
40. 2 Eo of SHE is zero.

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CHAPTER - 09

HYDROGEN

INTRODUCTION
Hydrogen means water producer. It is the most abundant element in the universe and the third most abundant
on the surface of the earth. In elemental form it exists as a diatomic molecule and is called dihydrogen. Its
formation was first recognised by Paracelus from the reaction between iron and sulphuric acid. In 1766,
English scientist, Henry Cavandish produced it in pure state by the action of acids on certain metals. The name
hydrogen was proposed by Lavoisier. Hydrogen forms more compounds than any other element. It has great
industrial importance and is being visualised as the major future source of energy.
9.1 POSITION OF HYDROGEN IN THE PERIODIC TABLE
Hydrogen is the first element in terms of atomic number; however, it is difficult to assign a place to it in the
periodic table because of its resemblance with halogens as well as alkali metals.
Resemblance with Alkali Metals: Electronic configuration; Hydrogen contains only one electron in the
valence shell, 1s1. Electropositive character; During electrolysis of water, hydrogen is liberated at the cathode;
(similar to alkali metals); Oxidation state. Hydrogen exhibits +1 oxidation state. Reducing agent; Hydrogen
is a very good reducing agent; Combination with electronegative elements; Hydrogen combines with
halogens, oxygen, sulphur, etc.
Difference from Alkali Metals: Ionisation enthalpy; Ionisation enthalpy (1312 kJ mol–1) is very high,
therefore, it has lesser tendency to form H+ ions. Size and existence of H+; H+ ion does not exist freely in
aqueous solution because of very small size (1.510-3 pm). It exists in aqueous solution as hydrated proton,
H3O+ , called hydronium ion. Halides; Hydrogen halides differ from halides of alkali metals. Pure HCl is
covalent while NaCl is ionic. HCl is a gas while NaCl is a solid at ordinary temperature.
Resemblance with Halogens: Electronic configuration; Hydrogen needs one more electron to attain
noble gas configuration. Atomicity and nonmetallic character; The atomicity of hydrogen is two. It is a
typical nonmetal. Electrochemical nature; During electrolysis of molten LiH, CaH2, etc., hydrogen is evolved
at the anode indicating electronegative nature. Oxidation state; Hydrogen exhibit -1 oxidation state in metal
hydrides. Combination with alkali metals; Hydrogen combines with alkali metals to form hydrides (salts).
Combination with non-metals; Hydrogen reacts with non-metals to form covalent compounds. Ionisation
enthalpy; Ionisation enthalpy of hydrogen is comparable to those of halogens.
Difference from Halogens: Lesser tendency for hydride-formation; Halogens form halides with a very
large number of metals, but hydrogen form hydrides only with a small number of metals like Na, Ca, etc.
Absence of unshared electrons; There is no unshared pair of electrons in hydrogen molecule, whereas
halogen molecules have six unshared electron pairs. Nature of oxides; The oxides of halogens are acidic
(Cl2O7, I2O5), whereas water is neutral.
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Thus, there is marked resemblance in the properties of hydrogen with alkali metals as well as halogens, therefore,
it is difficult to place it either in group 1 or in group 17. Thus, hydrogen is called rogue element.
Mendeleev placed hydrogen in isolation from other elements due to its unique properties. In the long form of
periodic table, hydrogen is placed along with the elements of group1, but slightly separated to indicate its
distinctive character.
9.2 DIHYDROGEN OR H2 MOLECULE
Hydrogen exists as diatomic covalent molecule. Molecular hydrogen is called dihydrogen. The term dihydrogen
is used for the mixture of diatomic molecules such as H2, D2 and HD. Diatomic molecules containing only
protium atoms (H2) are called diprotium. Diatomic molecules containing only deuterium atoms (D2) are called
dideuterium. Unipositive protium ion (H+) is called proton. An isotopic mixture of proton (H+) and deutron
(D+) with respect to their natural abundance is called hydron.
9.2.1 Occurrence. Hydrogen is the most abundant element in the universe (about 70% of total mass) and the
principal element in the solar atmosphere. On the earth, it occurs in the free state in traces in volcanic and
natural gases. In the combined state, it is widely distributed as water. It is an important constituent of organic
compounds, and animal and vegetable matter. It constitutes about 15.4% by mass of the earth’s crust and
oceans. It is the tenth element in the order of abundance and 0.15% by weight.
9.2.2 Isotopes of hydrogen. Hydrogen has three isotopes; protium, deuterium, and tritium.
i. Protium or hydrogen (H or 11 H ). It has one proton in its nucles and one electron in 1s orbital (nuclear
spin = ½). Naturally occuring hydrogen has 99.985% of this isotope.
ii. Deuterium or heavy hydrogen (D or 12 H ). It has one proton and one neutron and one electron in 1s
orbital (nuclear spin = 1). Naturally occuring hydrogen has 0.015% of this isotope, mostly in the form of
HD.
iii. Tritium (T or 13 H ). It has one proton and 2 neutrons in the nucleus and one electron in 1s orbital (nuclear
spin = ½). It is extremely rare (10-16 %) and is formed in the upper atmosphere by nuclear reactions
induced by cosmic rays. Tritium is radioactive and emits low energy   particles. Its half life (t1/2) is
12.33 years.
Isotopic effect. It is the difference in properties of an element arising from difference in mass numbers.
9.2.3 Preparation of Dihydrogen
1. By the action of water on metals. Metals react with water at different temperatures to displace hydrogen.
a) Very active metals like Na, K and Ca react at room temperature to form H2.
2Na  2H 2 O   2NaOH  H 2 ; Ca  2H 2 O   Ca  OH  2  H 2
b) Metals like Mg, Al, Zn, etc., react with water at boiling temperatures to form H2.
Mg  H 2O 
Heat
 MgO  H 2 ; Zn  H 2 O 
Heat
 ZnO  H 2 ; 2Al  3H2O 
Heat
 Al2O3  3H2
c) Metals like Fe, Ni, Sn, etc., react with steam at high temperatures.
1000K
3Fe  4H 2O  Fe3O 4  4H 2 ; Ni  H 2O 
 NiO  H 2
The reaction between iron and steam is used for the large scale preparation of dihydrogen by Lane’s process.
2. By the action of acids on metals (Laboratory preparation). Metals above hydrogen in the activity
series displace hydrogen from dilute HCl or dilute H2SO4.
Zn  H 2SO4 
 ZnSO4  H 2

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3. By the action of strong alkalies on metals. Metals like Be, Zn, Sn or Al react with strong alkalies such
as NaOH or KOH liberating H2 gas.
Heat
Heat
Zn  2NaOH   Na 2 ZnO2  H 2 ; 2Al  2NaOH  2H 2O   2NaAlO2  3H 2
Sodium zincate Sodium meta-aluminate
4. Electrolysis of Water. Electrolysis of water in presence of a small amount of acid or base using platinum
electrodes liberate dihydrogen at the cathode and dioxygen at the anode.
Electrolysis
2H 2O(l ) 
Traces of acid/base
 2H 2 (g)  O 2 (g)

At cathode: 2H 2O  2e 
 2H 2  2OH  ; At anode: 2OH 
 H2O  (1/ 2)O2  2e
9.2.3 Manufacture of Dihydrogen
1. Bosch’s Process: (a) Preparation of Water gas/Syngas. Water gas is a mixture of CO and H2. It is used
for the synthesis of methanol and hydrocarbons. Hence, it is also called synthesis gas or syngas.
(i) Coal gasification. It involves the action of steam with red hot coke
C  H 2O  g  
1270K
 CO  g   H 2  g 
Water gas

(ii) Steam reforming of hydrocarbons. Action of steam on hydrocarbons in the presence of catalyst.
Cn H 2n  2  nH 2O  g  
1270 K
Ni
 nCO   2n  1 H 2 ; C3H8  3H2O  g  
1270 K
3CO  g   7H2 (g)
Ni
Syngas

b) Separation of Hydrogen (water gas-shift reaction). When syngas mixed with steam is passed over
heated Fe2O3 and Cr2O3 at 770 K, CO is oxidised to CO2. The gaseous mixture of CO2 and H2 is then
bubbled into cold water under pressure when CO2 dissolves leaving behind dihydrogen gas.
H 2  CO  H 2 O  g  
Fe 2 O3
Cr2 O3
 CO 2  2H 2
2. Electrolysis of Water. Electrolysis using iron cathode, nickel coated iron anode and 15–20% NaOH
solution as electrolyte.
1270K
3. Thermal cracking of natural gas : CH 4   C  2H 2 (g)
Catalyst
4. As a byproduct in the manufacture of sodium hydroxide. Dihydrogen is obtained as a byproduct in the
manufacture of NaOH by the electrolysis of brine.
At anode: 2Cl (aq) 
 Cl 2 (g)  2e  ; At cathode: 2H 2 O(l )  2e  
 H 2 (g)  2OH  (aq)
Overall reaction: 2Na  (aq)  2Cl (aq)  2H 2 O(l ) 
 Cl 2 (g)  H 2 (g)  2Na  (aq)  2OH  (aq)
9.2.4 Physical Properties: (i) Dihydrogen is a colourless, odourless and tasteless combustible gas. (ii) It has
extremely low solubility in water. (iii) Its density is approximately 1/14 th of air. (iv) It can be liquified only at
high pressures and low temperature.
9.2.5 Chemical Properties: Dihydrogen is not very reactive at room temperature due to its high bond enthalpy
(435.9 kJ mol–1), which is the highest for a single bond between two atoms. It dissociates into hydrogen atoms
to the extent of only 0.081% even at 2000K. Dissociation becomes 95.5% at 5000K. Atomic hydrogen is
produced at high temperature in an electric arc or by ultraviolet radiations.
1. Combustibility. It is highly combustible. It burns in air with a pale blue flame to form water
2H 2  g   O 2  g  
 2H 2O  l 

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2. Reaction with metals. It reacts with metals like Na, Ca, Li, etc., to form salt-like hydrides.
525 K 573 K
2Na  H 2  2NaH Ca  H2  CaH2
Sodium hydride
; Calcium hydride

In such reactions, hydrogen aquires -1 oxidation state and acts as an oxidising agent.
Occlusion: With metals like Pt, Pd, Ni, etc., hydrogen forms interstitial hydrides in which hydrogen atoms
occupy interstitial holes in metallic crystals. This property is referred to as occlusion. The occluded hydrogen
can be liberated from the metal by strong heating.
Table - 01 Physical properties of isotopes of hydrogen.
Property Hydrogen Deuterium Tritium
-15
1.Relative abundance (%) 99.985 0.0156 10
-1
2.Relative atomic mass (g mol ) 1.008 2.014 3.016
-1
3. Density (g L ) 0.09 0.18 0.27
-1
4. Molar mass (g mol ) 2.016 4.028 6.03
5. Melting point (K) 13.96 18.73 20.62
6. Boiling point (K) 20.39 23.67 25
-1
7. Enthalpy of fusion (kJ mol ) 0.117 0.197 0.25
-1
8. Enthalpy of vaporisation (kJ mol ) 0.904 1.226 1.39
-1
9. Bond enthalpy (298 K) (kJ mol ) 435.88 443.35 446.9
-1
10. Ionisation enthalpy (kJ mol ) 1312
-1
11. Electron gain enthalpy (kJ mol ) -73
12. Covalent radius/pm 37
-
13. Ionic radius (H )/pm 210
14. Electronegativity 2.1
15. Internuclear distance (pm) 74.14 74.14
3. Reaction with non-metals:
i. With dioxygen : 2H 2  g   O 2  g  
970 K
 2H 2 O  g  ; H  285.9kJ mol1

ii. With dinitrogen (Haber’s process) : 3H 2  g   N 2  g  

Fe
 2NH3 (g); H  96.6 kJ
673K, 200atm.

iii. With sulphur : H 2  g   S  l  

Pr essure
 H 2S g 
710K
Hydrogen sulphide

 CH 4  g  ; H 2  g   2C  s    C2 H 2  g 
Electric arc
iv. With carbon : 2H 2  g   C  s  
1275 K
3300 K
Methane Acetylene

v. With halogens : H2  g   F2  g 


Dark
2HF ; H  g   Cl  g  
Sunlight
 2HCl  g 
2 2
Hydrogenfluoride Hydrogen chloride

H 2  g   Br2  g   2HBr  g 

670K
Catalyst
 ; H 2  g   I2 
670K, Pt
 2HI
Hydrogen bromide Hydrogen iodide

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The order of reactivity of halogens with hydrogen is : F2 > Cl2 > Br2 > I2
4. Reaction with metal oxides. Dihydrogen reduces metal oxides to metal.
CuO  s   H 2  g  
Heat
 Cu  s   H 2O  l  ; Fe3O 4  s   4H 2  g  
Heat
 3Fe  4H 2O

PbO  s   H 2  g  
Heat
 Pb  s   H 2O  l  ; ZnO  s   H 2  g  
Heat
 Zn  s   H 2 O  l 
5. Reaction with metal ions. Dihydrogen reduces some metal ions (metals lying below hydrogen in activity
series) in aqueous solution to the metallic state.
H 2  g   Pd 2  aq  
 Pd  s   2H   aq 
6. Reaction with carbon monoxide. Dihydrogen react with carbon monoxide at 700 K in the presence of a
catalyst ZnO/Cr2O3 to produce methanol.
CO  g   2H 2  g  
700K,200atm
ZnO,Cr2 O3
 CH3OH
Methanol

7. Hydrogenation of unsaturated hydrocarbons. Unsaturated hydrocarbons like ethene and ethyne

react with dihydrogen in presence of Ni, Pt or Pd as catalyst to form saturated hydrocarbons.
CH 2  CH 2  H 2 
Ni or Pt
473K
CH3  CH 3 ; CH  CH  2H 2 
Ni or Pt
473K
CH 3  CH3
Ethene Ethene Ethene Ethene

8. Hydrogenation (Hardening) of vegetable oils. When dihydrogen under pressure is passed through
vegetable oils such as ground-nut oil or cotton seed oil at about 473 K in presence of finely divided nickel
(Raney nickel) catalyst undergo hardening to edible fats known as margarine or vanaspati.
Vegetable oil  H 2 
473K
Ni
 Semisolid fat
 Vanaspati ghee 

The process is known as hydrogenation or hardening of oils.Vegetable oils contain unsaturated compounds
with C=C bonds, which undergo oxidation and develop unpleasant taste and smell. This phenomenon is
referred to as oxidative rancidity. Hydrogenation reduces double bonds and helps to control rancidity.
9. Hydroformylation of alkenes (Oxo Process). Hydroformylation of alkenes give aldehydes which further
get reduced to alcohols.
CH 3CH  CH 2  H 2  CO 
Catalyst
CH 3CH 2 CHO ; CH 3CH 2CHO  H 2 
Catalyst
 CH3CH 2 CH 2 OH
Propene Propanal Propanol

9.2.6 Uses of Dihydrogen: (i) Manufacture of ammonia which is a starting material for the maufacture of various
fertilizers like urea, ammonium sulphate, calcium ammonium nitrate, etc. (ii) Hydrogenation of oils. (iii)
Manufacture of methanol, synthetic petrol, etc. (iv) Manufacture of metal hydrides. (v) Reducing heavy metal
oxides to metals (in metallurgy). (vi) Oxy-hydrogen torch for welding at about 2500oC. (vii) Atomic hydrogen
torch for welding around 4000oC. (viii) As rocket fuel. (ix) Generating electrical energy in fuel cells.
9.3 HYDRIDES
Dihydrogen combines with metals as well as non-metals to form hydrides. Hydrides are binary compounds of
hydrogen with other elements. They have the general formula, EHx (e.g., MgH2) or EmHn (e.g., B2H6), where
E is symbol of the element. Hydrides are classified into three main classes, viz., ionic hydrides, covalent
hydrides, and metallic hydrides.
9.3.1 Ionic Hydrides (Saline or Salt-like hydrides). These are binary compounds of hydrogen with more
electropositive elements such as alkali and alkaline earth metals (except Be and Mg which form covalent
polymeric hydrides). Some common examples are : LiH, NaH, CaH2, SrH2, etc.

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a) General Characteristics of Ionic Hydrides. (i) They are crystalline solids having white or greyish colour.
(ii) They have high melting and boiling points. (iii) They have high density, generally higher than that of the
metals from which they are formed. This is because H– ions occupy the voids of metal atoms and thus do not
distrub their crystal structure. (iv) They have high values of  f H  . (v) Their melts conduct electricity liberating
dihydrogen at anode, which confirms the presence of H– in them.

2H  (melt) 
anode
 H 2 (g)  2e
vi) They react vigorously with water and other protonic solvents like ethanol and ammonia to liberate H2.

 LiOCH 3  H 2 ; NaH  NH3 

 NaOH  H 2 ; LiH  CH 3OH 
NaH  H 2 O   NaNH3  H 2
Lithium methoxide Sodamide

vii) Hydrides (except LiH) undergo combustion when heated strongly in air.
675  775K
CaH 2   Ca  H 2
viii) They act as powerful reducing agents especially at high temperatures.
2CO  NaH 
Heat
 HCOONa  C ; CO 2  NaH 
Heat
 HCOONa
Sodium formate Sodium formate

ix) LiH is rather unreactive at moderate temperatures towards O2 or Cl2, but reacts with metal halides to give
complex metal hydrides which are used as reducing agents in organic synthesis.
LiH  Al2 Cl6 
Dry ether
 2LiAlH 4  6LiCl ; LiH  B2 H 6 
Dry ether
 2LiBH 4
Lithium alu minium hydride Lithium borohydride

Similarly, NaH reacts with diborane to give sodium borohydride (NaBH4).

NaH  B2 H 6 
Dry ether
 2NaBH 4
9.2.2 Covalent or Molecular Hydrides. These are binary compounds of hydrogen and elements of comparatively
high electronegativity such as p-block elements, e.g., HCl, H2O, CH4, PH3, NH3, etc. The general formula of
covalent hydrides is XH(8–n) where n is the number of outershell electrons of X atom. However, group 13
elements such as B, Ga form polynuclear hydrides which are electron deficient, e.g., B2H6, Ga3H2, etc.
1. Classification. These are classified into three different categories depending upon the number of electrons
and bonds present in their Lewis structures.
a) Electron deficient hydrides. They do not have sufficient number of electrons to form normal covalent
bonds, e.g., elements of group 13 such as BH3, AlH3, etc. They generally exist in polymeric forms like
B2H6, B4H10, (AlH3)n, etc.
b) Electron precise hydrides. They have exact number of electrons to form normal covalent bonds, e.g.,
hydrides of elements of group 14 such as CH4, SiH4, GeH4, etc.
c) Electron rich hydrides. They have more electrons than required for forming normal covalent bonds.
The excess electrons are present as lone pairs, e.g., hydrides of elements of groups 15, 16, and 17 such as
 , HCl  :, H  
NH 2S :, H 2 O : ,etc.

3

2. Nomenclature. The systamatic names of molecular hydrides are derived from the name of the element by
attaching the suffix ‘ane’.
NH3 PH3 H2S H 2O SiH4 GeH4
Azane Phosphane Sulphane Oxidane Silane Germane

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Hydrides of the first elements of group 15, 16 and 17 (i.e., NH3, H2O, HF) have abnormally high boiling points
due to intermolecular hydrogen bonding. The intermolecular forces in hydrides of the rest of the elements of
each group are relatively weaker dipole-dipole interactions.
3. General Characteristics of Covalent Hydrides: (i) soft with low melting and boiling points; (ii) poor
conductors of electricity; (iii) more soluble in organic solvents; (iv) undergo thermal decomposition to the
respective elements; (v) strong reducing agents; (vi) some of them react with water to liberate dihydrogen;
(vii) become increasingly acidic in a period.
9.3.3 Metallic or Non-stoichiometric (or Interstitial ) Hydrides. These are binary compounds of hydrogen and
transition elements of groups 3, 4, 5 and Cr of group 6 and f-block elements. Metals of groups 7, 8, and 9 of
d-block do not form hydrides; this is referred to as hydride gap of d-block.
These hydrides conduct heat and electricity though not as efficiently as their parent metals. They are
nonstoichiometric, being deficient in hydrogen, e.g., LaH2.87, YbH2.55, TiH1.5–1.8, ZrH1.3–1.75, VH0.56, NiH0.6–0.7,
PdH0.6–0.8, etc. In these hydrides, the law of constant composition does not hold good.
Except for hydrides of Ni, Pd, Ce and Ac, these hydrides have lattice different from that of the parent metal.
Many transition metals like (e.g., Pt, Pd) absorb large quantities of hydrogen forming interstitial hydrides, e.g.,
red hot Pd, cooled in an atmosphere of H2, adsorbs 935 times its own volume of H2 gas. This property of
adsorption of H2 gas by metal (called occlusion) has high potential for hydrogen storage and as a source of
energy. The occluded H2 is liberated on strong heating
Red hot metal 
Cool
occlusion
 Interstitial hydride 
Strong heating
 Metal  H 2 (g)
9.3.4 Polymeric and Complex Hydrides. Polymeric hydrides are formed by elements having electronegativity
between 1.4 and 2.0. Molecules are held together in two or three dimensions by hydrogen bridges.
 BeH2 n ,  AlH3 n ,  InH3 n ,  SiH4 n , etc.
These are amorphous solids, stable up to 525 K. Above this temperature they decompose and evolve H2 gas.
Complex hydrides contain hydride (H-) ions coordinated to metal atoms/ions, e.g., LiAlH4 (lithium aluminium
hydride), NaBH4 (sodium borohydride), etc. These are generally very good reducing agents.
9.3.5 Uses of Hydrides: (i) Metal hydrides formed as a result of occlusion can be used as hydrogen storage media.
(ii) Occlusion can be used for separation of H2 and D2. (iii) Interstitial hydrides are used in catalytic reduction
and hydrogenation for preparing large number of useful compounds.
9.4 WATER
Water (H2O) is the most important hydride of oxygen. It is a principal constituent of the earth’s surface (about
75%). It constitutes about 65% of the human body and about 95% by weight of some plants.
9.4.1 Structure of Water Molecule. In the water molecule, two H atoms are bonded to an O atom by two
covalent bonds. The oxygen atom is in sp3 hybrid state. Each of the covalent bonds is formed by the axial
overlap of 1s-orbital of H atom and sp3 hybrid orbital of O atom. The O-H bond length is 95.7 pm. The two
bond pairs and two lone pairs around oxygen atom assume tetrahedral arrangement, therefore, H2O molecule
has a bent structure. Due to relatively greater repulsive interactions of lone pairs, the H–O–H bond angle
decreases from 109o.28´ to 104.5o.

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Polar nature of H2O. Since water molecule has bent shape, the bond moments of the two O–H bonds make
the molecule a permanent electrical dipole. A dipole moment of 1.84 D confirms its polar nature.
9.4.2 Aggregation of Water Molecules. In the gaseous state, H2O molecules exist as such. In the liquid state,
aggregates of varying number (H2O)n are formed due to association through intermolecular hydrogen bonds.
Thus, in liquid water, free and associated H2O molecules exist in a state of dynamic equilibrium. Intermolecular
hydrogen bonding is responsible for the abnormally high freezing point, boiling point, heat of fusion and heat of
vapourisation of water compared to the hydrides of other elements of oxygen family.

(a) (b)

Figure - 01 (a) Association of water molecules by H-bonding (b) Open cage like structure of ice.
9.4.3 Structure of Ice. In ice, each H2O molecule is tetrahedrally surrounded by four H2O molecules with their
oxygen atoms occupying the corners of the tetrahedron (distance between oxygen atoms = 276 pm).
There are four H atoms around each O atom; two of them bonded by covalent bonds (bond length 100 pm)
and the other two through hydrogen bonds (176 pm). This gives highly ordered three-dimensional structure
with large vacant spaces which may be compared to an open cage. Due to open-cage structure, ice has larger
volume and higher lower than liquid water.
Table - 02 Physical properties of water and heavy water at 298 K.
Prope rty H 2O D 2O
-1
1. Molar mass (g mol ) 18.015 20.028
2. Melting Point (K) 273.2 276.8
3. Boiling Point (K) 373.2 374.4
4. Tempertaure of Maximum Density (K) 276.98 284.4
-3
5. Density (g cm ) 1.000 1.1059
-1
6. Enthalpy of formation, DHf° (kJ mol ) -285.9 -294.6
-1
7. Enthalpy of Vaporisation at 373K (kJ mol ) 40.66 41.61
-1
8. Enthalpy of Fusion (kJ mol ) 6.01 6.28
--1 -1
9. Specific Heat (J g K ) 4.177 -
2 -2 -14 -15
10. Ionisation Constant, (mol L ) 1.008 × 10 1.95 × 10
2 2
11. Dielectric constant (C /Nm ) 78.39 78.06
12. Viscosity at 293 K (centipoise) 0.8903 1.107
-1 -1 -8
13. Electrical Conductivity at 293 K(ohm cm ) 5.7 × 10 -
Water has maximum density at 277 K. This is a boon for the survival of aquatic animals during winter when the
upper layer of sea water freezes. Frozen water does not sink to the bottom but keeps floating on the surface
which provides thermal insulation to the water below it.
At one bar pressure, ice has normal hexagonal form; however, at low temperatures, it adopts cubic form.
9.4.4 Physical Propertiesof Water
Water has higher specific heat, thermal conductivity and surface tension than many other liquids. Due to these
properties, water play a vital role in the biosphere. The high heat of vaporisation and high specific heat are
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responsible for the moderating influence of water on the climate and body temperature of living organisms.
Water is also an excellent solvent for transporting ions and molecules needed by plant and animal metabolism.
9.4.5 Chemical Properties
1. Amphoteric nature. Water is amphoteric because it can act as acid as well as base.
 NH4  OH ; HCl  H2O 
H2O  NH3   H3O  Cl
Acid Base Acid Base
The self ionization (autoprotolysis)of water can be represented as :
 H3O  (aq) 
H 2O(l )  H 2 O(l )  OH  (aq)
Acid 1 Base  2 Acid  2 Base 1
Acid Base Conjugate Base Conjugate Acid

However, the pH of water at 25oC is 7 and it is netural towards litmus.

2. Oxidation-reduction reactions. Water can undergo reduction as well as oxidation.
(i) As oxidising agent: Water reacts with metals having electrode potentials less than -0.83 V, liberating
dihydrogen. Thus, it reacts with active metals like Na, Ca, etc.
2H 2O  l   2e 
 2OH   aq   H 2  g  ; E o  0.83V

 2NaOH  H 2 ; C s   2H 2 O g  
2Na  2H 2 O   CO g   2H 2 g 
Oxidi sin g agent Oxidisingagent

(ii) As reducing agent: Water reacts with highly electronegative elements having electrode potentials higher
than +1.23 V, liberating dioxygen and trioxygen.
 O 2 g   4H   4e  ; E ox
2H 2O l   
 0.82 V

2F2  2H 2 O 
 4HF  O 2 ; 3F2 g   3H 2 Ol  
icecold water
6HF  O3
Reducing agent
Reducing agent

During photosynthesis, water is oxidised to dioxygen.

x CO 2 + y H 2O 
Sunlight
Chlorophyll
 Cx (H 2 O) y  x O 2
3. Hydrolysis reactions. Due to high dielectric constant, water has very strong hydrating tendency. It dissolves
many ionic compounds and hydrolyses certain covalent ionic compounds.
P4O10 + 6 H 2O 
 4H 3PO 4 ; SiCl4  2H 2O 
 SiO2  4HCl
Phosphoric acid Silicon tetrachloride Silicon dioxide

CaC 2  Ca  OH 2  C 2 H 2 ;
 2H 2 O  AlN  Al  OH 3  NH3
 3H 2O 
Calcium carbide Acetylene Aluminium nitride
4. Hydration reactions. Water combines with metal salts to form three categories of hydrates.
i) Coordinated water. Water molecules are coordinated to the central metal ion to form complex ions.
 Ni  H 2O 6   NO3  ;  Fe  H 2 O 6  3Cl  ; Cr  H 2 O 6  3Cl 
2 3+ 3+

ii) Hydrogen-bonded water. Water molecules are hydrogen bonded to certain oxygen containing anions, e.g.,
in CuSO4.5H2O, four water molecules are co-ordinated to the central Cu2+ ion while the fifth water molecule
is hydrogen bonded to sulphate groups. Thus, it can also be represented as [Cu(H2O)4] SO4.H2O.
iii) Interstitial water. Water molecules may occupy voids in the crystal lattice, e.g., BaCl2.2H2O.

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5. Water as solvent. Water has high dielectric constant due to polar character of its molecule. It is an
excellent solvant for ionic as well as covalent compounds. Dissolution of ionic compounds is due to ion-dipole
interactions. Dissolution of molecular compounds such as alcohols, amides, urea, sugar, glucose, etc., in water
is due to the formation hydrogen bonds with water molecules.
9.4.6 Hard water and Soft Water
a) Soft water. Water which readily produces lather with soap solution is called soft water, e.g.,distilled water,
demineralised water and rain water.
b) Hard water. Water which does not produce lather with soap solution readily is called hard water, e.g., river
water, sea water, lake water, water from borewells, etc.
Cause of Hardness of Water. Hardness of water is due to dissolved bicarbonates, chlorides and sulphates
of calcium and magnesium. Hard water does not produce lather with soap solution readily because the cations
(Ca+ and Mg2+) present in hard water react with soap (which is a mixture of sodium salts of higher fatty acids
like stearic acid, palmitic acid, oleic acid, etc.) to form a white curdy precipitate (scum) of Ca/Mg stearate.

M 2   C17 H35COO 2 M 2Na

 2C17 H35COONa 
(from hard water) Sodium stearate
(M = Ca or Mg)
Metal stearate

Thus, no lather is produced until all the calcium and magnesium ions are precipitated. This leads to wastage of
soap. Hard water is, therefore, not fit for washing purposes. It is also unfit for use in boilers because of
deposition of salts in the form of scales. Scale formation reduces efficiency and may cause boiler explosions.
9.4.7 Types of Hardness
i) Temporary hardness (carbonate hardness). This is due to the presence of soluble bicarbonates of
calcium and magnesium, i.e., Ca(HCO3)2 and Mg(HCO3)2.
ii) Permanent hardness (non-carbonate hardness). It is due to the presence of chlorides and sulphates of
calcium and magnesium, i.e., CaCl2, CaSO4, MgCl2, and MgSO4.
9.4.8 Softening of hard water. It is the process of removal of metallic ions responsible for hardness of water.
A. Removal of Temporary Hardness
i) Boiling. On boiling, soluble Mg(HCO3)2 and Ca(HCO3)2 are converted to insoluble Mg(OH)2 and CaCO3
respectively. These precipitates can be removed by filtration to get soft water.
Ca  HCO3 2 
Heat
 CaCO3  H2O  CO2 ; Mg  HCO3 2 
Heat
 Mg(OH)3  CO2
It is because of high solubility product of Mg(OH)2 as compared to MgCO3 that Mg(OH)2 is precipitated.
ii) Clark’s Process. Calculated quantity of lime is added to hard water and the precipitated calcium carbonate
and magnesium hydroxide are filtered off.

Ca  HCO3 2  Ca  OH 2 
 2CaCO3  2H 2 O

Mg  HCO3 2  2Ca  OH 2 
 MgCO3  2CaCO3  2H 2O

If excess lime is added, water will again become hard due to absorption of CO2 from the atmosphere.

Ca(OH) 2  2CO 2 
 Ca(HCO3 ) 2

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B. Removal of Permanent Hardness

1. Addition of washing soda. Ca2+ and Mg2+ ions can be precipitated by the addition of calculated amounts
of washing soda (Na2CO3.10H2O).

MCl2  Na 2CO3 
 MCO 3  2 NaCl  M  Ca or Mg 
MSO4  Na 2CO3 
 MCO3   Na 2 SO4  M  Ca or Mg 
2. Calgon Process. Ca2+ and Mg2+ ions are rendered ineffective (sequestration of ions) by the addition of
sodium hexametaphosphate (Na6P6O18), commercially called ‘calgon’.
[Na 4 P6 O18 ]2   2Na 
Na 6 P6 O18 

M 2  [Na 4 P6 O18 ]2 

 [Na 2 MP6 O18 ]2   2 Na  (M = Ca or Mg)
3. Ion Exchange Method. Ca2+ and Mg2+ ions present in hard water are exchanged for less damaging ions
from the exchangers.
a) Zeolite/permutit process. Permutit is hydrated sodium aluminium silicate Na2Al2SiO8.xH2O (or NaAlSiO4,
represented as NaZ) which can exchange cations like Ca2+ and Mg2+ in hard water for sodium ions.
2NaZ(s)  M 2  (aq) 
 MZ 2 (s)  2Na  (aq) (M = Ca or Mg)
The exhausted permutit/zeolite is regenerated for further use by treating with aqueous NaCl solution.
MZ 2 (s)  2NaCl(aq) 
 2NaZ(s)  MCl 2 (aq) (M = Ca or Mg)
The method can remove temporary as well as permanent hardness.
b) Synthetic resins method. Synthetic resins are complex organic molecules having either acidic group (e.g.,
R–SO3H or R–COOH) or basic group (OH– in the form of RH3N+OH-) on them.
Resins with acidic groups are called cation exchange resins or cation exchangers and those with basic
groups are called anion exchange resins or anion exchangers.
i) Removal of hardness. Ion exchange resin, RH is converted to RNa by treating it with NaCl. When hard
water is passed through cation exchange resin, Ca2+ and Mg2+ are exchanged with Na+ ions of the resin.
2RNa(s)  M 2  (aq) 
 R 2 M(s)  2Na  (aq) (M = Ca or Mg)
The exhausted resin can be regenerated for further use by treating it with aqueous NaCl solution.
9.5 HYROGEN PEROXIDE (H2O2)
Hydrogen peroxide was discovered by French chemist, J J Thernard in 1918. It is highly unstable, and therefore,
does not occur in nature. It is used in pollution control treatment of domestic and industrial effluents.
9.5.1 Methods of Preparation
a) From sodium peroxide (Merck’s process). A 30% solution of H2O2 is obtained by the action of calculated
amount of sodium peroxide on ice-cold dilute (20%) sulphuric acid.
Na 2 O 2  H 2 SO 4  
 Na 2 SO 4  H 2 O 2
b) From barium peroxide. Hydrated barium peroxide BaO2.8H2O reacts with cold dilute sulphric acid or
phosphoric acid. A five per cent dilute solution of hydrogen peroxide is obtained.
 Ba 3  PO 4 2  3H 2 O 2
 BaSO 4  8H 2 O  H 2O 2 ; 3BaO2  2H3PO 4 
BaO 2 .8H 2 O  H 2SO 4 

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9.5.2 Manufacture of Hydrogen Peroxide

a) By electrolysis of 50% H2SO4. In this method electrolytic reduction of HSO4– ion to H2S2O8 takes
place. The electrolysis is carried out in a cell with platinum anode and graphite cathode.

2H2SO4   
 2H  2HSO4

At cathode : 2H   2e  
 H2 ;  HO3S  O  O  SO3H  2e
At anode : 2HSO4 
Peroxodisulphuric acid

Peroxodisulphric acid is distilled with water under reduced pressure to get hydrogen peroxide.
H 2S2 O8  2H 2 O 
 H 2 O 2  2H 2SO 4
ii) A similar method is employed for the laboratory preparation of deutroperoxide (D2O2). Potassium
peroxodisulphate (K2S2O8) is treated with deuterium oxide (D2O).
K 2S2 O 8  2D 2 O 
 2KDSO 4  D 2 O 2
b) From 2-Ethylanthraquinol. Air is passed through a 10% solution of 2-Ethylanthraquinol dissolved in a
mixture of benzene and cyclohexane when H2O2 and 2-ethylanthraquinone are produced.

2-Ethylanthraquinone is reduced back to 2-ethylanthraquinol by hydrogenation using palladium catalyst.

9.5.3 Concentration of Hydrogen Peroxide solution
Hydrogen peroxide solution cannot be concentrated by boiling because it decomposes below its boiling point.
The 1% H2O2 obtained by the above method is extracted with water and concentrated to ~30% (by mass) by
distillation under reduced pressure. It is further concentrated to ~85% by distillation under low pressure. The
remaining water is frozen out to obtain pure H2O2.
Storage of Hydrogen Peroxide. Decomposition of H2O2 is catalysed by traces of metal impurities, strong
bases and light. Concentrated solution of H2O2 can be dangerous as rapid decomposition can result in explosion.
Therefore, it is stored in wax lined, coloured bottles as rough glass surface also causes its decomposition. A
small amount of stabilizer like phosphoric acid, glycerol or acetanilide is added to retard decomposition.
9.5.4 Structure of Hydrogen Peroxide: Hydrogen peroxide molecule has a non-planar structure. In the gas phase,
the dihedral angle is111.5o. In the solid phase it reduces to 90.2o on account of hydrogen bonding. The two
oxygen atoms are joined by a single bond. Its dipole moment (2.1D) is greater than that of water (1.84D).

(a) (b)

Figure - 02 (a) Gas phase (b) Solid phase.

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9.5.5 Pysical Properties of Hydrogen Peroxide: Pure H2O2 is a colourless/pale blue, syrupy liquid. It has odour
like that of nitric acid. Its aqueous solution has a bitter taste. It is miscible with water, alcohol and ether in all
proportions and forms a hydrate H2O2.H2O (m.p. 221K). The high density (1.44 g cm–3) is due to association
of molecules by intermolecular hydrogen bonds.
-3
Melting Point (K) 272.4 Density (liquid at 298 K) (g cm ) 1.44
Boiling point (extrapolated) (K) 423 Viscosity (290 K) (centipoise) 1.25
2 2
Vapour pressure at 298 K (mmHg) 1.9 Dielectric constant (298 K) (C /N.m ) 70.7
-3 -1 -1 -8
Density (solid at 268.5 K) (g cm ) 1.64 Electrical conductivity (ohm cm ) 5.1×10
9.5.6 Strength of H2O2 solutions: Percentage strength. A 30% aqueous solution (w/v) of H2O2 implies that 30 g
H2O2 is present in 100 mL of the solution.
Volume strength. It is the volume of O2 liberated at STP by the decomposition of 1 mL of H2O2. Thus, ‘100
volume’ H2O2, means that 1 mL of the solution will give 100 mL of oxygen at STP.
Commercially, H2O2 is marketed as 10 V, which means that it contains 3% H2O2.
Table - 03 Relationship between volume strength and other concentration units of H2O2.
-1
Volume strength Molarity (M) Normality (N) Strength (g L )
10 Volume 0.892 1.78 30.357
20 Volume 1.785 3.57 60.71
30 Volume 2.67 5.35 91.07
40 Volume 3.57 7.14 121.42
50 Volume 4.46 8.92 151.78
9.5.7 Chemical Properties
a) Decomposition. H2O2 readily decomposes on heating or on long standing to give water and dioxygen.
2H 2O2   2H 2O  O2 ; H  196.0 kJ
The decomposition is accelerated by finely divided platinum, manganese dioxide, carbon, blood, dust particles,
organic tissue and rough surfaces. It is an example of auto-oxidation and auto-reduction (disproportionation).
b) Acidic nature. Pure hydrogen peroxide is a weak acid (Ka = 1.55 × 10–12 at 298 K). It ionizes in water
 H   HO2 (hydroperoxide ion) ; HO 2 
H 2 O 2   H   O22 (peroxide ion)
It neutralises bases such as NaOH, Na2CO3, etc., to form corresponding peroxides
2NaOH  H 2 O 2   Na 2 O 2  2H 2 O ; Na 2 CO3  H 2 O 2   Na 2 O 2  H 2 O  CO 2
c) Oxidising and reducing nature. It acts as oxidising as well as reducing agent in both acidic and alkaline
media. The oxidation state of oxygen in hydrogen peroxide is –1. It can, therefore, be oxidised to dioxygen.
H2O2 can be reduced to H2O or OH– (oxidation number = –2). This accounts for its oxidising nature.
Oxidising nature. H2O2 can act as oxidising agent in acidic as well as basic medium.
Oxidising action in acidic medium: H 2 O 2 (aq)  2H  (aq)  2e  
 2H 2 O(l )
i) It oxidises black lead sulphide to white lead sulphate.
PbS  4H 2 O 2   PbSO 4  4H 2 O
This reaction is used for restoring the white colour of lead paintings.
ii) It oxidises ferrous salts to ferric salts in acidic medium
 Fe2 SO4 3  2H 2 O or
2FeSO 4  H 2SO 4  H 2 O2  2Fe 2   H 2 O 2  2H  
 2Fe3  2H 2 O

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iii) It oxidises iodides to iodine in acidic medium

 2KCl  2H 2O  I 2 or 2I  H 2O2  2H  
2KI  H 2 O 2  2HCl   I2  2H 2O
iv) It oxidises H2S to sulphur and sulphurous acid to sulphuric acid
H 2S  H 2 O 2 
 2H 2 O  S ; H 2SO 3  H 2O 2 
 H 2SO 4  H 2 O
Oxidising action in basic medium: H 2 O 2 (aq)  2e  
 2OH  (aq)
i) It oxidises nitrities to nitrates.
NaNO 2  H 2 O 2   NaNO3  H 2 O or NO 2  H 2 O 2 
 NO3  H 2 O
ii) It oxidises sulphites to sulphates.
Na 2SO 3  H 2 O 2 
 Na 2SO 4  H 2 O or SO32   H 2 O 2 
 SO 42  H 2 O
iii) It oxidises manganese salts to MnO2.
MnSO 4  H 2 O 2  2NaOH   Na 2SO 4  MnO 2  2H 2 O or Mn 2  H 2 O 2 
 Mn 4  2OH 
iv) It oxidises ferrous salts to ferric salts
FeSO 4  4NaOH  H 2 O 2   2Fe(OH)3  2Na 2SO 4 or 2Fe 2   H 2 O 2   2Fe3  2OH 
Reducing nature. H2O2 can act as reducing agent both in acidic as well as alkaline medium.
 2H  (aq)  O 2 (g)  2e 
Reducing action in acidic medium: H 2 O 2 (aq) 
i) It reduces ozone to dioxygen.
O3  H 2 O 2   H 2 O  2O 2
ii) It reduces acidified KMnO4 to colourless MnSO4.
2KMnO 4  3H 2SO 4  5H 2 O 2 
 K 2SO 4  2MnSO 4  8H 2 O  5O 2
or MnO 4  6H   5H 2 O 2   2Mn 2  8H 2 O  5O 2
iii) It reduces acidified K2Cr2O7 to chromium sulphate; orange colour changes to green.
K 2 Cr2 O 7  4H 2SO 4  3H 2 O 2 
 K 2SO 4  Cr2 (SO 4 )3  7 H 2 O  3O 2
v) It reduces chlorine and bromine to HCl and HBr respectively.
Cl2  H 2 O 2  2HCl  O 2 ; Br2  H 2 O 2 
 2HBr  O 2
vi) It reduces hypohalous acid to halide ion.
HOCl  H 2 O 2   H 3O   Cl  O 2
 HCl  H 2 O  O 2 or HOCl  H 2 O 2 
Reducing action in basic medium: H 2 O 2 (aq)  2OH  (aq)   2H 2 O(l )  O 2 (g)  2e 
i) It reduces moist silver oxide to silver and lead dioxide to lead monoxide.
Ag 2 O  H 2O 2 
 2Ag  H 2O  O 2 ; PbO 2  H 2 O 2 
 PbO  H 2 O  O 2
ii) It reduces KMnO4 to MnO2 in alkaline medium.
2MnO 4  3H 2 O 2   2MnO 2  2H 2 O  3O 2  2OH 
iii) It reduces halogens to halide ions.
I 2  H 2 O 2  2NaOH   2NaI  2H 2 O  O 2
iv) It reduces ferric salts to ferrous salts.
2Fe3  H 2 O 2  2OH  
 2Fe 2  2H 2 O  O 2
v) It reduces bleaching powder to calcium chloride.
 CaCl2  H 2 O  O 2 or OCl  H 2 O 2 
CaOCl 2  H 2 O 2   Cl  H 2 O  O 2
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e) Formation of addition compounds.

Hydrogen peroxide reacts with ethylene (CH2 = CH2) to give ethylene glycol (HOCH2CH2OH).
H 2 C  CH 2   HO  CH 2  CH 2  OH
f) Bleaching action. H2O2 is used to bleach delicate articles like ivory, silk, feather, wool, etc.
H 2 O 2   H 2 O  O ; Colouring matter + (O)   Oxidised matter (colourless)
Thus, bleaching action of hydrogen peroxide is due to nascent oxygen.
9.5.8 Uses of Hydrogen Peroxide: (i) For bleaching delicate articles such as silk, wool, ivory, paper pulp, leather,
oils, etc. (ii) Dilute solution of hydrogen peroxide under the name perhydrol is used in medicine as antiseptic
for washing wounds, teeth, ears, etc. (iii) For restoring the colour of old lead paintings. (iv) For the production
of epoxides, propylene oxide, and polyurethanes. (v) Synthesis of hydroquinone, pharmaceuticals and food
products like tartaric acid. (vi) In pollution control-treatment of domestic and industrial effluents. (vii) As
antichlor (to remove chlorine) in textile industry. (viii) Detection of metals such as Cr, Ti and V with which it
gives peroxides having characteristic colours. (ix) A 90% solution is used as oxidant for rocket fuel
(NaNH2+H2O2) and as propellant for torpedoes.
9.6 HEAVY WATER
Heavy water or deuterium oxide, D2O is the oxide of heavy hydrogen (deuterium). Heavy water was discovered
by American scientist, Harold C Urey (1932) who observed that 6000 parts of ordinary water contains about
1 part of heavy water. Lewis and Donald (1933) isolated a few ml of D2O by prolonged electrolysis of water.
9.6.1 Preparation of Heavy Water
1. Prolonged electrolysis of ordinary water. Hydrogen is liberated much more readily than deuterium on
electrolysis. This method involves multistage electrolysis of ordinary water containing NaOH using nickel
anode and steel cathode. The water left in the cell gets enriched with D2O at each stage till finally 100% D2O
is obtained.
2. Fractional distillation of ordinary water. Advantage is taken of the small difference in the boiling points
of protium oxide (373.2 K) and deuterium oxide (374.3 K). Since the difference in boiling points is very small,
a long fractionating column (about 13 m) is used for distillation and the process is repeated several times. The
lighter fraction (H2O) is distilled first while the heavier fraction (D2O) is left behind.
9.6.2 Physical properties: Heavy water is a colourless, tasteless and odourless liquid. All physical constants of
heavy water are higher than the corresponding values for ordinary water. Since dielectric constant of heavy
water is less than H2O, ionic compounds are less solule in D2O.
9.6.3 Chemical Properties. Heavy water is chemically similar to ordinary water, but reactions are slower.
i. Reaction with metals: 2Na  2D2O   Ca  OD 2  D2
 2NaOD  D2 ; Ca  2D2O 

ii. Reaction with metal oxides: Na 2O  D2O 

 2NaOD  Ca  OD 2
; CaO  D2 O 
Sodium deuteroxide
Calcium deuteroxide

iii. Reaction with non-metal oxides. Deutroacids are formed: SO3  D2 O 

 D SO
2 4
Deuterosulphuric acid

iv. Reaction with carbides :

 Ca  OD  2 
CaC 2  2D 2O  C2 D 2  4Al  OD 3  3CD 4
; Al 4 C3  12D2O 
Deuteroacetylene Deuteromethane

v. Electrolysis: 2D 2 O 
Electricity
 2D 2  O 2
 Cathode   anode 

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vi. Exchange reactions: NaOH  D 2 O 

 NaOD  HDO

HCl  D 2 O 
 DCl  HDO ; NH 4 Cl  4D 2 O 
 ND 4 Cl  4HDO

 Al  OD 3  3DCl ; BaS  2D2 O 

vii. Deutrolysis: AlCl3  3D2 O   Ba  OD  2  D2S
9.6.4 Uses of Heavy Water: (i) Heavy water is used as a moderator (substances used to slow down neutrons in
fission reactions) in nuclear reactors. (ii) Preparation of deuterium by electrolysis or decomposition with
metals produces deuterium. (iii) As a tracer for studying various reaction mechanisms.
Biological and physiological effects. Heavy water retards the growth of plants and animals. It has germicidal
and bactericidal properties. Water containing small qantity of D2O stimulates vegetable growth.
9.7 HYDROGEN AS FUEL
Dihydrogen is a proposed alternative for fossil fuels: (i) It is abundant in the combined state as water. (ii)
Hydrogen as fuel is pollution free because its combustion product is only water. (iii) Automobile engines using
hydrogen are 25 - 50% more efficient than conventional engine, burning gasoline. (iv) Heat of combustion per
gram of hydrogen is more than twice that of jet fuel. (v) Hydrogen-oxygen fuel cells may power motor
vehicles. (vi) Hydrogen is an excellent reducing agent and can replace coal in many industrial processes
because of reduced atmospheric pollution.
Table - 04 The energy released by combustion of fuels in moles, mass and volume.
Energy released on Dihydrogen Dihydrogen
LPG CH4 gas Octane (liquid)
combustion in kJ (gas) (liquid)
kJ mol-1 286 285 2220 880 5511
kJ g-1 143 142 50 53 47
kJ L-1 12 9968 25590 35 34005
9.7.1 Hydrogen Economy. Hydrogen economy refers to the use of dihydrogen as an alternative source of energy.
The basic principle is storage and transportation of energy in the form of dihydrogen instead of fossil fuels or
electric power. Cheap production of dihydrogen is the basic requirement of hydrogen economy.
Obstacles of Hydrogen Economy. (i) Availability of dihydrogen. Dihydrogen does not occur in the free
state in nature. The most likely future source of hydrogen is water. Hydrogen might be generated using nuclear
energy or solar energy. (ii) Storage and transportation. Hydrogen gas has explosive flammability which
causes problems in its storage and transportation. Hydrogen can be stored in vacuum insulated cryogenic
tanks (for space programmes). It can also be stored in underground tanks and transported by pipelines.
Another method is the use of Fe-Ti alloy which act like a sponge to absorb hydrogen and results in the
formation of fine silvery powder. Heating the powder safely releases hydrogen gas. Other small storage units
are alloys (interstitial hydrides) like LaNi5, Mg–MgH2, Ti–TiH2, etc. (iii) Platinum scarcity. In oxygen-
hydrogen fuel cells, platinum is required as catalyst. The demand for platinum exceeds supply. This may cause
problems in the manufacture of fuel cells.

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QUESTIONS
LEVEL - I
1. Hydrogen has the tendency to gain one electron to acquire helium configuration. In this respect, it resembles:
1) alkali metals 2) carbon 3) alkaline earth metals 4) halogens
2. Which of the following statement about hydrogen is incorrect?
1) Hydrogen has three isotopes of which tritium is the most common.
2) Hydrogen never acts as cation in ionic salts.
3) Hydronium ion, H 3O+ exists freely in solution.
4) Dihydrogen acts as reducing agent.
3. Ortho-hydrogen and para-hydrogen resemble in which of the following property?
1) Themal conductivity 2) Magnetic properties
3) Chemical properties 4) Heat capacity
4. Very pure hydrogen (99.9%) can be made by which of the following processes?
1) Reaction of methane with steam.
2) Mixing natural hydrocarbons of high molecular weight.
3) Electrolysis of water.
4) Reaction of salt like hydrides with water.
5. Which of the following reactions increases production of dihydrogen from synthesis gas?

1) CH 4 (g)  H 2 O(g) 

1270K
Ni
 CO(g)  3H 2 (g)

2) C(s)  H 2 O(g) 

1270K
 CO(g)  H 2 (g)

3) CO(g)  H 2O(g)   CO 2 (g)  H 2 (g)

673K
Catalyst

4) C2 H 6  2H 2 O 
1270K
Ni
 2CO  5H 2
6. Which of the following pairs of substances on reaction will not evolve H2 gas?
1) Fe and H2SO4 (aqueous) 2) Copper and HCl (aqueous)
3) Sodium and ethyl alcohol 4) Iron and steam
7. Which of the following produces hydrolith with dihydrogen?
1) Mg 2) Al 3) Cu 4) Ca
8. Of the following statements regarding dihydrogen, identify the statement which is not correct?
1) It is a colourless, odourless, tasteless gas
2) It has very low solubility in water
3) It forms more compounds than any other element
4) It is a highly reactive gas.

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9. The conversion of atomic hydrogen into ordinary hydrogen is:

1) exothermic change 2) endothermic change
3) nuclear change 4) photochemical change
10. In which of the following compounds does hydrogen exhibit a negative oxidation state:
1) LiH 2) H2O 3) HCl 4) NaOH
11. The combination of hydrogen and carbon monoxide in the presence of the catalyst ZnO and Cu gives:
1) C + H2O 2) CH4 + H2O 3) HCHO 4) CH3OH
12. Which of the following hydrides is electron precise hydride?
1) B2H6 2) NH3 3) HF 4) CH4
13. Which of the properties of interstitial hydrides is correct?
1) They generally form non-stoichiometric species
2) The hydrogen dissolved in titanium improves its mechanical properties
3) They give rise to metals fit for fabrication
4) On thermal decomposition, they afford a source of pure hydrogen
14. Hydrides of elements of Groups 3–5 are generally called:
1) interstitial hydrides 2) ionic hydrides
3) polymeric hydrides 4) complex hydrides
15. When electric current is passed through an ionic hydride in molten state:
1) hydrogen is liberated at anode 2) hydrogen is liberated at cathode
3) hydride ion migrates towards cathode 4) hydride ion remains in solution
16. An element that does not form stable hydride is:
1) Co 2) Ca 3) Li 4) Na
17. Which of the following has the highest extent of hydrogen bonding?
1) NH3 2) H2O 3) HF 4) Same in all
18. The boiling point of water is high because:
1) water molecule is linear
2) water molecule is not linear
3) water molecules possess covalent bond between H and O
4) water molecules associate due to H-bonding
19. The bond angle and dipole moment of water, respectively, are:
1) 109.5o, 1.84 D 2) 107.5o, 1.56 D 3) 104.5o, 1.84 D 4) 102.5o, 1.56 D
20. Which of the following statements regarding water is not correct?
1) High dielectric constant and strong solvating power make water an excellent solvent
2) Water expands on freezing
3) The density of water is maximum at 4oC
4) On cooling, density of water decreases upto 4oC followed by increase up to 0oC

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21. Which one of the following processes will produce hard water?
1) Saturation of water with CaCO3 2) Saturation of water with MgCO3
3) Saturation of water with CaSO4 4) Addition of Na2SO4 to water
22. Water softening by Clarke’s process uses:
1) Calcium bicarbonate 2) Sodium bicarbonate 3) Potash alum 4) Calcium hydroxide
23. When hard water is passed through permutit, which ions are exchanged with Ca2+ and Mg2+?
1) Na+ 2) Al3+ 3) H+ 4) K+
24. Hard water is not fit for washing clothes because:
1) it contains Na2SO4 and KCl 2) it gives precipitate with soap
3) it contains impurities 4) it is acidic in nature
25. Both temporary and permanent hardness is removed on boiling with:
1) Ca (OH)2 2) Na2CO3 3) CaCO3 4) CaO
26. Calgon used as a water softener is
1) Na2[Na4(PO3)6] 2) Na4[Na2(PO3)6] 3) Na2[Na4(PO4)5] 4) Na4[Na4(PO4)6]
27. Hydrogen peroxide is prepared in the laboratory by:
1) Adding MnO2 to dil. H2SO4 2) Passing CO2 into BaO2
3) Adding Na2O2 to cold water 4) Adding PbO2 to KMnO4
28. From the following statements regarding H2O2, choose the incorrect statement.
1) It decomposes on exposure to light
2) It has to be stored in plastic or wax lined glass bottles in dark
3) It has to be kept away from dust
4) It can act only as an oxidizing agent
29. The O  O  H bond angle in H2O2 in the gas phase is:
1) 106o 2) 109o28’ 3) 120o 4) 94.8o
30. Hydrogen peroxide is reduced by:
1) Ozone 2) Barium peroxide
3) Acidic solution of KMnO4 4) Lead sulphide suspension
31. The mass per cent of H2O2 in ‘30 volume H2O2’ is:
1) 4.56% 2) 9.11% 3) 11.39% 4) 13.67%
32. The bleaching property of hydrogen peroxide is due to its:
1) acidic nature 2) ability to liberate nascent oxygen
3) reducing nature 4) ability to liberate nascent hydrogen
33. Hydrogen peroxide acts both as an oxidizing and as a reducing agent depending upon the nature of the
reacting species. In which of the following cases H2O2 acts as a reducing agent in alkaline medium?
1) MnO 4 2) Cr2 O 72  3) SO32  4) KI

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34. The atom of oxygen lost by H2O2 molecule during oxidation reaction is that which is linked through:
1) an electrovalent bond 2) a covalent bond
3) a coordinate bond 4) a hydrogen bond
35. In which of the following reactions H2O2 acts as a reducing agent?
i) H 2 O 2  2H   2e   2H 2 O ii) H 2 O 2  2e   O 2  2H 
iii) H 2 O 2  2e   2OH  iv) H 2 O 2  2OH   2e   O 2  2H 2 O
1) i), ii) 2) iii), iv) 3) i), iii) 4) ii), iv)
36. Bond angles H–O–H and H–O–O– in water and hydrogen peroxide repectively are:
1) 104.5o in both 2) 94.8o in both 3) 104.5o, 94.8o 4) 94.8o and104.5o
37. In which reaction, hydrogen peroxide neither acts as an oxidising agent nor as a reducing agent?
1) PbS  H 2 O 2 
 2) SO3  H 2 O 2 


3) PbO 2  H 2 O 2 
 4) Na 2 CO3  H 2 O 2 

38. Decomposition of H2O2 is prevented by:
1) NaOH 2) MnO2 3) acetanilide 4) oxalic acid
39. Which of the following has lower value for D2O than for H2O?
1) Molecular mass 2) Boiling point
3) Viscosity 4) Ionization constant
40. Which of the following hydrides is used for the storage of hydrogen and serve as a source of energy?
1) Ionic hydride 2) Covalent hydride 3) Metallic hydride 4) Polymeric hydride
LEVEL - II
1. Hydrogen molecule differs from chlorine molecule in the following respect:
1) Hydrogen molecule is non-polar but chlorine molecule is polar
2) Hydrogen molecule is polar while chlorine molecule is non-polar
3) Hydrogen molecule can form intermolecular hydrogen bonds but chlorine molecule does not
4) Hydrogen molecule cannot participate in coordination bond formation but chlorine molecule can
2. Select incorrect statement.
1) Ortho and para hydrogen are different due to difference in their nuclear spins
2) Ortho and para hydrogen are different due to difference in their electron spins
3) Para hydrogen has a lower internal energy than that of ortho hydrogen
4) Para hydrogen is more stable at lower temperature
3. Out of the following metals which will give H2 on reaction with NaOH:
I : Zn II : Mg III : Al IV : Be
1) I, II, III, IV 2) I, III, IV 3) II, IV 4) I, III

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4. Zn gives H2 gas with H2SO4 and HCl but not with HNO3 because:
1) Zn acts as an oxidising agent when reacts with HNO3
2) HNO3 is weaker acid than H2SO4 and HCl
3) in electrochemical series Zn is above hydrogen
4) NO 3 is reduced in preference to hydronium ion
5. 2 g of aluminium is treated separately with excess of dilute H2SO4 and excess of NaOH. The ratio of volumes
of hydrogen evolved under similar conditions of pressure and temperature is:
1) 2 : 3 2) 1 : 1 3) 2 : 1 4) 1 : 2
6. In context with the industrial preparation of hydrogen from water gas (CO + H2), which of the following is the
correct statement?
1) CO and H2 are fractionally separated using differences in their densities.
2) CO is removed by absorption in aqueous Cu2Cl2 solution.
3) H2 is removed through occlusion with Pd.
4) CO is oxidized to CO2 with steam in the presence of a catalyst followed by absorption of CO2 in alkali.
7. The correct order of the thermal stability of hydrogen halides (H–X) is:
1) HI > HBr > HCl > HF 2) HF > HCl > HBr > HI
3) HCl < HF > HBr < HI 4) HI > HCl < HF > HBr
8. Under what conditions of temperature and pressure the formation of atomic hydrogen from molecular hydrogen
will be favoured most?
1) high temperature and high pressure 2) low temperature and low pressure
3) high temperature and low pressure 4) low temperature and high pressure
9. In which of the following reactions does hydrogen act as an oxidising agent?
1) H 2  F2 
 2) H 2  SiCl4 
 3) Na  H 2 
 4) CuO  H 2 

10. Electron-deficient, electron-precise and electron-rich hydrides are types of:
1) ionic hydrides 2) interstitial hydrides 3) covalent hydrides 4) metallic hydrides
11. Among CaH2, NH3, NaH and B2H6 which are covalent hydrides?
1) NH3 and B2H6 2) NaH and CaH2 3) NaH and NH3 4) CaH2 and B2H6
12. Which statements is/are correct?
1) Boiling point of H2O, NH3, HF are maximum in their respective group due to intermolecular H-bonding
2) Boiling point of CH4 out of CH4, SiH4, GeH4 and SnH4.
3) Formic acid forms dimer by H–bonding
4) All the above are correct statements
13. Consider following statements :
I : AlH 3  H  
 AlH 4 II : H 2 O  H  
 H 2  OH 
Select correct statements based on these reactions.
1) H– is a Lewis acid in I and Lewis base in II 2) H– is a Lewis base in I and Bronsted base in II
3) H– is a Lewis acid in I and Bronsted acid in II 4) H– is a Lewis base in I and II
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14. The hydride ion H– is stronger base than its hydroxide ion OH–. Which of the following reactions will occur if
sodium hydride (NaH) is dissolved in water?
1) 2H   aq  
 H 2  2e 2) H   aq   H 2O  l  
 OH   H 2

3) H   H 2O  l  
 No reaction 4) None of the above
15. Metallic hydrides are useful for hydrogen storage because:
1) they reacts with hydrogen and form stable compound
2) they absorb H-atoms
3) they adsorb H-atoms
4) they form unstable compounds with hydrogen
16. The correct statements among the following are:
i) Saline hydrides produce H2 gas when reacted with H2O.
ii) Reaction of LiAlH4 with BF3 leads to B2H6.
iii) PH3 and CH4 are electron – rich and electron – precise hydrides, respectively.
iv) HF and CH4 are called as molecular hydrides.
1) (iii) and (iv) only 2) (i), (ii) and (iii) only
3) (i), (ii), (iii) and (iv) 4) (i), (iii) and (iv) only
17. What is true about saline hydrides?
a) They are binary compounds of hydrogen and metallic elements
b) They are crystalline solids
c) They are generally very soft
d) Their common examples are, SiH4, CH4 etc.
1) a and b 2) b and c 3) c and d 4) a, b and c
18. In which of the following hydrides, hydrogen exists in negative oxidation state?
a) HCl b) NaH c) CaH2 d) HI
1) a and b 2) b and c 3) a, b and d 4) b, c and d
19. High dipole moment of water justifies that:
1) it is not linear molecule 2) it is a universal solvent
3) it has higher density than ice 4) it is neutral toward litmus
20. The melting points of most of the solid substances increase with an increase of pressure acting on them.
However, ice melts at a temperature lower than its usual melting point when the pressure is increased. This is
because:
1) ice is less dense than water 2) it generates heat
3) chemical bonds break under pressure 4) none

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21. Hydrogen bonding plays a central role in the following phenomena.

a) Ice floats in water.
b) Higher Lewis basicity of primary amines than tertiary amines in aqueous solutions.
c) Formic acid is more acidic than acetic acid.
d) Dimerization of acetic acid in benzene.
1) a and d 2) a,b and c 3) b, c and d 4) a, b and d
22. Which of the properties of water given below is/are false?
a) Water is a universal solvent.
b) Hydrogen bonding is present to a large extent in liquid water.
c) There is no hydrogen bonding in the frozen state of water.
d) Frozen water is heavier than liquid water.
1) a and b 2) b and c 3) c and d 4) a and d
23. The degree of hardness of water is usually expressed in terms of:
1) ppm by weight of MgSO4
2) g/L of CaCO3 and MgCO3 present
3) ppm by weight of CaCO3 irrespective of whether it is actually present
4) ppm of CaCO3 actually present is water
24. 10 L of hard water required 0.56 g of lime (CaO) for removing hardness. Hence, temporary hardness in ppm
(part per million, 106) of CaCO3 is:
1) 100 2) 200 3) 10 4) 20
25. The temporary hardness of a water sample is due to compound X. Boiling this sample converts X to
compound Y. X and Y, respectively, are :
1) Ca(HCO3)2 and CaO 2) Mg(HCO3)2 and MgCO3
3) Mg(HCO3)2 and Mg(OH)2 4) Ca(HCO3)2 and Ca(OH)2
26. Match List I with List II and select the correct answer using the codes given below in the lists:
List I List II
1. Heavy water A. Bicarbonates of Mg and Ca in water
2. Temporary hard water B. No foreign ions in water
3. Soft water C. D2O
4. Permanent hard water D. Sulphates and chlorides of Mg and Ca in water
Codes:
1) 1–C, 2–D, 3–B, 4–A 2) 1–B, 2–A, 3–C, 4–D
3) 1–B, 2–D, 3–C, 4–A 4) 1–C, 2–A, 3–B, 4–D
27. When 50% solution of H2SO4 is electrolysed by passing a current of high density at low temperature the main
products of electrolysis are:
1) oxygen and hydrogen 2) H2 and peroxo disulphuric acid
3) H2 and SO2 4) O2 and peroxy disulphuric acid

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28. 100 mL of 0.01 M KMnO4 oxidises 100 mL H2O2 in acidic medium. Volume of the same KMnO4 required
in alkaline medium to oxidise 100 mL of the same H2O2 will be: ( MnO 4 changes to Mn2+ in acidic medium
and to MnO2 in alkaline medium)
100 500 300 100
1) mL 2) mL 3) mL 4) mL
3 3 5 5
29. When H2O2 is added to ice cold solution of acidified potassium dichromate in ether and the contents are
shaken and allowed to stand :
1) a blue colour is obtained in ether due to formation of Cr2(SO4)3
2) a blue colour is obtained in ether due to formation of CrO5
3) a blue colour is obtained in ether due to formation of CrO3
4) chromyl chloride is formed
30. Select correct statement (s).
1) H2O2 reduces MnO 4 to Mn2+ in acidic medium
2) H2O2 reduces MnO 4 to MnO2 in basic medium
3) H2O2 can be used to bleach blackened oil paintings
4) All the above are correct statement
31. 10 mL of H2O2 solution (volume strength = x) requires 10 mL of N/0.56 MnO 4 solution in acidic medium.
Hence x is:
1) 0.56 2) 5.6 3) 0.1 4) 10
32. Which one of the following reactions does not form gaseous product?
1) PbO 2  H 2 O 2 
 2) PbS  H 2 O 2 

3) Cl2  H 2 O 2   4) Na 2 CO3  H 2 O 2  
33. Consider the reaction:
i) H 2 O 2  O3  2H 2 O  2O2 ii) PbS  4O3  PbSO 4  4O2
1) O3 is reduced both in (i) and (ii) 2) O3 is oxidized both in (i) and (ii)
3) O3 is oxidized in (i) and reduced in (ii) 4) O3 is reduced in (i) and oxidized in (ii)
34. The oxides which give H2O2 on treatment with dilute acid are:
a) PbO2 b) MnO2 c) Na2O2 d) BaO2
1) a only 2) a and b 3) c and d 4) a, c and d
35. Which of the following statements are correct?
a) H2O2 reduces MnO 4 both in acidic and basic media
b) H2O2 oxidises Fe2+ ions both in acidic and basic media
c) H2O2 oxidises Mn2+ to Mn4+ ions in basic medium
d) H2O2 liberates I2 from acidified KI solution and reduces I2 to I– ions in basic medium
1) a and b 2) c and d 3) a, b, and c 4) a, b, c and d
36. H2O2 is always stored in black bottles because:
1) it is highly unstable
2) its enthalpy of decomposition is high
3) it undergoes auto-oxidation on prolonged standing
4) Both 1 & 3
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37. Which of the following is not correct for D2O?

1) Boiling point is higher than H2O 2) D2O reacts slowly than H2O
3) Viscosity is higher than H2O at 25oC 4) Solubility of NaCl in D2O is more than H2O
38. Which of the following is not true?
1) D2O freezes at lower temperature than H2O
2) Reaction between H2and Cl2 is much faster than D2 and Cl2
3) Ordinary water electrolysed more rapidly than D2O
4) Bond dissociation energy of D2 is greater than H2
39. Which of the following is correct about heavy water?
1) Water at 4oC having maximum density is known as heavy water
2) It is heavier than water
3) It is formed by the combination of heavier isotope of hydrogen and oxygen
4) Both B and C
40. Which of the following statements are correct in the case of heavy water?
a) Heavy water is used as a moderator in nuclear reactor
b) Heavy water is used as a tracer for studying reaction mechanisms
c) Heavy water has lower boiling point than ordinary water
d) Enthalpy of vaporisation of heavy water is more than that of ordinary water
1) a only 2) a and b 3) a, b and d 4) a, b and d

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SOLUTIONS
LEVEL - I
1. 4 Just like halogens, hydrogen needs one electron to attain the configuration of nearest noble gas.
2. 1 Hydrogen has three isotopes of which tritium is radioactive and rare.
3. 3 The two forms differ in physical properties such as specific heat, thermal conductivity, boiling point,
and others, but their chemical properties are the same.
o
800  900 C
4. 3 CH 4  H 2O   CO  3H 2
In this method mixture of CO and H2 are obtained. High molecular weight hydrocarbon on cracking
gives H2 but this mixture also contain small amount of hydrocarbon.
Electrolysis of H2O is best method formation of 99.97% pure H2.
At the cathode : 2H 2 O  2e  2OH   H 2
1
At the anode : 2OH   2H 2 O  O 2  2e 
2
1
Overall reaction : H 2O  H 2  O2
2
5. 3
6. 2
o
300  400 C
7. 4 Calcium reacts with dihydrogen to produce hydrolith. Ca  H 2   CaH 2 .
8. 4 In dihydrogen gas, the hydrogen atoms have already shared their electrons to form the molecule, so H2
molecule is not very reactive.
9. 1 Atomic hydrogen combines to form ordinary hydrogen with the evolution of 104.2 kcal of heat
(exothermic process). This recombination is catalyzed by certain metals.
H  H  H 2 ; H  104.2 kcal

10. 1 LiH is an ionic hydride in which an electron is transferred from Li hence Li  and H  are formed.
11. 4 2H 2  g   CO g  
ZnO

 CH 3OH  l 
12. 4
13. 1 The interstitial hydrides are generally non-stoichiometric and their compositions vary with temperature
and pressure. For example, TiH1.73, CeH2.7, LaH2.8, etc.
14. 1 Hydrides of elements of Group 3–5 are generally called interstitial hydrides because in these hydrides,
hydrogen atoms, being small in size, occupy some interstitial sites in the metallic lattice producing distortion
without any change in its type.
15. 1 At anode : 2H  
 H 2  2e 
16. 1
17. 2 Among nitrogen, fluorine and oxygen, the increasing order of electronegativities are N < O < F. Hence,
HF > H2O > NH3. But, the actual order is H2O > HF > NH3. Although fluorine is more electronegative
than oxygen, the extent of hydrogen bonding is higher in water.
18. 4
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19. 3 Dipole moment of water is 1.84 D. Its bond angle is 104.5o.

20. 4 The density of water at 4oC is maximum.
21. 3 Alkaline earth metal salts cause hardness. Temporary hardness is caused by soluble Ca and Mg hydrogen
carbonates. Soluble sulphates and chlorides of calcium and magnesium cause permanent hardness.
22. 4 In Clarke’s process, slaked lime is added to remove the temporary hardness of water and insoluble
calcium carbonate precipitates out. Ca  OH  2  Ca  HCO3 2  2CaCO3  2H 2 O (slaked lime).
23. 1 For the softening of water by permutit process, the water is percolated through the bed of zeolite
packed in a vertical cylinder. The zeolite may be represented as Na2Z and the following reactions occur
in the zeolite bed:
Na 2 Z  CaSO 4  or CaCl 2   CaZ  Na 2SO4  or 2NaCl 

Na 2 Z  MgCl2  or MgSO4   MgZ  2NaCl  or Na 2SO4 

Na 2 Z  Ca  HCO3 2  or Mg  HCO3 2   CaZ  or MgZ   2NaHCO3

24. 2   RCOO  2 Mg  2Na 
2RCOONa  Mg 2  
Insoluble

2RCOONa  Ca 2
  RCOO  2 Ca  2Na 

Insoluble

25. 2 Na 2 CO 3  MgSO 4 
 Na 2SO 4  MgCO3 
26. 1 Calgon is sodium hexametaphosphate Na2[Na4(PO3)6]. It is used as water softener.
27. 2 CO 2  H 2 O  H 2 CO3  H   HCO3
2H + +Ba O 2  Ba 2   H 2 O 2
28. 4 The oxidation state of oxygen in H2O2 is –1 (peroxide). So, oxygen can increase and decrease its
oxidation number which means it can act as a reducing as well as an oxidizing agent.

29. 4 The structure of H2O2 in the gaseous phase is

Hydrogen peroxide is a dihydroxy compound (H–O–O–H) and the O–O linkage is known as a peroxide
linkage. It is a non-linear molecule as the two O–H bonds are in different planes. The interplanar
(dihedral) angle is 111.5o in the gaseous phase, but it is reduced to 90.2o in the crystalline state because
of hydrogen bonding.
30. 4 PbS  4H 2O 2  PbSO 4  4H 2 O
 black   white 

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31. 2 Strength of ‘30 volume H2O2’ = (molarity) (molar mass) =  2  30 / 22.4  mol L1 34 g mol 1 
= 91.1 g L–1
Mass per cent of H2O2 = 9.11%
32. 2 The bleaching property of hydrogen peroxide is due to its ability to liberate nascent oxygen. Bleaching
nature of hydrogen peroxide is due to its oxidation.

H 2O2  H 2 O   O Coloured matter   O   Colourless

Nascent ; Nascent  Bleached 
oxygen oxygen

33. 1 2MnO 4  6H   5H 2 O 2  2Mn 2  SO 2  8H 2 O

In this reaction KMnO4 is reduced from +7 to +2 oxidation state. Thus, it acts an oxidizing agent and
H2O2 act as a reducing agent in acidic medium.
34. 2
35. 4 The oxidation states of the species are as follows:
1 2 1 0
i) H 2 O 2  2H   2e   2H 2 O ii) H 2 O 2  2e   O 2  2H 
1 2 1 2 0 2
iii) H 2 O 2  2e   2 O H  iv) H 2 O 2  2 O H   2e   O 2  2H 2 O
In (ii) and (iv) H2O2 acts as a reducing agent as it loses electrons to reduce the other species and itself
undergoes oxidation.
36. 3
37. 4 H2O2 behaves as an acid when reacts with Na2CO3. Na 2 CO3  H 2 O 2  Na 2 O 2  CO 2  H 2O
38. 3 A small amount of phosphoric acid or glycerol or acetanilide is added in H2O2 during its storage. These
additives act as negative catalyst for the decomposition of H2O2.
39. 4 Ionic product of H2O 1.008 × 10–14 is higher than that of D2O 1.95×10–15. This is because of
strong association (H-bonding) in D2O molecule than water molecule.
40. 3 Metallic hydrides are hydrogen-deficient, that is, they do not hold the law of constant composition.
Metals like Pd, Pt have the capacity to accommodate a large volume of hydrogen. Therefore, they are
used for the storage of hydrogen and serve as a source of energy.
LEVEL - II
1. 4 Hydrogen molecule cannot participate in coordination bond formation but chlorine molecule can because
there is no unshared pair of electrons in hydrogen molecule (H2) whereas chlorine molecule has six
unshared electron pairs as shown below.
HH Cl  Cl
Coordinate bond is formed by sharing of an electron pair between the atoms where the shared pair of
electrons is contributed by only one of the atom.
2. 2
3. 2
4. 4

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5. 2  Al 2  SO4 3  3H 2 ; 2Al  2NaOH  2H 2 O 

2Al  3H 2SO4   2NaAlO 2  3H 2
Thus, ratio of volumes of hydrogen evolved is 1:1
6. 4 Hydrogen is made in large amounts by the steam reformer process. When water gas is passed over iron
or copper catalyst at 400oC, CO gets converted into CO2. Finally CO2 is absorbed in a solution of
alkali.
CO  H 2 
H 2O
 CO 2  Acidic oxide   2H 2
 KOH (base)
Water gas

K2CO3
7. 2 Thermal stability  bond dissociation energy. Bond dissociation energy of HF is the highest and HI is
the least. So, HF is the most stable halogen acid while HI is the least stable.
8. 3 Endothermic and increase of volume.
9. 3 The reaction in which hydrogen accepts the electron or its oxidation number changes from 0 to –1, it
 2Na  H  .
acts as an oxidising agent. 2Na  H 2 
10. 3 These are the types of molecular/covalent hydrides.
11. 1 N and B belong to p-block and since the electronegativity difference between these elements and H is
very less, they form covalent hydrides.
12. 4
13. 2
14. 2
15. 3 In metallic hydrides, hydrogen is adsorbed as H-atoms. This property of adsorption of hydrogen on
transition metals is widely used as its storage media. Some of the metals such as Pd, Pt can accommodate
a very large volume of hydrogen. This property has high potential for hydrogen storage and as a source
of energy. Metallic hydrides on heating decompose to form hydrogen and very finely divided metal.
16. 3
17. 1
18. 2 NaH and CaH2 are ionic hydrides, and Na  and Ca2+ lose electron which is gained by H to form
H ion.
19. 1 High dipole moment of water suggests that it is not a linear molecule.
20. 1
21. 4 Due to hydrogen bonding, ice has cage-like structure. In the case of amines, the order is
1o > 2o > 3o as primary amine forms maximum hydrogen bonds compared to secondary and tertiary.
Acetic acid dimerize in benzene due to hydrogen bonding.
22. 3 Hydrogen bonding is present in frozen state of water. Frozen water is lighter than liquid water due to its
lesser density.
23. 3 Degree of hardness is the number of parts of CaCO3 equivalent to various calcium and magnesium salts
present in a million parts of water by mass (ppm).
24. 1 Temporary hardness is due to HCO3 of Ca2+ and Mg2+ ; Ca  HCO3 2  CaO 
 2CaCO3  H 2 O

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25. 3 Temporary hardness is due to soluble Mg(HCO3)2 and Ca(HCO3)2.

Mg(HCO3 ) 2 
Boil
Mg(OH) 2  2CO 2 ;
Ca(HCO 3 ) 2 
Boil
CaCO3  H 2 O  CO 2
26. 4
27. 2 On electrolysis of cold 50% soultion of H2SO4
 H 2 g   Reduction 
At cathode: 2H   2e  

 H 2S2 O8  2e  Oxidation 

At anode: 2HSO4 
28. 2 MnO4  5e 
 Mn 2  acidic  MnO4  3e 
 MnO2  basic 
500
100 mL H2O2  100  5N MnO 4  V  3N MnO 2 , N = mL
3
29. 2 Cr2 O 72  2H   4H 2 O 2 
 2CrO5  5H 2O
Bluecolour
in ether

30. 4
31. 4 N of H2O2 = Volume strength/5.6 = x/5.6
mEq. of H2O2  mEq. of MnO 4  N1V1  N 2 V2
x 1
 10 mL   10 mL  x  '10 ' volume strength
5.6 0.56
32. 2 PbS  4H 2 O 2   PbSO 4  4H 2 O
 Solid   Liquid 
33. 1
1
34. 3 PbO 2  H 2SO 4 
 PbSO 4  H 2 O  O 2 ;
2
1
MnO 2  H 2SO 4 
 MnSO4  H 2O  O2
2
Na 2 O 2  H 2SO 4 
 NaSO 4  H 2 O 2 ;
BaO 2  H 2SO 4   BaSO 4  H 2 O 2
35. 4 All are correct statements.
36. 4 H2O2 readily undergoes decomposition on prolonged standing due to its unstable nature. The
decomposition or auto-oxidation is further accelerated by sunlight and rough surface. In order to check
it, hydrogen peroxide is stored in black bottles.
37. 4
38. 1 D2O actually has higher freezing point (3.8oC) than water H2O (0oC).
39. 4 Heavy water is formed by the combination of heavier isotope of hydrogen and oxygen.
2D  O2  2D 2O
2
Deutirium Heavy water

40. 4

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CHAPTER - 10

S-BLOCK ELEMENTS

INTRODUCTION
The s-block of the Periodic Table consists of elements of groups 1 and 2 in which the last electron enters the
outermost s-orbital. The general electronic configuration is [noble gas] ns1 for group 1 and [noble gas] ns2 for
group 2 metals. Although hydrogen (1s1) and helium (1s2) may be considered as s-block elements, they are
usually disscussed separately because of peculiar behaviour.
Group 1 elements are collectively known as alkali metals since they form strongly alkaline hydroxides.
Group 2 elements except beryllium are known as alkaline earth metals because their oxides and hydroxides
are alkaline and the oxides are found in the earth’s crust. These metals and their compounds are of utmost
importance in industry and in daily life. Sodium, potassium, magnesium and calcium ions are found in large
proportions in biological fluids.
10.1 GROUP 1 ELEMENTS: ALKALI METALS
Group 1 of the Periodic Table consists of six elements, namely lithium (Li), sodium (Na), potassium (K),
rubidium (Rb), caesium (Cs) and francium (Fr). Among alkali metals, sodium (7th) and potassium (8th) are
abundant and lithium (35th), rubidium (23rd) and caesium (46th) have much lower abundance. Francium is
highly radioactive (half-life of 223Fr is 21 minutes).
Occurance. Alkali metals are never found in the free state in nature; they occur only in the combined state as
halides, oxides, silicates, borates and nitrates.
Lithium as : Spodumene, LiAl(SiO3)2, Lepidolite, Li2Al2(SiO3)(F,OH)2. Sodium and potassium as : NaCl,
and KCl in sea water, Rock salt (NaCl), Borax (Na2B4O7.10H2O), Chile salt petre (NaNO3), Mirabilite
(Na2SO4), Trona (Na2CO3.NaHCO3.2H2O), Sylvite (KCl), Carnallite (KCl.MgCl2.6H2O), Feldspar
(K2O.Al2O3.6SiO2), etc.
10.1.1 Atomic and Physical Properties
1. Electronic configuration. General electronic configuration is ns1 outside the noble gas core. The losely
held s-electron makes them the most electropositive metals.
2. Atomic and ionic radii. In any period, the alkali metal atom has the largest size (except noble gases). The
atomic and ionic radii of alkali metals increase from Li to Cs. This is due to screening effect of inner filled
shells on the valence electron. The monovalent ions (M+) are smaller than the parent atom.
3. Ionization enthalpy. The ionization enthalpies are low and decrease down the group. This is because the
effect of increasing size outweighs increasing nuclear charge.
The second ionisation enthalpies (i are very high because the second electron has to be removed
from a unipositive ion which has noble gas configuration.
4. Hydration enthalpy. Hydration enthalpies decrease with increase in ionic size.
Li+> Na+ > K+ > Rb+ > Cs+.
Due to small size, Li+ has maximum degree of hydrationv(LiCl· 2H2O).

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5. Oxidation state. Alkali metals exhibit +1 oxidation state in their compounds and are strongly electropositive
in character. Electropositive character increases down the group due to larger size of the atoms which
facilitates easier removal of the outer electron.
6. Standard electrode potential. Alkali metals have very low standard reduction potentials and therefore
they are good reducing agents. Li has the lowest value (-3.04 eV) due to very high hydration energy.
Element/At. No. Li Na K Rb Cs Fr
Electronic 1 1 1 1 1 1
configuration [He] 2s [Ne] 3s [Ar] 4s [Kr] 5s [Xe] 6s [Rn] s

10.1.2 Physical Properties

i. Colour, appearance, hardness. Alkali metals are silvery white, soft and light metals. They can be easily
cut with a knife. They are soft because of large atomic size, loose packing and weak metallic bond.
Table - 02 Atomic and physical properties of alkali metals
Property Li Na K Rb Cs Fr
Atomic number 3 11 19 37 55 87
-1
Atomic mass (g mol ) 6.94 22.99 39.1 85.47 132.91 223
Metallic radius (pm) 152 186 227 248 265 375
+
Ionic radius (M /pm) 76 102 138 152 167 180
-1
Ionisation enthalpy,  i H1 (kJ mol ) 520 496 419 403 376 375
-1
Ionisation enthalpy, DiH2 (kJ mol ) 7268 4562 3051 2633 2230
-1
Hydration enthalpy (kJ mol ) -506 -406 -330 -310 -276
Electronegativity (Pauling scale) 0.98 0.93 0.82 0.82 0.79
-3
Density (g cm ) at 293 K 0.53 0.97 0.86 1.53 1.9
Melting point (K) 454 371 336 312 302
Boiling point (K) 1615 1156 1032 961 944
+ -
E° V at 298 K for M (aq)+e M(s) -3.04 -2.714 -2.925 -2.93 -2.927
* ** ** * * -18*
Occurrence in Lithosphere 18 2.27 1.84 78.12 2.6 ~10
*ppm, ** percentage by weight

ii. Density. Because of large size, these elements have low density which increases down the group from Li
to Cs. However, potassium is lighter than sodium.
iii. Melting and boiling points. Melting and boiling points are low indicating weak metallic bonding due to
the presence of only a single valence electron.
iv. Flame colour - flame test. Alkali metals and their salts impart characteristic colour to an oxidising flame.
This is because heat from the flame excites the outermost electron to a higher energy level and when the
excited electron returns to the ground state, radiation is emitted in the visible region.

Metal Li Na K Rb Cs
Colour Crimson Yellow lilac Red violet Blue
l/nm 670.8 589.2 766.5 780 455.5
Detection and determination: Alkali metals can be detected by flame test and estimated by flame
photometry or atomic absorption spectrometry.

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v. Photoelectric effect. Due to low ionisation potentials, these atoms eject electrons when irradiated with
light, therefore, caesium and potassium are used in photoelectric cells. Li does not exhibit photoelectric
effect because it has the highest ionisation enthalpy among alkali metals.
10.1.3 Chemical Properties
Alkali metals are highly reactive due to large size and low ionisation enthalpy.
(i) Reactivity towards air or oxygen: Alkali metals tarnish in dry air due to the formation of oxides. They
burn vigorously in oxygen forming oxides. Li forms monoxide, Na forms peroxide, and the other metals form
superoxides. In oxides, the oxidation state of the metal is +1. Reactivity increases down the group.
4Li  O 2 
 2Li 2 O (oxide) : 2Na  O2 
572K
 Na 2O2 (peroxide)

 MO 2 (super oxide) where M = K, Rb, Cs. ; K  O 2 

M  O 2   KO 2
Potassium superoxide

The superoxide ion (O2–) is stable only in the presence of large cations such as K, Rb, Cs. These are coloured
and paramagnetic because O2– has a three-electron bond in it.
Li does not form superoxide because Li+ has a strong positive field which attracts the negative charge so
strongly that it does not permit the oxygen to combine with another oxygen to form O2–.
Alkali metals except Li form ozonides, which decomposes to the superoxide and oxygen on standing.
M  O3  
 MO 3 ; 2MO 3 
 2MO 2  O 2

Lithium shows exceptional behaviour in reacting with N2 of air to form the nitride: 6Li  N 2  
 2Li 3 N
Because of high reactivity towards air and water, alkali metals are kept in kerosene.
(ii) Reactivity towards water: Alkali metals react with water to form the hydroxide and dihydrogen.
 2M   2OH   H 2
2M  2H 2 O  : 2Na  2H 2 O 
 2NaOH  H 2
Although lithium has the least E0 value, it reacts less vigorously with water than sodium which has the highest
E0 value among alkali metals. The reaction of sodium with water produces enough heat to melt it and the
hydrogen produced ignites in air (m.p. of Li >Na). Other metals of the group react explosively with water.
They also react with proton donors like alcohols, gaseous ammonia and alkynes to form alkali metal ethoxide,
alkali metal amide and alkali metal acetylide respectively.
2M  2C 2 H 5OH 
 2C 2 H 5 OM  H 2 : 2M  2NH 3 
 2MNH 2  H 2

2M  H  C  C  H 
 M  C  C  M  H2
(iii) Reactivity towards dihydrogen: Alkali metals react with dihydrogen at about 673K (lithium at 1073K)
to form hydrides. Thse hydrides are ionic solids with high melting points.
 2M  H  : 2 Li  H 2 1073
2M  H 2   K
 2LiH ; 2Na  H 2 
673K
 2NaH
Reactivity towards hydrogen decreases down the group. This is because lattice enthalpies of hydrides decrease
with increase in size of the metal atom (stability of the hydrides decreases down the group).
(iv) Reactivity towards halogens: They react vigorously with halogens to form ionic halides, M+X–. Lithium
halides are somewhat covalent due to high polarising power of lithium ion. Since anions with large size can be
easily distorted, lithium iodide is the most covalent among halides.
2M  X2 
2MX (where X = F, Cl, Br, and I)

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Reactivity of halogens towards a particular alkali metal decreases from F2 to I2.

(v) Reducing nature: Alkali metals are strong reducing agents, lithium being the most and sodium the least
powerful. Electrode potential depends on sublimation enthalpy, ionization enthalpy and hydration enthalpy :
(i) M(s) 
Sublimation enthalpy
 M(g) (ii) M(g) 
Ionization enthalpy
 M  (g)  e
(iii) M  (g)  H 2 O 
 M  (aq) + Enthalpy of hydration
With small size of its ion, lithium has the highest hydration enthalpy which accounts for its high negative E0
value and high reducing power.
(vi) Solutions in liquid ammonia: Alkali metals dissolve in liquid ammonia giving deep blue solutions (when
dilute) which are conducting and paramagnetic. This is due to the formation of ammoniated electrons.
[M (NH 3 ) x ]  [e(NH 3 ) y ]
M  (x  y)NH 3 
On standing, the solution slowly liberates hydrogen and forms amide.
M  (am)  e  NH3 (l ) 
 MNH 2(am)  1 2 H2 (g) (‘am’ denotes solution in ammonia)
The solubility may be as high as 5M. Concentrated solutions are bronze coloured (with metallic lustre) and
diamagnetic. Conductivity decreases with concentration since the ammoniated metal cations are bound by
the free unpaired electrons which have been described as expanded metals.
10.1.4 Uses of Alkali metals
Lithium: Lithium metal to make alloys; with lead (Li/Pb) in ‘white metal’ bearings for motor engines, with
aluminium (Li/Al) for aircraft parts, and with magnesium (Li/Mg) in armour plates.
In thermonuclear reactions and in electrochemical cells.
Lithium aluminium hydride (LiAlH4) as reducig agent in the laboratory.
Sodium: In Lassaigne’s test for the detection of elements (N, S and halogens) in organic compounds. As
reagent in Wurtz reaction. Sodium amalgam (Na/Hg) in the reduction of organic compounds.
In Na/Pb alloy to make PbEt4 and PbMe4 (earlier used as anti-knocks).
Production of artificial rubber (Buna-N, Buna-S), chemicals (NaOH, NaHCO3, Na2CO3, Na2O2, NaCN,
NaNH2), dyes, drugs, etc. As reducing agent in the extraction of B and Si.
In sodium vapour lamps. Liquid sodium as coolant in fast breeder nuclear reactors.
Potassium: Potassium has a vital role in biological systems. KCl is used as fertilizer. KOH is used in the
manufacture of soft soap and as absorbent for CO2.
Caesium: In photoelectric cells.
10.2 GENERAL CHARACTERISTICS OF THE COMPOUNDS OF ALKALI METALS
Most of the alkali metal compounds are ionic in nature.
1. Oxides and Hydroxides. On combustion in excess of air, lithium forms Li2O and some peroxide Li2O2,
sodium forms the peroxide, Na2O2 and some superoxide NaO2 whilst potassium, rubidium and caesium form
the superoxides, MO2. The increasing stability of the peroxide or superoxide, as the size of the metal ion
increases is due to the stabilisation of large anions by larger cations through lattice energy effects.
The oxides and peroxides are colourless when pure, but the superoxides are yellow or orange in colour. The
normal oxides are diamagnetic, whereas superoxides are paramagnetic (due to three-electron bond).
Sodium peroxide is widely used as an oxidising agent in inorganic chemistry.
These oxides are easily hydrolysed by water to form the hydroxides
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 2M   2OH  ; M2O2  2H 2O 
M 2O  H2 O   2M  2OH   H 2O2

 2M   2OH   H2O 2  O2
2MO2  2H2 O 
Alkali metal hydroxides are all white crystalline solids. They are the strongest bases. They dissolve freely in
water with the evolution of much heat on account of intense hydration.
2. Halides. Alkali metal halides, MX, (X = F, Cl, Br, I) are colourless crystalline high-melting solids. They
can be prepared by the reaction of the oxide, hydroxide or carbonate with aqueous hydrohalic acid (HX).
M 2 O  2HX 
 2MX  H 2 O ; MOH  HX 
 MX  H 2 O

M 2 CO3  2HX 
 2MX  H 2 O  CO 2

All halides have high negative enthalpies of formation. The fH values for fluorides become less negative as
we descend the group, whilst the reverse is true for chlorides, bromides and iodides. For a given metal
fH always becomes less negative from fluoride to iodide. The melting and boiling points follow the trend:
fluoride > chloride > bromide > iodide
All halides are soluble in water. The low solubility of LiF in water is due to its high lattice enthalpy, whereas the
low solubility of CsI is due to smaller hydration enthalpy of its ions. Other halides of lithium are soluble in
ethanol, acetone and ethylacetate. LiCl is soluble in pyridine.
Covalent character of halides decreases down the group; LiCl > NaCl > KCl > RbCl > CsCl (smaller the
cation, greater the polarising power and hence larger the covalent character).
Covalent character of lithium halides increases in the order; LiI > LiBr > LiCl > LiF (larger the anion greater
is its polarisability.)
3. Salts of Oxo-Acids. Oxo-acids are those in which the acidic proton is on a hydroxyl group with an oxo
group attached to the same atom, e.g., carbonic acid, H2CO3 (OC(OH)2), sulphuric acid, H2SO4 (O2S(OH)2),
H3PO4, HNO2, HNO3, etc.
Alkali metals form salts with all oxo-acids which are soluble in water and thermally stable. Their carbonates
(M2CO3) and most hydrogencarbonates (MHCO3) are highly stable towards heat. As the electropositive
character increases down the group, the stability of the carbonates and hydorgencarbonates increases.
Lithium carbonate is not very stable to heat; lithium being very small in size polarises the large CO32– ion
leading to the formation of more stable Li2O and CO2. In other words, the higher lattice energy of Li2O over
Li2CO3 favours its decomposition (LiCO3 does not exist as solid).

Li 2 CO3   Li 2 O  CO 2
Alkali metals form solid bicarbonates which decompose on gentle heating to carbonates and CO2.

2MHCO3   M 2CO3  CO 2  H 2 O
The hydrogencarbonate of lithium does not exist as a solid.
10.3 ANOMALOUS PROPERTIES OF LITHIUM
The anomalous behaviour of lithium is due to :
(i) Exceptionally small size of the atom and ion and high polarising power (charge/radius ratio). Thus, lithium
compounds have greater covalent character which is responsible for their solubility in organic solvents.
(ii) Comparatively high ionisation enthalpy and low electropositive character
(iii) Non-availability of d-orbitals and strong intermetallic bonding.

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Lithium exhibits diagonal relationship to magnesium.

10.3.1 Difference between Lithium and other Alkali Metals
(i) Lithium is much harder and its melting and boiling points are higher than other alkali metals.
(ii) Lithium is least reactive but the strongest reducing agent among alkali metals. Unlike other alkali metals, it
forms mainly monoxide, Li2O and the nitride, Li3N on combustion in air. LiOH is a weak base while the
hydroxides of other alkali metals are strong bases.
(iii) LiCl is deliquescent and crystallises as a hydrate, LiCl.2H2O, whereas other alkali metal chlorides do not
form hydrates.
(iv) Lithium hydrogencarbonate is not a solid while all other elements form solid hydrogencarbonates.
(v) Unlike other alkali metals, Li forms no ethynide on reaction with ethyne.
(vi) Lithium nitrate when heated gives lithium oxide, Li2O, whereas other alkali metal nitrates decompose to
give the corresponding nitrite.
 
4LiNO3   2Li2O  4NO2  O2 ; 2NaNO 3   2NaNO 2  O 2
(vii) LiF and Li2O are much less soluble in water than the corresponding compounds of other alkali metals.
10.3.2 Similarities between Lithium and Magnesium (Diagonal relationship). There is great similarity between
lithium and magnesium which arises from their similar sizes.
Both lithium and magnesium have similar atomic radii, ionic radii and electronegativity (Atomic radii : Li = 152
pm, Mg = 160 pm; Ionic radii : Li+ = 76 pm, Mg2+= 72 pm; Electronegativity : Li = 0.98, Mg = 1.2).
(i) Both lithium and magnesium are harder and lighter than other elements of the respective groups.
(ii) Lithium and magnesium react slowly with water.
(iii) Their oxides and hydroxides are much less soluble and their hydroxides decompose on heating.
(iv) Both form nitrides, Li3N and Mg3N2, by direct combination with nitrogen.
(v) Oxides, Li2O and MgO do not combine with excess oxygen to give any superoxide.
(vi) Both LiOH and Mg(OH)2 are weak bases.
(vii) Carbonates of lithium and magnesium decompose easily on heating to form the oxides and CO2.
(viii) Solid hydrogencarbonates are not formed by lithium and magnesium.
(ix) Both LiCl and MgCl2 are deliquescent and soluble in ethanol.
(x) Both LiCl and MgCl2 crystallise from aqueous solution as hydrates, LiCl·2H2O and MgCl2·8H2O.
10.4 SOME IMPORTANT COMPOUNDS OF SODIUM
10.4.1 Sodium Carbonate (Washing Soda), Na2CO3·10H2O.
(a) Solvay (ammonia-soda) process. When CO2 is passed through brine (NaCl) solution saturated with
ammonia, sodium hydrogencarbonate, being sparingly soluble, crystallises out. The function of ammonia is to
produce excess of HCO3- ions which precipitates NaHCO3 from the reaction mixture.
2NH 3  H 2 O  CO 2 
(NH 4 ) 2 CO3

(NH 4 ) 2 CO 3  H 2 O  CO 2 
 2NH 4 HCO3

NH 4 HCO3  NaCl 
 NH 4 Cl  NaHCO3
Sodium hydrogencarbonate crystals are heated to get sodium carbonate.

2NaHCO3   Na 2CO3  CO2  H 2O

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Ammonia is recovered when the solution containing NH4Cl is treated with Ca(OH)2.
Solvay process cannot be extended to the manufacture of potassium carbonate (K2CO3) because potassium
hydrogencarbonate (KHCO3) is too soluble to be precipitated by NH4HCO3 formed by passing CO2 through
ammoniated KCl solution.
Properties : Sodium carbonate is a white crystalline solid which exists as a decahydrate, Na2CO3·10H2O. It
is also called washing soda. It is readily soluble in water.
i. Action of heat. On heating, the decahydrate loses water of crystallisation to form monohydrate. Above
373K, the monohydrate becomes an anhydrous white powder called soda ash.
 373 K
Na 2 CO 3 .10H 2O  Na 2 CO 3 .H 2 O  9H 2 O
Na 2 CO3 .H 2 O 
> 373 K
Na 2 CO3  H 2 O
ii. Hydrolysis. Being a salt of a strong base (NaOH) and weak acid (H2CO3), sodium carbonate is hydrolysed
by water to form an alkaline solution.
CO32  H 2 O 
 HCO 3  OH 
iii. Action of acids. It reacts with dilute mineral acids evolving CO2 gas.
Na 2 CO3  2HCl 
 2NaCl  H 2 O  CO 2
iv. Reaction with milk of lime. Sodium hydroxide is formed.
Ca(OH) 2  Na 2 CO 3 
 CaCO 3  2NaOH
Uses: It is used in water softening, laundering and cleaning, in the manufacture of glass, soap, borax and
caustic soda, in paper, paints and textile industries, in petroleum refining and metal refining, as a laboratory
reagent in qualitative and quantitative analysis and a mixture of Na2CO3 and K2CO3 is used as fusion mixture.
10.4.2 Sodium Chloride, NaCl
The most abundant source of sodium chloride is sea water which contains 2.7-2.9% by mass of the salt. In
tropical countries like India, common salt is obtained by the evaporation of sea water.
Crude sodium chloride contains sodium sulphate, calcium sulphate, calcium chloride and magnesium chloride
as impurities (calcium chloride and magnesium chloride are deliquescent).
Preparation of pure sodium chloride. On saturating a concentrated solution of sodium chloride with hydrogen
chloride gas, crystals of pure sodium chloride separate out (due to common ion effect). Calcium and magnesium
chlorides, being more soluble, remain in solution.
Properties. Sodium chloride melts at 1081K.
It has a solubility of 36.0 g in 100 g of water at 273 K. The solubility does not increase appreciably with
increase in temperature.
Uses : (i) It is used as common salt or table salt for domestic purpose. (ii) It is used for the preparation of
Na2O2, NaOH and Na2CO3.
10.4.3 Sodium Hydroxide (Caustic Soda), NaOH
Sodium hydroxide is commercially prepared by the electrolysis of sodium chloride in a Castner-Kellner cell
or Mercury cathode cell using mercury cathode and carbon anode. Sodium metal discharged at the cathode
combines with mercury to form sodium amalgam. Chlorine is evolved at the anode.
Cathode: Na   e  
Hg
 Na  ama lg am
Anode : Cl 
Hg
1 2 Cl 2  e 
Sodium amalgam reacts with water to give sodium hydroxide and hydrogen gas.
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2(Na / Hg)  2H 2 O 
 2NaOH  2Hg  H 2
Now a days the use of mercury is avoided by using asbestoes diaphragm or Nafion (sulphonated
tetrafluoro ethylene based polymer) membrane.
Properties. Sodium hydroxide is a white, translucent solid. It melts at 591 K. It is readily soluble in water to
give strongly alkaline solution. Sodium hydroxide is deliquescent. NaOH solution reacts with CO2 of the
atmosphere to form Na2CO3.
Uses: It is used in the manufacture of soap, paper, artificial silk and a number of chemicals, in petroleum
refining, in the purification of bauxite, in textile industry for mercerising cotton fabrics, for the preparation of
pure fats and oils and as a laboratory reagent.
10.4.4 Sodium Hydrogencarbonate (Baking Soda), NaHCO3
Sodium hydrogencarbonate is known as baking soda because it decomposes on heating to generate bubbles
of carbon dioxide (leaving holes in cakes, pastries, etc., making them light and fluffy).
Preparation. Sodium hydrogencarbonate is prepared by saturating a solution of sodium carbonate with
carbon dioxide. Sodium hydrogencarbonate, being less soluble, separates out.
Na 2 CO3  CO 2  H 2 O   2NaHCO 3
Properties. White crystalline powder. It reacts with acids to evolve CO2.
NaHCO3  HCl   2NaCl  H 2 O  CO 2
Uses. It is used in the preparation of baking powder, as a mild antiseptic for skin infections, for generating
CO2 in fire extingushers and as an antacid.
10.5 BIOLOGICAL IMPORTANCE OF SODIUM AND POTASSIUM
A 70 kg human body contains about 90 g of Na and 170 g of K (5 g of iron and 0.06 g of copper). Sodium
ions are found primarily on the outside of cells - in blood plasma and the interstitial fluid which surrounds the
cells. These ions participate in the transmission of nerve signals, in regulating the flow of water across cell
membranes and in the transport of sugars and amino acids into cells.
Sodium and potassium, although chemically similar, differ in their ability to penetrate cell membranes, in their
transport mechanisms and in their efficiency to activate enzymes.
Potassium ions are the most abundant cations within cell fluids, where they activate many enzymes, participate
in the oxidation of glucose to ATP and with sodium, are responsible for the transmission of nerve signals.
There is considerable variation in the concentration of sodium and potassium ions on the opposite sides of the
cell membrane. E.g., in blood plasma, sodium level is 143 mmol L–1 and potassium level is only 5 mmol L–1.
Within the cells, these concentrations are 10 mmol L–1 (Na+) and 105 mmol L–1 (K+). These ionic gradients
indicate that a discriminatory mechanism, called the sodium-potassium pump, operates across cell membranes.
10.6 GROUP 2 ELEMENTS : ALKALINE EARTH METALS
The group 2 of the Periodic Table consists of beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr),
barium (Ba) and radium (Ra). These (except Be) are known as alkaline earth metals. Like alkali metals,
alkaline earth metals are also highly reactive and hence do not occur in the free state. Ca and Mg rank fifth and
sixth in abundance in the earth’s crust. Sr (15th) and Ba (14th) have much lower abundance. Be (51th) is rare
and Ra is the rarest (radioactive). Be and Ra together constitute 10-10 % of igneous rocks.
Occurance. Beryllium occurs as silicate minerals, beryl, Be3Al2Si6O18 and phenacite, Be2SiO4. Magnesium
exists mainly as carnallite, KCl.MgCl2.6H2O and magnesite, MgCO3 and in sea water (0.13%) as chloride
and sulphate. Calcium occurs as CaCO3 in the form of limestone, marble and chalk. It also occurs as fluorite
or fluorspar, CaF2, fluoropatite, [3Ca3(PO4)2.CaF2], gypsm, CaSO4.2H2O and anhydrite, CaSO4.
Strontium and barium are much less abundant. Strontium is present as celestite, SrSO4 and strontianite,

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SrCO3 and barium is present as barytes, BaSO4. Radium is found in minute amounts in pitchblende or
uraninite UO2.
10.6.1 Atomic properties
1. Electronic Configuration. These elements have two electrons in the s-orbital of the valence shell. The
general electronic configuration is [noble gas] ns2 outside the noble gas core.
Element/At. No. Be Mg Ca Sr Ba Ra
Electronic 2 2 2 2 2 2
configuration [He] 2s [Ne] 3s [Ar] 4s [Kr] 5s [Xe] 6s [Rn] s

2. Atomic and Ionic Radii. The atomic and ionic radii of alkaline earth metals are lower than those of the
alkali metals. This is due to increased nuclear charge of these elements. Within the group, the atomic and ionic
radii increase with increase in atomic number.
3. Ionization Enthalpies. Alkaline earth metals have low ionization enthalpies due to fairly large size of the
atoms. Since the atomic size increases down the group, ionization enthalpy decreases. The first ionisation
enthalpies of the alkaline earth metals are higher than those of the corresponding alkali metals. This is due to
smaller size of the atoms. The second ionisation enthalpies are also lower than those of alkali metals.
Table - 04 Atomic and physical properties of alkaline earth metals.
Property Be Mg Ca Sr Ba Ra
Atomic number 4 12 20 38 56 88
-1
Atomic mass (g mol ) 9.01 24.31 40.08 87.62 137.33 226.03
Metallic radius (pm) 112 160 197 215 222
+
Ionic radius (M /pm) 31 72 100 118 135 148
-1
Ionisation enthalpy,  i H1 (kJ mol ) 899 737 590 549 503 509
-1
 i H2 (kJ mol ) 1757 1450 1145 1064 965 979
-1
Hydration enthalpy (kJ mol ) -2494 -1921 -1577 -1443 -1305
Electronegativity (Pauling scale) 1.57 1.31 1 0.95 0.89 0.9
-3
Density (g cm ) at 293 K 1.84 1.74 1.55 2.63 3.59 5.5
Melting point (K) 1560 924 1124 1062 1002 973
Boiling point (K) 2745 1363 1767 1655 2078 1973
2+
E°/ V at 298 K for M /M -1.97 -2.36 -2.84 -2.89 -2.92 -2.92
-6*
Occurrence in Lithosphere 2* 2.76** 4.6** 384* 390* 10
*ppm (parts per million), ** percentage by weight.

4. Hydration Enthalpies. Like alkali metal ions, hydration enthalpies of alkaline earth metal ions decrease
with increase in ionic size.
Be2+> Mg2+ > Ca2+ > Sr2+ > Ba2+
The hydration enthalpies are greater than those of alkali metal ions. Thus, compounds of alkaline earth metals
are more extensively hydrated than those of alkali metals, e.g., MgCl2 and CaCl2 exist as MgCl2.6H2O and
CaCl2· 6H2O while NaCl and KCl do not form hydrates.
The existence of divalent ions in aqueous solution is due to greater enthalpy of hydration which counterbalances
the higher value of second ionisation enthalpy.
5. Reactivity and electrode potential. Alkaline earth metals are highly reactive due to strong tendency to

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lose two valence electrons and attain inert gas configuration. Thus, they have low ionisation enthalpies and
high negative standard electrode potentials. Chemical reactivity increases on descending the group.
10.6.2 Physical Properties
i. Colour and apparance. The alkaline earth metals are silvery white, lustrous and relatively soft, but
harder than the alkali metals. Beryllium and magnesium are greyish.
ii. Melting and boiling points. Due to smaller size and more close packed crystal lattice, the melting and
boiling points are higher than those of the corresponding alkali metals. The trend is, however, not systematic.
iii. Density. Alkaline earth metals are heavier than alkali metals due to smaller size and more close packed
crystal lattice. In the group, density first decreases fron Be to Ca and then increases.
iv. Electropositive or metallic character. Because of low ionisation enthalpies, they are strongly
electropositive in nature. The electropositive character increases down the group.
v. Nature of bonds. Alkaline earth metals form ionic bonds which are less ionic than alkali metals.
Due to smaller size of cations and greater number of valence electrons, metallic bonding in alkaline earth
metals is stronger than in alkali metals.
vi. Flame colour. Like alkali metals, alkaline earth metal salts also impart characteristic colour to the flame.
Metal Calcum Strontium Barium Radium
Colour Brick red Crimson Apple green Crimson
The electrons in beryllium and magnesium are too strongly bound to become excited by flame, therefore,
they do not impart colour to the flame.
Detection and estimation. The flame test for Ca, Sr and Ba is helpful in their detection in qualitative
analysis and estimation by flame photometry.
vii. Electrical and thermal conductivity. Alkaline earth metals, like alkali metals, have high electrical and
thermal conductivities.
10.6.3 Chemical Properties.
1. Reactivity towards air and water: Beryllium and magnesium are kinetically inert to oxygen and water
because of the formation of an oxide film on their surface. However, powdered beryllium burns brilliantly on
ignition in air to form BeO and Be3N2.
 
2Be  O2 (air)   2BeO ; 3Be  N 2 (air)   Be 3 N 2
Magnesium is more electropositive and burns with dazzling brilliance in air to form MgO and Mg3N2.
 
2Mg  O2 (air)   2MgO ; 3Mg  N2 (air)   Mg3 N2
Calcium, strontium and barium are readily attacked by air to form the oxide and nitride.
 
2M  O 2 (air)   2MO ; 3M  N 2 (air)   M3N2
They also react with water even in the cold to form hydroxides.
M  H 2O  
 M O  H 2 or M  2H 2 O  
 M (OH) 2  H 2
2. Reaction with halogens: These metals combine with halogens at elevated temperatures forming halides.
M  X 2 
 MX 2 (where X = F, Cl, Br, I)
Thermal decomposition of (NH4)2BeF4 is the best route for the preparation of BeF2.

(NH 4 ) 2 BeF4   BeF2  2NH 4 F
BeCl2 is conveniently made from the oxide by heating BeO with Cl2 in presence of carbon..

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
600 800K
BeO  C  Cl 2  BeCl 2  CO
3. Reactivity towards hydrogen: Alkaline earth metals except beryllium combine with hydrogen on heating
to form their hydrides, MH2.

M  H 2   MH 2 ( M = Mg, Ca, Sr, Ba)
BeH2 can be prepared by the reaction of BeCl2 with lithium aluminium hydride, LiAlH4.
2BeCl2  LiAlH 4 
 2BeH 2  LiCl  AlCl3
Both BeH2 and MgH2 are covalent compounds. Both have only four electrons in the valence shell, therefore,
they are electron deficient molecules. To make up for electron deficiency, each Be or Mg atom forms four
three centre - two electron (3c-2e) bonds or banana bonds. Thus, due to electron deficiency, BeH2 and
MgH2 have polymeric structures.

The stability of alkaline earth metal hydrides decreases down the group from Be to Ba since their lattice
enthalpies decrease with increase in size of the metal cations.
Hydrides of alkaline earth metals react with water liberating H2 gas and thus act as reducing agents.
MH 2  2H 2O 
 M(OH) 2  2H 2 ; CaH2  2H2O 
 Ca(OH)2  2H2
CaH2 is called hydrolith and is used for large scale production of H2 by the action of water on it.
5. Reactivity towards acids: The alkaline earth metals readily react with acids liberating dihydrogen.
M  2HCl 
 MCl2  H 2 (M = Be, Mg, Ca, Sr, Ba)
Being amphoteric, Be reacts with alkali forming sodium beryllate and H2.
Be  2NaOH  2H 2 O 
 Na 2 [Be(OH) 4 ]  H 2
6. Reducing nature: The alkaline earth metals are strong reducing agents. This is indicated by large negative
values of reduction potentials. However, their reducing power is less than alkali metals. Beryllium has less
negative value compared to other alkaline earth metals. Its reducing nature is due to large hydration energy
associated with small size of Be2+ ion and relatively large value of atomisation enthalpy of the metal.
7. Solutions in liquid ammonia: Like alkali metals, the alkaline earth metals dissolve in liquid ammonia to
give deep blue black solutions containing solvated (ammoniated) electrons (described as expanded metals).
However, concentrated solutions are bronze coloured due to the formation of metal clusters. The blue solutions
are paramagnetic, whereas bronze-coloured solutions are diamagnetic.
[M(NH3 ) x ]  [e (NH3 ) y ]
M  (x  y)NH3 
From these solutions, the ammoniates, [M(NH3)6]2+ can be recovered.
8. Formation of complexes. Be and Mg have the maximum tendency to form complexes, e.g., chlorophyll
is a complex of Mg. BeF2 combines with F- to form the tetrahedral complex ion, [BeF4]2-.
10.6.4 Uses of alkaline earth metals
Beryllium. It is used in the manufacture of alloys. Copper-beryllium alloys are used in the preparation of high
strength springs. Metallic beryllium is used for making windows of X-ray tubes.
Magnesium. It forms alloys with Al, Zn, Mn and Sn. Magnesium-aluminium alloys being light are used in air-

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craft manufacture. Duralumin is an alloy of Al = 95%, Cu = 4%, Mn = 0.5%, and Mg = 0.5%.

Magnesium (powder/ribbon) is used in flash powders and bulbs, incendiary bombs, and signals. It is also
used for preparing Grignard reagents (RMgX) and for igniting thermite charge in aluminothermy.
A suspension of magnesium hydroxide in water (milk of magnesia) is used as antacid in medicine. Magnesium
hydroxide and magnesium carbonate are ingredient of toothpaste.
Calcium. It is used in the extraction of metals from oxides which are difficult to reduce with carbon.
Calcium and barium, owing to their reactivity with oxygen and nitrogen at elevated temperatures, are used to
remove air from vacuum tubes.
Radium. Radium salts are used in radiotherapy (treatment of cancer).
10.7 GENERAL CHARACTERISTICS OF COMPOUNDS OF ALKALINE EARTH METALS
The predominant valence of Group-2 elements is the dipositive oxidation state (M2+). Alkaline earth metal
compounds are ionic, but less ionic than alkali metal compounds. This is due to increased nuclear charge and
smaller size. Compounds of beryllium and magnesium are more covalent than those of the heavier members.
(i) Oxides and Hydroxides. Alkaline earth metals burn in oxygen to form the monoxide, MO. Except for
BeO, all oxides have rock-salt structure. BeO is covalent.

2M  O 2   2MO (where M = Be, Mg, Ca)
The enthalpies of formation of these oxides are quite high, therefore, they are very stable towards heat. BeO
is amphoteric while oxides of other metals are ionic. Oxides, except BeO are basic and react with water to
form sparingly soluble hydroxides (sometimes called slaking).
MO  H 2 O 
 M(OH) 2 (where M = Ca, Sr, Ba)
Solubility, thermal stability and basic character of hydroxides increase with increasing atomic number. However,
alkaline earth hydroxides are less basic and less stable than alkali metal hydroxides. Be(OH)2 is amphoteric.

[Be  H 2O 4 ]Cl2
Be(OH)2  2HCl  2H 2O 

 Na 2 Be  OH 4
Be(OH) 2  2NaOH 
Sodium beryllate

This is because both lattice enthalpy and hydration enthalpy decreases as the size of the atom increases, but
lattice enthalpy decreases more rapidly than hydration enthalpy and hence solubility increases down the group.
(ii) Halides. Halides except beryllium halides are ionic. Beryllium halides are covalent and soluble in organic
solvents. Beryllium chloride has chain structure in the solid state. In the vapour phase BeCl2 forms a chloro-
bridged dimer which dissociates into the linear monomer at high temperatures (1200 K).

The tendency to form halide hydrates decreases (e.g., MgCl2·8H2O, CaCl2·6H2O, SrCl2·6H2O and
BaCl2·2H2O) down the group. The chlorides, bromides and iodides of Ca, Sr and Ba can be dehydrated on
heating. However, the hydrated halides of Be and Mg on heating undergo hydrolysis. The fluorides are relatively
less soluble than the chlorides due to high lattice energies.

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Melting points and boiling points of the halides follow the trend, fluoride > chloride > bromide > iodide.
Low solubility of LiCl is due to high lattice energy and that of CsCl is due to lower hydration energy of its ions.
(iii) Salts of Oxoacids. Group 2 metals form carbonates, bicarbonates, sulphates and nitrates.
Carbonates. Carbonates of alkaline earth metals are obtained by passing CO2 through hydroxide solutions.
M(OH) 2 (aq)  CO 2 (g) 
 MCO3 (s)  H 2 O(l )
Carbonates are sparingly soluble in water (solubility decreases as the atomic number of the metal ion increases)
and can be precipitated by the addition of sodium or ammonium carbonate solution.
Solubility of carbonates increases in presence of CO2 due to the formation of bicarbonates.
MCO 3  H 2 O  CO 2 
 M(HCO3 ) 2
All the carbonates decompose on heating to give carbon dioxide and the oxide.

MCO 3   MO  CO 2
Beryllium carbonate is unstable and can be kept only in an atmosphere of CO2. Thermal stability increases
with increasing cationic size :
BeCO3 (< 373 K), MgCO3 (813 K), CaCO3 (1173 K), SrCO3 (1563 K), BaCO3 (1633 K).
Bicarbonates. These are prepared by passing CO2 through a suspension of the metal carbonate in water.
MCO 3  H 2 O  CO 2 
 M(HCO3 ) 2
Bicarbonates of alkaline earth metals are stable only in solution and have not been isolated in the pure state.
Sulphates. Sulphates of alkaline earth metals are prepared by the action of sulphuric acid on metals, metal
oxides, metal carbonates, etc.
M  H 2SO 4 
 MSO 4  H 2 ; MO  H 2SO 4 
 MSO 4  H 2 O

M(OH)2  H2SO4 
 MSO4  2H2 O ; MCO 3  H 2SO 4 
 MSO 4  H 2 O  CO 2
Sulphates are white solids which are stable to heat. Solubility decreases from CaSO4 to BaSO4. BeSO4, and
MgSO4 are readily soluble in water since the greater hydration enthalpies of Be2+ and Mg2+ ions overcome the
lattice enthalpy factor. In the group, solubility of sulphates decreases in the order ; Be > Mg >> Ca > Sr > Ba.
Being insoluble in water and opaque to X-rays, BaSO4 is used for obtaining radiographs of stomach (for
diagnosing ulcers, etc.) under the name barium meal.
Nitrates. Nitrates of alkaline earth metals are prepared by treating the oxides, hydroxides or carbonates with
dilute nitric acid.
MO  2HNO 3 
 M(NO 3 ) 2  H 2 O ; M(OH) 2  2HNO 3 
 M(NO 3 )2  2H 2 O

MCO 3  2HNO 3 
 M(NO3 ) 2  H 2 O  CO 2
Magnesium nitrate crystallises with six molecules of water, whereas barium nitrate crystallises as anhydrous
salt. This shows a decreasing tendency to form hydrates with increasing size and decreasing hydration enthalpy.
All nitrates are soluble in water. Nitrates decompose on heating to give the oxide (like lithium nitrate).

2M(NO3 ) 2   2MO  4NO 2  O 2 (M = Be, Mg, Ca, Sr, Ba)
10.8 ANOMALOUS BEHAVIOUR OF BERYLLIUM
Beryllium, the first member of the alkaline earth metals, behaves differently from other members of the group
due to smaller atomic and ionic sizes, high ionisation enthalpy and absence of d-orbitals in its valence shell.
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(i) Beryllium is harder than other members of the group. It has higher melting and boiling points.
(ii) Beryllium has exceptionally small atomic and ionic sizes. Because of high ionisation enthalpy and small size
it forms covalent compounds which are easily hydrolysed.
(iii) Beryllium does not exhibit coordination number more than four as it has only four orbitals in its valence
shell. The other members of the group can have a coordination number of six by making use of d-orbitals.
(iv) The oxide and hydroxide of beryllium, unlike the hydroxides of other elements in the group, are amphoteric.
(v) Be does not react with water even at high temperatures while other metals do.
(vi) Beryllium carbide reacts with water to form methane; carbides of other metals give acetylene gas.
Be 2 C  4H 2 O 
 2Be(OH) 2  CH 4 ; CaC 2  2H 2 O 
 Ca(OH) 2  HC  CH
10.8.1 Diagonal Relationship between Beryllium and Aluminium.
Beryllium shows diagonal relationship with aluminium of group 13. This may be because the charge/radius
ratio of Be2+ (2/31 = 0.064) is nearly the same as that of the Al3+ ion (3/50 = 0.060).
Similarities : (i) Like aluminium, beryllium is not readily attacked by acids because of the presence/formation
of an oxide film on the surface of the metal.
(ii) Beryllium hydroxide dissolves in excess of alkali to give beryllate ion, [Be(OH)4]2– just as aluminium
hydroxide gives aluminate ion, [Al(OH)4]–.
(iii) The chlorides of both beryllium and aluminium have Cl– bridged chloride structure in vapour phase. Both
are soluble in organic solvents and are strong Lewis acids. They are used as Friedel Craft catalysts.

(iv) Beryllium and aluminium ions have strong tendency to form complexes, BeF42–, AlF63–.
(v) The oxides of both metals are hard, high-melting, insoluble solids.
(vi) Carbides of both metals react with water liberating methane gas
10.9 SOME IMPORTANT COMPOUNDS OF CALCIUM
Important compounds of calcium are calcium oxide, calcium hydroxide, calcium carbonate, calcium sulphate
and cement. These are also industrially important compounds.
10.9.1 Calcium Oxide or Quick Lime, CaO
It is prepared on a commercial scale by heating limestone (CaCO3) in a rotary kiln at 1070-1270 K.

1070 1270K
CaCO 3  
 CaO  CO 2 ; H  179.9 kJ
The carbon dioxide is removed as soon as it is formed to enable the reaction to proceed to completion.
Properties: CaO is a white amorphous solid with a melting point of 2870 K. On exposure to the atmosphere,
it absorbs moisture and carbon dioxide. On strong heating, it emits brilliant white light called limelight.
CaO  CO 2 
 CaCO3 ; CaO  H 2 O 
 Ca(OH) 2
Slaked lim e

The addition of limited amount of water breaks the lump of lime. This process, called slaking of lime is highly
exothermic.
 Ca(OH)2 ; H  64.5kJ mol1
CaO  H 2O 
Slaked lim e

Quick lime slaked with caustic soda gives solid soda-lime (CaO + NaOH).
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Action of acids and acidic oxides. Being a basic oxide, it combines with acids.

CaO  2HCl 
 CaCl 2  H 2 O ; 6CaO  P4 O10   2Ca 3 (PO 4 ) 2
1270K
It reacts with acidic oxides at high temperature : CaO  SiO2  CaSiO3
Calcium silicate

Reaction with coke. When heated with coke in an electric furnace at 2273K, calcium carbide is obtained.
CaO  3C 
2273K
 CaC 2  CO
Reaction with ammonium salts. On heating with ammonium salts, it liberates ammonia gas.

CaO  2NH 4 Cl   CaCl2  2NH 3  H 2 O
Uses: (i) It is an important primary material for manufacturing cement and is the cheapest form of alkali. (ii) In
the manufacture of sodium carbonate from caustic soda. (iii) In the purification of sugar and in the manufacture
of dye stuffs. (iii) In metallurgy, as a flux and as basic lining in furnaces. (iv) For drying alcohols and non-acidic
gases.(v) In the preparation of mortar, calcium carbide, ammonia, soda-lime and sodium carbonate.
10.9.2 Calcium Hydroxide (Slaked lime), Ca(OH)2.
Calcium hydroxide is prepared by adding water to quick lime, CaO.
CaO  H 2 O 
 Ca(OH) 2
Properties. It is a white amorphous powder, sparingly soluble in water. The aqueous solution is known as
lime water and a suspension of slaked lime in water is known as milk of lime.
Reaction with CO2 (Test for CO2). When carbon dioxide is passed through lime water, it turns milky due
to the formation of calcium carbonate.
Ca(OH) 2  CO 2 
 CaCO3   H 2 O
On passing excess of carbon dioxide, the precipitate dissolves to form calcium hydrogencarbonate (milkiness
disappears).
CaCO3  H 2O  CO2 
 Ca(HCO3 ) 2
so lub le

Action of heat. On heating above 700 K, slaked lime loses water to form CaO.
 700K
Ca(OH)2   CaO  H 2 O
Reaction with chlorine. Slaked lime reacts with chlorine to form hypochlorite, a constituent of bleaching
powder.
2Ca(OH) 2  2Cl2 
 CaCl2  Ca(OCl) 2  2H 2 O
Bleaching powder

Reaction with acids. Being basic, slaked lime reacts with acids and acidic oxides forming salts.
Ca(OH) 2  2HCl 
 CaCl2  2H 2O ; Ca(OH) 2  SO3 
 CaSO 4  H 2O
Uses: (i) In the preparation of mortar (a building material). (ii) In white wash due to its disinfectant nature. (iii)
In glass making and in tanning. (iv) For the preparation of bleaching powder and for purification of sugar. (v)
For absorbing acidic gases. (vi) Lime water is used for detecting CO2.
10.9.3 Calcium Carbonate, CaCO3.
Calcium carbonate occurs in nature in several forms such as limestone, chalk, marble etc.
Preparation. It can be prepared by passing carbon dioxide through slaked lime or by the addition of sodium
carbonate to calcium chloride.
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Ca(OH) 2  CO 2 
 CaCO3   H 2 O ; CaCl2  Na 2 CO 3 
 CaCO3  2NaCl
Excess of CO2 should be avoided since it leads to the formation of water soluble CaHCO3.
Properties. Calcium carbonate is a white fluffy powder, almost insoluble in water. When heated to 1200 K,
it decomposes to evolve carbon dioxide.
1070 1270K
CaCO3  CaO  CO 2
It reacts with dilute acids to liberate carbon dioxide.
CaCO 3  2HCl 
 CaCl2  H 2 O  CO 2
Uses: (i) It is used as a building material in the form of marble and in the manufacture of quick lime. (ii)
Calcium carbonate along with magnesium carbonate is used as a flux in the extraction of metals such as iron.
(iii) Specially precipitated CaCO3 is extensively used in the manufacture of high quality paper. (iv) As an
antacid, mild abrasive in tooth paste, a constituent of chewing gum, and a filler in cosmetics.
10.9.4 Calcium Sulphate (Gypsum) CaSO4.2H2O
It occurs in nature as Anhydrite (CaSO4) and Gypsum (CaSO4.2H2O). CaSO4.2H2O is also known as
alabaster.
Preparation: By action of H2SO4 or soluble sulphates on any soluble calcium salt.
CaCl2  H 2SO 4  CaSO 4  2HCl ; CaCl2  Na 2SO 4  CaSO 4  2NaCl
Properties: (i) Calcium sulphate is a white crystalline solid, which is sparingly soluble in water.

(ii) Dilute acids and ammonium sulphate dissolve CaSO4 due to the formation of  NH 4 2 SO4 .CaSO 4 .H 2O
(double sulphate).
(iii) Action of heat: Following changes are observed at different temperatures.

2  CaSO 4 .2H 2 O     2CaSO 4 .H 2 O  3H 2 O

o
120 C

Plaster of Paris

 2CaSO4 .H 2O 

o
200 C
 2CaSO  H 2 O ; 2CaSO 
4
Heated
 2CaO  2SO 2  O 2
4 strongly
Dead burnt plaster

Uses: In the manufacture of plaster of paris, cement, ammonium sulphate, blackboard chalks, etc.
10.9.5 Plaster of Paris (CaSO4·½ H2O).
It is hemihydrate of calcium sulphate, obtained when gypsum, CaSO4·2H2O, is heated to 393 K.

2 CaSO 4 .2H 2O 

393K
 2 CaSO 4 .1 2 H 2 O 3H 2 O
Gypsum Plaster of Paris

Above 393 K, anhydrous calcium sulphate, CaSO4 known as ‘dead burnt plaster’is formed.
 393K
2CaSO4 .1 2 H 2 O   2CaSO4  H 2O
Plaster of Paris Dead burnt Plaster

Properties: It is a white powder. On mixing with water it forms a plastic mass that sets into a hard solid in 5
to 15 minutes (reason for calling it plaster). Setting of Plaster of Paris is due to rehydration and conversion to
gypsm. During setting, slight expansion in volume occurs. As a result, it can take up the shape and impression
of the mould in which it is put.

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Uses: (i) In the building industry as well as manufacture of plasters. (ii) It is used for immobilising the affected
part of organ where there is a bone fracture or sprain. (iii) In dentistry. (iv) In ornamental work and for making
casts of statues and busts. (v) For making black-board chalks.
10.10 CEMENT (PORTLAND CEMENT )
Cement is an important building material, first introduced in England in 1824 by Joseph Aspdin. It is also
called Portland cement since it resembles the natural limestone quarried in the Isle of Portland, England.
Average composition of Portland cement. CaO, 50-60%; SiO2, 20-25%; Al2O3, 5-10%; MgO, 2-3%;
Fe2O3, 1-2% and SO3, 1-2%. The ratio of silica (SiO2) to alumina (Al2O3) should be between 2.5 and 4 and
the ratio of lime (CaO) to the total of the oxides of silicon (SiO2) aluminium (Al2O3) and iron (Fe2O3) should
be about 2.
Manufacture of cement. The raw materials for the manufacture of cement are limestone and clay. When
clay and lime are strongly heated together, they fuse and react to form cement clinker. Clinker is mixed with
2-3% by weight of gypsum (CaSO4·2H2O) to form cement. Important ingredients of cement are dicalcium
silicate (Ca2SiO4) 26%, tricalciumsilicate (Ca3SiO5) 51% and tricalcium aluminate (Ca3Al2O6) 11%.
Setting of Cement: When mixed with water, cement sets into a hard mass having three-dimensional network
structure involving -Si-O-Si- and -Si-O-Al- chains. This is due to hydration of the molecules of the constituents
and their rearrangement. The purpose of adding gypsum is only to slow down the process of setting of cement
so that it gets sufficiently hardened. Chemical reactions between components of cement and water produce
new solid phases; crystalline calcium hydroxide and ‘cement gel’ which consists of 3CaO.2SiO2,.3H2O.
Uses: Cement is a commodity of national necessity next to iron and steel. It is used in concrete, reinforced
concrete, and in plastering for the construction of bridges, dams, buildings, roads, etc.
10.11 BIOLOGICAL IMPORTANCE OF MAGNESIUM AND CALCIUM
Magnesium. An adult body contains about 25 g of Mg and 1200 g of Ca. The daily human requirement is
200 – 300 mg. All enzymes that utilise ATP in phosphate transfer require magnesium as the cofactor. Chlorophyll
contains magnesium.
Calcium. About 99 % of body calcium is present in bones as hydroxyapatite, Ca5(PO4)3OH and in the
enamel of the teeth as fluoroapatite [3Ca3(PO4)2.CaF2]. It also plays important roles in neuromuscular function,
interneuronal transmission, cell membrane integrity and blood coagulation. The calcium concentration in plasma
of about 100 mg L–1. It is maintained by two hormones, calcitonin and parathyroid hormone. A net loss of
bone calcium in elderly people, especially women, is called osteoporosis.

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QUESTIONS
LEVEL - I
1. Which one of the following represents the composition of carnallite mineral?
1) K 2 O.Al 2O3 .6SiO 2 2) KNO3

3) K 2SO 4 .MgSO 4 .MgCl2 .6H 2 O 4) KCl.MgCl2 .6H 2O

2. The similarity in the properties of alkali metals is due to:
1) Their same atomicity 2) Similar outer shell configuration
3) Same energy of outer shell 4) Same principal quantum number of outer shell
3. Saltpetre is:
1) KNO3 2) NaNO3 3) NaCl 4) Na2CO3
4. A solid compound of group I element and it gives a bright red colour in the flame test. The solid is:
1) LiBr 2) CsCl 3) KCl 4) NaCl
5. The alkali metals are low melting. Which of the following alkali metal is expected to melt if the room temperature
rises to 30oC?
1) Na 2) K 3) Rb 4) Cs
6. Which of the following statements is false regarding alkali metals?
1) Alkali metals are soft and can be cut with the help of knife
2) Alkali metals do not occur in free state in nature
3) Alkali metals are highly electropositive 4) Alkali metal hydrides are covalent in character
7. Alkali metals can be extracted from their salts by:
1) Reduction with carbon 2) Electrolysis of aqueous solution of their halides
3) Electrolysis of fused halides 4) Reduction with aluminium
8. Sodium hydride (NaH) when dissolved in water, produces:
1) Acidic solution 2) Basic solution 3) Neutral solution 4) Cannot be predicted
9. When sodium is added in scanty water, it catches fire. In this process which one of the following burns?
1) Na 2) H2O 3) CO 4) H2
10. Sodium peroxide which is a yellow solid, when exposed to air becomes white due to the formation of :
1) H2O2 2) Na2O 3) Na2O and O3 4) NaOH and Na2CO3
11. The alkali halide which is soluble in pyridine is
1) NaCl 2) LiCl 3) KCl 4) CsI
12. Sodium peroxide in contact with moist air turns white mainly due to the formation of:
1) Na2O 2) Na2CO3 3) NaHCO3 4) NaOH
13. There is loss in mass when mixture of Li2CO3 and Na2CO3.10H2O is heated strongly. The loss is due to:
1) Li2CO3 only 2) Na 2 CO3 .10H 2 O only

3) both Li 2 CO 3 and Na 2CO 3 .10H 2 O 4) CO2

14. Prepartion of which of the following substance does not involve electrolysis of NaCl as its one step?
1) Na metal 2) NaOH 3) Na2O2 4) Na2CO3

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15. Which of the following decomposes on heating?

1) LiOH 2) NaOH 3) KOH 4) CsOH
16. The stability of the following alkali metal chlorides follows the order:
1) KCl > CsCl > NaCl > LiCl 2) LiCl > KCl > NaCl > CsCl
3) CsCl > KCl > NaCl > LiCl 4) NaCl > KCl > LiCl > CsCl
17. The salt of an alkali metal gives yellow colour in the flame test. Also its aqueous solution gives an insoluble
white precipitate with barium chloride in acid medium. The salt is:
1) NaCl 2) K2SO4 3) Na2SO4 4) Li2SO4
18. Which of the following compound is used in gun powder?
1) LiNO3 2) NaNO3 3) Pb(NO3)2 4) KNO3
19. Which metal bicarbonates does not exist in solid state?
i) LiHCO3 ii) Ca(HCO3)2 iii) Zn(HCO3)2 iv) NaHCO3
v) AgHCO3
1) (i), (ii), (iii), (v) 2) (i), (ii), (iii) 3) (i), (ii), (v) 4) (ii), (iii), (iv)
20. Which of the following does not illustrate the anomalous properties of lithium?
1) Li is much softer than the other group first metals
2) The melting point and boiling point of Li are comparatively high
3) Li forms a nitride Li3N unlike group first metals
4) The ion of Li and its compounds are more heavily hydrated than those of the rest of the group elements
21. Na2CO3 can be manufactured by Solvay process but K2CO3 cannot be prepared because:
1) K2CO3 is more soluble 2) K2CO3 is less soluble
3) KHCO3 is more soluble than NaHCO3 4) KHCO3 is less soluble than NaHCO3
22. A pair of substances which cannot exist together in solution is:
1) NaHCO3 + NaOH 2) NaHCO3 + Na2CO3 3) Na2CO3 + NaOH 4) NaOH + NaCl
23. The raw materials in Solvay process are:
1) NaOH, CaO and NH3 2) Na2CO3, CaCO3 and NH3
3) Na2SO4, CaCO3 and NH3 4) NaCl, NH3, CaCO3
24. The alkali metal which acts as a nutrient for plants is:
1) Na 2) K 3) Li 4) Rb
25. Property of alkaline earth metals that increases with their atomic number is:
1) Ionization energy 2) Solubility of their hydroxides
3) Solubility of their sulphates 4) Electronegativity
26. Magnesium metal is prepared by:
1) reduction of MgO by coke 2) electrolysis of aqueous solution of Mg(NO3)2
3) displacement of Mg by iron from magnesium sulphate solution
4) electrolysis of molten magnesium chloride
27. The group-2-element which does not directly combine with hydrogen is:
1) Be 2) Ca 3) Sr 4) Ba
28 Among LiCl, RbCl, BeCl2 and MgCl2 the compounds with greatest and least ionic character respectively are:
1) LiCl, RbCl 2) RbCl, BeCl2 3) RbCl, MgCl2 4) MgCl2, BeCl2
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29. The metal X is prepared by the electrolysis of fused chloride. It reacts with hydrogen to form a colourless
solid from which hydrogen is released on treatment with water. The metal is:
1) Al 2) Ca 3) Cu 4) Zn
30. Which of the following pairs of substances gives same gaseous product on reaction with water?
1) Na and Na2O2 2) Ca and CaH2 3) Ca and CaO 4) Ba and BaO2
31. A piece of magnesium ribbon was heated to redness in an atmosphere of N2 and then treated with H2O, the
gas evolved is:
1) ammonia 2) hydrogen 3) nitrogen 4) oxygen
32. The solubilities of carbonates decrease down the magnesium group due to a decrease in:
1) Entropy of solution formation 2) Lattice energies of solids
3) Hydration energies of cations 4) Interionic attraction
33. Which of the following is acidic in nature?
1) Be(OH)2 2) Mg(OH)2 3) B(OH)3 4) Al(OH)3
34. Which of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy?
1) BaSO4 2) SrSO4 3) CaSO4 4) BeSO4
35. Which of the following on thermal decomposition yields a basic as well as an acidic oxide?
1) KClO3 2) CaCO3 3) NH4NO3 4) Na2CO3
36. Which one of the following order represents the correct sequence of the increasing basic nature of the given
oxides?
1) MgO  K 2 O  Al 2O3  Na 2 O 2) Na 2 O  K 2 O  MgO  Al2 O3
3) K 2 O  Na 2 O  Al2 O3  MgO 4) Al 2 O3  MgO  Na 2 O  K 2 O
37. Which one of the following is present as an active ingredient in bleaching powder for bleaching action?
1) CaOCl2 2) Ca(OCl)2 3) CaO2Cl 4) CaCl2
38. Beryllium and aluminium exhibit many properties which are similar. But the two elements differ in:
1) exhibiting maximum covalency in compounds 2) forming polymeric hydrides
3) forming covalent halides 4) exhibiting amphoteric nature in their oxides
39. Dead burnt plaster is:
1
1) CaSO4 2) CaSO4 . HO 3) CaSO 4 .H 2O 4) CaSO 4 .2H 2O
2 2
40. Portland cement does not contain:
1) CaSiO4 2) Ca2SiO5 3) Ca2Al2O6 4) Ca3(PO4)2
LEVEL - II
1. For which one of the following minerals, the composition given is incorrect?
1) Soda ash - (Na2CO3) 2) Carnallite - (KCl . MgCl2.6H2O)
3) Borax-(Na2B4O7 . 7H2O) 4) Glauber’s salt - (Na2SO4.10H2O)
2. The second ionization enthalpy of which of the following alkaline earth metals is the highest?
1) Ba 2) Mg 3) Ca 4) Be
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3. Which one of the following statements is true for all the alkali metals?
1) Their nitrates decompose on heating to give NO2 and O2
2) Their carbonates decompose on heating to give CO2 and the metal oxide
3) They react with oxygen to give mainly the oxide M2O
4) They react with halogens to give the halides MX

4. The mobility of metal ions in aqueous medium  Li  , Na  , K  , Rb   in the electric field, follows the order:

1) Li   Na   K   Rb  2) Rb   Na   K   Li 
3) Li   Na   K   Rb  4) Na   K   Rb   Li 
5. Sodium chloride imparts a golden yellow colour to the Bunsen flame. This can be interpreted due to :
1) low ionization potential of sodium 2) photosensitivity of sodium
3) sublimation of metallic sodium to give yellow vapour
4) emission of excess of energy absorbed as a radiation in the visible region.
6. The reaction between sodium and water can be made less vigorous by
1) Adding a little alcohol 2) Amalgamating sodium
3) Adding a little acetic acid 4) Lowering the temperature
7. In Down’s process, for manufacture of sodium metal, CaCl2 is added to NaCl in order to
1) Increase ionisation of NaCl 2) Increase the melting point of NaCl
3) Decrease the melting point of NaCl 4) Increase conductance of electrolyte
8. Which pair of the following chlorides do not impart colour to the flame?
1) BeCl2 and SrCl2 2) BeCl2 and MgCl2 3) CaCl2 and BaCl2 4) BaCl2 and SrCl2
9. A colourless salt gives violet colour to Bunsen flame and also turns moistured litmus paper blue. It is :
1) Na 2 CO3 2) KNO3 3) K 2 CO3 4) Cu  OH 2
10. “Pearl ash” and caustic potash are chemically:
1) K 2 CO3 and KOH 2) KOH and K 2 CO3 3) Na2CO3 and KOH 4) Na2CO3 and NaOH
11. Sodium sulphate is soluble in water whereas barium sulphate is sparingly soluble because:
1) the hydration energy of sodium sulphate is more than its lattice energy
2) the lattice energy has no role to play in solubility
3) the hydration energy of sodium sulphate is less than its lattice energy 4) None of the above
12. Sodium is heated in air at 350 C to form X. X absorbs CO2 and forms sodium carbonate and Y. Which of the
o

following is Y?
1) H2 2) O2 3) H2O2 4) O3
13. “Lithia water” used for the treatment of gout is
1) LiHCO3 2) Li2CO3 3) Li2SO4 4) LiOH
14. The solubility of metal halides depends on their nature, lattice enthalpy and hydration enthalpy of the individual
ions. Amongst fluorides of alkali metals, the lowest solubility of LiF in water is due to :
1) Ionic nature of lithium fluoride 2) High lattice enthalpy
3) High hydration enthalpy for lithium ion 4) Low ionisation enthalpy of lithium atom
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15. In the case of alkali metals, the covalent character decreases in the order:
1) MF > MCl > MBr > MI 2) MF > MCl > MI > MBr
3) MI > MBr > MCl > MF 4) MCl > MI > MBr > MF
16. The product of electrolysis of an aqueous solution of K2SO4 using inert electrodes, at anode and cathode
respectively:
1) O2 and H2 2) O2 and K 3) O2 and SO2 4) O2 and SO3
17. A and B are two salts. A with dilute HCl and B with conc. H2SO4 react to give reddish brown vapours, hence
A and B respectively are:
1) NaBr, NaNO3 2) NaNO3 , NaBr 3) NaBr, NaNO 2 4) NaNO 2 , NaBr
18. On heating a mixture containing 1 mole each of Li2CO3 and K2CO3 ..... is/are formed:
1) 2 moles of CO2 2) 1 mole of CO2 3) 1.5 moles of CO2 4) no carbon dioxide
19. Soda lime is
1) Na 2 CO3  CaO 2) NaOH + NaHCO3 3) NaOH + CaO 4) NaH + Na2CO3
20. Microcosmic salt is
1) Na  NH 4  HPO 4 .4H 2 O 2) Na  NH 4  .H 2 O
3) Na  NH3  HPO4 .4H 2O 4) K  NH 4  HPO4 .2H 2O
21. Sodium thiosulphate, Na2S2O3.5H2O is used in photography to:
1) Reduce the AgBr grains to metallic Ag
2) Convert metallic Ag to Ag salt
3) Remove undecomposed AgBr as soluble silver thiosulphate complex
4) Removed reduced silver
22. Celestine is an ore of:
1) Ba 2) Ca 3) Sr 4) Mg
23. A chloride dissolves appreciably in cold water. When placed on a platinum wire in Bunsen flame, no distinctive
colour is noticed. Which one is cation?
1) Mg2+ 2) Ba2+ 3) Pb2+ 4) Ca2+
24. The correct order of ability to form complexes is:
1) Be 2  Mg 2  Ca 2  Ba 2  2) Be 2  Mg 2   Ca 2   Ba 2 
3) Be 2  Mg 2   Ca 2  Ba 2  4) Ba 2   Mg 2   Ca 2   Be 2 
25. For two ionic solids CaO and KI, identify the wrong statement among the following?
1) Lattice energy of CaO is much higher than that of KI
2) KI is soluble in benzene
3) CaO has high melting point
4) KI has high melting point
26. Solubility of alkaline earth metal hydroxides increases from Be(OH)2 to Ba(OH)2 because:
1) hydration energy > lattice energy 2) lattice energy > hydration energy
3) hydration energy is equal to lattice energy 4) none of the above
27. A substance which gives a brick red flame and breaks down on heating giving oxygen and a brown gas is:
1) calcium carbonate 2) magnesium carbonate 3) magnesium nitrate 4) calcium nitrate
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28. Metal carbonates decompose on heating to give metal oxide and carbon dioxide. Which of the metal carbonates
is most stable thermally?
1) MgCO3 2) CaCO3 3) SrCO3 4) BaCO3
29. Which of the following Group 2 metal hydroxides is soluble in sodium hydroxide?
1) Be  OH 2 2) Mg  OH 2 3) Ca  OH  2 4) Ba  OH  2
30. A chemical A is used for the preparation of washing soda to recover ammonia. When CO2 is bubbled through
an aqueous solution of A, the solution turns milky. It is used in white washing due to its disinfectant nature.
What is the chemical formula of A?
1) Ca  HCO3  2 2) CaO 3) Ca  OH  2 4) CaCO3
31. Choose the incorrect statement in the following:
1) BeO is almost insoluble but BeSO4 is soluble in water
2) BaO is soluble but BaSO4 is insoluble in water
3) LiI is more soluble than KI in ethanol 4) Both Li and Mg form solid hydrogen carbonates
32. Which category of salts of alkaline earth metals is not found in solid state, but found in solution state?
1) Carbonates 2) Bicarbonates 3) Hydroxides 4) Sulphates
33. Correct order of increasing thermal stabilities of alkaline earth metal sulphates is:
1) SrSO 4  CaSO 4  MgSO 4  BeSO 4 2) BeSO 4  MgSO 4  CaSO 4  SrSO 4
3) CaSO 4  BeSO 4  MgSO 4  SrSO 4 4) MgSO 4  BeSO 4  CaSO 4  SrSO 4
34. Which salt does not show hydrolysis?
1) Mg(NO3)2 2) Be(NO3)2 3) Ca(NO3)2 4) Ba(NO3)2
35. Bleaching powder loses its power on keeping for a long time because:
1) it changes into calcium hypochlorate 2) it changes into CaCl2 and Ca(OH)2
3) it absorbs moisture 4) it changes into calcium chloride and calcium chlorate
36. The main constituent of egg-shells is:
1
1) CaCO3 2) CaSiO3 3) CaSO 4 . H 2 O 4) CaSO 4 .2H 2O
2
37. Superphosphate of lime is a mixture of:
1) primary calcium phosphate and epsom 2) primary magnesium phosphate and epsom
3) primary magnesium phosphate and gypsum 4) primary calcium phosphate and gypsum
38. Gypsum is added to clinker during cement manufacture to:
1) decrease the rate of setting of cement 2) make the cement impervious
3) bind the particles of calcium silicate 4) to facilitate the formation of colloidal gel
39. Identify the correct statement:
1) The percentage of calcium is lower in gypsum in comparison to plaster of Paris.
2) Gypsum is not a natural product. It is obtained by heating of plaster of Paris.
3) Plaster of Paris is obtained by hydration of gypsum.
4) Plaster of Paris is formed by oxidation of gypsum.
40. Which of the following statements is false?
1) Ca2+ ions are not important in maintaining the regular beating of the heart
2) Mg2+ ions are important in the green parts of the plants
3) Mg2+ ions form a complex with ATP
4) Ca2+ ions are important in blood cloating.
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SOLUTIONS
LEVEL - I
1. 4
2. 2 Outer electronic configuration : ns1
3. 1
4. 1 Lithium salts impart bright red colour to the flame.
5. 4 Cs has the lowest melting point and melts at 30oC.
6. 4 Alkali metal hydrides are ionic in character.
7. 3
8. 2 NaH  H 2 O 
 NaOH  H 2

9. 4 Na  s   H 2 O  NaOH  H 2
10. 4 In presence of moisture and CO2 present in air, Na2O2 is converted into NaOH and Na2CO3 respectively
both are white
11. 2 LiCl being covalent is soluble in organic solvents such as pyridine.
1
12. 4 Na 2O 2  H 2 O  2NaOH  O 2
2
13. 3
14. 4 Sodium metal on oxidation gives Na2O2.
15. 1 2LiOH  Li 2 O  H 2 O

16. 1 The stability of the compound depends on f H . Evidently, more negative the value of f H , more is
the stability.
fH of KCl = –428 k J mol–1, fH of CsCl = –424 k J mol–1

fH of NaCl = –400 k J mol–1, fH of LiCl = –398 k J mol–1

17. 3 it is be sodium sulphate, Na2SO4.
18. 4 Gun powder is a mixture of 75% KNO3, 13% charcoal and 12% sulphur.
19. 1 Bicarbonates of alkali metal except LiHCO3 only are found in solid state.
20. 1 Li is harder than other alkali metals.
21. 3
22. 1 NaHCO3 is a sodium salt of weak acid, while NaOH is a strong base, they cannot exist together in
solution.
23. 4 Solvay process is based on electrolysis of brine (NaCl) solution.
24. 2 K is used as fertilizer (NPK) for nutrition of plants.
25. 2 Down the group, the solubility of M(OH)2 increases down the group.
26. 4 Alkali and alkaline earth metals are extracted by the electrolysis of their fused salt.

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27. 1 BeH2 can’t be prepared by direct action of H2 on Be. BeH2 is prepared by the action of LiAlH4 on
BeCl2.
28. 2
29. 2 Ca reacts with water to give H2.
30. 2 Ca  2H 2O  Ca  OH 2  H 2
CaH 2  2H 2O  Ca  OH 2  2H 2
31. 1 3Mg  N 2  Mg3 N 2 
6H 2O
 3Mg  OH  2  2NH3
32. 3
33. 3 B is a non metal and B(OH)3 is a weak Lewis acid.
34. 4 Among alkaline earth metal sulphates, lattice enthalpy remains almost constant but hydration enthalpy
decreases down the group. BeSO4 has higher hydration enthalpy as compared to lattice enthalpy.

35. 2 CaCO3   CaO  CO 2
(basic) (acidic)

36. 4 The increasing order of basic nature is

Al 2 O3  MgO  Na 2 O  K 2 O
37. 2 Bleaching powder is a mixture of calcium hypochlorite, Ca(OCl)2 and basic calcium chloride, CaCl2,
Ca(OH)2.H2O. The active ingredient in bleaching powder is Ca(OCl)2.
38. A Be (Z = 4) has maximum covalency of 4 while Al (Z = 13) has maximum covalency of 6.
39. 1
40. 4 Portland cement is a mixture of Ca silicates and aluminates.
LEVEL - II
1. 3 Borax is Na 2 B4 O7 .10H 2 O
2. 4 Due to small size, Be has highest 1st and 2nd ionization energies amongst alkaline earth metals.
3. 4 Alkali metals react with halogen to give halides.
4. 3 Down the group with decrease in positive charge density of M  ion, tendency to get hydrated decreases
and hence the size of hydrated ion decreases.
5. 4
6. 2
7. 3 The addition of calcium chloride lowers the melting point of sodium chloride to 850-875 K
8. 2 BeCl2 and MgCl2 do not impart colour to the flame.
9. 3 Aqueous solution of K2CO3 is alkaline.
10. 1 Pearl ash is K2CO3; caustic potash is KOH.
11. 1 For an ionic compound, if lattice energy is less than its hydration energy, it is water soluble.
12. 2 2Na 
 air   Na 2 O 2 ; 2Na 2 O 2  2CO 2  2Na 2 CO 3  O 2
O2

13. 1 Lithia water (LiHCO3) is used for treatment of gout.

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14. 2 The lattice energy of LiF is much higher due to their comparable size.
15. 3 As the size of halide ion decreases, I > Br > Cl > F , covalent character decreases.
16. 1
17. 4 NaNO2 gives NO2 (brown) with dil. acids, whereas NaBr and NaNO2 both give brown vapours of
Br2 and NO2 respectively with conc. acids.
18. 2 Li2CO3 decomposes while K2CO3 is stable and does not decompose.
19. 3 Soda lime is NaOH + CaO.
20. 1 Microcosmic salt is Na(NH4) HPO4.4H2O.
3
21. 3  Ag  S2 O3 2  is formed.

22. 3 Celestine is SrSO4.

23. 1 PbCl2 is insoluble in cold water. Mg2+ and Pb2+ do not show flame colour.
24. 3 Complexing tendency depends upon charge/size ratio.
25. 4 Due to lower lattice energy of KI as compared to that of CaO, the m.p. of KI is much lower than that
of CaO.
26. 1 For an ionic compound to be soluble in water, its hydration energy should be more than its lattice
energy.
27. 4 2Ca  NO3 2 
Heat
 2CaO  4NO 2  O2
28. 4 Thermal stability of metal carbonates increases from Be(OH)2 to Ba(OH)2. Thus, BaCO3 is the most
stable.
29. 1 Be(OH)2 being amphoteric dissolves in NaOH.
30. 3
31. 4 Both Li and Mg do not form solid hydrogen carbonates.
32. 2 Bicarbonates of alkaline earth metals exist only in solution.
33. 2 In the order of ionic character
34. 4 The decreasing tendency for hydration down the group is due to decreasing hydration enthalpy with
increasing size of the cation, from Be2+ to Ba2+.
35. 4 6Ca  OCl  2 
Stan ding for
long time
5CaCl2  Ca  ClO3 2
36. 1 Egg-shells are made up of CaCO3.
37. 4 It is a mixture of calcium dihydrogen phosphate and gypsum and is obtained by treating phosphatic
rock with conc. H2SO4.
38. 1
1
39. 1 The formulae of gypsum and plaster of paris are CaSO 4 .2H 2O and CaSO 4 . H 2 O respectively..
2
Hence, molecular mass of gypsum is higher than molecular mass of plaster of Paris. Therefore, the
percentage of calcium is lower in gypsum in comparison to plaster of Paris.
40. 1 Ca2+ maintains the regular beating of the heart.

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CHAPTER - 11
P-BLOCK ELEMENTS

INTRODUCTION
The p-block of the Periodic Table consists of six groups (groups 13 to 18) of 36 elements in which the last
electron enters the outermost p-orbital (except He). All non-metals and metalloids belong to this block. The
most electronegative elements also belong to p- block. Boron, carbon, nitrogen, oxygen, fluorine and helium
head the groups.
11.1 GENERAL CHARACTERISTICS OF P-BLOCK ELEMENTS
The valence shell electronic configuration of p-block elements is ns2 np1-6 (except He). The first members
differ from the remaining members of their respective groups in size and absence of d-orbitals. The difference
in the inner core of electrons greatly influences the properties of these elements.
11.1.1 Oxidation state. The maximum oxidation state of p-block elements is equal to the total number of valence
electrons. In boron, carbon and nitrogen families, the group oxidation state is most stable for lighter elements.
For heavier elements, an oxidation state two units less than the group oxidation state is more stable; this is
called inert pair effect.
Group 13 14 15 16 17 18
2 1 2 2 2 3 2 4 2 5
Electronic configuration ns np ns np ns np ns np ns np ns np6 (1s 2 for He)
2

First member of the group B C N O F He

Group oxidation state 3 4 5 6 7 8
Other oxidation states 1 2, 4 3, 3 4, 2, 2 5, 3, 1, 1 6, 4, 2
11.1.2 Metallic character. Metallic character increases down the group. In general, non-metals have higher
ionisation enthalpies and higher electronegativities than metals. Hence, in contrast to metals which readily form
cations, non-metals readily form anions.
The compounds of highly reactive non-metals with highly reactive metals are generally ionic because of large
differences in electronegativities. Compounds formed between non-metals are largely covalent because of
small differences in their electronegativities.
11.1.3 Nature of the oxides. Oxides of non-metals are acidic or neutral, whereas metal oxides are basic in nature.
The second period elements of p-block, with electronic configuration 2s2 2pn, starting with boron are restricted
to a maximum covalence of four (using 2s and three 2p orbitals).
The third period elements, with electronic configuration 3s2 3pn, have vacant 3d orbitals lying between the 3p
and 4s energy levels. Therefore, these elements can expand their covalence above four, e.g., while boron
forms only [BF4]–, aluminium forms [AlF6]3– ion.
The combined effect of size and availability of d-orbitals influences the ability of these elements to form
 -bonds.
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The first member of a group differs from the heavier members in its ability to form p - p multiple bonds to
itself ( e.g., C  C, C  C, N  N ) and to other second row elements (e.g., C  O, C  N, C  N, N  O ).
This type of  -bonding is not particularly strong in heavier p-block elements.
The heavier elements form  -bonds involving d-orbitals (d  – p  or d  –d  ). However, the coordination
number in species of heavier elements may be higher than for the first element in the same oxidation state. For
example, in +5 oxidation state both N and P form oxoanions, NO3– (three-coordination with  – bond
involving one nitrogen p-orbital) and PO43- (four-coordination involving s, p and d orbitals contributing
to the  – bond).
11.2 GROUP 13 ELEMENTS: THE BORON FAMILY
Group 13 comprise Boron (B), Aluminium (Al), Gallium (Ga), Indium (In), Thallium (Tl) and Nihonium (Nh).
Boron is a typical non-metal, aluminium is a metal but shows many chemical similarities to boron. Gallium,
indium and thallium are almost exclusively metallic in character.
Nihonium is a transactinide element which is extremely radioactive. Its most stable isotope, Nh286 has a half-
life of about 10 seconds.
Boron is a fairly rare element; its abundance in earth crust is less than 0.0001% by mass. There are two
isotopic forms of boron 10B (19%) and 11B (81%). It mainly occurs as orthoboric acid, (H3BO3), borax,
Na2B4O7·10H2O, kernite, Na2B4O7·4H2O and colemanite, Ca2B6O11.5H2O.
Aluminium is the most abundant metal and the third most abundant element in the earth’s crust (8.3% by mass)
after oxygen (45.5%) and Si (27.7%). The important minerals are bauxite, Al2O3. 2H2O, cryolite, Na3AlF6,
corundum (anhydrous alumina), Al2O3 and beryl, 3BeO.Al2O3.6SiO2 or Be3Al2Si6O18.
Gallium, indium and thallium are less abundant. Gallium (0.1-1.0%) is found in germanite, a complex sulphide
of Zn, Cu, Ge and As.
11.2.1 Electronic Configuration. The general outer electronic configuration of these elements is ns2 np1. While
boron and aluminium have noble gas core, gallium and indium have noble gas plus 10 d-electrons, and thallium
has noble gas plus 14 f- electrons plus 10 d-electrons cores. Thus, the electronic structures of these elements
are more complex than for the first two groups of elements. This difference in electronic structures affects the
properties of these elements.
Element 5B 13 Al 31 Ga 49 In 81 Tl
Electronic
2 1 2 1 10 2 1 10 2 1 14 10 2 1
configuration [He] 2s 2p [Ne] 3s 3 p [Ar] 3d 4s 4p [Kr] 4d 5s 5p [Xn] 4f 5d 6s 6p
11.2.2 Physical Properties
1. Atomic and ionic radii. In the group, atomic radius increases. However, atomic radius of gallium (135
pm) is less than that of aluminium (143 pm). This is due to the presence of ten 3d-electrons which offer
only poor shielding for the outer electrons from the increased nuclear charge in gallium. The atomic radii
increase in the order : B < Ga < Al < In < Tl.
The ionic radii, however, follow the regular trend, i.e., increases from B to Tl.
2. Ionization Enthalpy. The first ionisation enthalpies of group 13 elements are lower than those of group 2
because in group 13 elements, electron is removed from a p-orbital while in group 2 elements, electron is
removed from a filled s-orbital which has lower energy.
The ionisation enthalpies do not decrease regularly down the group. The first ionisation enthalpies are in
the order : B > Tl > Ga > Al > In. The decrease from B to Al is associated with increase in size. The
discontinuity between Al and Ga, and between In and Tl are due to low screening effect of d- and f-
electrons.
The order of ionisation enthalpies is  i H1   i H 2   i H 3 . The sum of the first three ionisation enthalpies
for each element is very high.
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3. Electronegativity. Group 13 elements are more electronegative than alkali and alkaline earth elements.
Down the group, electronegativity decreases from B to Al due to increase in atomic size and then increases
slightly because effective nuclear charge increases due to poor shielding by inner d- and f-electrons.
4. Electropositive or metallic character. Boron group elements are less electropositive or metallic than
alkali and alkaline earth elements.
Down the group, electropositive character increases from B to Al and then decreases from Al to Tl.
Boron with the highest sum of first ionisation enthalpies   i H1   i H 2  i H3  is the least electropositive.
Due to increase in atomic size, the value is low for aluminium, therefore, it is highly electropositive.
Table - 01 Atomic and physical properties of group 13 elements
Property B Al Ga In Tl
Atomic number 5 13 31 49 81
-1
Atomic mass (g mol ) 10.81 26.98 69.72 114.82 204.38
Atomic radius (pm) 88 143 135 167 170
3+
Ionic radius (M /pm) 27 53.5 62 80 88.5
+
Ionic radius (M /pm) 120 140 150
-1
Ionisation enthalpy, i H1 (kJ mol ) 801 577 579 558 589
-1
i H2 (kJ mol ) 2427 1816 1979 1820 1971
-1
i H3 (kJ mol ) 2659 2744 2962 2704 2877
Electronegativity (Pauling scale) 2 1.5 1.6 1.7 1.8
-3
Density (g cm ) at 293 K 2.35 2.7 5.9 7.31 11.85
Melting point (K) 2453 933 303 430 576
Boiling point (K) 3923 2740 2676 2353 1730
3+ -
E° V at 298 K for M (aq)+3e M(s) -0.87 -1.66 -0.56 -0.34 1.26
+ -
E° V at 298 K for M (aq)+e M(s) 0.55 -1.39 (alkali) -0.18 -0.34
0.79 (acid)
5. Colour, appearance, and hardness. Boron is a non-metallic extremely hard black solid which exists in
many allotropic forms.
6. Melting and boiling points. The melting points decrease sharply from B to Ga and then increases from
Ga to Tl. Due to very strong crystalline lattice, boron has unusually high melting point (3923 K).
Other members are soft metals with low melting points and high electrical conductivity. Gallium (m. p.
303K) exist in the liquid state during summer. Its high boiling point (2676K) makes it useful for measuring
high temperatures.
7. Density. Density increases down the group due to smaller atomic and ionic radii.
11.2.3 Chemical Properties
Oxidation state and trends in chemical reactivity. Down the group, increased effective nuclear charge
holds the ns electrons tightly, restricting their participation in bonding. As a result, only the p-orbital electron
involves in bonding. The reluctance of ns2 electrons to participate in bonding is called intert pair effect.
In Ga, In and Tl, both +1 and +3 oxidation states are observed. The relative stability of +1 oxidation state
increases from Al to Tl, i.e., Al < Ga < In < Tl. In thallium, +1 oxidation state is predominant whereas +3
oxidation state is highly oxidising in character. The compounds in +1 oxidation state are more ionic than those
in +3 oxidation state.

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In the trivalent state, the number of electrons around the central atom in a molecule of a compound of these
elements will be only six. Such electron deficient molecules accept a pair of electrons to achieve stable
electronic configuration and thus, behave as Lewis acids. The tendency to behave as Lewis acid decreases
with increase in size. Thus, BCl3 easily accepts a lone pair of electrons from ammonia to form BCl3·NH3.
AlCl3 achieves stability by forming a dimer. Both boron and aluminium halides are Lewis acids, but only
aluminium halides form dimers because boron atom is so small that it cannot accomodate four halogen atoms
around it.

In the trivalent state, most of the compounds being covalent are hydrolysed by water. The trichlorides on
hyrolysis form tetrahedral [M(OH)4]- species in which element M is in sp3 hybridised state.
Aluminium chloride in acidified aqueous solution forms octahedral [Al (H2O]6]3+ ion. In this complex ion, the
3d orbitals of Al are involved and the hybridisation is sp3d2.
1. Reactivity towards air. Crystalline form of boron is unreactive. Aluminium forms a very thin oxide layer
on its surface which protects the metal from further attack. Amorphous boron and aluminium metal on heating
in air form B2O3 and Al2O3 respectively. With dinitrogen at high temperature they form nitrides.
 
2B(s)  N 2 (g)   2BN(s) ; 2Al(s)  N 2 (g)   2AlN(s)
Boron nitride Alu min ium nitride


4B(s)  3O 2 (g) 
973 K
 2B2 O3 (s) ; 4Al(s)  3O2 (g)   2Al2O3 (s)
Boron trioxide Alu min ium oxide

Boron nitride is a white slippery solid having a layered structure similar to that of graphite. Therefore, it is
called inorganic graphite.
2. Acid-base character of oxides. The trioxides of group 13 reacts with water to form hydroxides.
M 2 O 3  3H 2 O 
 2M(OH)3
The basic nature of oxides increases down the group. Boron trioxide is acidic. It reacts with basic
oxides forming metal borates. Aluminium and gallium oxides are amphoteric and indium and thallium oxides
are basic.
673823K
B2 O3  2NaOH   2NaBO2  H 2O
Sodium metaborate

Being amphoteric, alumina reacts with both acids and bases forming salts.
Al 2 O3  2NaOH 
Fuse
 2NaAlO 2  H 2 O
3. Reactivity towards acids and alkalies. Boron does not react with acids or alkalies even at moderate
temperature. At high temperature it reacts with oxidising acids (e.g., mixture of conc. H2SO4 and conc.
HNO3) to form boric acid.
B(s) + 3HNO 3 (aq) 
conc. H 2SO 4
 H 3 BO 3 (aq) +3NO 2 (g)
Aluminium dissolves in mineral acids and aqueous alkalies and thus exhibits amphoteric character.
 2Al3+ (aq) + 6Cl  +3H 2 (g)
2Al(s) + 6HCl (aq) 

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Concentrated nitric acid renders aluminium passive by forming a protective oxide layer on the surface.
2Al(s) + 6HNO3 (conc.) 
 Al 2 O3 (aq) + 6NO 2 +3H 2 O(g)
Aluminium also reacts with aqueous alkali to form sodium tetrahydroxoaluminate(III) and liberates dihydrogen.

 2Na + [Al(OH)4 ] +3H 2 (g)

2Al(s) + 2NaOH(aq) + 6H 2O(l ) 
Boron reacts with alkalies above 773K.

2B(s) + 6KOH(s) 
 2K 3BO3 +3H 2 (g)
Tl does not react with alkalies.
4. Reactivity towards halogens. Group 3 elements react with halogens to form trihalides.

2M(s) + 3X 2 (g)   2MX 3 (X = F, Cl, Br, I)
11.2.4 Important trends and anomalous properties of boron
The trihalides of group 3 elements, being covalent, undergo hydrolysis. Except for boron, tetrahedral [M(OH)4]–
and octahedral [M(H2O)6]3+ species exist in aqueous medium.
The monomeric trihalides, being electron deficient, are strong Lewis acids.
Boron trifluoride reacts with Lewis bases such as NH3 to complete octet around boron.
F3B + :NH3 (aq) 
 F3B  NH3
The maximum covalence of boron is 4 because of the absence of d-orbitals. Since d-orbitals are available
with Al and other elements, their maximum covalence can be more than 4. Most of the other metal halides
(e.g., AlCl3) are dimerised through halogen bridging (e.g., Al2Cl6). The metal species completes its octet by
accepting electrons from halogen.
11.3 SOME IMPORTANT COMPOUNDS OF BORON
11.3.1 Borax. Borax or sodim tetraborate decahydrate is a white crystalline solid of formula Na2B4O7·10H2O. It
contains tetranuclear units [B4O5(OH)4]2-. The correct formula is Na2[B4O5 (OH)4].8H2O.
Preparation. (i) From boric acid. Boric acid is neutralised with sodium carbonate.

4H 3 BO3 + Na 2 CO3   Na 2 B4 O 7  6H 2 O  CO 2
Boric acid Borax

Properties. It is a white crystalline solid. Borax dissolves in water to give an alkaline solution due to hydrolysis.
Na 2 B4O7 + 7H 2 O 
 2NaOH  4H3BO3
Orthoboric acid

Action of heat. On heating, borax first loses water molecules and swells up. On further heating it turns into a
transparent liquid, which solidifies into glass like material known as borax bead.

Na 2 B4O7 .10H 2 O 
Heat
10H 2O
Na 2 B4O7   2NaBO 2  B2 O3
Borax
  
Sodium metaborate Boric anhydride

Transparent glassy bead

Borax bead test. The metaborates of certain transition metals (Ni2+, Co2+, Cr3+,Cu2+, Mn2+, etc) have
characteristic colours, therefore, borax bead test is used to identify them in the laboratory. E.g., when borax
is heated in a Bunsen burner flame with CoO on a loop of platinum wire, a blue coloured Co(BO2)2 bead is
formed.

CoSO 4   CoO  SO3 ; 
CoO +B2O3   Co(BO 2 ) 2
Cobalt sulphate Cobalt oxide Cobalt oxide Cobalt metaborate

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Uses. (i) In the manufacture of heat resistant borosilicate (pyrex) glass. (ii) In the preparation of medicinal
soaps, due to its antiseptic properties. (iii) As a flux in soldering. (iv) In the manufacture of enamels and glazes
for earthen wares (tiles, pottery, etc.) (v) In the laboratory for borax bead test. (vi) Softening of water.
11.3.2 Orthoboric acid (Boric acid) H3BO3 or B(OH)3.
Orthoboric acid, H3BO3 is a white crystalline solid with soapy touch. It is sparingly soluble in water but highly
soluble in hot water.
Preparation. It is prepared by acidifying an aqueous solution of borax with HCl or H2SO4.
Na 2 B 4 O 7 + 2HCl + 5H 2 O 
 4H 3 BO 3  2NaCl

It is also formed by the hydrolysis of most boron compounds (halides, hydrides, etc.) with water or acids.
BC l 3 + 3H 2 O 
 H 3 BO 3  3H C l ; B 2 H 6 + 6H 2 O 
 2H 3 BO 3  6H 2
Properties. It has a layer structure in which planar BO3 units are joined by hydrogen bonds.
Boric acid is a weak monobasic acid. It is not a protonic acid but a Lewis acid accepting electrons from a
hydroxyl ion:
[B(OH) 4 ]  H 3O 
B(OH)3 +2H 2 O 
On heating, orthoboric acid above 370K gives metaboric acid, HBO2 which on further heating yields boric
oxide, B2O3.
H3BO3 
370K
 HBO2  H 2 O ; 4HBO2  410K
 H 2O
 H 2 B4 O7  red heat
2B2 O3  H 2O
Metaboric acid Metaboric acid Tetraboric acid Boron trioxide

Reaction with ethyl alcohol. Orthoboric acid reacts with ethyl alcohol in presence of conc. H2SO4 to form
triethyl borate. These vapours burn with green edged flame.
H 3 BO3 + 3C2 H 5OH 
conc.H 2SO 4
B(OC2 H5 )3  3H 2O
Orthoboric acid Triethyl borate

This reaction is used as a test for borates and boric acid.

Structure. Boron in borate ion (BO33-) is sp2 hybridised. In boric acid, planar BO33- units are joined by
hydrogen bonds to give a layered structure. The adjacent layers (318 pm apart) are held by weak forces of
attraction. The layers can slide over one another, and therefore, boric acid is soft and soapy to touch.
Uses. (i) In the manufacure of heat resistant borosilicate (pyrex) glass. (ii) As preservative. (iii) In the manufacture
of enamels and glazes for earthen wares (tiles, pottery, etc). (iv) Aqueous solution is used as mild antiseptic for
eye ailments (boric lotion).

(a) (b)

Figure - 01 (a) Boric acid (dotted lines represent H- bonds), (b) Borazine (Inorganic Benzene).

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11.3.3 Boron Hydrides (Boranes)

Hydrides of boron are collectively called boranes in analogy with alkanes.
1. Diborane, B2H6. It is the simplest boron hydride.
Preparation. Diborane is prepared by treating boron trifluoride with LiAlH4 in diethyl ether.
4BF3 + 3LiAlH 4 
 2B 2 H 6  3LiF  3AlF3
Laboratory preparation: By the oxidation of sodium borohydride with iodine.
2NaBH 4 + I2 
diglyme
 B2 H6  2NaI  H 2
Industrial preparation: Diborane is produced by the reaction of BF3 with sodium hydride.
2BF3 + 6NaH 
450 K
 B 2 H 6  6NaF
Properties. Diborane is a colourless, highly toxic gas, b.p.180 K. On exposure to air, it catches fire
spontaneously. It burns in oxygen releasing enormous amount of energy.
 B2 O3  3H 2 O ;  c H   1976 kJ mol 1
B2 H 6 + 3O 2 
Most of the higher boranes also are spontaneously flammable in air.
Hydrolysis. Boranes are readily hydrolysed by water to give boric acid.
B2 H 6 (g) + 6H 2 O(l ) 
 2B(OH)3 (aq)  6H 2 (g)
Cleavage reactions. Diborane undergoes cleavage reactions with Lewis bases to give borane adducts.
B2 H 6 + 2NMe3 
 2BH 3 .NMe3 ; B2 H 6 + 2CO 
 2BH 3 .CO
Reaction with ammonia. With ammonia, diborane gives initially B2H6.2NH3, i.e., [BH2(NH3)2]+ [BH4]–.
Further heating gives borazine (borazole), B3N3H6, known as inorganic benzene because of its ring structure
with alternate BH and NH groups.
3B2 H 6 + 6NH 3 
low temp.
 3[BH 2 (NH3 ) 2 ] [BH 4 ] 
473 K
 2B3 N 3 H 6  12H 2
Formation of complex borohydrides. Lithium and sodium tetrahydridoborates, also known as borohydrides,
are prepared by the reaction of metal hydrides with B2H6 in diethyl ether.
2LiH  B2 H 6 
Diethyl ether
2Li  [BH 4 ] ; 2NaH  B2 H 6 
Diethyl ether
2Na  [BH 4 ]
Lithium borohydride Sodium borohydride

LiBH4 and NaBH4 are used as reducing agents in organic synthesis. They are used as starting materials for
preparing other metal borohydrides.
Uses.i) For preparing borohydrides such as LiBH4, NaBH4, etc.
ii) As a reducing agent in organic reactions.
iii) As a fuel for rockets.
Structure of diborane. Boron atoms in diborane are sp3 hybridised. Out of the four sp3 hybrid orbitals on
each B atom, one is without an electron (shown in broken lines). The terminal B-H bonds are normal 2-
centre-2electron bonds, but the two bridge bonds (B-H-B) are 3-centre-2-electron bonds referred to as
banana bonds.The four terminal hydrogen atoms and the two boron atoms lie in one plane. Above and
below this plane, there are two bridging hydrogen atoms.

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(a) (b)

Figure - 02 (a) The structure of diborane, B2H6 (b) Bonding in diborane.

11.4 Uses of boron and aluminium and their compounds
11.4.1 Uses of Boron. Boron being an extremely hard refractory solid of high melting point, low density and very
low electrical conductivity, finds many applications.
Boron fibres are used in making bullet-proof vests and light composite material for aircraft.
The boron-10 isotope has high ability to absorb neutrons and, therefore, metal borides are used in nuclear
industry as protective shields and control rods.
Borax and boric acid are used in the manufacture of heat resistant glasses (e.g., Pyrex), glass-wool and
fibreglass.
Borax is also used as a flux for soldering metals, for heat, scratch and stain resistant glazed coating to
earthenwares.
Borax is used as constituent of medicinal soaps.
An aqueous solution of orthoboric acid is generally used as a mild antiseptic.
11.4.2 Uses of aluminium. Being a bright silvery-white metal, with high tensile strength, and high electrical and
thermal conductivity. On a weight-to-weight basis, the electrical conductivity of aluminium is twice that of
copper.
Aluminium is used extensively in industry and every day life. It forms alloys with Cu, Mn, Mg, Si and Zn.
Aluminium and its alloys can be shaped into pipes, tubes, rods, wires, plates or foils and, therefore, find uses
in packing, utensil making, construction, aeroplane and transportation industry.
Al powder is used in flash light bulbs for indoor photography.
Al powder is used in aluminothermic process for the extraction of metals like chromium and manganese.
The use of aluminium and its compounds for domestic purposes is now reduced considerably because
of their toxic nature.
11.4.3 Alloys of aluminium
(i) Aluminium bronze (Al 95%, Cu 5%). Light, strong alloy with golden lustre, resistant to corrosion,
Used for making coins, utensils, jewellery, picture frames etc.
(ii) Magnalium (Al 95%, Mg 5%). Light, tough and strong. Used for making light instruments, balance
beams, pressure cookers, etc.
(iii) Duralumin (Al 95%, Cu 4%, Mg 0.5%, Mn 0.5%). Light, tough, ductile, and resistant to corrosion.
Used for making aeroplanes, automobile parts, pressure cookers, etc.
11.4.4 Compounds of aluminium
(i) Al(OH)3 is used as antacid
(ii) Anhydrous AlCl3 is used as catalyst in Friedel-crafts reactions and in cracking of petroleum.
(iii) Hydrated AlCl3 is used as a mordant in dyeing.
(iv) Potash alum, K2SO4.Al2(SO4)3.24H2O is used for purification of water, as styptic for stopping bleeding,
in foam type fire-extinguishers, as mordant in dyeing, for tanning of leather, in calico printing and sizing
of paper.

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11.5 GROUP 14 ELEMENTS: THE CARBON FAMILY

Group 14 of the periodic table comprises Carbon (C), silicon (Si), germanium (Ge), tin (Sn), lead (Pb), and
Flerovium (Fl). Carbon is the seventeenth most abundant element by mass in the earth’s crust. In elemental
state, it is found as coal, graphite and diamond. In the combined state it is present as carbon dioxide gas
(0.03%) in air, metal carbonates, hydrocarbons, etc.
Carbon is the most versatile of all elements. It combines with other elements such as dihydrogen, dioxygen,
chlorine and sulphur to form numerous materials ranging from living tissues to drugs and plastics.
It is an essential constituent of all living organisms. Organic chemistry is the chemistry of carbon compounds.
Carbon has two stable isotopes; 12C and 13C. 14C is a radioactive isotope with half-life 5770 years and used
in radiocarbon dating.
Silicon is the second most abundant element on the earth’s crust (27.7 % by mass) and is present in the form
of silica and silicates. Silicon is a very important component of ceramics, glass and cement.
Germanium exists only in traces. Germanium is transparent to infrared light and is, therefore, used for making
prisms, lenses and windows of infrared spectrometers. Ultrapure germanium and silicon are used in transistors
and semiconductor devices.
Tin occurs mainly as cassiterite, SnO2 and lead as galena, PbS. Other ores of lead are anglesite, PbSO4 and
cerussite, PbCO3.
Flerovium is a synthetic super heavy transactinide element which is extremely radioactive. Its longest-lived
isotope, Fl289 has half-life of 19 seconds.
11.5.1 Electronic Configuration
The valence shell electronic configuration of these elements is ns2np2.

Element 6C 14 Si 32 Ge 50 Sn 82 Pb
2 2 2 2 10 2 2 10 2 2 14 10 2 2
Electronic configuration [He] 2s 2p [Ne] 3s 3 p [Ar] 3d 4s 4p [Kr] 4d 5s 5p [Xn] 4f 5d 6s 6p

11.5.2 Atomic and physical properties

1. Covalent Radius. There is considerable increase in covalent radius from C to Si but only a small increase
from Si to Pb. This is due to the presence of completely filled d and f orbitals in heavier members.
2. Ionisation Enthalpy. The first ionisation enthalpy of group 14 elements is higher than those of group 13.
In general, ionisation enthalpy decreases down the group. Decrease in iH from Si to Sn and slight
increase in iH from Sn to Pb is due to poor shielding effect of intervening d and f orbitals and increase in
size of the atom.
3. Electronegativity. Due to small size, elements of this group are slightly more electronegative than group
13 elements. The electronegativity values for Si to Pb are almost the same.
4. Melting and boiling points. All group 14 slements are solids. Carbon and silicon are non-metals,
germanium is a metalloid, and tin and lead are soft metals with low melting points. Melting points and
boiling points are much higher than those of corresponding elements of group 13.

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Table - 02 Atomic and physical properties of group 13 elements.

Property C Si Ge Sn Pb
Atomic number 6 14 32 50 82
-1
Atomic mass (g mol ) 12.01 28.09 72.60 118.71 207.2
Covalent radius (pm) 77 118 122 140 146
4+
Ionic radius (M /pm) 40 53 69 78
2+
Ionic radius (M /pm) 73 118 119
-1
Ionisation enthalpy, i H1 (kJ mol ) 1086 786 761 708 715
-1
i H2 (kJ mol ) 2352 1577 1537 1411 1450
-1
i H3 (kJ mol ) 4620 3228 3300 2942 3081
-1
i H4 (kJ mol ) 6220 4354 4409 3929 4082
Electronegativity (Pauling scale) 2.5 1.8 1.8 1.8 1.9
-3
Density (g cm ) at 293 K 3.51*2.22** 2.34 5.32 7.26 11.34
Melting point (K) 4373 1693 1218 505 600
Boiling point (K) 3550 3123 2896 2024
-1 14 16 -5 -5
Electrical resistivity (ohm cm ) 298K 10 -10 50 50 10 210
* diamond, ** graphite
11.5.3 Chemical Properties.
a. Oxidation states and trends in chemical reactivity. The common oxidation states Group 14 elements
are +4 and +2. Carbon also exhibits negative oxidation states. Since the sum of the first four ionization
enthalpies is very high, compounds with +4 oxidation state are generally covalent. The tendency to show +2
oxidation state increases from Ge to Sn (inert-pair effect).
Carbon and silicon mostly show +4 oxidation state. Germanium forms stable compounds in +4 state and
some compounds in +2 state. Tin forms compounds in both oxidation states (Sn in +2 state is a reducing
agent). Lead compounds in +2 state are stable and in +4 state are strong oxidising agents.
In tetravalent state, the number of electrons around the central atom in a molecule (e.g., carbon in CCl4) is
eight. Being electron-precise molecules, they are not electron acceptors or electron donors.
Trends in oxidising and reducing character. Since Ge, Sn and Pb show +2 and +4 oxidation states, these
elements in +2 oxidation state are good reducing agents while in +4 oxidation state are good oxidising agents.
Thus, SnCl2 is widely used as a reducing agent.
2FeCl3 + SnCl2 
 2FeCl 2  SnCl 4
b Tendency to form p-p bonds. Due to small size and high electronegativity, carbon forms p-p
multiple bonds either with itself (C=C, CC) and with other atoms of similar size (C=O, C=N, CN). On
descending the group, this tendency decreases due to increase in size and decrease in electronegativity.
c. Tendency to form d-p bonds. Due to the absence of d-orbitals, carbon do not form d-p bonds.
d. Maximum covalency and tendency to form complexes. Carbon cannot exceed its covalence beyond
4, but other elements of the group can do so because of the presence of d-orbitals. Therefore, their halides
undergo hydrolysis and have a tendency to form complexes by accepting electron pairs from donor species.
In species like, [SiF5]-, [SiF6]2-, [GeCl6]2-, [Sn(OH)6]2-, [Pb(OH)6]2-, [PbCl6]2-, etc., the hybridisation of the
central atom is sp3d2.

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1. Reactivity towards oxygen. Group 14 elements form two types of oxides; monoxide (MO) and dioxide
(MO2). Oxides in higher oxidation states are generally more acidic than those in lower oxidation states. Thus,
CO2, SiO2 and GeO2 are acidic, while SnO2 and PbO2 are amphoteric. Among monoxides, CO is neutral,
GeO is acidic and SnO and PbO are amphoteric. Besides PbO and PbO2, lead forms trilead tetroxide,
Pb3O4. It is obtained by heating PbO (litharge) with excess of air or oxygen.
6PbO  O 2 
673K
 2Pb3O 4
2. Reactivity towards water. Carbon, silicon and germanium do not react with water. Tin decomposes
steam to form the dioxide and dihydrogen gas.
Sn  2H 2 O  
 SnO 2  2H 2 O
Lead is unaffected by water because of protective oxide film formation.
3. Reactivity towards halogens. Group 14 elements except carbon react directly with halogens to form
halides of formula MX2 and MX4 (X = F, Cl, Br, I). Most MX4 are covalent. The central metal atoms in these
halides are sp3 hybridised and the molecules are tetrahedral in shape (except SnF4 and PbF4, which are ionic).
PbI4 does not exist because the lead- iodine bond does not release enough energy to unpair 6s2 electrons and
excite one of them to a higher orbital to have four unpaired electrons around lead atom.
Ge, Sn and Pb form halides of formula MX2. Stability of dihalides increases down the group. Thermal and
chemical stability of GeX4 is more than GeX2, whereas PbX2 is more stable than PbX4.
Tetrachlorides except CCl4 are easily hydrolysed by water because the central atom can accommodate the
lone pair of electrons from oxygen atom of water molecule in d-orbital. Thus, SiCl4 undergoes hydrolysis by
accepting a lone-pair of electrons from water in the d-orbitals of Si and forming silicic acid, Si(OH)4.

11.5.4 Important trends and anomalous behaviour of carbon

Like the first members of other groups, carbon also differs from the rest of its group due to smaller size, higher
electronegativity, higher ionisation enthalpy and unavailability of d-orbitals.
Carbon has only s- and p-orbitals available for bonding, therefore, its covalency is limited to four. Other
members of the group can expand their covalence due to the presence of d-orbitals.
Carbon has the unique ability to form p – p multiple bonds with itself and with other atoms of small size and
high electronegativity, e.g., C=C, C  C, C = O, C = S, and C = N. Heavier elements do not form p – p
bonds because their atomic orbitals are too large and diffuse to have effective overlap.
Carbon atoms link with one another through covalent bonds forming chains and rings; a property called
catenation. This is because C–C bonds are very strong. Down the group, atomic size increases and
electronegativity decreases, therefore, tendency for catenation decreases. This is evident from the bond
enthalpies.
-1 -1
Bond Bond enthalpy (kJ mol ) Bond Bond enthalpy (kJ mol )
C-C 348 Ge-Ge 260
Si-Si 297 Sn-Sn 240
The order of tendency for catenation is C > > Si > Ge  Sn. Lead does not exhibit catenation. Due to
catenation and p – p bond formation, carbon is able to show allotropic forms.

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11.5.5 Allotropes of carbon

Carbon exists in many allotropic forms; both crystalline and amorphous. Diamond and graphite are two well-
known crystalline forms of carbon. A third allotrope of carbon called fullerene was discovered by H.W.Kroto,
E.Smalley and R.F.Curl in 1985 (Nobel Prize in 1996). Another allotropic form is carbon nanotubes.
Due to different structures, they have different properties.
1. Diamond. It has a crystalline lattice. Each carbon atom is sp3 hybridised and linked to four other carbon
atoms in tetrahedral fashion through covalent bonds (C–C bond length =154 pm). The structure
produces a rigid three-dimensional network of carbon atoms. It is very difficult to break extended covalent
bonding, therefore, diamond is the hardest substance on earth.
It has high density (3.51 g cm-3). It does not melt at 1atm. It’s m.p. at 10 GPa is 5000 K.
Since all valence electrons are held firmly in C–C,  bonds, there are no free electrons, therefore, diamond
is a bad conductor of electricity. However it has very high thermal conductivity.
Because of high refractive index (2.5), diamond can reflect and refract light. Hence, it is transparent.
2. Graphite. It occurs in nature and can be manufactured by heating coke at 3273-3300 K in an electric
furnace. Graphite has layered structure. Carbon is sp2 hybridised and forms three sigma bonds with three
neighbouring carbon atoms forming hexagonal rings. CC bond length is 141.5 pm. Thus graphite has two-
dimensional sheet like (layered) structure consisting of a number of benzene-like rings. The layers are held
together by van der Waals forces and the distance between two layers is 340 pm.
The  electrons are delocalised over the whole sheet. Electrons are mobile and, therefore, graphite conducts
electricity along the sheet. Its thermal conductivity is lower than that of diamond.
The layers of graphite can slip over one another, therefore, it is very soft and slippery. Thus, it is used as a dry
lubricant in machines running at high temperature, where oil cannot be used.
Unlike diamond, graphite is black and has metallic lustre. Its sublimation temperature is between 3895 and
4020 K at atmospheric pressure.

(a) (b) (c) (d)

Figure - 03 The structure of (a) diamond (b) graphite (c) Buckminsterfullerene(C60) (d) carbon nanotube.
3. Fullerenes. Fullerenes are prepared by heating graphite in an electric arc in an atmosphere of inert gases
such as helium or argon. The soot consists mainly of C60 with small quantity of C70 and traces of fullerenes of
even number of carbon atoms up to 350 or above.
 Etectric arc
Graphite   Carbon vapour 
Condensation
 Fullerene soot(C60  little C70 )
Unlike graphite or diamond, fullerenes are soluble in organic solvents. A solution of C60 fullerene in toluene is
purple, whereas that of C70 fullerene is orange red. Fullerenes are the only forms of pure carbon because they
have smooth structure without ‘dangling’ bonds which attract other atoms.
Fullerenes are cage like molecules. C60 molecule has a shape like a soccer ball (football) and is called
Buckminster fullerene. Spherical fullerenes are also called bucky balls.
C60 fullerene contains 20 six- membered rings and 12 five-membered rings. A six membered ring is fused with
six or five membered rings but a five membered ring can only fuse with six membered rings.
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All the carbon atoms are sp2 hybridised. Each carbon atom forms three sigma bonds with neighbouring
carbon atoms. The remaining electron (in the unhybridised p-orbital) on each carbon is delocalised in molecular
orbitals to give aromatic character to the molecule. This ball shaped molecule has 60 vertices, occupied by
one carbon atom each. It also contains both single and double bonds with C–C distances of 143.5 pm and
138.3 pm respectively.
Graphite is the most thermodynamically stable allotrope of carbon and, therefore, f  of graphite is taken as
zero. f  values of diamond and fullerene, C60 are 1.90 and 38.1 kJ mol–1, respectively.
4. Carbon nanotubes. Carbon nanotubes (CNTs) are allotropes of carbon with a cylindrical nanostructure.
It consists of a two dimensional array of hexagonal rings of carbon just as in a layer of graphite (graphene
sheet) rolled into a cylinder and capped on both ends with C60 hemispheres.
5. Other forms of elemental carbon. Other forms of elemental carbon like carbon black, coke, and charcoal
are all impure forms of graphite or fullerenes. Amorphous carbons also contain fine crystals, therefore, these
allotropes are called microcrystalline carbons.
Carbon black or lamp black is obtained by burning hydrocarbons in a limited supply of air. Charcoal and
coke are obtained by heating wood or coal respectively at high temperatures in the absence of air. Animal
charcoal (bone black) is obtained by the destructive distillation of bones (10% carbon and the rest is
calcium phosphate). Sugar charcoal is the purest form of amorphous carbon and is obtained by the action of
conc. H2SO4 on cane sugar.

CH 4  O 2   C  2H 2 O ; C12 H 22 O11 C o nc.H 2SO 4
12C  11H 2 O
Activated charcoal is charcoal whose adsorption capacity is enhanced by heating at 1273 K in a current of
steam. This process removes impurities like residual hydrocarbons, oxygen, etc., from the capillary pores.
11.5.6 Chemical properties
Charcoal. Among the allotropes of carbon, charcoal is the most reactive. It readily burns in air to give CO2.
It reduces conc.HNO3 to NO2 and conc. H2SO4 to SO2. It combines with metals to form carbides, e.g.
CaC2, Al4C3, etc. It forms CS2 when heated with sulphur.
Diamond. It is stable in vacuum up to 1773K. It burns in air at 1173 K to form CO2. Diamond changes to
graphite at 2073 K and more rapidly at 2273 K. It is resistant to most chemical reagents, but slowly oxidised
by a mixture of (K2Cr2O7 and conc.H2SO4,.i.e., chromic acid, H2CrO4) at 473 K to give CO2.
Graphite. It is not attacked by dilute acids. It is slowly oxidised by chromic acid to CO2, by conc. HNO3 to
graphitic acid, C11H4O5 and by alkaline KMnO4 to mellitic acid (benzene hexacarboxylic acid), C6 (COOH)6.
11.5.7 Uses of Carbon.
Diamond. i) As abrasive for sharpening hard tools and in making dies. ii) For cutting glass and making borers
for rock drilling. iii) Diamond is a precious stone. Cut and polished diamond is used in jewellery. Tt is measured
in carats (1 carat = 200 mg). iv) For making dies for drawing thin wires from metals. v) For grinding and
polishing hard materials.
Graphite. i) As a reducing agent in steel manufacture. ii) For making electrodes of dry cells. iii) For making
cores of lead pencils (called black lead). iv) As a moderator in atomic reactors. v) As a solid lubricant for
heavy machinery at high temperatures. vi) Crucibles made of graphite are inert to dilute acids and alkalies. vii)
Graphite fibres embedded in plastic form high strength, lightweight composites which are used in products
such as tennis rackets, fishing rods, aircrafts and canoes.
Activated charcoal. Being highly porous, is used in adsorbing poisonous gases. It is also used in water filters
to remove organic contaminants and in airconditioning systems to control odour.
Carbon black. It is used as black pigment in black ink and as filler in automobile tyres.
Coke. It is used as a fuel and largely as a reducing agent in metallurgy.

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11.6 SOME IMPORTANT COMPOUNDS OF CARBON AND SILICON

Oxides of Carbon. Two important oxides of carbon are carbon monoxide, CO and carbon dioxide, CO2.
11.6.1 Carbon Monoxide. It is prepared by the direct oxidation of carbon in a limited supply of oxygen or air.

2C(s)  O 2 (g)   2CO(g)
Pure CO is prepared by the dehydration of formic acid with concentrated H2SO4 at 373 K.
HCOOH  O 2 (g) 
373K
Conc.H 2SO4
H 2 O  CO(g)

It is commercially prepared by passing steam over hot coke. The mixture of CO and H2 thus produced is
known as water gas or synthesis gas or simply syngas.
4731273K
C(s)  H 2O(g)   CO(g)  H 2 (g)
 
Water gas

When air is used instead of steam, a mixture of CO and N2 is produced, which is called producer gas.

2C(s)  O 2 (g)  4N 2 (g) 

1273K
 2CO(g)  4N 2 (g)
   
Air Pr oducer gas

Water gas and producer gas are important industrial fuels. Carbon monoxide in water gas or producer gas
can undergo further combustion to form carbon dioxide with the liberation of heat.
 2CO 2 (g) ;  r H  566 kJ mol1
CO(g)  O 2 (g) 
Properties. Carbon monoxide is a colourless, odourless, and almost water insoluble gas. It is a neutral
oxide. It is a powerful reducing agent and reduces almost all metal oxides except those of alkali and alkaline
earth metals, aluminium and a few transition metals. Therefore, CO is used in the extraction of many metals
from their oxide ores.
 
Fe 2 O 3 (s)  3CO(g)   2Fe(s)  3CO 2 (g) ; ZnO(s)  CO(g)   Zn(s)  CO 2 (g)
Because of the presence of a lone-pair on carbon, CO molecule acts as donor and reacts with certain metals
when heated to form metal carbonyls.
330  350K
Ni  4CO   Ni(CO) 4 ; Fe  5CO 
473K ,100 atm.
 Fe(CO)5
Ni(CO)4, when heated to 440-450 K, dissociates giving Ni metal (purification of Ni by Mond’s process).
CO combines with chlorine in presence of sunlight to form carbonyl chloride or phosgene which is an extremely
poisonous gas.
h
CO(g)  Cl 2 (g)   CoCl 2 (g)
Phosgene

Toxicity. CO is highly poisonous nature because of its ability to form a complex with haemoglobin which is
about 300 times more stable than the oxygen-haemoglobin complex.
Haemoglobin  CO 
 Carboxyhaemoglobin
This prevents haemoglobin from carrying oxygen from the lungs to the various parts of the body which may
result in suffocation and finally death.

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Structure. In CO molecule, both C and O atoms are sp-hybridised. The sp-hybridised orbitals overlap to
form a C-O  -bond while the other sp-hybridised orbital contains the lone pair of electrons. The two
 
unhybridised p-orbitals of C and two p-orbitals of O form two p  -p  bonds, : C  O : .
Due to the presence of lone pair on the C atom, CO acts as a Lewis base or ligand which coordinates to

metals to form metal carbonyls, M  : C  O : .
11.6.2 Carbon Dioxide. It is prepared by the complete combustion of carbon or carbon containing fuels in excess
of air.
 
C(s)  O 2 (g)   CO 2 (g) ; CH 4 (s)  2O 2 (g)   2CO 2 (g)  2H 2 O
In the laboratory it is prepared by the action of dilute HCl on calcium carbonate.

CaCO3 (s)  2HCl(aq)   CaCl2 (aq)  CO 2 (g)  H 2 O(l )
It is prepared on a commercial scale by heating limestone.
1070 1270K
CaCO3 (s)  CaO(s)  CO 2 (g)
Properties. It is a colourless and odourless gas about 1.5 times heavier than air. It is not poisonous but it can
cause suffocation and eventually death due to lack of oxygen.
Solubility in water. It has only low solubility which increases with pressure. On dissolution, it forms carbonic
acid, H2CO3 which is a weak dibasic acid.

H 2 O  CO 2 
 H 2 CO3
Carbonic acid

Soda water and other aerated soft drinks are solutions of CO2 in water under pressure.
Carbonic acid dissociates in two steps:
 HCO3 (aq)  H 3O  (aq)
H 2 CO3 (aq)  H 2 O(l ) 

 CO32 (aq)  H3O (aq)

H 2 CO3 (aq)  H 2 O(l ) 
H2CO3/HCO3– buffer system helps to maintain pH of blood between 7.26 to 7.42.
Being acidic in nature, it combines with alkalies to form metal carbonates.
2NaOH  CO2 
 Na 2CO3  H2O
When CO2 is passed into lime water it turns milky due to the formation of insoluble calcium carbonate.
Ca(OH)2  CO2 
 CaCO3  H 2 O
Lime water

This is used as a test for carbonate in the laboratory.

Carbon dioxide, normally present to the extent of ~ 0.03 % by volume in the atmosphere, is consumed by
green plants in the synthesis of carbohydrates such as glucose (Photosynthesis).

6CO 2  12H 2 O   C 6 H12 O 6  6O 2  6H 2 O
CO2 is a major greenhouse gas. Use of fossil fuels and decomposition of limestone for cement manufacture
increase the CO2 content of the atmosphere. This leads to increased green house effect and thus, raise the
temperature of the atmosphere (global warming) which might have serious consequences.
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Solid carbon dioxide (dry ice). Carbon dioxide can be obtained in the solid form by allowing liquified CO2
to expand rapidly or when CO2 is cooled under 50-60 atm pressure. It is also called dry ice or cardice.
Uses. i) It is extensively used to carbonate soft drinks. ii) Being heavy and non-supporter of combustion it is
used as fire extinguisher. iii) In the manufacture urea. iv) Dry ice is used as a refrigerant for ice-cream and
frozen food. v) Super critical CO2 is used as a solvent to extract organic compounds from their natural
sources such as caffeine from coffee beans and perfume from flowers. vi) For artificial respiration (for victims
of CO poisoning) as a mixture of 95% O2 and 5% CO2 under the name carbogen.
Structure. In CO2 molecule carbon atom is in sp hybridised state. Two sp hybridised orbitals of carbon
overlap with two p-orbitals of oxygen atoms to form two sigma bonds while other two electrons of carbon are
involved in p  – p  bonding with oxygen atoms. This results in a linear molecule having both carbon-oxygen
bonds of equal length (115 pm) and no dipole moment. The resonance structures of CO2 are:
.. .. .. ..
:O.. – C O: :O = C = O: :O C – O:
11.6.3 Silicon Dioxide or Silica (SiO2). Silica and silicates constitute 95% of the earth’s crust. Silicon dioxide,
commonly known as silica, exists in several crystallographic forms such as quartz, cristobalite and tridymite.
The crystalline forms of silica are interconvertable at suitable temperatures.
Silicon dioxide is a covalent, three-dimensional network solid in which each Si atom is covalently bonded in
a tetrahedral fashion to four O atoms. Each O atom in turn is covalently bonded to two Si atoms. Each corner
is shared with another tetrahedron. The crystal may be regarded as a giant molecule formed by eight membered
rings with alternate Si and O atoms.

Figure - 04 Three dimensional structure of SiO2.

Silica is almost non reactive due to very high Si-O bond enthalpy (368 kJ mol-1). It resists attack by halogens,
dihydrogen, most acids and metals even at high temperatures. However, it is attacked by NaOH, HF and F2.

SiO 2  2NaOH 
Fuse
 Na 2SiO3  H 2O ; SiO 2  4HF   SiF4  2H 2O
Sodium silicate Silicon tetrafluoride

SiF4 dissolves in HF forming hydrofluorosilicic acid.


SiF4  2HF   H SiF
2 6
Hydrofluorosilicic acid

Due to the formation of H2SiF6, SiO2 of glass dissolves, therefore, HF cannot be stored in glass bottles.
Quartz is piezoelectric material; it is used in extremely accurate clocks, modern radio and television
broadcasting and mobile radio communication. Silica gel is used as a drying agent and support material in
chromatography and in catalysis (Kieselghur an amorphous form of silica is used in filtration plants).
11.6.4 Silicones. These are organosilicon polymers having general formula, (R2SiO)n where R is alkyl or aryl
group. Silicones are manufactured from alkyl or aryl substituted silicon chlorides, RnSiCl(4–n). When methyl
chloride reacts with silicon in presence of copper catalyst at 573K, methyl substituted chlorosilanes of formula
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MeSiCl 3, Me2SiCl 2, and Me3SiCl with a small amount of Me4Si are formed. Hydrolysis of
dimethyldichlorosilane, (CH3)2SiCl2 followed by condensation polymerisation yields straight chain polymers.
2CH 3Cl  Si 
Cu powder/570K
 (CH 3 ) 2 SiCl 2
Dichlorodimethylsilane

The chain length of the polymer can be controlled by adding (CH3)3SiCl which blocks the ends :

Short chain silicones are oily liquids, medium chain length silicones are viscous oils, jellies and greases and
those with very long chains are rubbery elastomers and resins.
Being surrounded by non-polar alkyl groups, silicones are water repelling in nature. In general, they have high
thermal stability, high dielectric strength and resistance to oxidation and chemicals.
Uses. Silicones are used as sealant, electrical insulator and for water proofing of fabrics. As lubricants at high
as well as low temperatures since their viscosities do not change with temperature. Being biocompatible they
are also used in surgical and cosmetic implants.
11.6.5 Silicates. Silicates are compounds in which the anions are either discrete (SiO4)4- tetrahedra or a number of
such units joined together through the corners.

(a) (b)

Figure - 05 (a) Tetrahedral structure of SiO44– anion (b) Representation of SiO44–unit

When silicate units are linked together, they form chain, ring, sheet or three-dimensional structures. If all the
four corners are shared with other tetrahedral units, a three-dimensional network is formed. Negative charge
on silicate structure is neutralized by positively charged metal ions.
A large number of silicate minerals exist in nature, e.g., feldspar, zeolites, mica and asbestos. Two important
man-made silicates are glass and cement.
Types of silicates. Silicates are classified into different types depending on the number of corners of the
SiO44– tetrahedron shared with other tetrahedra through oxygen atoms.
a. Orthosilicates. These are simple silicates containing discrete SiO44– tetrahedra (no corners are shared),
e.g., Zircon(ZrSiO4), Forestrite (Mg2SiO4), Phenacite (Be2SiO4), Wellimite (Zn2SiO4), etc.
b. Pyrosilicates. In pyrosilicates, two SiO44– tetrahedra share one corner oxygen (the shared oxygen is
common to the two tetrahedra) resulting in the formation of Si2O76– units, e.g., Thortveitite, Sc2(Si2O7),
and Hemimorphite, Zn3(Si2O7).Zn(OH)2.H2O.
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c. Cyclic or ring silicates. If two oxygen atoms per SiO44– are shared, closed structures with general
formula, (SiO32–)n or (SiO3)n2n– are formed. Examples of such anions are Si3O96– and Si6O1812–.
d. Chain silicates. Two oxygen atoms per SiO44– tetrahedron are shared such that a linear chain of (SiO32–)n
or (SiO3)n2n– is formed, e.g., Pyroxenes such as Spodumene, LiAl(SiO3)2 and Diopside, CaMg(SiO3)2. If
two chains are cross-linked, the double stranded chain silicates having formula [(Si4O11)6-]n are called
amphiboles, e.g., Tremolite, Ca2Mg5(Si4O11)2(OH)2, Asbestos, CaMg3O(Si4O11) etc.
e. Sheet silicates. The sharing of three corners of each tetrahedron results in two-dimensional sheet structure
of general formula (Si2O5)n2n– or (Si2O52–)n, e.g., Kaolinite, [Al(OH)4Si2O5], Talc, Mg3(OH)3Si4O11, etc.
f. Three-dimensional silicates. If all the four corner oxygens are shared between tetrahedra, three
dimensional network structure is obtained, e.g., different forms of silica such as quartz, tridymite and
crystobalite.
11.6.6 Zeolites. If some of the silicon atoms in three-dimensional network of silicon dioxide are replaced by
aluminium atoms, overall structure known as aluminosilicate, acquires a negative charge. Cations such as Na+,
K+ or Ca2+ balance the negative charge. Examples are feldspar (KAlSi3O8) and zeolites (NaAlSi2O6.H2O).
Zeolites are widely used as catalyst in petrochemical industries for cracking of hydrocarbons and isomerisation,
e.g., ZSM-5 is used to convert alcohols directly into gasoline. Hydrated zeolites (permutit) are used as ion
exchangers in softening of hard water.

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QUESTIONS
LEVEL - I
1. Which of the following is not a mineral of boron?
1) Colemanite 2) Kernite 3) Boric anhydride 4) Borax
2. Which of the following is used in artificial gems?
1) SnO2 2) B2O3 3) Al2O3 4) PbO2
3. Mineral of aluminium that does not contain oxygen is:
1) corundum 2) diaspore 3) bauxite 4) cryolite
4. The element which exists in liquid state for a wide range of temperature and can be used for measuring high
temperature is:
1) B 2) Al 3) Ga 4) In
5. Al and Ga have the same covalent radii because of :
1) greater shielding power of s-electrons of Ga atoms
2) poor shielding power of s-electrons of Ga atoms
3) poor shielding power of d-electrons of Ga atoms
4) greater shielding power of d-electrons of Ga atoms
6. Which one of the following is correct statement?
1) The hydroxide of aluminium is more acidic than that of boron
2) The hydroxide of boron is basic, while that of aluminium is amphoteric
3) The hydroxide of boron is acidic, while that of aluminium is amphoteric
4) The hydroxides of boron and aluminium are amphoteric
7. The shapes of BF3 and  BH 4  are respectively:
1) planar, tetrahedral 2) tetrahedral, planar 3) planar, planar 4) tetrahedral, tetrahedral
8. Boron can’t form which one of the following anions?
1) BF63 2) BH 4 3) B(OH) 4 4) BO 2

9. Ionisation enthalpy   i H1kJ mol1  for the elements of Group 13 follows the order:
1) B > Al > Ga > In > Tl 2) B < Al < Ga < In < Tl
3) B < Al > Ga < In > Tl 4) B > T1 > Ga > Al > In
10. The stability of +1 oxidation state increases in the sequence:
1) Tl < In < Ga < Al 2) In < Tl < Ga < Al 3) Ga < In < Al < Tl 4) Al < Ga < In < Tl
11. Which gas is liberated when Al4C3 is hydrolysed?
1) CH4 2) C2H2 3) C2H6 4) CO2
12. BCl3 exists as monomer whereas AlCl3 dimerises through halogen bridging. This is because of the
1) small size of B atom as compare to Al 2) absence of d orbital in B atom
3) p  p back bonding in AlCl3 4) p  p back bonding in BCl3
13. Aluminium chloride exits as a dimer because aluminium has:
1) Greater ionisation enthalpy 2) Incomplete p-orbital
3) High nuclear charge 4) Larger radius

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14. Boron nitride is isoelectronic with:

1) C2 2) B2 3) N2 4) O2
15. When NaOH is slowly added to AlCl3(aq). It gives a white precipitate which dissolves as more NaOH is
added. This is due to:
1) the formation of Al(OH)3 2) the formation of sodium aluminate which is soluble
3) the formation of soluble Al2O3 4) the formation of Al+3 ion
16. When excess of NaOH solution is added in potash alum the product is:
1) A bluish precipitate 2) Clear solution 3) White precipitate 4) Greenish precipitate
17. Aluminium is more reactive than iron. But aluminium is less easily corroded than iron because:
1) Iron forms mono and divalent ions
2) Iron undergoes reaction easily with water
3) Oxygen forms a protective oxide layer on aluminium
4) Aluminium is a noble metal
18. In Al2Cl6, which statement is incorrect?
1) Four Al–Cl bonds are of 206 pm length and four of 221 pm length
2) Six Al–Cl bonds are of same length and two of different length
3) The angle Cl–Al–Cl is 118o and 79o
4) The angle Al–Cl–Al is 101o
19. Which of the following statements is correct?
1) Aluminium chloride (AlCl3) is a Lewis acid because it can donate electrons
2) All the Al–Cl bonds in Al2Cl6 are equivalent
3) Aluminium exists in two polymorphic forms, namely,   Al 2 O3 and   Al2 O3
4) Anhydrous aluminium chloride can be prepared by heating hydrated salt
20. Among group 13 elements, the one forming an amphoteric oxide is:
1) Tl 2) Al 3) B 4) In
21. Boric acid is polymeric due to:
1) Its monobasic nature 2) Its acidic nature 3) The presence of hydrogen bonds 4) Its geometry
22. Borax bead test is responded by:
1) Divalent metals 2) Heavy metals
3) Light metals 4) Metals which form coloured metaborates
23. H3BO3 is
1) a monobasic acid and weak Lewis acid 2) a monobasic and weak Bronsted acid
3) a monobasic and strong Lewis acid 4) a tribasic and weak Bronsted acid
24. Na 2 B4 O7 .10H 2 O is correctly represented as:

1) 2NaBO 2 .Na 2 B2 O3 .10H 2 O 2) Na 2  B4 O5  OH 4  .8H 2 O

3) Na 2  B4  H 2 O 4 O 7  .6H 2 O 4) All of the above

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25. The correct Lewis acid order for boron halides is:
1) BF3  BCl3  BBr3  BI3 2) BCl3  BF3  BBr3  BI3
3) BI3  BBr3  BCl3  BF3 4) BBr3  BCl3  BI3  BF3
26. Boron trichloride on reaction with water produces ‘X’ along with HCl.’X’ is:
1) BOCl3 2) B2H6 3) B(OH)3 4) B2O3
27. When orthoboric acid is heated to red heat the residue is
1) Boron 2) Boron trioxide 3) Pyroboric acid 4) Metaboric acid
28. On the addition of mineral acid to an aqueous solution of borax, the compound formed is:
1) borodihydride 2) orthoboric acid 3) metaboric acid 4) pyroboric acid
29. The name and formula of the compound of boron which is called ‘inorganic benzene’ are:
1) Borazole, B6H6 2) Borazine, B6N6 3) Borazine, B3N3H6 4) Borazine, B6N3H3
30. Which is correct for the structure of diborane?
1) It contains four (2c, 2e) covalent bonds and two (3c, 2e) covalent bonds.
2) It contains three (2c, 2e) covalent bonds and three (3c, 2e) covalent bonds
3) It contains two (2c, 2e) covalent bonds and four (3c, 2e) covalent bond
4) It contains six (2c, 2e) covalent bonds
31. Which of the following is most basic?
1) CO 2) GeO 3) SnO 4) PbO
32. Which of the following halide of carbon is used as refrigerant?
1) CCl4 2) CF4 3) CH2Cl2 4) CCl2F2
33. Which of the following statements is correct with respect to the property of elements in the carbon family with
an increase in the atomic number? Their
1) atomic size decreases 2) stability of +2 oxidation state increases
3) metallic character decreases 4) ionization energy increases
34. Which of the following statements is not correct?
1) Silicon is extensively used as a semiconductor 2) Carborundum is SiC
3) Silicon occurs in free state in nature 4) Mica contains the element silicon
35. Which of the following statements is correct?
1) The electronegativity of Si is more than that of C 2) Si–C bonds are as strong as C–C bonds
3) Both C and Si can form p  p bond 4) Both CO2 and SiO2 has linear structures
36. Inert pair effect causes:
1) increase in stability of the (+IV) oxidation state on descending 14 group
2) decrease in stability of the (+IV) oxidation state on descending 14 group
3) decrease in stability of the (+II) oxidation state on descending 14 group
4) the decrease in tendency to form dimer in group 14.
37. CCl4 is inert towards hydrolysis but SiCl4 is readily hydrolysed because:
1) Carbon cannot expand its octet but silicon can expand its octet
2) Ionisation enthalpy of carbon is higher than silicon
3) Electronegativity of carbon is higher than that of silicon
4) Carbon forms double and triple bonds

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38. What is the number of free electrons present on each carbon atom in graphite?
1) Zero 2) 3 3) 2 4) 1
39. Graphite is soft solid lubricant extremely difficult to melt. The reason for this anomalous behaviour is that
graphite:
1) has carbon atoms arranged in large planes of rings of strongly bound carbon atoms with weak interplanar bonds
2) is a non-crystalline substance
3) is an allotropic form of carbon
4) has molecules of variable molecular masses like polymers
40. Me2SiCl2 on hydrolysis will produce:

1) Me2Si  OH  2 2) Me2Si  O \

3) [ O – (Me)2 Si – O ]n 4) Me 2SiClOH
LEVEL - II
1. The stability of monohalides of group 13 elements:
1) Increases down the group 2) Decreases down the group
3) First increases and then decreases 4) First decreases and then increases
2. Thallium shows different oxidation states because:
1) it is a transition metal 2) of inert pair effect 3) of its amphoteric character 4) of its high reactivity
3. For the properties mentioned, the correct trend for the different species is in
1) strength as Lewis acids – BCl3 > AlCl3 > GaCl3 2) inert pair effect – Al > Ga > In
3) oxidising property – Al > In > Tl
3+ 3+ 3+
4) first ionization enthalpy – B > Al > Tl
4. Which one of the following compounds does not exist?

1) B2 H 4  CH3 2 2) B2 H 3  CH 3 3 3) B2 H 2  CH3 4 4) B2 H  CH3 5

5. Which one of the following is hardest compound of boron?
1) Boron carbide 2) Silicon carbide 3) Magnesium boride 4) Silicon boride
6. Heating an aqueous solution of aluminium chloride to dryness will give:
1) AlCl3 2) Al2Cl6 3) Al2O3 4) Al(OH)Cl2
7. Which of the following statements regarding aluminium chloride is not correct?
1) Fused aluminium chloride does not conduct electricity
2) Aluminium chloride exists as dimer in organic solvents such as benzene and ether
3) Aluminium chloride exists as monomer in organic solvents such as benzene and ether
4) Aluminium chloride forms double salt AlCl3.6NH3 with ammonia
8. The unexpected order of acidic strength of the trihalides of boron can best be explained by:
1) p  p back bonding 2) Hybridisation
3) Trigonal planar structure 4) None of the above

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9. The B–F bond length in BF3 is shorter than that in BF4 . This is because of:

1) resonance in BF3 but not in BF4 2) p  p back bonding in BF4 but not in BF3

3) p  p back bonding in BF3 but not in BF4 4) p  d back bonding in BF3 but not is BF4
10. AlCl3 is an electron-deficient compound but AlF3 is not, due to:
1) Atomic size of F is smaller than Cl, which makes AlF3 more covalent
2) AlCl3 is a covalent compound while AlF3 is an ionic compound
3) Al in AlCl3 is sp3 hybridised but in AlF3, Al is sp2 hybridised.
4) AlCl3 exists as a dimer but AlF3 does not
11. BCl3 does not exist as a dimer but BH3 exists as B2H6 because:
1) Chlorine is more electronegative than hydrogen
2) Large size of chlorine atom does not fit between small, sized boron atoms, while small-sized hydrogen
atoms occupy the space between boron atoms.
3) There is p  p back bonding in BCl3.
4) Both (2) and (3)
12. What is not true about borax?
1) Molecular formula is Na 2 B4 O7 .10H 2 O
2
2) Crystalline borax contains tetranuclear unit of  B4 O5  OH 4 
3) It hydrolyses to give an acidic solution 4) White crystalline solid
o o
13. H 3 BO3 
100 C
 X 
160 C
 Y 
Red hot
 B2 O3 ; X and Y respectively are:
1) X = metaboric acid ; Y = Tetraboric acid 2) X = Borax; Y = Metaboric acid
3) X = Tetraboric acid; Y = Metaboric acid 4) X = Tetraboric acid; Y = Borax
14. Aqueous solution of borax acts as a buffer because:
1) it contains weak acid and its salt with strong base 2) it contains tribasic acid and strong base
3) it contains number of neutral water molecules 4) none of these
15. Reactivity of borazole is greater than that of benzene because:
1) borazole is non-polar compound 2) borazole is polar compound
3) borazole is electron deficient compound 4) of localized electrons in it
16. In the structure of diborane
1) All hydrogen atoms lie in one plane and boron atoms lie in a plane perpendicular to this plane
2) 2 boron atoms and 4 terminal hydrogen atoms lie in the same plane and 2 bridging hydrogen atoms lie in
the perpendicular plane
3) 4 bridging hydrogen atoms and boron atoms lie in one plane and two terminal hydrogen atoms lie in a
plane perpendicular to this plane
4) All the atoms are in the same plane
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17. A compound X of boron, reacts with NH3 on heating to give another compound Y which is called inorganic
benzene. The compound X can be prepared by treating BF3 with Lithium aluminium hydride. The compounds
X and Y are represented by the formulas
1) B2 H 6 , B3 N 3 H 6 2) B2 O3 , B3 N 3 H 6 3) BF3 , B3 N3 H 6 4) B3 N 3 H 6 , B2 H 6
18. An alkali metal hydride (NaH) reacts with diborane in ‘A’ to give a tetrahedral complex ‘B’ which is extensively
used as reducing agent in organic synthesis. The compounds ‘A’ and ‘B’ respectively are:
1) CH 3COCH 3 and B3 N3 H 6 2)  C2 H5  2 O and NaBH 4

3) C2 H 6 and C2 H 5 Na 4) C6 H 6 and NaBH 4

19. Which of the following statements is incorrect regarding B2H6?
1) Banana bonds are longer but stronger than normal B–H bonds
2) B2H6 is also known as 3c–2e compound
3) The hybridization of B in B2H6 is sp3 while that of sp2 in BF3.
4) It cannot be prepared by reacting BF3 with LiBH3 in the presence of dry ether.
20. Which statement is not true about potash alum?
1) Its empirical formula is KAl(SO4)2.12H2O
2) Its aqueous solution is basic in nature
3) It is used in dyeing industries
4) On heating it melts and loses its water of crystallization
21. Which tetrahalide does not act as Lewis acid?
1) CCl4 2) SiF4 3) GeCl4 4) SnCl4
22. Which of the following lead oxide is called ‘red lead’?
1) PbO 2) PbO2 3) Pb2O3 4) Pb3O4
23. A tetravalent element forms monoxide and dioxide with oxygen. When air is passed over heated element
(1273 K), producer gas is obtained. Monoxide of the element is a powerful reducing agent and reduces ferric
oxide to iron. Identify the element.
1) Lead 2) Carbon 3) Tin 4) Silicon
24. The stability of hydrides of carbon family is in the order:
1) CH4 > SiH4 > GeH4 > SnH4 > PbH4 2) CH4 < SiH4 < GeH4 < SnH4 < PbH4
3) CH4 > SnH4 > GeH4 > SiH4 > PbH4 4) None of the above
25. Which of the following statements is not correct?
1) Sn(II) salts or lead (II) compounds are essentially ionic whereas tetravalent are generally covalent
2) Lead (II) compounds are more stable than lead (IV) compounds
3) Sn (II) are good oxidizing agent
4) Sn (II) salts are more stable than Sn (IV) salts
26. PbCl2 is more ionic than PbO2 because
1) the radius of Pb2+ is more than that of Pb4+ 2) of inert pair effect
3) chlorine is more electronegative than oxygen 4) chlorine atom is smaller than oxygen atom
27. Lead does not dissolve in concentrated HCl because
1) A surface coating of PbO is formed 2) The reaction is kinetically unfavourable
3) A surface coating of PbCl2 is formed 4) A protective layer of PbCl4 is formed
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28. Which gas is evolved when PbO2 is treated with concentrated HNO3?
1) NO2 2) O2 3) N2 4) N2O
29. Which of the following statements regarding graphite is not correct?
1) Graphite is a good conductor of electricity 2) Graphite is less dense than diamond
3) The bond length of  bonded carbon-carbon bond is 154 pm
4) Graphite is thermodynamically more stable than diamond
30. Which of the following is known as pyrene?
1) CS2 2) Al4C3 3) CCl4 4) Solid CO2
31. Carbon suboxide, C3O2 has:
1) bent structure 2) trigonal planar sturucture
3) linear structrue 4) distorted tetrahedral structure
32. Which of the following statements about SiO2 is NOT correct?
1) SiO2 dissolves rapidly in fused NaOH forming silicates
2) SiO2 reacts only with F2 amongst halogens
3) Kieselguhr is form of SiO2, used as an abrasive
4) SiO2 reacts with hydrochloric acid to form silicon tetrachloride
33. In silicon dioxide:
1) each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atom
2) each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms
3) silicon atom is bonded to two oxygen atoms.
4) there are double bonds between silicon and oxygen atoms
34. Which of the following oxide is used to make scratch resistant glass?
1) Cr2O3 2) PbO2 3) SnO2 4) MgO
35. Among the following substituted silanes, the one which will give rise to crosslinked silicone polymer on hydrolysis
is:
1) R4Si 2) RSiCl3 3) R2SiCl2 4) R3SiCl
36. Which of the following statements is not correct?
1) Organosilicon polymers are known as silicones
2) Silicones have the general formula (R2SiO)n where R = –CH3, –C2H5, –C6H5, etc.
3) Hydrolysis of dialkyldichlorosilane produces cross-linked silicon polymer
4) Hydrolysis of alkyltrichlorosilane produces cross-linked silicon polymer
37. ‘Butter of tin’ refers to:
1) SnCl2.2H2O 2) SnCl4.5H2O 3) SnBr2.5H2O 4) SnBr4.5H2O
38. The name of the structure of silicates in which three oxygen atoms of [SiO4] are shared is:
4–

1) pyrosilicate 2) sheet silicate

3) linear-chain silicate 4) three-dimensional silicate
39. On heating Pb(NO3)2, the products formed are:
1) PbO, N2, O2 2) Pb(NO2)2, O2 3) PbO, NO2, O2 4) Pb, N2, O2
40. Name the type of the structure of silicate in which one oxygen atom of [SiO4] is shared?
4–

1) Linear chain silicate 2) Sheet silicate 3) Pyrosilicate 4) Three dimensional

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SOLUTIONS
LEVEL - I
1. 3
2. 3 Al2O3 is used in the preparation of artificial gems.
3. 4 Cryolite is Na3AlF6
4. 3 Gallium consists of only Ga2 molecule. It exists as liquid up to 2000oC and hence used in high temperature
thermometry.
5. 3
6. 3 Boric acid behaves as a weak monobasic acid. [Al(OH)3] is amphoteric.
7. 1 [BH4]– species is tetrahedral in shape.
8. 1 Due to absence of 3d-orbital, maximum covalency is four. Thus, BF63 is not formed.
9. 4
10. 4 Lewis acid strength of boron trihalides : BI3 > BBr3 > BCl3 > BF3.

11. 1 Al4C3  12H 2 O  3CH 4  4Al  OH 3

12. 2 Due to the absence of d-orbitals, BCl3 does not dimerise but in AlCl3, Al contains empty d-orbital so it
is capable to undergo dimerisation.
13. 2
14. 1 BN and C2 are isoelectronic.

15. 2 This is due to the formation of aluminate ion AlCl3  4NaOH  Na  Al  OH 4   3NaCl
so luble

16. 2 Potash alum is K 2SO 4 .Al2  SO4 3 .24H 2O

Al2 SO4 3 reacts with NaOH to form NaAlO 2 which is soluble.

17. 3

18. 1 Al2Cl6 has the structure given:

19. 3
20. 2 Oxides of Al and Ga, ie, Al2O3 and Ga2O3 respectively are amphoteric.
21. 3 B(OH)3 units are joined together by hydrogen bonds.
22. 4
23. 1 B  OH 3  H 2 O 
 [B(OH)4] + H
H3BO3 is a monobasic acid and acts as a weak Lewis acid by accepting OH ions.
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24. 2 Borax molecule is actually made of two tetrahedra and two triangular units joined as shown.

25. 3 The Lewis acid order for boron halides are explained in terms of back-bonding.
26. 3  B  OH 3  3HCl; X is B(OH)3
BCl3  H 2O 
27. 2 2H 3 BO 3 
Re d heat
 B2 O 3  3H 2 O
28. 2 Na 2 B4 O 7  H 2SO 4  5H 2 O 
 Na 2SO 4  4H 3 BO3
29. 3 Borazine is isoelectronic and isostructural with benzene and hence is also known as inorganic benzene.
30. 1

31. 4
32. 4 Dichloro difluoro methane (Freon-12)
33. 2
34. 3 Si occurs as silicates in nature.
35. 2 Si–C bond is also strong. Si does not form p-p bond due to its larger size. Carbon dioxide molecules
is linear while SiO2 is a network of Si and O atoms bonded by a single bond.
36. 2 Inert pair effect represents the inactiveness of pair of electrons and thus the stability of M(+II) is increased
in comparison to M(+IV)
37. 1 Since carbon has no d-orbital it cannot extend its co-ordination number beyond four, so its halides are
not hydrolysed by water.
38. 4 In graphite one electron is free on each carbon atom.
39. 1 Since the layers of C-atoms are held together by weak van der Waals forces of attraction, the layers can
slip over one another. Thus graphite is soft and acts as a good lubricating agent.
40. 3 Linear silicons can be obtained by the hydrolysis and subsequent condensation of dialkyl or diaryl silicon
chloride, R2SiCl2.
LEVEL - II
1. 1 Due to inert pair effect, stability of lower oxidation state, ie, +1 increases down the group.
2. 2 The inert pair effect begins when n  4 and increases with increasing value of n.
3. 1 Strength of Lewis acids decreases in the order : BCl3  AlCl3  GaCl3 :
4. 4 Not more than four hydrogen atoms can be substituted by methyl groups in the molecules of B2H6. The
bridge hydrogen atoms are not to be substituted.
5. 1
6. 3 On heating AlCl3(aq) to dryness, Al2O3 is formed.

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7. 3 In organic solvents, aluminium chloride exists as dimer.

8. 1 The tendency of p  p bond formation decreases in the order: BF3 > BCl3 > BBr3
9. 3 Because of back bonding, B–F bond has some double bond character. All the three boron-fluorine
bonds are shorter than the usual single boron-fluorine bond.
10. 2
11. 3
12. 3  2NaOH  4B  OH 3 . On hydrolysis, an alkaline solution is formed.
Na 2 B4 O7  7H 2 O 
Strong base Weak acid
o
 HBO 2 (Metaboric acid) ; 4HBO 2   H 2 B4 O7 (Tetraboric acid)
o 160 C
13. 1 H 3 BO3 
100 C
 H 2O  H 2O
(X) (Y)
14. 1
15. 2 Borazole is polar due to B–N bond; benzene is non-polar due to C–C bonds.

16. 2

17. 1 3B2 H 6  6NH3 

473K
 2B3 N3 H 6  12H 2
Diborane(X) Inorganicbenzene (Y)

4BF3  3LiAlH 4 
 2B 2 H 6  3LiF  3AlF3

 2 5  
2Na   BH 4 
C H O A 
18. 2 2NaOH  B2 H 6  
Diethyl ether

Sod.hydride Sod.borohydride(B)(Tetrahedral complex)

19. 4 B-H bond enthalpy is 381kJ mol–1, B-H-B bond enthalpy is 441kJ mol–1.
20. 2 Alum form acidic solution on dissolution in water due to hydrolysis of Al3+ ions
21. 1 CCl4 does not form hexachloro complex as d-orbitals are not present in carbon.
22. 4 Pb3O4 is called ‘red oxide’ it is a mixed oxide.
23. 2 Producer gas is a mixture of CO and N2.
The carbon monoxide is a strong reducing agent and reduces ferric oxide to iron.
24. 1 The stability of hydrides of carbon family decreases down the group, hence order is
25. 3 Sn (II) is not an oxidizing agent. It is more stable than Sn (IV)
26. 1
27. 3 Pb does not dissolve in conc. HCl due to the formation of surface coating of PbCl2.
28. 2  2Pb  NO3  2  2H 2 O  O2
The reaction is 2PbO2  4HNO3 

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29. 3 The bond length of  bonded carbon-carbon bond in graphite is 142 pm

30. 3 CCl4 used as fire extinguisher.
31. 3 Carbon suboxide has linear structure. O  C  C  C  O
32. 3
33. 1 Si is tetrahedrally surrounded by four O atoms. Each corner is shared with another tetrahedron, thus
giving an infinite array.
34. 3 Casseterite or tin stone, SnO2 is added to glass to make it scratch resistant.
35. 2 Hydrolysis of alkyl trichlorosilane RSiCl3 gives complex cross linked polymer.

36. 3 The hydrolysis of dialkyldichlorosilane produces linear silicon.

37. 2 Butter of tin is SnCl4.5H2O.

38. 2 In sheet silicates, three oxygen atoms of silicate are shared.

1
39. 3 Pb  NO3 2 

 PbO  2NO 2  O 2
2
Pyrosilicate contains two units of SiO 44 joined together through one oxygen atom given  Si 2 O 7 
6
40. 3
anion.

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CHAPTER - 12
ORGANIC CHEMISTRY : BASIC PRINCIPLES - I
NOMENCLATURE

INTRODUCTION
Organic chemistry is the study of the composition, structure, synthesis, properties and reactions of organic
compounds. Organic compounds are compounds of carbon containing usually hydrogen or one or more
additional elements like nitrogen, oxygen, halogens, sulphur, phosphorus, etc. Organic compounds form the
basis for all earthly life constitute the majority of known chemicals.
In 1815, Berzelius, a Swedish chemist proposed that organic compounds could be produced only by living
matter because they require the presence of a mysterious force called vital force which exists only in living
organisms. However this theory was rejected in 1828 when German Chemist, Friedrich Wohler synthesized
urea by heating ammonium cyanate obtained by heating a mixture of ammonium chloride and potassium cyanate.

NH 4 Cl  KCNO 
 NH 4 CNO  KCl ; NH 4 CNO 
Heat
 NH 2 CONH 2
Ammonium cyanate Ammonium cyanate Urea

The synthesis of acetic acid by Kolbe (1845) and of methane by Berthelot (1856) showed conclusively that
organic compounds could be synthesised in the laboratory from inorganic sources.
12.1 CHARACTERISTICS OF ORGANIC COMPOUNDS
Organic compounds are covalent while inorganic compounds are electrovalent. Being non-polar, they are
usually insoluble in water while inorganic compounds being polar are soluble in water. The reactions of organic
compounds are molecular in nature while those of inorganic compounds are ionic in nature.
Catenation. Carbon has the remarkable property of self-linking through covalent bonds to form long straight
or branched chains and rings of different sizes. This property is called catenation.
Isomerism. Many organic compounds exhibit the phenomenon of isomerism by virtue of which a single
molecular formula may represent two or more structures.
Due to the property of catenation and the phenomenon of isomerism, carbon forms a very large number of
compounds. About 5 million organic compounds are known to date.
12.2 STRUCTURAL REPRESENTATION OF ORGANIC COMPOUNDS
Structure of organic compounds may be represented by Lewis dot structure, dash structure, condensed structure
and bond line structural formulae. Lewis structures can be simplified by representing the covalent bond by a
dash (–). A single dash represents a single bond, double dash () is used for double bond and a triple dash ()
represents triple bond. Lone-pairs of electrons on heteroatoms (e.g.,O, N, S, halogens etc.) may or may not
be shown. Thus, ethane (C2H6), ethene (C2H4), ethyne (C2H2) and methanol (CH3OH) can be represented by
the following structural formulae. Such structural representations are called complete structural formulae.

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These structural formulae can be abbreviated by omitting some or all of the dashes representing covalent
bonds and indicating the number of identical groups attached to an atom by a subscript. This representation is
called condensed structural formula. Thus, ethane, ethene, ethyne and methanol can be written as:

CH3CH3 H 2C=CH 2 HC  CH CH3OH

Ethane Ethene Ethyne Methanol

Similarly, CH3CH2CH2CH2CH2CH2CH2CH3 can be condensed to CH3(CH2)6CH3.

Further simplification is possible by bond-line structural representation in which lines representing carbon-
carbon bonds are drawn in a zig-zag fashion. The only atoms specifically written are oxygen, chlorine, nitrogen
etc. The terminals denote methyl (–CH3) groups (unless otherwise indicated by a functional group), while the
line junctions denote carbon atoms bonded to appropriate number of hydrogens required to satisfy its
tetracovalency. 3-methyloctane can be represented in various forms as:

2-bromobutane can be represented as:

In cyclic compounds, the bond-line formulas may be as follows:

12.2.1 Three-Dimensional Representation of Organic Molecules. The three-dimensional (3-D) structure of

organic molecules can be represented on paper using solid and dashed wedge formula. Solid-wedge is used to
indicate a bond projecting out of the plane of paper, towards the observer. Dashed-wedge is used to depict
the bond projecting out of the plane of the paper, away from the observer. Wedges are shown in such a way
that the broad end of the wedge is towards the observer. The bonds lying in the plane of the paper are depicted
using a normal lines. 3-D representation of methane molecule on paper is shown below:

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Molecular Models. Molecular models (made of wood, plastic or metal) are used for visualisation and
perception of three-dimensional shapes of organic molecules. Commonly used molecular models are: (1)
Framework model, (2) Ball-and-stick model, and (3) Space filling model.

12.2 CLASSIFICATION OF ORGANIC COMPOUNDS

Organic compounds are broadly divided into the following classes.

12.2.1 Acyclic or open chain compounds. These compounds are also called aliphatic compounds and consist of
straight or branched chain compounds, for example:

12.2.2 Alicyclic or closed chain or ring compounds. Aliphatic cyclic compounds are called alicyclic compounds.
Alicyclic (aliphatic cyclic) compounds contain carbon atoms joined in the form of a ring (homocyclic or
carbocyclic). Sometimes atoms other than carbon are also present in the ring (heterocylic).

12.2.3 Aromatic compounds. Completely conjugated cyclic polyenes which obey Huckel’s rule, i.e., contain
(4n + 2)  -electrons, where n = 0, 1, 2, 3, .....etc. are called aromatic compounds. Benzene is the simplest
aromatic hydrocarbon. These are of two types:
i) Benzenoids. Aromatic compounds and their alkyl, alkenyl and alkynyl derivatives which contain one or
more benzene rings either fused or isolated in their molecules are called benzenoids or arenes.

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ii) Non-benzenoids. Aromatic compounds which do not contain a benzene ring but instead contain other
highly unsaturated rings are called non-benzenoids. For example,

2. Heterocyclic compounds. Cyclic compounds containing one or more heteroatoms (e.g., O, N, S, etc.) in
their rings are called heterocyclic compouds. These are of two types:
i) Alicyclic heterocyclic compounds. Heterocyclic compounds which resemble aliphatic compounds in their
properties are called alicyclic heterocyclic compounds. For example,

ii) Aromatic heterocyclic compounds. Heterocyclic compounds which resemble benzene and other aromatic
compounds in most of their properties are called aromatic heterocyclic compounds. For example,

12.4 FUNCTIONAL GROUPS

A functional group is an atom or a group of atoms joined in a specific manner which is responsible for the
characteristic properties of an organic compound, e.g., –OH (hydroxy), –CHO (aldehydic), –COOH
(carboxylic), -NO2 (nitro), etc.
12.5 HOMOLOGOUS SERIES
Homologous series is a family of structurally similar organic compounds, all the members of which contain the
same functional group, show gradation in physical properties and similarity in chemical properties and any two
consecutive members of which differ by a –CH2 group. Each homologous series can be represented by a
general formula, e.g, ROH (alcohols), R–X (alkyl halides), RCOOH (carboxylic acids), etc., where R is any
alkyl group. The different members of a homologus series are called homologues and the phenomenon is
called homology.
12.6 NOMENCLATURE OF ORGANIC COMPOUNDS
1. Common or Trivial Names. Organic compounds were assigned names based on their origin or certain
properties. For instance, citric acid is named so because it is found in citrus fruits. The acid found in red ant is
named formic acid since the Latin word for ant is formica. Some common names are followed even today,
e.g., Buckminsterfullerene. Common names are useful when the systematic names are lengthy and complicated.

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Table - 01 Common or Trivial Names of Some Organic Compounds

Compound Common name Compound Common name

CH4 Methane CHCl3 Chloroform
H3CCH2CH2CH3 n-Butane H3CCOOH Acetic acid
(H3C)2CHCH3 Isobutane C6H6 Benzene
(H3C)4C Neopentane C6H5OCH3 Anisole
H3CCH2CH2OH n-Propyl alcohol C6H5NH2 Aniline
HCHO Formaldehyde C6H5COCH3 Acetophenone
(H3C)2CO Acetone H3COCH2CH3 Ethyl methyl ether
12.6.2 The IUPAC System of Nomenclature
Organic chemistry deals with millions of compounds. In order to identify them clearly, a systematic method of
naming has been developed by the International Union of Pure and Applied Chemistry (IUPAC). In this
system of nomenclature, the names are correlated with the structure such that the structure can be deduced
from the name. The systematic name of an organic compound is derived by identifying the parent hydrocarbon
and the functional group(s) attached to it. Using prefixes and suffixes, the parent name can be modified to
obtain the actual name.

Compounds containing only carbon and hydrogen are called hydrocarbons. A hydrocarbon is termed saturated
if it contains only carbon-carbon single bonds. The IUPAC name for a homologous series of saturated
hydrocarbons is alkane. Unsaturated hydrocarbons contain at least one carbon-carbon double bond (alkenes)
or triple bond (alkynes).

Type of carbon chain Primary suffix General name

Saturated -ane Alkane
Unsaturated with double bond -ene Alkene
Unsaturated with triple bond -yne Alkyne
12.6.3 IUPAC Nomenclature of aliphatic hydrocarbons
I. Nomenclature of Straight chain Alkanes. The names of alkanes are based on their chain structure, and
end with suffix ‘-ane’ and carry a prefix (word root) indicating the number of carbon atoms in the chain (except
CH4 to C4H10, where the prefixes are derived from trivial names).

Chain le ngth Word root Chain length Word root Chain le ngth Word root
C1 Meth- C5 Pent (a)- C9 Non (a)
C2 Eth- C6 Hex (a)- C10 Dec (a)
C3 Prop (a)- C7 Hept (a)- C11 Undec (a)
C4 But (a)- C8 Oct (a)- C12 Dodec (a)

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Formula IUPAC name Formula IUPAC name Formula IUPAC name

CH4 Methane C8H18 Octane C15H32 Pentadecane
C2 H 6 Ethane C9H20 Nonane C16H34 Hexadecane
C3 H 8 Propane C10H22 Decane C17H36 Heptadecane
C4H10 Butane C11H24 Undecane C18H38 Octadecane
C5H12 Pentane C12H26 Dodecane C19H40 Nonadecane
C6H14 Hexane C13H28 Tridecane C20H42 Eicosane
C7H16 Heptane C14H30 Tetradecane C30H62 Triacontane

Molecular Name of alkane Structural formula of Name of alkyl group

formula of alkane alkyl group
CH4 Methane -CH3 Methyl
C2H6 Ethane -CH2CH3 Ethyl
C3H8 Propane -CH2CH2CH3 Propyl
C4H10 Butane -CH2CH2CH2CH3 Butyl
C10H22 Decane -CH2(CH2)8CH3 Decyl
Abbreviations are used for some alkyl groups, e.g., methyl as Me, ethyl as Et, propyl as Pr and butyl as Bu.
Branched alkyl groups. Alkyl groups can be branched also. Common branched groups have specific trivial
names, e.g., the propyl groups can either be n-propyl group or isopropyl group. The branched butyl groups
are called sec-butyl, isobutyl and tert-butyl group. –CH2C(CH3)3 is called neopentyl group.

Types of carbon and hydrogen atoms. Based on the number of carbon atoms to which it is bonded, a
carbon atom can be called primary (1o), secondary (2o), tertiary (3o) or quaternary (4o). The hydrogens
attached to 1o, 2o and 3o carbon atoms are called 1o, 2o and 3o hydrogen atoms respectively.

II. Nomenclature of Branched chain Alkanes.

1. Longest continuous chain or the parent or root chain. The longest carbon chain in the molecule is
identified. It is considered as the parent or root chain and the rest are side chains or substituents. In the
examples given below, (a) with eight carbons is the longest chain

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2. Lowest locant rule. The carbon atoms of the parent chain are numbered to identify the parent alkane and
to locate the positions of the carbon atoms at which branching takes place. The numbering is done in such a
way that the branched carbon atoms get the lowest possible numbers (locants).
3. The names of alkyl groups attached (branches) are prefixed to the name of the parent alkane and positions
of the substituents are indicated by the numbers. If different alkyl groups are present, they are listed in alphabetical
order. Thus, the name for the compound given above is 5-ethyl-2-methyloctane. The numbers are separated
from the groups by hyphens and there is no break between methyl and nonane.

4. If two or more identical substituent groups are present then the numbers are separated by commas. Prefixes
such as di (for 2), tri (for 3), tetra (for 4), penta (for 5), hexa (for 6) etc. are used to indicate their numbers.
However, these prefixes are not considered while writing the name of the substituents in alphabetical order.

5. If the two substituents are present in equivalent positions, the lower number is given to the one coming first
in alphabetical listing. Thus, the following compound is 3-ethyl-6-methyloctane and not 6-ethyl-3-methyloctane.

6. Branched alkyl groups can be named by the above procedure. However, the carbon atom of the branch
that attaches to the root alkane is numbered 1. The name of the branch alkyl group is placed in parenthesis.

For writing the trivial names of substituents in alphabetical order, the prefixes iso- and neo- are considered to
be part of the name of the alkyl group, but prefixes sec- and tert- are not considered to be part of the name.
The use of common prefixes for naming alkyl groups is allowed as long as these are not further substituted.
In case of multisubstituted compounds, the following rules should also be followed:
i. If there are two chains of equal size, the chain which contains more side chains should be selected.
ii. Numbering of the chain should be done from the end closer to the substituent.

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III. Nomenclature of Unsaturated Hydrocarbons ( Alkenes and Alkynes)

The parent chain must contain the multiple bond regardless of whether it denotes the longest chain of carbon
atoms or not. For example, in structure (a), the parent chain consists of five carbon atoms while the longest
continuous chain contains six carbon atoms (b).

If the carbon chain contains two, three, four or more double or triple bonds, numerical prefixes such as di, tri,
tetra, etc., are added before the primary suffix. E.g., diene, triene, tetraene, etc.
If both double and triple bonds are present, the numbering of the parent chain should be done from that end
which is nearer to the double/triple bond, i.e., the lowest locant rule for multiple bonds must be followed.

However, if there is a choice in numbering, the double bond is always given preference over the triple bond.
While naming unsaturated hydrocarbons, the locant for the double bond or triple bond is placed immediately
before the suffix ‘ene’ or ‘yne’ and not followed eariler.

If however, both double and triple bonds are present in the compound, their locants are written immediately
before their suffixes and the terminal ‘e’ from the suffix ‘ene’ is dropped while writing its complete name.

IV. Nomenclature of Cyclic Compounds: An alicyclic compound is named by prefixing ‘cyclo’ to the
corresponding straight chain alkane. The suffixes ‘ane’, ‘ene’, or ‘yne’ are written depending upon the saturation
or unsaturation of the compound. If side chains are present, then the rules given above are applied.

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If two or more alkyl groups or other substituent groups are present in the ring, their positions are indicated by
arabic numerals, i.e., 1,2,3.....etc. While numbering the carbon atoms of the ring, the substituent which comes
first in the alphabetical order is given the lowest number provided it does not violate the lowest locant rule.

V. Nomenclature of Compounds Containing Functional Groups

1. Select the longest chain of carbon atoms containing the functional group (carbon atoms involved in the
functional groups like –COOH, –CHO, –CN etc., or those which carry functional groups like –OH,
–NH2, –Cl, –NO2, etc.). The number of carbon atoms decides the word root.
2. The primary prefix is decided on the basis of the presence of carbon-carbon multiple bonds in the chain.
3. Secondary suffix or prefix is selected based on the functional group present.
4. The chain is numbered in such a way that the carbon atom of the functional group or the carbon to which
the functional group is attached gets the lowest number.
In the case of polyfunctional compounds, one of the functional groups is chosen as the principal functional
group and the compound is named on that basis. The remaining functional groups are considered as subordinate
functional groups and named as substituents using prefixes. The choice of principal functional group is based
on the following order of preference: -COOH, –SO3H, -COOR (R = alkyl group), COCl, -CONH2, -CN,-
HC=O, >C=O, -OH, -NH2, >C=C<, C  C  .
The –R, C6H5-, halogens (F, Cl, Br, I), –NO2, alkoxy (–OR) etc., are always prefix substituents.

VI. Nomenclature of Compounds with more than one Similar Functional Groups.
If the organic molecule contains more than one similar functional group, then in addition to various rules the
prefixes di, tri, tetra, etc. are added before the secondary suffix which indicates the functional group. While
adding such words, the vowel ‘e’ of the primary suffix is retained.

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Table - 02 Functional Groups and Classes of Compounds

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12.6.4 Nomenclature of Substituted Benzene Compounds. The benzene ring is called the nucleus and the alkyl
groups attached to it are called the side chains. Benzene forms only one monosubstituted derivative. In
IUPAC nomenclature, the substituent is placed as prefix to the word benzene. However, common names
(given in bracket) of many substituted benzenes are also used.

In disubstituted benzene, the positions of substituents are defined by numbering the carbon atoms of the ring in
such a way that the substituents are located at the lowest numbers possible. E.g., compound (b) is named as
1,3-dichlorobenzene and not as 1,5-dichlorobenzene.

In the trivial system, the prefixes ortho (o), meta (m) and para (p) are used to indicate the relative positions
1,2- ;1,3- and 1,4- respectively. Thus, 1,2-dichlorobenzene is named o-dichlorobenzene, 1,3-dichlorobenzene
is named m-dichlorobenzene and 1,4-dichlorobenzene is named p-dichlorobenzene, respectively.
When an aromatic compound contains two or more functional groups, it is named as a derivative of the
compound with the principal functional group at position 1.

Tri and polysubstituted derivatives are named by numbering the chain in such a way that the parent group gets
the lowest number and the lowest locant rule is obeyed. In case three or more functional groups are present
they are named in alphabetic order. For example,

If the functional groups present in the benzene ring are such which are normally treated as substituent groups,
the various groups are arranged in alphabetical order with the group named first in the alphabetical order
getting the lowest locant provided it does not violate the lowest locant rule for all the substituents. For example,
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12.7 IUPAC Names of Compounds Represented by Bond-Line Notation

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QUESTIONS
LEVEL - I
1. Which of the following is the correct bond line formula for the given compound? (CH3)2C = CHCH2CH3?

1) 2) 3) 4)

2. Which of the following has only 1o and 2o C atoms?

1) 2-Methyl butane 2) Butane 3) 2,2-Dimethyl butane 4) 2,2,3,3-Tetramethyl pentane
3. In 2-Chloro-3-methyl hexanoic acid, the primary suffix is:
1) 2-Chloro- 2) -3-Methyl 3) an(e) 4) oic acid
4. The number of  -and  -bonds in 1-buten-3-yne is:
1) 5  and 5  2) 7  and 3  3) 8  and 2  4) 6  and 4 
5. Which of the following is not a cumulated diene?
1) Hexa-1,2-diene 2) Hexa-2, 3-diene 3) Penta-2,3-diene 4) Penta-1,3-diene
6. Which of the following functional groups will provide secondary suffix if all are present in one molecule?
1) –OH 2) –CN 3) –COOH 4) –CHO
7. Choose the option which show correct preferential order of groups among the following:
1) –COOH, –CHO, –OH, –NH2 2) –NH2, –OH, –CHO, –COOH
3) –COOH, –OH, –NH2, –CHO 4) –COOH, –NH2, –CHO, –OH
8. Which of the following statements is wrong for a homologous series?
1) All members have a general formula 2) All members have the same functional group
3) All members have the same chemical properties4) All members have the same physical properties
9. The family to which methoxyethene belongs is:
1) hydrocarbon 2) ketone 3) unsaturated ether 4) ester
10. The correct decreasing order or priority for the functional groups of organic compounds in the IUPAC system
of nomenclature is:
1) –CONH2, –CHO, –SO3H, –COOH 2) –COOH, –SO3H, –CONH2, –CHO
3) –SO3H, –COOH, –CONH2, –CHO 4) –CHO, –COOH, –SO3H, –CONH2

11. The number of functional groups in aspartame is:

1) 4 2) 6 3) 5 4) 7
H C4H9

12. The IUPAC name of CH3–– C –– C––CH3 is :

C2H5 CH3
1) 2-Butyl-2-methyl-2-ethylbutane 2) 2-Ethyl-3-3-dimethylheptane
3) 3, 4, 4- Trimethylheptane 4) 3, 4, 4- Trimethyloctane

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13. The IUPAC name of vinyl acetylene is:

1) Pent-1-en-4-yne 2) Pent-4-yn-1-ene 3) But-1-en-3-yne 4) But-1-yn-3-ene
14. The IUPAC name of (CH3)3C–CH=CH2 is:
1) 2,2-Dimethylbut-3-ene 2) 2,2-Dimethylpent-4-ene
3) 3,3-Dimethylbut-1-ene 4) Hex-1-ene
15. Which of the following compounds has isopropyl group?
1) 2,2,3,3-Tetramethylpentane 2) 2,2-Dimethylpentane
3) 2,2,3-Trimethylpentane 4) 2-Methylpentane
16. The structural formula of 2, 2, 4-trimethyl pentane is:
CH3 CH3 CH3 CH3

1) CH 3  C  CH  CH 2  CH3 2) CH 3  C  CH 2  CH  CH3
CH3 CH3

CH3 CH3 CH3

3) CH 3  C  CH  CH 2  CH 2  CH 3 4) CH 3  CH  CH 2  CH 2  C  CH 3

CH3 CH3 CH3

17. Which of the following IUPAC names is correct?
1) 2-Methyl-3-ethylpentane 2) 2-Ethyl-3-methylpentane
3) 3-Ethyl-2-methylpentane 4) 3-Methyl-2-ethylpentane
18. The IUPAC name of neopentane is
1) 2-Methylbutane 2) 2,2-Dimethylpropane3) 2-Methylpropane 4) 2,2-Dimethylbutane
19. Which compound is 2,2,3-trimethylhexane?
CH3 CH3 CH3 CH3

1) CH 3  C  CH  CH 2  CH 3 2) CH 3  C  CH 2  CH  CH 3
CH3 CH3

CH3 CH3 CH3

3) CH 3  CH  CH 2  CH 2  C  CH 3 4) CH 3  C  CH  CH 2  CH 2  CH 3
CH3 CH3 CH3
1
2

20. The correct IUPAC name of the compound, 3 4 is:

5
6
1) 3-(1-Ethyl propyl) hex-1-ene 2) 4-Ethyl-3-propylhex-1-ene
3) 3-Ethyl-4-ethenyl heptane 4) 3-Ethyl-4-propylhex-5-ene
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21. The IUPAC name of the compound is:

1) 2,2,4,4-Tetramethylpentane 2) 2,2-Dimethylpropane
3) 3-Methyl-4-ethylhexane 4) 3-Ethyl-4-methylhexane

CN
N
22. The IUPAC name of the compound, N  is :
1) Hexane-1, 2, 5-tricarbonitrile 2) Hexane-1, 3, 6-tricarbonitrile
3) Butane-1, 2, 4-tricarbonitrile 4) Butane-1, 3, 4-tricarbonitrile
23. The IUPAC name of CH3COCH(CH3)2 is :
1) iso-Propylmethyl ketone 2) 2-Methyl-3-butanone
3) 4-Methyl iso-propyl ketone 4) 3-Methyl-2-butanone
OH
24. IUPAC name of is :
OH
1) But-2-ene-2,3-diol 2) Pent-2-ene-2,3-diol
3) 2-Methylbut-2-ene-2,3-diol 4) Hex-2-ene-2,3-diol
Cl

25. The IUPAC name of the compound shown is:

Br
1) 2-Bromo-6-chlorocyclohex-1-ene 2) 6-Bromo-2-chlorocyclohexene
3) 3-Bromo-1-chlorocyclohexene 4) 1-Bromo-3-chlorocyclohexene
26. The general formula CnH2nO2 could be for open chain:
1) diketones 2) carboxylic acids 3) diols 4) dialdehydes
27. The structural formula of cyclohexyl alcohol is
CH2OH
OH OH CH2OH

1) 2) 3) 4)

28. The correct IUPAC name of the compound is:

1) 1,1-Dimethyl-3-hydroxycyclohexane 2) 3, 3-Dimethyl-1-hydroxycyclohexane
3) 3,3-Dimethyl-1-cyclohexanol 4) 1,1-Dimethyl-3-cyclohexanol
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H
29. IUPAC name of following compound, CH 3  C  CH 2  CH 3 is:

1) 2-Cyclohexylbutane 2) 2-Phenylbutane 3) 3-Cyclohexylbutane 4) 3-Phenylbutane

30. Name the following substance:
COOH
Cl

CH2CH3
1) 1-Ethyl-3-chlorobenz-4-oic acid 2) 6-Chloro-4-ethyl benzoic acid
3) 4-Ethyl-2-chlorobenzoic acid 4) 2-Chloro-4-ethyl benzoic acid
LEVEL - II
1. The number of 1 , 2 , and 3 H atoms in 3-ethyl-5-methyl heptane, respectively, is:
o o o

1) 12, 8, 1 2) 14, 4, 2 3) 12, 6, 2 4) 12, 8, 2

2. Which alkane would have only the primary and tertiary carbon?
1) pentane 2) 2-methyl butane 3) 2,2-dimethyl propane
4) 2,3-dimethyl butane
3. The decreasing order of priority for the following functional group is:
I) C  N II) –CONH2 III) C O IV) –CHO
1) II > I > IV > III 2) III > IV > I > II 3) I > II > IV > III 4) I > II > III > IV

4. The correct IUPAC name of the compound, CH3 –– CH –– CH–– CH –– CH2 –– CH3 is:

Cl Br I
1) 4-bromo-5-chloro-3-iodohexane 2) 3-bromo-2-chloro-4-iodohexane
3) 3-bromo-4-iodo-2-chlorohexane 4) 2-chloro-3-bromo-4-iodohexane
5. Which is the structure of compound 2-(1-Cyclobutenyl)-1-hexene?

1) 2)

3) 4)

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6. In which of the following cases does the suggested name not follow the IUPAC system of nomenclature?
CH3 CH2CH3
1) CH ––CH––CH––CH 2) CH3–– C C–– CH(CH3)2
3 3 4-Methylpent-2-yne
2, 3-Dimethylpentane
CH3––CH––CO––CH3 CH3––CH––CH––COOH
3) CH3 4) CH3 CH3
2-Methyl-3-butanone 2, 3-Dimethyl butanoic acid

7. Give the IUPAC name of

1) 2,2-Dimethyl-3-propyl-4-isopropyl heptane 2) 4-Isopropyl-5-t-butyl octane

3) 4-t-Butyl-5-isopropyl octane 4) 2-Methyl-3-propyl-4-isopropyl heptane
8. Which is the simplest alkane, that is, the one with the smallest molecular weight, which possesses primary,
secondary and tertiary carbon atoms?
1) 2-Methylpropane 2) 2-Methylbutane 3) 2-Methylpentane 4) 3-Methylpentane
9. The IUPAC name of the compound CH 3CH=CHCH=CHC  CCH3 is :
1) 4,6-Octadien-2-yne 2) 2,4-Octadien-6-yne 3) 2-Octyn-4,6-dienes 4) 6-Octyn-2,4-diene
CH3

10. The IUPAC name for CH3 –– CH –– CH2 –– C –– CH3 is:

OH OH
1) 1, 1-Dimethyl-1, 3-butanediol 2) 4-Methyl-2, 4-pentanediol
3) 2-Methyl-2, 4-pentanediol 4) 1, 3, 3-Trimethyl-1, 3-propanediol
11. The structure of 4-methyl-2-penten-1-ol is
1) CH3CH2CH=CHCH2OH 2) (CH3)2C=CHCH2CH2OH
3) (CH3)2CHCH=CHCH2OH 4) CH3CHOH–CH=C(CH3)2
12. The IUPAC name of CH3CH = CHCOOC2H5 is
1) Ethylbut-1-enoate 2) Ethylbut-2-enoate 3) Ethylprop-2-enoate 4) Ethylprop-1-enoate
O

13. The IUPAC name of the compound CH3––CH2––CH––CH2 is:

1) 2-Ethyl oxirane 2) Ethyl methyl ether 3) Ketopentanone 4) Ketobutanone
14. IUPAC name of CH3CH(OH)CH2CH2COOH is:
1) 4-Hydroxypentanoic acid 2) 1-Carboxy-3-butanoic acid
3) 1-Carboxy-4-butanol 4) 4-Carboxy-2-butanol

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15. The correct IUPAC name of (C2H5)4C is:

1) Tetraethyl methane 2) 2-Ethylpentane 3) 3, 3-Diethylpentane 4) 2, 2-Diethylbutane
16. The prefix name of –SH group in IUPAC system is:
1) mercapto 2) thiol 3) sulphide 4) sulphonic acid
17. Which of the following represents neopentyl alcohol?
1) CH3CH(CH3)CH2CH2OH 2) (CH3)3CCH2OH
3) CH3(CH2)3OH 4) CH3CH2CH(OH)CH3

18. The IUPAC name of the compound is

1) 1-Iodo-2-methyl-3-ethylcyclohexane 2) 1-Methyl-2-ethyl-6-iodocyclohexane
3) 1-Iodo-2-methyl-1-ethylcyclohexane 4) 1-Ethyl-3-iodo-2-methylcyclohexane

19. The compound Et have its IUPAC name as:

HO C

O
1) Octadec-9-enoic acid 2) Oleic acid
3) Ethylhexadec-9-enoic acid 4) Ethylpentadec-9-enoic acid
20. Which of the following compounds has incorrect IUPAC nomenclature?
O

1) CH3CH2CH2COC2H5 2) CH3CHCH2CHO
Ethyl butanoate
O CH3
3-Methyl butanal

3) CH3CHCCH2CH3 4) CH3CHCHCH3

CH3 H3C OH
2-Methyl-3-pentanone 2-Methyl-3-butanol
CH3
21. IUPAC name of CH 3  C  C  CH  CH  CH 2 is:

CH2

CH3
1) 2-Ethyl-3-methyl-hexa-l-en-4-yne 2) 5-Ethyl-4-methyl-hexa-2-yn-5-yne
3) 3-Methylene-4-methylhepta-5-yne 4) 5-Methylene-5-ethyl-4-methylhepta-2-yne
O

CH 2  C  OH
OH
22. The IUPAC name of compound C is :
COOH
CH2 COOH
1) 1,2,3-Tricarboxy-2, 1-propane 2) 3-Carboxy-3-hydroxy-1,5-pentanedioic acid
3) 3-Hydroxy-3-carboxy-1,5-pentanedioic acid 4) none of the above

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23. Which is wrong IUPAC name?

1) CH3CH2CH2COOCH2CH3 (Ethyl butanoate) 2) CH 3  CHCH 2 CHO (3-Methyl butanal)
CH3

3) CH3CH CHCH3 (2-Methyl butanal) 4) CH3CHCOCH2CH3 (2-Methyl-3-pentan-3-one)

OH CH3 CH3

Cl
24. The IUPAC name of is :
O
1) 2-Ethyl-3-methylbutanoyl chloride 2) 2,3-Dimethylpentanoyl chloride
3) 3,4-Dimethylpentanoyl chloride 4) 1-Chloro-l-oxo-2,3-dimethylpentane
25. Structure of the compound whose IUPAC name is 3-Ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid is:
OH
OH
COOH COOH
1) 2)

OH

COOH COOH
3) 4)
OH
O CN

26. The IUPAC name of the compound CH 3  C  CH 2  C  CH3 is:

CH3
1) 4-Cyano-4-methyl-2-oxopentane 2) 2-Cyano-2-methyl-4-oxopentane
3) 2,2-Dimethyl-4-oxopentanenitrile 4) 4-Cyano-4-methyl-2-pentanone
27. Which of the following structures represents 2-Bromo-6-isobutyl 4-methyldec-3-en-1-ol?

1) 2)

3) 4)

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28. Structure of the compound whose IUPAC name is 3-Ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid is:

1) 2) 3) 4)

29. IUPAC name of 4-isopropyl-m-xylene is:

1) 1-isopropyl-2,4-dimethylbenzene 2) 4-isopropyl-m-xylene
3) 4-isopropyl-3,5-dimethylbenzene 4) 4-isopropyl-3,5-dimethylbenzene
OH

30. The IUPAC name of the following compound is:

CN
Br
1) 4-Bromo-3-cyanophenol 2) 2-Bromo-5-hydroxy benzonitrile
3) 2-Cyano-4-hydroxy bromo benzene 4) 6-Bromo-3-hydroxy benzonitrile

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SOLUTIONS
LEVEL - I
1. 2 The carbon skeleton and the bond line formula are represented as
H3C H
C C CH3
H2
CH3

2. 2

3. 3

4. 2  7,3 

5. 4 Cumulated diene is (C=C=C).

A) B) C)
(All cumulated diene)

D) It is not a cumulated diene .

6. 3
7. 1 The choice of principal functional group is made on the basis of the following order.
Carboxylic acid > sulphonic acid > anhydrides > esters > acid halides > acid amides > nitriles >
aldehydes > ketones > alcohols > amines
8. 4 The physical properties such as density, melting point, boiling point of the members of a homologous
series show a regular gradation with rise in molecular mass.
9. 3 Methoxy ethene is CH3O – CH = CH2; an unsaturated ether.
10. 2 The priority order is: –COOH > –SO3H > –CONH2 > –CHO.
11. 3 It has phenyl (C6H5), carboxyl (–COOH), amide (–CONH2), ester (–COOCH3) and primary amine
(–NH2) as functional groups.
5 6 7 8
CH2 –– CH 2 –– CH 2 –– CH3
3 4
12. 4 CH3 –– C –– C –– CH 3
2
CH2 CH3
1
CH3
3, 4, 4-Trimethyloctane

13. 3 (Vinylacetylene) (But-1-en-3-yne)

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14. 3 (3,3-Dimethyl but-1-ene)

15. 4 2-Methyl pentane:

16. 2
17. 3
CH3
3 2 1
18. 2 The structure is of Neopentane is H 3 C  C  C H 3

CH3
The IUPAC name is 2,2-Dimethylpropane.

19. 4 Write the structure of 2,2,3-Trimethyl hexane.

20. 2
21. 4
1 2 3 4
22. 3 CH2 –– CH –– CH2 –– CH2

CN CN CN
Butane-1, 2, 4-tricarbonitrile

O O

23. 4 CH3 –– C –– CH –– CH3 , keto (–– C ––) group is given priority..

1 2 3 4

CH3
3-Methyl butan-2-one
24. 2 The terminal ‘e’ of ene is retained if suffix name starts with consonant.
25. 3
26. 2 This is general formula for carboxylic acids. For example, for n = 2, we have CH3COOH.
27. 1
28. 3
29. 2
30. 4 The name of the compound is 2-Chloro-4-ethyl benzoic acid.

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LEVEL - II

1. 4

2. 4
3. 1
4. 2 The decreasing order of priority of prefix in numbering the carbon chain of an organic compound is
1 2 3 4 5 6
bromo > chloro > iodo. CH3 –– CH –– CH –– CH –– CH2 –– CH3

Cl Br I
3-bromo-2-chloro-4-iodohexane

5. 2
2-(1-Cyclobutenyl)-1-hexene

4 3 2 1
6. 3 CH3 –– CH –– COCH3

CH3
3-Methyl-2-butanone

7. 3 . Numbering is started from that side of the chain where the complex substituent is

at a lower position.
8. 2 2-Methylbutane is the alkane with smallest molecular weight which possesses primary, secondary and
tertiary carbon atoms.
9. 2 The preference goes to a triple bond, but counting is done from the double bond end. The correct name
is 2,4-Octadien-6-yne.
CH3
5 4 3 2 1
10. 3 CH3–– CH –– CH2 –– C –– CH3

OH OH
2-Methyl pentane –2, –4 – diol

11. 3 4-Methyl-2-penten-1-ol is CH3 –– CH –– CH CH –– CH2OH

CH3
O

12. 2 CH3 –– CH –– CH –– C –– O –– CH2CH3

ester
Ethyl but-2-enoate

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13. 1 CH3––CH2––CH––CH2
2-Ethyl oxirane

14. 1
15. 3
16. 1
17. 2 Neopentyl alcohol is (CH3)3CCH2OH.
18. 4 Substituted cycloalkanes are named in the same way as alkanes.
15 14 11 10
17 16 13 12
Et 9
19. 1 HO C1 2 5 6
3 4 7 8
O Octadec-9-enoic acid

Incorrect 4 3 2 1
20. 4 numbering CH3 –– 2CH –– CH –– CH3
1 3 4

CH3 OH Correct numbering

Correct name is 3-Methylbutan-2-ol functional group
should be given priority

21. 1
22. 2
23. 3 It is 3-Methyl butan-2-ol.
24. 2
OH
2
3 1
COOH
25. 1
4
5

26. 3

27. 4

28. 4

29. 1

30. 2 CN– group is given the highest priority. (2-Bromo-5-hydroxy benzonitrile)

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CHAPTER - 13
ORGANIC CHEMISTRY : BASIC PRINCIPLES - II
ISOMERISM AND REACTION MECHANISM

PART - I ISOMERISM
Compounds having the same molecular formula but different physical and chemical properties are called
isomers and the phenomenon is called isomerism. Isomerism is mainly of two types, structural isomerism
and stereoisomerism. Each one of these is further divided into different types.

13.1 STRUCTURAL ISOMERISM

Compounds which have the same molecular formula but different arrangement of atoms within the molecule is
called structural isomers and the phenomenon is called structural isomerism.
13.1.1 Chain or Nuclear Isomerism. Compounds which have the same molecular formula but different skeletons
of the carbon chain (straight or branched) are called chain or nuclear isomerism. For example,
(i) Butane (C4H10) has two chain isomers:
CH3
|
CH3CH 2CH 2CH3 CH3  CH  CH3
Butane( n Bu tan e) 2-Methylpropane (Isobutane)

(ii) Pentane (C5H12) has three isomers:

CH3
CH 3 |
| CH3  C  CH3
CH3CH 2CH 2CH 2CH3 CH3  CH  CH 2CH3 |
Pentane ( n  Pen tan e) 2-Methylbutane (Isopentane)
CH3
2,2-Dimethylpropane (Neopentane)

Similarly, hexane (C6H14) has 5, heptane (C7H16) has 9, octane (C8H18) has 18, nonane (C9H20) has 35 and
decane (C10H22) has 75 chain isomers.
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13.1.2 Position isomerism. Compounds having the same carbon chain structure, but differ in the position of the
multiple (double or triple) bond or the functional group are called position isomers and the phenomenon is
called position isomerism.

OH
|
(i) CH3  CH 2  CH 2  OH CH3  CH  CH3
Pr opan 1ol ( n Pr opyl alcohol) Propan-2-ol (Isopropyl alcohol)

(ii) CH3  CH 2  CH  CH 2 CH 3  CH  CH  CH 3
But  2 ene
But 1ene

13.1.3 Functional group isomerism. Compounds having the same molecular formula but different functional groups
are called functional isomers. The phenomenon is called functional isomerism. The following classes of
compounds show functional isomerism.
(i) Aldehydes and ketones (CnH2nO). C3H6O represents:

CH3  CO  CH3 CH3CH 2  CHO

Propanone (Acetone ) Propionaldehyde

(ii) Alcohols and ethers (CnH2n+2O). C2H6O represents:

CH 3CH 2  OH CH 3  O  CH 3
Ethanol ( Ethyl alcohol ) Methoxymethane ( Dimethyl ether )

(iii) Carboxylic acids and esters (CnH2nO2). C3H6O2 represents:

CH 3CH 2  COOH CH3  CO  OCH3

Propanoic acid (Propionic acid ) Methyl acetate

(iv) Nitroalkanes and alkyl nitrites (CnH2n+1NO2). C2H5NO2 represents:

CH 3CH 2  NO 2 CH3CH 2 O  N  O
Nitroethane Ethy l n itrite

13.1.4 Metamerism. Compounds having the same molecular formula, but different number of carbon atoms or
different alkyl groups on either side of the functional group ( –O–, –S– , –NH–) are called metamers and the
phenomenon is called metamerism. Metamerism occurs among members of the same homologus series.

(i) CH3CH 2  O  CH 2CH3 CH3  O  CH 2CH 2CH3 CH3  O  CH  CH3 2

Ethoxyethane 1-Methoxypropane 2-Methoxypropane
(Diethyl ether) (Methyl n -propyl ether) (Isopropyl methyl ether)

Metamers may also be position isomers, e.g., pentan-3-one and pentan-2-one may be regarded as position
isomers as well as metamers.

(ii) CH3COCH 2CH 2CH3 CH3CH 2  CO  CH 2CH3 CH3  CO  CH  (CH3 )2

Pentan-2-one Petan-3-one 3-Methylbutan-2-one
(Methyl n -propyl ketone) (Diethyl ketone) (Isopropyl methyl ketone)

(iii) CH3CH 2  NH  CH 2CH3 CH3  NH  CH 2CH 2CH3 CH3  NH  CH(CH3 )2

Diethyl amine Methyl n -propylamine Isopropylmethylamine

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O O
|| ||
(iv) CH 3  C  OCH 2CH3 CH 3CH 2  C  OCH 3
Ethylethanoate(Ethyl acetate) Methylpropanoate

O O O
|| || ||
(v) CH3CH 2  C  OCH 2CH3 CH3  C  OCH 2CH 2CH3 CH3  C  O  CH  CH3 2
Ethyl propanoate n -Propyl acetate
Isopropyl acetate
13.1.5 Tautomerism. It is a type of functional isomerism in which the isomers exist in dynamic equilibrium with each
other. It arises due to migration of a hydrogen atom from one polyvalent atom to the other within the same
molecule. The isomers are called tautomers and the phenomenon is called tautomerism or desmotropism.
(i) 1, 2-migration of a proton from one polyvalent atom to the other within the same molecule.
 
H  C  N  HN C
Hydrogen cyanide Hydrogen isocyanide
The alkyl derivatives of these tautomers are called cyanides and isocyanides respectively.
(ii) Hydrogen migrates from one polyvalent atom to the third polyvalent atom within the same molecule. The
most important example is keto-enol tautomerism in which one tautomer contains the keto group while the
other contains the enolic group.
O OH
 

 
CH3  C  H CH 2  C  H
Acetaldehyde (keto form ) Vinyl alcohol (enolic form )

The greater stability of the keto-form than the enol form is due to the greater strength of the carbon-oxygen -
bond (364 kJ mol–1) as compared to carbon-carbon -bond (254 kJ mol–1).
13.1.6 Ring Chain Isomerism. Compounds having the same molecular formula but possessing open chain and
cyclic structures are called ring chain isomers and the phenomenon is called ring-chain isomerism.

(i) C3H6 represents : CH3CH  CH 2 and

Pr opene

13.2 STEREOISOMERISM
Isomers having the same structural formula but different relative arrangement of atoms or groups in space are
called stereoisomers and the phenomenon is called stereoisomerism.
13.2.1 Conformational Isomers. The different relative arrangements of atoms or groups resulting from free rotation
about a single bond are called conformations and each possible structure is called a conformer or rotamer.

13.2.2 Configurational Isomers. The spatial arrangement of atoms that characterises a particular stereoisomer is
called its configuration. Configurational isomers differ in their configuration.
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1. Geometrical isomerism. It is the phenomenon by which compounds with same molecular formula and
structural formula posses different arrangement of atoms or groups in space due to hindered rotation about a
double bond or a single bond in a ring system. The isomer with identical groups on the same side are called
cis- isomer and the one with identical groups on opposite sides is called trans- isomer.

2. Optical Isomerism. It is the phenomenon by which a compound posseses two spatial isomers which are
non-superimposable mirror images of each other (called enantiomers). Objects which are non-superimposable
on their mirror images are said to be chiral and the property is called chirality.

A carbon atom connected to four different types of atoms/groups is called asymmetric carbon or chiral
carbon or stereocentre. A molecule containing asymmetric carbon has no elements of symmetry and hence
asymmetric molecule. Molecular asymmetry responsible for optical activity. Optically active substances
rotate the plane of plane polarised light. Substances which rotate the plane of polarised light are said to be
optically active and the property is called optical activity.

PART - II FUNDAMENTAL CONCEPTS IN REACTION MECHANISM

13.3 Reaction mechanism
Reaction mechanism is a detailed description of a possible pathway for a chemical reaction at the molecular
level. It is a sequential account of each step describing electron movement, energetics of bond cleavage and
bond formation, and the rates of transformation of reactants into products.
Knowledge of reaction mechanism may make it possible to choose reaction conditions that suits the formation
of maximum amount of desired products, and to find correlations between systems not otherwise related.
Products

Organic molecule 

Attacking
reagent
  Intermediate 
 substrate  Biproducts

The organic molecule that reacts with a reagent is called substrate. It supplies carbon to the new bond.
13.3.1 Types of Reagents
1. Electrophiles: Electrophiles are electron loving chemical species. Their attraction for electrons is due to
the presence of an electron-deficient atom in them. Electrophiles may be either positively charged or neutral.
  
(i) Positive electrophiles: H  , H 3O  , Cl , Br  , I  , N O 2 , N O,  N  N, R  (carbocation), etc.

(ii) Neutral electrophiles : R (free radicals), : CR2 (carbenes), : NR

 (nitrenes), BF3, AlCl3, ZnCl2, FeCl3, SnCl4,
etc.

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Since both positively charged and neutral electrophiles are short by a pair of electrons, they act as Lewis
acids. Electrophiles always attack the substrate molecule at the site of highest electron density.
2. Nucleophiles: Nucleophiles are nucleus loving chemical species. Since the nucleus of any atom is positively
charged, nucleophiles must be electron rich containing at least one lone pair of electrons. They may be either
negatively charged or neutral chemical species.
(i) Negative nucleophiles: H– (hydride ion), Cl– , Br–, I–, R– (carbanion), RC  C  (acetylide ion),
OH  , OR  , SR  , C 6 H 5 O  , NH 2 , NO 2 , CN  , RCOO  , etc.
.. .. .. .. .. ..
(ii) Neutral nucleophiles:H2O:, NH3, RNH2, ROH, .. RSH,
.. ROR, .. R-MgX, R-Li, LiAlH4, etc,
Since both negatively charged and neutral nucleophiles contain at least one unshared pair of electrons, they
have a strong tendency to donate a pair of electrons to electron deficient species and hence behave as Lewis
bases. Nucleophiles always attack the substrate molecule at the site of lowest electron density.
13.3.2 Fission of a Covalent Bond
1. Homolytic (symmetrical) fission or Homolysis. If a covalent bond breaks in such a way that each atom
takes away one electron of the shared pair, it is called homolytic or symmetrical fission or homolysis. Homolytic
fission is usually indicated by a fish hook arrow which denotes one-electron displacement.
. .
A . . B Homolytic fission A + B
Heat or light Free radicals
The neutral chemical species ( A  and B ) containing an odd or unpaired electron, produced by homolytic
fission of covalent bonds is called free radical. Homolytic fission usually occurs in non-polar bonds and is
favoured by high temperature, ultraviolet (UV) radiations and the presence of radical initiators such as peroxides.
2. Heterolytic (unsymmetrical) fission or Heterolysis. When a covalent bond between two atoms A and
B breaks in such a way that both the electrons of the covalent bond (i.e. shared pair) are taken away by one
of the bonded atoms, the mode of bond cleavage is called heterolytic fission or heterolysis. Heterolytic
fission is usually indicated by a curved arrow which denotes two-electron displacement.

Heterolytic fission A+ + : B– (B is more electronegative than A)

A: B

A : B Heterolytic fission A:– + B+ (A is more electronegative than B)

Thus, heterolytic fission results in the formation of charged species, i.e., cations and anions. It usually occurs
in polar covalent bonds and is favoured by polar solvents.
3. Homolytic vs heterolytic fission. Homolytic fission requires much less energy than heterolytic fission:
Homolytic fission . .
CH3CH2 – Br CH3CH2 + Br; H = 289 kJ mol–1

Heterolytic fission +
CH3CH2 – Br CH3 CH2 + Br–; H = 720 kJ mol–1
13.3.3 Types of electron displacements in covalent bonds
I. Inductive effect. It involves permanent displacement of -electrons along a saturated carbon chain towards
the more electronegative substituent (atom or group). The effect weakens steadily with increasing distance
from the substitutent and becomes negligible after two carbon atoms.
+ + + –
C C C C Cl

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Groups or atoms which withdraw electrons of the -bond towards themselves as compared to hydrogen are
said to have electron - withdrawing or electron -attracting or –I-effect. The decreasing order of
–I- effect of some substitutents relative to hydrogen is:
  
R 3 N    S R 2   N H 3   NO 2  SO 2 R  CN  COOH   F  Cl   Br   I  OAr 
COOR  OR  COR  SH  SR  OH  NH2  C  CR  Ar  CH  CR 2
Groups or atoms which donate electrons to the carbon chain are said to have electron-releasing or electron-
donating or + I-effect. Examples are:
O  COO    CH3 3 C   CH3  2 CH    CH 3CH 2   CH 3   D  H
Inductive effect is responsible for high melting points, boiling points and dipole moments of polar molecules.
II. Electromeric effect or E-effect or Polarisability effect. It involves the complete transfer of electrons
of a multiple bond to one of the bonded atoms (usually more electronegative) in presence of an attacking
reagent.
Reagent added
C=O C – O–
Reagent removed
This effect is temporary and takes place only in the presence of a reagent. When the reagent is removed, the
molecule reverts back to its original position.
+E-effect: If the electrons of the -bond are transferred to that atom of the double bond to which the reagent
gets finally attached, the effect is called +E-effect. For example, addition of acids to alkenes.

C=C + H+ C C
Electrophile
H
–E-effect : If the electrons of the double bond are transferred to an atom of the double bond other than the
one to which the reagent gets finally attached, the effect is called -E-effect. For example, the addition of
cyanide ion to the carbonyl group.

weakly C – O–
C = O + CN– basic
Nucleophile CN
Wherever inductive effect and electromeric effect operate simultaneously, electromeric effect
predominates.
III. Resonance or Mesomerism. The phenomenon of resonance occur when a molecule can have two or
more Lewis structures which differ in the position of electrons, but not in the relative position of atoms. The
various Lewis structures are called canonical or resonance or contributing structures.
The real structure of the molecule is not represented by any of the canonical structures, but is a resonance
hybrid of all Lewis structures. The resonance structures are separated by a double headed arrow (  ). Thus,
benzene can be represented as a resonance hybrid of the following two Kekule (Lewis) structures, I and II.

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Adjacent carbon atoms in benzene are neither joined by pure single bond nor by pure double bond. As a
result, all the carbon-carbon bond lengths are equal (1.39 Å) and lie in between carbon-carbon double bond
length (1.34 Å) and carbon-carbon single bond length (1.54 Å). The resonance hybrid or actual molecule is
usually represented by formula III. The circle inside the ring denotes completely delocalized six -electrons.
1. Resonance stabilisation energy or Resonance energy or Delocalization energy. It is defined as the
difference in internal energy of the resonance hybrid and the most stable canonical structure. Greater the
number of resonance structures, greater the delocalization of electrons, larger the resonance energy and more
stable the compound. In case of benzene, resonance energy is 150.62 kJ (or 36.0 kcal)mol–1.
2. Rules for writing Resonance Structures : (i) The resonance structures should differ only in the position
of electrons and not in the position of atoms or nuclei.
(ii) All resonance structures should have the same number of unpaired electrons.
(iii) As far as possible, resonance structures should have nearly the same energy.
(iv) In case of elements of second period (no d-orbitals), resonance structures should not violate octet rule.
3. Relative contribution of resonance structures. When the resonance structures differ in energy content
or relative stabilities, the more stable structure makes greater contribution towards the resonance hybrid.
(i) Structures which are indistinguishable are of equal energy and hence contribute equally towards the
resonance hybrid.
 
and H 2 C  CH  CH 2 
 H 2 C  CH  C H 2

Thus, the various resonance structures of a molecule should not differ much in energy.
(ii) Structures with greater number of covalent bonds contribute more towards the resonance hybrid.

.. .. CH – CH = CH – CH
CH 2 = CH – CH = CH 2 CH 2 – CH = CH – CH 2 2 2
I II III
Since formation of a bond is accompanied by release of energy, structure (I) with two -bonds is more
stable than structures (II and III) which contain one -bond each.
(iii) Structures which involve separation of positive and negative charges are of higher energy and hence
contribute little towards the resonance hybrid.
(iv) Lesser the separation of positive and negative charges, more stable is the resonance structure.
(v) When atoms of different electronegativities are involved, the structure with a negative charge on the more
electronegative atom and positive charge on the less electronegative atom is of lower energy and hence
contributes more towards the resonance hybrid.
(vi) Structures with like charges on adjaent atoms are highly unstable and do not contribute towards the
resonance hybrid.
(vii) Structures which help to delocalise the positive charge make significant contribution towards the resonance
hybrid regardless of whether the positive charge is on a less electronegative or more electronegative atom.
(viii)Resonance structures in which all atoms have octet of electrons make greater contribution towards the
resonance hybrid.
IV. Resonance effect or mesomeric effect. In conjugated systems (having alternate -and -bonds),
-electrons can flow from one part of the molecule to the other due to resonance. As a result, centres of low
and high electron density are created in such molecules. This flow of electrons in conjugated systems as a result
of resonance is called resonance effect (R-effect) or mesomeric effect (M-effect). It is of two types:

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(i) +R Effect. Groups which donate electrons to the double bond or a conjugated system are said to have
+R-effect, e.g., – OH, – OR, – SH, – SR, – NH2, – NHR, – NR2, – X (halogens), etc. Thus,
.. . ..CH – CH = Cl .. (+ R-effect)
CH2 = CH – Cl.. . 2 ..

(ii) -R Effect. Groups which withdraw electrons from the double bond or from a conjugated system towards
themselves due to resonance are said to have –R -effect, e.g., >C = O, –CHO, –COOR, –CN, –NO2, etc.
..
CH2 = CH – C .. N.. CH2 – CH = C = N ..– (– R-effect)

V. Hyperconjugation effect or Baker - Nathan Effect or no-bond resonance. When an alkyl group is
attached to an unsaturated system such as a double bond or a benzene ring, the order of inductive effect is
usually reversed. This effect is called hyperconjugation effect or Baker-Nathan effect. It involves
delocalisation of -electrons through overlap of p-orbitals of a double bond and - orbital of the adjacent
single bond (i.e., conjugation) as shown.

Figure - 01 Orbital representation of hyperconjugation --conjugation) in propene

Hyperconjugation may be represented as follows:

Structures I, II and III are called hyperconjugative structures. Since there is no bond between carbon and
hydrogen atoms in these structures, hyperconjugation is also called no-bond resonance. The magnitude of
hyperconjugation depends on the number of -hydrogen atoms. Hyperconjugation decreases in the order :
CH3   CH 3CH 2    CH3 2 CH    CH3 3 C 
Significance of the hyperconjugation effect. Although hyperconjugation is a much weaker effect than
resonance effect, it is useful in explaining some of the physical and chemical properties of organic molecules
such as o, p -directive influence of alkyl groups, shortening of carbon-carbon single bonds adjacent to multiple
bonds, relative stability of alkenes and relative stability of carbocations and free radicals. .
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13.3.4 Reactive Intermediates. The highly reactive short-lived (10–6 s to a few seconds) chemical species through
which majority of the organic reactions occur are called reactive intermediates, e.g., carbocations, carbanions,
free-radicals, carbenes, etc.

1. Carbocations (carbonium ions) or carbenium ions. These are chemical species having a positive charge
on carbon and carrying six electrons (sextet) in its valence shell. These are formed by heterolytic cleavage of
covalent bonds. Carbocations can be classified as primary (1°), secondary (2°) and tertiary (3°).
These are planar chemical species (sp2 hybridised carbon). The relative stability of simple alkyl carbocations
follows the sequence : 3° > 2° > 1o > methyl,
   
C H 3  CH3 C H 2   CH 3 2 C H  C  CH3 3
This order of stability is due to + I-effect and hyperconjugation effect of the alkyl groups.
2. Carbanions. Chemical species in which the carbon atom bears a negative charge and is surrounded by
octet of electrons are called carbanions. These are produced by heterolytic cleavage of covalent bonds.
Simple alkyl carbanions are of three types; primary, secondary and tertiary depending on the nature of the
carbon bearing the negative charge.
The stability of alkyl carbanions follows a sequence which is exactly the reverse of that of alkyl free radicals
and carbocations, i.e., methyl > primary > secondary > tertiary. This due to + I-effect of the alkyl groups.
Alkyl carbanions are pyramidal (sp3 hybridised carbon). In carbanions which are resonance stabilised, the
negative charge is on an sp2 hybridised carbon. Thus (CH3)3C- is pyramidal while allyl carbanion is planar.
3. Carbenes. Neutral divalent carbon species in which the carbon atom is bonded to two monovalent atoms
or groups and surrounded by a sextet of electrons are called carbenes, e.g., :CH2 (methylene), :CC12
(dichlorocarbene), etc. Like in carbocations, the central carbon atom in carbenes has only a sextet of electrons
in its valence shell Carbenes, therefore behave as Lewis acids or electrophiles.
Carbenes are of two types, i.e., singlet and triplet. In singlet carbenes, the central carbon atom is sp2-
hybridised, whereas in triplet carbenes, the central carbon is sp- hybridised. Triplet carbene is generally more
stable than a singlet carbene.

4. Free Radicals. Free radicals are neutral chemical species (atom/group) containing an odd or unpaired
electron. It is obtained by homolytic cleavage of a covalent bond.

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Free radicals are of three types; primary, secondary and tertiary.

Their order of stability is : Tertiary > Secondary > Primary > Methyl. This order of stability is due to
hyperconjugation effect of the alkyl groups.
Free radicals are planar chemical species. They are very short-lived, highly reactive and paramagnetic (due
to the presence of unpaired electron).
13.4 TYPES OF ORGANIC REACTIONS
Common types of organic reactions are (i) substitution (ii) addition (iii) elimination (iv) rearrangement (v)
condensation (vi) isomerisation (vii) pericyclic reactions and (viii) polymerisation.
13.4.1 Substitution reactions. These reactions involve direct replacement of atom or group in the organic molecule
by another atom or group without any change in the remaining part of the molecule. The product is called
substitution product and the atom or group which gets attached to the molecule is called substituent.
Substitution may be free radical, nucleophilic or electrophilic.
(i) Free radical substitution reactions. These are initiated by free radicals, e.g., halogenation of alkanes in
presence of heat or light.
hv or 
CH 4  Cl2  CH 3Cl  HCl
Methane Methyl chloride

The reaction proceeds by the following mechanism :

Initiation :

Propagation : ;

Termination :
(ii) Nucleophilic substitution reactions. These are brought about by nucleophiles, e.g., hydrolysis of alkyl
halides.
R  X  KOH  aq  
 R  OH  KX

(iii) Electrophilic substitution reactions are brought about by electrophiles, e.g., halogenation, nitration,
sulphonation and Friedel-Crafts reactions of aromatic compounds.
C6 H 6  Cl2 
Anh. AlCl3
 C 6 H 5 Cl  HCl

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13.4.2. Addition reactions. These reactions involve the addition of a reagent across a double or triple bond of an
unsaturated compound. There are three types of addition reactions:
(i) Free radical addition reactions. These reactions are initiated by free radicals.
CH 3CH  CH 2  HBr 
Peroxides
 CH 3CH 2CH 2 Br
Pr opene n  Pr opyl bromide

The reaction proceeds by the following mechanism :

Initiation : ;

Propagation :

Termination :
(ii) Electrophilic addition reactions. These reactions are initiated by electrophiles.
Br

CH 3CH  CH 2  HBr 

Absence of peroxides
CH3 – CH – CH3
Pr opene
Isopropyl bromide

(iii) Nucleophilic addition reactions. These are initiated by nucleophiles.

13.4.3 Elimination reactions. Elimination reaction involves the loss of two atoms or groups from the same or
adjacent atoms of a substance leading to the formation of a multiple (double or triple) bond. Depending on the
relative positions of the atoms or groups eliminated, these are divided into three types :
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(i) -Elimination reactions. In these reactions, the loss of two atoms or groups occurs from the same atom
of the substrate molecule. For example,

Dichlorocarbene is the reactive intermediate involved in carbylamine reaction and Reimer- Tiemann reaction.
(ii) -Elimination reactions. In these reactions, the loss of two atoms or groups occur from the adjacent
atoms of the substrate molecule. For example, acid-catalysed dehydration of alcohols and base-catalysed
dehydrohalogenation of alkyl halides :
(a) Acid-catalysed dehydration of alcohols :

 
Conc. H2SO4
H – CH2 – CH2 – OH CH2 = CH2 + H2O
Ethanol
 Ethene

(b) Base-catalysed dehydrohalogenation of alkyl halides :

(iii) -Elimination reactions. In these reactions, loss of two atoms or groups occur from -and -positions
(i.e., three bonds away) of the molecule leading to the formation of three-membered rings.

BrH 2 C  CH 2  CH 2 Br 
Zn dust

  ZnBr2
1,3 Dibromopropane Cyclopropane

This reaction, called Freund reaction is used in the synthesis of three-membered rings.
13.4.4 Condensation Reactions. These are reactions in which two or more molecules of the same or different
reactants combine to form product with or without the elimination of a small molecule like H2O, NH3, HCl,
etc. For example, in aldol condensation, two molecules of acetaldehyde, in presence of dilute alkali, condenses
to form 3-hydroxybutanal.

In the condensation of benzaldehyde with aniline to form benzylideneaniline, a molecule of water is eliminated.

13.4.5 Rearrangement Reactions. These are reactions involving migration of an atom or group from one atom to
another within the same molecule. For example, dehydration of 2, 2-dimethylpropan-1-ol to 2-methyl but-2-
ene occurs through 1, 2-migration of the methyl group.

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13.4.6 Isomerisation Reactions. These reactions involve the interconversion of one isomer to another, e. g.,
1-bromobutane isomerises to 2-bromobutane in presence of anhydrous aluminium chloride.

13.4.7 Pericyclic Reactions. These reactions take place in a single step via a cyclic transition state; bond breaking
and bond making occurs simultaneously. They do not involve ionic or free radical intermediates and do not
require any catalyst or initiator (such as heat or light).

13.4.8 Polymerisation Reactions. These are reactions which involve the union of large number of small molecules,
called monomers, to form a large molecule having very high molecular mass, called polymer.

There are two basic types of polymerisation, chain-reaction (or addition) and step-reaction (or condensation)
polymerisation.
ADDITIONAL INFORMATION
13.5.1 Nucleophilic substitution reaction (SN): Replacement (displacement) of an atom or group in molecule by
another atom or group is known as substitution reaction. If substitution reaction is brought about by a nucleophile,
it is known as nucleophilic substitution. Generally substitution takes place at sp3 hybridised carbon atom.
R  lg   R  Nu  lg 
Unimolecular nucleophilic substitution reaction (SN1): Nucleophilic substitution which involves two step
process.
(a) Step I : Slow step involves ionisation to form carbocation (lg = leaving group)
R  lg   R  +lg 
(b) Step II : Fast attack of nucleophile on carbocation to form product.
SN1 Reaction of Alkyl halide - Mechanism:

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Characteristics of SN1 reactions: i) It is unimolecular, two step process. ii) Carbocation intermediate is
formed, therefore, rearrangement is possible in SN1 reaction. iii) It is first order reaction.

Rate   Alkyl halide  Rate = k[(CH3)3 C – X]

Stereochemistry of SN1 reactions. In SN1 mechanism, the carbocation intermediate is sp2 hybridised and
planar, therefore, nucleophile can attack the carbocation from either side. If reactant is chiral, both enantiomers
of the product are formed. This is called racemisation.
Mechanism of racemisation:

13.5.2 Bimolecular nucleophilic substitution reaction (SN2). Nucleophilic substitution in which incoming group
replaces leaving group in one step. No intermediates are formed. This mode of attack of the nucleophile from
the back side causes inversion of configuration (walden inversion).

Characteristic of SN2. (i) It is biomolecular, one step process. (ii) It is second order reaction because in the
rate determining step two species are involved.
Rate  [alkyl halide] [nucleophile]  rate = k[alkyl halide] [nucleophile]
13.5.3 Elimination reactions. It is a type of reaction in which two substitutents from a molecule are removed in
either a one-step (E1) or two-step (E2) mechanism.
13.5.4 E1 reaction: Proton and leaving group depart in two different steps.
(i) Step 1 : Slow step - involves ionisation to form carbocation.
(ii) Step 2 : Abstraction of proton.

Step 1: ; Step 2 :

It is unimolecular, two step process (first order reaction). Reaction intermediate is carbocation, therefore,
rearrangement is possible. In the second step, a base abstracts a proton from the carbon adjacent to the
carbocation and forms alkene.
Rate  [Alkylhalide]  Rate = k [Alkylhalide]

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13.5.5 E2 Reaction: The one-step mechanism is known as E2 reaction.

E2 Reaction of Alkyl halide : Dehydrohalogenation is the elimination of a hydrogen and a halogen from an
alkyl halide to form an alkene. Dehydrohalogenation can take place by E1 and E2 mechanisms and is brought
about by:
(i) Hot alcoholic solution KOH or EtO– / EtOH (ii) NaNH2 (iii) t-BuO–K+ in t-BuOH

Characteristics of E2 reaction: i) This is a single step, bimolecular (second order) reaction.

Rate  [R – X] [Base]  Rate  k[R  X][B ]
Rearrangement is not possible. For the lower energy of activation, transition state must be stable. E2 follows
a concerted mechanism. The orientation of proton and leaving group should be antiperiplanar. Here –H is
eliminted by base, hence called  elimination.
Positional orientation of elimination In most E1 and E2 eliminations where there are two or more possible
elimination products, the product with the most highly substituted double bond will predominate. This rule is
called saytzeff or zaitsev rule.
13.5.6 Elimination Unimolecular conjugate Base or E1cB Reaction. In E1cB, H leaves first and then the X.
This is a two step process, the intermediate is a carbanion.

Step 1 : ; Step 2 :
The first step consists of the removal of a proton by the base, generating a carbanion. In second step, the
carbanion looses the leaving group to form alkene. For E1cB, the substrate must contain acidic hydrogens and
poor leaving groups.

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QUESTIONS
LEVEL - I

1. What type of isomers are 3-hydroxy propanol and 1-hydroxy propanone?

1) chain isomers 2) position isomers 3) functional group isomers 4) metamers
2. Which among the following pairs of compounds are chain isomers?
1) n-propylchloride and isopropylchloride 2) sec. butyl chloride and tert. butylchloride
3) n-butylchloride and sec. butylchloride 4) isobutylchloride and tert. butylchloride
3. Total number of isomers (including stereoisomers) having molecular formula C 4 H10 O is:
1) 4 2) 5 3) 7 4) 8
4. The dihedral angle of the most stable conformation of n-butane is:
1) 60o 2) 120o 3) 180o 4) 270o
5. Which among the following compounds will exhibit geometrical isomerism?
1) 1-Phenylbut-2-ene 2) 3-Phenylbut-1-ene 3) 2-Phenylbut-1-ene 4) 1, 1-Diphenyl prop-1-ene
6. Which of the following has a pyramidal shape?
1) CH 3 2) CH 3 3) CH3 4) H 2 C :
7. Electrophilic addition reactions proceed in two steps. The first step involves the addition of an electrophile.
Name the type of intermediate formed in the first step of the following addition reaction.

H 3C  HC  CH 2  H  
?
o o o o
1) 2 carbanion 2) 1 carbocation 3) 2 carbocation 4) 1 carbanion
8. Which of the following is a free radical substitution reaction?
1) Benzene  Me  Cl 
Anhyd.
AlCl3
 Ph  CH 3 2) Ph Cl  AgNO 2  Ph NO 2

Me Me OMe
H
3) Ph  CH 3  Cl 2  Ph
hv
Cl 4)  O  MeOH 
H H OH
9. The stability of a carbonium ion depends upon:
1) the bond angle of the attached group 2) the substrate with which it reacts
3) the inductive effect and hyper-conjugative effect of the attached group 4) None of the above
10. Most stable carbocation is:

CH2 CH2 CH2 CH 2

1) 2) 3) 4)

Cl NO2 OCH3

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CH2 CH2

11. is more stable than because

NO2
NO2
(I) (II)
1) 180o location of NO2 and CH2 in (I) 2) – NO2 operate both –I and –M in (I)
3) –I in (II) is weaker than (I) 4) Due to steric repulsion
12. Which one of the following is an intermediate in the reaction of benzene with CH3Cl in the presence of
anhydrous AlCl3?
+

1) Cl+ 2) Cl– 3) CH 3 4)

13. Most stable radical among the following is:

1) 2) 3) 4)

14. The order of decreasing reactivity towards an electrophilic reagent, for the following:
I. Benzene II. Toluene III. Chlorobenzene IV. Phenol
1) I > II > III > IV 2) II > IV > I > III 3) IV > III > II > I 4) IV > II > I > III
15. Consider the following carbanions

I) CH3O CH2 II) O2N CH2 III) CH2

Correct order of stability is

1) I > II > III 2) III > II > I 3) II > III > I 4) I > III > II
16. Which among the following statements are true with respect to electronic displacement in a covalent bond?
I. Inductive effect operates through  -bond II. Resonance effect operates through  -bond
III. Inductive effect operates through  -bond IV. Resonance effect operates through  -bond
V. Resonance and inductive effects operate through  -bond
1) III and IV 2) I and II 3) II and IV 4) I and III
17. The arrangement of  CH3 3 C,  CH3  2 CH , CH3CH 2  when attached to benzene or unsaturated group
in increasing order of inductive effect is:
1)  CH3 3 C  (CH3 ) 2 CH   CH3CH 2  2) CH3CH 2    CH3 2 CH    CH 3 3 C 
3)  CH 3 2 CH    CH3 3 C  CH3CH 2  4)  CH 3 3 C  CH3CH 2    CH3 2 CH 

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18. Among the following, the least stable resonance structure is:
O O
1) N 2) N

O O
O O
3) N 4) N
O O
19. An organic molecule has 5 C = C bonds (heat of hydrogenation for C = C bond is 28.8 kJ mol–1) and
experimental value for heat of hydrogenation is 99 kJ mol–1. The resonance energy in kJ mol–1 is :
1) 45 2) 90 3) 70 4) 140
20. Ionic species are stabilised by the dispersal of charge. Which of the following carboxylate ion is the most
stable?
O O O O
F
1) CH3 – C – O– 2) Cl – CH2 – C – O– 3) F – CH2 – C – O– 4) CH – C – O–
F
21. For 1-methoxy-1, 3-butadiene, which of the following resonating structures is the least stable?

1) H2C = CH– CH – CH – O – CH3 2) H2C = CH – CH – CH = O – CH3

3) H2C – CH – CH = CH – O – CH3 4) H2C – CH = CH – CH = O – CH3

22. Which of the following statements regarding resonance is not correct?
1) The different resonating structures of a molecule have fixed arrangement of atoms
2) The different resonating structures differ in the arrangement of electrons
3) All resonating structures are equally probable
4) The characteristics of a molecule exhibiting resonance cannot be explained on the basis of one resonating
structure.
23. Which of the following does not show electromeric effect?
1) Alkenes 2) Ethers 3) Aldehyde 4) Ketones
24. Which of the following alkenes will show maximum number of hyperconjugation forms?

1) 2) CH 3  CH  CH  CH 3

CH3 CH3
H CH3
C=C CH3 – C – CH = CH – C – CH3
3) 4)
H3C CH3 CH3
CH3

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25. In which of the following molecules all the effects namely inductive, mesomeric and hyperconjugation operate?

1) Cl 2) CH3 3) COCH3 4)

CH3 CH3

26. The stability of Me2C = CH2 is more than that of MeCH2CH = CH2 due to:
1) inductive effect of the Me group 2) resonance effect of the Me group
3) hyperconjugative effect of the Me group 4) resonance as well as inductive effect of the group
27. Which of the following compounds is the most likely to undergo a bimolecular nucleophilic substitution reaction
with aqueous NaOH?

CH3O Br
Br
Br CH3 O2N Br
1) 2) 3) 4)
NO2
OCH3
CH3
28. Which of the following cannot show SN1 reaction?
CH2X

X
1) X 2) 3) 4)
X

29. The conversion of 2,3 -Dibromobutane to 2-butene with Zn is:

1) Redox reaction 2) -Elimination

3) -Elimination 4) Both -elimination and redox reaction

For the reaction PhCH 2CH 3   the major product formed is:
NBS
30. CCl4 , heat

1) PhCHBrCH3 2) PhCH2CH2Br 3) PhCH = CH2 4) PhCHBrCH 2 Br

(racemic) (racemic)

31. Which of the following has maximum number of -hydrogen?

1) 2) 3) 4)

32. Which among the following is the strongest nucleophile?

1) NH2 2) CH3 3) OH 4) F

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33. Which will undergo fastest SN2 substitution reaction when treated with NaOH?

CH3 CH 3 CH 3 H
| | | |
1) H 5 C2  C  Br 2) CH 3  C  Br 3) H  C  Br 4) H  C  CH 2  CH 2  CH3
| | | |
H CH 3 C2 H 5 Br
34. Which of the following carbocations will not rearrange?

1) 2) 3) 4)

Et
Dil. H2SO4
35. The major product in the reaction is: Ph
Me
Et Et Et
Me OH Me
1) Ph 2) Ph 3) Ph 4) Both (2) and (3)
OH Me Me Me OH
LEVEL - II
1. The isomerism exhibited by the pair of compounds HCOOC2H5 and C2H5COOH is:
1) functional group isomerism 2) metamerism 3) tautomerism 4) position isomerism

2. Functional group isomers of allyl alcohol CH 2  CH  CH 2OH are:

O
O
(i) (ii) (iii) (iv) O

1) (i) and (iii) 2) (ii) and (iv) 3) (ii), (iii) and (iv) 4) (i), (ii), (iii) and (iv)
3. How many cyclic constitutional isomers have molecular formula C5H10?
1) 3 2) 4 3) 5 4) 6
4. Which among the following compounds will have the highest percentage of enol content in pure liquid state?

1) C6 H 5COCH 2COC6 H5 2) CH 3COCH 2 COOC2 H 5

3) CH 3COCH 2 COCH 3 4) CH 3COCH 3

5. Which among the following pair of compound represent metamers?

1) 1-methoxypropane and 2-methoxy propane 2) 1-methoxypropane and ethoxy ethane
3) propanal and propanone 4) butanone and but-2-en-2-ol

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6. In which of the following pairs A is more stable than B?

7. (CH 3)3CCH2COOH is more acidic than (CH3)3SiCH2COOH because:

1) Size of Si is more than that of carbon
2) Electronegativity of carbon is less than that of silicon
3) Silicon is more electropositive than carbon due to which (CH3)3SiCH2COO– becomes less stable
4) None of the above
8. Compare relative stability of following carbocation:

(i) ii) iii)

1) i > ii > iii 2) iii > i > ii 3) i > iii > ii 4) iii > ii > i
9. The stability of carbanions is in the order:
  
I. RC  C II. III. R 2 C  C H IV. R 3C  C H 2
1) I > II > III > IV 2) II > III > IV > I 3) IV > II > III > I 4) I > III > II > IV
10. The effective electrophile in aromatic sulphonation is:

1) HSO 4 2) SO2 3) +SO2 4) SO3

11. Which of the following explain why propene undergo electrophilic addition with HBr, but not with HCN?
1) Br– is better nucleophile than CN– 2) HBr being better source of proton as it is stronger
acid than HCN 3) HCN attacks preferentially via lone pair of nitrogen
4) The C–Br bond being stronger is formed easily as compared to C–CN bond
12. p-Chlorophenol is a stronger acid than phenol because:
1) Cl is less electronegative than oxygen atom
2) of the –I effect of a halogen, which is greater than its +R effect
3) of +R effect of Cl, which is stronger than its –I effect
4) of +R effect of Cl.

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13. Which order is true for resonance energy?

1) < 2) <
OH CH2
O

3) < 4) >
NO2
NO2 N N

H H
14. The stability of carbanions in the following:

(i) RC C (ii) (iii) R2C = CH (iv) R3C – CH2 in the order of :

1) (iv) > (ii) > (iii) > (i) 2) (i) > (iii) > (ii) > (iv) 3) (i) > (ii) > (iii) > (iv) 4) (ii) > (iii) > (iv) > (i)
15. Among the following structure the one which is not a resonating structure of other is:

16. In which of the following molecules, the resonance effect is not present?

1) NH2 2) NH3 3) OH 4) Cl

17. Which one of the following substitutents at para-position is most effective in stabilizing the phenoxide
O–
ion?

1) –CH3 2) –OCH3 3) –COCH3 4) –CH2OH

18. The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to:
1)    (empty) and    * electron delocalizations
2)    * and    electron delocalizations 3)    (filled) and    electron delocalizations
4)  (filled)   * and    * electron delocalizations

19. Consider the following compounds:

Hyperconjugation occurs in:

1) III only 2) I and II 3) I only 4) II only

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20. Rearrangement reactions are mainly shown by:

1) Carbanion 2) Free radical 3) Carbene 4) Carbocation
21. Arrange the following in correct order of acidic strength:
I) CH 3  NO 2 II) NO2  CH 2  NO 2 III) CH 3  CH 2  NO2 IV) NO2 – CH – NO2

NO2
1) IV > II > I > III 2) IV > II > III > I 3) III > I > II > IV 4) III > I > IV > II
22. Arrange the following compounds in the order of decreasing reactivity towards electrophilic substitution.
OH OCH3 CH3 Cl

(I) (II) (III) (IV) (V)

1) V > IV > III > II > I 2) I > II > III > V > IV 3) I > II > IV > III > V 4) I > III > IV > II > V
23. Give the stability order of the following compounds.
OCH3 OCH3 NH2 NH2

y H
H y H y H y
(I) (II) (III) (IV)
1) I > II > III > IV 2) IV > I > II > III 3) IV > I > III > II 4) IV > II > I > III
24. A mong the following free radical bromination reactions, select those in which 2o halide is the major product.
CH2 – CH3

Br2/hv Br2/hv Br2/hv

(P) (Q) (R)

Br2/hv Br2/hv Br2/hv

(S) (T) (U)

1) P, Q, R, S 2) P, R, U 3) P, R, S, T 4) P, Q, R, S, T
25. Which is the correct reaction coordinate diagram for the following solvolysis reaction?

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26. Arrange the following in decreasing order of reactivity towards EAS (electrophilic aromatic substitution).
CH3 CD3 CT3

(a) (b) (c)

1) a > b > c 2) c > b > a 3) a > c > b 4) c > a > b
CH3 CH2
alc. KOH
27. CH3 – C – X CH3 – C + HX . The reaction is:

CH3 CH3
1) SN1 reaction 2) SN2 reaction 3) E1 reaction 4) E2 reaction
28. Which of the following alkyl benzene cannot be oxidised to benzoic acid?
CH3
CH3 CH2CH3 H3C — CH — CH3
H3C — C — CH3

1) 2) 3) 4)

29. Examine the following statements regarding SN2 reaction:

1. The rate of reaction is independent of concentration of nucleophile
2. The nucleophile attacks the carbon atom on the side of molecule opposite to the group being displaced
3. The reaction proceeds with simultaneous bond formation and rupture
Which of the above written statements are correct?
1) 1, 2 2) 1, 3 3) 1, 2, 3 4) 2, 3
30 The substitution reaction among the following is:
O O
OH
1) + O O 2) C = O + NaHSO3 C
SO3Na
O O
Br
H3C H3C
3) H3C COH 
Dry HCl/ Anhy.ZnCl2
Lucas reagent
H3C CCl 4)  CH3 2 C  CH 2  BrCl 
  CH3 2 C  CH 2
H3C H3C
Cl
31. Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is:
1) CH3COCl 2) CH 3COOCH 3 3) CH 3CONH 2 4) CH 3COOCOCH 3

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32. An incorrect statement with respect to SN1 and SN2 mechanism for alkyl halide is:
1) A strong nucleophile in an aprotic solvent increases the rate or favours SN2 reaction
2) Competing reaction for an SN2 reaction is rearrangement
3) SN1 reaction can be catalysed by some Lewis acids
4) A weak nucleophile and a protic solvent increases the rate or favours SN1 reaction
33. The order of decreasing ease of abstraction of hydrogen atoms in the following molecule is:
Ha Me
Hb

Hc
1) Ha > Hb > Hc 2) Ha > Hc > Hb 3) Hb > Ha > Hc 4) Hc > Hb > Ha
34. In the following carbocation, H/CH3 that is most likely to migrate to the positively charged carbon is:

1) CH 3 at C4 2) H at C 4 3) CH 3 at C2 4) H at C2

35. If the carbocation rearranges to gain stability, it will rearrange to:

+
1) 2) 3) CH2 4)

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SOLUTIONS
LEVEL - I
1. 3 They are functional group isomers.
2. 2 CH3CH(Cl) CH2CH3 and (CH3)3C–Cl are chain isomers.
3. 4 C4H9OH has 4 structural isomers of which sec. butyl alcohol is chiral.
1-methoxy propane, 2-methoxy propane and ethoxy ethane.
4. 3 The ‘anti’ conformation is the most stable for n-butane.
5. 1 C6 H 5  CH 2  CH  CH  CH3 exhibit geometrical isomerism.
6. 3 Methyl carbanion is pyramidal

7. 3 H 3C  HC  CH 2  H  
 H 3C  HC   CH 3
2o carbocation
(morestable)

8. 3 Free radical substitution. In the presence of light, free radicals are generated.
9. 3
10. 3 EDG (+ M increases stability)
11. 2 Mesomeric effect is applicable at ortho and para position.

12. 3 C H 3 group acts as an intermediate in Friedal-Craft alkylation.

13. 3
14. 4 Benzene having any activating group, ie, OH, R etc. undergoes electrophilic substitution very easily as
compared to benzene itself. Chlorine with +E and +M effect deactivates the ring due to strong –I effect.
Correct order is phenol > toluene > benzene > chlorobenzene
IV II I III
15. 3 –NO2 group shows –M effect while CH3O – group shows +M effect (–M effect stabilises an anion).
16. 1 Inductive effect is a permanent effect operates on  -electrons while resonance involves delocalisation
of  -electrons.
17. 1 The inductive effect of the alkyl group is attached to an unsaturated system such as double bond or a
benzene ring, the order of inductive effect is actually reversed. This effect is called hyperconjugation
effect or Baker-Nathan effect. Now, the order is  CH 3 3 C   CH 3  2 CH   CH3  CH 2   CH 3 

18. 1 Two positive charges present at the adjacent place, elevates the energy, thus lowers the stability most.
19. 1 Resonance energy = ( Hydrogenation) Theoretical – Experimental
20. 4 Factors which decrease the –ve charge, increase the stability of anion.

21. 1 The structure is expected to be least stable.

22. 3

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23. 2 Simple ethers do not contain a multiple bond, they do not show electromeric effect.
24. 1 More is the number, of H, more is hyperconjugative structures.
O (–m)
25. 3 C — CH3

(H. C. > + I)
Me MeCH2 H
26. 3 C = C has 6-hyperconjugate forms; C=C has 2-hyperconjugate forms.
Me H H

O2N Br
27. 4 is activated for nucleophilic substitution by electron withdrawing group present at
NO2
ortho-position.
28. 3 Substrates that give stable carbocation generally undergo SN1 reaction.
Br Br
Zn / CH 3OH
CH 3 – C – C – CH 3 CH 3 – CH = CH – CH 3
29. 3 
H H
30. 1 It is benzylic bromination.

31. 1
32. 2 CH3 is the strongest nucelophile compared to others.
33. 4 Primary alkyl halides undergo bimolecular nucleophilic substitution most readily.
34. 3 The rearrangement of a carbocation is possible only if it can form a more stable carbocation.

35. 1 The product is obtained by methyl shift.

LEVEL - II
1. 1 Ethyl ethanoate and propanoic acid are functional group isomers.
2. 4 All are functional group isomers of allyl alcohol C3H6O.

3. 3 & have molecular formula C5H10.

4. 1 C6H5COCH2COC6H6 exist in almost 100% enol form

5. 2 Methyl propyl ether and diethyl ether are metamers.
6. 4 Stability of radicals can be compared on the basis of electronic effects and angle strain.

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only +I
CH3 CH3
effect
CH3 — C — CH2 — C — OH CH3 — Si — CH2 — C — OH
7. 3
CH3 O CH3 O
8. 3
9. 1 The stability order of carbanions is:
RC  C  C 6 H 5  R 2 C  CH  R 3C  CH 2
10. 4 SO3 is Electrophile of aromaticsulphonation.
11. 2 HBr is a better source of proton. It gives H+ and Br– ions.
12. 2 Electron-withdrawing groups that stabilize the phenoxide ion increases the acid strength of phenols.
13. 4 I structure is aromatic in all these examples and II structure is not aromatic and resonance energy of
atomatic compound is higher than non-aromatic compound.
14. 3 The overall stability decreases in the order : (i) > (ii) > (iii) > (iv).
15. 4 Resonating structure involve movement of electrons, but they do not involve movement of atom or

group of atoms.


16. 2 Resonance effect is not present in C6 H 5 N H 3

17. 3 The substituent –COCH3 is electron withdrawing.

18. 1 In tert-butyl cation, the carbon bearing positive charge has an empty p-orbital and hence hyperconjugation
here involves  -p (empty) electron delocalisation.

In 2-butene,  -bond contains two electrons, one electron from each of the two carbon atoms, ie,
 -MO contains a pair of electrons, but  * MO is empty and hence electrons of the  C – H bond can
move only into the  * orbital.
19. 1 Hyperconjugation in free radicals occurs only when the  -carbon to the carbon atom carrying the odd
electron has a hydrogen atom.
20. 4 Only carbocation shows rearrangement (but in some cases radical also rearranges).
21. 1 More is number of EWG
22. 2 Rate of electrophilic substitution reaction  Stability of arenium ion.
23. 3 More is +I effect more is stability.

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Br Br
24. 2 In ‘Q’ it is . In ‘S’ it is and in ‘T’ it is 3o halide is major..

25. 2  Highly endothermic so has high G of activation.

3o carbocation is stabilised by H2O. This SN1 reaction has three transition states and two intermediates.
26. 1
27. 3
28. 4
29. 4
H3C H3C
30. 3 H3C COH 
Dry HCl/ Anhy.ZnCl2
Lucas reagent
H3C CCl is nucleophilic substitution reaction.
H3C H3C

31. 1 Cl– is a weakest base, therefore good leaving group.

32. 2 Rearrangement occur in SN1 reaction and not in SN2 reaction.
33. 2 The ease of abstraction of different H atoms depends upon the stability of free radicals formed after the
abstraction of the H atom. The relative stability of the free radicals formed after the abstraction of Ha,
Hc and Hb atoms respectively follows the order:

34. 4 Here the driving force is conjugation with oxygen.

35. 3

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CHAPTER - 14
ORGANIC CHEMISTRY: SOME BASIC PRINCIPLES - III
PURIFICATION AND CHARACTERISATION OF ORGANIC COMPOUNDS

INTRODUCTION
Organic compounds either obtained from natural sources or synthesised in the laboratory may not be pure,
therefore, it is essential to purify them. Various methods used for the purification of organic compounds are
based on the nature of the organic compound and the impurity present in it. The various techniques are
sublimation, crystallisation, distillation, differential extraction, and chromatography.
Most of the pure organic compounds have sharp melting and boiling points, therefore, the purity of orgainc
compounds is finally checked by determining its melting or boiling point. Nowadays, chromatographic and
spectroscopic techniques are also used to ascertain the purity of organic compounds.
14.1 METHODS FOR PURIFICATION OF SOLIDS
a. Sublimation. Some substances change from solid to vapour state without passing through the liquid state on
heating. This property is useful for separating sublimable compounds from non-sublimable impurities.
The impure substance is heated in a dish covered with a perforated asbestos sheet on which an inverted funnel
is placed (Fig.01a). The funnel is kept cool by wrapping it with a wet filter paper or wet cloth. Vapours of the
sublimable substance condense on the cooler walls of the funnel.

(a) (b)

Figure - 01 (a) Sublimation (b) Hot water funnel.

b. Crystallisation. This method is based on the difference in the solubilities of the compound and the impurities
in a suitable solvent. The impure compound is dissolved in a solvent in which it is sparingly soluble at room
temperature but appreciably soluble at a higher temperature. The solution is heated to get a saturated solution.
On cooling the solution, pure compound crystallises out and is removed by filtration (Fig.01b) or filtration under
suction. E.g., benzoic acid is purified by this method.
If the compound is highly soluble in one solvent and sparingly soluble in another solvent, crystallisation can be
carried out in a mixture of these solvents.
A compound containing impurities of comparable solubilities can be purified by repeated crystallisation.
c. Adsorption. Coloured impurities are removed by adsorption over-activated charcoal.
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14.2 METHODS FOR PURIFICATION OF LIQUIDS

a. Simple distillation. This method is used to separate volatile liquids from non-volatile impurities having
sufficient difference in their boiling points.
Liquids having different boiling points vaporise at different temperatures. The vapours are cooled and liquids so
formed are collected separately. For example, chloroform (b.p. 60oC) and aniline (b.p.189oC) can be separated
by simple distillation due to large difference in their boiling points.
The liquid mixture of aniline and chloroform is taken in a round-bottom (R.B.) flask and heated. The vapours of
CHCl3 (low-boiling liquid) are formed first, which are condensed by using a condenser and liquid CHCl3 is
collected in a conical flask (Fig.02a). The vapours of aniline (high-boiling liquid) are formed at 189oC and are
collected separately.
b. Fractional distillation. This method is used to separate two liquids when the difference in their boiling
points is small. The vapours of such liquids are formed within the same temperature range and are condensed
simultaneously. In this method, vapours of the liquid mixture are passed through a fractionating column, fitted to
the mouth of the R.B. flask (Fig.02b).

(a) (b) (c)

Figure - 02 (a) Simple distillation (b) Fractional distillation, (c) Different types of fractionating columns.

Fractionating columns are of various sizes and designs (Fig.02c). They provide many surfaces for heat exchange
between the ascending vapours and the descending condensed liquid. The vapours of lower boiling fraction
reach the top of the column first followed by the vapours of higher boiling fractions.

The vapours of the liquid with higher boiling point condense before the vapours of the liquid with lower boiling
point. Thus, the vapours rising in the fractionating column become richer in more volatile liquid. By the time the
vapours reach the top of the fractionating column (F.C.), these are rich in the more volatile liquid. In the F.C.,
some of the condensed liquid gets heat from the ascending vapours and revaporises.

The vapours thus become richer in low-boiling component. The vapours of the low-boiling liquid ascend to the
top of the F.C., and become pure and pass through the condenser and are collected in a receiver. After a
number of successive distillations, the remaining liquid in the distillation flask becomes richer in high boiling
liquid. (Each sucessive condensation and vaporisation unit in the F.C. is called a theoretical plate.)

This technique is widely used to separate different fractions of crude oil in petroleum refineries.

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(a) (b)

Figure - 03 (a) Distillation under reduced pressure (b) Steam distillation.

c. Distillation under reduced pressure. This technique is used to purify liquids having very high boiling points
and those which decompose at or below their boiling points. Such liquids are made to boil at a temperature
below their normal boiling point by reducing the pressure on their surface. A water pump or vacuum pump is
used to reduce the pressure (Fig.03a). This method is widely used in the separation of glycerol from spent lye
in soap manufacture.
d. Steam distillation. This method is used to separate compounds which are steam volatile and insoluble in
water. Steam from a steam generator is passed through a heated flask containing the liquid to be distilled (Fig.
03b). The liquid boils when the sum of the vapour pressure of liquid (p1) and that of steam (p2) becomes equal
to the atmospheric pressure, p = p1+ p2.
Since p1 is lower than p, the liquid vaporises at a temperature lower than its boiling point (the principle
of this method is similar to that of distillation under reduced pressure).
The mixture of steam and the volatile organic compound is condensed and collected in a receiver. Water and the
water-insoluble substance will form separate layers which can be separated using a separating funnel (Fig. 04).
E.g., aniline can be separated from aniline-water mixture by this technique.

Figure - 04 Separating funnel.

14.3 Differential extraction. When an organic compound is present as an aqueous solution, it can be separated
by shaking it with an organic solvent in which it is more soluble than in water. The organic solvent and aqueous
solution should be immiscible so that they can form separate layers in a separatory funnel. The solution
layer containing the organic compound is separated (using a separating funnel) and collected and the solvent is
removed by distillation or evaporation to get the compound. E.g., benzoic acid is extracted from its water
solution using benzene.

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If the organic compound is less soluble in the organic solvent, a very large quantity of the solvent would be
required to extract a small quantity of the compound. In such cases, the technique of continuous extraction is
employed, where the same solvent is repeatedly used for extraction of the compound.
14.4 Chromatography. Chromatography is a technique used to separate mixtures into their components, purify
compounds and to test the purity of compounds. The method was first used to separate coloured substances
obtained from plants by Tswett, a Russian botanist in 1906. It is the technique of separating the components
of a mixture based on the differential movement of individual components through a stationary phase
under the influence of a mobile phase.
(i) Stationary phase. The stationary phase may be a solid or a liquid, e.g., alumina, silica gel, water held on an
inert material, etc.
(ii) Mobile phase. The mobile phase may be a pure solvent, a mixture of solvents or a gas which is allowed to
move slowly over the stationary phase.
Based on the principle involved, chromatography is classified into different categories. Two of these are
(i) Adsorption chromatography, and (ii) Partition chromatography.
14.4.1 Adsorption Chromatography. Adsorption chromatography is based on the fact that different compounds
are adsorbed on an adsorbent to different degrees. Commonly used adsorbents are silica gel and alumina.
When a mobile phase is allowed to move over a stationary phase (adsorbent), the components of the mixture
move by varying distances over the stationary phase. The following are two main types of chromatographic
techniques based on the principle of differential adsorption:
(a) Column Chromatography. It involves separation of a mixture over a column of adsorbent (stationary
phase) packed in a glass tube. The glass tube is fitted with a stopcock at its lower end.
The mixture is placed at the top of the adsorbent column packed inside a glass tube. A suitable solvent (called
eluant) is allowed to flow slowly down the column. The mixture is separated out depending on the degree of
adsorption. The most readily adsorbed constituents are retained near the top, whereas others come down to
various distances in the column. The eluant is run continuously and each component is collected separately. The
solvent is removed by evaporation to get the compound.

Figure - 05 Column chromatography (different stages of separation of components of a mixture)

(b) Thin layer chromatography (TLC). It is another type of adsorption chromatography in which a mixture
of substances is separated over a thin layer (0.2 mm thick) of an adsorbent (silica gel or alumina) which is
spread over a glass plate (called chromatography plate or chromaplate). The solution of the mixture to be
separated is placed as a small spot at about 2 cm above one end (base line) of the TLC plate.
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The glass plate (chromaplate) is then placed in a closed jar containing the solvent (eluant). With the rise of the
solvent, the constituents of the mixture move up along with the solvent to different distances depending on their
differential adsorption, and separation occurs. The relative adsorption of each constituent of the mixture is
expressed in terms of its retardation factor, i.e., Rf value.
Distance moved by thesubstance from base line (x) x
Rf  or Rf  .
Distance moved by thesolvent from base line (y) y
The Rf value of a substance depends on the: (i) nature of the substance, (ii) nature of the solvent,(iii) nature of the
filter paper used, and (iv) temperature.
The spots of coloured components are visible on the TLC plate. Spots of colourless components can be made
visible by (i) using uv light (e.g., uv light of 254 nm causes flourescence), (ii) action of iodine vapours (spots
absorb iodine vapours and become brown) or (iii) chemical methods (e.g., amino acids can be detected by
spraying ninhydrin and aldehydes and ketones by 2,4-dinitrophenylhydrazine ).

Figure - 06 Thin Layer Chromatography (TLC) experimental setup and chromaplate

14.4.2 Partition chromatography. It is based on continuous differential partitioning (distribution) of components
of a mixture between stationary and mobile phases.
(a) Paper chromatography. Paper chromatography is the type of partition chromtograpy in which a special
quality paper, called chromatographic paper is used. The paper can hold water (trapped in it), which acts as the
stationary phase.
In ascending paper chromatography (solvent is kept at the bottom), the solution of the mixture is spotted at the
base of a strip of chromatography paper and suspended in a suitable solvent or a mixture of solvents. This
solvent, which acts as mobile phase, rises up the paper by capillary action and flows over the spot. Different
constituents are separated according to their differing partitions in the two phases. The paper strip so developed
is called chromatogram. The spots of different coloured compounds are visible at different distances. The
spots of the separated colourless compounds may be observed under uv light or by using appropriate reagent
spray to produce different colours.

Figure - 07 Paper Chromatography (chromatographic paper in two different shapes)

This type of chromatography is mainly used for the separation of amino acids and sugars. It is further classified
into (i) radial chromatography and (ii) ascending and descending paper chromatography.
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In descending chromtography, the solvent is kept in a trough at the top of the chamber with the spotted end of
the paper dipped in it nd the solvent is allowed to flow downwards.
(b) Gas chromatography. It is a chromatographic technique using gas as the mobile phase. If the stationary
phase is a solid, it is called gas solid chromatography (GSC). If the stationary phase is a high-boiling liquid
deposited on a suitable support, it is called gas liquid chromatography (GLC) and is widely used.
14.5 QUALITATIVE ANALYSIS OF ORGANIC COMPOUNDS
The elements present in organic compounds are carbon and hydrogen. They may also contain oxygen, nitrogen,
halogens, sulphur and phosphorous.
14.5.1 Detection of C and H. Many organic compounds burn with a sooty flame or char when strongly heated. C
and H are detected by heating the compound with CuO (cupric oxide) in a dry test tube. They are oxidised to
CO2 and H2O, respectively. CO2 turns lime water milky and H2O turns anhydrous CuSO4 to hydrated CuSO4
which is blue in colour.
 
i. C  2CuO   2Cu  CO 2 ; ii. 2H  CuO   Cu  H 2 O

iii. CO 2  Ca(OH) 2 
 CaCO3   H 2 O ; iv. CuSO 4  5H 2O 
 CuSO 4 .5H 2 O
White Blue

14.5.2 Detection of other Elements

Lassaigne’s test. Nitrogen, sulphur, and halogens present in organic compounds can be detected by this test.
The given organic compound is fused with metallic sodium, which converts the elements present in the compound
from covalent to ionic form.


Na  C  N   NaCN ; 2Na  S   Na 2S

Na  X   NaX  X  Cl, Br, or I 
NaCN, Na2S, and NaX are extracted from the fused mass by boiling it with disilled water. This extract is
known as sodium fusion extract or Lassaigne’s extract (L.E.).
a. Test for nitrogen. The sodium fusion extract or L.E. is boiled with FeSO4 and then acidified with con.
H2SO4. The appearance of Prussian blue colour confirms the presence of N (due to the formation of iron(III)
hexacyanoferrate or ferriferrocyanide, Fe4III[FeII(CN)6]3).

  Fe(CN) 6 
4
Fe 2  6CN  
ferrocyanideion
4 3
3 [Fe(CN) 6 ]  4Fe 
xH 2 O
 Fe 4 [Fe(CN) 6 ]3 .xH 2 O
Iron(III)hexacyanoferrate(II)
(Prussian blue colour)

b. Test for sulphur. The sodium fusion extract or L.E. is treated with sodium nitroprusside. The appearance of
violet colour indicates the presence of sulphur.
Na 2S  Na 2 [Fe(CN)5 NO] 
 Na 4 [Fe(CN)5 NOS]
Sodium nitroprusside (Violet colour )

The L.E. is acidified with acetic acid and lead acetate is added to it. A black precipitate of lead sulphide
indicates the presence of sulphur.
Na 2S  (CH3COO) 2 Pb 
 PbS  2CH3COONa
(Black ppt.)

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c. Test for halogens

i. Organic layer test. L.E. is mixed with CS2 or CCl2 and then with Cl2 water or KMnO4, shaken well and
keep for some time. The appearance of orange colour in the organic layer confirms the presence of Br and
violet colour confirms the presence of iodine.
2Br   Cl 2 
 Br2  2Cl ; 2I   Cl 2 
 I 2  2Cl 
16H   2MnO 4   10Br  
 5Br2  2Mn 2  8H 2 O
16H   2MnO 4   10I   5I 2  2Mn 2   8H 2O
ii. Beilstein test. The organic compound is heated on a clean copper wire in Bunsen flame. A green or blue
colour due to the formation of volatile copper halides indicate the presence of halogens.
iii. L.E. is acidified with HNO3 and then treated with AgNO3. A white precipitate soluble in NH4OH indicates
the presence of Cl, a yellowish precipitate sparingly soluble in NH4OH indicates the presence of Br, and a
yellow precipitate insoluble in NH4OH shows the presence of I.
X   Ag    AgX [X=Cl, Br or I]
d. Test for phosphorous. The organic compound is heated with an oxidising agent (sodium peroxide).
Phosphorus is oxidised to phosphate. The solution is then boiled with conc. HNO3 and treated with ammonium
molybdate. A yellow precipitate confirms the presence of phosphorous.
Na 3 PO 4  3HNO3 
 H 3 PO 4  3NaNO3
H 3PO 4  12  NH 4 2 MoO4  21HNO3 
  NH 4 3 PO .12MoO
4 3  21NH 4 NO3  12 H 2O
Ammonium molybdate Ammonium phosphomolybdate(Yellowppt.)

14.6 QUANTITATIVE ANALYSIS

a. Estimation of C and H. The percentage composition of elements present in an organic compound is
determined by the following methods.
Liebig’s combustion method. A known mass of the compound is heated with CuO. Carbon is oxidised to
CO2 and hydrogen is oxidised to H2O.
 
C  2CuO   CO 2  2Cu ; 2H  CuO   H 2 O  Cu

or Cx H y  (x  y / 4)O2 
 xCO 2  (y / 2)H 2 O
Water and carbon dioxide are absorbed in anydrous calcium chloride and potassium hydroxide solutions
contained in U -tubes. The increase in mass of the tubes give the masses of CO2 and H2O formed.
12 Mass of CO 2
Percentage of C   100
44 Mass of compound

2 Mass of H 2 O
Percentage of H    100
18 Mass of compound

Figure - 08 Estimation of carbon and hydrogen

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b. Estimation of halogens
(i) Carius method. A known mass of compound is heated with conc.HNO3 in the presence of AgNO3 contained
in a hard glass tube known as Carius tube (Fig.09) in a furnace. C and H are oxidised to CO2 and H2O. The
halogen forms the corresponding AgX which is filtered, dried, and weighed.
Atomic mass of X Mass of AgX
i. Percentage of X   100
Molecular mass of AgX Mass of compound

35.5 Mass of AgCl

ii. Percentage of Cl   100
143.5 Mass of compound

80 Mass of AgBr
iii. Percentage of Br   100
188 Mass of compound

127 Mass of AgI

iv. Percentage of I    100
235 Mass of compound

Figure - 09 Carius method

c. Estimation of sulphur (Carius method). A known mass of compound is heated with fuming HNO3 or
sodium peroxide (Na2O2) in the presence of BaCl2 solution in a Carius tube. Sulphur is oxidised to H2SO4 and
precipitated as BaSO4. It is filtered, dried, and weighed.
Atomicmassof S Massof BaSO4 32 Mass of BaSO 4  100
Percentage of S   100  
Molecular massof BaSO4 Massof compound 233 Mass of compound
d. Estimation of phosphorous (Carius method):
Method -I. A known mass of compound is heated with fuming HNO3 in Carius tube which converts
phosphorous to phosphoric acid. It is precipitated as ammonium phospho molybdate [(NH4)3PO4. 12MoO3]
by adding NH3 and ammonium molybdate [(NH4)2MoO4]. It is filtered, dried, and weighed.
Atomic mass of P Mass of Amm.phosphomolybdate
Percentage of P    100
Mol.mass of Amm. phospho molybdate Mass of compound

31 Mass of  NH 4 3 .PO 4 .12MoO3

   100
1877 Mass of compound
Method II. A known mass of compound is heated with fuming HNO3 or sodium peroxide (Na2O2) which
converts phosphorous to H3PO4. Magnesia mixture (MgCl2+NH4Cl) is then added, which gives the precipitate
of magnesium ammonium phosphate (MgNH4.PO4) which on heating gives magnesium pyrophosphate
(Mg2P2O7), which is weighed.
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MgCl2  NH 4 Cl  H 3 PO 4 
 MgNH 4 PO 4  3HCl

2MgNH 4 PO 4   Mg 2 P2 O 7  2NH 3  H 2 O

Atomic mass of P Mass of Mg 2 P2 O7

Percentage of P   100
Molecular mass of Mg 2 P2O7 Mass of compound
62 Mass of Mg2 P2O7
  100
222 Mass of compound
e. Estimation of nitrogen. There are two methods for the estimation of nitrogen: (i) Dumas method and
(ii) Kjeldahl’s method.
i. Dumas method . A known mass of compound is heated with copper oxide (CuO) in an atmosphere of CO2,
which gives free nitrogen along with CO2 and H2O.
Cx H y N z   2x  y / 2  CuO 
 xCO 2  y / 2(H 2O)  z / 2  N 2    2x  y / 2  Cu
The gaseous mixture is passed over a heated copper gauze which converts traces of nitrogen oxides formed to
N2. The gaseous mixture is collected over an aqueous solution of KOH which absorbs CO2, and nitrogen is
collected in the upper part of the graduated tube (Fig.11).
Let the volume of N2 collected be V1 ml
p1V1  273
Volume of N2 at STP   Vml
760  T1
where p1 and V1 are the pressure and volume of N2, and p1 = atmospheric pressure – aqueous tension.
22400 ml of N2 at STP weights 28 gm
28  V
V ml of N2 at STP weighs  gm
22400
Molecular mass of N 2 Volume of N 2 at STP
Percentage of N =   100
22400 ml Mass of compound
28 Vml
   100
22400 Mass of compound

Figure - 10 Dumas method

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ii. Kjeldahl’s method. A known mass of an organic compound (0.5 gm) is heated with K2SO4 (10 gm),
CuSO4 (1.0 gm) or a drop of mercury and conc. H2SO4 (25 ml), in Kjeldahl’s flask. CuSO4 or Hg acts as
catalyst, while K2SO4 raises the boiling point of H2SO4 (Fig.12). The nitrogen in the organic compound is
quantitatively converted to ammonium sulphate. The resulting mixture is then distilled with excess of NaOH
solution and the NH3 evolved is passed into a known excess volume of standard HCl or H2SO4. The acid left
unused is estimated by titration with standard alkali. The amount of acid used against NH3 can thus be known
and from this the percentage of nitrogen is calculated.
C  H  S 
Conc.
H 2SO 4
 CO 2  H 2O  SO 2

N 
Conc.
H 2SO4
  NH 4  2 SO 4

 NH4 2 SO4  2NaOH 

 Na 2SO4  2NH 3  2H 2O

  NH 4  2 SO4
2NH3  H 2SO4 
Calculation of percentage of N
Let the mass of organic compound be m gm.
Volume of H2SO4 of molarity M [or (2M) normality] taken = V ml
Volume of NaOH of molarity M (or M normality) used for titration of excess H2SO4 = V1 ml

Fig. 11 Kjeldahl’s method

V1mL of NaOH of molarity M = V1/2 mL of H2SO4 of molarity M
Volume of H2SO4 of molarity M unused = (V-V1/2) mL.
(V-V1/2) mL of H2SO4 of molarity M = 2(V-V1/2) mL of NH3 solution of molarity M.
1000 mL of 1 M NH3 solution contains 17 g NH3 or 14 g of N
2(V-V1/2) mL of NH3 solution of molarity M contains:
14  M  2  V  V1 / 2 
= gN
1000
14  M  2  V  V1 / 2  100 1.4  M  2(V  V1 / 2)
Percentage of N   
1000 m m
This method is not applicable to compounds containing N in nitro and azo groups, and N present in the ring
(e.g., pyridine) as N of these compounds does not change to (NH4)2SO4 (ammonium sulphate) under reaction
conditions.

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f. Estimation of oxygen. It is usually found by the difference between the total percentage composition (100)
and sum of the percentages of all the other elements.
Percentage of O = 100 – (Percentage of C + Percentage of H + Percentage of N)
i) Aluise’s method. A known mass of compound is decomposed by heating in presence of N2 gas. The mixture
of gaseous products containing O2 is passed over red hot coke when all O2 is converted to CO. This mixture is
heated with I2O5 (iodine pentoxide) when CO is oxidised to CO2 liberating I2.
Organic compound 

 Other gaseous products + O2
[2C + O 2 
1373K
 2CO]  5
[I 2 O5  5CO 
 5CO 2  I 2 ]  2
10CO 2  2I 2 or 5O 2  10CO 2  2I 2 or O  CO 2  (1/ 5)I 2
10C  5O 2  2I 2 O5 
Each mole of oxygen liberated from the compound will produce two moles of CO2. Thus, 88g carbon dioxide
is obtained if 32 g oxygen is liberated.
32 Mass of CO 2
Percentage of O  88  Mass of compound  100
ii) The percentage of oxygen can also be derived from the amount of iodine produced.
16 5 Mass of I 2 40 Mass of I 2
Percentage of O  127  2 Mass of compound  100  127 Mass of compound
14.7. CHN elemental analyser. Nowadays, the estimation of elements in an organic compound is carried out with
automatic experimental techniques using micro quantities of the compound. The elements C, H, and N present
in an organic compound are determined by an automatic instrument called CHN elemental analyser using a very
small amount of the compound (1–3 mg), which displays the result within a very short time.

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QUESTIONS
LEVEL - I
1. Benzoic acid is soluble in hot water but sparingly soluble in cold water. An impure sample of benzoic acid can
be purified by:
1) distillation 2) steam distillation 3) crystallisation 4) solvent extraction
2. A mixture of iodine and sodium chloride can be easily separated by:
1) fractional distillation 2) steam distillation 3) chromatography 4) sublimation
3. The distillation technique most suited for separating glycerol from spent-lye in the soap industry is:
1) simple distillation 2) fractional distillation
3) steam distillation 4) distillation under reduced pressure
4. Steam distillation is based on the fact that vaporisation of organic liquid takes place at:
1) lower temperature than its boiling point 2) higher temperature than its boiling point
3) its boiling point 4) water and organic liquid both undergo distillation
5. In steam distillation of toluene, the pressure of toluene in vapour is:
1) equal to pressure of barometer 2) less than pressure of barometer
3) equal to vapour pressure of toluene in simple distillation
4) more than the vapour pressure of toluene in simple distillation.
6. Which of the following can be purified from nonvolatile impurities by steam distillation?
1) benzamide 2) benzoic acid 3) aniline 4) cinnamic acid
7. Some organic compounds are purified by distillation at low pressure because the compounds are:
1) low boiling liquids 2) high boiling liquids
3) highly volatile 4) dissociated before reaching their boiling points.
8. A mixture of benzene (b.p. 80oC) and toluene (b.p. 111oC) are best separated into pure compounds by:
1) steam distillation 2) simple distillation
3) fractional distillation 4) differential extraction
9. The separation of the constituents of a mixture by column chromatography depends upon their:
1) different solubilities 2) different boiling points
3) different refractive indices 4) differential adsorption
10. In paper chromatography:
1) mobile phase is liquid and stationary phase is solid
2) mobile phase is solid and stationary phase is liquid
3) both phases are solids
4) both phases are liquids
11. A compound migrates 7.6 cm from its point of application on a thin layer chromatographic plate where as
during the same time the solvent migrates 16.2 cm from point of application. In another similar experiment on
relevant plate if the solvent migrates 14.3 cm beyond the point of sample application. The distance moved by
the same compound from the origin will be:
1) 7.6 cm 2) 6.7 cm 3) 4.5 cm 4) 5.8 cm
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12. Which among the following is not correctly matched with their colour?
Compound Colour

1) Na 4  Fe  CN 5 NOS purple

2) Fe4  Fe  CN 6  blue
3

3) Fe  CNS3 blood red

4) AgCl light yellow
13. When thiourea is heated with metallic sodium, the compound which can’t be formed is:
1) NaCNS 2) NaCN 3) Na2SO4 4) Na2S
14. The Lassaigne’s extract is boiled with con. HNO3 while testing for halogens. By doing so it:
1) decomposes Na2S and NaCN, if formed 2) helps in the precipitation of AgCl
3) increases the solubility product of AgCl 4) increase the concentration of NO 3 ions
15. The presence of halogen, in an organic compound, is detected by:
1) iodoform test 2) silver nitrate test 3) Beilstein’s test 4) Millon’s test
16. Sodium fusion extract of an organic compound gives a blood red colouration with few drops of FeCl3 solution.
This indicates the presence of :
1) nitrogen 2) sulphur
3) both nitrogen and sulphur 4) both sulphur and chlorine
17. The sodium extract on acidification with acetic acid and then adding lead acetate solution gives a black precipitate.
The organic compound contains:
1) nitrogen 2) halogen 3) sulphur 4) phosphorous
18. An organic compound is fused with fusion mixture and extracted with HNO3. The extract gives yellow precipitate
with ammonium molybdate. It shows the presence of which element?
1) P 2) Br 3) S 4) both Br and P
19. Sodium nitroprusside reacts with sulphide ion to give a purple colour due to the formation of:
3  4 3
1)  Fe  CN 5 NO  2)  Fe  NO 5 CN  3)  Fe  CN 5 NOS 4)  Fe  CN 5 NOS

20. In Dumas method for the estimation of nitrogen in an organic compound, nitrogen is determined in the form of:
1) Gaseous nitrogen 2) Sodium cyanide 3) Ammonium sulphate 4) Gaseous ammonia
21. In Kjeldahl’s method, 29.5 mg of an organic compound containing nitrogen was digested and the evolved
ammonia was absorbed in 20 mL of 0.1 N HCl solution. The excess of the acid required 15 mL of 0.1 N
NaOH solution for complete neutralization. The percentage of nitrogen in the compound is:
1) 29.5 2) 59.0 3) 47.4 4) 23.7
22. An organic compound containing sulphur is estimated by Carius method in which fuming HNO3 is used to
convert S into:
1) SO32  2) SO 24  3) SO3 4) SO2
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23. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The
percentage of bromine in the compound is (at mass Ag = 108; Br = 80)
1) 36 2) 48 3) 60 4) 24
24. Analysis of organic compound (0.36 g) containing phosphorus gave 0.66 g of Mg2P2O7 when treated with
concentrated nitric acid followed by magnesia mixture. Calculate the amount of phosphorus present in the
compound.
1) 51.20% 2) 61.20% 3) 73.5% 4) 68.3%
25. Tyrosine is one of the amino acids present in protein. Its content in protein is 0.22% and its molecular weight
is 181 g mol–1. Lowest molecular mass of protein is:
1) 18100 2) 2200 3) 82273 4) 18132
26. 15 mL of a gaseous hydrocarbon required 45 mL of oxygen for complete combustion. 30 mL of CO2 is
formed. The formula of hydrocarbon is:
1) C2H6 2) C2H4 3) C3H6 4) C2H2
27. 0.22 g of organic compound CxHyO which occupied 112 mL at NTP and on combustion gave 0.44 g CO2. If
the percentage of oxygen is 36.45%, then the ratio of x to y in the compound is:
1) 1 : 1 2) 1 : 2 3) 1 : 3 4) 1 : 4
28. 4g of hydrocarbon on complete combustion gave 12.571g of CO2 and 5.143g of water. What is the empirical
formula of the hydrocarbon?
1) CH 2) C2H3 3) CH2 4) CH3
29. 0.0833 mole of a carbohydrate of empirical formula CH2O contains 1.00 g of hydrogen. The molecular
formula of the carbohydrate is:
1) C5H10O5 2) C3H4O3 3) C12H22O11 4) C6H12O6
30. In a compound C, H and N are present in 9 : 1 : 3.5 by mass. If molar mass of the compound is 108, the
molecular formula of compound is:
1) C2H6N2 2) C3H4N 3) C6H8N2 4) C9H12N3

LEVEL - II
1. Two solids which are soluble in the same liquid to different extents may be separated by:
1) crystallization 2) sublimation
3) evaporation 4) fractional crystallization
2. For a compound to be purified by steam distillation:
1) impurities must be non-volatile
2) the liquid must be completely immiscible with water
3) the vapour pressure of the liquid must be sufficiently high
4) all of the above are correct.
3. An organic compound present in dissolved state in large volume of water is generally isolated by:
1) crystallisation 2) distillation 3) sublimation 4) solvent extraction
4. The speparation of mixture of organic compounds into individual components by column chromatography is
based on the difference in their:
1) solubilities 2) adsorptivities 3) melting points 4) partition co-efficients

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5. The most commonly used abdsorbent in column chromatography is:

1) sand 2) calcium carbonate 3) charcoal 4) silica gel
6. The identity of two solid organic samples having identical melting points can be established by:
1) undipressed mixed melting point
2) identical RF value in thin layer chromatogram
3) identical RF value in paper chromatogram
4) All the above
7. For which of the following compound Lassaigne’s test will fail?
1) NH 2 CONH 2 2) NH 2 CONH NH 2 .HCl

3) NH 2 NH 2 .HCl 4) C6 H 5 NH 2 NH 2 .2HCl
8. Which of the following statements is not applicable to Beilstein test?
1) Green or bluish green flame is due to the formation of volatile cupric halides
2) It does not tell us to which halogen is present in the organic compound
3) It is a very sensitive test that can be easily performed
4) It is a sure test for the presence of halogen
9. An organic compound which produces a bluish green coloured flame on heating in presence of copper is:
1) chlorobenzene 2) benzaldehyde 3) aniline 4) benzoic acid
10. Which of the following will give blood red colour while doing Lassaigne’s test for nitrogen?
1) H2N SO3H 2)  NH 2  2 C  0 3) C6 H 5SO3H 4)  NH 4  2 SO 4
11. When one of the following hydrocarbons is burnt in excess of oxygen, the volume of CO2 evolved is just
double to that of hydrocarbon taken. The hydrocarbon is:
1) CH4 2) C2H6 3) C3H8 4) C3H6
12. The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was
passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide
solution for complete neutralization. The organic compound is:
1) acetamide 2) thiourea 3) urea 4) benzamide
5
13. g of an organic compound gave 22.4 cm3 of moist nitrogen measured at 280 K and 732.7 mm pressure.
19
The percentage of nitrogen in the substance is approximately: (Aqueous tension at 280 K = 12.7 mm)
1) 9.8 2) 19.6 3) 4.9 4) 9.0
14. In duma’s method of estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected
at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of
nitrogen in the compound is:
1) 16.76 2) 15.76 3) 17.36 4) 18.20
15. 0.156 g of an organic compound on heating with fuming HNO3 and AgNO3 gives 0.235 g of AgI. Calculate
the percentage of iodine in the compound.
1) 81.41% 2) 68.32% 3) 52.75% 4) 79.68%
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16. ClCH2COOH is heated with fuming HNO3 in the presence of AgNO3 in Carius tube. After filtration and
washing the precipitate obtained is:
1) AgNO3 2) AgCl 3) Ag2SO4 ClCH2COOAg
17. 20 mL of CH4 is burnt with 60 mL of O2. If all measurements are made at the same P and T, what is the volume
of unreacted oxygen?
1) 10 mL 2) 20 mL 3) 30 mL 4) 40 mL

18. Liquid benzene (C6H6) burns in oxygen according to 2C6 H6  l   15O2  g  

12CO2  g   6H 2O  g 

How many litres of O2 at STP are needed for complete combustion of 39 gm of liquid benzene?
1) 11.2 litres 2) 74 litres 3) 84 litres 4) 22.4 litres.
19. 9.9 g amide with molecular formula C4H5 Nx Oy on heating with alkali liberated 1.7 g of ammonia. If the
percentage of oxygen is 33.33% then the ratio of ‘N’ and ‘O atoms in the compound is:
1) 1 : 1 2) 1 : 2 3) 2 : 3 4) 3 : 2
20. 0.75 g platinic chloride of a mono-acidic base on ignition gave 0.245 g platinum. The molar mass of the base
is:
1) 75.0 2) 93.5 3) 100 4) 80.0
21. The silver salt of a monobasic acid on ignition gave 60% of Ag. The molecular weight of the acid is:
1) 37 2) 33 3) 73 4) 77
22. A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2. The empirical formula of
the hydrocarbon is:
1) C7H8 2) C2H4 3) C3H4 4) C6H5
23. Ten millilitre of a gaseous hydrocarbon was burnt completely in 80 ml of O2 at STP. The volume of the remaining
gas is 70 ml. The volume became 50 ml on treatment with NaOH. The formula of the hydrocarbon is:
1) C2H6 2) C2H4 3) C3H8 4) C3H6
24. 0.24 gm of a volatile liquid upon vaporisation gives 45 ml of vapours at STP. What will be the vapour density
of the substance? (Density of H2 = 0.089 gm litre–1)
1) 9.539 2) 59.7 3) 5.993 4) 95.39
25. A Duma’s bulb full of air weighs 22.567 g at 20oC and 755 mm pressure. Full of vapours of a substance at
120oC and the same pressure, it weighs 22.8617 g. The capacity of the bulb is 200 mL. Find out the molecular
mass of the substance. [Density of air = 0.00129 g/mL at STP)
1) 86.59 2) 48.32 3) 63.27 4) 36.49
26. Nine volumes of a gaseous mixture consisting of gaseous organic compound A and just sufficient amount of
oxygen required for complete combustion yielded on burning four volumes of CO2, six volumes of water
vapour, and two volumes of N2, all volumes measured at the same temperature and pressure. If the compound
contains C, H, and N only, the molecular formula of the compound A is:
1) C2H3N2 2) C2H6N2 3) C3H6N2 4) C3H6N

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27. A mixture of ethylene and excess of H2 had a pressure of 600 mm Hg. The mixture was passed over nickel
catalyst to convert ethylene to ethane. The pressure of the resultant mixture at the similar conditions of temperature
and volume dropped to 400 mm Hg. The fraction of C2H4 by volume in the original mixture is:
1) 1/3rd of the total volume 2) 1/4th of the total volume
3) 2/3rd of the total volume 4) 1/2 of the total volume
28. A hydrocarbon contains 10.5 g of carbon for each one gram of hydrogen. The mass of 1 litre of hydrocarbon
vapours at 127oC and 1 atmospheric pressure is 2.8 g. Find out the molecular formula.
1) C6H12 2) C4H8 3) C7H8 4) C4H10
29. A substance on analysis gave the following data: 0.3112 g gave 0.4291 g of CO2 and 0.0585 g of water.
0.2293 g of the substance when heated with nitric acid and silver nitrate gave 0.3969 g of bromide and
chloride of silver. 0.2202 g of this mixture of halides was found to contain 0.1435 g of silver. Find the empirical
formula of the original substance.
1) C6H4ClBr 2) C6H3Cl2Br 3) C6H3ClBr2 4) C6H4Cl2
30. An organic diacid base was found to contain 40% carbon, 13.33% hydrogen and 46.67% Nitrogen. 0.94 g of
Chloroplatinate salt of the base gave 0.39 g platinum. Molecular formula of the base is:
1) C2H7N 2) C3H10N2 3) C2H8N2 4) CH4N

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SOLUTIONS
LEVEL - I
1. 3 Crystallisation is the preferred method of purification.
2. 4 Sublimation is the process employed.
3. 4 Glycerol decomposes at its b.p., it is separated from spent lye by distillation under reduced pressure.
4. 1 In steam can be purified by steam distillation. Vaporisation of organic liquid takes place at lower
temperature than its boiling point.
5. 2
6. 3 aniline is steam volatile.
7. 4
8. 3 fractional distillation since the b.ps are not widely differing.
9. 4
10. 1
distance moved by solute
11. 2 R F value 
distance moved by solvent
12. 4
13. 3 The chemical formula of thiourea is NH2CSNH2 so here Na2S, NaCN and NaCNS will be formed but
not Na2SO4.
14. 1 HNO3 decomposes Na2S and NaCN, if formed.
15. 3
16. 3 The presence of both nitrogen and sulphur in the Lassaigne’s test in an organic compound shows a
blood red colouration with FeCl3 solution.
17. 3 A black precipitate of PbS is formed.
18. 1 The formation of canary yellow precipitate with am. molybdate confirms the presence of P.
19. 3
20. 1 N present in the organic compound is converted into N2 gas by heating the compound with CuO.
1.4×Volume ×Normality 1.4  5 103 L 1gL1
21. 4 =
% Nitrogen Weight of substance taken ×100 = 100  23.73 %
29.5 103 g
22. 2 S is converted to H2SO4
Atomic mass of Br×Wt.of AgBr×100 80  141 100
23. 4 Percentage of Br = =  24%
Mol. mass of AgBr×Wt.of org. substance 188  250
24. 1 The percentage of phosphorus in an organic compound when estimated as magnesium pyrophosphate
62 Mass of Mg 2 P2 O7 formed  m1  100 62  0.66  100
is given by  %= %  51.20%
222 Mass of compound taken  m  222  0.36

100
25. 3 Molecular mass of protein =  181 = 82272.727
0.22
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 y y
26. B C x H y   x   O 2  xCO 2  H 2 O ; 15 x = 30 x2  y4
 4 2
15 45 0 0
15y
0 0 15x
2

wRT
27. 2 Molar mass of compound   44
PV
% of C in the compound  54.54%
% of H = 9.01%
Then, the ratio of number of atoms of ‘C’ and ‘H’ = 1 : 2
12 12.571
28. 3 %C    100  85.7
44 4.0
2 5.143
%H    100  14.3
18 4.0
The mole ratio of C to H is = 1: 2
1
29. 4 1 mole of carbohydrate will contain g of hydrogen, ie, 12 g hydrogen.
0.0833
 The molecular formula of the compound will be C6H12O6.
9 108 1108 3.5  108
30. 3 13.5g compound  9 g C  1g H  3.5 g N  108g  gC g H gN
13.5 13.5 13.5
9  108 1 108 3.5  108
 mol C  mol H  mol N  Molecular formula = C6H8N2
13.5  12 13.5  1 13.5  14
LEVEL - II
1. 4 The solid with lesser solubility crystalizes out first followed by the other.
2. 4 These are characteristics of steam distillation.
3. 4 Solvent extraction is used to separate the dissolved organic compounds in water.
4. 2 Column chromatography is based on difference in adsorptivities.
5. 4 Silicagel and aluminia are the commonly used adosrbants.
6. 4 Add all these are methods to identify the two samples to be the same.
7. 3 For Lassaigne’s test of N, compound must contain C in addition to nitrogen.
8. 4 Compounds like urea, thiourea etc. also give blue or green colour in this test.
9. 1 Copper produces a bluish green flame due to formation of volatile cupric chloride.
10. 1
7
11. 2 C 2 H 6  O 2 
 2CO 2  3H 2 O
2
1.4  2  50  0.1
12. 3 Percentage of nitrogen present in the given organic compound   46.6%
0.3
Nitrogen present in urea (NH2CONH2) = 46.6%

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13. 1 Pressure of dry nitrogen = (732.7 – 12.7) mm = 720 mm

28 720  22.4 273  1 

% of N       100  9.828%
22400 280 760  5 / 19 
14. 1 P1 = 725 – 25 = 700 mm, P2 = 760 mm ; T1 = 300 K, T2 = 273 K
P1V1T2 28 33.52
 V2   33.52 mL ; % N   100  16.76
P2 T1 22400 0.25
127 0.235
15. 1 Percentage of iodine =  100  81.41%
235 0.156
16. 2 Chlorine compounds  AgCl
(Cl)

17. 2 CH 4  2O 2 
 CO 2  2H 2 O
20 mL of CH4 will react with 40 mL of O2.
18. 3 2C6 H 6  15O 2  12CO 2  6H 2O
15  22.4  39
2  78 g  15  22.4 litre of O 2 at STP ; 39 g   84 litre
2  78
9.9
19. 2 Amide C4H5 NxOy will give x mole NH3.  Molecular mass of amide   17x  99x
1.7
14x
% Nitrogen in the amide =  100  14.14 ; % Oxygen in the amide = 33.33
99x
14.14 33.33
Ratio of number of atoms of ‘N’ and ‘O’ = :  1: 2
14 16
Mass of B2 H 2 PtCl6 Mass of Pt
20. 2 
2B  410 195
0.75 0.245
   B  93.5
2B  410 195
Eq. mass of base = 93.5; since it is monoacidic.  Molar mass of base = 93.5 × 1 = 93.5
Weight of silver
21. 3 % of Ag  Weight of silver salt  100  60

Weight of silver salt 100

Equivalent weight of acid   108  107 =  108  107  73
Weight of silver 60
Molecular weight of acid = 73 × 1 = 73.
22. 1 Let the mass of gaseous hydrocarbon taken = w gm
12 3.08 84 2 0.72 8
 % C   100  ; % H   100 
44 w w 18 w w
84 1 8 1
Rate of C : H atoms =  :   7 :8 Thus, E.F = C7H8.
w 12 w 1
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 y y
23. 2 Cx H y   x   O2  x CO 2  H 2O
 4 2
Volume of CO2 = (70 – 50) = 20 ml
Volume of O2 (used) = 80 – 50 = 30 ml; x = 2, y = 4; the formula of hydrocarbon is C2H4.
24. 2 It is an example of Victor Meyer’s method
Mass of compound  22400 0.24  22400
Mw    119.46
Volume of vapours at STP 45
Mw 119.46
Vapour density =   59.7
2 2
25. 1 Given, V1 = volume of bulb = 200 mL

200  755 273

V2 = Volume of gas at STP =   185.122 mL
293 760
Mass of air = V2 × 0.00129 = 185.122 × 0.00129 = 0.2388 g
Mass of empty bulb = (22.567 – 0.2388) = 22.3282 g
Mass of vapours = (22.8617 – 22.3282) = 0.5335 g

200  755 273

Volume of vapours at STP    138 mL
393 760

mass of vapours
Molecular mass of the substance   22400
volume of vapours at NTP
0.5335
  22400  86.59
138
26. 2 2Cx H y N z  7O 2 
 4CO2  6H 2 O  2N 2

2x = 4  x = 2, 2y = 12  y = 6, 2z = 4 z = 2
Formula = C 2 H 6 N 2
27. 1 Let n mol of (C2H4 + H2) and x mol of C2H4. H2 = (n – x) mol

C2 H 4  H 2  C 2 H 6
x x x mol
After reaction (C2H6 + H2 left) x+n–x–x=n–x
Total H2 = (n – x), H2 reacted = x H2 left = (n – x – x) n = 600. n – x = 400
n 600

n  x 400
n 1
x volume of C2H4  rd of total volume
3 3
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28. 3 Carbon : Hydrogen : : 10.5 : 1

Empirical formula = C7H8 Empirical formula mass = (12 × 7) + (1 × 8) = 92

wRT
Molar mass M   92 g mol 1
PV

Molecular mass 92
n  1
Empirical formula mass 92

Molecular formula = Empirical formula = C7H8

35.5 0.1720
29. 1 Percentage of Cl   100  18.55
143.5 0.2293
80 0.2249
and percentage of Br    100  41.73
188 0.2293
12 0.4291
Percentage of carbon    100  37.6%
44 0.3112
2 0.0585
Percentage of hydrogen   100  2.09%
18 0.3112
EF = C6H4ClBr
30. 3 E.F. of the compound = CH4N
Chloroplatinate salt = B(H2PtCl6) Since a diacid base.
0.94
Mol. mass of B.H2PtCl6 = 195  470
0.39
Mol. mass of base = 60
Hence base = C2H8N2

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CHAPTER - 15

HYDROCARBONS

INTRODUCTION
Hydrocarbons are compounds of carbon and hydrogen only. These are widely distributed in nature in the form
of petroleum, natural gas, etc. Hydrocarbons are considered to be the parent organic compounds while all
others are thought to have been derived by replacing one or more hydrogen atoms by other atoms or groups.
Hydrocarbons play a key role in our daily life. LPG (Liquefied Petroleum Gas), CNG (Compressed Natural
Gas) and LNG (Liquefied Natural Gas) are used as fuels. Coal gas, obtained by destructive distillation of coal
is also used as fuel. Petrol, diesel and kerosene are obtained by the fractional distillation of petroleum. Kerosene
is used as a domestic fuel. Automobiles use petrol, diesel and CNG as fuel.
Hydrocarbons are extensively used in the manufacture of polymers such as polythene, polypropene, polystyrene,
etc. Higher hydrocarbons are used as solvents for paints. These are also used as the starting materials for
manufacture of dyes and drugs.
15.1 CLASSIFICATION OF HYDROCARBONS
Hydrocarbons are classified into (i) Acyclic or open chain hydrocarbons and (ii) Cyclic or closed chain
hydrocarbons based on carbon structure.
15.1.1 Acyclic or open chain hydrocarbons. These are compounds contain open chains of carbon atoms in their
molecules. They are also called aliphatic hydrocarbons. These are further classified into alkanes, alkenes,
and alkynes based on the type of bond between carbon atoms.
i) Alkanes. Alkanes have only carbon-carbon single bonds :
CH3  CH3 CH3  CH 2  CH3 CH3  CH 2  CH 2  CH3
Ethane Propane Butane

ii) Alkenes. Alkenes have one carbon-carbon double bond :

CH 2  CH 2 CH3  CH  CH 2 CH3  CH  CH  CH 3
Ethene Propene But-2-ene

iii) Alkynes. Alkynes have one carbon-carbon triple bond :

CH  CH CH3  C  CH CH3CH 2  C  CH CH3  C  C  CH 3
Ethyne Pr opyne But 1 yne But  2 yne

15.1.2 Cyclic or closed chain hydrocarbons. These are compounds containing closed chains or rings of carbon
atoms. They are divided into two classes : (a) Alicyclic hydrocarbons, and (b) Aromatic hydrocarbons
a) Alicyclic hydrocarbons. These are hydrocarbons containing a ring of three or more carbon atoms and
having properties similar to aliphatic hydrocarbons. These are further divided into three categories :
(i) Cycloalkanes (ii) Cycloalkenes and (iii) Cycloalkynes
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(i) Cycloalkanes. Saturated alicyclic hydrocarbons in which all the carbon atoms are joined by single covalent
bonds are called cycloalkanes

(ii) Cycloalkenes. Unsaturated alicyclic hydrocarbons which contain one carbon-carbon double bond are
called cycloalkenes.

iii) Cycloalkynes. Unsaturated alicyclic hydrocarbons containing one carbon-carbon triple bond are called
cycloalkynes, e.g., cyclopentyne, cyclohexyne, cycloheptyne, cyclooctyne, cyclononyne, etc.
b) Aromatic hydrocarbons. Aromatic hydrocarbons are a special type of cyclic compounds possessing
aromaticity. These are of two types:
i) Benzenoid aromatic compounds. Hydrocarbons and their alkyl, alkenyl and alkynyl derivatives which
contain one or more benzene rings, either fused or isolated in their molecules are called benzenoid aromatic
hydrocarbons or arenes (aromatic alkenes).

ii) Non-benzenoid aromatic compounds. Aromatic hydrocarbons which do not contain a benzene ring
but other highly unsaturated rings are called non-benzenoid aromatic compounds, e.g., azulene, etc.

PART - I ALKANES
15.2 3Alkanes. Alkanes are saturated open chain hydrocarbons containing carbon-carbon single bonds. They are
also called paraffins (Latin : parum = little, affinis = affinity). They are unreactive towards most reagents.
15.2.1 Nomenclature of Alkanes
In the IUPAC system, saturated hydrocarbons are called alkanes. The general formula is CnH2n+2 where n = 1,
2, 3,.... etc.
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Table - 01 IUPAC names of some straight chain alkanes.

No.of carbon Formula IUPAC name No.of carbon Formula IUPAC name
atoms atoms
1 CH4 Methane 11 C11H24 Undecane
2 C2H 6 Ethane 12 C12H26 Dodecane
3 C3H 8 Propane 13 C13H28 Tridecane
4 C4H 10 Butane 14 C14H30 Tetradecane
5 C5H 12 Pentane 15 C15H32 Pentadecane
6 C6H 14 Hexane 20 C20H42 Eicosane
10 C10H 22 Decane 30 C30H62 Triacontane

(Longest chain rule and lowest loctant rule and alphabetical order of substituents are followed.)

(Numbering in the complex substituent begins with the carbon atom attached to the main chain.)

(The prefix iso is considered to be part of the fundamental name of the alkyl group while considering the
alphabetical order of substituents.)

(The prefix sec is not considered while deciding alphabetical order of substituents and isopropyl is taken as
one word.)
15.2.2 Isomerism in Alkanes
Alkanes exhibit structural isomerism (chain isomerism). Structural isomers which differ in the arrangement of
carbon chain are called chain or nuclear isomers. Methane, ethane and propane have only one structure
each, but butane has two isomers since four carbon atoms can be joined in two different ways:

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Pentane has three isomers:

Hexane has five isomers:

Heptane (C7H16) has 9, octane (C8H18) has 18, nonane (C9H20) has 35 and decane (C10H22) has 75 isomers.
15.2.3 Structure of alkanes
Carbon atoms in alkanes are sp3-hybridized, therefore, they have tetrahedral structures. The four bonds are
directed towards the corners of a regular tetrahedron. The angle between any two adjacent bonds is
109o 28 (tetrahedral angle).
Alkanes contain only carbon-carbon and carbon-hydrogen single bonds with average bond lengths of 154 pm
and 110 pm respectively.
Structure of methane (CH4). In methane, sp3 hybridised carbon lies at the centre of the tetrahedron while the
four hydrogens are present at the vertices of the regular tetrahedron. Thus, methane has tetrahedral structure.

(a) (b) (c) (d) (e)

Figure-01 Representation of methane molecule (a,b) tetrahedral (c,d) three-dimensional and (e) two-
dimensional or Fisher Projection Formula.
Three-Dimensional (3-D) Structure of Alkanes. Three-dimensional structure of organic molecule can be
represented by wedge and dash formula (Fig.01c). Another method of representing 3-D structures on a 2-D
surface was introduced by Emil Fischer (German chemist, 1891) called Fischer Projection Formula.

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15.2.4 Methods of Preparation of Alkanes

1. From unsaturated hydrocarbons (Hydrogenation). Alkenes and alkynes add hydrogen in presence of
a catalyst such as Raney nickel powder, platinum or palladium to form alkanes.
Pt/Pd/ Ni
Pt/Pd/ Ni
CH2  CH2  H2   CH3  CH3 ; CH 3  C  CH  2H 2   CH 3  CH 2  CH 3
Ethene Ethane Pr opene Propane

Pt and Pd catalyse the reaction at room temperature. Hydrogenation of unsaturated hydrocarbons using nickel
at 523–573 K is known as Sabatier and Sendern’s reaction.
2. From alkyl halides
i) Reduction of alkyl halides. Alkanes can be prepared by the reduction of alkyl halides (except fluorides)
using Zn and CH3COOH, Zn and HCl, Zn and NaOH or Zn-Cu couple and alcohol.
 
Zn, H Zn, H
CH 3Cl  H 2  CH 4  HCl ; CH 3CH 2 CH 2Cl  H 2  CH 3CH 2CH 3  HCl
ii) By Wurtz reaction. When an alkyl halide (preferably bromide or iodide) is treated with metallic sodium in
dry diethyl ether, symmetrical alkane containing even number of carbon atoms is formed.
Dryether
Dryether
2R  X  2Na   R  R  2 NaX ; 2 CH 3  Br  2 Na   CH 3  CH3  2 NaBr
Alkyl halide Alkane Bromomethane Ethane

With two different alkyl halides, a mixture of alkanes is obtained.

Dry ether
3CH 3  I  6Na  3CH 3 CH 2  I   CH 3CH 2  CH 3  CH 3CH 3  CH 3 CH 2 CH 2 CH 3  6NaI
Iodomethane Iodoethane Pr opane Ethane Butane

iii) Reduction of alkyl halide with HI and red P. Alkyl iodides can be reduced to the corresponding
alkanes by heating with conc. HI in presence of red P at 423 K.
423K/redP
CH 3CH 2  I HI   CH 3  CH 3  I 2
Iodoethane Ethane
iv) Through Grignard reagents. Grignard reagents (alkyl magnesium halides) readily react with compounds
containing active hydrogen such as water, alcohols, ammonia, amines, etc., to form alkanes.
CH 3CH 2  MgBr  H 2 O   CH 3  CH 3  Mg  OH  Br
Ethylmagnesiumbromide Ethane

3. From carboxylic acids

i) Decarboxylation. Decarboxylation is the process of removal of CO2 from an organic compound.
When a carboxylic acid is heated with soda-lime (NaOH + CaO in the ratio 3 : 1) at 630 K, a molecule of
CO2 is lost and an alkane with one carbon atom less the carboxylic acid is formed.
CaO,630K
 R  COONa  H 2O ; CH3COONa  NaOH  CH 4  Na 2CO3
R  COOH  NaOH 
Carboxylicacid Sod.acetate Methane

CaO keeps NaOH dry since it is hygroscopic (absorbs mositure).

ii) Kolbe’s electrolytic method. When a concentrated aqueous solution of the sodium or potassium salt of
monocarboxylic acid is electrolysed, alkane containing even number of carbon atoms is produced.
Electrolysis
2R  COONa  2H 2 O   R  R  2NaOH  H 2  2CO 2
Alkane
Electrolysis
2CH 3COONa  2 H 2 O   CH3  CH3  2NaOH  H 2  2 CO 2
Sodium ethanoate Ethane

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2e
At anode: 2CH 3COO    2CHCOO  
 2H C 
3  2CO 2  ; 2H3C  
 H3C  CH3 
Acetate ion Acetate free radical Methyl free radical

Electrolysis
At cathode: H 2 O  e   OH   H  ; 2H  
 H2 
4. From beryllium and aluminium carbides. Both these carbides on treatment with water yield methane.
Be2 C  CH 4  2 Be  OH 2 ;
 4 H 2 O  Al4 C3  12 H 2O 
 3CH 4  4 Al (OH)3
Berylliumcarbide Methane Aluminiumcarbide Methane
15.2.5 Physical properties
1. Boiling points. Amongst straight chain alkanes, the first four members (C1–C4) are gases, the next thirteen
(C5–C17) are liquids and the higher members (C18 onwards) are colourless waxy solids.
The boiling points of straight chain alkanes increase by 20–30 K for the addition of each CH2 group to the
chain. This is due to increase in van der Waals forces of attraction with molecular size/surface area.
Among isomeric alkanes, branched chain isomers have lower boiling points than the straight chain alkane, i.e.,
greater the branching, lower the boiling point of the isomer. This is due to the fact that with branching the shape
of the molecule tends to approach that of a sphere. As a result, area of contact decreases and van der Waals
forces of attraction becomes weaker. E.g., n-Pentane has higher boiling point than its isomers, Isopentane (2-
methyl butane) or Neopentane (2,2-dimethyl propane).

2. Melting points. The melting points of alkanes increase with increase in carbon content but the variation is
not regular due to alternation effect.
Table - 02 Melting points of some n-alkanes
No.of carbon Formula IUPAC name Molecular m.p./(K) b.p./(K)
atoms mass/u
1 CH4 Methane 16 90.5 111
2 C2H6 Ethane 30 101 184
3 C3H8 Propane 44 85.3 230.9
4 C4H10 Butane 58 134.6 272.4
4 C4H10 2-Methylpropane 58 114.7 261
5 C5H12 Pentane 72 143.3 309.1
5 C5H12 2-Methylbutane 72 113.1 300.9
5 C5H12 2,2-Dimethylpropane 72 256.4 282.5
6 C6H14 Hexane 86 178.5 341.9
8 C8H18 Octane 114 216.2 398.7
Alkanes with even number of carbon atoms have higher melting points than those with odd number of
carbon atoms. This property is known as alternation effect.
3. Solubility. ‘Like dissolves like’ is the rule of solubility. Alkanes being non-polar are insoluble in polar
solvents such as water, alcohol, etc., but highly soluble in non-polar solvents such as petroleum ether, benzene,
carbon tetrachloride, etc.

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4. Density. The densities increase with increase in molecular masses till the limiting value of about 0.8 g cm–3.
Thus all alkanes are lighter than water.
15.2.6 Chemical Properties
Alkanes contain only strong  -bonds, therefore they are the least reactive of all hydrocarbons.
1. Substitution reactions
i. Halogenation. Halogenation is carried out by treating alkane with halogen in presence of ultraviolet light or
by heating the reaction mixture to 520–670 K. The order of reactivity of halogens is F2 > Cl2 > Br2 > I2.
(i) Chlorination. During chlorination of methane, all the four hydrogen atoms are replaced one by one to form
a mixture of products. For example,
h or h or
CH 4  Cl2 520  670K
 CH 3Cl  HCl CH 3Cl  Cl2  520  670K
 CH 2 Cl2  HCl
Methane Chloromethane ; Chloromethane Dischloromethane
(Methyl chloride) (Methylene chloride)
h or
CH 2 Cl 2 h or
 Cl2   CHCl3  HCl CHCl3  Cl 2 
520 670 K  CCl4  HCl
520 670K
Dichloromethane ; Trichloromethane
Trichloromethane Tetrachloromethane
(Carbon tetrachloride)
(Chloroform)
(ii) Bromination. Bromine reacts with alkanes in a similar manner but less readily.
hvor
CH3  CH 3  Br2 
520 670K  CH 3  CH 2  Br  HBr
Ethane Bromoethane

(iii) Iodination. The reaction of iodine with alkane is reversible because hydrogen iodide formed as a
by-product is a moderate reducing agent and hence reduces the iodoalkane back to alkane.
 CH 3  I  HI
CH 4  I2 
Iodomethane

Iodination can be carried out in presence of an oxidising agent such as iodic acid (HIO3) or nitric acid (HNO3)
which oxidises HI to I2.
5HI  HIO 3 
 3I 2  3 H 2O ; 2HNO3  2 HI 
 2 H 2 O  2 NO 2  I 2
Mechanism of halogenation. Halogenation occurs by free radical mechanism consisting of three steps; (a)
initiation (b) propagation and (c) termination.
a) Initiation. When a mixture of CH4 and Cl2 is heated to 520-670 K in the dark or subjected to uv light at
room temperature, Cl2 absorbs energy and undergoes homolytic fission producing chlorine free radicals.
hν
Cl  Cl 
homolysis  Cl  + Cl 
chlorine free radicals
.
b) Propagation. This step consists of two reactions. In the first reaction, the Cl attacks the CH4
molecule and abstracts a hydrogen atom forming .CH3 and a molecule of HCl. In the second reaction,.CH3
reacts with a molecule of Cl2 forming methyl chloride and another .Cl. These reactions (i) and (ii) are repeated
over and over again and the chain gets propagated.
hν hν
CH 4  Cl     CH 3  HCl ; CH 3  Cl  Cl   CH 3Cl  Cl
Highly halogenated products are also formed.
CH 3Cl  Cl 
  CH 2 Cl  HCl ; CH2Cl  Cl  Cl 
 CH2Cl2  Cl
c) Termination. The chain reactions come to a halt if two free radicals combine amongst themselves.
.Cl  .Cl   CH 3  CH 3 ; .CH 3  .Cl 
 Cl  Cl ; .CH 3  .CH 3   CH 3  Cl
The above mechanism can explain the formation of ethane as a by product during chlorination of methane.
ii. Nitration. Replacement of hydrogen by nitro (–NO2) group is called nitration. Alkanes react with fuming
HNO3 at 423-673K under pressure forming a mixture of nitro alkanes (vapour phase nitration).
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Fuming HNO3
CH 3  CH 3 
673K  CH 3CH 2 NO 2  CH 3 NO 2
Nitroethane(80%) Nitromethane(20%)

iii. Sulphonation. Substitution of hydrogen atom of an alkane by sulphonic acid group (–SO3H) is called
sulphonation. It is carried out by heating an alkane with fuming sulphuric acid (H2SO4 + SO3) at 675–725 K.
The ease of substitution of hydrogens is 3o > 2o > 1o.
CH 3 (CH 2 ) 4 CH 3  H 2SO 4 
SO3
675K
 CH 3 (CH 2 ) 4 CH 2  SO3 H  H 2 O
n  Hexane n  Hexanesulphonic acid

2. Combustion. Alkanes readily burn in air/oxygen producing CO2 and H2O. The process is highly exothermic.
 nCO2   n  1 H 2 O
Cn H 2n  2  [(3n  1) / 2]O2 

CH 4  g   2O 2  g  
 CO2  g   2 H 2 O  l  ;  c H o  890 kJ mol1

C4 H10  g   (13 / 2) O 2 
 4CO 2  g   5 H 2 O  l  ;  c H o  2875.8kJ mol 1
b) Incomplete combustion. Combustion in limited supply of air/oxygen gives CO and carbon black.
2 CH 4  3O 2 
 2 CO  4 H 2 O ; CH 4  O 2 
 C  2 H 2O
(limited) (limited) Carbon black

c) Controlled (Catalytic) oxidation. i) When a mixture of methane and oxygen (9:1 by volume) at a pressure
of 100 atmospheres is passed through a copper tube at 523 K, methanol is obtained.
100atm,523K
2CH 4  O2 
Cu tube  2 CH 3OH
Methane Methanol

ii) Methane oxygen mixture under pressure, passed over heated molybdenum oxide forms methanal.
Mo2 O3
CH 4  O 2 
 ,pressure  HCHO  H 2O
Methane Methanal

iii) Higher alkanes on oxidation in presence of manganese acetate give carboxylic acids.
 CH COO Mn
2CH3  CH3  3O2 
3

2
2CH3COOH  2H2O
Ethane Ethanoic acid

iv) Oxidising agents such as KMnO4 and K2Cr2O7 have no effect on alkanes. However, alkanes containing
tertiary hydrogen are oxidised to the corresponding alcohols.
KMnO4
(CH 3 )3 CH   (CH 3 )3 C  OH
2  Methylpropane 2  Methylpropan  2  ol

3. Isomerisation. When n-alkanes are heated with anhydrous aluminium chloride and hydrogen chloride at
573 K under pressure (about 35 atm), these are converted into brached chain alkanes.
CH3

CH3  CH 2  CH 2  CH3 

AlCl3 / HCl
573K
 CH3  CH  CH3
n-Butane Isobutane

Isomerisation is useful in increasing the octane number of petroleum fractions.

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4. Aromatisation. n-Alkanes containing six to eight carbon atoms when heated to about 773 K and
10-20 atm pressure in presence of oxides of vanadium (V2O5), molybdenum (MoO3) or chromium (Cr2O3)
supported over alumina undergo dehydrogenation to form aromatic hydrocarbons. This process which involves
cyclisation, isomerisation and dehydrogenation is called aromatization or reforming.
Cr2O3 or V2O5 or Mo2O3
CH3  (CH2 )4  CH3 
773K,1020atm  C6H6
n-Hexane Benzene
Cr2O3 or V2O5 or Mo2O3
CH3  (CH 2 )5  CH3 
773K,10 20atm  C 7 H8
n-Heptane Toluene

5. Pyrolysis. Decomposition of a compound by heating is called pyrolysis. Pyrolysis of higher alkanes to give
a mixture of lower alkanes, alkenes, etc., is called cracking. It is carried out by heating the higher alkanes to
high temperatures (773-973 K) under a pressure of 6-7 atmospheres in presence or absence of a catalyst.
773K
C6 H14 
6 7atm  C 6 H12  H 2  C 4 H 8  C 2 H 6  C3 H 6  C 2 H 4  CH 4
Hexane      
Hexene Hydrogen Butene Ethane Pr opene Ethene Methane

Pyrolysis of alkanes involves breaking of C–C and C–H bonds and occurs by a free radical mechanism.
Preparation of oil gas from kerosene oil and petrol gas from petrol is based on the process of pyrolysis.
Pt or Pd or Ni
C12 H 26  973K  C 7 H16  C5 H10  Other products
Dodecane Heptane Pentene

6. Reaction with steam. Methane reacts with steam at 1273 K in presence of nickel catalyst forming a
mixture of CO and H2, called syngas.
1273K, Ni
CH 4 + H 2 O   CO  3H 2
(Steam) 
Syngas

15.2.7 Uses of Alkanes

(i) Methane (natural gas) is used as fuel in automobiles. LPG (mixture of butane and isobutane) is used as fuel
in homes and in industry. (ii) Methane is used to make carbon black used in the manfacture of printing inks,
paints and automobile tyres. (iii) Alkanes on catalytic oxidation give alcohols, aldehydes and carboxylic acids.
(iv) Higher alkanes in form of gasoline, kerosene oil, diesel, lubricating oils and paraffin wax are widely used.
15.2.8 Conformations
The infinite number of momentary arrangement of atoms in space which result through rotation about a single
bond are called conformations or conformers or rotational isomers or rotamers.
Conformations of Ethane. In ethane, the two carbon atoms are connected by a  -bond. If one of the
methyl groups in ethane molecule is kept fixed and the other is rotated about the C–C bond, a large number of
arrangements of the hydrogen atoms on one carbon atom with respect to the hydrogen atoms on the other
carbon atom are obtained. Out of the infinite number of possible conformations, only two extreme conformations,
i.e., staggered and eclipsed are important. All other conformations in between these two are known as
gauche or skew conformations.
Two-dimensional representation of Conformations. Three-dimensional structures can be represented on
a two-dimensional (2-D) surface with the help of Fischer Projections.
a) Sawhorse Projections. The molecule is viewed slightly from above and from the right and projected on
paper. The bond between the two carbon atoms is drawn diagonally and slightly elongated. The lower left
hand carbon is considered to be towards the front and the upper right hand carbon towards the back. The
lines are inclined at an angle 120 to each other..

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(a) (b)

Figure - 02 (a) Sawhorse projections of ethane (b) Newman’s projection of ethane (The bonds of the rear
carbon in the eclipsed form are shown slightly displaced to make them visible)
b) Newman Projections. These projection formulae are obtained by viewing the molecule along the bond
joining the two carbon atoms. The carbon atom near the eye is represented by a point and the three atoms or
groups attached to it by three equally spaced (120o) radii. The carbon atom farther from the eye is designated
by a circle and the three atoms or groups attached to it by three equally spaced radial extensions.
Relative stability of Staggered and Eclipsed conformations of Ethane. In staggered conformation, each
hydrogen atom on the front carbon lies exactly between the hydrogens on the rear carbon. In this conformation,
any two hydrogen atoms on adjacent carbon atoms are as far apart as possible (H to H distance = 310 pm).
As a result, repulsion between the electron clouds of  -bonds of two hydrogen atoms is minimum.
In eclipsed conformation, hydrogen atoms on the rear carbon lies exactly behind the hydrogen atoms on the
front carbon. (H to H distance = 229 pm). As a result, the electron clouds of the hydrogen atoms repel each
other. Thus, the energy of the eclipsed conformation is greater than that of the staggered conformation. This
repulsive interaction between the electron clouds, which affects the stability of a conformation is called torsional
strain. The magnitude of the strain, depends on the angle of rotation about C–C bond. This angle is also called
dihedral angle or torsional angle.
The staggered conformation has minimum torsional strain and the eclipsed form has maximum torsional strain.
The staggered conformation of ethane is about 12.55 kJ mol–1 more stable than the eclipsed conformation.
This energy barrier is not large enough to prevent rotation. Thus, the two conformers are readily interconvertible
and therefore, it is not possible to separate the two conformers. At any moment, most of the ethane molecules
exist in the staggered conformation due to minimum energy and maximum stability.
PART - II ALKENES
15.3 Alkenes. Alkenes are acyclic unsaturated hydrocarbons containing a carbon-carbon double bond. They are
also called olefins (Greek: olefiant = oil forming) since the lower members of this class such as ethene (ethylene),
propene (propylene), etc. produce oily products on reaction with halogens.Their general formula is CnH2n
where n = 2,3,4... etc.
15.3.1 Structure of Double Bond
The double bond in alkenes consists of one carbon-carbon  -bond formed by head-on overlap of sp2-
hybridised orbitals (bond enthlpy, 397 kJ mol-1) and a  -bond (bond enthlpy, 284 kJ mol–1) formed by lateral
or sideways overlap of the two 2p-orbitals of the two carbon atoms.

(a) (b) (c) (d)

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Figure-03 Orbital picture of ethene showing (a)  -bonds only, (b) π -bond, (c) π -cloud and (c) bond angles
and bond lengths
To have effective overlap, the p-orbitals move closer, therefore, a double bond is shorter (134 pm) than a
single bond (154 pm). The presence of  -bond makes alkenes less stable and more reactive than
alkanes.Therefore, alkenes are readily attacked by (electrophiles) electrophilic reagents.
15.3.2 Nomenclature of Alkenes
In the IUPAC system, the longest carbon chain containing the double bond is selected. The name of alkene is
obtained by replacing the suffix ane of the corresponding alkane by suffix ene.
The first member of the series, :CH2, known as methylene in the common system and methene in the IUPAC
system is highly unstable. Therefore, the first stable member of the series is C2H4 known as ethylene or ethene.

15.3.3 Isomerism
Alkenes exhibit structural isomerism and stereoisomerism.
a) Structural isomerism. Ethene and propene have only one structure, but alkenes containing four or more
carbon atoms exhibit both position and chain isomerism.E.g., the formula C4H8 represent the three isomers :

Structures I and II are position isomers, whereas structures I and III, and II and III are chain isomers.
b) Geometrical Isomerism-Hindered rotation around carbon-carbon double bond. A double bond consists
of a  -bond and a  -bond. The breaking of a  -bond requires 251 kJ mol–1 of energy, therefore, rotation
about carbon-carbon double bond is strongly hindered or restricted.
Many substituted alkenes have two distinct isomers which differ only in the relative positions of atoms or
groups in space around the double bond, e.g, but-2-ene.

These isomers have the same structural formula, but different relative spatial arrangement of hydrogen atoms
and methyl groups around the double bond. Such isomers which have the same structural formula but differ in
the relative spatial arrangement of atoms or groups around the double bond are called geometrical isomers
and the phenomenon is called geometrical isomerism. This is a type of space or stereoisomerism.
Isomer I, in which similar atoms or groups lie on the same side of the double bond is called
cis-isomer and isomer II, in which similar atoms or groups lie on opposite sides is called the trans-isomer.
Thus, geometrical isomerism is also called cis-trans isomerism.

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i. Properties of geometrical isomers. Due to different arrangement of atoms or groups in space, these
isomers differ in their physical properties such as m.p., b.p., dipole moment, solubility, etc. Dipole moment of
cis-but-2-ene is 0.33 D while that of trans-but-2-ene is zero; trans-but-2-ene is non-polar.

In the case of solids, the trans isomer has higher melting point than the cis isomer.
ii. Conditions for Geometrical Isomerism. (i) The molecule must contain a double bond. (ii) Each of the
two carbon atoms of the double bond must have different substituents which may be same or different.
Geometrical or cis-trans isomerism is also shown by alkenes of the types XYC = CXZ and XYC = CZW.

However, geometrical isomers are not possible if one or both the doubly bonded carbon atoms carry two
similar substituents. This is because the two possible configurations are identical as shown below:

Thus, terminal alkenes such as propene, but-1-ene, 2-methylprop-1-ene, etc. and alkenes carrying identical
substituents on one of the doubly bonded carbon atoms such as 2-methylbut-2-ene and 2, 3-dimethylbut-2-
ene, etc. do not show geometrical isomerism.

15.3.4 Stability of Alkenes. Alkenes have high heats of hydrogenation; unsubstituted alkenes such as ethene has
the highest value. Greater the number of alkyl groups attached to the doubly bonded carbon, more stable is the
alkene : R2C=CR2 > R2C=CHR > R2C=CH2 > RCH=CHR > RCH=CH2 > H2C=CH2.
The relative stability of alkenes may be due to hyperconjugation. trans-2-butene is more stable than cis-2-
butene due to absence of seric hindrance.
15.3.5 General Methods of Preparation
1. By partial reduction of alkynes. Catalytic hydrogenation of alkynes to alkenes is faster than alkenes to
alkanes. Alkynes can be reduced to give cis-or trans-alkenes depending on the nature of the catalyst used.
E.g., catalytic reduction of alkynes in presence of palladised charcoal, partially deactivated by sulphur or
quinoline (Lindlar’s catalyst) gives cis-alkenes. When alkynes are reduced with sodium in liquid ammonia
(Birch reduction), trans-alkenes are the major products.

Pd /C Pd /C
CH  CH  H 2  CH 2  CH 2 ; CH 3  C  CH  H 2  CH3  CH  CH 2

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2. From alkyl halides or haloalkanes (dehydrohalogenation). Alkyl halides on heating with a strong base
such as sodium ethoxide or a conc. alcoholic solution of KOH undergo dehydrohalogenation to give alkenes.

Dehydrohalogenation is an example of elimination reaction. Since, a hydrogen is removed from  -carbon and
a halogen from the  -carbon, it is called  -elimination reaction. The ease of dehydrohalogenation of alkyl
halides having the same alkyl group but different halogens is Iodides > Bromides > Chlorides.
The ease of dehydrohalogenation of isomeric alkyl halides having the same halogen, but different structures is
tertiary > secondary > primary. Thus, a tertiary alkyl iodide is most reactive.
Saytzeff rule. Depending on the structure, alkyl halides may give one or more isomeric alkenes. E.g.,
dehydrohalogenation of 1-chlorobutane gives only one alkene (but-1-ene) since only one type of  - hydrogen
is available in the molecule.

If, however, the alkyl halide has  -hydrogens on either side of the carbon atom carrying the halogen, it can
undergo elimination in two different ways giving two alkenes. The relative amounts of these two alkenes is
governed by Saytzeff rule. A ccording to this rule, whenever two alkenes are theoretically possible during
a dehydrohalogenation reaction, it is always the more highly substituted alkene (i.e., having lesser
number of hydrogen atoms on the double bond) which predominates.

3. From vicinal dihalides or 1,2- dihaloalkanes (dehalogenation). Alkenes can be prepared by heating
a suitable vicinal or 1,2 dihaloalkane with zinc dust in methanol or ethanol.
CH OH
Br  CH 2  CH 2  Br  Zn 
3

 CH 2  CH 2  ZnBr2
1,2 Dibromoethane(Ethylene dibromide) Ethene(Ethylene)

CH OH
CH3  CHBr  CH 2 Br  Zn 
3

 CH3  CH  CH 2  ZnBr2
1,2 Dibromopropane(Pr opylene dibromide) Pr opene

4. From monohydric alcohols or alkanols by acidic dehydration. Monohydric alcohols containing a  -

hydrogen on heating with a mineral acid such as conc H2SO4 or H3PO4 or on passing their vapours over
heated alumina at 623-633 K eliminate a molecule of water to form alkenes.

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conc.H SO ,443 4443K

CH3  CH 2  OH 
2 4
or Al O ,623633K
 CH 2  CH 2  H 2O
Ethanol 2 3
Ethene (Ethylene)
The order of dehydration of alcohols is : 30 > 20 > 10.
5. Kolbe’s electrolytic reaction. Electrolysis of sodium or potassium salts of saturated dicarboxylic acids
give alkenes.

15.3.6 Physical Properties. The first three members i.e., ethene, propene and butenes are colourless gases; the
next fourteen (C5–C18) are liquids while the higher ones are solids. Except ethene which has pleasant smell, all
other alkenes are colourless and odourless gases. They are insoluble in water, but fairly soluble in non-polar
solvents such as benzene, petroleum ether, etc.
Boiling points : Their boiling points increase regularly with increase in molecular mass. The boiling points
generally increase by 20-30 K for the addition of each CH2 group to the chain. Like alkanes, straight chain
alkenes have higher boiling points than isomeric branched chain alkenes.
15.3.7 Chemical Properties
I. Addition Reactions. Alkenes contain loosely held  -electrons, therefore, they undergo addition reactions
in which electrophiles add on to the carbon-carbon double bond to form the addition products.
1. Addition of dihydrogen (Catalytic hydrogenation). Alkenes readily add dihydrogen in presence of
Raney nickel (an active form of nickel), platinum or palladium as catalyst at room temperature or ordinary
nickel at 523 - 573 K to form alkanes (Sabatier and Senderen’s reduction)
Ni Ni
CH 2  CH 2  H 2 
or Pt or Pd
 CH3  CH3 ; CH 3CH  CH 2  H 2 
523573K
 CH3CH 2CH 3
2. Addition of halogens. Halogens readily add to alkenes to form 1,2- dihaloalkanes.
CCl
CH 2  CH 2  Br2 
4
 CH 2 Br  CH 2 Br
Ethene 1,2 Dibromoethane

Ether
CH3  CH  CH 2  Cl 2  CH3  CHCl  CH 2Cl
Pr opene 1,2 Dichloropropane

During the addition of bromine to alkenes, the orange red colour of bromine is discharged since the dibromide
formed is colourless. This reaction is, therefore, used as a test for unsaturation in organic compounds.
Fluorine reacts violently with alkenes, while iodine does not react at ordinary temperatures. The order of
reactivity of addition of halogens to alkenes is F2 > Cl2 > Br2 > I2.
3. Addition of hydrogen halides. Alkenes react with hydrogen halides (HCl, HBr, HI) to form monohaloalkanes
called alkyl halides. The order of reactivity of hydrogen halides in this reacation is HI > HBr > HCl.
Lower the bond dissociation energy, more reactive is the hydrogen halide; HI (300kJ mol-1) > HBr (360kJ
mol-1) > HI (430kJ mol-1). The product formed, depends on whether the alkene is symmetrical or unsymmetrical.
(i) Addition of HBr to symmetrical alkenes. When the alkene is symmetrical only one product is possible.

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(ii) Addition of HBr to unsymmetrical alkenes. When the alkene is unsymmetrical, two products are
possible. E.g., the addition of HBr to propene in the dark and in the absence of peroxides give two products,
1-bromopropane and 2-bromopropane, but 2-bromopropane is the major product.

This may be explained on the basis of Markovnikov’s rule.

Markonikov’s rule. Markovnikov’s rule (Russian chemist, 1869) states that in the addition of unsymmetrical
reagents to unsymmetrical alkenes, the negative part of the addendum goes to the carbon atom bearing
the lesser number of hydrogen atoms.

The secondary carbocation, being more stable than the primary carbocation is formed at a faster rate. It is
attacked by Br – to form 2-bromopropane as expected (major product).

Anti Markonikov Addition or Peroxide effect. In presence of peroxides such as benzoyl peroxide, addition
of HBr (but not HCl or HI) to unsymmetrical alkenes takes place contrary to Markovnikov’s rule. This is
known as Peroxide effect or Kharasch effect.
(C H COO) , 
CH 3CH  CH 2  HBr 
6 5 2
Anti.Mark.addn.
 CH 3CH 2 CH 2 Br .
Pr opene 1 Bromopropane

Mechanism : The reaction proceeds via free radical chain mechanism as given below:

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The formation of 1-bromopropane as the major product is because the secondary free radical (step iii) is more
stable than the primary free radical.
Peroxide effect is not observed in the case of HCl or HI. This may be becaause the H–Cl bond being stronger
(430.5 kJ mol–1) than H–Br bond (363.7 kJ mol–1) is not cleaved by the free radical, whereas the H–I bond
is weaker (296.8 kJ mol–1) and iodine free radicals combine to form iodine molecules instead of adding to the
double bond.
4. Addition of sulphuric acid (Indirect hydration of alkenes). Cold conc. H2SO4 adds to alkene to form
alkyl hydrogen sulphate which on boiling with water, undergoes hydrolysis to produce alcohol. In case of
unsymmetrical alkenes, addition occurs in accordance with Markovnikov’s rule.

 
(i)Conc.H SO
 CH3CH 2  OSO2 OH ; CH 2  CH 2 
CH 2  CH 2  H OSO2OH  2
(ii) H 2O,
4
 CH3CH 2OH
Ethene Sulphuric acid Ethyl hydrogen sulphate Ethene Ethanol

5. Addition of water (Direct hydration of alkenes): Alkenes add water in presence of mineral acids to
form alcohols. Addition occurs in accordance with Markovnikov’s rule.

II. Oxidation Reactions

1. Combustion. Alkenes burn in air or oxygen to form CO2 and H2O. All combustion reactions are highly
exothermic.
 2CO 2  2H 2 O;  C H  1411kJmol 1
CH 2  CH 2  3O 2 
2. Oxidation with cold dilute neutral or alkaline KMnO4 solution. Alkenes are readily oxidised by cold
dilute neutral or alkaline KMnO4 solution to give vicinal or 1, 2 -diols or 1, 2-glycols. This reaction is called
hydroxylation since two hydroxyl groups are added across the double bond.
dil.KMnO
CH 2  CH 2  H 2 O  O 
273K
4
 HO  CH 2  CH 2  OH
Ethene Ethane1,2diol (Glycol )

dil.KMnO
CH 3CH  CH 2  H 2O  O 
273K
4
 CH 3CH(OH)  CH 2 (OH)
Pr opene Pr opan e1,2diol

The pink colour of KMnO4 solution is discharged with the formation of a brown precipitate of manganese
dioxide.Therefore, this reaction is used as a test for unsaturation (Baeyer’s test)

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(b) Oxidation with acidic KMnO4 or K2Cr2O7 solution. Acidic potassium permanganate or acidic potassium
dichromate oxidises alkenes to ketones and/or acids depending on the nature of the alkene and the experimental
conditions. With terminal alkenes, one of the products is always methanoic acid. With non-terminal alkenes,
carboxylic acids or ketones or both are obtained.
KMnO ,H  KMnO4 ,H 
(CH 3 ) 2 C  CH 2 
4
 (CH 3 )2 C  O ; C H3CH  CHCH3   2CH3COOH
2 Methylpropene Pr opan 2 one But 2ene Ethanoic acid

KMnO ,H 
CH3CH 2CH  CHCH 3  4[O] 
4
 CH 3CH 2COOH  CH3COOH
2 Pentene Pr opanoic acid Ethanoic acid
(ii) Ozonolysis. When ozone is passed through a solution of alkene in an inert solvent such as CH2Cl2, CHCl3
or CCl4 at low temperature (196-200K), the corresponding ozonide is formed. Ozonides are unstable explosive
compounds, therefore, they are reduced, in situ, with zinc dust and water or H2/Pd to give aldehydes or
ketones or a mixture of these depending on the structure of the alkene (reductive cleavage).

This reaction is useful for detecting the position of the double bond in alkenes and other unsaturated compounds.
III. Polymerisation Reactions. Polymerisation is a process by which a large number of simple molecules
combine to form a large molecule. The simple molecules are called monomers while the large molecule is
called a polymer or macromolecule. Alkenes readily undergo polymerisation in presence of catalysts.
(a) Polymerisation of ethene. When ethene is heated to 473K at 1500 atmosphere pressure in presence of
traces of oxygen (0.001-0.1%) polythene (polyethylene) is obtained.

Polythene is widely used as packaging material and insulation for electric wires and cables. It is also used for
making squeeze bottles, refrigerator trays, toys, pipes, radio and T.V. cabinets, etc.
15.3.8 Uses of alkenes. (i) Lower members are used as fuels and illuminants. (ii) Alkenes and substituted alkenes
on polymerization, form a number of useful polymers such as polythene, PVC, teflon, orlon etc. (iii) Ethene is
used for the preparation of ethyl alcohol and ethylene glycol (anti-freeze). (iv) Ethylene is also used for artificial
ripening of green fruits.

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PART - III - ALKYNES

15.4 Alkynes. Acyclic unsaturated hydrocarbons containing a carbon-carbon triple bond are called alkynes or
acetylenes. Their general formula is CnH2n-2 where n = 2, 3, 4, etc.
15.4.1 Nomenclature
Alkynes are named as derivatives of acetylene, e.g., methyl acetylene (propylene or propene). In IUPAC
system, they are called alkynes. The name of an alkyne is obtained by replacing the terminal ‘ane’ of the
corresponding alkane by the suffix ‘yne’. The position of triple bond is indicated by arabic numerals; the
numbering starting from that end which is closer to the triple bond.
Table - 03 Common and IUPAC names of alkynes (Cn H2n-2)
Value of n Formula Structure Common name IUPAC name
2 C2 H 2 HC  CH Acetylene Ethyne
3 C3 H 4 CH 3  C  CH Methyl acetylene Propyne
4 C4 H 6 CH 3CH 2 C  CH Ethyl acetylene But  1  yne
4 C4 H 6 CH 3C  CCH 3 Dimethyl acetylene But  2  yne

(i) If both double and triple bonds are present, the double bond is always given preference. (ii) If there is no
choice in numbering, lowest set of locants rule is followed.
1 2 3 4 5 6
C H2  C H C H2 C H2 C  C H
Hex 1en 5 yne

15.4.2 Isomerism in Alkynes

(i) Position isomerism. The first two members, i.e., ethyne and propyne exist in one form only. But butyne
and higher alkynes exhibit position isomerism due to different positions of the triple bond on the carbon chain.

CH  CCH 2CH 2CH 2CH3 CH 3 C  CCH 2 CH 2 CH 3 CH 3 CH 2 C  CCH 2 CH 3

Hex 1 yne Hex  2  yne Hex  3 yne
(ii) Chain isomerism. Alkynes having five or more carbon atoms show chain isomerism due to different
structures of the carbon chain.

15.4.3 Classification of Alkynes. Alkynes are classified as terminal or non-terminal alkynes according as the triple
bond is present at the end of the carbon chain or within the carbon chain.

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Terminal alkynes :

Non-terminal alkynes :

15.4.4 Structure of Triple bond

Alkynes contain a carbon-carbon triple bond which consists of a  -bond and two  -bonds. Each carbon
atom in ethyne is sp-hybridised. sp-hybridised orbitals of carbon atoms undergo head on overlap to form the
carbon-carbon  -bond. The second sp-hybridised orbital on each carbon overlaps with s-orbital of hydrogen
forming sp-s (C–H)  -bond. The two unhybridised p-orbitals (2px and 2py) on each carbon atom, which are
perpendicular to each other as well as to the C–C  -bond, overlap side ways to form two  -bonds.
Since a p-orbital has two lobes, the electron cloud of a  -bond has two halves. The four halves of the electron
clouds of two -bonds merge together to form a single electron cloud which has cylindrical symmetry about the
internuclear axis. Because of cylindrical symmetry of the electron cloud between carbon atoms, ethyne is a
linear molecule with H–C–C bond angle of 180o.
The strength of C  C bond of ethyne (823 kJ mol–1) is greater than C=C bond of ethene (599 kJ mol–1) and
C–C bond of ethane (348 kJ mol–1). Further, due to the smaller size of sp-orbitals (as compared to sp2 and
sp3) and sideways overlap of p-orbitals, the carbon-carbon bond length in ethyne is shorter (120pm) than
those of C=C (134pm) and C–C (154pm).

(a) (b)

Figure - 04 Orbital picture of ethyne showing (a)  -bond and (b) π -bond
In spite of the presence of two  -bonds, alkynes are less reactive than alkenes towards addition reactions.
Further, due to linear structure, alkynes (unlike alkenes) do not exhibit geometrical isomerism.
15.4.5 Methods of Preparation
1. By the action of water on calcium carbide. Ethyne (acetylene) is prepared in the laboratory as well as
on a commercial scale by the action of water on calcium carbide.
 2275 K
CaCO3 
 CaO  CO 2 ; CaO  3C   CaC2  CO
Calcium carbonate Calcium oxide Calcium carbide

CaC2  2H 2O 
 HC  CH  Ca(OH) 2
Calcium carbide Acetylene

2. By dehydrohalogenation of dihaloalkanes.
i) Dehydrohalogenation of vic-dihaloalkanes by heating with alcoholic KOH.

CH 2 Br  CH 2 Br  2KOH  alc. 
 HC  CH  2KBr  2H 2O
1, 2Dibromoethane Acetylene

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ii) Alkynes can be prepared from gem-dihalides by the action of alcoholic KOH followed by treatment with
NaNH2 in liquid NH3 or sodamide in liquid NH3.
KOH(alc.) NaNH
CH 3CHBr2  2KOH  alc. 
 HBr, H O
 CH 2CHBr  2
 NaBr,  NH
 HC  CH
1,1Dibromoethane 2
Vinyl bromide 3 Ethyne

3. By dehalogenation of tetrahalides. Tetrahaloalkanes when heated with zinc dust in methanol undergo
dehalogenation to yield alkynes.
CH OH
CHBr2  CHBr2  2Zn 
3

 HC  CH  ZnBr2
1,1,2,2Tetrabromoethane Acetylene

4. Synthesis from carbon and hydrogen. Acetylene is obtained on passing a steram of hydrogen through
an electric arc struck between carbon electrodes.
Electric arc
2C  H 2 
 HC  CH
3270K Acetylene
5. Kolbe’s electrolytic reaction. Acetylene is obtained on electrolysis of concentrated solution of sodium
or potassium salt of maleic acid or fumaric acid.
Electrolysis
KOOC  CH  CH  COOK  2H 2O 
 HC  CH  2CO2  H 2  2KOH
Potassium fumarate
15.4.6 Physical Properties
i) Physical state. The first three members (ethyne, propyne, and butyne) are colourless odourless gases, the
next eight are liquids and the higher ones are solids.
ii) Melting and boiling points. The melting and boiling points of alkynes are slightly higher than those of the
corresponding alkenes and alkanes. This may be because alkynes have linear structures and hence their molecules
can be more closely packed in the crystal lattice.
iii) Solubility. Alkynes, being non-polar, are insoluble in water, but readily soluble in organic solvents such as
petroleum ether, benzene, carbon tetrachloride, etc.
iv) Density. Densities of alkynes, like those of alkenes and alkanes, increase with molecular size. They are all
lighter than water since their densities lie in the range 0.69 – 0.77 g/cm3.
15.4.7 Reactivity of Alkynes versus Alkenes. Although alkynes contain two  -bonds, they are less reactive
than alkenes towards electrophilic addition reactions. This is due to two reasons :
(i) Due to greater electronegativity of sp-hybridised carbon atoms (50% s-character) of a triple bond than sp2-
hybridised carbon atoms of a double bond, the  -electrons of alkynes are more tightly held by the carbon
atoms than  -electrons of alkenes and hence are less easily available for reaction with electrophiles.
(ii) Due to cylindrical nature of the  -electron cloud of alkynes,  -electrons of a triple bond are more delocalised
than  -electrons of a double bond. Consequently, the  -electrons of a triple bond are less readily available
for addition reactions than those of the double bond.
15.4.8 Chemical Properties
Alkynes show acidic character, undergo electrophilic and nucleophilic addition reactions, reduction, oxidation
and polymerisation reactions.
I. Acidic character of Alkynes. Unlike alkanes and alkenes, the hydrogen atoms of alkynes, i.e., acetylenic
hydrogens are acidic in nature.
(i) Formation of alkali metal acetylides. Ethyne and other terminal alkynes or 1-alkynes react with strong
bases such as sodium metal at 475 K or sodamide in liquid ammonia at 196 K to form sodium acetylides.

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475K 2Na
2HC  CH  2Na  2HC  C Na  
 2 Na  C  C  Na
Ethyne H2 Monosodium ethynide H 2 Disodium ethynide

liq.NH
CH3  C  CH  Na  NH 2 
3
 CH 3  C  C  Na   NH3
Pr opyne Sodium propynide

These reactions are not given by alkenes and alkynes, hence used to distinguish between alkynes alkenes and
alkanes. Acidic behaviour of alkynes alkenes and alkanes : HC  CH  H 2C  CH 2  H 3C  CH3

Acidic behaviour of alkynes follow the order : HC  CH  H3C  C  CH  H 3C  C  C  CH3 .

II. Electrophilic Addition Reactions. Due to the presence of a triple bond, they can add two molecules of
dihydrogen, halogens, hydrogen halides, etc. Most of the reactions are electrophilic additions. The product
depends on the stability of the vinylic cation. Addition to unsymmetrical alkynes obey Markovnikov’s rule.

1. Addition of dihydrogen. Alkynes add two molecules of hydrogen in presence of Pt, Pd or Ni catalyst to
form the corresponding alkane.

  H 2 C  CH 2  
Pt/Pd/ Ni H
HC  CH  H 2  2
CH 3  CH 3
Ethyne Ethane

  CH3  HC  CH 2  
Pt/Pd/ Ni H
CH3  C  CH  H 2  2
CH3  CH 2  CH3
Pr opyne Pr opane

2. Addition of halogens. Chlorine and bromine readily add to alkynes first forming 1, 2-dihaloalkenes and
then 1, 1, 2, 2-tetrahaloalkanes.
Cl Cl
HC  CH 2
CCl
 Cl  CH  CH  Cl 2
CCl
 Cl2CH  CHCl2
Ethyne 4 1,2 Dichloroethene 4
1,1,2,2Tetrachloroethane (Westron)

Br Br
CH3  C  CH 
2
 CH3  CBr  CHBr 
2
 CH3  CBr2  CHBr2
CCl4 CCl4
Pr opyne 1,2 Dibromopropene 1,1,2,2Tetrabromopropane

During this reaction, the reddish brown colour of Br2 is decolourised and hence this reaction is used as a test
for unsaturation (for double and triple bonds). The order of reactivity of halogens is Cl2 > Br2 > I2.
3. Addition of hydrogen halides. HF adds to alkynes only under pressure. The addition of halogen acids can
take place in accordance with Markovnikov’s rule to form vinyl halides and then alkylidene halides.
Their order of reactivity being HI > HBr > HCl > HF.
HBr HBr
HC  CH 
 CH2  CHBr  CH3  CHBr2
Ethyne Mark. addn.
1Bromoethene (Vinylbromide) 1,1Dibromoethane

HBr HBr
CH3  C  CH  CH3CBr  CH2  CH3  CBr2  CH3
Mark. addn. Mark. addn.
Pr opyne 2Bromopropene 2,2Dibromopropane

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However, in presence of peroxides, antiMarkonikov addition occurs,

HBr,RCOOR HBr,RCOOR
CH3  C  CH 
 CH3CH  CHBr 
 CH3  CH 2  CHBr2
AntiMark. addn. AntiMark. addn.
Pr opyne 1Bromopropene 1,1Dibromopropane

When acetylene is passed through dil. HCl in presence of Hg2+ ions as catalyst, vinyl chloride is formed.
398K
HC  CH  HCl 
2  CH 2  CHCl
Ethyne Hg
1Chloroethene (Vinylchloride)

4. Addition of H2O (Hydration of alkynes). Alkenes undergo hydration in the presence of 60% H2SO4 and
Hg2+ ions at 333 K. Under these conditions, H2O adds to the triple bond to form an enol which tautomerises
to the carbonyl compound.

In case of unsymmetrical terminal alkynes, addition occurs in accordance with Markovnikov’s rule.

III. Oxidation Reactions

1. Oxidation with air or oxygen - Combustion. Acetylene burns with a luminous yellow sooty flame due to
the presence of higher carbon content.

2CH  CH  5O 2  4CO 2  2H 2 O; H  1300 kJ mol1

Acetylene

Oxyacetylene flame. With air or oxygen under high pressure, it burns with a blue flame producing high
temperature of the order of 3000 K which is used for cutting and welding of metals.
2. Oxidation with cold dilute potassium permanganate. Alkynes are readily oxidised by cold dilute alkaline
KMnO4 solution. Acetylene, gives oxalic acid due to further oxidation of the initially formed glyoxal

During the reaction, the pink colour of KMnO4 solution is discharged, therefore, this reaction is, used as a test
for unsaturation (Baeyer’s test).
3. Oxidation with ozone . Alkynes react with ozone in presence of inert solvents such as CH2Cl2, CHCl3 or
CCl4 at low temperature (196-200 K) to form ozonides. These ozonides on decomposition with Zn dust and
water or H2/Pd (reductive cleavage) give 1,2-dicarbonyl compounds.

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IV. Polymerisation
(a) Linear polymerisation. In this type of polymerisation, two or more molcules of ethyne combine to form
products which have linear structures.
(i) In presence of CuCl / NH4Cl, acetylene first gives vinylacetylene and then divinylacetylene.

CuCl,NH Cl HCCH
2CH  CH 
4
 HC  C  CH  CH 2 
CuCl,NH Cl
 H 2 C  CH  C  C  CH  CH 2
Acetylene Vinylacetylene 4
Divinylacetylene

(ii) Under suitable conditions, polymerisation of acetylene produces the linear polymer, polyacetylene. It is a
high moleular weight conjugated polyene containing the repeating unit (CH = CH – CH = CH)n. Under proper
conditions, this material conducts electricity. Polyacetylene film is used as electrodes in batteries; it is lighter
and cheaper than metal conductors.
(b) Cyclic polymerisation. In this type of polymerisation, three or more molecules of alkyne combine to form
products having ring structures, e.g., when ethyne is passed through a red hot iron tube at 873 K, it trimerises
to give benzene.

Propyne trimerises under similar conditions to form mesitylene (1, 3, 5-trimethyl benzene).
15.4.9 Uses of Alkynes. (i) Acetylene is used as illuminant in hawker’s lamp and in light houses. (ii) It is used for
ripening of fruits and vegetables. (iii) It is also used for the manufacture of ethyl alcohol, acetaldehyde, acetic
acid, vinyl plastics, synthetic rubbers such as Buna N and synthetic fibres such as Orlon. (iv) Oxyacetylene
flame is used for cutting and welding of metals. (v) Acetylene and its derivatives are used for the synthesis of
cis- and trans-alkenes, methyl ketones, etc. (vi) Lewisite (ClHC = CHAsCl2), a war gas used in World War-
II, is obtained by the action of arsenic trichloride (AsCl3) on acetylene in presence of anhydrous AlCl3.

PART - IV AROMATIC HYDROCARBONS OR ARENES

15.5 Arenes. Hydrocarbons and their alkyl, alkenyl and alkynyl derivatives containing one or more benzene rings,
isolated or fused, in their molecules are called aromatic hydrocarbons or arenes (aromatic alkenes). Since
most of them have pleasant odour, these are called aromatic compounds.
Arenes containing benzene ring are called benzenoids and those not containing benzene ring are known as
non-benzenoids, e.g., azulene, tropolone, cyclopentadienyl anion, cycloheptatrienyl cation, etc.

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15.5.1 Classification and Nomenclature

Arenes are classified into different series depending on the number of fused benzene rings present in their
molecules. The general formula is CnH2n-6m where n is the number of carbon atoms and m is the number of
rings. For monocyclic arenes, m  1 and n  6 , or more.

For bicyclic arenes, n  10 and n  2 . The first member of this series is naphthalene. For tricyclic arenes,
n  14 and m  3 , e.g., anthracene and phenanthrene.

Certain arenes contain two or more isolated benzene rings.

15.5.2 Isomerism in Arenes

Benzene is a symmetrical molecule. All the six hydrogen atoms of benzene are equivalent, therefore, replacement
of any one hydrogen atom of benzene by a substituent will give only a single product. Thus, monosubstitution
products of benzene does not exhibit isomerism.
When two hydrogen atoms of benzene ring are replaced by same or different substituents, three isomers which
differ in the position of the substituents are possible. Thus, disubstitution products of benzene show position
isomerism. The three isomers are called ortho (o-) , meta (m-) and para (p-) according as the relative
positions of the two substituents are 1,2-; 1,3- and 1,4- respectively. Thus, the three position isomers of
dimethylbenzene (xylene) are:

(Besides the three dimethylbenzenes, the fourth isomer is ethylbenzene.)

15.5.3 Structure of Benzene
Benzene was isolated by Michael Faraday in 1825. The molecular formula of benzene, C6H6, indicated highly
degree of unsaturation, but the molecule was found to be stable. It was found to form a triozonide which
indicates the presence of three double bonds. It was further found to produce only one monosubstituted
derivative which indicates that all the six carbon and six hydrogen atoms are identical.
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Kekule structure. The first insight into the structure of benzene was given by Friedrich August Kekule
(1865). He proposed that the six carbon atoms of benzene are joined to each other by alternate single and
double bonds to form a hexagonal ring. Each carbon is further connected to a hydrogen atom. Benzene,
according to this proposal, is 1, 3, 5-cyclo octatetraene.

Figure - 05 Kekule structure of benzene

Kekule structure indicates the possibility of two isomeric 1, 2-dibromobenzenes. In one of the isomers, the
bromine atoms are attached to doubly bonded carbon atoms, whereas in the other, they are attached to singly
bonded carbons. However, benzene was found to form only one ortho disubstituted product. In order to
overcome this problem, Kekule suggested the concept of oscillating nature of the double bonds, i.e., the
positions of single and double bonds are not fixed, but oscillate back and forth between adjacent positions.
Molecular orbital structure of benzene. Since all the bond angles in benzene are 1200, all the carbon atoms
of benzene are sp2 -hybridised. Each carbon atom forms two C-C,  -bonds with neighbouring carbon atoms
and one C-H  -bond with a hydrogen atom. Thus, in all there are six C-C,  -bonds and six C-H,  -bonds
which lie in one plane.
Each carbon has an unhybridised p-orbital. The six unhybridised p-orbitals have one electron each and are not
only parallel to one another but also perpendicular to the plane of  -bonded carbon skeleton. Any two
p-orbitals on adjacent carbon atoms can overlap sideways to form a  -bond.

(a) (b) (c)

(d) (e)

Figure - 06 (a)  -bonded hexagonal carbon skeleton with unhybridised p-orbitals. (b, c) Two possible
sideways overlap of six unhybridised p-orbitals to form three  -bonds. (d)  -electron clouds lying above
and below the plane of the hexagonal ring. (e) Modern picture of benzene molecule.
In fact, each 2p-orbital overlaps equally well with 2p-orbitals on adjacent carbon atoms on either side to form
two doughnut shaped  -electron clouds; one of which lies above and other below the plane of the carbon
skeleton (Fig. 06d).
The participation of  -electrons in more than one bond is called delocalisation of electrons. As a result, all the
carbon- carbon bond lengths in benzene are equal (139 pm), all the carbon-hydrogen bonds are equivalent
(110 pm), its dipole moment is zero.

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Due to delocalization of electrons, benzene is shown by a circle within a hexagon. The hexagon represents the
six carbon atoms and the circle represents the six completely delocalised  -electrons (Fig. 06e). The
delocalisation of  -electrons makes benzene more stable than the hypothetical cyclohexatriene molecule.
15.5.4 Resonance and Stability of Benzene. Benzene can be represented as a resonance hybrid of the following
Kekule structures (I and II). Structures (I) and (II) are the canonical forms of benzene, but the actual structure
is a resonance hybrid of these two, represented as (III).

Since the canonical forms are of equal energy, they contribute equally (50%) towards the resonance hybrid.
Any two adjacent carbon atoms in benzene molecule are neither joined by a pure single bond nor by a pure
double bond. All carbon-carbon bond lengths are equal (139 pm) and lie between C=C bond length (133 pm)
and C-C bond length (154 pm). Thus, benzene is more stable than either of the two Kekule structures; the
resonance energy of benzene is 150.62 kJ mol-1.

15.5.5 Aromaticity - Huckel (4n+2) Rule. According to this rule, a system possessing the following characteristics
would behave as an aromatic compound.
(i) The molecule should be planar.
(i) It should have complete delocalization of  -electrons in the ring.
(iii) It should contain (4n+2)  -electrons in the ring where n = 0,1, 2, 3...etc.
A molecule which does not satisfy any one or more these conditions is said to be non-aromatic.
Applications of Huckel rule. A planar cyclic system containing 2, 6, 10, 14, 18 etc.  -electrons and having
a single cyclic  -electron cloud encompassing all the carbon atoms in the ring is aromatic. For example,

15.5.6 Methods of preparation

Benzene (phene), C6H6, was first isolated by Michael Faraday (1825). In 1845, benzene was found in coal-
tar by Hofmann. Arenes are mainly present in the light oil fraction of coal tar distillation.
1. Cyclic polymerization of ethyne. Benzene (b.p.353K) was first synthesized by Berthelot (1870) by
passing ethyne through a red-hot iron tube at 873K. Under similar conditions, propyne gives mesitylene.

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2. Decarboxylation. Benzene is obtained when sodium salts of benzoic acid is heated with soda-lime.

3. Reduction of phenol with zinc dust. Phenol is reduced to benzene on passing its vapours over Zn dust.

4. From chlorobenzene. Benzene is prepared by reduction of chlorobenzene with Ni-Al alloy/NaOH.

15.5.7 Physical Properties

Benzene and its homologues containing upto eight carbon atoms are colourless liquids while the higher ones
are solids with characteristic smell. They burn with a sooty flame due to high carbon content. They are
insoluble in water, but soluble in organic solvents.
Melting and boiling points. Boiling points of arenes increase with increase in molecular size
due to increase in magnitude of van der Waals forces of attraction. Amongst isomeric arenes, (o-, m– and
p-xylenes) the p-isomer, which is most symmetrical has the highest melting point.
15.5.8 Chemical Properties
Arenes are highly unsaturated but behave like saturated hydrocarbons. They are more reactive than alkanes
but less reactive than alkenes and alkynes. The lower reactivity is due to extra stability of delocalised
  electrons.
I. Electrophilic Substitution Reactions. Common electrophilic substitution reactions are halogenation,
nitration, sulphonation and Friedel-Crafts alkylation and acylation.
1. Halogenation. Benzene reacts with chlorine and bromine in presence of Lewis acids such as AlCl3, FeCl3,
FeBr3, etc. as catalyst and in absence of light to form chlorobenzene and bromobenzene respectively.

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The function of the Lewis acid is to carry the halogen to the aromatic hydrocarbon (halogen carrier).
Fluorination of arenes is too vigorous to be of practical use.
Two or more or even all the hydrogen atoms of benzene may be successively replaced by the electrophile if
excess of reagent is used. Thus, benzene on treatment with excess Cl2 in presence of anhydrous AlCl3 in the
dark yields hexachlorobenzene (C6Cl6).

2. Sulphonation. The process of replacement of a hydrogen atom by a sulphonic acid group (-SO3H) is
called sulphonation. Arene treated with fuming sulphuric acid or oleum (concentrated sulphuric acid containing
dissolved sulphur trioxide) or chlorosulphonic acid gives benzene sulphonic acid.

3. Nitration. It is the process of replacement of a hydrogen atom by nitro group (–NO2). Nitration is carried
out by treating an arene with a mixture of conc. HNO3 and conc. H2SO4 (nitrating mixture).

4. Friedel - Crafts reaction. This is a convenient method for the introduction of an alkyl or an acyl (RCO-)
group into an arene. It is of two types;
(i) Friedel - Crafts alkylation. When benzene or its homologue is treated with an alkyl halide in presence of
anhydrous aluminium chloride as catalyst, alkylbenzene is formed.

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(ii) Friedel - Crafts acylation. With acid chloride or acid anhydride in presence of anhydrous aluminium
chloride, benzene gives an acylbenzene.

Mechanism of Electrophilic Substitution Reactions

Benzene ring acts as a source of electrons and attracts electrophiles. Thus, the reactions of benzene and other
arenes are initiated by electrophiles. Whereas, arenes can undergo electrophilic addition as well as substitution
reactions, they preferably undergo electrophilic substitution (SE).
Electrophilic substitution reactions proceed via three steps: (i) Generation of the eletrophile, (ii) Formation of
carbocation intermediate and (iii) Removal of proton from carbocation intermediate.
Step 1 : Generation of electrophile, E . During chlorination, alkylation and acylation of benzene, anhydrous
AlCl3, being a Lewis acid, helps in the generation of the elctrophiles, Cl (chloronium ion), R  , RC O
(acylium ion) respectively by combining with the attacking reagent :

 SO3  HSO 4  H3O

; 2H 2SO 4 

Sulphur trioxide which acts as the electrophile in sulphonation is generated by the acid- base equilibrium
between two molecules of H2SO4.
Nitration is carried out using a mixture of conc.HNO3 and conc.H2SO4, called nitrating mixture. Nitronium
ion, NO2+ is produced by the transfer of a proton from sulphuric acid to nitric acid :

(a) (b)
Step 2 : Formation of carbocation intermediate. Attack of the electrophile on arene results in the formation
of a  -complex or arenium ion in which a carbon atom becomes sp3 hybridised.

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The arenium ion is resonance stabilised

The sigma complex loses its aromatic character because delocalisation of electrons is not possible at the sp3
hybridised carbon.
Step 3 : Removal of proton. The  -complex releases a proton from the sp3 hybridised carbon and regains
aromatic character on attack by [A lCl 4]–in case of halogenation, alkylation and acylation and [HSO4]– in case
of nitration.

E.g., Mechanism of halogenation. Halogenation is carried out at low temperatures (310-320 K), in the
absence of sunlight and in presence of a Lewis acid such as anhydrous ferric or aluminium chloride as catalyst.
Step 1 : Generation of electrophile.

Step 2 : Attack of electrophile. The positive end of the polarised chlorine molecule accepts the  -electron
cloud of the benzene ring to form a  -complex or the carbocation intermediate which is stabilized by resonance.

This step is slow and hence is the rate-determining step of the reaction.
 

Step 3 : Abstraction of proton. The base AlCl 4 present in the reaction mixture then abstracts a proton to
form chlorobenzene.

This step is fast and hence does not affect the rate of the reaction.
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II. Addition Reactions. Benzene and its homologues usually do not undergo addition reactions. However, at
high temperature and pressure, they undergo certain addition reactions.
(a) Addition of dihydrogen (Hydrogenation ): Benzene adds three molecules of dihydrogen in presence of
Raney nickel or platinum at 473 - 523K to form cyclohexane.

(b) Addition of halogens : In presence of uv light and in the absence of halogen carrier (eg. AlCl3, FeBr3,
etc.), benzene adds three molecules of chlorine to form benzene hexachloride (BHC) or Gammaxane.

In the absence of sunlight and in the presence of halogen carriers, benzene undergoes substitution
reactions but in presence of sunlight and absence of halogen carriers, it undergoes addition reactions.
III. Oxidation (Combustion). Arenes readily burn in air or oxygen with a luminous yellow sooty flame
producing CO2 and H2O.

2C 6 H 6  15O 2  12CO 2  6H 2 O;  C H o  3000kJ mol 1

The sooty flame is due to higher carbon content. This is a qualitative test to distinguish aromatic compounds
from aliphatic compounds.
15.5.9 Uses of Arenes
(i) Lower arenes such as benzene, toluene, xylenes, etc. are used as solvents for oils, fats, rubbers,etc. (ii)
Arenes are extensively used in the manufacture of dyes, drugs, perfumes, explosives (TNT) and insecticides
(DDT, BHC). (iii) Benzene is blended with petrol to improve its octane number. (iv) p-Xylene is used to
prepare terephthalic acid and its ester for manufacture of synthetic fibre called Terylene or Dacron.
15.5.10 Directive Influence of Substituents and their Effect on Reactivity
When a monosubstituted benzene is subjected to further substitution, the substituent already present determines
the position of the incoming group. Either ortho and para products or meta product is predominently formed.
The ability of a substituent already present in the benzene ring to direct the incoming group to a particular
position is called directive influence of substituents.
I. o- and p-Directing Groups. These are substituents which direct the incoming group to ortho and para
positions, e.g., CH 3 , CH 2 CH 3 , C6 H 5 , Cl,  Br,  I,  OH, OCH 3,  OCOCH 3 ,  NH 2 ,  NHCH 3 ,

 NHCOCH 3 , etc. In general, all electron - donating groups are o- and p-directing.

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Nitration of toluene gives a mixture of ortho and para - nitrotoluene

Although two o-positions and one p-position are available for disubstitution, it is usually the p-isomer which
predominates. This is due to steric hindrance by the group at o-position.
Groups having at least one lone pair of electrons can involve in resonance with the   electrons of the benzene
ring. As a result, electron density increases at all the nuclear positions of the benzene ring, but the increase is
much more at o- and p-positions (structures II-IV) than at m-positions.

In electrophilic substitution reactions, the electrophile attacks the benzene ring at a position where the electron
density is high. Since the electron density is high at o- and p- positions, the electrophile will attack preferentially
at o- and p-positions. Thus, –OH group is o- and p- directing.
i. Directive influence of halogens on reactivity of arenes. Halogens, though deactivating are o-
and p-directing. Halogens are deactivating due to-I-effect. However, because of the +R-effect, electron density
increases at o-and p-postions than at m-positions.

As a result, halogens are o- and p-directing. The combined effect of +R- effect and -I effect is that, halogens
are somewhat deactivating but o- and p-directing.
iii. Effect of o- and p-directing substituents on reactivity. Since o- and p- directing groups increase the
electron density in the benzene ring, the ring gets activated and further electrophilic substitution becomes
easier. Thus, all o- and p-directing groups except halogens are called activating groups. Because of the
ability of these groups to donate electrons to the benzene ring, o- and p-directing groups are also called
electron-repelling or electron-donating groups (EDG).
(i) o- and p- Directing groups facilitate further electrophilic substitution reaction. E.g, nitration of toluene
occurs faster than that of benzene because electron density in toluene ring is higher than that in benzene ring
due to hyperconjugation effect of the CH3 group.

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(ii) Higher the electron-donating ability of the substituent, the more facile the reaction. E.g., bromination
of aniline is faster than phenol because of the electron-donating ability of -NH2 group.

The electron -donating ability of substituents follows the order ; O   NH 2   NR 2  OH   OCH 3 ,

 NHCOCH3  CH3  X(F  Cl  Br  I) .

II. m-Directing groups. These are groups which direct the incoming group to the meta position e.g;
 CH3 3 N ,  NO2 , CN, CF3 , CHO, COR, COOH, COOR,SO3H,etc. In general, all electron-
withdrawing groups are m- directing.
Nitration of nitrobenzene mainly gives m- dinitrobenzene.

Groups in which the atom directly attached to benzene ring has one more electronegative atom linked to it by
a multiple bond pulls the electrons of the multiple bond towards it which, in turn, withdraws electrons from all
the nuclear positions. As a result, electron density falls at all the nuclear positions, but the decrease is more at
o- and p-positions than at m-positions.
The meta- directing influence of –NO2 group, is due to electron-withdrawing resonance effect (-R-effect) of
the –NO2 group.

i. Effect of m-directing substituents on reactivity. Since m- directing groups decrease the electron-density
in the benzene ring, the ring gets deactivated and further electrophilic substitution becomes difficult. Therefore,
meta-directing groups are deactivating groups. These are also called electron-attracting or electron-
withdrawing groups (EWG).
(i) m-Directing substituents make further electrophilic substitution more difficult. For example, further
nitration of nitrobenzene is difficult because of electron-withdrawing -I and -R effects of –NO2 group.
(ii) Higher the electron-withdrawing ability of the substituent, the more difficult is the reaction. Nitration
of nitrobenzene occurs much slower than benzoic acid since –NO2 is much more electron-withdrawing
than –COOH.

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15.6 CARCINOGENICITY AND TOXICITY

Benzene and polynuclear aromatic hydrocarbons, (PAHs) containing more than two benzene rings fused
together are toxic and cancer-causing (carcinogenic), e.g., 1, 2-benzanthracene, 3-methylcholanthrene, 1, 2-
benzpyrene, 9, 10-dimethyl -1, 2-benzanthracene and 1, 2, 5, 6-dibenzanthracene.

Polynuclear aromatic hydrocarbons are formed by incomplete combustion of organic matter such as coal,
petroleum, tobacco, etc. They are widely present in the environment and are the major cause of cancer.
It is belived that when PAHs enter the human body, they are converted into dihydroxy epoxides which react
with purine bases such as guanine present in DNA and RNA of the cells. This causes mutations and ultimately
leads to cancer.

PAH 
O2
 PAH epoxide 
 PAH dihydroxy epoxide 
DNA or RNA
 Mutations 
 Cancer

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QUESTIONS
LEVEL - I
1. The structure of alkane or cycloalkane with molecular formula C8H18 that has only 1o H atoms is:
1) 2, 2, 3, 3 -Tetramethylbutane 2) 2, 2, 3-Trimethylpentane
3) 2, 2, 4-Trimethylpentane 4) 2, 3,3-Trimethylpentane
2. Which of the following can be used for the preparation of propane?
1) CH 3CH  CH 2 (i) B2 H 6
(ii) H 2 O 2 / NaOH
2) CH 3CH 2 CH 2 Cl 
(i) Mg/ether
(ii) H 2 O

NaOH  CaO 
3) CH3CH 2CH 2 I 
HI./Heat
150o C
 4) CH 3CH 2 CH 2 COONa 
Heat

3. Wurtz reaction converts alkyl halide into alkane when it is made to react with:
1) Na in alcohol 2) Na in dry ether 3) Zn in alcohol 4) Zn in dry ether
4. Which one of the following is expected to have minimum boiling point?
1) n-Butane 2) n-Pentane 3) 2-Methylbutane 4) 2,2-Dimethylpropane
5. Which statement is correct?
1) Low chemical reactivity of alkanes is due to strong C – C and C – H bonds.
2) Alkanes show characteristic substitution reactions because they are saturated
3) Reaction of alkanes with fluorine is explosive even in dark
4) All of the above
6. On mixing a certain alkane with chlorine, and irradiating it with ultraviolet light, it forms only one
monochloroalkane. This alkane could be
1) Neopentane 2) Propane 3) Pentane 4) Isopentane
7. Which of the following product is obtained at cathode during Kolbe’s electrolysis of RCOO–Na+(aq.)?
1) Alkane 2) Sodium 3) Hydrogen 4) Sodium hydroxide
8. The product obtained on heating n-heptane with Cr2O3-Al2O3 at 600 C is:
o

1) cyclohexane 2) cyclohexene 3) benzene 4) toluene

9. Chlorination of n-butane produces:
1) 1-chlorobutane as the chief product 2) 2-chlorobutane as the chief product
3) 1-chlorobutane more than 2-chlorobutane 4) 2-chlorobutane more than 1-chlorobutane
10. In the complete combustion of CnH2n+2, the number of moles of oxygen required is:
n  n 1  3n  1  n2
1)   O 2 2)   O2 3)   O2 4)   O2
2  2   2   2 
11. The relative stability of the three isomers of pentane, namely, n-pentane, isopentane and neopentane follows
the order
1) n-pentane > isopentane > neopentane 2) n-pentane > neopentane > isopentane
3) neopentane > isopentane > n-pentane 4) neopentane > n-pentane > isopentane
12. The compound 1, 2-butadiene has:
1) only sp-hybridised carbon atoms 2) only sp2-hybridised carbon atoms
3) both sp- and sp2-hybridised carbon atoms 4) sp-, sp2- and sp3- hybridised carbon atoms
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13. n-Propyl bromide on treating with ethanolic potassium hydroxide produces:

1) propane 2) propene 3) propyne 4) propanol
14. The dehydration of 2-methylbutanol with concentrated H2SO4 produces:
1) 2-methylbutene as the major product 2) 2-methylbut-2-ene as the major product
3) 1-pentene 4) pent-2-ene
15. Which of the following yields both alkane and alkene?
1) Williamson’s synthesis 2) Kolbe’s electrolytic method
3) Wittig reaction 4) Sandmeyer’s reaction
16. Which of the following hydrocarbons has the lowest dipole moment?
H3C CH3

1) C=C 2) CH 3C  CCH 3 3) CH 3CH 2C  CH 4) CH 2  CH  C  CH

H H
17. Ozonolysis of 2, 3-dimethyl-1-butene followed by reduction with zinc and water gives:
1) Methanoic acid and 3-methyl-2-butanone 2) Methanal and 3-methyl-2-butanone
3) Methanal and 2- methyl-3-butanone 4) None of these
18. Position of double bond in alkenes is identified by:
1) Bromine water 2) Ammoniacal silver nitrate solution
3) Ozonolysis 4) Baeyer’s reagent
19. The addition of HCl in the presence of peroxide does not follow anti-Markovnikov’s rule because:
1) HCl bond is too strong to be broken homolytically
2) Cl atom is not reactive enough to add on to a double bond
3) Cl combines with H to give back HCl 4) HCl is a reducing agent
20. The treatment of CH 3C  CH 2 with NaIO4 or boiling KMnO4 produces:
|
CH 3
1) CH 3COCH3  CH 2O 2) CH 3CHO  CH 3CHO
3) CH 3COCH3  CO2 4) CH 3COCH 3  HCOOH
CH2OH
21. In reaction, CH 2  CH 2 
Hypochlorous acid
 A 
B
 , then A and B are:
CH2OH
1) CH3CH2Cl and NaOH 2) CH3– CH3 and KOH
3) CH3CH2OH and HCl 4) CH2OH – CH2Cl and aq. NaHCO3
22. Which of the following is the predominate product in the reaction of HOBr with propene?
1) 2–Bromopropan–1–ol 2) 3–Bromopropan–1–ol
3) 2–Bromopropan–2–ol 4) 1–Bromopropan–2–ol
23. An alkene, obtained by the dehydration of an alcohol (A), on ozonolysis gives two molecules of acetaldehyde
for every molecule of alkene. The alcohol (A) is:
1) CH 3CH 2 CH 2 OH 2) CH 3CH 2 OH 3) CH 3CH  CHCH 2OH 4) CH 3CH 2 CHCH3
|
OH
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Br2/hv Alc. KOH HBr./Peroxide

24. X Y Z. The compound Z is:

CH2
Br Br Br
1) 2) 3) 4)

25. The reaction of C 6 H 5 CH  CHCH 3 with HBr produces:

CH = CHCH3

1) C6 H5 CH 2 CHCH 3 2) C 6 H 5CH 2 CH 2 CH 2 Br 3) 4) C6 H5 CHCH 2 CH 3

| |
Br Br
Br
CH 3
|
(i) Hg  CH3COO 2 ; THF
26. The product of following reaction is CH 3  C  CH  CH 2 
 ii  NaBH 4  NaOH 
:
|
CH 3
CH 3 CH3
| |
1) CH 3  C  CH  CH3 2) CH 3  C  CH 2  CH 2 OH
| | |
CH 3 OH CH3
OH CH 3 CH3
| | |
3) CH 3  C  CH  CH3 4) HOCH 2  C  CH 2  CH 3
| |
CH 3 CH3
27. 2-Phenyl propene on acidic hydration gives:
1) 2-Phenyl-2-propanol 2) 2-Phenyl-1-propanol
3) 3-Phenyl-1-propanol 4) 1-Phenyl-2-propanol
28. Identify a reagent from the following list which can easily distinguish between 1-butyne and 2-butyne:
1) bromine, CCl4 2) H2, Lindlar catalyst
3) dilute H2SO4, HgSO4 4) ammoniacal Cu2Cl2 solution
29. The addition of water to propyne in the presence of HgSO4 – H2SO4 produces:
1) CH3CH = CHOH 2) CH 3COCH 3 3) CH 3CH 2 CH 2 OH 4) CH3CHO

30. The reduction of an alkyne to alkene using Lindlar catalyst results into:
1) cis addition of hydrogen atoms
2) trans addition of hydrogen atoms
3) a mixture obtained by cis and trans additions of hydrogen in equimolar amounts.
4) a mixture obtained by cis and trans additions of hydrogen atoms in unequal amounts.
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31. The treatment of CH 3CH 2 C  CCH 2 CH 2 CH 3 with KMnO4 under neutral conditions at room temperature gives

1) CH3CH 2 CH  CHCH 2 CH 2 CH3 2) CH3CH 2CH  C  CH 2CH 2CH3

| | | ||
OH OH OH O
3) CH 3CH 2  C  C  CH 2 CH 2 CH 3 4) CH 3CH 2 COOH  CH 3CH 2 CH 2 COOH
|| ||
O O
32. The hydrocarbon which decolourises alkaline KMnO4 solution, but does not give any precipitate with
ammoniacal silver nitrate is:
1) benzene 2) acetlyene 3) propyne 4) butyne-2
33. The compound (i) decolourises KMnO4 (ii) forms ozonide with ozone and (iii) undergoes polymerization. It
will be:
1) C6H6 2) C3H8 3) C2H4 4) C2H6
34. Which of the following species is aromatic ?
1) cyclopropenyl cation 2) cyclobutadiene 3) cyclopentadiene 4) cyclopropane
35. Which of the following is non-aromatic in nature?

1) 2) 3) 4)

36. Which of the following has maximum resonance energy?

1) Diphenyl 2) Benzene 3) Naphthalene 4) Phenanthrene
37. The reaction of toluene with chlorine in presence of ferric chloride gives predominantly
1) benzoyl chloride 2) m-chlorotoluene 3) benzyl chloride 4) o-and p-chlorotoluene
38. In the reaction, C6 H 5 CH 3 
Oxidation
 A 
NaOH
 B 
Soda lim e

 C the product C is:

1) C6 H 5OH 2) C6 H 6 3) C6 H 5COONa 4) C6 H 5ONa

39. The correct sequence of activating power of a group in benzene is
1) – NH2 > –NHCOCH3 > – CH3 2) – NH2 < – NHCOCH3 < – CH3
3) – NH2 > – NHCOCH3 < – CH3 4) – NH < – NHCOCH3 > – CH3
40. Which product is formed when the following compound is treated with Br2 in the presence of FeBr3?
CH3

CH3
CH3 CH3 CH3 CH3
Br Br
1) 2) 3) 4)
CH3 CH3 CH3 Br CH3
Br

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LEVEL - II
1. Out of the five isomeric hexanes, the isomer that can give two monochlorinated compounds is:
1) 2,3-Dimethyl butane 2) 2, 2-Dimethyl butane 3) 2, 2-Dimethyl pentane 4) n-Hexane
o o o o
2. The compound which contains all the four 1 , 2 , 3 , 4 carbon atoms is:
1) 2, 3-Dimethylpentane 2) 2, 2, 4-trimethylpentane
3) 2, 3, 4-Trimethylpentane 4) 3, 3-Dimethylpentane
3. What volume of CH4 at NTP is formed when 20.5 g of CH3COONa is treated with sodalime?
1) 4.4 litre 2) 2.2 litre 3) 3.2 litre 4) 5.6 litre
CH 2 Br
|
Zn / 
4. Identify ‘X’ in the reaction: Br  CH 2  C  CH 2 Br   X ; X is:
|
CH 2 Br
CH 3  C  CH 2
1) | 2) 3) 4) CH 3C  C  CH  CH 2
CH  CH 2
5. Which of the following reactions will not give propane?
1) Cl 
Mg/ether
H2O
 2) 
B2 H 6 /ether
CH3COOH

Cl O
3) P  HI

 4) 
CH3 MgX
H2O

Me Me
6. What would be the product formed when 1-bromo-3-chlorocyclobutane reacts with two equivalents of
metallic sodium in water?

1) 2) Cl Br

3) Cl Cl 4) Br Cl

7. Which of the following molecules has the minimum bond energy of the indicated C—H bond?
1) CH3 – H 2) CH3CH2 – H 3) (CH3)2CH – H 4) (CH3)3C – H
8. Out of the following compounds, I) Pent-1-ene II) Pent-2-ene III) 2-Methyl but-1-ene
IV) 2-Methyl but-2-ene. Which pair has the lowest and the highest heats of combustion, respectively?
1) (IV) and (I), respectively 2) (I) and (IV), respectively
3) (II) and (III), respectively 4) (III) and (II), respectively
9. 10 ml of a certain hydrocarbon require 25 mL of oxygen for complete combustion and the volume of CO2
produced is 20 mL. What is the formula of hydrocarbon?
1) C2H2 2) C2H4 3) CH4 4) C2H6
10. What product would you expect from addition of deuterium chloride to 2-cyclohexyl-4-methyl-2-pentene?

1) 2) 3) 4)

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11. Which of the following conformations of cyclohexane is most stable?

1) Chair 2) Boat 3) Twist-boat 4) Half-chair
12. The most stable conformation of the product of following reaction is:

HBr/H2O2 HBr/Dark
C CH (1 equivalent) X Product

H Ph Ph H
Ph Br H Br H H Br H

1) H H 2) H H 3) H Br 4) Ph Br

Br Br Br H
13. When alcoholic solution of ethylene dibromide is heated with granulated zinc, the compound formed is:
1) ethane 2) ethylene 3) butane 4) isobutane
14. When ethene reacts with bromine in aqueous sodium chloride solution, the product(s) obtained is (are)
1) ethylene dibromide only 2) ethylene dibromide and 1-bromo-2-chloroethane
3) 1-bromo-2-chloroethane only 4) ethylene dichloride only
15. An alkene on oxidative ozonolysis gives adipic acid. The alkene is:
1) cyclohexene 2) 1-methylcyclopentene 3) 1,2-dimethylcyclobutene 4) 3-hexene
16. Which of the following will have lower pK a value?

Ha
I. II. III. Hc IV.
Hd
Hb
1) Ha 2) Hb 3) Hc 4) Hd
17. . Which reagent cannot be used for the above conversion?

1) 2) Et3N 3) POCl3 4) NH3

18. Which one of the following alkenes will react fastest with H2 under catalytic hydrogenation condition?
R R R H R R R R
1) 2) 3) 4)
H H R H R H R R

O OH
19. Alkene (A)  KMnO 4

. ‘A’ is
O

1) 2)

3) 4)

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20. Propene, CH 3  CH  CH 2 can be converted into 1-propanol by oxidation. Which set of reagents among
the following is ideal to effect the conversion?
1) Alkaline KMnO4 2) B2H6 and alk. H2O2 3) O3/zinc dust 4) OsO4/CHCl3
21. Which compound on reductive ozonolysis forms only glyoxal?
1) Ethyne 2) Ethene 3) Ethane 4) 1, 3-butadiene

22. In the following reaction, the major product is :

1) 2) 3) 4)

23. During ozonolysis of CH2 = CH2 if hydrolysis is made in absence of Zn dust the products formed are:
1) HCHO 2) HCOOH 3) CH3OH 4) CH2OHCH2OH
24. A hydrocarbon of formula C6H10 absorbs only one molecule of H2 upon catalytic hydrogenation. Upon
ozonolysis the hydrocarbon yields,

H H
| |
O  C – CH 2 – CH 2  CH 2  CH 2  C  O . The hydrocarbon is:

1) cyclohexane 2) benzene 3) cyclohexene 4) cyclobutane

25. Hydroboration oxidation and acid hydration will yield the same product in case of :
Me
Me Me Ph Ph
(i) (ii) Me—  —Me (iii) (iv)
Me
1) (i) and (ii) 2) (ii) and (iv) 3) (i) and (iv) 4) (i), (ii), (iii) and (iv)
26. Which of the following will undergo faster dehydrobromination?
Br Br Br Br

1) 2) 3) 4)

27. The synthesis of 3-Octyne is achieved by adding a bromoalkane into a mixture of sodium amide and alkyne.
The bromoalkane and alkyne, respectively, are

1) BrCH 2 CH 2 CH 2CH 2 CH 3 and CH 3CH 2C  CH 2) BrCH 2CH 2 CH3 and CH 3CH 2 CH 2C  CH

3) BrCH 2CH 2 CH 2CH 2CH 3 and CH3C  CH 4) BrCH 2CH 2 CH 2CH 3 and CH 3CH 2 C  CH

28. 2-Hexyne gives trans-2-hexene on treatment with:

1) Li/NH3 2) Pd/BasO4 3) LiAlH4 4) Pt/H2

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29. Addition of HBr on CH  C  CH 2  CH  CH 2 and CH  C  CH  CH 2 separately gives:

1) CH  C  CH 2  CHBr  CH 3 and CH 2  C  CH  CH 2
|
Br
2) CH 2  C  CH 2  CHBr  CH 2 and CH 2  C  CH  CH 2
| |
Br Br
3) CH  C  CH 2  CH  CH 3 and CH  C  CH  CH 3 4) None of the above
| |
Br Br
30. The compound X (C5H8) reacts with ammoniacal AgNO3 to give a white precipitate and reacts with excess of
KMnO4 to give the acid,  CH3  2 CH  COOH . Therefore, X is:

1) CH 2  CH  CH  CH  CH 3 2) CH 3  CH 2 2 C  CH

3)  CH3  2 CHC  CH 4)  CH3 2 C  C  CH 2

31. H–   H (i) O3
(ii) H 2 O/ Zn
(X) 
Zn /CH3COOH
  Y  . Compound (Y):

O O OAC
OH
1) 2) Me – COOH 3) HO 4) Me
H H OAC
32. Which statement is correct about cyclopentadienyl anion (I) and benzene (II)?
1) Both (I) and (II) are aromatic but (II) is more stable than (I)
2) Both (I) and (II) are aromatic and have the same stability
3) (II) is more aromatic and more stable than (I) and it is non-aromatic
4) (I) is more stable than (II) though both are aromatic.
33. Which among the following is aromatic?
– 2–

1) 2) 3) 4)

34. Which of the following molecules/species are aromatic in character?

35. On passing benzene vapour through a tube at 700-800oC or through molten lead we get:
1) diphenyl 2) phenol 3) toluene 4) benzaldehyde

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36. Benzene contains double bonds but does not give addition reactions because:
1) double bonds in benzene are strong
2) double bonds change their position rapidly
3) resonance lowers the energy of benzene molecule and leads to greater stabilisation
4) none of the above
37. Oxidation of naphthalene by acidic KMnO4 gives:
1) toluene 2) benzaldehyde 3) phthalic acid 4) benzoic acid

(i) BH 3 , THF HF
38. Identify ‘ C’ in the following reaction: + CH2 = CH– CH2Cl 
AlCl3
A 
B C
(ii) H 2 O 2 / OH Heat

1) 2) 3) 4)
OH
CH2CH2CH3

Br2
39. Provide the structure of the major product(s) from the following reaction. :
hv

CH2CH2CH3
CH2CH2CH2Br CH2CHBrCH3 CHBrCH2CH 3

1) 2) 3) 4)

Br


CH2  CH  CH3 , H
40. The major product ‘U’ in the following reactions is: 
hight pressure, heat
 T 
radical initiator, O2
U

1) 2) 3) 4)

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SOLUTIONS
LEVEL - I

o
1. 1 (2, 2, 3, 3-Tetramethyl butane contains only 1 H atoms)

2. 4
3. 2 The reagent used in Wurtz reaction is Na in dry ether.
4. 4 2,2-Dimethylpropane is expected to have minimum boiling point. The boiling point decreases with
increasing branching of carbon chain
5. 4
6. 1 As only one monochloro alkane is formed, it means that the alkane contains only one set of hydrogen.
CH3

This is present only in neopentane. CH3 C CH3

CH3
7. 3 Alkane is obtained at anode while H2 gas liberates at cathode.

8. 4 Cr2 O3  Al 2 O 3
CH 3CH 2 CH 2 CH 2 CH 2 CH 2 CH 3   CH3

9. 4 2-chlorobutane is produced more than 1-chlorobutane.

 3n  1 
10. 3 Cn H 2n  2    O 2  nCO 2   n  1 H 2 O
 2 
11. 3 The stability of structural isomers generally increases with increased branching.
sp2 sp sp2 sp3
12. 4 CH2 = C = CH – CH3
1,2-Butadine

13. 2 CH3CH 2CH 2 Br 

ethanolic
KOH
 CH3CH  CH 2 .
14. 2 The dehydration involves rearrangement of carbocation.

CH3CH 2CHCH 2 OH 
 CH3CH  C  CH3
| |
CH3 CH3
15. 2 A concentrated aqueous solution of Na/K salts of saturated monocarboxylic acid on electrolysis yields
alkane. A concentrated aqueous solution of Na/K salt of saturated dicarboxylic acid on electrolysis
gives an alkene.
16. 2 The dipole moment of CH 3C  CCH 3 is zero. It is a linear molecule and has symmetrical arrangement
of methyl substituents.

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 Brilliant STUDY CENTRE

CH3 O
(i) O3
17. 2 CH2 = C – CH – CH3  HCHO + CH3 — C — CH –– CH
(ii) Zn/H 2 O
3
Methanal
CH3 CH3
3-Methyl 2- butanone
18. 3 The position of the double bond in alkene is identified by ozonolysis. Bromine water is used to detect
the presence of  -bond whereas ammoniacal AgNO3 is used to detect the presence of terminal alkynes
or –CHO group.
19. 1 The addition of HCl in the presence of peroxide does not follow anti-Markovnikov’s rule.
20. 3 CH3  C  CH 2 
NaIO 4
 CH 3COCH3  CO 2
|
CH3

CH2 CH2OH CH2OH

21. 4 

HOCl 
aq.NaHCO3
or aq.KOH
CH2 CH2Cl CH2OH
Chlorohydrin

22. 4
23. 4 Alkene is CH3CH = CHCH3, a symmetrical alkene and therefore alcohol is, CH3CH 2 CHCH3 which
|
will give alkene-2 as major product. OH

24. 3

25. 4

Hg  OOC.CH 
26. 1  CH 3 3 CCH  CH 2   CH3 3 CCHCH 2 HgOOCCH 3 
3 2 NaBH
NaOH
  CH3 3 CCHOHCH 3
4

|
OOCCH 3
27. 1 Acidic hydration of 2-phenyl propene follows electrophilic reaction mechanism forming an intermediate
3o carbocation (more stable), thereby forming 2-phenylpropan-2-ol.

28. 4 1-Butyne contains acetylinic hydrogen atom. It will form insoluble cuprous acetylides.

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29. 2 The product is CH 3C  CH 2 which rearranges to CH3COCH3.

|
OH
30. 1 Lindlar catalyst causes syn adiditon. Hence, cis addition of hydrogen atoms occur.

31. 3 At neutral conditions and at room temperature, C  C  is oxidized to C  C  .

|| ||
O O
32. 4 Non-terminal alkynes do not have acidic H-atom.
33. 3 These are characteristics of C2H4.
34. 1 Cyclopropenyl cation is the smallest aromatic compound which satisfies 4n + 2 rule with n = 0.
35. 2 10  e ’s in delocalisation, but is non-aromatic because ring is non-planar, as the H atoms that point to
the interior to the ring interfere with each other.
36. 4 Due to more canonical forms.
37. 4 Methyl group is ortho and para directing.
38. 2  
O CaO  NaOH
C 6 H 5 CH 3  C 6 H 5 COOH 
NaOH
 C6 H 5 COONa  C6 H 6
39. 1 The correct sequence of activating power of a group in benzene is – NH2 > – NHCOCH3 > – CH3
40. 3 –CH3 group is o, p-directing.
LEVEL - II
1. 1 The isomeric hexane should have two different types of H atoms and four similar types of H atoms to

give two monochlorinated products.

2. 2 2, 2, 4-trimethyl pentane contain 1o, 2o, 3o and 4o carbon atoms.

3. 4 82 g CH3COONa gives 22.4 litre CH4.

4. 2 Product will be spiropentane.

5. 4

6. 3 (C – Br) bond is weaker than (C – Cl) bond; hence, Wurtz reaction will take place with (C – Br) bond.

7. 4  CH3 3 C  H has a minimum bond energy..

8. 1 More-stable, more-substituted alkene has the lowest (more negative) H oC .
The order of stability of alkene : IV > III > II > I.
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 Brilliant STUDY CENTRE

5
9. 1 C2 H 2  O 2 
 2CO 2  H 2 O
2
10. 2 According to Markovnikov’s rule, deuterium (D) will get attached to the carbon with fewer alkyl
substituents and the chloride (Cl) will get attached to the carbon with more alkyl substituents.
11. 1

12. 3

13. 2 CH 2 BrCH 2 Br 

Zn
dust
 CH 2  CH 2  ZnBr2
14. 2
15. 1 O3 COOH
COOH

Cyclohexene Adipic acid

16. 1

The order of Aromatic stability is : Aromatic > Non-aromatic > Anti-aromatic

(I) > (II) > (IV) > (III)

 H a in (I) is more acidic since it will give H faster to become a stable aromatic anion.
17. 3 POCl3 is used for the dehydration of alcohol to alkene, others (A), (B) and (C) are used for the
dehydrohalogenation of alkyl halide to alkene.
1
18. 1 Stability of alkene  Heat of hydrogenation

Hence, the reactivity of alkene  heat of hydrogenation of alkenes.

In these, cis-alkenes are more reactive because of greater strain at the double bond.
O OH
KMnO4
19. 1 +
O

20. 2 CH3CH  CH 2 

B2 H 6
  CH3CH 2 CH 2 3 B 
H 2 O2
 CH 3CH 2 CH 2 OH  H3 BO3 ;
CHO
21. 1 CH  CH 
O3
2H
CHO
22. 4 The 1,4-addition is the major product as it is the result of the formation of a stable allylic carbocation.

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23. 2 O3 HOH
CH2 = CH2 CH2 – O – CH2 2HCHO + H2O2 
 2HCOOH

O O

24. 3 
H2
 Also, 
 ii  H  OHCCH 2 CH 2 CH 2 CH 2 CHO
(i) O3

25. 4 Same product is obtained with HBO and catalytic hydration reaction with symmetrical alkene or alkyne.

26. 4

27. 2

28. 1 2-Hexyne gives trans-2-hexene on treatment with Li/NH3.

29. 1 The first addition will occur on double bond. Thus,

CH  C  CH 2  CH  CH 2 
HBr
 CH  C  CH 2  CHBr  CH3 but in
CH  C  CH  CH 2 , the addtion will occur at CH  C  because the product formed is conjugated
alkadiene, which is more stable.
CH  C  CH  CH 2 
HBr
 CH 2  C  CH  CH 2
|
Br
30. 3 It should be terminal alkyne as it gives white ppt. with amm. AgNO3.

31. 3

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32. 1 Both 6 e 's and in delocalisation, both aromatic, but (II) is more stable than (I) because (II) is
uncharged structure.
2–

33. 4 Compound has 8 + 2 = 10  -electrons, hence is aromatic.

has 4  e– , has 8  e– , while has 8 + 1 = 9  e– ,

hence, all these species are not aromatic.

34. 3 ie, cyclopropenyl carbocation has two  -electron or 4n + 2 = 2.  n = 0, and it is a planar

molecule. Thus, it is aromatic in nature according to Huckel rule.
35. 1 2C6 H 6 
Molten lead
700o C
 C6 H5  C6 H5
36. 3
[O] COOH
37. 3
COOH
38. 2
39. 4 The major product is (1-bromopropyl) benzene as it involve the formation of stable benzyl cation as an
intermediate.
CH2CH2CH3 CHBrCH2CH 3

Br2
hv

40. 2

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CHAPTER - 16
ENVIRONMENTAL CHEMISTRY

INTRODUCTION
Environmental chemistry is the branch of science which deals with the chemical phenomena occuring in the
environment. It is the study of the origin, transport, reactions, effects and fates of chemical species in the
environment.
16.1 ENVIRONMENTAL POLLUTION
The addition of undersirable material to air, water and soil by natural source or due to human activity to such
a level of concentration that adversely affects the life on the earth is called environmental pollution.
16.1.1 Environmental Segments
Environment is nature which supports and affects various forms of life. The four environmental segments are
the atmosphere, hydrosphere, lithosphere and the biosphere. The atmosphere consists of gaseous layers
around the earth which protects the earth from the environment of outer space. It absorbs harmful radiations
like cosmic-rays, ultra-violet rays, etc., and maintains the heat balance of the earth. The hydrosphere comprises
all sources of water on earth; oceans, lakes, rivers, ground water, glaciers, polar ice caps, etc. It covers about
75% of earth’s surface. About 96% of the earth’s water is present in the oceans. The lithosphere comprises
earth’s crust up to a depth of about 16 km. It contains soil, water, minerals, and some organic matter. The
upper part consisting of weathered rocks and organic matter is called soil. The lower 8 - 40 km thick part is
called crust.
The biosphere refers to the domain of living organisms in contact with atmosphere, hydrosphere, and lithosphere.
16.1.2 Pollutant and Contaminant
1. Pollutant. A substance which causes pollution is known as pollutant. Pollutants can be solids, liquids or
gases present in greater concentration than in natural abundance and are produced due to human activities
or natural happenings.
i) Primary and secondary pollutants. Primary pollutants are those which after their formation enter
into the environment directly, e.g., NO formed from N2 and O2. Secondary pollutants are those
which are formed from primary pollutants, e.g., peroxyacylnitrate (PAN) from oxides of nitrogen and
hydrocarbons.
ii) Biodegradable and non-biodegradable pollutants. Biodegradable pollutants are those which
are decomposed by micro organisms (e.g., vegetable matter, domestic sewage, etc.) and thus may not
be harmful.
Non-biodegradable pollutants are those which are not easily decomposed and hence are harmful,
e.g., mercury, DDT, plastics, heavy metals, nuclear wastes etc.
2. Contaminant. A substance which does not occur in nature, but is introduced by human activity is called
conatminant, e.g., methyl isocyanate (MIC) which caused Bhopal Gas Tragedy.

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3. Source. The site from which pollutants or contaminants originate is called source.
4. Receptor. Any organism that is affected by a pollutant is called receptor. Humans are the receptor of
smog, polluted water, etc.
5. Sink. The material or medium which consumes or interacts with a long-lived pollutant is called sink.
Subsoil and ground water act as sink for pesticides. Oceans act as sink for atmospheric gases like CO2,
SO2, etc.
6. Threshold limiting value (TLV). It is the permissible limit of a pollutant to which a healthy person can
be exposed during 8 hours a day or 50 hours a week for lifetime without any adverse effect on health.
TLV of CO is 40 ppm while that of CO2 is 5000 ppm.
16.2 ATMOSPHERIC POLLUTION
Air pollution or atmospheric pollution is the addition of undesirable materials into the atmosphere from natural
phenomena or human activity which adversely affects the quality of air and hence life on the earth.
16.2.1 Composition and Structure of the Atmosphere. The atmosphere is a cover of gases upto a height of
about 1600 km from the sea level. Gases mainly present are O2, N2, O3, CO2 and H2O vapour. The atmosphere
is divided into four regions :
a) Troposphere. It is the lowest region of the atmosphere (0–11 km) and domain of all living organisms. With
altitude, temperature decreases from 15oC to – 56oC. Troposphere is a turbulent, dusty zone containing air,
water vapour and clouds. This is the region of strong air movement and cloud formation. It is greatly affected
by air pollution.
b) Stratosphere. Above the troposphere, between 11 and 50 km above sea level lies stratosphere. It contains
ozone and hence is called ozonosphere. It protects the living organisms from harmful ultraviolet radiations.
The temperature increases from –56oC to – 2oC.
c) Mesosphere. It contains gases in the ionized form and (O2+, NO+, etc.) is called ionosphere (50–85 km).
Its temperature decreases from –2oC to – 92oC.
d) Thermosphere. It is also ionosphere (85–500 km). Its temperature increases from –92oC to 1200oC.
Altitude range Temperature range
Region Important chemical constituents
(km) (°C)
1. Troposphere 0 - 11 15 to -56 NO2 , O2, CO2 , H2 O, NO, NH3 , SO2, H2 S,
hydrocarbons
2. Stratosphere 11-50 -56 to -2 O3
+ +
3. Mesosphere 50 - 85 -2 to -92 O2 , NO , N2 , O2
+ + +
5. Thermosphere 85 - 500 -92 to 1200 O2 , O , NO
16.2.2 Tropospheric Pollution. Tropospheric pollution may be due to (i) gaseous pollutants or by (ii) particulate
pollutants in air. Even small amounts of pollutants in air is significant compared to similar levels in food
because human beings need 12 to 15 times more air than food.
1. Gaseous air pollutants. The most important part of the atmosphere is the troposphere. The main pollutants
of the troposphere are the following:
(a) Carbon monoxide (CO). Carbon monoxide enters into the atmosphere mainly due to incomplete
combustion of fossil fuels or hydrocarbon fuels in automobiles. It is poisonous because it combines with
haemoglobin to form carboxyhaemoglobin which is much more stable (about 300 times) than oxyhaemoglobin.

Haemoglobin  CO 
 Carboxyhaemoglobin or  Hb  CO  HbCO 
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Thus transport oxygen to different parts of the body is hindered and the body becomes oxygen-starved. The
condition is called anoxia or asphyxiation. If CO exceeds 100 ppm, person suffers headache, dizziness, low
eyesight, nervousness and cardiovascular disorder. If CO exceeds 750 ppm, it leads to coma and death. In
pregnant women, increased CO levels in blood may induce premature birth, abortions and deformed babies.
(b) Carbon dioxide (CO2). Carbon dioxide is released into the atmosphere by respiration, burning of fossil
fuels, and decomposition of limestone during manufacture of cement, and volcanic eruptions. Normally it
forms about 0.03 per cent by volume of the atmosphere. With the increased use of fossil fuels, a large amount
of CO2 gets released into the atmosphere.
CO2 in air is used by green plants during photosynthesis and this maintains an appropriate CO2 level in the
atmosphere. Deforestation and burning of fossil fuels increases the CO2 level and disturb the balance in the
atmosphere. Increased level of CO2 in air is mainly responsible for global warming, i.e., CO2 is a major
greenhouse gas.
(c) Oxides of Nitrogen. At high altitudes when lightning strikes, oxygen and nitrogen combine to form oxides
of nitrogen. NO2 is oxidised to nitrate ion, NO3 which is washed down to soil (during rain), where it serves as
a fertilizer. When fossil fuels are burnt in an automobile engine (at high temperature), dinitrogen and dioxygen
combine to yield significant quantities of nitric oxide (NO) and nitrogen dioxide ( NO2).

N 2 (g)  O 2 (g) 

1483K
 2NO(g)
NO reacts instantly with oxygen to give NO2
2NO(g)  O 2 (g) 
 2NO 2 (g)
Rate of production of NO2 is faster when nitric oxide reacts with ozone in the stratosphere.
NO(g)  O 3 (g)  
 NO 2 (g)  O 2 (g)
Nitrogen dioxide (NO2) can combine with moisture to form HNO3.
The irritant reddish haze in traffic congested areas is due to oxides of nitrogen. Higher concentrations of NO2
damage leaves of plants and retard rate of photosynthesis. Nitrogen dioxide is a lung irritant that can lead to
respiratory diseases in children. It is toxic to living tissues also. Nitrogen dioxide is harmful to various textile
fibres and metals. Oxides of nitrogen in air cause acid rain and photochemical smog.
(d) Oxides of sulphur (SO2 and SO3). Oxides of sulphur enter into atmosphere mainly due to volcanic
eruptions. They are also produced by the combustion of sulphur containing coal and fuel oil in thermal power
plants and due to roasting of sulphide ores (FeS2, CuFeS2, CuS, ZnS, PbS, etc.). Particulate matter in polluted
air catalyses the oxidation of sulphur dioxide to sulphur trioxide.
2SO 2 (g)  O 2 (g) 
 2SO 3 (g)
The reaction can also be promoted by ozone and hydrogen peroxide.
SO 2 (g)  O3 (g) 
 SO3 (g)  O 2 (g)

SO 2 (g)  H 2 O 2 (l ) 
 H 2SO 4
Both SO2 and SO3 are strongly irritating to the respiratory tract. SO2 causes throat and eye irritation. Even at
low concentration, it causes respiratory diseases like asthma, bronchitis, and emphysema. High concentration
of SO2 leads to stiffness of flower buds which eventually fall off from plants.

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(e) Hydrocarbons. Hydrocarbons are produced by the incomplete combustion of fuel in automobiles. Many
hydrocarbons are carcinogenic, i.e., they may cause cancer. They harm plants by causing ageing, breakdown
of tissues and shedding of leaves, flowers and twigs.
Methane, released in huge quantities by decaying organic matter is a greenhouse gas. It is produced naturally
when vegetation is burnt, digested or rotted in the absence of oxygen. Large amounts of methane are released
from paddy fields, coal mines, from rotting garbage dumps, by cattle and by fossil fuels.
Global Warming and Greenhouse Effect
The temperature on the earth is a balanced equilibrium between the energy received from the sun and the
energy radiated back into space. Solar flux or the incident energy from the sun on earth is 1.36 kilowatt/m2.
About 66 % of the solar energy reaching the earth is absorbed while the rest is radiated back into space. The
earth absorbs radiations in the visible region (short wavelengths) and reflects those in the infrared (I.R.) region
(long wavelengths). Infrared are heat radiations.
Carbon dioxide in the atmosphere acts as a barrier against the flow of heat from the earth into space. Thus,
they add to the heating of the atmosphere. This is called greenhouse effect because, like the glass of a
greenhouse, CO2 permits the passage of short wavelength radiations and impedes the passage of long wavelength
(heat) radiations. This results in global warming.
Carbon dioxide is the major contributor to global warming. Other greenhouse gases are methane, water
vapour, nitrous oxide, chlorofluorocarbons (CFCs), and ozone.
Effects. (i) If the amount of carbon dioxide crosses the 0.03 percent, the natural greenhouse balance may get
disturbed. With a doubling of CO2 concentration in the atmosphere, the temperature on the earth’s surface will
increase by 1.9°C.
(ii) Global warming may lead to melting of glaciers and polar ice caps and flooding of low lying areas all over
the earth. Global warming could make changes in climate.
(iii) Increase in global temperature will increases the incidence of infectious diseases like dengue, malaria,
yellow fever, sleeping sickness etc.
(iv) Summers will be more scorching and winters more freezing. Rains will be more intensive.
Acid rain
Oxides of nitrogen and sulphur along with solid particles in the atmosphere settle down on the ground either as
dry deposition or in water, fog and snow as wet deposition.
Normal pH of rain water is 5.6 due to the presence of H+ ions formed by the dissolution of atmospheric CO2.
 H CO (aq), H CO (aq) 
H2O(l) + CO2(g)   H+(aq) + HCO - (aq)
2 3 2 3 3

When the pH of the rain water is below 5.6, it is called acid rain.
Thus, acid rain is rain water containing H2SO4 and HNO3 along with small amounts of HCl which are formed
from the oxides of sulphur and nitrogen present in air as pollutants by the following reactions:
2SO 2 (g) + O 2 (g) + 2H 2 O (l ) 
 2H 2SO 4 (aq)

4NO 2 (g) + O 2 (g) + 2H 2 O (l ) 

 4HNO3 (aq)
HNO3 and H2SO4 formed above combine with HCl present in air and come down to earth along with rain.
Effects. (i) Acid rain can cause damage to buildings made of marble (like The Taj Mahal of Agra), marble
statues (stone leprosy), and iron and steel structures.

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(ii) It is harmful for agriculture, trees and plants as it dissolves and washes away nutrients.
(iii) It causes respiratory ailments in human beings and animals.
(iv) When acid rain reaches rivers, lakes, etc., it affects plants and animal life in aquatic ecosystem. Increase
of acid in lakes causes fish to die.
(v) It corrodes water pipes resulting in the leaching of heavy metals such as iron, lead and copper into the
drinking water.
Control. Acid rain can be reduced by limiting the use of vehicles driven by fossil fuels, using less sulphur
content fossil fuels for power plants and industries and using catalytic convertors in internal combustion engineer
so that NO2 are converted into N2.
Particulate Pollutants
Particulates are small solid particles and liquid droplets suspended in air. They are present in air in the form of
soot (produced by incomplete combustion of fossil fuels), fly ash (coming with the furance flue gases), organic
particles (paraffins, oleffins, polycyclic aromatic hydrocarbons, i.e., PAH, etc.).
Viable and Non-viable particulates. Particulates may be classified as viable and non-viable. Viable
particulates are small size living organisms such as bacteria, fungi, moulds, algae, etc. Non-viable particulates
are formed by disintegration of large size materials or condensation of small size particles or droplets. These
include mist, smoke, fumes and dust.
Effects. Particulates enter the lungs, where they act as sites for adsorption of carcinogenic compounds causing
lung cancer and bronchital asthma. This disease is common among industrial workers and is called
pneumoconiosis (asbestosis due to asbestos, silicosis due to silica, etc).
Human beings are allergic to some of the fungi found in air.
Particulate pollutants bigger than 5 microns are likely to lodge in the nasal passage, whereas particles of about
1.0 micron enter into lungs easily. Lead used to be a major air pollutant emitted by vehicles.
Leaded petrol used to be the primary source of air-borne lead emission in Indian cities. This has now been
overcome by using unleaded petrol. Lead interferes with the development and maturation of red blood cells.
Particulates scatter light and hence affect visibility on foggy nights.
They counteract the greenhouse effect as they reflect back heat of sunlight.
Smog
The word smog is derived from smoke and fog because earlier, it was found to be formed by condensation of
fog on carbon particles. This is the most common example of air pollution that occurs in many cities throughout
the world. There are two types of smog:
(i) Classical smog (London smog or Sulphurous smog). Classical smog occurs in cool humid climate. This
type of smog was first observed in London in 1952. It is formed due to presence of SO2 and humidity in the air
which combine to form H2SO4 fog which deposits on particulates. It involves smoke and fog. It is formed in
the winter months, particularly in the morning hours when the temperature is low. As it contains C and SO2, it
is a reducing mixture and so it is also called as reducing smog. It causes bronchitis (irriation in the lungs).
(ii) Photochemical smog (Los Angeles smog). It was first observed in Los Angeles in 1950. The common
components of photochemical smog are ozone, nitric oxide, acrolein, formaldehyde and peroxyacetyl nitrate
(PAN). It is formed in the summer months, during afternoon when there is bright sunlight so that photochemical
reactions can take place.
It is formed due to photochemical reactions taking place when air contain NO and hydrocarbons. When these
pollutants build up to sufficiently high levels, a chain reaction occurs from their interaction with sunlight in which
NO is converted into nitrogen dioxide (NO2). This NO2 in turn absorbs energy from sunlight and breaks up
into nitric oxide and free oxygen atom.
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2NO + O 2 
 2NO 2 , NO 2 
hv
 NO + O ..... (i)
Oxygen atoms are very reactive and combine with the O2 in air to produce ozone.
O 2  O 
 O3 ..... (ii)
The ozone formed in the above reaction (ii) reacts rapidly with the NO(g) formed in the reaction (i) to regenerate
NO2. NO2 is a brown gas and at sufficiently high levels can contribute to haze.
NO + O3 
 NO2 + O2 ,
Ozone is a toxic gas and both NO2 and O3 are strong oxidising agents and can react with the unburnt
hydrocarbons in the polluted air to produce chemicals such as formaldehyde (HCHO), acrolein
(CH2=CHCH=O) and peroxyacetyl nitrate (PAN).
Hydrocarbons + O   RCO (free radicals) , RCO + O   RCO . ,2 3

RCO3.  Hydrocarbons 
 RCHO, R 2 C=O ,

RCO3.  NO 
 RCO 2  NO 2 , RCO3.  O 2 
 O3  RCO 2 ,

RCO3.  NO2 
 RCO NO 3 2 , 3CH 4 + 2O3 
 3CH 2 = O + 3H 2 O
Peroxyacetyl nitrate (PAN) Formaldehyde

Thus, concentration of O3, PAN, RCHO and R2CO builds up in the atmosphere.
Photochemical smog has high concentration of oxidising agents (O3, NO2) and, therefore, an oxidising smog.
Effects of photochemical smog. (i) Photochemical smog causes serious health problems. Both ozone and
PAN act as powerful eye irritants. Ozone and nitric oxide irritate the nose and throat and their high concentration
causes headache, chest pain, dryness of the throat, cough and difficulty in breathing. (ii) It leads to cracking of
rubber and extensive damage to plant life. (iii) It also causes corrosion of metals, stones, building materials,
rubber and painted surfaces.
Control of photochemical smog. (i) Use of catalytic converters in automobiles prevent the release of nitrogen
oxide and hydrocarbons into the atmosphere. (ii) Planting trees like Pinus, Juniparus, Quercus, Pyrus and Vitis
which metabolise nitrogen oxide.
16.2.2 Stratospheric Pollution
1. Ozone layer (Ozone Umbrella). The upper stratosphere (altitude 25–30 km) consists of considerable
amount of ozone (1-5 ppm), which protects life on earth from harmful ultraviolet (UV) radiations ( 255 nm)
from the sun. These radiations cause skin cancer (melanoma) in humans.
Formation and Breakdown of Ozone. Ozone in the stratosphere is a product of UV radiation acting on O2.
UV radiations split molecular oxygen into oxygen atoms, which combine with molecular oxygen to form ozone.
O 2 (g) 
uv
 O(g) + O(g) , O(g) + O 2 (g) 
 O3 (g)
O3 (g) 
uv
 O 2 (g)  O(g)  Heat
Ozone is unstable and decomposes to molecular oxygen; the heat liberated raises the temperature of stratosphere.
Thus, a dynamic equilibrium exists between the production and decomposition of ozone molecules.
In recent years, there have been reports of depletion of ozone layer. The main reason for ozone layer depletion
is believed to be the release of chlorofluorocarbons (CFCs), also known as freons into the atmosphere.

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CFCs are non reactive, non flammable, non toxic organic molecules used as refrigerants, propellants in spray
cans, in the production of plastic foam, for cleaning computer parts, etc. Once released into the atmosphere,
CFCs mix with atmospheric gases and eventually reach the stratosphere, where they are broken down by UV
radiations, releasing chlorine free radical.
 
CF2 Cl 2 (g) 
uv
 Cl (g) + C F2 Cl (g) ..... (i)
The chlorine radical reacts with stratospheric ozone to form chlorine monoxide free radicals and molecular
oxygen.
 
Cl (g) + O3 (g) 
 Cl O (g) + O 2 (g) ..... (ii)
Reaction of chlorine monoxide free radical with atomic oxygen produces more chlorine free radicals.
 
Cl O (g) + O(g) 
 Cl (g) + O 2 (g) ..... (iii)
Chlorine free radicals are continuously regenerated and cause the breakdown of a large number of ozone
molecules.
2. The Ozone Hole. The Ozone hole over the South Pole (Antarctica) was first reported in the 1980s. A
unique set of conditions is responsible for the ozone hole. In the summer season, nitrogen dioxide and methane
react with chlorine monoxide and chlorine atoms forming chlorine sinks, preventing ozone depletion.
 
 ClONO 2 (g) ; Cl (g) + CH 4 (g) 
Cl O (g) + NO 2 (g)   CH 3 (g) + HCl(g)
In winter, polar stratospheric clouds provide surface on which chlorine nitrate gets hydrolysed to form
hypochlorous acid. It also reacts with hydrogen chloride to give molecular chlorine.
ClONO 2 (g) + H 2 O (g) 
 HOCl(g) + HNO3 (g)
ClONO 2 (g) + HCl (g) 
 Cl 2 (g) + HNO3 (g)
When sunlight returns to the Antarctica in the spring, the sun’s warmth breaks up the clouds and HOCl and Cl2
are photolysed by sunlight to give chlorine radicals.
  
h
HOCl(g)   OH (g) + Cl (g) ; Cl 2 (g) 
 2Cl (g)
The chlorine free radicals, initiate the chain reaction for ozone depletion as described in Eqns. (ii) and (iii).
Effects of Depletion of the Ozone Layer. With the depletion of ozone layer, more UV radiation reach the
troposphere. UV radiations cause ageing of skin, cataract, sunburn, skin cancer, death of many phytoplanktons,
damage to fish productivity etc. Plant proteins are affected by UV radiations which leads to the harmful
mutation of cells. It also increases evaporation of surface water through the leaves of plants and decreases the
moisture content of the soil. Increase in UV radiations damage paints and fibres, causing them to fade faster.
16.3 WATER POLLUTION
Water pollution is the presence of undesirable substances in water in undesirable levels, which makes it harmful
to living beings and unfit for domestic, industrial and agricultural use.
Point source and non-point source. Easily identified source or place of pollution is called point source,
e.g., place where municipal and industrial discharge enter the water-source. Non-point sources are those
where a source of pollution cannot be easily identified, e.g., agricultural run off, acid rain, drainage, etc.
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16.3.1 Types of water pollution

(i) Ground water pollution. Ground water (water present below the surface of the earth) gets contaminated
due to industrial effluents, fertilizers, pesticides, etc.
(ii) Lake pollution occurs due to inflow of inorganic nutrients from agricultural lands and toxic materials from
urban areas or dumping of industrial effluents.
(iii) River pollution occurs due to discharge of industrial wastes and domestic sewage.
(iv) Sea water pollution or marine pollution occurs due to wreckage of oil tankers or oil leakage from pipe
lines or refineries or deliberate marine pollution by crude oil. The spreading of oil into sea is called oil-spill
and the thick layer of oil on the surface is called oil slick. It causes heavy damage to fishery, sea birds and
aquatic animals.
16.3.2 Causes of water pollution
Major sources of water pollution are sewage and domestic wastes, industrial effluents, agricultural nutrients,
pesticides and insecticides, polychlorinated biphenyls (PCBs), etc.
Pathogens (disease causing agents) are serious water pollutants. These include bacteria and other organisms
that enter water from domestic sewage and animal excreta. Human excreta contain bacteria such as Escherichia
coli and Streptococcus faecalis which cause gastrointestinal diseases.
Excessive phytoplankton growth in water also causes of water pollution. Organic wastes are biodegradable
by bacteria, but they consume dissolved oxygen in water.
Table - 01 Major water pollutants and their sources

Pollutant Source
Micro-organisms Domestic sewage
Organic wastes Domestic sewage, animal excreta and waste, decaying animals and plants, discharge
from food processing units
Plant nutrients Chemical fertilisers
Heavy metals Industries and chemical factories
Sediments Erosion of soil by agriculture and mining, deforestation.
Pesticides Chemicals used for killing insects, pests, rodents, fungi, weeds, etc.
Radioactive materials Mining of minerals of uranium and other radioactive metals.
Heat Water used as coolant in industries, power plants, etc.
Chemical Pollutants. Soluble inorganic chemicals that contain heavy metals such as cadmium, mercury,
nickel, etc., constitute an important class of pollutants. These metals damage the kidneys, central nervous
system, liver, etc. Acids like H2SO4 from mine drainage and salts from various sources are water pollutants.
Petroleum products may cause major oil spills in oceans. Pesticides like DDT, BHC and various industrial
chemicals like polychlorinated biphenyls (PCBs) which are used as cleansing solvents, detergents and fertilizers
are water pollutants. PCBs are suspected to be carcinogenic.

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Nowadays most of the detergents available are biodegradable, but their use can also create problems. Bacteria
responsible for degrading these detergents can feed on them, grow rapidly and use up the dissolved oxygen in
water. Lack of oxygen can kill all other forms of aquatic life such as fish and plants.
Eutrophication. The process in which nutrient (such as phosphates from fertiliser run off) enriched water
bodies support excessive algal growth which kills animal life by depriving oxygen and results in loss of biodiversity
is known as eutrophication.16.3.3 Water Quality Parameters
1. Dissolved oxygen (DO). The amount of oxygen that water can hold in solution is limited (up to 10 ppm).
Dissolved oxygen in water is very important for aquatic life. Decomposition of organic matter depletes DO. If
DO of water is below 6 ppm, the growth of fish and other aquatic organisms gets inhibited.
The dissolved oxygen is used by microorganisms to oxidise organic matter. If too much of organic matter is
present in water, all the available oxygen is used up. This causes oxygen dependent aquatic life to die. Thus,
anaerobic bacteria begin to break down the organic waste and produce foul smelling harmful chemicals.
Aerobic bacteria degrade these organic wastes, thereby depleting dissolved oxygen.
2. Biochemical Oxygen Demand (BOD). It is the total amount of oxygen consumed by bacteria in
decomposing the organic matter present in a certain volume of a sample of water. BOD5 is a measure of
dissolved oxygen consumed in 5 days. Clean water has BOD less than 5 ppm, whereas highly polluted water
has BOD greater than 17 ppm.
3. Chemical Oxygen Demand (COD). It is the total amount of oxygen required to oxidise all the oxidisable
pollutants present in a sample of water. The water sample is treated with an oxidising agent (K2Cr2O7+H2SO4)
and the amount of oxidant consumed is expressed in terms of oxygen equivalence.
16.3.4 International Standards for Drinking Water. Water quality parameters suggested by WHO:
1. pH. The pH of drinking water should be 5.5 - 9.2.
2. Fluoride. The deficiency of fluoride in drinking water can cause tooth decay, etc. Soluble fluoride in
drinking water should be upto 1 ppm (1 mg dm–3). Fluoride makes teeth harder by converting hydroxyapatite,
[3(Ca3(PO4)2.Ca(OH)2], the enamel, into fluorapatite, [3(Ca3(PO4)2.CaF2]. However, fluoride
concentration above 2 ppm causes brown mottling of teeth and over 10 ppm causes harmful effects on
bones and teeth.
3. Lead. Drinking water gets contaminated with lead when lead pipes are used for water supply. The prescribed
upper limit concentration of lead in drinking water is about 50 ppb. Lead poisoning can damage kidney,
liver, reproductive system, etc.
4. Sulphate. Excessive sulphate (>500 ppm) in drinking water causes laxative effect.
5. Nitrate. Excess nitrate (>50 ppm) in drinking water can cause a disease called methemoglobinemia
(‘blue baby’ syndrome).
6. Metals. The recommended maximum concentration of some metals in drinking water are given below:

Metal Fe Mn Al Cu Zn Cd
-1
Permissible le ve l (mg L ) 0.2 0.05 0.2 3 5 0.005

16.4 SOIL POLLUTION

Soil pollution is defined as the presence of toxic pollutants or contaminants in soil, in undesirable concentrations
to pose a risk to human health and the ecosystem. The main soil pollutants are, (i) industrial wastes, (ii) urban
wastes, (iii) radioactive wastes and (iv) fertilizers containing excess of nitrates and phosphates and (v) pesticides.
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Pesticides. After World War II, DDT was put to use in agriculture to control insects, rodents, weeds and
various crop diseases. However, due to adverse effects, its use has been banned in India.
The repeated use of the same or similar pesticides give rise to pests that are resistant to that group of pesticides.
As insect resistance to DDT increased, other organic pesticides such as Aldrin and Dieldrin were introduced.

Most of the organic toxins are water insoluble and nonbiodegradable. These highly persistent toxins are
transferred from lower trophic level to higher trophic level through food chain. At each trophic level, the
pollutant becomes 10 times concentrated. Over time, the concentration of toxins in higher animals reach a level
which causes serious metabolic and physiological disorders.

Figure - 03 At each trophic level, the pollutant gets 10 times concentrated

Due to persistence of chlorinated organic toxins, less persistent, biodegradable organophosphates and
carbamates have been introduced, but these are severe nerve toxins and hence more harmful to humans.
Insects have become resistant to these insecticides also.
Herbicides such as sodium chlorate (NaClO3), sodium arsinite (Na3AsO3) , etc., caused a shift from mechanical
to chemical weed control, but these are also not environment friendly. Most herbicides are toxic to mammals,
but are not as persistent as organochlorides (decompose in a few months). Like organochlorides, these too
become concentrated in the food web. Some herbicides cause birth defects.
16.5 INDUSTRIAL WASTE
Non-biodegradable wastes are generated by thermal power plants which produce fly ash, integrated iron and
steel plants which produce blast furnace slag and steel melting slag, industries producing metals such as aluminium,
zinc and copper produce mud and tailings and fertilizer industries produce gypsum. Hazardous wastes such as
inflammables, composite explosives or highly reactive substances are produced by industries dealing in metals,
chemicals, drugs, pharmaceuticals, dyes, pesticides, rubber goods, etc.
16.6 STRATEGIES TO CONTROL ENVIRONMENTAL POLLUTION
Control/Management of environmental pollution. Two main sources of environmental pollution are (i)
household wastes and (ii) industrial wastes.

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Household waste can be managed using separate garbage bins for biodegradable and non-biodegradable
(recyclable) wastes like polythene bags, etc. which choke the sewers. Biodegradable wastes can be deposited
in land fills or converted into compost. Non-biodegradable materials such as plastic, glass, metal scraps, etc.
can be sent for recycling.
Industrial waste can be managed by: (i) recycling wastes, e.g., glass, plastic, etc., (ii) burning and incineraton,
e.g., plant leaves, etc., (iii) sewage treatment before disposing it off, (iv) digesting, i.e., degradation of toxic
organic waste by micro organisms, and (v) dumping of sewage sludge into land.
New innovations have led to different uses of waste material. Nowadays, fly ash and slag from the steel
industry are utilised for manufacture of cement. Plastic waste is being convertd into fuel which has high octane
number; it is a green fuel. Plastic waste is used to make clothes by textile industries.
16.7 GREEN CHEMISTRY
Green chemistry deals with producing chemicals of our daily needs using such reactions and chemical processes
which neither use toxic chemicals nor emit such chemicals into the atmosphere. The techniques generally used
in green chemistry are: (i) Use of sunlight (photochemistry) and microwaves; (ii) Use of sound waves
(sonochemistry); (iii) Use of enzymes.
In a chemical reaction, if reactants are fully converted into useful environment friendly products by using an
environment friendly medium then there would be no pollution. Synthetic reactions in aqueous medium should
be promoted since water is cost effective, noninflammable and devoid of carcinogenic effects.
During synthesis, care must be taken to choose starting materials and methods that can provide maximum
(upto 100 per cent) yield.
Metathesis reactions. Yves Chauvin, Robert H. Grubbs and Richard R. Schrock (2005 Nobel Prize in
chemistry) developed the metathesis method in organic synthesis. It is a way to rearrange groups of atoms
within molecules to get product(s). Metathesis has applications in pharmaceuticals, biotechnology and food
production industries. It is also used in the development of environment-friendly polymers
Green Chemistry in day-to-day Life
(i) Dry Cleaning of Clothes. Tetra chloroethene (Cl2C = CCl2), used as solvent for dry cleaning, contaminates
ground water and is a suspected carcinogen. The process has been replaced by one in which liquid
carbondioxide, with a suitable detergent is used.
Nowadays hydrogen peroxide (H2O2) is used for bleaching clothes, which gives better results and use
lesser amount of water.
(ii) Bleaching of Paper. Chlorine gas used for bleaching paper has been replaced by hydrogen peroxide with
suitable catalyst.
(iii) Synthesis of Chemicals. Ethanal (CH3CHO) is produced by one step oxidation of ethene in presence of
ionic catalyst in aqueous medium with an yield of 90%.
CH 2  CH 2 + O 2 
Pd (II)/Cu (II)
 CH 3CHO
Thus, green chemistry is a cost effective approach which involves reduction in material, energy consumption
and waste generation.

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QUESTIONS
LEVEL - I
1. Co-existence of biotic and abiotic components of the environment along with climatic factors such as
temperature, humidity, etc. is called:
1) eutrophication 2) ecosystem 3) atmosphere 4) halomorphism
2. Pollutant is a chemical substance or factor which disturbs:
1) our balanced environment 2) geochemical cycles
3) flora of any region 4) fauna of any region
3. Which of the following is not a major constituent of air pollutants?
1) Oxides of sulphur 2) Oxides of nitrogen 3) Carbon monoxide 4) Hydrogen sulphide
4. The pollutants released by jet aeoplanes in the atmosphere are called:
1) Photochemical oxidants 2) Photochemical reductants
3) Aerosols 4) Physical pollutants
5. Acid rain water principally contains sulphuric acid produced by oxidation of:
1) Hydrogen sulphide 2) Iron pyrites 3) Sulphur rich coal 4) Metal sulphates
6. The non-viable particulate is
1) bacteria 2) moulds 3) fungi 4) dust
7. Presence of high concentration of ozone and smog in atmospheric air causes:
1) embrittlement and decreases of folding resistance of paper
2) cracking of rubber products
3) fading of dye on textiles
4) damage of electrical insulator on high tension power line
8. When rain is accompanied by a thunderstorm, the collected rain water will have a pH value:
1) slightly higher than that when the thunderstorm is not there
2) uninfluenced by occurrence of thunderstorm
3) which depends upon the amount of dust in air
4) slightly lower than that of rain water without thunderstorm
9. ‘White lung cancer’ is caused by:
1) asbestos 2) silica 3) textiles 4) paper
10. Which one of the following is not a common component of photochemical smog?
1) Ozone 2) Acrolein
3) Peroxyacetyl nitrate 4) Chloroflurocarbons
11. All are primary pollutants except:
1) SO2 2) H2SO4 3) NO2 4) Particulate matter
12. London smog is found in:
1) Summer during day time 2) Summer during morning time
3) Winter during morning time 4) Winter during day time
13. Among the following, the one which is not a ‘greenhouse gas’ is:
1) N2O 2) CO2 3) CH4 4) O2
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14. The pollutants which come directly in the air from sources are called primary pollutants. Primary pollutants
are sometimes converted into secondary pollutants. Which of the following belongs to secondary air pollutants?
1) CO 2) Hydrocarbon 3) Peroxyacetyl nitrate 4) NO
15. Incomplete combustion of petrol or diesel in automobile engines can be best detected by testing the fuel gases
for the presence of:
1) carbon monoxide and water vapour 2) carbon monoxide
3) nitrogen dioxide 4) sulphur dioxide
16. Man dies in the atmosphere of CO because it:
1) dries up the blood 2) combines with O2 present in the body
3) reduces the organic matter of tissues
4) combines with the haemoglobin of blood, thereby making the latter incapable of absorbing O2
17. Lung diseases are four times more in urban areas than rural areas. This is due to the presence of:
1) SO2 2) CO2 3) N2 4) Water-vapour
18. SO2 and NO2 causes pollution by increasing:
1) acidity 2) alkanity 3) buffer action 4) none of these
19. The following process does not occur in the trophosphere:
1) photosynthesis 2) combustion 3) greenhouse effect 4) ozone depletion
20. Which of the following does not contribute towards the formation of photochemical smog?
1) NO 2) SO2 3) O3 4) Hydrocarbons
21. Particulate matter with size less than one micron that remain suspended in air indefinitely and transported by
wind currents are called:
1) fumes 2) mist 3) aerosols 4) soot
22. Particulates are added to the atmosphere by:
1) industrial processes 2) combustion of fuels 3) agriculture burning 4) All of these
23. Which of the following is responsible for depletion of the ozone layer in the upper strata of the atmosphere?
1) Freons 2) Polyhalogens 3) Ferrocene 4) Fullerenes
24. Depletion of ozone layer causes:
1) blood cancer 2) lung cancer 3) skin cancer 4) breast cancer
25. In the upper layer of the atmosphere, ozone is formed by the:
1) action of UV rays on oxygen 2) combination of oxygen molecules
3) action of electric discharge of oxygen molecules 4) effect of high pressures on oxygen
26. Freon is not recommended to be used in refrigerators because they:
1) increase temperature 2) deplete ozone 3) affect environment 4) affect human body
27. Peeling of ozone umbrella, which protects us from UV rays is caused by:
1) PAN 2) CO2 3) CFCs 4) Coal burning
28. Main source of lead pollution is from:
1) sewage 2) leaded gasoline 3) tobacco 4) insecticide
29. Excess nitrate in drinking water can cause:
1) methemoglobinemia 2) kidney damage 3) liver damage 4) laxative effect

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30. Biochemical Oxygen Demand, (BOD) is a measure of organic material present in water. BOD value less than
5 ppm indicates a water sample to be :
1) rich in dissolved oxygen 2) poor in dissolved oxygen
3) highly polluted 4) not suitable for aquatic life
31. Addition of phosphate fertilizers into water leads to:
1) increased growth of decomposers 2) reduced algal growth
3) increased algal growth 4) nutrient enrichment (eutrophication)
32. The process of ‘eutrophication’ is due to:
1) increase in concentration of insecticide in water
2) increase in concentration of fluoride ion in water
3) the reduction in concentration of the dissolved oxygen in water due to phosphate pollution in water
4) attack of younger leaves of a plant by peroxyacetyl nitrate
33. If BOD of a river is high, it means that the river is:
1) not polluted
2) very much polluted with inorganic chemicals
3) very much polluted with organic chemicals which are decomposed by micro-organisms
4) polluted with pesticides
34. Photochemical smog formed in congested metropolitan cities mainly consists of :
1) hydrocarbons, SO2 and CO2 2) hydrocarbons, ozone and SO2
3) ozone, peroxyacetyl nitrate and NOx 4) smoke, peroxyacetyl nitrate and SO2
35. Green chemistry deals with:
1) study of plant physiology
2) study of extraction of natural products from plants
3) detailed study of reactions involved in the synthesis of chlorophyll
4) utilization of existing knowledge base for reduction the chemical hazards along with developmental activities
LEVEL - II
1. Persistent pesticides such as DDT pass into food chain and increase in amount per unit weight of organisms
due to their accumulation in fat. This phenomenon is called:
1) biodegradation 2) biomagnification 3) biosynthesis 4) decomposition
2. Two important sinks of CO2 are:
1) plants, vehicular exhaust 2) oceans, plants
3) oceans, soil 4) plants, limestone
3. Black lung disease is caused by inhalation of:
1) coal dust 2) silica dust
3) cotton fiber dust 4) asbestos dust
4. Pneumoconiosis is caused by inhalation of :
1) CO 2) Particulate matter 3) SO2 4) CFCs

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5. The major products of the digestion of sewage sludge are:

1) CO2 + H2O 2) CO + H2O 3) CO2 + CH4 4) CH4 + H2O
6. The smog is essentially caused by the presence of :
1) O2 and O3 2) O2 and N2
3) Oxides of sulphur and nitrogen 4) O3 and N2
7. Which of the following is present in maximum amount in acid rain?
1) HNO3 2) H2SO4 3) HCl 4) H2CO3
8. Which statement among the following is not correct?
1) Primary pollutants are those which are emitted directly from the source.
2) Secondary pollutants are those which are formed in the atomosphere by chemical interactions among
primary pollutants and atmospheric constituents.
3) Acid rain is a primary pollutant.
4) Particulates refer to all atomospheric substances that are not gases.
9. Photochemical oxidant PAN is formed:
1) By action of oxides of nitrogen on hydrocarbons in presence of sunlight
2) By action of carbon dioxide on hydrocarbons in presence of sunlight
3) By action of hydrogen sulphide on hydrocarbons in presence of sunlight
4) By the action of SO2 and hydrocarbons
10. Which of the following statements is not true about classical smog?
1) Its main components are produced by the action of sunlight on emissions of automobiles and factories
2) Produced in cold and humid climate
3) It contains compounds of reducing nature
4) It contains smoke, fog and sulphur dioxide
11. Photo-chemical oxidants such as PAN or PBN are formed:
1) by action of nitrogen oxides on hydrocarbons in presence of sunlight
2) by action of carbon dioxide on hydrocarbons in presence of sunlight
3) by action of hydrogen sulphide on hydrocarbons in presence of sunlight
4) none of the above
12. Pick up the correct statement.
1) CO which is major pollutant resulting from the combustion of fuels in automobiles plays a major role in
photochemical smog
2) Classical smog has an oxidizing character while the photochemical smog is reducing in character
3) Photochemical smog occurs in day time whereas the classical smog occurs in early morning hours
4) During formation of smog the level of ozone in the atmosphere goes down
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13. Which statement among the following is correct?

1) NO is more harmful than NO2
2) SO2 is more harmful than SO3
3) Acid rain contains mainly HNO3
4) Acid rain contains mainly H2SO4 and lesser concentrations of HNO3 and HCl
14. Photochemical transformation of the automobile exhaust emission in UV wavelength of sunlight results into:
1) CH4 and C6H6 2) O3 and PAN 3) CO2 and NO2 4) CO and CO2
15. The gas emitted by supersonic jet planes that slowly depletes the ozone layer is:
1) CO 2) NO 3) SO2 4) O2
16. Identify the wrong statement in the following:
1) Ozone layer does not permit infrared radiation from the sun to reach the earth
2) Acid rain is mostly because of oxides of nitrogen and sulphur
3) Chlorofluorocarbons are responsible for ozone layer depletion.
4) Green house effect is responsible for global warning
17. Which of the following statements is wrong?
1) Ozone is not responsible for greenhouse effect
2) Ozone can oxidise nitrogen monoxide present in the atmosphere to nitrogen dioxide
3) Ozone hole is thinning of ozone layer present in stratosphere
4) Ozone is produced in upper stratosphere by the action of UV rays on oxygen
18. Which of the following statements about the depletion of ozone layer is correct?
1) The problem of ozone depletion is less serious at poles because NO2 solidifies and is not available for
consuming ClO* radicals.
2) The problem of ozone depletion is more serious at poles because ice crystals in the clouds over poles act
as catalyst for photochemical reactions involving the decomposition of ozone by Cl* and ClO* radicals.
3) Freons, chlorofluorocarbons, are inert chemically, they do not react with ozone in stratosphere.
4) Oxides of nitrogen also do not react with ozone in stratosphere.
19. ‘Fluorosis’ disease is caused due to the reaction of ....... with excess of fluoride in the body.
1) Ca 2) Mg 3) Fe 4) K
20. Oxidizing agent usually used in determination of COD is:
1) H2O2 2) K2Cr2O7 3) KMnO4 4) O3
21. The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to
be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm respectively. This water is unsuitable for drinking due to high
concentration of:
1) fluoride 2) lead 3) nitrate 4) iron

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22. High concentration of fluoride is poisonous and harmful to bones and teeth at levels over:
1) 1 ppm 2) 3 ppm 3) 5 ppm 4) 10 ppm
23. Growth of fish is not as healthy in warm water as in cold water because:
1) the amount of D.O. in warm water is higher than in cold water
2) warm water is not liked by fish 3) cold water contains more marine plants
4) the amount of D.O. in warm water is less than in cold water
24. Which of the following statements is false?
1) The main reason for river water pollution is industrial and domestic sewage discharge
2) Surface water contains a lot of organic matter, mineral nutrients and radioactive materials
3) Oil spill in sea water causes heavy damage to fishery
4) Oil slick in water increases D.O. value
25. Lead in water can cause:
1) eye disease 2) arthritis 3) kidney damage 4) hair falling
26. Which is incorrectly matched?
Disease Cause
1) Fluorosis - Fluorides in water
2) Minamata - Mercury in fishing water
3) White lung disease - Chemical from petroleum industry
4) Pneumoconiosis - Particulate pollutants from industry
27. Select the incorrect statement.
1) Water is considered pure if it has BOD less than 5 ppm.
2) In COD determination, the pollutants resistant to microbial oxidation are not oxidized by oxidizing agent
like K2Cr2O7.
3) The lower the concentration of DO, the more polluted is the water sample
4) The tolerable limit of lead in drinking water is 50 ppb.
28. A fertile soil is likely to have a pH of:
1) 3 2) 9 3) 6-7 4) 14
29. DDT and BHC may act as :
1) allergens 2) carcinogens
3) asthmatic agents 4) none of these
30. Growing more trees help to:
1) reduce oxygen in the environment 2) increase carbon dioxide in the environment
3) reduce carbon dioxide only in the environment 4) reduce CO2 and increase O2 in the environment

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31. Modes of controlling pollution in large cities include:

1) less use of insecticides
2) proper disposal of organic wastes, sewage and industrial effluents
3) shifting of factories out of the residential area
4) all the above
32. Which of the following is incorrect regarding green chemistry?
1) It involves forestation of wastelands.
2) It involves utilization of information regarding toxicity and hazardous nature of chemicals.
3) It involves development of new synthetic methods and analytical tools.
4) It follows the same tradition of scientific discovery and understanding that has characterized chemistry from
its origins.
33. Green chemistry has helped in development of process for manufacture of polystyrene foam sheet packaging
material by using........ as the blowing agent.
1) Steam 2) CO2 3) CFCs 4) Methane
34. Which one is not correct? Green house effect:
1) is due to high concentration of CO2 in atmosphere
2) is influenced by gases such as CH4, O3 and chlorofluorocarbons
3) would result in the warming up of the earth
4) would result in lowering the level of oceans due to high evaporation
35. Which of the following practices will not come under green chemistry?
1) If possible, making use of soap made of vegetable oils instead of using synthetic detergents
2) Using H2O2 for bleaching purpose instead of using chlorine based bleaching agents
3) Using bicycle for travelling small distances instead of using petrol/diesel based vehicles
4) Using plastic cans for neatly strong substances

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SOLUTIONS
LEVEL - I
1. 2 Ecosystem is defined as the co-existence of biotic and abiotic components along with climatic factors.
2. 1
3. 4
4. 3
5. 2
6. 4 Dust is the non-viable particulate (mist, smoke, fumes and dust).
7. 2
8. 4 Normal rain water has a pH value 5.6. Thunderstorm forms NO which after oxidation gives HNO3.
9. 3
10. 4
11. 2
12. 3
13. 4
14. 3 Peroxyacetylnitrate (PAN) formed from the primary pollutants NO2, O3 and hydrocarbons is the
secondary pollutant.
15. 2 Incomplete combustion of petrol or diesel in automobile engines produces CO.
16. 4
17. 1
18. 1
19. 4
20. 2 SO2 has no contribution in photochemical smog.
21. 3
22. 4
23. 1
24. 3
25. 1
26. 2
27. 3
28. 2
29. 1 Excess nitrates in drinking water causes methemoglobinemia (blue baby syndrome) which results in a
blue colour skin and a congenital heart defect in infants.
30. 1 BOD less than 5 ppm shows that water is pure, ie, it is rich in dissolved oxygen.
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31. 4 Addition of phosphate fertilizers into water leads to nutrient enrichment (eutrophication)
32. 3 Eutrophication process is due to the reduction in concentration of D.O. as a result of phosphate pollution
in water.
33. 3
34. 3
35. 4
LEVEL - II
1. 2 Biomagnification is a process of increasing the amount of persistent pesticides per unit weight of organisms
due to their accumulation in fat.
2. 2 Oceans and plants are the two important sinks of CO2
3. 1 Black lung disease is caused by the inahalation of coal dust
4. 2

5. 3 The main gas evolved from sludge digestion tank is CH4; 2  CH 2 O  CO 2  CH 4

6. 3
7. 2 H2SO4 is the main contributor (60-70%), next is HNO3 (30-40%) and least is HCl.
8. 3
9. 1
10. 1 Photochemical smog and not the classical smog is produced by the action of sunlight on emissions of
automobiles and factories.
11. 1
12. 3 Photochemical smog occurs in day time whereas the classical smog occurs in early morning hours.
13. 4 NO2 is more harmful than NO, SO3 is more harmful than SO2 and acid rain contains mainly H2SO4 and
lesser amounts of HNO3.
14. 2
15. 2 NO  O3  NO 2  O 2 .
16. 1 Ozone layer does not permit ultraviolet radiation from the sun to reach the earth.
17. 1 Ozone is one of the green house gases.
 .
18. 2 CFCl3  UV light  CFCl 2  Cl
. . .   O
Cl + O3  ClO + O2 ; ClO  O  Cl 2
. 
Reaction without polar stratospheric clouds: CFCl3  UV light  Cl  CFCl 2
. . .
Cl + O3  ClO + O2; ClO  NO 2  ClONO 2 (Reservoirs)

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With polar stratospheric clouds:

The main reason for ozone depletion in polar region is that ice crystals catalyze ozone depletion reaction.
19. 1
20. 2
21. 3 In drinking water, maximum permissible concentration (tolerable limits) of different chemicals are
fluoride (1 ppm), lead (50 ppb), nitrate (50 ppm) and iron (0.2 ppm). As the given sample of water
contain higher concentration of nitrate ions (100 ppm) than the permissible limit of 50 ppm.
22. 4
23. 4 Due to lesser amount of D.O. in warm water, the growth of fish in it is not as healthy as in cold water.
24. 4 Oil slick (layer) in sea water decreases D.O. value.
25. 3 Lead in water can cause kidney damage.
26. 3 White lung disease is due to textile industry and not due to chemicals from petroleum industry.
27. 2 In COD determination the pollutants resistant to microbial oxidation are oxidized by oxidizing agent like
K2Cr2O7.
28. 3
29. 2
30. 4
31. 4
32. 1
33. 2
34. 4
35. 4

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