Page | 1

Page | 2
Page | 3
Page | 4
Page | 5
 STUDY MATERIAL OF CLASS XI CHEMISTRY: SESSION 2023-24
1. Some basic concepts of chemistry
What is chemistry?
Chemistry is the science of molecules and their properties which deals with the
study of matter, its composition, the changes that matter undergoes and the relation
between changes in composition and changes in energy..
Importance of Chemistry:
Principles of chemistry are applicable in diverse areas, such as weather patterns,
functioning of brain and operation of a computer, production in chemical industries,
manufacturing fertilisers, alkalis, acids, salts, dyes, polymers, drugs, soaps,
detergents, metals, alloys, etc., including new material. It also plays an important
role in meeting human needs for food, healthcare products and other material
aimed at improving the quality of life.
Few such examples: Cisplatin and taxol are effective against cancer therapy
whereas AZT (Azydithymidine) used against AIDS.
Matter and its nature:
Anything that occupies space, has mass, and can be felt directly by our senses are
all matters.

Physical Classification of Matter :

Properties Solid Liquid Gas

1. volume Definite Definite Indefinite

2. Shape Definite Indefinite Indefinite

Page | 6
 3. Very high force of Moderate Negligible / Very low
Intermolecular attraction
force of
attraction

4. Orderly arranged Free To move Free to move

Arrangement of within the volume every where
Molecules

5. Inter Very small Space Slightly greater Very great

molecular
Space

7. Not compressible Not compressible Highly

Compressibility compressible

8. Expansion on Very little Very little Highly expand

Heating

9. Rigidity Very rigid Not rigid Not rigid

known as fluid known as fluid

9. Fluidity Can’t flow Can flow Can flow

10. Diffusion Hardly Can diffuse Can diffuse

rate of diffusion is rate rate of diffusion is rate
very fast very fast

Chemical Classification of matter---

Elements

An element is the simplest form of matter that cannot be split into simpler substances
or built from simpler substances by any ordinary chemical or physical method.
There are 118 elements known to us, out of which 92 are naturally occurring. After
atomic no. 92, elements are known as synthetic elements, or trans-uranic
elements. Elements are further classified into metals, non-metals and metalloids.

Page | 7
 Compounds

A compound is a pure substance made up of two or more elements combined in

a definite proportion by mass, which could be split by suitable chemical methods.

Characteristics of compound

∙ Compounds always contain a definite proportion of the elements by mass.

∙ The properties of compounds are totally different from the elements from which they
are formed.

∙ Compounds are homogeneous.

∙ Compounds are broadly classified into inorganic and organic compounds. Inorganic
compounds are those, which are obtained from non-living sources such as minerals.
For example, common salt, marble and limestone. Organic compounds are those,
which occur in living sources such as plants and animals. They all contain carbon.
Common organic compounds are oils, wax, fats etc.

Mixtures

A mixture is a combination of two or more elements or compounds in any proportion so

that the components do not lose their identity. Air is an example of a mixture. Mixtures
are of two types, homogeneous and heterogeneous.

Homogeneous mixtures have the same composition throughout the sample. The
components of such mixtures cannot be seen under a powerful microscope. They are
also called solutions. Examples of homogeneous mixtures are air, sea water, gasoline,
brass etc.

Heterogeneous mixtures consist of two or more parts (phases), which have different
compositions. These mixtures have visible boundaries of separation between the
different constituents and can be seen with the naked eye e.g., sand and salt, chalk
powder in water etc.

LAWS OF CHEMICAL COMBINATIONS

Law of Conservation of Mass (Given by Antoine Lavoisier in 1789). It states that

matter (mass) can neither be created nor destroyed.

Law of Definite Proportions or Law of Constant Composition: This law was

proposed by Louis Proust in 1799, which states that: 'A chemical compound always
consists of the same elements combined together in the same ratio, irrespective of
the method of preparation or the source from where it is taken'.

Law of Multiple Proportions Proposed by Dalton in 1803, this law states that:

'When two elements combine to form two or more compounds, then the
different masses of one element, which combine with a fixed mass of the other, bear
a simple ratio to one another'.

Page | 8
 Gay Lussac’s Law of Gaseous Volumes (Given by Gay Lussac in 1808.) It states
when gases combine or are produced in a chemical reaction they do so in a simple
ratio by volume provided all gases are at same temperature and pressure.

e.g.H2(g) + Cl2(g) →2HCl(g)

1vol 1vol 2 vol

All reactants and products have simple ratio 1 : 1 : 2 by volume.

Avogadro Law (In 1811, Given by Avogadro) :According to this law equal volumes
of gases at the same temperature and pressure should contain equal number of
molecules.

Dalton's Atomic Theory: Postulates

*All substances are made up of tiny, indivisible particles called atoms. *Atoms of the
same element are identical in shape, size, mass and other properties.
*Atoms of different elements are different in all respects.
*Atom is the smallest unit that takes part in chemical reactions.
*Atoms combine with each other in simple whole number ratios to form molecules.
*Atoms cannot be created, divided or destroyed during any chemical or
physical change.

Atoms and Molecules

The smallest particle of an element, which may or may not have
independent existence, is called an atom, while the smallest particle of a substance
which is capable of independent existence is called a molecule.
Molecules are classified as homoatomic and heteroatomic. Homoatomic
molecules are made up of the atoms of the same element and heteroatomic molecules
are made up of the atoms of the different element having different atomicity (number
of atoms in a molecule of an element) like monoatomic, diatomic, triatomic and
polyatomic.
Atomic Mass Unit
One atomic mass unit is defined as a mass exactly equal to one twelfth the mass
of one carbon -12 atom. And 1 amu = 1.66056×10 g. –24

Today, amu ‘has been replaced by u’ which is known as unified mass.

Atomic Mass
Atomic mass of an element is defined as the average relative mass of an atom of
an element as compared to the mass of an atom of carbon -12 taken as 12.

Gram Atomic Mass

The quantity of an element whose mass in grams is numerically equal to its
atomic mass. In simple terms, atomic mass of an element expressed in grams is the
gram atomic mass or gram atom.
For example, the atomic mass of oxygen = 16 amu (1 atom)
Therefore gram atomic mass of oxygen = 16 g ( 1 mole of atoms)

Molecular Mass

Page | 9
 Molecular mass of a substance is defined as the average relative mass of its
molecule as compared to the mass of an atom of C-12 taken as 12. It expresses as to
how many times the molecule of a substance is heavier than 1/12th of the mass of an
atom of carbon.
For example, a molecule of carbon dioxide is 44 times heavier than 1/12th of
the mass of an atom of carbon. Therefore the molecular mass of CO 2 is 44 amu. It is
obtained by adding the atomic masses of all the atoms present in one molecule.
Gram Molecular Mass
A quantity of substance whose mass in grams is numerically equal to its
molecular mass is called gram molecular mass. In simple terms, molecular mass of a
substance expressed in grams is called gram molecular mass.
e.g., the molecular mass of oxygen = 32 amu
Therefore, gram molecular mass of oxygen = 32 g
Formula Mass
Sum of atomic masses of the elements present in one formula unit of a compound.
It is used for the ionic compounds.
Mole Concept
Mole is defined as the amount of a substance, which contains the same number
of chemical units (atoms, molecules, ions or electrons) as there are atoms in exactly
12 grams of pure carbon-12.
A mole represents a collection of 6.022 x1023(Avogadro's number) chemical units.
The mass of one mole of a substance in grams is called its molar mass.
Molar Volume
The volume occupied by one mole of any substance is called its molar volume. It
is denoted by V. One mole of all gaseous substances at 273 K and 1 atm
pressure occupies a volume equal to 22.4 litre or 22,400 mL. The unit of molar volume
is litre per mol or millilitre per mol.

PERCENTAGE COMPOSITION—
The mass percentage of each constituent element present in any compound is
called its percentage composition

Mass % of the element = Mass of element in 1 molecule of compound X100

Molecular mass of the compound
Empirical Formula and Molecular Formula:
An empirical formula represents the simplest whole number ratio of various atoms
present in a compound. E.g. CH is the empirical formula of benzene C6H6.
The molecular formula shows the exact number of different types of atoms
present in a molecule of a compound. E.g. C6H6 is the molecular formula of benzene.
Relationship between empirical and molecular formulae
The two formulas are related as Molecular formula = n x empirical formula
Limiting Reagent- The reactant which gets consumed first or limits the amount of
product formed is known as limiting reagent.
Reactions in Solutions-- The concentration of a solution can be expressed in
any of the following ways.
1. Mass Percent is the mass of the solute in grams per 100 grams of the solution.

A 5 % solution of sodium chloride means that 5 g of NaCl is present in 100g of the

solution.

Page | 10
2. Volume percent is the number of units of volume of the solute per 100 units of the
volume of solution.

A 5 % (v/v) solution of ethyl alcohol contains 5 cm of alcohol in 100 cm of the

3 3

solution.
3. Molarity of the solution is defined as the number of moles of solute dissolved per litre
(dm ) of the solution. It is denoted by the symbol M. Measurements in Molarity can
3

change with the change in temperature because solutions expand or contract

accordingly.
Molarity of the solution = No. of moles of the solute = n
Volume of the solution in litre V
In terms of weight, molarity of the substance can be expressed as:

Molarity equation
To calculate the volume of a definite solution required to prepare solution of
other molarity, the following equation is used:
M1V1 = M2V2,
where M1= initial molarity, M2= molarity of the new solution, V1= initial volume and
V2= volume of the new solution.
4. Molality- Molality is defined as the number of moles of solute dissolved per 1000 g (1
kg) of solvent. Molality is expressed as 'm'.

5. Mole Fraction is the ratio of number of moles of one component to the total number of
moles (solute and solvents) present in the solution. It is expressed as 'x'.
Mole fraction of the solute = Moles of the solute
Moles of solute + Moles of solvent
Mole fraction of the solvent = Moles of the solvent
Moles of solute + Moles of solvent
Mole fraction of the solute + Mole fraction of solvent = 1

ASSERTION-REASON TYPE QUESTIONS OF 1 MARK EACH

In the following questions, a statement of Assertion (A) followed by a statement of
Reason (R) is given. Choose the correct option out of the choices given below each
question:
(a) Assertion and Reason both are correct and Reason is the correct explanation of
Assertion.
(b) Assertion and Reason both are correct statements but Reason is not the correct
explanation of Assertion.
(c) Assertion is correct but Reason is wrong.
(d) Assertion is wrong but Reason is correct.

(1) Assertion: 1 mole O and N contain have same volume at a same temperature
2 2

and pressure.
Reason: 1 mole of any gas at STP occupy 22.4 L at STP.
Correct Answer: (b)
(2) Assertion: Equal moles of different substances contain same number of
constituent particles.
Reason:. Equal weights of different substances contain the same number of
constituent particles.

Page | 11
 Correct Answer: (c)
(3) Assertion: The empirical formula of Ethanoic acid is CH O.
2

Reason: Empirical formula gives us exact no. of atoms present in a molecule.

Correct Answer: (c)
(4) Assertion: When 2 g hydrogen reacts with 32 g Oxygen to form water, Oxygen is
the limiting reagent.
Reason: Limiting reagent is present in lesser amount than stoichiometric amount.
Correct Answer: (d)
(5) Assertion: One atomic mass unit is defined as one-twelfth of the mass of one
carbon-12 atom.
Reason: Carbon-12 isotope is the most abundant isotope of carbon and has been
chosen as the standard.
Correct Answer: (a)

VERY SHORT ANSWER TYPE QUESTIONS OF 1 MARK EACH

(1) Define one mole?
Ans. One mole is the amount of a substance that contains as many particles as there
are atoms in exactly 12 g of the carbon-12.
(2) What is the limiting reagent?
Ans. The reactant which gets consumed first or limits the amount of product formed
is known as limiting reagent.
(3) Write the empirical formula of the following:
(a)N2O4 (b)6H12O6 (c) H2O (d) H2O2
Ans. (a)NO2 (b) CH2O (c) H2O (d) HO
(4) Why the atomic masses of most of the elements are fractional?
Hints: Due to isotopic abundance/
(5) Which of the following have largest no. of atoms?
(i) 1 g Au (ii) 1 g Na (iii) 1 g Li (iv) 1 g Cl2
Hints: 1 g Li as its gram atomic mass is least.
(6) Calculate the molecular mass of the of the following (i)H SO (ii) NaOH
2 4

Ans: Molar mass of H SO = 2 + 32 + 4x16 = 98 amu

2 4

Molar mass of NaOH = 23 + 16 + 1= 40amu

(7) What do mean by Mole fraction?
Ans. Mole Fraction is the ratio of number of moles of one component to the total
number of moles (solute and solvents) present in the solution. It is
expressed as 'x'
(8) Define the law of multiple proportions. Explain it with one example.
Ans.When two elements combine to form two or more compounds, then the different
masses of one element, which combine with a fixed mass of the other, bear a
simple ratio to one another. For example- carbon combines with oxygen to form two
compounds CO and CO2.
Compound CO CO2
Mass of C 12 : 12 So, Mass of O = 16 : 32
Masses of oxygen which combine with a fixed mass of carbon (12g) bear a simple
ratio of 16 : 32 or 1 : 2.
(9) How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?
Ans. Molar mass of Na2CO3= 2 x 23 +12 + 3 x 16 = 106 g / mol
0.50 mol Na2CO3 and 0.50 x 106 = 53 g

Page | 12
 0.50 M Na2CO3 means 0.50 mol i.e. 53 g of Na2CO3 are present in I L of the solution.
(10) What is the law called which deals with the ratios of the volumes of the gaseous
reactants and products?
Ans. Gay Lussac‘s law of gaseous volumes.
(11) What is the difference between molality and molarity?
Ans: Molality is the number of moles of solute present in 1 kg of solvent, whereas
molarity is the number of moles of solute dissolved in 1 litre of a solution.
Molality is independent of temperature, whereas molarity depends on temperature
(12) State Avogadro’s law.
Ans: According to Avogadro’s law, equal volumes of gases at the same temperature and
pressure should contain equal number of molecules.
(13) Define one atomic mass unit (amu).
Ans: One atomic mass unit (amu) is defined as a mass exactly equal to one – twelfth the
mass of one carbon – 12 atom

(14) At NTP, what will be the volume of molecules of H?

2

Ans: 22.4 Litres

(15) Calculate the number of molecules present in 0.5 moles of CO 2?
Ans: The number of molecules in 0.5 moles of O2 is

MULTIPLE CHOICE QUESTIONS OF ONE MARK EACH

Question No.
1) A chemical equation is balanced according to the law
(a) Definite proportion (b) Multiple proportion
(b) Avogadro Law (d) Conservation of mass [Correct Response (d)]
2) 1 amu is equal to
(a) 1.008 g (b) 0.00059 g
(c) 1.66 X 10-24 g (d) 6.023X1023 g [Correct Response (c)]
3) Which of the following contain highest number of molecules
(a) 7 g N 2(b) 4 g H 2 (c) 8 g O 2 (d) 71 g Cl [Correct Response (b)]
2

4) The empirical formula and molecular mass of a compound are CH O and 180 g
2

respectively. What will be the molecular formula of the compound?

(a) CH3COOH (b) HCHO (c) CH3OH (d) C6H12O6 [Correct Response (d)]
5) Which of the following is dependent on temperature?
(a) Molarity (b) Molality
(c) Mole fraction (d) Mass percentage [Correct Response (a)]
6) Which one of the following is not equal with 1 mol O 2

(a) 16 g Oxygen (b) N no of molecules of O

A 2

(c) 22.4 L of Oxygen gas (d) None of these [Correct Response (a)]
7) One Gram Molecule of Benzene is equal to
(a) 72 g Benzene (b) 18 g Benzene
(c) 76 g of Benzene (d) 78 g Benzene [Correct Response (d)]
8) A compound was found to contain Nitrogen and Oxygen in the ratio 28 g and 80 g.
The formula of the compound is
(a) NO (b) NO2 (c) N2O5 (d) N2O3 Correct Response (c)
9) If the concentration of glucose (C H O ) in blood is 0.9 g/L, molarity of glucose in blood?
6 12 6

Page | 13
 a) 5M b) 50 M c) 0.005 M d) 0.5 M [Correct Response(c)]
10) What is the mass percent of carbon in carbon dioxide?

SHORT ANSWER TYPE QUESTIONS OF 2 MARKS EACH

(1) Differentiate between molarity and molality.
(Hints: Any two differences (i) Unit of Molarity is mol L whereas molality is mol kg
-1 -1

(ii) Molarity is temperature dependent whereas molality is independent of temperature

(iii) Definition of molarity and molality
(2) Convert into mole.
(i) 5.6 L of CO2 at STP (ii) 1023 atoms of Na
(Hints (i) 5.6/22.4 (ii) 1023/6.02x1023)
(3) Give one example each of a molecule in which empirical formula
and molecular formula are (i) same (ii) Different.
Ans:(i) Same molecular formula and empirical formula. Carbon dioxide, both is CO2.
(ii) When molecular formula and empirical formula are different –
Hydrogen peroxide: molecular formula is H 2O2 and empirical formula is HO.
(4) How many moles of methane are required to produce 22g CO 2 (g) after
combustion?
Ans: According to the chemical equation, CH4 (g) + 2O2 (g) → CO2(g) + 2H2O (g)
44g CO2 (g) is obtained from 16 g CH4 (g).
[∴ 1 mol CO2(g) is obtained from 1 mol of CH4(g)]
Number of moles of CO2 (g) = 22 g CO2 (g) × 1 /44
= 1/2 mol CO g = 0.5 mol CO2 (g).
Hence, 0.5 mol CO2 (g) would be obtained from 0.5 mol CH4 (g) or 0.5 mol of CH4 (g)
would be required to produce 22 g CO2 (g).
(5) Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in
enough water to form 250 mL of the solution.
Ans: M = 3 mol L–1 Mass of NaCl in 1 L solution = 3 × 58.5 = 175.5 g
Mass of 1L solution = 1000 × 1.25 = 1250 g (since density = 1.25 g mL –1)
Mass of water in solution = 1250 –75.5 = 1074.5 g
Molality = No. of moles of solute/Mass of solvent in kg = 3 mol/1.0745 kg = 2.79 m

LONG ANSWER TYPE QUESTIONS CARRYING THREE MARKS EACH

(1) A compound on analysis was found to contain C = 34.6%, H = 3.85% and O =
61.55%. Calculate the empirical formula.
(Hints: Step 1 Calculating relative no. in moles for each atom e.g.,C=34.6/12=2.88
H = 3.85 / 1 = 3.85 O = 61.55 / 16 = 3.85
Step-2 Relative whole No. of atoms C:H:O = 2.88:3.85:3.85 = 1:1.33:1.33=3:4:4
Hence Empirical formula of the compound C3H4O4
(2) What is the percentage of carbon, hydrogen and oxygen in ethanol?
Ans. Molecular formula of ethanol is : C H OH
2 5

Molar mass of ethanol is : (212.01 + 61.008 + 16.00) g = 46.068 g Mass per cent of
carbon = (24.02g / 46.068g) ×100 = 52.14%
Mass per cent of hydrogen = (6.048g / 46.068 g) ×100 = 13.13%
Mass per cent of oxygen = (16.00 g / 46.068g)×100 = 34.73%
(3) What do mean by molarity. Calculate the molarity of NaOH in the solution prepared
by dissolving its 4 g in enough water to form 250 mL of the solution.
Ans. The number of moles of solute dissolved per litre (dm3) of the solution is called
molarity.

Page | 14
Since molarity (M) = No. of moles of solute /Volume of solution in litres
= (Mass of NaOH / Molar mass of NaOH) / 0.250 L
= (4g / 40g 0.1 mol)/0.250L =0.1 mol / 0.250 L = 0.4 mol L-1 = 0.4 M

2. STRUCTURE OF ATOM
Atomic Number: The number of protons present in the nucleus is equal to the
atomic number (z). In order to keep the electrical neutrality, the number of electrons
in an atom is equal to the number of protons (atomic number, z)
Atomic Number (z) = Number of protons in the nucleus of an atom= Number of
electrons in a neutral atom
Mass Number: Number of protons and neutrons present in the nucleus are
collectively known as nucleons. The total number of nucleons is termed as mass
number (A) of the atom.
Mass Number (A) = Number of protons (p) + Number of neutrons (n).
Isotopes: Atoms with identical atomic number but different atomic mass number are
known as Isotopes. Isotopes of Hydrogen: protium 11𝐻 , deuterium 21𝐻 and tritium 13𝐻 .
Isobars: Isobars are the atoms with same mass number but different atomic
number for Example: 146𝐶 , 146𝑁
(i) Dual character of the electromagnetic radiation which means that radiations
possess both wave like and particle like properties.
(ii) Experimental results regarding atomic spectra which can be explained only by
assuming quantized electronic energy levels in atoms.
Characteristics of a Wave
(i) Wavelength (𝛌): It is defined as the distance between any two consecutive
crests or troughs. It is represented by X and its S.I. unit is metre (m).
(ii) Frequency (ν): Frequency of a wave is defined as the number of waves passing
through a point in one second. It is represented by v (nu) and is expressed in Hertz
(Hz). 1 Hz = 1 cycle/sec.
(iii)Velocity: Velocity of a wave is defined as the linear distance travelled by the
wave in one second. It is represented by c and is expressed in cm/sec or m/sec.
c = v𝛌
(v) Amplitude: Amplitude of a wave is the height of the crest or the depth of the
through. It is represented by V and is expressed in the units of length.
(vi) Wave Number : It is defined as the number of waves present
in 1 metre length. Evidently it will be equal to the reciprocal of the
wavelength. It is represented by bar v (read as nu bar).
Electromagnetic Spectrum: When electromagnetic radiations are arranged in order
of their increasing wavelengths or decreasing frequencies, the complete spectrum
obtained is called electromagnetic spectrum.
Limitations of Electromagnetic Wave Theory Electromagnetic wave theory was
successful in explaining properties of light such as interference diffraction etc; but it
could not explain the following:
(i) The phenomenon of black body radiation.
(ii) The photoelectric effect.
(iii) The variation of heat capacity of solids as a function of temperature.

Page | 15
Atomic spectrum: When a ray of white light is passed through a prism, we get a
series of colored bands called spectrum. This spectrum is called continuous
spectrum, because here violet merges into blue, blue into green and so on.
Emission and Absorption Spectra
Emission Spectra: Emission Spectra is noticed when the radiations emitted from a
source are passed through a prism and then received on the photographic plate.
Radiations can be emitted in a number of ways such as: (i) from sun or glowing
electric bulb. (ii) by passing electric discharge through a gas at low pressure. (iii) by
heating a substance to high temperature. The emission spectra of atoms in the gas
phase do not form a continuous spectrum.
Absorption Spectra: When white light is passed through the vapours of a
substance and the transmitted light is then allowed to strike a prism, dark lines
appear in the otherwise continuous spectrum. The dark lines indicate that the
radiations corresponding to them were absorbed by the substance from the white
light. This spectrum is called absorption spectrum. Dark lines appear exactly at the
same positions where the lines in the emission spectra appear.
Each element has a unique line emission spectrum. So, line emission spectra are
also called finger print of atoms.
Line Spectrum of Hydrogen: When electric discharge is passed through hydrogen
gas enclosed in discharge tube under low pressure and the emitted light is analyzed
by a spectroscope, the spectrum consists of a large number of lines which are
grouped into different series. The complete spectrum is known as hydrogen
spectrum.

On the basis of experimental observations, Johannes Rydberg noted that all series
of lines in the hydrogen spectrum could be described by the following expression:
where Z is the atomic number of the species.
Here RH = constant, called Rydberg constant for hydrogen and n1, n2 are integers
from which electrons jumps (n2 > n1) For any particular series, the value of n 1 is
constant while that of n2 changes. For example,
Name of series Region n1 n2
Lyman series Ultra violet 1 2, 3, 4, 5………..∞
Balmer series, Visible 2 3, 4, 5, 6………..∞
Paschen series Infrared 3 4, 5, 6, 7………..∞
Brackett series Infrared 4 5, 6, 7, 8………..∞
Pfund series Infrared 5 6, 7, 8, ………..∞
Thus, by substituting the values of n1 and n2 in the above equation, wavelengths and
wave number of different spectral lines can be calculated. When n 1 = 2, the

Page | 16
expression given above is called Balmer’s formula.

n (n−1)
the maximum no. of lines of H atom from n th orbit = (For n = 6 no. of H lines = 15)
2
hc
E = h𝜈 = ( for 1 quantum or 1 photon)
λ
n hc
& E = n h𝜈 = ( for n quanta or n photons) where h = Planck’s constant = 6.6 x 10-31 Js
λ

Bohr’s Model of Atom

Niels Bohr in 1913, proposed a new model of atom on the basis of Planck’s Quantum
Theory.
The main postulates of this model are as follows:
(i) In an atom, the electrons revolve around the nucleus in certain definite circular paths
called orbits.
(ii) Each orbit is associated with definite energy and therefore these are known as energy
levels or energy shells. These are numbered as 1, 2, 3, 4……….. or K, L, M, N………..
(iii) As long as electron is present in a particular orbit, it neither absorbs nor loses energy
and its energy, therefore, remains constant.
(iv) When energy is supplied to an electron, it absorbs energy only in fixed amounts as
quanta and jumps to higher energy state away from the nucleus known as excited state. The
excited state is unstable, the electron may jump back to the lower energy state and in doing
so, it emits the same amount of energy.
hc hc
The energy change ∆E = E2 – E1 = h𝜈 = or, 𝜆 =
λ ΔE
(v) Those orbits are permitted for the electron in which angular momentum of the electron is
ℎ 𝑛ℎ
a whole number multiple of . Angular momentum of electron, mvr = 2𝜋 (n = 1, 2, 3, 4 etc).
2𝜋

Dual Behavior of Matter (de Broglie Equation) de Broglie in 1924, proposed that
matter, like radiation, should also exhibit dual behavior i.e., both particle like and
wave like properties. This means that like photons, electrons also have momentum
as well as wavelength.

Heisenberg’s Uncertainty Principle It states that,

“It is impossible to determine simultaneously, the exact position and exact
momentum (or velocity) of an electron”.

Significance of Uncertainty Principle (i) It rules out existence of definite paths or

trajectories of electrons and other similar particles.
(ii) The effect of Heisenberg’s uncertainty principle is significant only for microscopic
objects and is negligible for macroscopic objects.

Page | 17
Reasons for the Failure of Bohr Model (i) The wave character of the electron is
not considered in Bohr Model. (ii) According to Bohr Model an orbit is a clearly
defined path and this path can completely be defined only if both the position and the
velocity of the electron are known exactly at the same time. This is not possible
according to the Heisenberg’s uncertainty principle.
Quantum Mechanical Model of Atom Quantum mechanics: Quantum mechanics is
a theoretical science that deals with the study of the motions of the microscopic
objects that have both observable wave like and particle like properties.
Quantum Numbers Atomic orbitals can be specified by giving their corresponding
energies and angular momentums which are quantized (i.e., they have specific
values). The quantized values can be expressed in terms of quantum number. These
are used to get complete information about electron i.e., its location, energy, spin
etc.
The principal quantum number (n) gives us the following information:
1. It gives the size the orbit.
2. It gives the energy of electron in an orbit.
3. It gives the shell in which the electron is found.
4. It also gives the average distance between the electron and the nucleus. As the
value of n increases, the distance between the electron and the nucleus also
increases. The possible values of n are 1, 2, 3, 4, 5 etc. If n = 1 the electron is in K
shell n = 2 the electron is in L shell n = 3 the electron is in M shell and so on.
5. The maximum number of electrons in the shell with principal quantum number n is
equal to 2n2 .
Azimuthal or Subsidiary or Orbital Angular Quantum Number (l) The following
information’s are obtained from Ɩ.
1. It gives the shape of the orbital.
2. It gives the sub shell or sub level in which the electron is located.
3. It also gives the orbital angular momentum of the electron.
4. the values of l = 0 to (n-1) The number of sub shells in a principal shell is equal to
the value of n. For example,
When n = 1, Ɩ = 0. i.e. K shell contains only one sub shell : 1s sub shell
when n = 2, Ɩ = 0 and1. i.e. L shell contains two sub shells: 2s and 2p sub shells
when n = 3, Ɩ = 0, 1 and 2. i.e. M shell contains three sub shells: 3s, 3p and 3d sub
shells
when n = 4, Ɩ = 0, 1, 2 and 3. i.e. N shell contains four sub shells : 4s, 4p, 4d and 4f
sub shells
5. the number of electrons that can be accommodated in sub-energy level is equal to
2(2l+1).
If l = 0 for s orbital then = 2(2x1 + 1) = 2 electrons
If l = 1 for p orbital then = 2(2x2 + 1) = 6 electrons
If l = 2 for d orbital then = 2(2x3 + 1) = 10 electrons
if l = 3 for f orbital then = 2(2x3 + 1) = 14 electrons
Magnetic Orbital Quantum Number (m or ml)
1. The magnetic orbital quantum number determines the number of preferred
orientations of the electrons present in a sub-shell.
2. The magnetic quantum number is denoted by letter m and for a given value of l, it
can have all the values ranging from – l to + l including zero. Thus, for energy value
of l, m has 2l + 1 values. For example, For, l = 0 (s-sub-shell), m can have only one
value i.e., m = 0. This means that s-sub-shell has only one orientation in space.
In other words, s-subshell has only one orbital called s-orbital.

Page | 18
Spin Quantum Number (S or ms) This quantum number helps to explain the
magnetic properties of the substances. A spinning electron behaves like a micro-
magnet with a definite magnetic moment. If an orbital contains two electrons, the two
magnetic moments oppose and cancel each other.
• Shapes of s-orbitals s-orbital with only one orientation has a spherical shape with
uniform electron density along all the three axes.

Shapes of p-orbitals p-orbitals are present in the p-subshell for which l = 1 and ml
can have three possible orientations – 1, 0, + 1. Thus, there are three orbitals in the
p-subshell which are designated as p x, py and pz orbitals depending upon the axis
along which they are directed. The general shape of a p-orbital is dumb-bell
consisting of two portions known as lobes. Moreover, there is a plane passing
through the nucleus along which finding of the electron density is almost nil. This is
known as nodal plane as shown in the fig.

• Shapes of d-orbitals For d-orbitals, Ɩ = 2 and mƖ = -2, -1, 0, +1 and +2. i.e., there are
five possible orientations for d orbitals. So there are 5 types of d-orbitals. They are
dxy, dxz, dyz, dx2 - y2 and dz2 . . The shapes of the first four d orbitals are double dumb-
bell and that of the fifth one, dz2 , is dumb-bell having a circular collar in the XYplane.
The five d-orbitals have equivalent energies. For d orbitals the number of radial
nodes is 2 and the total number of nodes is n-1. Boundary surface diagrams for d-
orbitals are as follows.

Number of radial nodes = n - Ɩ – 1 Number of angular nodes = Ɩ

Total number of nodes = n -1

Electronic Configuration of Atoms The distribution of electrons into orbitals of an

atom is called its electronic configuration. The electronic configurationis explained

Page | 19
 based on Aufbau principle, Hund’s rule, Pauli’s exclusion principle and stability
concept.
1. Aufbau principle: The principle states: In the ground state of the atoms, the orbitals
are filled in order of their increasing energies. In other words, electrons first occupy
the lowest energy orbital available to them and enter into higher energy orbitals only
after the lower energy orbitals are filled. This rule has two sub rules: a) The various
orbitals are filled in the increasing order of their (n+l) value. b) If two orbitals have the
same (n+l) values, the orbital with the lower n value is filled first.The increasing order
of orbitals is as follows.
The order in which the energies of the orbitals increase and
hence the order in which the orbitals are filled is as follows:
Is, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, id, 5p, 6s, if, 3d, 6p, 7s, 5f 6d,
7p…

(2) Pauli Exclusion Principle: According to this principle, no two electrons in the
orbitals of an atom can have the same set of four quantum numbers.
Pauli exclusion principle can also be stated as: “Only two electrons may exist in the
same orbital and these electrons must have opposite spins”.

Example: Let e1 and e2 are two atoms of an atom

Let, for e1, n = 4, l = 0, ml = 0 & ms = +½
For e2, n = 4, l = 0, ml = 0 & ms = - ½. i.e., here ms cannot be + ½ otherwise the two
electrons will be of same spin & will repel each other & cannot exist in an orbital.

(3) Hund’s rule of maximum multiplicity: It states that: pairing of electrons in the
orbitals belonging to the same sub-shell (p, d or f) does not take place until each
orbital belonging to that sub-shell has got one electron each with parallel spin. For
example the electronic configuration of N is 1s2 2s2 2px1 py1 pz1 and not 1s2 2s2 2px2
2py1 . Example:

Stability of Completely Filled and Half-Filled Subshells: For atoms having half-
filled or completely filled electronic configurations have extra stability compared to
other atoms. This is due to their symmetrical distribution of electrons and greater
exchange energy. For example, the electronic configuration of Cr is [Ar] 3d5 4s1 and
not 3d4 4s2 . This is because d5 represents a half filled configuration and has extra
stability. Similarly for Cu the electronic configuration is [Ar] 3d 104s1 and not 3d9 4s2

Example: Copper: 29Cu (29e) → Expected Elect. Conf. 1s2 2s2 2p6 3s2 3p6 4s2 3d9
(Partially filled 3d9)

Page | 20
Copper: Cu (Correct Elect. Conf.) 1s2 2s2 2p6 3s2 3p6 4s1 3d10 (Fully filled elec.
conf.) → more stable.
So, the correct electronic configuration: 1s2 2s2 2p6 3s2 3p6 3d10 4s1

Assertion - Reason Questions

In the following questions (1 to 5), a statement of assertion is followed by a corresponding
statement of reason. Of the following statements, choose the correct one:
A. Both assertion and reason are correct statements and reason is the correct
explanation of the assertion.
B. Both assertion and reason are correct statements and reason is not the correct
explanation of the assertion.
C. Assertion is correct but reason is incorrect statement.
D. Assertion is incorrect but reason is correct statement.

1 Assertion: Bohr model of atom fails in case of multi electron species.

Reason: It does not mention electron-electron repulsion.
Answer B
2 Assertion: Both position and momentum of an electron cannot be determined
simultaneously with maximum accuracy.
Reason: The path of an electron in an atom is clearly defined.
Answer C
3 Assertion: According to Thomson model of atom, mass of the atom is
assumed to be uniformly distributed over the atom.
Reason: According to Thomson model of atom , positive charge is
concentrated in the centre of the atom.
Answer C
4 Assertion: Electronic configuration of helium is 1S2.
Reason: Hund’s rule demands that the configuration should display maximum
stability.
Answer B

5 Assertion: The outer electronic configuration of Cr and Cu are 3d54s1 and

10 1
3d 4s respectively.
Reason: Electrons are filled in orbitals in order of increasing energies given by
(n+l) rule.
Answer B

Very Short Answer Questions(one word or one sentence)(One mark questions)

1 In the multi-electronic atoms, indicate the orbital which has lower
energy? 4s or 3d?
Answer: 4s
2 How many unpaired electrons are present in zinc?
Answer: It has no unpaired electrons.
3 Write the electronic configuration of an element having atomic number 25.
Answer Mn- [Ar]184s23d5
4 Write de Broglie relation.
Answer: λ = h/mv (symbols have their usual significance).
5 Write the mathematical form of Heisenberg Uncertainty Principle.
Answer: ∆x. ∆p ≥ h/4π (symbols have their usual significance).
6 State one defect of Bohr model of atom.

Page | 21
Answer: It can not explain Stark effect.
7 Is an electron stationary in stationary energy state?
Answer: No, the electron is not stationary. Only the energy associated with it is
stationary.
8: State one point of Planck’s Quantum theory.
Answer: Atoms and molecules could emit or absorb energy only in discrete
quantities and not in a continuous manner.
OR
The energy (E) of a quantum of radiation is proportional to its frequency and is given
by : E = hv ( h= planck constant = 6.626 x 10-34Js), v(neu) = frequency.
9 State Pauli’s exclusion principle.
Answer: No two electrons can have all the four quantum numbers same.
10 State all the four quantum numbers.
Answer: Principal quantum numbers, Azimuthal quantum numbers, magnetic
quantum number and spin quantum number.
11 Explain why atoms with half-filled and completely filled orbitals have extra
stability?
Answer: Due to exchange energy and symmetry.
12 Which quantum number specifies the shape of an orbital?
Answer: Azimuthal quantum number.
13 Which of the following orbitals are not possible?
2s,2p,3f,3d.
Answer: 3f
14 What do you understand by Hund’s rule?
Answer: For the orbitals like p,d,f which have degenerate states, the pairing of
electrons take pace after all the orbitals getting singly filled.
15 What are the shapes of the following orbitals: d and d .
xy z2

Answer: dxy has clover leaf shape (double dumb-bell) and d has dough-nut shape.
z2

Multiple Choice Questions (Single correct option)

1 A p-orbital can accommodate up to :
A) four electrons B) six electrons
C) two electrons with parallel spin D) two electrons with opposite spins
Answer: B
2 Principal quantum number of an atom is related to the:
A) size of the orbital B) spin angular momentum
C) orientation of the orbital in space D) orbital angular momentum
Answer: A
3 Quantum numbers n = 2, l = 1 represent:
A) 1s orbital B) 2s orbital C) 2p orbital D) 3d orbital
Answer: C
4 For principal quantum number n = 4, the total number of orbitals having l = 3
is : A) 3 B) 7 C) 5 D) 9
Answer: B
5 The electronic configuration of titanium is :
A) [Ar]184s23d1 B) [Ar]184s23d3 C) [Ar]184s23d2 D) [Ar]184s24p2
Answer C

Page | 22
6 Which expression represents de Broglie relationship:
A) h/mv = p B) λ = h/mv C) λ = h/mp D) λm = v/p
Answer: B
7 Which of the following series of transitions in the spectrum of hydrogen falls in
visible region ?
A) Paschen series B) Lyman series C) Balmer series D) Brackett series
Answer: C
8 Which one of the following statements rules out the existence of definite paths
of electrons and other similar particles?
A) Pauli’s exclusion principle B) de Broglie relationship
C) Aufbau principle D) Heisenberg Uncertainty Principle
Answer: D
9 Rutherford’s gold foil experiment drew the following conclusion:
A) mass of the atom is assumed to be uniformly distributed over the atom
B) an atom possesses a spherical shape in which the positive charge is
uniformly distributed
C) most of the space in the atom is empty
D) none of these
Answer: C
10 Half-filled and completely filled shells have extra stability due to :
A) Pauli’s exclusion principle B) Hund’s rule
C) Symmetrical distribution of electrons and exchange energy
D) Aufbau principle
Answer: C
Section-C: Two Marks Questions
1 State two postulates of Bohr model of atom.
Answer: Bohr’s postulates of hydrogen atom:
a. The electrons revolve around the nucleus in some selected permissible orbits
called stationary orbits.
b. The angular momentum of an electron in a stationary state is an integral multiple
of h/2π ( mvr = nh/2π).The symbols have their usual significances.

2 Explain Pauli’s exclusion principle with an example.

Answer: Pauli’s exclusion principle: No two electrons can have all the four quantum
numbers same.
Example: For Helium the three quantum numbers are same but fourth quantum
number (spin quantum number) is different.
E.C. of He: 1s2
For 1st electron , n=1,l=0,m=0 and s=+1/2.
For 2nd electron n=1,l=0,m=0 and s= -1/2.
3 Which one of the following electronic configurations is correct for
chromium? a) [Ar]18 4s23d4 or b) [Ar]18 4s13d5?
Justify your answer.
Answer: b) [Ar]18 4s13d5 is correct as half-filled orbitals are more stable than nearly
half-filled orbitals due to exchange energy and more symmetry.
4 Out of the following, which electronic configurations is correct for nitrogen?
Give reason in support of your answer.
a) 1s2 2s2 2px1 2py1 2pz 1 b) 1s2 2s2 2px2 2py1 2pz 0
c) 1s2 2s2 2px1 2py2 2pz 0
Answer: a) 1s2 2s2 2px1 2py1 2pz 1 is correct.

Page | 23
Reason: According to Hund’s rule, for the orbitals like p,d,f which have degenerate
states, the pairing of electrons take pace after all the orbitals getting singly filled.
5 Designate the orbitals using s, p, d and f notations which are applicable:
a) n=4, l=2 b) n=5, l=1
Answer a) 4d b) 5p.
6 Write the electronic configuration of the following ions?
a) Na+ b) O2-
Answer: a) Configuration of Na =1s2 2s2 2p6 3s1 and Na+ =1s2 2s2 2p6
b) Configuration of O=1s2 2s2 2p4 and O2- =1s2 2s2 2p6

Three marks Questions

1a) Write de Broglie equation.
b) What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10
meter per second. Given : Planck’s constant (h) = 6.626 x10-34 Js.
Answer a): de Broglie equation: λ= h/mv (symbols have usual significances).
b) According to de Broglie equation: λ= h/mv
Given: h= 6.626 x 10-34 Js , m=0.1 kg , v= 10 m s-1
λ= h/mv = (6.626 x 10-34) / (0.1 x 10) = 6.626 x 10-34 m.
2 Write the significance (one point) of each of the following:
a) Principal quantum number b) Azimuthal quantum number
c) Magnetic quantum number.
Answer: Significance of principal Quantum number(n): The principal quantum
number determines the size and to large extent the energy of the orbital. It also
identifies the shell.
Significance of azimuthal Quantum number(l): It defines the 3-dimensional shape of
the orbital.
Significance of magnetic Quantum number(m): It gives information about the spatial
orientation of the orbital with respect to standard set of co-ordinate axis.
3 State the following: a) Aufbau principle, b) Hund’s rule of maximum
multiplicity c)Heisenberg Uncertainty principle.
Answer: a) Aufbau principle: In the ground state of the atoms, the orbitals are filled
with electrons in order of their increasing energies.
It means the lower energy orbitals are filled first followed by higher energy orbitals.
b) Hund’s rule of maximum multiplicity:
In the orbitals like p, d, f etc which have degenerate states, the pairing of electrons
takes place after all the orbitals getting singly filled.
c) Heisenberg Uncertainty principle:It is impossible to determine simultaneously the
exact position and exact momentum or velocity of subatomic particles like electrons.

4a) The uncertainty in the position of a moving bullet of mass 10 g is 10 m.-5

Calculate the uncertainty in its velocity.

(Given : h = 6.626x10 Js).
-34

Answer: According to Heisenberg Uncertainty principle :

∆x. ∆p ≥ h/4π
∆x. m(∆v) ≥ h/4π
Given: h= 6.626 x 10-34 Js , ∆x=10-5 m , m=10 g =10-2 kg .
∆v= h / 4πm(∆x)
∆v= (6.626 x 10-34)/(4 x 3.14 x 10-2 x 10-5) = 5.27 x 10-28 ms-1 (Ans)

Page | 24
 3: CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

1. INTRODUCTION
‘Periodic table may be defined as the table which classifies all the known
elements in accordance with their properties in such a way that elements with similar
properties are grouped together in the same vertical column and dissimilar elements
are separated from one another’.
2. DEVELOPMENT OF THE PERIODIC TABLE
All the earlier attempts of classification of elements were based upon their atomic
weights.
J.W. Dobernier in 1829 classified the known elements in groups of three on the basis
of similarities in their properties. These group ware called triads. In the triads:
➢ Properties of elements in each triad were similar
➢ Elements in the triad were arranged in the order of increasingatomic weights such
that the atomic-mass of middle elementwas approximately equal to the arithmetic
mean of the othertwo elements
Elements Li Na K
At.mass 7 23 39 7 + 39 / 2 = 23
New Lands Law of octaves: When elements were arranged in order of their
increasing relative atomic masses. The properties of every eight elements were
similar to the first one, like the eighth note of a musical scale. This repetition in the
properties of elements is just like the repetition of eighth node in anoctave of music.
This arrangement of elements is called as Newland’s law of Octave.
Li Be B C N O F
Na Mg Al Si P S Cl
K Ca

Page | 25
This classification worked quite well for the lighter elements but it failed in case of
heavier elements and therefore, discarded.
MENDELEEV’S PERIODIC LAW:
The law states that “The properties of elements are periodic function of their atomic
weights.”
Features of Mendeleev’s periodic law are as follows:
➢ All 63 elements then known were arranged in an increasing order of their atomic
masses; in such a way that elements with similar
properties come under the same vertical column.
Horizontal rows were called as Periods and vertical columns as groups.
➢ There were eight groups and a zero group and seven periods.
➢ Group I to group VII were further divided into two sub-groups A and B. Group VIII
had nine elements in three sets containing three elements each.
➢ There were 7 periods in the table which were numbered as 1,2, 3, 4, 5, 6 and 7.
➢ The first 6 periods were complete while the seventh was incomplete.
• Merits of Mendeleev’s periodic classification:-
➢ Properties of elements could be easily studied because elements present in same
sub-group showed similar properties.
➢ Mendeleev considered similarities in properties of elements rather than atomic
masses. So he corrected the atomic mass of some elements.
➢ Prediction of the existence and properties of new elements –
Mendeleev left some gaps in his periodic table and he also predicted the properties
of new elements which were undiscovered. E.g. Gaps were left below Aluminium
and Silicon for the undiscovered elements. They were named as Eka-Aluminium and
Eka-Silicon. These elements were further discovered as Gallium and Germanium.
➢ The inert gases when discovered were placed as a separate group without
upsetting the order of other elements.
Demerits of Mendeleev’s periodic Classification:-
➢ Position of Hydrogen – Position of Hydrogen was not clear in the Mendeleev’s
periodic table, because it has properties similar to alkali metals of first group and
halogens of seventh group.
➢Position of Isotopes – Isotopes of an element have different atomic masses, so
according to increasing order of atomic masses,they should be placed in different
groups .
➢ Some elements with higher atomic masses were placed before elements with
lower atomic masses. e.g. (i) Ar40, K39 (ii) Co59.9, Ni58.6
All these pairs of elements got placed in wrong order of their atomic masses.
MODERN PERIODIC LAW: The modern periodic law states that “The physical and
chemical properties of the elements are periodic functions of their atomic umbers”.
MODERN PERIODIC TABLE
▪ The present form of the periodic table is the modern periodic table. It is based on
the modern periodic law.
▪ It is also known as the Long Form of Periodic Table
▪ The elements are arranged in horizontal rows referred as ‘periods’ and vertical
columns known as ‘groups’.
▪ There are seven periods and 18 groups in the long form of the periodic table.
IUPAC NOMENCLATURE- Nomenclature of elements with Atomic Nos. > 100

Page | 26
The IUPAC has made recommendation that until a new elements discovery is
proved, and its name is officially recognized, a systematic nomenclature is derived
from the atomic number of the elements Z> 100. The nomenclature is as follows:

Example: At. no. Z = 109 → Un + nil + enn + ium = Unnilennium Sumbol= Une
3. ELECTRONIC CONFIGURATION OF ELEMENTS & THE PERIODIC TABLE
• There is a close relationship between electronic configuration of elements and their
position in the periodic table.
• Each period begins with the filling up of a new principle quantum no; ‘n’.
• The period number indicates the value of n (principle quantum number) for the
valence shell.
The elements are further classified into 4 blocks.
s–block elements (Groups 1 & 2)
Elements in which the last electron enters the s-orbital of their outermost energy
level are called s-block elements. Their general valance shell electronic configuration
is ns1-2
. It consists of Group –1 (ns1) and Group – 2 (ns2) elements.
p-block elements
The elements in which the last electron enters the p-orbital of their outermost energy
level, are called p-block elements. Their general valance shell electronic
configuration is ns2 np1-6
. It consists of Groups13 to 18 of periodic table.
“The elements of group 1, 2 and 13 to 18 are called representative elements.”
d-block elements or transition metals:
The elements consist of group 3 to 12 and have general valance shell electronic
configuration (n-1) d1-10 ns1or2 . The valence electron enters the d – orbital of
penultimate shell (n-1).
f-block elements or inner transition elements
The elements in which last electron enters f orbital the anti-penultimate energy level
and have general valance shell electronic configuration (n – 2)f0 - 14(n-1)d0 or 1 ns2
PERIODIC TRENDS IN PROPERTIES
1. ATOMIC RADII
Atomic radius is defined as the distance from the center of nucleus of the atom to the
outer most shell of electrons.
a)Covalent Radius
It is defined as one half of the distance between the centers of the nuclei of two
similar atoms in a molecule bonded by a single covalent bond.

Page | 27
 b) Metallic radius: It is defined as half inter nuclear distance between two adjacent
atoms in metallic lattice.

c) Van Der Waal’s Radius: It is defined as one half of the inter nuclear distance
between two adjacent atoms of the same element belonging to two nearest
neighbouring molecules of the same substance in solid state.

VARIATION IN ATOMIC RADII

a) Along the period:
➢ Atomic radii decrease from left to right in a period.
➢ Reason: Along a period, nuclear charge increases progressively by one unit but
the electrons are added to the same principal shell. As are result the electron cloud
is pulled closer to the nucleus and atomic size decreases.
b) Down the group:
➢ The atomic radii increase from top to bottom in-group.
➢ Reason: On moving down the group, nuclear charge increases and also there is a
progressive increase in the principal energy level. The increase in number of
principal shell is more pronounced than the increase in nuclear charge. So the
distance of the outer most electrons from the nucleus gradually increases down the
group & the size increases.
2. IONIC RADII
• Radius of cation is smaller than radius of parent atom. In a cation, the number of
electrons is less but nuclear charge remain the same. The effective nuclear charge
per electron increases which pulls the election cloud more strongly towards the
nucleus & hence the size decreases.
• Radius of anion is larger than the parent atom In an anion, number of electrons
increases, and the effective nuclear charge per electron decreases & the size
increases.
3. ISOELECTRONIC IONS

Page | 28
Isoelectronic ions are ions of different elements which have same number of electrons
but differ in magnitude of nuclear charge. E.g. Na+, Mg2+, Al3+, F-, O2-, N3-. All have 10
electrons.
The size of iso electronic ions decreases with increases in magnitude of nuclear
charge.
➢ The increasing order of size of the following isoelectronic species are as follows:

IONIZATION ENTHALPY (∆iH)

It is defined as the amount of energy required to remove the most loosely bound
electron from an isolated gaseous atom.
The amount of energies required to remove most loosely bound electron from
unipositive, dipositive, tripositive ... ions of the element in gaseous state are called
2nd, 3rd ..... Ionization enthalpies respectively. Or successive ionization enthalpy.

2. FACTORS GOVERNING IONIZATION ENTHALPY

(i) Size of atom – Smaller the size, greater is the I.E. (electrons are closer to the
nucleus and so more energy is required to remove it)
(ii) Charge on nucleus – Greater the charge on nucleus, greater will be I.E.
(Greater the nuclear pull more energy needed to remove the electron)
(iii) Screening effect of inner electron – In multi electron atoms, the outer most
electrons are shielded from the nucleus by the inner electrons.
The effective nuclear charge felt on these electrons is less than the total nuclear
charge. This is known as screening effect.
VARIATION IN IONIZATION ENTHALPY
Along a group
• The I.E decreases along a group as we move from top to bottom.
• Reason: In a group atomic size increases and also the shielding effect on the
outermost electrons increases. As a result, electrons become less & less firmly held
to the nucleus as we move down the group & the I.E. decreases, i.e. tendency to
lose electron increases.

Along a period: The I.E increases along a period

• Along a period nuclear charge increases and the atomic size decreases, as a result
the electrons are more tightly held to the nucleus & the I.E. increases, i.e, tendency
to lose electron decreases.

Page | 29
IRREGULARITIES IN IONIZATION ENTHALPIES OF DIFFERENT ELEMENTS:
(I) Be & B: B is having less I.E. than Be.
Reason: (a) The electronic configuration of B (1s2 2s2 2p1) is less stable than Be(1s2
2s2 ) which has completely filled orbital.
(b) In B, last electron enters the 2p orbital which has lesser penetrating power than
2s electron of Be and so it can be easily removed.
ELECTRON GAIN ENTHALPY (∆egH)
It is amount of energy associated when an electron is added to a neutral isolated
gaseous atom to convert it into a
negative ion (anion).
X(g) + 1e → X-(g) ∆egH

If an atom has high tendency to add

an electron, energy is released &
electron gain enthalpy is negative,
exothermic.
If an atom has no tendency to add
an electron, energy is required for
this process & its electron gain
enthalpy is positive, endothermic.

ELECTRONEGATIVITY: “Electro-negativity is the ability of an atom in a chemical

compound to attract the shared pair of electrons to itself.”
FACTORS AFFECTING MAGNITUDE OF ELECTRONEGATIVITY
➢ Atomic radius: As the atomic radius of the element increases the electro-
negativity value decreases.
➢ Effective nuclear charge: The electro-negativity value increases as the
effective nuclear charge on the atomic nucleus increases.
ANOMALOUS BEHAVIOR OF FIRST ELEMENT OF EACH GROUP OF 2ND
PERIOD:
The first element of each group shows property different from the rest of the
elements in a group.
• This is due to (i) their small size (ii) large charge/radius ratio (iii) high ionization
enthalpy. (iv) Absence of d orbital.
(The first member of each group has only 2s and 2p orbitals available for bonding
whereas the rest of the elements also have d orbitals for bonding.)
DIAGONAL RELATIONSHIP
It has been observed that some elements of the second period show similarities in
properties with the elements of the third period placed diagonally to each other,
though belonging to different groups. This similarity in properties of elements placed
diagonally to each other is called diagonal relationship.
• This is because diagonally situated elements have almost same size, ionization
enthalpy, electronegativity etc.
For example, lithium (of group 1) resembles magnesium (of group 2) and

Page | 30
beryllium (of group 2) resembles aluminium (of group 13) and so as.

ASSERTION REASON QUESTIONS

For questions below, two statements are given one labelled Assertion (A) and the
other labelled Reason (R). Select the correct answer to these questions from the
codes (a), (b), (c) and (d) given below :
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.

1.Assertion. Helium and beryllium have similar outer electronic configuration of the
type ns .
2

Reason. Both are chemically inert.

2.Assertion: Electron gain enthalpy becomes less negative as we go down a group.
Reason: Size of the atom increases on going down the group and the added electron
would be farther from the nucleus.
3.Assertion: Generally, ionization enthalpy increases from left to right in a period.
Reason: When successive electrons are added to the orbitals in the same principal
quantum level, the shielding effect of inner core of electrons does not increase very
much to compensate for the increased attraction of the electron to the nucleus.
4. Assertion. Zinc is a s-block element.
Reason. The last electron in zinc goes to s-orbital.
5. Assertion: Of all the halogens, fluorine is most electronegative element.
Reason. Fluorine has the most negative electron gain enthalpy
Answer: 1-c, 2-a, 3- a, 4-d, 5-c

VERY SHORT ANSWER QUESTIONS 1 MARK EACH

1.What is meant by periodicity in properties of elements?
Ans: The recurrence of similar properties of the elements after certain regular
intervals when they are arranged in the order of increasing atomic numbers in called
periodicity.
2. What is the basic difference in approach between Mendeleev’s Periodic Law and
the Modern Periodic Law?
Answer: The basic difference in approach between Mendeleev’s Periodic Law and
Modern Periodic Law is the change in basis of classification of elements from atomic
weight to atomic number.
3 Arrange (Increasing order of electron gain enthalpy) .
F, Cl, Br, I
Ans: I < Br < F< Cl .
4.Write the IUPAC name & symbol of an element having atomic no. 109.
Ans: Unnilennium Symbol: Une

Page | 31
5.) Consider the following species: N , O , F , Na , Mg and Al What is common in
3- 2- - + 2+ 3+.

them?
Ans: They are isoelectronic, having same number of electrons.
6. Arrange the following elements in the increasing order of their metallic character
Mg, Ca, K, Ge, Ga
Ans: Ge< Ga< Mg< Ca< K.
7. Where would you locate the following element in the periodic table?
A soft metal stored under kerosene
Ans: Group -1
8. Alkali metals do not form dipositive ions. Why?
Ans: They have only one valency electron.
9. Oxygen has lower ionization enthalpy than nitrogen. Give reason.
Ans: In Oxygen by removing one electron from 2p orbital, it acquire stable
configuration whereas nitrogen has half-filled 2p orbitals, it difficult to remove
electron.
10. Write the atomic number of the element present in the third period and
seventeenth group of the periodic table.
Answer: The element is chlorine (Cl) with atomic number (Z) = 17.
11. Chlorine can be converted to its chloride more easily when compared to
conversion of Fluorine to fluoride ion. Give reason.
Ans: Chlorine has more negative electron gain enthalpy than Fluorine, more energy
is released when Chlorine gains electron to form Chloride ion, than flourine.
12. In terms of period and group where will you locate the element with z = 114?
Answer: Period – 7 and Group -14, Block-p.
13. Why do elements in the same group have similar physical and chemical
properties?
Ans: Elements in the same group have similar physical and chemical properties
same electronic configuration of outer shell.
14 How do atomic radius vary in a period and in a group? How do you explain the
variation?
Ans: In group- atomic size increases on moving from top to bottom. In period- atomic
size decreases on moving left to right in a period.
15. Which of the following pairs of elements would have a more negative electron
gain enthalpy? (i) O or F (ii) F or Cl
Ans: (i) F (ii) Cl.
16.Write the general electronic configuration of d , and f-block elements?
–

Answer: d-Block elements:(n – 1) d1-10 ns 0-2 where n = 4-7.

f-Block elements: (n – 2) f0-14 (n -1) d0-1 ns2where n = 6 – 7
17.Assign the position of the element having outer electronic configuration,
(i) ns2 np4 for n = 3 (ii) (n – 1) d2 ns2 for n = 4 and (iii) (n – 2) f7 (n – 1) d1 ns2 for n = 6
in the periodic table?
Answer: (i) n = 3 Thus element belong to 3rd period, p-block element.
Since the valence shell contains = 6 electrons, group No = 10 + 6 = 16 configuration
=1s2 2s2 2p6 3s2 3p4 element name is sulphur.
Multiple Choice Questons 1 MARK EACH

1. Which set of triads doesn't fit into Dobereiner’s Law of Triads?

a) lithium, sodium, potassium b) calcium, strontium, barium

Page | 32
c) phosphorus, arsenic, antimony d) iron, cobalt, nickel
Ans: (d)
2.In Mendeleev ’s Periodic Table, gaps were left for the elements to be discovered
later. Which of the following elements found a place in the periodic table later?
a) Germanium b) Chlorine c) Oxygen d) Silicon
Ans: (a)
3. Which of the following statements about the Modern Periodic Table is correct:
(a) It has 18 horizontal rows known as Periods
(b) It has 7 vertical columns known as Periods
(c) It has 18 vertical columns known as Groups
(d) It has 7 horizontal rows known as Groups
Ans:(c)
4. An element which is an essential constituent of all organic compounds belongs to
(a) group 1 (b) group 14 (c) group 15 (d) group 16
Ans: (b)
5. Which among the following elements has the largest atomic radii?
(a) Na (b) Mg (c) K (d) Ca
Ans: (c)
6.The anion O is isoelectronic with
-

(a) Ne (b) N2- (c) N3- (d) F-

Ans: (b)
7.The correct order of radii of three species Ca, Ca and Ca is
+ 2+

(a) Ca > Ca > Ca

+ 2+ (b) Ca > Ca > Ca
2+ +

(c) Ca > Ca > Ca

+ 2+ (d) Ca > Ca > Ca
+ 2+

Ans: (a)
8.The family of elements with the highest ionization enthalpy
(a) alkaline earth metals (b) halogens
(c) noble gases (d) alkali metals
Ans: (c)
9.The correct order in which the first ionization enthalpy increases is
(a) K, Be, Na (b) Be, Na, K
(c) Na, K, Be (d) K, Na, Be
Ans: (d)
10.Which of the following property increases down the group?
a) electronegativity b) electropositive nature of element
c) atomic size d) both b and c
Ans: (d)
11. The size of isoelectronic species-F , Ne and Na is affected by
– +

(a) nuclear charge (Z) (b) valence principal quantum number (n)
(c) electron-electron interaction in the outer orbitals
(d) none of the factors because their size is the same
Answer: (a) Nuclear charge (Z).
12. Which of the following statements is incorrect in relation to ionization enthalpy?
(a)ionization enthalpy increases for each successive electron.
(b)The greatest increase in ionization enthalpy is experienced on removal of
electrons from core noble gas configuration.
(c)End of valence electrons is marked by a big jump in ionization enthalpy.
(d)Removal of electron from orbitals bearing lower n value is easier than from orbital

Page | 33
having higher n value.
Answer: (d) is incorrect.
13.The elements in which 4f orbitals are progressively filled are called
(a) Actinides (b) Transition elements (c) Lanthanides (d) Halogens.
Answer: (c)
14. Which of the following has least electron affinity?
(a) Oxygen (b) Argon (c) Nitrogen (d) Boron
Answer: (b)
15.The general outer electronic configuration of transition elements is
(a) ns²nd1-10 (b) ns²np1(n – 1) d1-10 (c) ns²np6(n – 1) d1-10 (d) ns0-2 (n – 1) d1-10
Answer: (d)
16.Total number of elements present in the 3rd period is
(a) 6 (b) 32 (c) 18 (d) 8
Answer: (d)
SHORT ANSWER QUESTION 2 MARKS EACH
1. Write 2 achievements of Mendeleev’s periodic table.
Ans:i) Mendeleev considered similarities in properties of elements rather than atomic
masses. So he corrected the atomic mass of some elements.
ii) Prediction of the existence and properties of new elements –Mendeleev left some
gaps in his periodic table and he also predicted the roperties of new elements which
were undiscovered. E.g. Gaps were left below Aluminium and Silicon for the
undiscovered elements. They were named as Eka-Aluminium and Eka-Silicon.
These elements were further discovered as Gallium and Germanium.
2.Why is atomic number a better basis for classification of the elements than atomic
mass?
Ans: The atomic mass of the element relates to the nucleus which is present in the
centre of the atom. But the properties of the elements depend upon the electronic
configuration which is linked with atomic number. As the atomic number or electronic
configuration atom changes, the properties of the element involved also change.
3.What are the differences between electron gain enthalpy and electro-negativity?
Ans: Electron gain enthalpy : It is amount of energy associated when an electron is
added to a neutral isolated gaseous atom to convert it into a negative ion (anion).
Electro-negativity :If an atom has high tendency to add an electron, energy is
released. If an atom has no tendency to add an electron, energy is required for this
process. X(g) + 1e → X-(g) ∆egH
4. Name the factors which affect the ionisation enthalpy of an element.
Ans: (i) Size of atom or ion (ii) Nuclear charge (iii) Electronic configuration (iv)
Screening effect (v)Penetration effect of the electrons.
5.What do you understand by the term diagonal relationship? Explain with the help of
an example.
Ans: some elements of the second period show similarities in properties with the
elements of the third period placed diagonally to each other, though belonging to
different groups. This similarity in properties of elements placed diagonally to each
other is called diagonal relationship.
For example, lithium (of group 1) resembles magnesium (of group 2) and beryllium
(of group 2) resembles aluminium (of group 13)

Short answer type questions:

Page | 34
6. Explain why cation are smaller and anions larger in radii than their parent atoms?
Answer: A cation is smaller than the parent atom because it has fewer electrons
while its nuclear
charge remains the same. The size of anion will be larger than that of parent atom
because the addition of one or more electrons would result in increased repulsion
among the electrons and a decrease in effective nuclear charge.
7.How would you explain the fact that the first ionization enthalpy of sodium is lower
than that of magnesium but its second ionization enthalpy is higher than that of
magnesium?
Answer: Electronic configuration of Na and Mg are
Na = 1s2 2s2 2p6 3s1
Mg = 1s2 2s2 2p6 3s2
First electron in both cases has to be removed from 3s-orbital but the nuclear charge
of Na (+ 11) is lower than that of Mg (+ 12) therefore first ionization energy of sodium
is lower than that of magnesium. After the loss of first electron, the electronic
configuration of Na+ = 1s2 2s2 2p6 Mg+ = 1s2 2s2 2p6 3s1
Here electron is to be removed from inert (neon) gas configuration which is very
stable and hence removal of second electron requires more energy in comparison to
Mg. Therefore, second ionization enthalpy of sodium is higher than that of
magnesium.
8).

Which of the above elements is likely to be:

(a) the least reactive element (b) the most reactive metal
(c) the most reactive non-metal (d) the least reactive non-metal
(e) the metal which can form a stable binary halide of the formula MX (X = halogen)
2

(f) the metal which can form a predominantly stable covalent halide of the formula
MX (X = halogen)?
Answer: (a) The element V has highest first ionization enthalpy (∆ H ) and positive
i 1

electron gain enthalpy (∆ H) and hence it is the least reactive element. Since inert
eg

gases have positive ∆ H, therefore, the element-V must be an inert gas. The values
eg

of ∆ H , ∆ H and ∆ H match that of He.

i 1 i 2 eg

(b) The element II which has the least first ionization enthalpy (∆ H ) and a low
i 1

negative electron gain enthalpy (∆ H) is the most reactive metal. The values
eg

of ∆ H , ∆ H and ∆ H match that of K (potassium).

i 1 i 2 eg

(c) The element III which has high first ionization enthalpy (∆ H ) and a very high
i 1

negative electron gain enthalpy (∆ H) is the most reactive non-metal. The values
eg

of ∆ H , ∆ H and ∆ H match that of F (fluorine).

i 1 i 2 eg

(d) The element IV has a high negative electron gain enthalpy (∆ H ) but not so high
eg

first ionization enthalpy (∆ H). Therefore, it is the least reactive non-metal. The
eg

values of ∆ H , ∆ H and ∆ H match that of I (Iodine).

i 1 i 2 eg

Page | 35
 (e) The element VI has low first ionization enthalpy (∆ H ) but higher than that of
i 1

alkali metals. Therefore, it appears that the element is an alkaline earth metal and
hence will form binary halide of the formula MX (where X = halogen). The values of
2

∆ H , ∆ H and ∆ H match that of Mg (magnesium).

i 1 i 2 eg

(f) The element I has low first ionization (∆ H ) but a very high second ionization
i 1

enthalpy (∆ H ), therefore, it must be an alkali metal. Since the metal forms a

i 2

pronominally stable covalent halide of the formula MX (X = halogen), therefore, the

alkali metal must be least reactive. The values of ∆ H , ∆ H and ∆ H match that of Li
i 1 i 2 eg

(lithium).

LONG ANSWER TYPE QUESTIONS

1. Nitrogen has positive electron gain enthalpy whereas oxygen has negative.
However, oxygen has lower ionization enthalpy than nitrogen explain.
Sol: The outermost electronic configuration of nitrogen is 2s 2 2p1x 2p1y 2p1z. It is stable
because it has exactly half filled 2p-subshell. Therefore, it has no tendency to accept
extra electron and energy has to be supplied to add additional electron. Thus, electron
gain enthalpy of nitrogen is slightly positive. On the other hand, the outermost
electronic configuration of O is 2s2 2p2x 2p1y 2p1z. It has higher positive charge (+8)
than nitrogen (+7) and lower atomic size than N. Therefore, it has a tendency to accept
an extra electron. Thus, electron gain enthalpy of O is negative. However, oxygen has
four electrons in the 2p subshell and can lose one electron to acquire stable half filled
configuration and therefore, it has low ionization enthalpy. Because of stable
configuration of N, it cannot readily lose electron and therefore, its ionization enthalpy
is higher than that of O.
2. Arrange the elements N, P, O and S in the order of
(i) increasing first ionisation enthalpy. (ii) increasing non-metallic character.
Give reason for the arrangement assigned.
Sol: (i) S < P < O < N
Ionisation enthalpy increases from left to right in a period and decreases down the
group. N has higher ionisation enthalpy than O due to extra stability of half-filled
orbitals. Similarly, P has higher ionisation enthalpy than S due to half-filled orbitals.
(ii) P < S < N < O
Non-metallic character decreases down the group and increases from left to right in
a period.
3. Explain the following:
(a) Electronegativity of elements increases on moving from left to right in the
periodic table.
(b) Ionisation enthalpy decreases in a group from top to bottom.
Sol: (a) The electronegativity generally increases on moving across a period from
left to right (e.g., from Li to F in the second period). This is due to decrease in atomic
size and increase in effective nuclear charge. As a result of increase in effective
nuclear charge, the attraction for the outer electron and the nucleus increases in a
period and therefore, electronegativity also increases.
(b) On moving down a group there is a gradual decrease in ionisation enthalpy. The
decrease in ionisation enthalpy down a group can be explained in terms of net effect
of the following factors:
(i) In going from top to bottom in a group, the nuclear charge increases.
(ii) There is a gradual increase in atomic size due to an additional main energy shell
(n).
(iii) There is increase in shielding effect on the outermost electron due to increase in
the number of inner electrons.

Page | 36
The effect of increase in atomic size and the shielding effect is much more than the
effect of increase in nuclear charge. As a result, the electron becomes less and less
firmly held to the nucleus as we move down the group. Hence, there is a gradual
decrease in the ionization enthalpies in a group.

3. CHEMICAL BONDING AND MOLECULAR STRUCTURE

CHEMICAL BOND: The attractive force which holds various constituents (atoms,
ions, etc.) together in different chemical species is called a chemical bond.
Lewis Symbols: A Lewis Symbol is constructed by placing dots representing
electrons in the outer energy around the symbol for the element

Octet Rule: The tendency of atoms to have eight electrons in the valence shell is
known as the "Octet rule”.
Formal Charge: The formal charge is the difference in the number of valence
electrons in the atom and the number of valence electrons in the Lewis structure.

Page | 37
Resonance: According to the concept of resonance, whenever a single Lewis
structure cannot describe a molecule accurately, a number of structures with similar
energy, positions of nuclei, bonding and non- bonding pairs of electrons are taken as
the canonical structures of the hybrid which describes the molecule accurately.

Dipole moment (µ) = charge (q) × distance of separation (d)

Unit: Debye units (D). The conversion factor is 1 D = 3.33564 × 10 –30 C m where C is
coulomb and m is meter.

Page | 38
THE VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY
Main Postulates are the following::
(i) The exact shape of molecule depends upon the number of electron pairs (bonded
or non-bonded) around the central atoms.
(ii) The electron pairs have a tendency to repel each other since they exist around
the central atom and the electron clouds are negatively charged.
(iii) Electron pairs try to take such position which can minimize the repulsion between
them.
(iv) The valence shell is taken as a sphere with the electron pairs placed at
maximum distance.
(v) A multiple bond is treated as if it is a single electron pair and the electron pairs
which constitute the bond as single pairs.
(vi) Where two or more resonance structures can represent a molecule, the VSEPR
theory is applicable to any such structure.
The repulsive interaction of electron pairs decreases in the order:
Lone pair (lp)-Lone pair (lp) > Lone pair (lp) – bond pair (bp) > Bond pair (bp) –
Bond pair (bp)

Page | 39
Types of Orbital Overlap:
Depending upon the type of overlapping, the covalent bonds are of two types, known as
sigma (σ ) and pi (π) bonds.
(i) Sigma (σ bond): Sigma bond is formed by the end to end (head-on) overlap of bonding
orbitals along the internuclear axis. The extent of overlap is more so, sigma bond is a strong
bond. The axial overlap involving these orbitals is of three types:
(a) s-s overlapping: In this case, there is overlap of two half-filled s-orbitals along the
internuclear axis as shown below:

(b) s-p overlapping: This type of overlapping occurs between half-filled s-orbitals of one
atom and half filled p-orbitals of another atoms.

(c) p-p overlapping: This type of overlapping takes place between half-filled p-orbitals of
the two approaching atoms. Generally, pz orbitals are taken in inter nuclear axis.

(ii) pi (π bond): π bond is formed by the atomic orbitals when they overlap in such a way
that their axes remain parallel to each other and perpendicular to the internuclear axis. The
orbital formed is due to lateral overlapping or side wise overlapping. The extent of overlap is
less, so pi bond is a weak bond. It can be pπ - pπ or pπ - dπ or dπ - dπ overlap.

Page | 40
Hybridisation: The process of intermixing of the orbitals of slightly different energies
so as to redistribute their energies, resulting in the formation of new set of orbitals of
equivalent energies and shape.

Types of Hybridisations: (sp, sp2, sp3, sp3d, sp3d2)

(i) sp hybridisation: When one s and one p-orbital hybridise to form two equivalent orbitals,
the orbital is known as sp hybrid orbitals, and the type of hybridisation is called sp
hybridisation.
Each of the hybrid orbitals formed has 50% s-character and 50%, p-character. This type of
hybridisation is also known as diagonal hybridisation having linear structure & bond angle =
180°.

Example: BeCl2, BeF2, C2H2 etc.

Page | 41
(ii) sp2 hybridisation: In this type, one s and two p-orbitals hybridise to form three
equivalent sp2 hybrid orbitals in the same plane making an angle of 120° & trigonal planar
shape. Each of the hybrid orbitals formed has 33.3 % s-character and 66.7 %, p-character.
Example: A few compounds in which sp 2 hybridisation takes place are BF3, BCl3 carbon
compounds containing double bond (C2H4) etc.

(iii) sp3 hybridisation: In this type, one s and three p-orbitals in the valence shell of an atom
get hybridised to form four equivalent sp3 hybrid orbitals. Each orbital has 25% s-character
and 75% p-character. With tetrahedral structure & bond angle 109.5° The four sp3 orbitals
are directed towards four corners of the tetrahedron.

A compound having sp3 hybridisation is CH4, CCl4, NH4+ etc.

The structures of NH3 and H2O molecules can also be explained with the help of
sp3 hybridisation.

Hybridisation in PCl5

Page | 42
Molecular Orbital Theory: According to this theory, the atomic orbitals combine to
form the molecular orbitals. The number of molecular orbitals formed is equal is the
number of atomic orbitals involved. According to this theory.
The molecular orbitals are formed by LCAO (Linear combination of atomic orbitals)
method, i.e., by addition or subtraction of wave functions of individual atoms,
thus Ψ MO = ΨA ± ΨB
Ψb = Ψ A + Ψ B
Ψa = Ψ A – Ψ B
2. Molecular orbital of lower energy is known as bonding molecular orbital and that of
higher energy is known as anti-bonding molecular orbital.
3. Aufbau rule, Pauli’s exclusion principle and Hund’s rule are all applicable for
molecular orbitals.
4. The shape is governed by the shape of atomic orbitals, e.g., s-s and p-p
overlapping.
(i) Combination between s-atomic orbitals (ii) Combination between p-atomic
orbitals (iii) Combination between 2p x and 2py atomic orbitals

Electronic Configuration and Bond Order (BO) of Molecular Species: The order
of energy of molecular orbitals has been determined experimentally by spectroscopy
for the elements of the second period. The increasing order of energies of the
molecular orbitals in homonuclear diatomic molecules is
(A) For (O2, F2 & Ne2 (hypothetical) :
Electronic configuration: σ1s σ*1s σ2s σ*2s σ2pz (π2px = π 2py) (π*2px = π* 2py)
σ*2pz
(B) For B2, C2, N2

Page | 43
Electronic configuration: σ1s σ*1s σ2s σ*2s (π2px = π 2py) σ2pz (π*2px = π* 2py)
σ*2pz
Bond Order (BO) : Bond order is defined as half of the difference between the number of
electrons present in bonding molecular orbitals (Nb) and antibonding molecular orbitals (Na).
Bond order (B.O.) = 1/2 [Nb-Na] where Nb & Na are no. of bonding electrons & anti bonding
electrons.
(i) Stability: If BO > 0 or positive, Nb > Na ➔ the molecular species is stable & exists in
nature.
If BO ≤ 0 (i.e. zero or negative) : Nb ≤ Na ➔ the molecular species is unstable & does not
exist in nature.
(ii) The integral value of BO corresponds to the no. of covalent bonds.
(iii) Higher the BO, the molecular species is more stable.
(iv) Bond order ∝ stability of molecular species
Bond order ∝ bond dissociation enthalpy of molecular species.
1
Bond order ∝ bond length of molecular species
(v) BO is fractional ➔ the molecular species is paramagnetic. (having one or more unpaired
electrons & is weakly attracted in a magnetic field.
Bonding in Homonuclear
(Diatomic) Molecules:
(1) Hydrogen molecule (H2): It is
formed by the combination of two
hydrogen atoms. Each hydrogen
atom has one electron in Is orbital,
so, the electronic configuration of
hydrogen molecule (2 e-) is σ 1s2
σ*1s
Nb − Na 2−0
BO = = = 1 Hence, two
2 2
hydrogen atoms are bonded by a
single covalent bond. Bond dissociation energy of hydrogen has been found = 438 kJ/mole.
Bond-Length = 74 pm
No unpaired electron is present therefore, it is diamagnetic.

(2) Helium molecule (He2): Each helium atom contains 2

electrons (1s2), thus in He2 molecule has 4 electrons.
He2: electronic configuration is σ1s2 σ*1s2
Nb − Na 2−2
BOND ORDER = = =0 So, He2 does not
2 2
exist.

Page | 44
(4) Explain: Be2 does not exist in nature:
Be2 has 8 electrons
Its electronic configuration: σ1s2 σ*1s2 σ2s2 σ*2s2
Nb − Na 4−4
So, bond order = = = 0 So, Be2 does not exist.
2 2

The energy level diagram of Be 2 is ➔

(5) Explain: Dioxygen molecule is paramagnetic.

Draw its energy level diagram.
Ans. Atomic no. of oxygen atom is 8 so, O2 has 16
electrons.
Molecular orbital electronic configuration of O 2 is
σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 (π2px2 = π 2py2) (π*2px1 =
π* 2py1)
Due to the presence of unpaired electrons in π*2px1 =
π* 2py1 O2 molecule is paramagnetic
Nb − Na 10−6
(Bond order = = = 2. It has 2 covalent
2 2
bonds.)

Hydrogen Bond
It is defined as the electrostatic force of attraction existing between hydrogen atom
covalently bonded to highly electronegative atom (N, O or F) and the electronegative
atom belonging to another molecule of the same or different substance. It is
represented by dotted lines. The chains possess a zig – zag structure.

Types of hydrogen bonds are

(i) Intermolecular H-bonding: H-bonding involving two or more molecules.
(ii) Intramolecular H-bonding: H-bonding within a molecule.

Page | 45
Water is a liquid due to inter molecular H bonds.

Applications of Intermolecular H-bonding

(i) Melting point and boiling point of water has the lowest molecular weight among
the hydrides of group 16 elements yet it has the highest melting and boiling points. It
is due to intermolecular H Bonding in H2O.
(ii) Melting point and boiling point of alcohols the marked difference between the
melting and boiling points of alcohols is also due to H-bonding.

Strength of bonds: Ionic bond > covalent bond > metallic bond > H-bond.

Assertion & Reason Questions

Directions: Each of these questions contain two statements, Assertion and Reason.
Each of these questions also has four alternative choices, only one of which is the
correct answer. You have to select one of the codes (a), (b), (c) and (d) given below.
(a) Assertion is correct, reason is correct; reason is a correct explanation for
assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for
assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.

Q.1. Assertion : The bond order of helium is always zero.

Reason : The number of electrons in bonding molecular orbital and antibonding
molecular orbital is equal.
Ans- A
Q.2. Assertion : The lesser the lattice enthalpy more stable is the ionic compound.
Reason : The lattice enthalpy is greater, for ions of highest charge and smaller radii.
Ans- D
Q.3. Assertion : pi bonds are weaker than σ bonds.
Reason : pi bonds are formed by the overlapping of p-p orbitals side wise.
Answer- A
Q.4. Assertion : BF molecule has zero dipole moment.
3

Reason : F is electronegative and B–F bonds are polar in nature.

Answer- B
Q.5. Assertion : CH2Cl2 is non-polar and CCl4 is polar molecule.
Reason : Molecule with zero dipole moment is non-polar in nature.
Answer- D
Q.6. Assertion : Lone pair-lone pair repulsive interactions are greater than lone pair-
bond pair and bond pair-bond pair interactions.
Reason : The space occupied by lone pair electrons is more as compared to bond
pair electrons.

Page | 46
Answer- A

Very Short Answer Type Questions

1. Define a chemical bond.
Ans - It is the attractive force which hold the constituent particles (atoms, ions etc)
together in different chemical species is called a chemical bond.
2. Give the main feature of Lewis approach of chemical bonding.
Ans - Lewis postulated that atoms achieve the stable octet when they are linked by
chemical bonds. In case of sodium and chlorine, both attain octet by complete
transfer of electron.
3. Write electron dot structure (Lewis structure) of Na, Ca, B, Br, Xe, As, Ge, N -3

4.Which one of the following has the highest bond order? N2, N2+ or N2-.
Ans - N2 (BO = 3)
5. Write the type of hybridisation of C is involved in CH4,C2H4 and C2H2.
Answer: CH4= sp3 C2H4 = sp2 C2H2 = sp
6. Predict the shapes of the following molecules using VSEPR theory?
(i) BeCl2 (ii) SiCl4
Ans. (i) Linear (ii) Tetrahedral.
7. Define bond order.
Ans - In the Lewis description of covalent bond, the Bond Order is given by the
number of bonds between the two atoms in a molecule.
8. What type of bond is formed when atoms have high difference of
electronegativity?
Ans - Electrovalent bond or Ionic bond.
9. Define dipole moment.
Ans. The product of the magnitude of the charge and the distance between the
centres of positive and negative charge.
10. Give the mathematical expression of dipole moment.
Ans. Dipole moment (µ) = charge (q) × distance of separation (d)
11. Why is dipole moment of BeF2, BF3 is zero?
Ans - The dipole moment in case of BeF 2 is zero. This is because the two
equal bond dipoles point in opposite directions and cancel the effect of
each other
In BF , the dipole moment is zero although the B – F bonds are
3

oriented at an angle of 120° to one another, the three bond

moments give a net sum of zero as the resultant of any two is
equal and opposite to the third.
12. How is bond order related to the stability of a molecule?
Answer: Higher the bond order, greater is the stability.
13. Write the resonating structure of O3 molecule.
Ans.

Page | 47
14. Arrange O2, O2–, O22-, O2+in increasing order of bond energy.
Answer: O22-< O2– < O22-< O2+
15. Arrange the following, according to increasing covalent nature.
NaCl, MgCl2, AlCl3
Answer: NaCl < MgCl2 < AlCl3
17. Out of sigma and Π bonds, which one is stronger and why?
Awer: sigma-bond is stronger. This is because sigma-bond is formed by head-on
overlapping of atomic orbitals and Π bond is formed by side wise overlapping.
18. What type of bond are formed due to orbital overlap?
Ans - Covalent bond
19. Define bonding & antibonding molecular orbital.
Ans - When two atomic orbitals combine, two molecular orbitals are formed. One is
known as bonding molecular orbital while the other is called antibonding molecular
orbital.
The bonding molecular orbital has lower energy and hence greater stability than the
corresponding antibonding molecular orbital.
20. Define bond order.
Ans - Bond order (B.O.) is defined as one half the difference between the number of
electrons present in the bonding and the antibonding Orbitals
Bond order: = ½( Nb –Na )
21. Why B2 is paramagnetic in nature while C2 is not?
Answer: The molecular orbital electronic configuration of both B 2 and C2 are-

Since, B2 has two impaired electrons,

B2 is paramagnetic. C2 has no unpaired electron. Thus, C2 is diamagnetic.
22.What are the types of H-bonding? Which of them is stronger?
Ans - Intramolecular H-bonding & Intramolecular H-bonding. Intermolecular H-
bonding is stronger.
Multiple Choice Questions (MCQs ) 1 mark each:
Question 1. A co-ordinate bond is formed by:
(a)sharing of electrons contributed by both the atoms
(b)complete transfer of electrons
(c)sharing of electrons contributed by one atom only (d) none of these.
Question 2. The species CO, CN and N are:
–
2

(a) isoelectronic (b) having coordinated bond

(c)having polar bond (d) having low bond energies.
Question 3. The axial overlap between the two orbitals leads to the formation of a:
(a) sigma bond (b) pi bond (c) multiple bond (d) none of these.

Page | 48
Question 4. In S0 molecule, S atom is:
2

(a)sp hybridized
3 (b) sp hybridized (c) sp hybridized
2 (d) d sp hybridized.
2

Question 5 MX is a molecule with octahedral geometry. How many X – M – X

6

bonds are at 180°?

(a)four (b)two (c)three (d)six.

Question 6. The molecule Ne2 does not exist because

(a) Nb > Na (b) Nb = Na (c) Nb < Na (d) None of these.

Question 7 Which of the molecules does not have a permanent dipole moment?
(a) SO3 (b) SO2 (c) H2S d) CS2

Question 8. In sp , sp and sp hybridized carbon atom, the p character is maximum

3 2

in: (a) sp3 (b) sp2 (c) sp (d) all of the above have same p-character.

Question 9. Out of the following, intramolecular hydrogen bonding exists in:

(a) water (b)H2S (c) 4-nitrophenol (d) 2-nitrophenol.

Question 10. The outer orbitals of C in ethene molecule can be considered to be

hybridized to give three equivalent sp² orbitals. The total number of sigma (s) and pi
(p) bonds in ethene molecule is
(a) 1 sigma (s) and 2 pi (p) bonds (b) 3 sigma (s) and 2 pi (p) bonds
(c) 4 sigma (s) and 1 pi (p) bonds (d) 5 sigma (s) and 1 pi (p) bonds.

MCQs Answer: 1. (c) 2. (a) 3. (a) 4. (c) 5. (c) 6. (b) 7. (d) 8. (a) 9. (d) 10. (d)

SA 2 marks
Q.1 A molecule of H exists while that of the He does not. Explain.
2 2

Ans-Hydrogen atom (Z = 1) has only one electron present in the s-orbital(1s1). The
two such half-filled atomic orbitals combine to form a molecular orbital which
contains both these electrons. But helium (Z = 2) has already a filled orbital (1s 2).
Therefore, the atomic orbitals of the helium atoms do not combine. Thus, a molecule
of H2 exists while that of He2 does not. (BO of H2 = 1 but BO of He2 = 0)
Q.2 Bond angle in NH3 is more than in PH3. Explain.
Ans- Both NH3 and PH3 are the hydrides of nitrogen and phosphorus which are
present in group 15. Nitrogen is more electronegative (3.0) as compared phosphorus
(2.1). As a result, the shared electron pairs lie closer to the nitrogen atom in
NH3 than to the phosphorus atom in PH3. Therefore, the electron density around the
nitrogen atom is more than around the phosphorus atom. This results in greater
repulsion in the electron pairs around nitrogen atom than around phosphorus atom.
As a result, bond angle in NH3 is more than in PH3.

Q. 3 A molecule of PCl5 exists while that of NCl 5 does not.

Page | 49
Ans-The valence shell electronic configurations of both N and P are as follows :

Since
nitrogen atom has no vacant 2d orbitals, it cannot extend its covalency to five.
Therefore, a molecule of NCl 5 does not exist. But in case of phosphorus, an electron
can be promoted from 3s filled orbital to 3d vacant orbitals. Therefore, phosphorus
can extend its covalency to five. Thus, a molecule of PCl5 can exist.
Q. 4 o-nitrophenol is steam volatile while p-nitrophenol is not. Discuss.
Answer: In o-nitrophenol, intra molecular hydrogen bonding is present while inter
molecular hydrogen is present in the molecules of p-nitrophenol. Energy is needed to
overcome attractive forces in the molecules of p-nitrophenol but no such energy is
required in case of molecules in o-nitrophenol. This means that the boiling point of o-
nitrophenol is less and is steam volatile while that of p-nitrophenol is more and is,
therefore, non-volatile in steam.

Q. 5 Draw the Lewis structures of the following molecules and ions and tell in which
case/cases the octet rule is violated CO2, SO2, BeCl2, NH3, AlCl3, PCl5, CO32−
Answer:

The octet rule is violated in case of BeCl 2, AlCl3 and PCl5.

Q.6 On the basis of VSEPR theory, predict the shapes of the following molecules /
ions? (i) SiF4 (ii) NH−2 (iii) NH4+ (iv) C2H2 (v) H3O+(vi) F2O (vii) PCl3 (viii) PF5.
Answer: (i) Tetrahedral (ii) V-shape (lii) Tetrahedral (iv) Cylindrical (v) Trigonal
pyramidal (vi) V-shape (vii) Trigonal pyramidal (viii) Trigonal bipyramidal.

Q. 7 Draw the shapes of the following hybrid orbitals ; sp,sp and sp

2 3

Answer:

All the hybrid orbitals have same shape. However, their sizes are in the order

Page | 50
sp < sp2 < sp3

Q. 8. Name the type of hybridisation of each C-atom in a molecule of (i) propylene

(propene) (ii) propyne. How many σ and π-bonds are present in each case?
Answer:

bonds =8,π-bonds = 1.

(ii)
C−2 and C−3 are sp), σ- bonds = 6, π-bonds = 2.

Q. 9. Which out of CH3F and CH3Cl has a higher dipole moment and why?
Answer: The dipole moment of CH3Cl is greater than that of CH3F. The C-F bond
length in CH3F is smaller than the C-Cl bond length in CH3Cl. The charge separation
in the C-F bond is more than in the Cl-C bond- fluoride being more electronegative
than chlorine. The bond length has a greater effect than the charge separation.
Hence the dipole moment of CH3Cl is greater than that of CH3F.
Q. 10. Why HCl is polar whereas the Cl 2 molecule is non-polar?
Answer: In Cl2 both atoms have the same electronegativity. Hence the shared pair of
electrons is attracted equally by both and remains exactly in the center.
In HCl, chlorine is more electronegative than H. Hence shared pair of electrons is
more attracted towards chlorine, which, therefore acquires a negative charge while H
acquires a positive charge.
SA 3MARKS
Q.1 Write the resonance structures for SO , NO and NO
3 2 3

Answer:

Q. 2 Discuss the shape of the following molecules using the VSEPR model:
BeCl2, BCl3 , SiCl4, AsF5, H2S,PH3
Answer:

Page | 51
Q.3 Compare the relative stability of the following species and indicate their
magnetic properties: O2, O2, O2– (Superoxide),O22- (peroxide)
Answer: O2 :Bond order = 2, paramagnetic
O2+ : Bond order = 2.5, paramagnetic
O2– : Bond order = 1.5, paramagnetic
O22-: Bond order = 1, diamagnetic
Order of relative stability is O2+ > O2 > O2– > O22- as bond order of O2+, O2 , O2–, O22-
are (2.5) (2.0) (1.5) (1.0) respectively.
Q.4 Account for the following:
(i) Water is a liquid while H2S is a gas
(ii) NH3 has higher boiling point than PH3.
Answer: (i) In case of water hydrogen bonding causes association of the H 2O
molecules. There is no such hydrogen bonding in H 2S, that’s why it is a gas.
(ii) There is hydrogen bonding in NH3 but not in PH3.
Q 5. (a) Define dipole moment. What are the units of dipole moment?
(b) Dipole moment values help in predicting the shapes of covalent molecules.
Explain.
Answer: (a)Dipole moment: In a polar molecule, one end bears a positive charge and
the other has a negative charge. Thus, the molecule has two poles with equal
magnitude of the charges. The molecule is known as dipolar molecule and
possesses dipole moment.
It is defined as the product of the magnitude of the positive or negative charge and
the distance between the charges. µ (dipole moment) = q x d
SI unit of dipole moment is coulomb metre (m) or Debye.
(b)The dipole moment values are quite helpful in determining the general shapes of
molecules.

Page | 52
For molecules with zero dipole moment, shapes will be either linear or symmetrical.
For Example. BeF2 CO2 etc. Molecules that possess dipole moments, their shape will
not be symmetrical.
Q 6. Discuss the orbital structures of the following molecules on the basis of
hybridisation, (i) BH3 (ii) C2H2
Answer:

B atom gets hybridised to form three equivalent hybrid orbitals directed towards
three comers of equilateral triangle with B atoms in the centre. Bond angle = 120°.

Both the carbon atoms are sp hybridised. Both the carbon atoms have also two
unhybridized orbitals which overlap sidewise with the similar orbitals of the other
carbon atom to form two pi bonds.

3) Which is most stable out of O2+, O2‾ & O22 ‾.? Explain. Arrange them in increasing
order of bond length & bond dissociation energy.
Ans. Molecular orbital electronic configurations are
O2+ = (15 e-) σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 (π2px2 = π 2py2) (π*2px1 = π* 2py0)
O2‾ = (17e-) σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 (π2px2 = π 2py2) (π*2px2 = π* 2py1)
O22 ‾ = (18 e-) σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 (π2px2 = π 2py2) (π*2px2 = π* 2py2)
Nb − Na 10−5 Nb − Na 10−7
Bond order of O2+ = = = 2.5 Bond order of O2‾ = = = 1.5
2 2 2 2
Nb − Na 10−8
Bond order of O22 ‾ = = =1
2 2
So, Bond order increases O22- < O2- < O2+ Hence, O2+ is most stable.
Bond length: O2+ < O2‾ < O22‾
Bond dissociation energy: O22 ‾ < O2‾ < O2+

4. Thermodynamics
[MLL: System, Surroundings, Types of system, State function, Extensive and
intensive properties, Isothermal and adiabatic process, Internal Energy and Internal
Energy change, First Law of thermodynamics and its mathematical formula, Enthalpy
and enthalpy change, relation between heat of reaction at const pressure and

Page | 53
constant volume (∆H = ∆ U + ∆n(g) RT), simple numerical based on this formula,
Heat capacity, Gibbs’ Free energy, Entropy, Second law of thermodynamics, Hess’s
law, numerical based on this law and simple numerical based on the formula
∆G° = – 2.303RT log K, ∆G = ∆H − T∆S, ∆S = qrev/T & NCERT Text Book
Question No. 6.1 to 6.6 MCQ, 6.14, 6.17, 6.19, 6.20]
MLL THERMODYNAMICS CLASS XI
System: The part of universe in which observations are made.
Surroundings: The part of universe other than the system is known as
surroundings.
Types of Systems:
Open system: The system in which energy and matter both can be exchanged with
the surroundings.
Closed system :The system in which only energy can be exchanged with the
surroundings.
Isolated system:The system in which neither energy nor matter can be exchanged
with the surroundings.
Intensive Properties: Properties which do not depend on the quantity of matter.
e.g., pressure, temperature, specific heat, molar heat capacity etc
Extensive Properties: Properties which are dependent on the quantity of matter
e.g., internal energy, volume, enthalpy, heat capacity etc.
State functions: Properties of a system which are independent of path followed
and depend only on initial and final state e.g., U, H, G etc.
Path functions: Properties which depend upon the path followed e.g., work, heat,
etc.
Thermodynamic Process: It is the operation which brings change in the state of
the system. Thermodynamic processes are as follows:-
Isothermal process: In which temperature remains constant, i.e.,(∆T = 0, ΔU = 0).
Isochoric process: In which volume remains constant, i.e., (Δ V = 0).
Isobaric process: In which pressure remains constant, i.e., (Δp = 0).
Adiabatic process: In which heat is not exchanged between the system and the
surroundings (q= 0).
First Law of Thermodynamics:
Energy can neither be created nor destroyed although it can be converted from
one form to the other. Mathematically, ΔU = q + W
Sign conventions: q is + ve = heat is supplied to the system
q is – ve = heat is lost by the system
W is + ve = work done on the system (eg during compression of a gas)
W is – ve = work done by the system (eg during expansion of a gas)
Enthalpy (H): H = U + PV
At constant volume ∆V = 0, q v =∆ U
So ∆H = ∆U + p∆V, qp = H2- H1 = ∆H = ∆U + P∆ V
Exothermic reaction: Reaction accompanied by release of energy. q = ∆H = -ve
Endothermic reactions: Reaction accompanied by release of energy. q = ∆H = +ve
Specific heat capacity: Specific heat, also called specific heat capacity is the
quantity of heat required to raise the temperature of unit mass of a substance by one
degree Celsius (or one kelvin).
Standard Enthalpy of Reactions:The standard enthalpy of reaction is the enthalpy
change for a reaction when all the participating substances are in their standard
states. The standard state of a substance at a specified temperature is its pure form
at 1 bar
Standard enthalpy of fusion: The enthalpy change that accompanies melting of
one mole of a solid substance in standard state.

Page | 54
Standard enthalpy of vaporization: The enthalpy change that accompanies
vaporization one mole of a liquid at constant temperature and under standard
pressure.
Standard enthalpy of sublimation: ∆subH0 is the change in enthalpy when one
mole of a solid substance sublimes at a constant temperature and under standard
pressure (1bar).
Standard enthalpy of formation: The enthalpy change for the formation of one
mole of a compound from its elements in their standard states.
Bond enthalpy (B.E. or ∆bondH: It is amount of energy released when gaseous
atoms combines to form one mole of bonds between them or heat absorbed when
one mole of bonds between them are broken to give free gaseous atoms. Further
∆rH = B.E.(Reactants) - B.E.(Products)
Lattice Enthalpy ∆latticeH: The lattice enthalpy of an ionic compound is the enthalpy
change which occurs when one mole of an ionic compound dissociates into its ions
in gaseous state.
Enthalpy of Solution (∆solH0): Enthalpy of solution of a substance is the enthalpy
change when one mole of it dissolves in a specified amount of solvent.
Hess’s Law of constant heat summation: The total enthalpy change in a reaction
is same whether the reaction takes place in one step or in several steps.
i.e. ∆H = H1 + H2 + H3 + --------
Entropy (S): It is measure of randomness or disorder in a system.
Entropy change ( ∆S) = q(rev.) /T
Order of entropy: Gas > Liquid > Solid.
Crystallization of a solid: Entropy decrease (∆S = -ve)
H2 → 2H : Entropy increases (∆S = +ve)
Melting of a solid : Entropy increases (∆S = +ve)
Vaporization of a liquid: Entropy increases (∆S = +ve)
Dissolution of a solid : Entropy increases (∆S = +ve)
Spontaneity in term of ( ∆S)
∆S(total) = ∆S(universe) = ∆S(system) + ∆S(surrounding)
If ∆S(total) is +ve, the process is spontaneous.
If ∆S(total) is –ve, the process is non spontaneous
Second Law of thermodynamics: In any spontaneous process, the entropy of the
universe always increases. ∆Stotal = ∆Suniverse = ∆Ssystem + ∆Ssurrounding > 0
Gibb’s energy (G): G = H – TS
∆G =∆H – T∆S (Gibb’s equation)
Spontaneity in term of ( ∆G) If ∆G = - ve the process is spontaneous.
If ∆G = + ve the process is non-spontaneous If ∆G = 0 for Equilibrium
ΔH ΔS ΔG Reaction will be spontaneous / non spontaneous
-ve +ve -ve Spontaneous at all temp
+ve -ve +ve Non spontaneous
-ve -ve -ve (at low temp) Spontaneous at low temp
+ve (at high temp) Non spontaneous at high temp
+ve +ve +ve (at low temp) Non spontaneous at low temp
-ve (at high temp) Spontaneous at high temp

SOME IMPORTANT FORMULAE

∆U=q+w

∆U=q v

Page | 55
∆H=q p

∆H=∆U+P∆V
∆H=∆U+∆n RT g

Q = c m ∆T (c is specific heat capacity)

Q = C ∆T (C is heat capacity)
Q = c n ∆T
m (c is molar heat capacity and n is no. of moles)
m

∆rH = B.E. (Reactants) - B.E. (Products)

∆r H = ∑ ∆f H (Products) - ∑ ∆f H (Reactants)

Hess’s Law ∆H=∆H1+∆H2+∆H3……

∆Hsol=∆H lattice + ∆H hydration

∆G = ∆H – T∆S

∆rG0 = - RT ln K
∆rG0 = - 2.303 RT ln K
Assertion Reasoning Type Question:
In the following questions a statement of Assertion (A) followed by a statement of
Reason (R) is given. Choose the correct option out of the choices given below each
question.
a. Both A and R are true and R is the correct explanation of A.
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
1. Assertion (A): Combustion of all organic compounds is an exothermic reaction.
Reason (R): The enthalpies of all elements in their standard state are zero.
(Answer) a) Both A and R are true but R is not the correct explanation of A.

2. Assertion (A): Spontaneous process is an irreversible process and may be

reversed by some external agency.
Reason (R): Decrease in enthalpy is a contributory factor for spontaneity
(Answer) (b) Both A and R are true but R is not the correct explanation of A.

3. Assertion (A) : A liquid crystallises into a solid and is accompanied by decrease in

entropy.
Reason (R) : In crystals, molecules organise in an ordered manner
(Answer) (a) Both A and R are true and R is the correct explanation of A.

4. Assertion (A): The thermodynamic factor which determines the spontaneity of a

process is the free energy. For a process to be spontaneous the free energy must be
-ve.
Reason (R) The change in free energy is related to the change in enthalpy and
change in entropy. The change in entropy for a process must always be positive if it
is spontaneous.
(Answer) (c) A is true but R is false.

Page | 56
5. Assertion: The sum of q + w is a state function.
Reason: Work and heat are state functions.
(Answer)(c) A is true but R is false.

6. Assertion: A reaction in which randomness of the system increases, will always be

spontaneous.
Reason: According to second law of thermodynamics, spontaneous reaction is
accompanied by increase in randomness of the universe.
(Answer): (d) A is false but R is true.

7. Assertion: Standard enthalpy of formation of diamond is zero.

Reason: Standard enthalpy of formation of substance is the enthalpy change for
formation of one mole of the substance from its elements in their standard states.
(Answer): (d) A is false but R is true

8. Assertion : The sum of q + w is a state function.

Reason : Work and heat are state functions.
(Answer): (c) A is true but R is false.

C. Short Answer Type : 1 mark each.

1. ∆G = – 2.303RT log K, for K = 1 what is the value of ∆G ?
0 0

Ans: Zero ∵ log 1 = 0

2. Identify intensive properties out of the following :Mass, density, heat capacity,
Temperature, specific heat capacity
Answer: density, Temperature, specific heat capacity
3 . At what condition of a chemical change, the value of ∆H and ∆U will be equal?
∆H = ∆U when ∆n(g) = 0 i.e. no of moles of reactants and products are equal
4. State Hess’s Law
Ans: The Enthalpy change for a particular reaction is always constant and is
independent of path in which the reaction takes place ΔH = ΔH +ΔH + ……..
1 2

5. What is the value of entropy change ( ∆S ) at equilibrium state?

Ans: ∆S = 0
6. For an isolated system ∆U = 0; what will be ∆S? Ans: ∆S > 0
7. Calculate the value of ∆n (g)
(i) C(s) + O 2(g) →CO2 (g) (ii) CO (g) + ½ O 2(g) → CO 2 (g)
Ans ; (i) ∆n (g) = 0 (ii) ∆n (g) = I – ( 1 + ½) = − ½
8. State 3rd law of thermodynamics.
Ans: The entropy of perfectly crystalline substances at absolute zero is taken as
zero.
9. State the second law of thermodynamics.
Ans: In any spontaneous process, the entropy of the universe always increases.
10. Define the term Heat Capacity.
Ans; The amount of heat required to raise the temperature of a substance by 1 0 is
called Heat Capacity.
11. For a reaction both ∆H and ∆S, under what condition the reaction will
be occurred spontaneously?
Ans; ∆G = ∆H − T∆S, If ∆G is −ve then the reaction will be spontaneous, at
higher temperature the term T∆S will be large and ∆G value will be –ve.

Page | 57
12. What is meant by an adiabatic process?
Answer: A process in which there is no transfer of heat between the system
and surroundings.
13. Define Lattice enthalpy.
Answer: Enthalpy change when one mole of an ionic solid changes into its ions in
gaseous state.
14. Define standard enthalpy of formation of a substance.
Answer: Enthalpy change when one mole of a substance is formed from its
elements in their standard states.
15. For the reaction, 2 Cl(g) → Cl (g), what are the signs of ΔH and ΔS ?
2

Answer: ΔH and ΔS are negative.

16. Identify state functions out of the following H, q, U , w
Answer: H, U.
17. Predict whether the entropy increases/decreases in the following changes-
(i) dissolution of a solid to form aqueous solution
(ii) Temperature of a crystallize solid raised from OK to 115K
Answer: i) increases ii) increases.
18. Given: N2(g) + 3H2(g) → 2NH3(g), ΔH0 = –92.4KJ.mol–1 What is the
standard Enthalpy of formation of NH3(g)?
Answer: standard enthalpy of formation of NH3(g) = -92.4/2 = -46.2 kJ/mol.
19. The enthalpy of atomisation for the reaction CH 4(g)→ C(g) + 4H (g) is 1665 kJ
mol–1. What is the bond energy of C–H bond?
Answer: The bond energy of C–H bond = 1665/4 = 416.2 kJ/mol
20. For the reaction CO(g) + ½ O2(g)→ CO2(g) Which out of Enthalpy change and
internal energy change will have a larger value?
Answer: Since ∆ng = - ½, using formula ∆H = ∆U + ∆ngRT
∆H < ∆U
Multiple Choice Questions Type: 1 mark each
1. Which of the following statements is correct?
(i) The presence of reacting species in a covered beaker is an example
of open system.
(ii) There is an exchange of energy as well as matter between the
system and the surroundings in a closed system.
(iii) The presence of reactants in a closed vessel made up of copper is
an example of a closed system
(iv) The presence of reactants in a thermos flask or any other closed
insulated vessel is an example of a closed system
Ans: (iii)
2. The volume of gas is reduced to half from its original volume. The specific heat will
be ______.
(i) reduce to half (ii) be doubled
(iii) remain constant (iv) increase four times
Ans: (iii)
3. The entropy change can be calculated by using the expression ∆S = q / T. When
rev

water freezes in a glass beaker, choose the correct statement amongst the following
(i) ∆S (system) decreases but ∆S (surroundings) remains the same.
(ii) ∆S (system) increases but ∆S (surroundings) decreases.
(iii) ∆S (system) decreases but ∆S (surroundings) increases.
(iv) ∆S (system) decreases and ∆S (surroundings) also decrease.

Page | 58
Ans: (iii)
4. The work done in case of isothermal free expansion is
a) maximum b) minimum c) zero d) positive
(Answer) c) zero.
5. Which of the following statement is not correct?
(i) ∆G is positive for a spontaneous reaction
(ii) ∆G is negative for a spontaneous reaction
(iii) ∆G is positive for a non-spontaneous reaction
(iv) ∆G is zero for a reversible reaction
Ans: (i)
6. ∆ U of formation of CH4 (g) at certain temperature is –393 kJ mol–1. The
f
0

value of ∆fH0 is (i) zero (ii) < ∆fU0 (iii) > ∆fU0 (iv) equal to ∆fU0
Ans: (ii)
7. Which of the following is an extensive property
i) Molar heat capacity ii) Temperature iii) Enthalpy iv) All of these
Answer: iii) Enthalpy.
8. The q is ……………. when heat is transferred from the surroundings to the system
and w is………………………..When a gas undergoes expansion
i) positive, negative ii) negative, positive
iii) positive, positive iv) negative, negative
Answer: i) positive, negative.
9. ΔH for the reaction N2(g) + 3H2(g) ⇋ 2NH3(g)
(i) ΔU – 2RT (ii) ΔU + 2RT (iii) ΔU – RT (iv) ΔU = RT
Ans: (i)
10. In bomb calorimeter which quantity is kept constant to measure ΔU?
(i)volume (ii) pressure (iii) temperature (iv) all of these
Ans: (i) volume

Short Answer Type : 2 mark each

1. State whether the entropy increases or decreases in the following
transformation
(i) 2H2O (l) → 2 H2 (g) + O2(g) (ii) CaCO3 (s) → CaO ( s) + CO2 (g)
(iii) 2SO2(g) + O2 (g) → 2 SO3(g) (iv) H2O (ice) → H2O (l)
Ans i) Entropy increases ii) Entropy increases

2. The equilibrium constant for the reaction is 10. Calculate the value of ∆G. Given R
= 8 J/k/mol ; T = 300K
Ans: ∆G 0 = – 2.303RT log K
= - 2.303 x 8 x 300 x log 10 = - 2.303 x 8 x 300 [∵ log 10=1 ]
= – 5.527kJ /mol

3. Identify the state functions and path functions out of the following: enthalpy,
entropy, heat, temperature, work, free energy.
Ans: State function: enthalpy, entropy, heat, temperature
Path function: work, free energy

4. Calculate the entropy change in surroundings when 1.0 mol of water H O is2 (l)

formed under standard condition. Given ∆H = -286 kJ/mol

0

Ans.

Page | 59
6.The heat of combustion of gaseous CH4 at const. volume is measured in a bomb
calorimeter at 298 K and it is found to be – 885.4 kJ/mol. Find the value of enthalpy
change.
Solution: CH4 (g) + 2O2 (g) → CO2(g) + H2O (l)
∆n(g) = 1 – ( 1+2) = -2 moles
∆ U = – 885.4 kJ / mol = - 885400 J / mol
∆H = ∆ U + ∆n(g) RT or, ∆H = - 885400 − 2 x 8.314 x 298 = - 890355 J
= - 890.355 kJ
6. For a reaction at 298 K, 2A + B ——-> C
∆H = 40 kJ / mol and ∆S = 0.2 kJ / K / mol. At what temperature will the reaction
become spontaneous considering ∆H and ∆S to be constant over the temperature
range?
Answer: As per the Gibbs Helmholtz equation: ΔG = Δ H − TΔ S
For equilibrium, ΔG = 0 ; ΔH = TΔS or T = ΔH / ΔS
T = (400 KJ mol-1) / (0.2 KJ K-1 mol-1) = 2000 k
Thus, reaction will be in a state of equilibrium at 2000 K and will be spontaneous
above this temperature.
7. During a change,5 kJ of heat is supplied to the system and 200J of work is done
by the system. Calculate the change in internal energy of the system.
Answer: △U = q + w
Heat is supplied to the system. So, q = +5kJ = +5000J
Work is done by the system. So, w = -200J
Putting values in the equation
△U = 5000J – 200J
△U = 4800 J
8. Calculate the heat necessary to raise the temperature of 54 g of aluminium
from 10°C to 30°C. Molar heat capacity of Al is 24 J mol –1 K–1
Answer: C = 24 J mol–1 K–1
△T = T2 – T1 = 30 – 10 = 20°C = 20 K
Number of mole of Aluminium =54/2 =2 mole
Q = n Cm △T
Q = 2 mole × 24 J mol –1 K–1 × 20 K
Q = 960 KJ
9.Enthalpy of fusion and enthalpy of vaporisation of sodium metal are 2.6 and
98.2kJ/mol respectively. What is the enthalpy of sublimation of sodium?
Answer: △subH = △fusH + △vapH = 2.6 + 98.2 = 100.8 kJ/mol
10. Given: N2(g) + 3H2(g) → 2NH3(g), ∆H0 = –92.4KJ.mol–1 What is the standard
enthalpy of formation of NH3(g).
Answer: The reaction for formation of one mole of ammonia is
½ N2(g) + 3/2 H2(g) → NH3(g)
The standard enthalpy of formation of ammonia is
∆fH0 NH3(g) = - 92.4= - 46.2 KJmol-1
2
Long Answer Type Questions:

Page | 60
1. For the reaction 2 A(g) + B(g) → 2D(g) ∆U 0 = –10.5 kJ and ∆S0 = –44.1 JK–1.
Calculate ∆G0 for the reaction, and predict whether the reaction may occur
spontaneously.
Ans. 2 A(g) + B(g) → 2D(g) ∆U 0 = –10.5 kJ and ∆S0 = – 44.1 J K–1
∆ng = 2 – (2 + 1) = –1 mole
∆Ho = ∆Uo + ∆ngRT
= (–10.5 kJ) – (–1) (8.314 × 10–3 kJ K–1 mol–1 ) (298 K)
= –10.5 kJ – 2.48 kJ
∆Ho= – 12.98 kJ
Substituting the values of ∆H o and ∆So in the expression of ∆Go
∆Go = ∆Ho– T∆So
= –12.98 kJ – (298 K) (–44.1 J K–1 )
= –12.98 kJ + 13.14 kJ
∆Go= + 0.16 kJ
Since ∆Go for the reaction is positive, the reaction will not occur spontaneous.

3. From the following bond energies:

H – H bond energy: 431.37 kJ mol –1
C = C bond energy: 606.10 kJ mol –1
C – C bond energy: 336.49 kJ mol –1
C – H bond energy: 410.50 kJ mol –1
Calculate Enthalpy for hydrogenation of Ethene involving the reaction,

Answer:

3. Enthalpies of formation of C2H4 (g) , CO2(g) and H2O(l) at 25°C and 1atm
pressure are 52, – 394 and – 286 kJ/mol respectively. Calculate Enthalpy of
combustion of Ethene.
Answer:
The reaction is C2H4 + 3O2 → 2CO2 + 2H2O.
Enthalpy of reaction △rH = △fH(products) - △fH(reactants) {△fH(O2) = 0}
= [2x(-394) + 2x(-286)] – [52 + 0]
= – 1412 kJ/ mol.

4. Define the following terms: i) Enthalpy (ii) Entropy (iii) Gibb’s Free energy
Ans: i) Enthalpy may be defined as the “ sum of the internal energy and pressure-
volume energy of the system” H = U + P.V
ii) Entropy is measure of randomness or disorder of a system.
i.e. Gas > Liquid > Solid.
(iii) Gibbs energy (G) It is a measure of capacity of a system to do maximum useful
work. G = H – T.S

Page | 61
 5. State the First Law of Thermodynamics. Established the mathematical formula of
First Law of Thermodynamics.
Ans: Energy can neither be created nor be destroyed; it may be converted from one
from into another.
Mathematically ∆U = q + w,
Let us consider a system whose internal energy is U , now if the system are supplied
1

q amount of heat, the energy of the system become U 1 + q.

Due to supply of heat the system do some work (expansion) i. e. w.
Hence, the final energy of the system U2 become U1 + q + w
Therefore, U2 = U1 + q + w
Or U2 – U1 = q + w
Or, ∆U = q + w
∆U = q + w
w = – P V∆ (work of expansion)

6. Calculate the standard enthalpy of formation of CH3OH. from the following

data: (i) CH3OH(l) + 3/2 02 (g) ———-> CO2 (g) + 2H20 (l); ∆rH– = – 726kj mol-1
(ii) C(s) + 02(g) —————>C02 (g); ∆cH– = -393 kj mol-1
(iii) H2(g) + 1/202(g) —————->H20 (l); ∆fH– = -286 kj mol-1
Answer:
The equation we aim at; C(s) + 2H2(g) + l/202(g) ———> CH3OH ∆fH– = ? …
Multiply eqn. (iii) by 2 and add to eqn. (ii)
C(s) + 2H2(g) + 2O2(g) ————->CO2(g) + 2H2O(l) ;
∆H = – (393 + 522) = – 965 kj mol…..(iv)
Subtract eqn. (iv) from eqn. (i)
CH3OH(l) + 3/2O2 (g) ————> CO2 (g) + 2H20 (l) ∆H = – 726 – (-965) kJ mol-1
C(s) + 2H2(g) + l/202(g) ———-> CH3OH(l) ; ∆fH = – 239 kj mol-1

8. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are

given in the following reactions. Calculate the Enthalpy of formation of CH 4 (g)
i) CH4 + 2O2 ══> CO2 + 2H2O ; ∆CH0 = − 890.3 kJ/mol
ii) C(s) + O2 ══> CO2 ; ∆CH0 = − 393.5 kJ/mol
iii) H2 + ½O2 ══> H2O ; ∆CH0 = − 285.8 kJ/mol
Answer:
Multiplying equation (iii) by 2 and add equation (ii) we get
C(s) + O2 + 2 H2 + O2 ══> CO2 + 2H2O ; ∆cH1 = − 393.5 + 2x − 285.8
Or, C(s) + 2O2 + 2 H2 ══> CO2 + 2H2O ∆cH2 = − 965.1 kJ/mol …….(iv)
Now, subtract eqn (iv) from (i)
CH4 + 2O2 − C(s) - 2O2 −2 H2 ══> CO2 + 2H2O ̶ CO2 − 2H2O
∆H0 = − 890.3 – ( − 965.1) kJ / mol
Or, CH4 + 2O2 − C(s) - 2O2 −2H2 = 0 ; ∆H0 = 74.8 kJ/mol
Or, − C (s) −2 H2 = − CH4 ; ∆H0 = 74.8 kJ/mol
Or C(s) +2H2 = CH4 ; ∆H0 = −74.8 kJ/mol

Equillibrium
KEY POINTS
• Equilibrium in physical processes:
• Solid-liquid equilibrium
• Liquid-vapour equilibrium
• Dynamic equilibrium
• law of chemical equilibrium and equilibrium constant

Page | 62
• homogeneous equilibrium
• Equilibrium constant in gaseous systems
• Factors affecting equilibrium-Le Chatteliers principle
• Ionic equilibrium-ionisation of acids and bases
• Strong and weak electrolytes
• Degree of ionisation
• Ionisation of polybasic acids
• Acid strength –pH
• Hydrolysis of salt, buffer solution
• Henderson equation
• Solubility product
• Common ion effect:
1.Equilibrium in chemical process – At equilibrium the concentration of each
reactant and product becomes constant.
At equilibrium the rate of forward reaction becomes equal to the rate of backward
reaction and hence dynamic in nature.
2. Equilibrium Constant – The product of molar concentration of products raised to
power their molar coefficient divided by molar concentration of the reactants raised
to the power their molar coefficient is constant at constant temperature and is called
equilibrium constant.
3. Relationship between Kp and Kc.
Kp = Kc (RT)∆n
4. Relationship between equilibrium constant and reaction quotient.
i) If Qc < Kc, the reaction proceeds in forward direction.
ii) If Qc > Kc, the reaction proceeds in backward direction.
iii) If Qc = K , the reaction is in equilibrium.
C

5. Different acid- base concepts.

According to Arrhenius an acid is a substance that gives H+ ion on dissociation and
base is a substance that gives OH- ion on dissociation.
6. According to Lowry Bronsted concept acid is an proton donor and base is an
proton acceptor.
7. According to Lewis concept a base is an electron donor and acid is an electron
acceptor.
8. Le Chatliers principle – When an equilibrium is subjected to change in
temperature pressure or concentration, the equilibrium will shift in that direction to
undo the effect.
9. (i) pH = - log [H+]
(ii) Kw = [H3O+] [OH-], where Kw = ionic product of water.

Page | 63
Ionic product of water increases
with increase in temperature
because dissociation of water into
H3O+ & OH- increases with increase
in the temperature.
11.Solubility product is the product
of molar concentration of ions in a
saturated solution but ionic product
is the product in any solution.
12.The common ion effect is an
effect that suppresses the ionization
of an electrolyte when another
electrolyte (which contains an ion
which is also present in the first
electrolyte, i.e. a common ion) is
added.

ASSERTION-REASON TYPE QUESTIONS

(A) Both Assertion and reason are true and reason is correct explanation of
assertion.
(B) Assertion and reason both are true but reason is not the correct explanation of
assertion.
(C) Assertion is true, reason is false.
(D) Assertion is false, reason is true

1. Assertion (A): For any chemical reaction at a particular temperature, the

equilibrium constant is fixed and is a characteristic property.
Reason (R) : Equilibrium constant is independent of temperature
Ans – C.
2. Assertion (A): Chemical equilibrium is in a state of dynamic equilibrium.
Reason (R): At equilibrium the forward and backward reactions are proceeding at the
same rate
and no macroscopic change is observed.
Ans – A.
3.Assertion (A): If Q (reaction quotient) < K (equilibrium constant) reaction moves in
c C

direction of reactants.
Reason (R): Reaction quotient is defined in the same way as equilibrium constant at
any stage of the reaction.
Ans – D.

Page | 64
4.Assertion: When Qc =Kc, reaction is at equilibrium.
Reason: At equilibrium, ΔG is 0.
Ans – B.
5. Assertion: An aqueous solution of ammonium acetate can act as buffer solution.
Reason: An aqueous solution of any pure liquid can act as buffer solution
Ans: C
SHORT QUESTIONS (1 Mark Each)
Q 1. Solubility of CO2 decreases with rise in temperature Why?
Ans - The dissolution of CO2 gas in water is exothermic in nature because the gas
contracts in volume. Thus, on increasing the temperature, the speed of the
endothermic or reverse process will increase according to Le Chatelier's principle.
Q 2. What is the relationship between Kp and Kc?
Ans - Kp = Kc (RT)∆n
Q 3. Define common ion effect.
Ans - The common ion effect is an effect that suppresses the ionization of an
electrolyte when another electrolyte (which contains an ion which is also present in
the first electrolyte, i.e. a common ion is added.
Q 4. Ammonia is termed as a base although it does not have OH– ions in it. Why?
Ans - It is a Lewis base, as it can donate its lone pair of electrons.
Q 5. How does a catalyst affect the equilibrium constant?
Ans - It helps to attain the equilibrium faster.
Q 6 Write conjugate acid and conjugate base of H 2O?
Ans - H3O+ & OH-.
Q 7. Name one natural buffer solution.
Ans - Blood is an example of natural buffer solution.
Q 8. What is the expression for Keq? for the reaction 2N20(g)+02(g)⇌4NO(g)
Ans - K = [NO]4 / [N2O]2 [O2]
Q 10. How is ionic product of a salt different from its solubility product?
Ans: -- Solubility product is the product of molar concentration of ions in a saturated
solution but ionic product is the product in any solution.
Q 11.When is a precipitate formed when solution of BaCl 2 and Na2SO4 are mixed?
Ans - A precipitate is formed when in the final solution after mixing, the ionic product
of [Ba2+][SO4 2-] > Ksp for BaSO4.
Q 12. Which of the following are Lewis acid ? H2O , BF3 , H+, NH4+
Ans - BF3, H+, NH4+.
Q 13. Explain why pure liquids and solids can be ignored while writing the value of
equilibrium constants.
Answer: This is because molar concentration of a pure solid or liquid is independent
of the amount present. Since density of pure liquid or solid is fixed and molar mass is
also fixed. Therefore, molar concentration are constant
Q 14. Give one example of everyday life in which there is gas solution equilibrium.
Ans - Soda - water bottle.
Q 15. What happens to the dissociation of PCl5 in a closed vessel if helium gas is
introduced into it at the same temperature?
Answer: Volume is constant, molar concentration of all reactant & products are
constant, so there is no change in the ttate of equilibrium.

Page | 65
MULTIPLE CHOICE QUESTIONS OF ONE MARK EACH
Q 1. According to Lewis concept, an acid is:
(a) proton donor (b) electron pair donor
(c) electron pair acceptor (d) proton acceptor
Ans: (c)
Q 2. Ostwald’s dilution law is applicable to:
(a) Strong electrolytes only (b) Weak electrolyte only
(c) Non-electrolytes (d) Strong and weak electrolytes
Ans: (b) Weak electrolyte only
Q 3. The pH of a solution of hydrochloric acid is 4. The molarity of the solution is:
(a) 4.0 (b) 0.4 (c) 0.0001 (d) 0.04
Ans: (c)

Q 4. The strong conjugate base is (a) NO3‐ (b) Cl‐ (c) SO42- (d) CH3COO ‐
Ans (d)
Q 5. Which of the following pairs constitutes a buffer?
(a) NaOH and HCl (b) HNO3 and NH4NO3 (c) HCl and KCl (d) HNO2 and NaNO2
Ans (d)
Q 6. Le Chatelier’s principle is applicable to:
(a) only homogeneous chemical reversible reactions
(b) only heterogeneous chemical reversible reactions
(c) only physical equilibria
(d) all systems, chemical or physical in equilibrium.
Ans (d)
Q 7. Which of the following is the weakest base?
(a) NaOH (b) Ca(OH)2 (c) NH4OH (d) KOH
Ans ( c )
Q 8. For the reversible reaction: N2 (g) + 3H2(g) ⇋ 2NH3 + Heat
The equilibrium shifts in forward direction
(a) by increasing the concentration of NH3(g) (b) by decreasing the pressure
(c) by decreasing the concentration of N2 and H2
(d) by increasing pressure and decreasing temperature
Ans (d)
Q 9. A base according to Bronsted concept is a substance which can:
(a) lose pair of electrons (b) donate protons
(c) gain a pair of electrons (d) accept protons
Ans (d)
Q 10. When NH4Cl is added to NH4OH solution the dissociation of ammonium
hydroxide is reduced. It is due to:
(a) common ion effect (b) hydrolysis (c) oxidation (d) reduction
Ans (a)

Q 11. In the chemical reaction A → B, the system will be known in equilibrium

when? (a) A completely changes to B (b) 50% of A changes to B
(c) The rate of change of A to B and B to A on both the sides are same
(d) Only 10% of A changes to B
Ans (c)
Q 12. The reaction A + B ⇌ C+ D+ heat has reached equilibrium. The reaction may
be made to proceed forward by:

Page | 66
(a) Adding more C (b) Adding more D (c) Decreasing the temperature
(d) Increasing the temperature
Ans (c)
Q 13. A reaction is A + B ⇌ C + D. Initially we start with equal concentration of A
and B. At equilibrium we find the moles of C are two times of A. What is the
equilibrium constant of the reaction?
a. 4 b. 2 c. 1 / 4 d. 1 / 2
Ans (a)
Q 14. An acidic buffer solution can be prepared by mixing solution of
(a) Ammonium acetate and acetic acid
(b) Ammonium chloride and hydrochloric acid
(c) Sulphuric acid and sodium sulphate
(d) Acetic acid and sulphuric acid
Ans (a)
Q 15. The expression for the solubility product of Al2(SO4)3 is
(a) K = [Al3+] (SO42-] (b) K = [Al3+]3[SO42-]2
(c) K =[Al3+]2 [SO42-]3 d. K= [Al3+]2[SO42-]2
Ans (c)

SHORT QUESTIONS (2 Marks each)

Q 1. Write down the characteristics of buffer solution
Ans. Characteristics of buffer solution
(i) It has a definite pH.
(ii) Its pH does not change on standing for long periods of time.
(iii) Its pH does not change on dilution.
(iv) Its pH is slightly changed by the addition of small quantity of an acid or base.

Q 2. What is the effect of temperature on ionic product of water and why?

Ans;- Ionic product of water increases with increase in temperature because
dissociation of water into H3O+ & OH- increases with increase in the temperature.

Q 3. What are pH & pOH values of a neutral solution at a temperature at which K w =

10 . What happens to the ionic product of water if some acid is added into water?
-13

Ans - pKw = pH + pOH. But pKw = 13

For neutral solution pH = pOH. Hence, pH = pOH = 6.5.
So, Ionic product of water remains unchanged.
Q 4. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10 M. What
-3

is its pH?
Ans - pH = - log [H+]
= - log (3.8 × 10-3) = 3 - 0.5798
= 2.42.
Q 5 How are Kp and Kc related with for the following equilibrium?
2A(g)+B(g) ⇄ C(g)
Ans. Δng = 1-3 = -2
Kp = Kc (RT) Δng
= Kc (RT)-2
Therefore, Kc= Kp (RT)2
Q 6 The species: H2O and HCO3–can act both as Bronsted acids and bases. For
each case give the corresponding conjugate acid and conjugate base.
Species Conjugate Conjugate
acid base

Page | 67
H2O H3O+ OH–
HCO3– H2CO3 CO32–

At 450 K, Kp = 2.0 ×1010 / bar for the given reaction at equilibrium.

2SO2(g) + O2(g) ⇌ 2SO3(g) What is Kc at this temperature?
Ans: Kp = Kc(RT)∆n
Kc = Kp/(RT)∆n = (2×10 10) /(.0831)(450) = 7.48×1011 L / mol

Short Answer type questions: 3 MARKS each.

Q 1. How can you predict the following stages of a reaction by comparing the value
of Kc and Qc?
(i) Net reaction proceeds in the forward direction.
(ii) Net reaction proceeds in the backward direction.
(iii) No net reaction occurs.
Ans: i) If Qc < Kc ii) If Qc > Kc iii) If Qc = KC.
Q 2. On the basis of Le-Chatelier principle explain how temperature and pressure
can be adjusted to increase the yield of ammonia in the following reaction.
N2(g) + 3H2(g) ⇌2NH3(g) ∆H = – 92.38 kJmol–1
What will be the effect of addition of argon to the above reaction mixture at constant
volume?
Ans - According to Le- Chateliers principle, decrease in temperature and increase in
pressure will increase the yeild of ammonia.
It is an exothermic process as ∆H is negative.

Effect of temperature According to Le-Chatelier’s principle, low temperature is

favourable for high yield of ammonia, but practically very low temperatures slow
down the reaction. So, optimum temperature 700 K is favourable in attainment of
equilibrium.
Effect of pressure Similarly, high pressure about 200 atm is favourable for high
yield of ammonia. On increasing pressure, reaction goes in the forward direction
because the number of moles decreases in the forward direction.
Addition of argon At constant volume addition of argon does not affect the
equilibrium because it does not change the partial pressures of the reactants or
products involved in the reaction and the equilibrium remains undisturbed.

Q 3. The value of Kc for the reaction 2A ⮀ B + C is 2 × 10–3. At a given time, the

composition of reaction mixture is [A] = [B] = [C] = 3 × 10–4 M. In which direction the
reaction will proceed?
Ans. For the reaction the reaction quotient Qc is given by, Qc = [B][C] / [A]2
as [A] = [B] = [C] = 3 × 10–4M
Qc = (3 ×10–4) (3 × 10–4) / (3 ×10–4)2 = 1
as Qc > Kc so the reaction will proceed in the reverse direction.
Q 4. What do you understand by buffer solutions? Classify them and give one
example of each.
Ans. A solution that resists change in its pH due to addition of small amount of strong
acid or small amount of strong base
They are classified in two categories: Acidic Buffer and Basic buffer solutions
(i) Acidic Buffer: CH3COOH + CH3COONa
(ii) Basic Buffer: NH4OH + NH4Cl

Page | 68
 REDOX REACTION
KEY POINTS :
The chemical changes that occur when electrons are transferred between reactants
are called oxidation – reduction reactions.

Oxidation number denotes the charge assigned to an atom of a compound or an

ion according to some arbitrary rules. It is equal to the number of electrons in the
valence shell of an atom, that are gained or lost while forming a bond in a
compound.
Calculation of oxidation number:
1.Oxidation number (O.N) of all the elements in their elemental form (in standard
state) is taken as zero. Ex: O. N. of elements in Cl2, F2, O2, P4, O3, Fe(s), H2, N2,
C(graphite) is zero.
2. Common O. N. of elements of group 1 is one. Common O.N. of elements of group
2 is two.
3. For ions composed of only one atom, the oxidation number is equal to the charge
on the ion.
4. The oxidation number of oxygen in most compounds is –2. While in peroxides
(e.g., H2O2, Na2O2), O.N. –1, in super oxides (e.g., KO2, RbO2) each oxygen atom
has O.N. of –(½).
5. In oxygen difluoride (OF2) and dioxygen difluoride (O2F2), the oxygen is assigned
an oxidation number of +2 and +1,respectively.
6. The oxidation number of hydrogen is +1 but in metal hydride its O.N. is–1.
7. In all its compounds, fluorine has an O.N. of –1.
8. The algebraic sum of the O.N. of all the atoms in a compound must be zero.
9. In polyatomic ion, the algebraic sum of all the O.N. of atoms of the ion must equal
the charge on the ion.
Oxidising agent: A reagent which can increase the oxidation number of an element in
a given substance. These reagents are also called as oxidants. It itself gets
reduced.
Reducing agent: A reagent which lowers the oxidation number of an element in a
given substance. These reagents are also called as reductants. It teself gets
oxidised.
CuO + C → Cu + CO, Oxidant = CuO & reductant is C.
Zn + Cu2+ → Zn2+ + Cu Oxidant = Cu2+ & reductant is Zn.
Types of redox reactions
1. Combination reactions: A reaction in which two atoms or molecules
combine together form molecule or compound.
For e.g., C (s) + O2 (g) → CO2 (g)
2. Decomposition reactions: They are the opposite reactions of combination
reactions. or breakdown of a compound into two or more components at least one of
which must be in the elemental state.
For e.g., 2H2O(l) → 2H2(g) + O2(g)

Page | 69
3. Displacement reactions: In a displacement reaction, an ion (or an atom) in a
compound is replaced by an ion (or an atom) of another element.
For e.g. (i) Metal displacement: CuSO4(aq.) + Zn(s) → Cu(s) + ZnSO4(aq.)
(ii) Non-metal displacement: 2Na(s) + 2H2O(l) → 2NaOH(aq.) + H2(g)
4. Disproportionation reactions: In a disproportionation reaction an element in
one oxidation state is simultaneously oxidized and reduced in the same reaction.
For e.g., P4(s) + 3OH- (aq.) +3H2O(l) → PH3(g) + 3H2PO2- (aq.)
2 H2O2 → H2O + O2.

SECTION A: ASSERION AND REASON (1 MARK)

In the following questions a statement of assertion (A) followed by a statement of
reason (R) is given. Choose the correct option out of the choices given below each
question.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.

Assertion (A) : Among halogens fluorine is the best oxidant.

Reason (R) : Fluorine is the most electronegative atom.
Ans: (ii) Both A and R are true but R is not the correct explanation of A.

2. Assertion (A): In the reaction between potassium permanganate and potassium

iodide, permanganate ions act as oxidising agent.
Reason (R) : Oxidation state of manganese changes from +2 to +7 during the
reaction.
Ans: (iii) A is true but R is false.
3. Assertion (A) : The decomposition of hydrogen peroxide to form water and oxygen
is an example of disproportionation reaction.
Reason (R) : The oxygen of peroxide is in –1 oxidation state and it is converted to
zero oxidation state in O2 and –2 oxidation state in H2O.
Ans: (i) Both A and R are true and R is the correct explanation of A.
4. Assertion (A) : Redox couple is the combination of oxidised and reduced form of a
substance involved in an oxidation or reduction half-cell.
Reason (R) : In the representation E 0 E
Fe
+3 , Fe / Fe and Cu / Cu are redox
/ Fe
+2 0
Cu2+ / Cu
+3 2+

couples.
Ans: (ii) Both A and R are true but R is not the correct explanation of A.
5. Assertion (A): HgCl and SnCl cannot stay together.
2 2

Reason (R): HgCl is an oxidising agent but SnCl is reducing agent.

2 2

Ans: (i) Both A and R are true and R is the correct explanation of A.

SECTION B: VERY SHORT ANSWER (1 MARK)

1. What is Oxidation in terms of electron transfer.
Ans: Loss of electrons.
2. What is Reduction in terms of electron transfer.
Ans: Gain of electrons.
3. What is Oxidant in terms of electron.
Ans: Substance which can accept one or more electron.
4. What is Reductant in terms of electron.

Page | 70
Ans. Substance which can donate one or more electron.
5. Give two examples of Oxidant.
Ans: KMnO4 and K2Cr2O7.
6. Give two examples of Reductant.
Ans. HCl and SnCl2.
7. Calculate Oxidation number of the metal in the compound BaO . 2

Ans: +2.
8. Give example of an element whose Oxidation number is zero.
Ans. Cl2 (Any Pure Element)
9. Give example of a compound of Nitrogen where its Oxidation number is negative.
Ans: NH3
10. Give example of a compound of Carbon where its Oxidation number is zero.
Ans: CH2Cl2 or C6H12O6
11. Write the Disproportionation Reaction of ClO -.
Ans: ClO– → 2Cl- + ClO3– (Any other example)
12. Iron undergo Oxidation more readily than Copper. Why?
Ans: Electrode potential of Iron is lower than that of copper.
13. What is the Oxidation number of Chlorine in Bleaching Powder.
Ans: Zero.
14. Identify the substances reduced in the following reaction:
Fe2O3(s) + 3CO(g) → 2Fe (s) + 3CO2 (g)
Ans: Substance reduced is Fe2O3.
15. Give an example of disproportionation reaction.
Ans: Cl2 + 2NaOH → NaCl + NaClO + H2O
0 –1 +1
Chlorine is getting oxidised as well as reduced.
∴ It is disproportionation reaction.
16. What is the oxidation number of Mn in KMnO 4
Ans: Let oxidation number of Mn be x.
∴ +1 + x – 8 = 0 ⇒ x = +7.
17. Which is best reducing agent and best oxidising agent?
Ans: (i) Li is best reducing agent due to lowest standard reduction potential.
(ii) F2 is best oxidising agent due to highest standard reduction potential.
18. Can we store copper sulphate in an iron vessel? Why?
Ans: No, because iron is more reactive than copper, and therefore, it will displace
copper from its salt solution.
19. An electrochemical cell consist of two electrodes i.e. anode and cathode. What
is direction of flow of electrons in the cell?
Ans: Electrons flow from anode to cathode because electron density is more at
anode due to loss of electron and less at cathode due to gain of electrons.
20. A freshly cut apple is almost white but turns brown after some time, why?
Ans: Apple contains Fe2+ which get oxidised to Fe3+ which is brown in colours. Apple
turns brown due to oxidation of Fe2+ to Fe3+
21. Represent a galvanic cell in electrode and ions contain Cu electrode dipped in 1
molar solution of copper sulphate and silver electrode dipped in 1 molar solution of
silver nitrate.

Page | 71
 Given = ECu2+/Cu(s)= + 0.34 V, EAg+/Ag(s) = + 0.80 V
Ans: Cu(s) | Cu2+(1 M) || Ag+(1 M) | Ag(s)

SECTION C: MULTIPLE CHOICE QUESTION (1 MARK)

1. Which of the following is not an example of redox reaction?
(i) CuO + H2 → Cu + H2O (ii) Fe2O3 + 3CO → 2Fe + 3CO2
(iii) 2K + F2 → 2KF (iv) BaCl2 + H2SO4 → BaSO4 + 2HCl
Ans: (iv) BaCl2 + H2SO4 → BaSO4 + 2HCl
2. The more positive the value of E0, the greater is the tendency of the species to get
reduced. Using the standard electrode potential of redox couples given below find
out which of the following is the strongest oxidising agent. The E0 values:
Fe3+/Fe2+ = + 0.77V; I2(s)/I– = +0.54V; Cu2+/Cu = +0.34V; Ag+/Ag +0.80V
(i) Fe3+ (ii) I2(s) (iii) Cu2+ (iv) Ag+
Ans: (iv) Ag+
3. E0 values of some redox couples are given below. On the basis of these values
choose the correct option. The E0 values: Br2/Br– = + 1.90V;
Ag+ /Ag(s) = +0.80V; Cu2+/Cu(s) = + 0.34V; I2(s)/I– = + 0.54V
(i) Cu will reduce Br– (ii) Cu will reduce Ag
–
(iii) Cu will reduce I (iv) Cu will reduce Br2
Ans: (iv) Cu will reduce Br2
4. Using the standard electrode potential, find out the pair between which redox
reaction is not feasible. E0 values: Fe3+/Fe2+ = + 0.77; I2/I– = + 0.54;
Cu2+/Cu = + 0.34; Ag+/Ag = + 0.80 V
(i) Fe3+ and I– (ii) Ag+ and Cu (iii) Fe3+ and Cu (iv) Ag and Fe3+
3+
Ans: (iv) Ag and Fe
5. The oxidation number of an element in a compound is evaluated on the basis of
certain rules. Which of the following rules is not correct in this respect?
(i) The oxidation number of hydrogen is always +1.
(ii) The algebraic sum of all the oxidation numbers in a compound is zero.
(iii) An element in the free or the uncombine state bears oxidation number zero.
(iv) In all its compounds, the oxidation number of fluorine is –1.
Ans: (i) The oxidation number of hydrogen is always +1.
6. Which of the following arrangements represent increasing oxidation number of the
central atom?
(i) CrO2– , ClO3– , CrO42–, MnO4–
(ii) ClO3– , CrO42– , MnO4– , CrO2–
(iii) CrO2– , ClO3– , MnO4– , CrO42–
(iv) CrO42–, MnO4– , CrO2– , ClO3–
Ans: (i) CrO2– , ClO3– , CrO42–, MnO4–
7. The largest oxidation number exhibited by an element depends on its outer
electronic configuration. With which of the following outer electronic configurations
the element will exhibit largest oxidation number?
(i) 3d14s2 (ii) 3d34s2 (iii) 3d54s1 (iv) 3d54s2
Ans: (iv) 3d54s2
8. Identify disproportionation reaction
(i) CH4 + 2O2 → CO2 + 2H2O (ii) CH4 + 4Cl2 → CCl4 + 4HCl
(iii) 2F2 + 2OH– → 2F– + OF2 + H2O (iv) 2NO2 + 2OH– → NO2– + NO3– + H2O
Ans: (iv) 2NO2 + 2OH– → NO2– + NO3– + H2O

9. The oxidation number of Cl in Cl 2O7 is

Page | 72
 (a) + 7 (b) +5 (c) +3 (d) – 7
Ans. (a)
10. The oxidation number of Mn is maximum in
(a) MnO2 (b) K2MnO4 (c) Mn3O4 (d) KMnO4
Ans. (d)

SECTION D: SHORT ANSWER (2 MARKS)

1. Calculate the oxidation amount of phosphorus in the following species.
(a) HPO32- and (b) PO43-
Ans: (a) +3 (b) +5.
2. Calculate the oxidation number of each sulphur atom in the following compounds:
(a) Na2S2O3 (b) Na2S4O6 (c) Na2SO3 (d) Na2SO4
Ans: (a) +2 (b) +5,0,0, +5 (c) +4 (d) +6
3. MnO4-2 undergoes disproportionation reaction in acidic medium but MnO 4– does
not. Give reason.
Ans: In MnO4–, Mn is in the highest oxidation state i.e. +7. Therefore, it does not
undergo disproportionation. MnO4–2 undergoes disproportionation as follows:
3MnO4–2 + 4H+ → 2MnO4– + MnO2 + 2H2O
4. Nitric acid is an oxidising agent and reacts with PbO but it does not react with
PbO2. Explain why?
Ans: PbO is a basic oxide and simple acid base reaction takes place between PbO
and HNO3. On the other hand, in PbO2 lead is in +4 oxidation state and cannot be
oxidised further. Therefore, no reaction takes place. Thus, PbO2 is passive, only
PbO reacts with HNO3.
2PbO + 4HNO3 → 2Pb (NO3)2 + 2H2O (Acid base reaction)
5. Using the standard electrode potentials given in the table, predict if the reaction
between the following is feasible: (i) Fe3+ (aq) and I– (aq) (ii) Ag+ (aq) and Cu (s)
(E°Fe3+/Fe2+ = +0.77V, E°I2/I-= +0.54V) (E°Ag+/Ag = +0.80V, E°Cu2-/Cu= +0.34V)

Ans. (i) (ii)

SECTION E: LONG ANSWER (3 MARKS)

1. Explain redox reactions on the basis of electron transfer. Give suitable examples.
Ans: 2Fe+3 + 2e → 2Fe+2 (Red)…………….(I)
Sn +2
→ Sn+4 + 2e (Ox)………………..(ii)
2Fe+3 + Sn+2 = 2Fe+2 + Sn+4
In the yellow colour Ferric Chloride solution white colour Stanous Chloride solution
drop wise added and the mixture was stirred continuously with a glass rod. The
yellow colour solution turns to light green colour because the above mentioned

Page | 73
chemical reaction occurs. Here Ferric Chloride is reduced to Ferrous Chloride by the
gain of electron and Stanous Chloride is oxidized to Stanic Chloride by the loss of
electron. So in the above process electron transfer is occurring.
2. Why does fluorine not show disproportionation reaction?
Ans: Disproportionation is a redox reaction in which one compound of intermediate
oxidation state converts to two compounds, one of higher and one of lower oxidation
states. So, to occur such type of redox reaction, the element should exist in at least
three oxidation states. So that element present in the intermediate state and it can
change to both higher and lower oxidation state during disproportionation reaction.
Fluorine is the most electronegative element and a strong oxidizing agent and is the
smallest in size of all the halogens. It does not show a positive oxidation state
(shows only −1 oxidation state) and hence, does not undergo disproportionation
reaction.
3. Write the balanced chemical equation corresponding to the reaction.
Cr2O7 2– (aq) + Fe2+ (aq) → Cr3+ (aq) + Fe3+ (aq)
Ans. Step 1:
Cr2O7 2– (aq) + Fe2+ (aq) Cr3+ (aq) + Fe3+ (aq)
+6 +2 +3 +3
Step 2: The oxidation state of Cr decreases by 3 per chromium atom, total
decrease is 6 for two chromium atoms, oxidation state of Fe changes from +2 to +3,
i.e., increases by 1, therefore, to equalize the increase and decrease, we
multiply Fe2+ by 6 and Cr2O72– by 1.
Step 3: Balancing ‘Cr’ and Fe on both sides:
Cr2O72– (aq) + 6 Fe2+ (aq) → 2Cr3+ (aq) + 6Fe3+ (aq)
Step 4: To balance oxygen, we add 7 molecules of H 2O on RHS.
Cr2O72– (aq) + 6 Fe2+ (aq) → 2Cr3+ (aq) + 6Fe3+ (aq) + 7H2O (l)
Step 5: To balance hydrogen, we add 14H+ on LHS and we get balanced
equation:
Cr2O72– (aq) + 6 Fe2+ (aq) + 14H+ (aq) = 2Cr3+ (aq) + 6Fe3+ (aq) + 7H2O (l)

4. Permanganate ion reacts with bromide ion in basic medium to give

manganese dioxide and bromate ion. Write the balanced chemical equation for the
reaction. 3
Ans: Step 1:
MnO4– (aq) + Br–(aq) → MnO2(aq) + BrO3– (aq)
Step 2: MnO4 is oxidant because its oxidation state is decreasing. Br – is
–

reductant because its oxidation state is increasing.

Step 3: Oxidation state of Mn is decreasing by 3. Oxidation state of Br is
increasing by 6.
To equalise increase and decrease, multiply MnO 4– by 2 and Br by 1 we get:
2MnO4– (aq) + Br–(aq) → 2MnO2(s) + BrO3– (aq)
Step 4: Now for balancing oxygen, we add 1 molecule of H 2O on RHS.
2MnO4– (aq) + Br–(aq) → 2MnO2(s) + BrO3– (aq) + H2O (l)
Step 5: As the reaction is taking place in basic medium to balance hydrogen,
add 2H2O molecule on LHS and 2OH– on RHS
2MnO4– (aq) + 2H2O(l) + Br–(aq) → 2MnO2(s) + BrO3– (aq) + H2O (l)
–
+ 2OH (aq)
It can be seen 1 molecule of H2O gets cancelled on both sides, we get:
2MnO4– (aq) + H2O(l) + Br–(aq) = 2MnO2(s) + BrO3– (aq) + 2OH– (aq)
is a balanced equation.

Page | 74
1. Consider the following table of standard reduction potentials:

Reaction E° (volts)
A3+ + 2e– → A +
1.36
B2+ + 2e– → B 0.72
C2+ + 2e– → C –0.28
D+ + e– → D –1.42

(i) Which substance is (a) strongest oxidizing agent? (b) Strongest reducing agent
(ii) Which substances can be oxidized by B2+?
(iii) Which substance can be reduced by C?
Ans. (i) A3+ (ii) (D) (iii) C & D (iv) A3+ & B2+

ORGANIC CHEMISTRY - SOME BASIC PRINCIPLES AND TECHNIQUES

Organic Chemistry Some basic principles and Techniques
Organic Chemistry – Organic chemistry is the study of carbon compounds that
always contain carbon and it is limited to other elements. Compounds obtained from
plants and animals were termed organic to indicate their ultimate source was a living
organism.
Organic chemistry is one of the most important disciplines of science which deals
with the study of carbon compounds especially hydrocarbons and their derivatives.
General Introduction:
1. Catenation – Catenation can be defined as the self-linking of atoms of an element to
form chains and rings. This definition can be extended to include the formation of
layers (two-dimensional catenation) and space lattices (three-dimensional
catenation).
2. Tetravalency and small size – Carbon exhibits’ tetravalency. The tetravalency of
carbon can be satisfied by forming bonds with carbon, hydrogen or other atoms. The
carbon atom has 4 electrons in its valence shell. In order to account tetravalency it is
believed during the process of bond formation which is energy-releasing process the
two electrons in the 2s orbital get unpaired and out of them, one is promoted to
empty orbital.
Structural Representations of Organic Compounds
1. Complete Structural Formula – Full structural equations show all the atoms in a
molecule, the types of bonds that bind them, and how they are interconnected.
2. Condensed Structural Formula – The Condensed structural formula is used to
save space, structural formulas are conveniently abbreviated as condensed
structural formulas.
3. Bond Line Structural Formula – A bond-line structure is a less cluttered drawing
than a condensed structural formula. However, to understand the simplified bond-
line structure, the reader has to mentally add many more features to comprehend the
overall structure.
Classification of Organic Compounds
1. Acyclic or Open Chain Compounds & Alicyclic or Closed Chain or Ring
Compounds – Organic compounds are classified as open-chain compounds and
closed chain compounds in terms of the carbon chain. Also termed as Organic

Page | 75
 Compounds Acyclic or Open Chain or Aliphatic Compounds, Cyclic or Closed Chain
or Ring Compounds.
2. Aromatic Compounds – Plants and micro-organisms have an exclusive route to
benzene-ring compounds. The great majority of aromatic compounds in nature,
therefore, are produced by plants and microorganisms, and animals are dependent
upon plants for many aromatic compounds either directly or Indirectly.
3. Heterocyclic Aromatic Compounds – In the twentieth century it is witnessed that
the first inorganic heteroaromatic compound produced in the laboratory. Some of
these heterocyclic aromatic compounds are very important in biochemical
processes, drugs, and agrochemicals.
Nomenclature: IUPAC Rules – Nomenclature of Organic Compounds follows
recommendations of IUPAC in naming organic compounds, carbocations, etc. The
International Union of Pure and Applied Chemistry (IUPAC) formulated rules for
naming organic compounds.
Isomerism
1. Structural Isomerism – Structural Isomerism arises due to different arrangement of
atoms within the molecule. Two molecules are structural isomers if they share the
same molecular formula.
2. Stereoisomerism – Stereo-isomers are isomeric molecules having the same
molecular formula and the same sequence of bonded atoms, but are only different in
the 3D orientations of their atoms in space. Stereoisomerism may be of two types
viz. geometrical (or cis-trans) isomerism and optical (or d-l or mirror-image)
isomerism.
Fundamental Concepts
1. The Shapes of Carbon Compounds – Tetravalent carbon atom is the building
block of structural organic chemistry. The four hydrogen atoms, with four carbon
atoms, form a structure known as a tetrahedron.
2. Functional Groups – Functional groups were introduced as a useful method for
organizing this vast number of compounds because chemical reactions occur at the
functional group and compounds with the same functional group undergo similar
reactions. Example: C2H60 has isomers: C2H5-OH & CH3 – O – CH3.
3. Homologous Series – A series of organic compounds in which every succeeding
member differs from the previous one called Homologous Series.
These differ by -CH2-
Methods of Purification
1. Simple crystallization – Crystallization is one of the most effective purification
techniques for solids. Simple crystallization involves the selection of the solvent and
preparation of the solution.
2. Fractional crystallization – Fractional crystallization is used for the purification of a
single substance contaminated with small quantities of impurities.
3. Sublimation – Sublimation is an excellent method for purifying relatively volatile
organic solids on scales ranging from a few milligrams to tens of grams.

Page | 76
4. Simple distillation – Simple distillation is the process of converting a liquid into its
vapour, transferring the vapour to another place, and recovering the liquid by
condensing the vapour.
5. Fractional distillation – Fractional distillation is the separation procedure of a
mixture into sections or fractions of its material. By heating them to a temperature at
which one or more parts of the mixture vaporize, chemical compounds are isolated.
6. Steam distillation – A steam distillation is simply a distillation in which steam is
involved as a process component. Steam distillation and organic solvent extraction
have both been widely used to extract compounds from spices.
(i) Ortho nitro phenol is separated from its mixture with p-nitro phenol by steam dist.
(ii) Aniline is separated from aniline-water mixture by steam distillation.
7. Azeotropic distillation – Azeotropic distillation is accomplished by adding to the
liquid phase, a volatile third component that changes the volatility of one of the two
components more than the other so that the components are separated by
distillation.
8. Distillation under reduces reduced pressure: Glycerol is separated from spent-lye
in soap industry by this technique.
9. Chromatography – The purpose of preparative chromatography is to separate the
components of a mixture for further use (and is thus a form of purification).
Quantitative Analysis:
The elementary organic analysis is also considered, along with the development of
the theory of analytical chemistry. Quantitative analysis refers to assessing how
much is present in a sample of a given item. The quantity of any or all components of
a sample can be expressed in terms of size, concentration, or relative abundance.
Nitrogen in organic compound is detected by Lassaigne’s test. The compound
formed id Prussian blue colouration; Fe4[Fe(CN)6]3.
ASSERTION AND REASONING QUESTIONS (Each question carry 1 mark)
In the following questions a statement of assertion (A) followed by a statement of
reason (R) is given. Choose the correct option out of the choices given below each
question.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
1. ASSERTION (A): benzene shows resonating structures
REASON (R): Benzene has delocalised pi electrons
Ans. (a) Both A and R are true and R is the correct explanation of A.
2. ASSERTION(A): H2O is a nucleophile
REASON (R): H2O is an electron rich species
Ans. (a) Both A and R are true and R is the correct explanation of A.
3. ASSERTION(A): CH3 CH2+ is less stable than (CH3)2CH+
REASON (R): CH3 CH2+ has more hyper conjugative forms than (CH3)2CH+
Ans. (c) A is true but R is false.
4. ASSERTION(A): both the carbon atoms of ethene are sp 3 hybridised
REASON (R): it has C=C bond
Ans. (d) A is false but R is true.

Page | 77
5. ASSERTION(A) carbon shows catenation property
REASON (R): C-C single bond is very weak
Ans. (c) A is true but R is false
VERY SHORT TYPE QUESTIONS (1 mark for each question)
1. Identify the nucleophiles from the following:
CN- , NO2+, OH-, Cl+, H2O
Ans. CN-, OH-, H2O
2. Write the IUPAC name of the following: CH3-CH2-COOH
Ans. Propanoic acid.
3.Define homolytic cleavage of covalent bond?
Ans. Breaking of covalent bond takes place in such a way, one of the electrons of
the shared pair in a covalent bond goes with each of the bonded atoms.
4.Which bond is more polar in the following pair of molecules?
H3C-NH2, H3C-OH
Ans. C-O bond is more polar than C-N bond as O is more electronegative than N.
5.Give two examples of organic compounds where resonance effect is possible.
Ans. a. Benzene b. chlorobenzene
6. Name the metal which is used in Lassaigne extract.
Ans. Sodium metal.
7.Name the chemicals which are used to detect nitrogen in organic compounds.
Ans. Freshly prepared ferrous sulphate and concentrated sulphuric acid.
8. How many sigma and pi bonds in the following molecule:
CH2=CHCN.
Ans. Sigma bonds = 6 and pi bonds = 3.
9. Name a suitable technique of separation of the components from a mixture of
calcium sulphate and camphor.
Ans. Sublimation process.
10. Arrange the following in order of decreasing priority in IUPAC system
-CONH2, -COOH, -SO3H, -CHO
Ans. -SO3H> -COOH> -CONH2> -CHO
11. Draw the structure of Pent-4-en-2-ol.
Ans. CH3-CH(OH) CH2-CH=CH2
12. Name the first synthesis organic compound.
Ans. Urea.
13.Write the IUPAC name of chloroform.
Ans. Trichloromethane
14.Name two methods for the estimation of nitrogen in organic compounds.
Ans. Dumas method and Kjeldahl method.
15.Identify the functional group present in the following compounds:
a. CH3-CO-C2H5 b. CH3-O-C2H5
Ans. a. ketonic functional group b. ether functional group

MCQ TYPE QUESTIONS (EACH CARRY 1 MARK)

1.The IUPAC name of CH3COCH(CH3)2 is
(a) 3-Methyl-2-butanone (b) Isopropyl methyl ketone
(c) 2-Methyl-3-butanone (d) 4-Methyl isopropyl ketone

Page | 78
2. What is the state of hybridization of carbon in carbanion
(a) sp (b) sp2 (c) sp3 (d) sp3d
3. Which of the following is the correct order regarding the –I effect of the
substituents?
(a) –NR2 < -OR < -F (b) –NR2 > -OR > -F (c) –NR2 < -OR > -F (d) –NR2 > -OR < -F
4.Which of the following carbocations has the least stability
(a) Methyl (b) Ethyl (c) tert-butyl (d) Isopropyl.
5.Which of the following is not an electrophile?
(a) :CH2 (b) + CH3 (c) SO3 (d) :SH-
6.Heterolytic cleavage of a C-X bond may result in the formation of
(a)a free radical (b) a carbocation (c) a carbene (d)a carbanion
7. Of the five isomeric hexanes, the isomer that can given two monochlorinated
compounds is (a) 2-methyl pentane (b) 2,2-dimethyl butane
(c) 2,3-Dimethyl butane (d) n-hexane.
8. Consider the following compounds: A. chloroethene B.benzene
C. 1,3-butadiene D.1,3,5-hexatriene.
All the carbon atoms are sp2-hybridised in
(a) A,C,D only (b)A,B only (c) B,C,D only (d)A,B,C and D.
9. The Lassaigne’s extract is boiled with dil.HNO3 before testing for halogens
because
(a) silver halides are soluble in HNO3
(b) Na2S and NaCN are decomposed by HNO3
(c) Ag2S is soluble in HNO3 (d) AgCN is soluble in HNO3.
10. Which of the following carboction is most stable?
(a)(CH3)3CC+H2 (b) (CH3)3C+ (c) CH3CH2C+H2 (d) CH3C+HCH2CH3.
Answer. 1.(a) 2.(c) 3.(a) 4.(a) 5.(d) 6.(a) 7.(c) 8.(d) 9.(b) 10.b

SHORT ANSWER TYPE QUESTIONS (2 MARKS FOR EACH QUESTION)

1. Explain, how is the electronegativity of carbon atoms related to their state of

hybridisation in an organic compound?
Ans. Since s-electrons are more firmly attached by the nucleus than p-electrons,
therefore, electronegativity increases as the s-character of the hybridised orbital
increases, i.e., in the order sp3 < sp2 < sp.
2. Identify the most stable species in the following set of ions giving reasons
CH3+, CH2+Br, CH+Br2, C+Br3
Ans. CH3+ is the most stable species because replacing H atom by Br (-Inductive
effect) increases +ve charge on carbon atom and destabilised the species.
3. Write structural formulae for compounds named as-
(i) 1-Chloroheptane
(ii) 5-Bromoheptanoic acid
Ans. (i) CH3−CH2−CH2−CH2−CH2−CH2−CH2Cl
(ii) CH3−CH2−CH(Br)−CH2−CH2−CH2−COOH
4. Arrange the following in order of increasing stability
(CH3)3C+,(CH3)2C+H, CH3+, CH3C+H2
Ans. CH3+< CH3C+H2< (CH3)2C+H < (CH3)3C+
5. Why does NH3 act as a nucleophile?

Page | 79
Ans. N in NH3 has loan pair electron. So, it is electron rich species and can donate
these electrons.
LONG ANSWER QUESTIONS (EACH CARRY 3 MARKS)
1. Two liquids (A) and (B) can be separated by the method of fractional distillation.
The boiling point of a liquid (A) is less than the boiling point of a liquid (B). Which of
the liquids do you expect to come out first in the distillate? Explain.
Ans. Fractional distillation is used to separate the components of a mixture if their
boiling points differ by 20∘ or less. In this method, a fractionating column is used in
between the flask and the condenser. The fractionating column aims to provide
hurdles for the ascending vapours and provide ample surface area for condensing
the high boiling liquid. Consequently, the low boiling liquid (A) vapours will move up
while those of the high boiling liquid will condense and fall back into the flask. As a
result of this fractionation, liquid A with a lower boiling point will distil first, and liquid
B with a higher boiling point afterwards.
2. Give three points of differences between inductive effect and resonance effect.
Ans. 1. Inductive Effect involves the displacement of electrons in saturated
compounds. Resonance Effect involves displacing % electrons or lone pairs of
electrons in unsaturated and conjugated compounds.
2. In the inductive Effect, a slight displacement of σ electrons and partial +ve or-ve
charge develops. In the resonance effect, there is a complete transfer of π electrons,
and as a result, a complete +ve or -ve charge develops.
3. The Inductive Effect can move only up to 3 to 4 carbons.In the resonance effect,
the movement of electrons all along the length of the conjugated system takes place.

HYDROCARBON
Minimum Level of Learning MLL):
Naming hydrocarbons according to IUPAC system of nomenclature;
Structures of isomers of alkanes, alkenes, alkynes and aromatic hydrocarbons,
Various methods of preparation of hydrocarbons;
Distinguish between alkanes, alkenes, alkynes and aromatic hydrocarbons on the
basis of physical and chemical properties;
Draw and differentiate between various conformations of ethane;
The formation of the addition products of unsymmetrical alkenes and alkynes on the
basis of electronic mechanism;
Structure of benzene, explain aromaticity and mechanism of electrophilic substitution
reactions of benzene, the directive influence of substituents in monosubstituted
benzene ring, carcinogenicity and toxicity.

ASSERTION AND REASON TYPE QUESTION (1M)

Choose the correct option from the following.
(a) Assertion and reason both are correct statements and reason is
correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct

Page | 80
explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

Q.1) Assertion: Boiling point of alkanes increases with increase in molecular weight.
Reason: van der Waal’s forces increase with increase in molecular weight.
Ans:(a) Assertion and reason both are correct statements and reason is
correct explanation for assertion.
Q.2) Assertion: Methane cannot be obtained by Wurtz reaction.
Reason: Wurtz reaction leads to the formation of symmetrical alkane having
an even number of carbon atoms.
Ans: a) Assertion and reason both are correct statements and reason is
correct explanation for assertion.
Q.3) Assertion: Trans-pent-2-ene is polar but trans-but-2-ene is non-polar.
Reason: The polarity of cis-isomer is more than trans which are either
non-polar or less polar.
Ans: (b) Assertion and reason both are correct statements but reason is not
correct explanation for assertion.
Q.4) Assertion: All the hydrogen atoms in CH2 =C= CH2 are attached to
sp2 hybridised carbon atom.
Reason: All the carbon atoms in its are sp 2 hybridized.
Ans: (c) Assertion is correct statement but reason is wrong statement.
Q.5) Assertion: Acetylene is acidic in nature.
Reason: Acetylene is sp hybridised.
Ans: (b) Assertion and reason both are correct statements but reason is not
correct explanation for assertion

MULTIPLE CHOICE QUESTION (1M)

Q.1) In the compound CH2 = CH – CH2 – CH2 CH3, the hybridisation of
C2 – C3 bond is of the type
(a) sp2 – sp² (b) sp³ – sp³ (c) sp – sp³ (d) sp² – sp³
Ans: (a) sp2 – sp³
Q.2) The number and type of bonds in benzene are
(a) twelve sigma (σ) and three pi (π) bonds (b) six sigma (σ) and two pi (π) bonds
(c) three sigma (σ) and one and a half pi (π) bonds
(d) One sigma (σ) bond

Page | 81
Ans: (a) twelve sigma (σ) and three pi (π) bonds.

Q.3) The Number of structural isomers for C 5H12 is

(a) 3 (b) 4 (c) 5 (d) 6
Ans: (a) 3
Q.6 Homologous series those __________
a) differ from each other by CH2 group (b) aliphatic compounds
(c) saturated compounds (d) hydrocarbon.
Ans: (a) differ from each other by CH2 group

Q.7) Sodium salt of carboxylic acids on heating with soda lime gives alkane
containing ___________than the carboxylic acid
(a) one carbon less (b) one carbon more (c) two carbon less (d) two carbon more
Ans: (a) one carbon less.
Q.8) Which of the following compounds reacts with sodium to liberate hydrogen gas?
(a) Ethane (b) Propene (c) Acetylene (d) Benzene
Ans: (c) Acetylene.
Q.9) What is a general formula of alkynes?
(a) CnH2n-2 (b) CnHn-2 (c) CnH2n (d) CnH2n+2
Ans: (b) CnHn-2
Q.10) An organic compound, ethane and ethene can be distinguish by
(a) Bromine water (b) Tollen reagent
(c) Fehling solution (d) Ammonical silver nitrate solution.
Ans: (a) Bromine water

Page | 82
Very short answer (1Mark Each)
Q.1) What effect does branching of an alkane chain on its boiling point?
Ans: Branching of carbon atom chain decreases the boiling point of alkane.
Q.2) Arrange the isomers of pentane in increasing order of their boiling points.
2-methylbutane, 2, 2-Dimethylpropane, pentane
Ans. 2, 2-Dimethylpropane < 2-methylbutane < pentane.
Q.3) What are conformations?
Ans: Conformations are spatial arrangements of atoms which are obtained by
rotation around single bond.
Q.4) what are the two-extreme type of conformation in Ethane?
Ans: 1) Staggered 2) Eclipsed.
Q.5) Why alkyne do not show geometrical isomerism?
Ans: Alkyne have linear shape hence do not show geometrical isomerism.
Q.6) How would you distinguish between but-1-yne and but-2-yne.
Ans: On a reaction with Sodium metal , but-1-yne liberate Hydrogen gas where
as but-2-yne does not reacts.
Q.7) How will you distinguish between propene and propyne?
Ans: Propyne react with sodium metal and liberate hydrogen gas. Propene do not
react with sodium metal.
Q.8) How will you distinguish between propene and propane?
Ans: When we pass them through Br2 in CCl4 solution. Propene will decolourise
the reddish orange colour of bromine solutions but propane does not react.
Q.9) Arrange the following in the increasing order of acidic character.
HC≡CH, CH3CH3, H2C=CH2
Ans: CH3CH3 < H2C=CH2 < HC≡CH
Q.10) Name the alkane that cannot be prepared by Wurtz reaction.
Ans: Methane.
Q.11) Arrange the following: HCl, HBr, HI, HF in order of decreasing reactivity
towards alkenes.
Ans: HI > HBr > HCl > HF
Q.12) What are ortho, para directing group in aromatic electrophilic substitution
reactions.
Ans: Those group which increases electron density at ortho and para position are
known as ortho-para directing group.
All the activating groups are ortho-para directing groups.
Q. 13. Show the canonical structures of phenol.

Q.14) Name the examples of ortho para directing group..

Ans: OH, -NH2 -NHR, -NHCOCH3, -OCH3 -CH3, -C2H5 etc.

Q.15) Write IUPAC names of the following compounds: CH3CH=C(CH3)2.

Page | 83
Ans.

Short answer (2M)

Q 1) What are the necessary conditions for any system to be aromatic?
Ans: i) It should be planar.
ii) Complete delocalization of the π electron in the ring
iii) Presence of (4n + 2) π electrons in the ring (where n = 0, 1, 2, 3 )
(This is according to Huckel Rule)
Q.2) Write IUPAC names of the products obtained by the ozonolysis of the following
compound Pent-2-ene.
Ans :

The IUPAC name of product (I) ethanal and product (II) propanal.

Q.3) Draw Newman and Sawhorse projections for the eclipsed and staggered
conformations of ethane. Which of these conformations is more stable and why?
Ans. In staggered form of ethane, the electron clouds of carbon-hydrogen bonds are
as far apart as possible (torsional angle = 60°). Thus, there are minimum repulsive
forces (torsional strain), minimum energy and maximum stability of the molecule.
On the other hand, when the staggered form changes into the eclipsed form, the
electron clouds of the carbon-hydrogen bonds come closer to each other (torsional
angle = 0°) resulting in increase in electron cloud repulsions. To minimise the
increased repulsive forces, molecule will have to possess more energy and thus has
lesser stability.

Page | 84
Q.4) Explain with example Wurtz reaction.
Ans: Wurtz reaction: In this reaction, sodium metal is reacted with two molecules
alkyl halides in the presence dry ether to form a higher alkane. Higher alkane.
R−X + 2 Na + R-X → R-R + 2NaX.
Q.5) How would you convert Acetylene to Benzene.
Ans: When acetylene(ethyne) gas is heated to 873 K in a red-hot iron tube,
benzene is produced.

Q.6) How will you convert benzene into (i) p-nitrobromobenzene (ii) m-
nitrobromobenzene?

Q27.
Long answer question (3M)
Q 1) (a) Define substitution reactions. Why do benzene undergo electrophilic
substitution reactions even though they contain double bonds?
(b) What happens when benzene is treated with
(i) Br2 in presence of anhydrous AlCl 3
(ii) Conc. H2SO4 at 330 K.

Page | 85
Ans: (a) Substitution reactions are those reactions in which an atom or group of
atoms directly attached to carbon in the substrate molecule is replaced by another
atom or group of atoms, for example,

Benzene undergoes electrophilic substitution reactions because benzene ring has

delocalized electrons is an electron-rich system. It is attacked by electrophiles giving
substitution products.

Q.2). Draw the cis and trans structures of hex-2-ene. Which isomer will have higher
boiling point and why?
Structural formula of Hex-2-ene

Geometrical isomers of hex-2-ene are

The dipole moment of cis-compound is a sum of the dipole moments of C–CH3 and
C–CH2CH2CH3 bonds acting in the same direction.
The dipole moment of trans-compound is the resultant of the dipole moments of C–
CH3 and C–CH2CH2CH3 bonds acting in opposite directions.
Hence, cis-isomer is more polar than trans-isomer.
The higher the polarity, the greater is the intermolecular dipole-dipole interaction and
the higher will be the boiling point. Hence, cis-isomer will have a higher boiling point
than trans-isomer.

Page | 86