SOME BASIC CONCEPTS OF CHEMISTRY 1

UNIT 1

SOME BASIC CONCEPTS OF CHEMISTRY

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Chemistry is the science of molecules and their

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transformations. It is the science not so much of the one

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hundred elements but of the infinite variety of molecules that
may be built from them ...
able to
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After studying this unit, you will be
Roald Hoffmann

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• understand and appreciate the
role of chemistry in different
spheres of life;
• explain the characteristics of
three states of matter;
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Chemistry deals with the composition, structure and

• classify different substances into
elements, compounds and
properties of matter. These aspects can be best described
mixtures; and understood in terms of basic constituents of matter:
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• define SI base units and list some atoms and molecules. That is why chemistry is called
commonly used prefixes; the science of atoms and molecules. Can we see, weigh
• use scientific notations and and perceive these entities? Is it possible to count the
perform simple mathematical number of atoms and molecules in a given mass of matter
operations on numbers; and have a quantitative relationship between the mass and
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• differentiate between precision and number of these particles (atoms and molecules)? We will
accuracy;
like to answer some of these questions in this Unit. We
• determine significant figures;
would further describe how physical properties of matter
• convert physical quantities from
one system of units to another;
can be quantitatively described using numerical values
• explain various laws of chemical
with suitable units.
combination; 1.1 IMPORTANCE OF CHEMISTRY
to

• appreciate significance of atomic

mass, average atomic mass,
Science can be viewed as a continuing human effort to
molecular mass and formula systematize knowledge for describing and understanding
mass; nature. For the sake of convenience science is sub-divided
• describe the terms – mole and into various disciplines: chemistry, physics, biology,
molar mass; geology etc. Chemistry is the branch of science that studies
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• calculate the mass per cent of the composition, properties and interaction of matter.
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different elements constituting a Chemists are interested in knowing how chemical

compound;
transformations occur. Chemistry plays a central role in
• determine empirical formula and
science and is often intertwined with other branches of
molecular formula for a compound
from the given experimental data; science like physics, biology, geology etc. Chemistry also
• perform the stoichiometric plays an important role in daily life.
calculations. Chemical principles are important in diverse areas, such
as: weather patterns, functioning of brain and operation
2 CHEMISTRY

of a computer. Chemical industries

manufacturing fertilizers, alkalis, acids, salts,
dyes, polymers, drugs, soaps, detergents,
metals, alloys and other inorganic and organic
chemicals, including new materials, contribute
in a big way to the national economy.
Chemistry plays an important role in meeting
human needs for food, health care products

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and other materials aimed at improving the
quality of life. This is exemplified by the large
scale production of a variety of fertilizers,
improved varieties of pesticides and
insecticides. Similarly many life saving drugs Fig. 1.1 Arrangement of particles in solid, liquid

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such as cisplatin and taxol, are effective in and gaseous state

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cancer therapy and AZT (Azidothymidine)
Everything around us, for example, book, pen,

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used for helping AIDS victims, have been
isolated from plant and animal sources or pencil, water, air, all living beings etc. are
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prepared by synthetic methods. composed of matter. You know that they have
mass and they occupy space.

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With a better understanding of chemical
principles it has now become possible to You are also aware that matter can exist in
design and synthesize new materials having three physical states viz. solid, liquid and gas.
specific magnetic, electric and optical The constituent particles of matter in these
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properties. This has lead to the production of three states can be represented as shown in
superconducting ceramics, conducting Fig. 1.1. In solids, these particles are held very
polymers, optical fibres and large scale close to each other in an orderly fashion and
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miniaturization of solid state devices. In recent there is not much freedom of movement. In
years chemistry has tackled with a fair degree liquids, the particles are close to each other
of success some of the pressing aspects of but they can move around. However, in gases,
environmental degradation. Safer alternatives the particles are far apart as compared to those
to environmentally hazardous refrigerants like present in solid or liquid states and their
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CFCs (chlorofluorocarbons), responsible for movement is easy and fast. Because of such
ozone depletion in the stratosphere, have been arrangement of particles, different states of
successfully synthesised. However, many big matter exhibit the following characteristics:
environmental problems continue to be (i) Solids have definite volume and definite
matters of grave concern to the chemists. One shape.
such problem is the management of the Green
House gases like methane, carbon dioxide etc. (ii) Liquids have definite volume but not the
to

Understanding of bio-chemical processes, use definite shape. They take the shape of the
of enzymes for large-scale production of container in which they are placed.
chemicals and synthesis of new exotic (iii) Gases have neither definite volume nor
materials are some of the intellectual challenges definite shape. They completely occupy the
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for the future generation of chemists. A container in which they are placed.
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developing country like India needs talented These three states of matter are
and creative chemists for accepting such interconvertible by changing the conditions of
challenges. temperature and pressure.
1.2 NATURE OF MATTER he at
Solid



⇀
heat

liquid



⇀

Gas
↽



cool
↽



cool

You are already familiar with the term matter
from your earlier classes. Anything which has On heating a solid usually changes to a
mass and occupies space is called matter. liquid and the liquid on further heating
 SOME BASIC CONCEPTS OF CHEMISTRY 3

changes to the gaseous ( or vapour) state. In composition is variable. Copper, silver, gold,
the reverse process, a gas on cooling liquifies water, glucose are some examples of pure
to the liquid and the liquid on further cooling substances. Glucose contains carbon,
freezes to the solid. hydrogen and oxygen in a fixed ratio and thus,
At the macroscopic or bulk level, matter like all other pure substances has a fixed
composition. Also, the constituents of pure
can be classified as mixtures or pure
substances cannot be separated by simple
substances. These can be further sub-divided
physical methods.
as shown in Fig. 1.2.

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Pure substances can be further classified
into elements and compounds. An element
consists of only one type of particles. These
particles may be atoms or molecules. You may
be familiar with atoms and molecules from the

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previous classes; however, you will be studying

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about them in detail in Unit 2. Sodium, copper,

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silver, hydrogen, oxygen etc. are some
examples of elements. They all contain atoms
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elements are different in nature. Some elements
such as sodium or copper, contain single
Fig. 1.2 Classification of matter atoms held together as their constituent
particles whereas in some others, two or more
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Many of the substances present around atoms combine to give molecules of the
you are mixtures. For example, sugar solution element. Thus, hydrogen, nitrogen and oxygen
in water, air, tea etc., are all mixtures. A mixture gases consist of molecules in which two atoms
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contains two or more substances present in it combine to give their respective molecules. This
(in any ratio) which are called its components. is illustrated in Fig. 1.3.
A mixture may be homogeneous or When two or more atoms of different
heterogeneous. In a homogeneous mixture, elements combine, the molecule of a
the components completely mix with each other
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compound is obtained. The examples of some

and its composition is uniform throughout. compounds are water, ammonia, carbon
Sugar solution, and air are thus, the examples
of homogeneous mixtures. In contrast to this,
in heterogeneous mixtures, the composition
is not uniform throughout and sometimes the
different components can be observed. For
example, the mixtures of salt and sugar, grains
to

and pulses along with some dirt (often stone)

pieces, are heterogeneous mixtures. You can
think of many more examples of mixtures
which you come across in the daily life. It is
worthwhile to mention here that the
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components of a mixture can be separated by

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using physical methods such as simple hand

picking, filtration, crystallisation, distillation
etc.
Pure substances have characteristics
different from the mixtures. They have fixed
composition, whereas mixtures may contain
the components in any ratio and their Fig. 1.3 A representation of atoms and molecules
4 CHEMISTRY

dioxide, sugar etc. The molecules of water and chemical properties are characteristic
carbon dioxide are represented in Fig 1.4. reactions of different substances; these include
acidity or basicity, combustibility etc.
Many properties of matter such as length,
area, volume, etc., are quantitative in nature.
Any quantitative observation or measurement
is represented by a number followed by units
Water molecule Carbon dioxide in which it is measured. For example length of

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(H2O) molecule (CO2 ) a room can be represented as 6 m; here 6 is
the number and m denotes metre – the unit in
Fig. 1.4 A depiction of molecules of water and which the length is measured.
carbon dioxide
Two different systems of measurement, i.e.

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You have seen above that a water molecule the English System and the Metric System

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comprises two hydrogen atoms and one were being used in different parts of the world.

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oxygen atom. Similarly, a molecule of carbon The metric system which originated in France
dioxide contains two oxygen atoms combined in late eighteenth century, was more
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with one carbon atom. Thus, the atoms of convenient as it was based on the decimal

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different elements are present in a compound system. The need of a common standard
in a fixed and definite ratio and this ratio is system was being felt by the scientific
characteristic of a particular compound. Also, community. Such a system was established
the properties of a compound are different in 1960 and is discussed below in detail.
from those of its constituent elements. For
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example, hydrogen and oxygen are gases 1.3.1 The International System of Units
whereas the compound formed by their (SI)
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combination i.e., water is a liquid. It is The International System of Units (in French
interesting to note that hydrogen burns with Le Systeme International d’Unités –
a pop sound and oxygen is a supporter of abbreviated as SI) was established by the 11th
combustion, but water is used as a fire General Conference on Weights and Measures
extinguisher. (CGPM from Conference Generale des Poids
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Moreover, the constituents of a compound et Measures). The CGPM is an inter

cannot be separated into simpler substances governmental treaty organization created by
by physical methods. They can be separated a diplomatic treaty known as Metre Convention
by chemical methods. which was signed in Paris in 1875.
The SI system has seven base units and
1.3 PROPERTIES OF MATTER AND
they are listed in Table 1.1. These units pertain
THEIR MEASUREMENT
to

to the seven fundamental scientific quantities.

Every substance has unique or characteristic The other physical quantities such as speed,
properties. These properties can be classified volume, density etc. can be derived from these
into two categories – physical properties and quantities.
chemical properties.
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The definitions of the SI base units are given

Physical properties are those properties
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in Table 1.2.
which can be measured or observed without
changing the identity or the composition of the The SI system allows the use of prefixes to
substance. Some examples of physical indicate the multiples or submultiples of a unit.
properties are colour, odour, melting point, These prefixes are listed in Table 1. 3.
boiling point, density etc. The measurement Let us now quickly go through some of the
or observation of chemical properties require quantities which you will be often using in this
a chemical change to occur. The examples of book.
 SOME BASIC CONCEPTS OF CHEMISTRY 5

Table 1.1 Base Physical Quantities and their Units

Base Physical Symbol Name of Symbol

Quantity for SI Unit for SI
Quantity Unit
Length l metre m
Mass m kilogram kg
T ime t second s

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Electric current I ampere A
Thermodynamic T kelvin K
temperature
Amount of substance n mole mol

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Luminous intensity Iv candela cd

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Table 1.2 Definitions of SI Base Units
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Unit of length metre The metre is the length of the path travelled

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by light in vacuum during a time interval of
1/299 792 458 of a second.
Unit of mass kilogram The kilogram is the unit of mass; it is equal
to the mass of the international prototype
of the kilogram.
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Unit of time second The second is the duration of 9 192 631 770
periods of the radiation corresponding to the
transition between the two hyperfine levels
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of the ground state of the caesium-133 atom.

Unit of electric current ampere The ampere is that constant current which,
if maintained in two straight parallel
conductors of infinite length, of negligible
circular cross-section, and placed 1 metre
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apart in vacuum, would produce between

these conductors a force equal to 2 × 10–7
newton per metre of length.
Unit of thermodynamic kelvin The kelvin, unit of thermodynamic
temperature temperature, is the fraction 1/273.16 of the
thermodynamic temperature of the triple
point of water.
to

Unit of amount of substance mole 1. The mole is the amount of substance of a

system which contains as many elementary
entities as there are atoms in 0.012
kilogram of carbon-12; its symbol is “mol.”
2. When the mole is used, the elementary
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entities must be specified and may be atoms,

molecules, ions, electrons, other particles,
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or specified groups of such particles.

Unit of luminous intensity candela The candela is the luminous intensity, in a
given direction, of a source that emits
monochromatic radiation of frequency
540 × 10 12 hertz and that has a radiant
intensity in that direction of 1/683 watt per
steradian.
6 CHEMISTRY

Table 1.3 Prefixes used in the SI System chemistry laboratories, smaller volumes are
used. Hence, volume is often denoted in cm3
Multiple Prefix Symbol
or dm3 units.
10–24 yocto y
–21
10 zepto z
–18
10 atto a
–15
10 femto f

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10–12 pico p
10–9 nano n
10–6
micro µ
–3
10 milli m

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–2
10 centi c

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–1
10 deci d

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10 deca da
102
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3
10 kilo k
6
10 mega M
9
10 giga G
12
10 tera T
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1015
peta P Fig. 1.5 Analytical balance
1018 exa E
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1021 zeta Z Maintaining the National

1024
yotta Y Standards of Measurement
The s ystem of units including unit
1.3.2 Mass and Weight definitions keeps on changing with time.
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Whenever the accuracy of measurement

Mass of a substance is the amount of matter of a particular unit was enhanced
present in it while weight is the force exerted substantially by adopting new principles,
by gravity on an object. The mass of a member nations of metre treaty (signed in
substance is constant whereas its weight may 1875), agreed to change the formal
vary from one place to another due to change definition of that unit. Each modern
in gravity. You should be careful in using these industrialized country including India has
a National Metrology Institute (NMI) which
to

terms.
maintains standards of measurements.
The mass of a substance can be determined
This responsibility has been given to the
very accurately in the laboratory by using an National Physical Laboratory (NPL),
analytical balance (Fig. 1.5). New Delhi. This laboratory establishes
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The SI unit of mass as given in Table 1.1 is experiments to realize the base units and
kilogram. However, its fraction gram derived units of measurement and
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(1 kg = 1000 g), is used in laboratories due to maintains National Standards of

Measurement. These standards are
the smaller amounts of chemicals used in
periodically inter -compared with
chemical reactions. standards maintained at other National
Volume Metrology Institutes in the world as well
Volume has the units of (length)3. So in SI as those established at the International
system, volume has units of m 3. But again, in Bureau of Standards in Paris.
 SOME BASIC CONCEPTS OF CHEMISTRY 7

A common unit, litre (L) which is not an SI

unit, is used for measurement of volume of
liquids.
1 L = 1000 mL , 1000 cm3 = 1 dm3
Fig. 1.6 helps to visualise these relations.
In the laboratory, volume of liquids or
solutions can be measured by graduated

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cylinder, burette, pipette etc. A volumetric flask
is used to prepare a known volume of a
solution. These measuring devices are shown
in Fig. 1.7.

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Density

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Fig. 1.6 Different units used to express volume

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Density of a substance is its amount of mass
per unit volume. So SI units of density can be
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obtained as follows:

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SI unit of mass
SI unit of density =
SI unit of volume
kg –3
= 3 or kg m
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m
This unit is quite large and a chemist often
expresses density in g cm–3, where mass is
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expressed in gram and volume is expressed in

cm3.
Temperature
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There are three common scales to measure Fig 1.7 Some volume measuring devices
temperature — °C (degree celsius), °F (degree
fahrenheit) and K (kelvin). Here, K is the SI unit. o o
373.15 K 100 C Boiling point 212 F
The thermometers based on these scales are of water
shown in Fig. 1.8. Generally, the thermometer
with celsius scale are calibrated from 0° to 100°
to

where these two temperatures are the freezing 310.15 K

o
37 C
Human body
temperature 98.6 F
o

point and the boiling point of water 298.15 K

o
25 C Room o
77 F
temperature
respectively. The fahrenheit scale is
represented between 32° to 212°. o Freezing point
273.15 K 0C 32 o F
of water
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The temperatures on two scales are related

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to each other by the following relationship:

9
°F = (°C) + 32
5 Kelvin Celsius Fahrenheit
The kelvin scale is related to celsius scale
as follows :
Fig. 1.8 Thermometers using different
K = °C + 273.15
temperature scales
8 CHEMISTRY

It is interesting to note that temperature 1.4 UNCERTAINTY IN MEASUREMENT

below 0 °C (i.e. negative values) are possible Many a times in the study of chemistry, one
in Celsius scale but in Kelvin scale, negative has to deal with experimental data as well as
temperature is not possible. theoretical calculations. There are meaningful
Reference Standard ways to handle the numbers conveniently and
After defining a unit of measurement such present the data realistically with certainty to
as the kilogram or the metre, scientists the extent possible. These ideas are discussed
agreed on reference standards that make below in detail.

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it possible to calibrate all measuring
devices. For getting reliable measurements, 1.4.1 Scientific Notation
all devices such as metre sticks and As chemistry is the study of atoms and
analytical balances have been calibrated by
their manufacturers to give correct
molecules which have extremely low masses
and are present in extremely large numbers,

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readings. However, each of these devices
is standardised or calibrated against some a chemist has to deal with numbers as large

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reference. The mass standard is the as 602, 200,000,000,000,000,000,000 for the

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kilogram since 1889. It has been defined molecules of 2 g of hydrogen gas or as small
as the mass of platinum-iridium (Pt-Ir) as 0.00000000000000000000000166 g
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cylinder that is stored in an airtight jar at mass of a H atom. Similarly other constants

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International Bur eau of Weights and
such as Planck’s constant, speed of light,
Measures in Sevres, France. Pt-Ir was
chosen for this standard because it is
charges on particles etc., involve numbers of
highly resistant to chemical attack and its the above magnitude.
mass will not change for an extremely long It may look funny for a moment to write or
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time. count numbers involving so many zeros but it

Scientists are in search of a new
offers a real challenge to do simple
standard for mass. This is being attempted
through accurate determination of mathematical operations of addition,
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Avogadr o constant. Work on this new subtraction, multiplication or division with

standard focuses on ways to measure such numbers. You can write any two
accurately the number of atoms in a well- numbers of the above type and try any one of
defined mass of sample. One such method, the operations you like to accept the challenge
which uses X-rays to determine the atomic
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and then you will really appreciate the difficulty

density of a crystal of ultrapure silicon, has
an accuracy of about 1 part in 106 but has
in handling such numbers.
not yet been adopted to serve as a This problem is solved by using scientific
standard. There are other methods but notation for such numbers, i.e., exponential
none of them are presently adequate to notation in which any number can be
replace the Pt-Ir cylinder. No doubt, represented in the form N × 10n where n is an
changes are expected within this decade. exponent having positive or negative values
The metre was originally defined as the
to

length between two marks on a Pt-Ir bar

and N is a number (called digit term) which
kept at a temperature of 0°C (273.15 K). In varies between 1.000... and 9.999....
1960 the length of the metre was defined Thus, we can write 232.508 as
as 1.65076373 ×106 times the wavelength 2.32508 ×102 in scientific notation. Note that
of light emitted by a krypton laser. while writing it, the decimal had to be moved
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Although this was a cumbersome number,

to the left by two places and same is the
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it preserved the length of the metre at its

agreed value. The metre was redefined in exponent (2) of 10 in the scientific notation.
1983 by CGPM as the length of path Similarly, 0.00016 can be written as
travelled by light in vacuum during a time 1.6 × 10–4. Here the decimal has to be moved
interval of 1/299 792 458 of a second. four places to the right and ( – 4) is the exponent
Similar to the length and the mass, there in the scientific notation.
are reference standards for other physical
quantities. Now, for performing mathematical
operations on numbers expressed in scientific
 SOME BASIC CONCEPTS OF CHEMISTRY 9

notations, the following points are to be kept accuracy is the agreement of a particular value
in mind. to the true value of the result. For example, if
Multiplication and Division the true value for a result is 2.00 g and a
student ‘A’ takes two measurements and
These two operations follow the same rules reports the results as 1.95 g and 1.93 g. These
which are there for exponential numbers, i.e. values are precise as they are close to each
(5.6 ×10 ) × (6.9 × 10 ) = (5.6 × 6.9 ) (10 )
5 8 5+ 8
other but are not accurate. Another student
repeats the experiment and obtains 1.94 g and
= ( 5.6 × 6.9) ×1013

ed
2.05 g as the results for two measurements.
= 38.64 × 1013 These observations are neither precise nor
accurate. When a third student repeats these
= 3.864 × 1014 measurements and reports 2.01g and 1.99 g
(9.8 × 10 ) × (2.5 ×10 ) = (9.8 × 2.5 ) (10 ( ) )
−2 −6 − 2+ − 6 as the result. These values are both precise and

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accurate. This can be more clearly understood
= ( 9.8 × 2.5) (10 )

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−2− 6
from the data given in Table 1.4

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Table 1.4 Data to Illustrate Precision and
= 24.50 × 10–8 Accuracy
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−3
2.7 × 10
= ( 2.7 ÷ 5.5 ) (10− 3− 4 ) = 0.4909 × 10 –7 1 2 Average (g)
5.5 ×104 Student A 1.95 1.93 1.940
= 4.909 × 10–8 Student B 1.94 2.05 1.995
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Addition and Subtraction Student C 2.01 1.99 2.000

For these two operations, first the numbers are
written in such a way that they have same The uncertainty in the experimental or the
N

exponent. After that, the coefficient are added calculated values is indicated by mentioning
or subtracted as the case may be. the number of significant figures. Significant
figures are meaningful digits which are known
Thus, for adding 6.65 × 104 and 8.95 × 103,
with certainty. The uncertainty is indicated by
6.65 × 104 + 0.895 × 104 exponent is made
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writing the certain digits and the last uncertain

same for both the numbers. digit. Thus, if we write a result as 11.2 mL, we
Then, these numbers can be added as follows say the 11 is certain and 2 is uncertain and
(6.65 + 0.895) × 104 = 7.545 × 104 the uncertainty would be +1 in the last digit.
Similarly, the subtraction of two numbers can Unless otherwise stated, an uncertainty of +1
be done as shown below : in the last digit is always understood.
2.5 × 10–2 – 4.8 × 10–3 There are certain rules for determining the
to

number of significant figures. These are stated

= (2.5 × 10–2) – (0.48 × 10–2) below:
= (2.5 – 0.48) × 10–2 = 2.02 × 10–2 (1) All non-zero digits are significant. For
1.4.2 Significant Figures example in 285 cm, there are three
significant figures and in 0.25 mL, there
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Every experimental measurement has some are two significant figures.

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amount of uncertainty associated with it.

However, one would always like the results to (2) Zeros preceding to first non-zero digit are
be precise and accurate. Precision and not significant. Such zero indicates the
accuracy are often referred to while we talk position of decimal point.
about the measurement. Thus, 0.03 has one significant figure and
Precision refers to the closeness of various 0.0052 has two significant figures.
measurements for the same quantity. However, (3) Zeros between two non-zero digits are
10 CHEMISTRY

significant. Thus, 2.005 has four significant Since 2.5 has two significant figures, the
figures. result should not have more than two
(4) Zeros at the end or right of a number are significant figures, thus, it is 3.1.
significant provided they are on the right While limiting the result to the required
side of the decimal point. For example, number of significant figures as done in the
0.200 g has three significant figures. above mathematical operation, one has to keep
But, if otherwise, the terminal zeros are not in mind the following points for rounding off
significant if there is no decimal point. For the numbers

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example, 100 has only one significant 1. If the rightmost digit to be removed is more
figure, but 100. has three significant than 5, the preceding number is increased
figures and 100.0 has four significant by one. for example, 1.386
figures. Such numbers are better If we have to remove 6, we have to round it

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represented in scientific notation. We can
to 1.39
express the number 100 as 1×102 for one

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significant figure, 1.0×10 2 for two 2. If the rightmost digit to be removed is less

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significant figures and 1.00×102 for three than 5, the preceding number is not changed.
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significant figures. For example, 4.334 if 4 is to be removed,
then the result is rounded upto 4.33.

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(5) Counting numbers of objects, for example,
2 balls or 20 eggs, have infinite significant 3. If the rightmost digit to be removed is 5,
figures as these are exact numbers and can then the preceding number is not changed
be represented by writing infinite number if it is an even number but it is increased
of zeros after placing a decimal i.e., by one if it is an odd number. For example,
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2 = 2.000000 or 20 = 20.000000 if 6.35 is to be rounded by removing 5, we

In numbers written in scientific notation, have to increase 3 to 4 giving 6.4 as the
result. However, if 6.25 is to be rounded
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all digits are significant e.g., 4.01×102 has three

significant figures, and 8.256 × 10–3 has four off it is rounded off to 6.2.
significant figures. 1.4.3 Dimensional Analysis
Addition and Subtraction of Significant Often while calculating, there is a need to
©

Figures convert units from one system to other. The

The result cannot have more digits to the right method used to accomplish this is called factor
of the decimal point than either of the original label method or unit factor method or
numbers. dimensional analysis. This is illustrated
12.11 below.
18.0 Example
1.012
to

A piece of metal is 3 inch (represented by in)

31.122
long. What is its length in cm?
Here, 18.0 has only one digit after the decimal
We know that 1 in = 2.54 cm
point and the result should be reported only
up to one digit after the decimal point which From this equivalence, we can write
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is 31.1. 1 in 2.54 cm
=1=
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Multiplication and Division of Significant 2.54 cm 1 in

Figures
1 in 2.54cm
In these operations, the result must be reported thus 2.54cm equals 1 and 1 in also
with no more significant figures as are there in
the measurement with the few significant equals 1. Both of these are called unit factors.
figures. If some number is multiplied by these unit
2.5×1.25 = 3.125 factors (i.e. 1), it will not be affected otherwise.
 SOME BASIC CONCEPTS OF CHEMISTRY 11

Say, the 3 in given above is multiplied by i.e., 2 days − −− − −− = −− − seconds

the unit factor. So,
The unit factors can be multiplied in
2.54cm series in one step only as follows:
3 in = 3 in × 1 in = 3 × 2.54 cm = 7.62 cm
24 h 60 min 60 s
2 day × × ×
Now the unit factor by which multiplication 1day 1h 1 min
2.54 cm = 2 × 24 × 60 × 60 s

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is to be done is that unit factor ( in = 172800 s
1 in
the above case) which gives the desired units 1.5 LAWS OF CHEMICAL COMBINATIONS
i.e., the numerator should have that part which The combination of elements
is required in the desired result. to form compounds is

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It should also be noted in the above governed by the following five
basic laws.

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example that units can be handled just like

is
other numerical part. It can be cancelled, 1.5.1 Law of Conservation
divided, multiplied, squared etc. Let us study of Mass
re ER
one more example for it. It states that matter can Antoine Lavoisier

bl
Example neither be created nor (1743—1794)

A jug contains 2 L of milk. Calculate the volume destroyed.

of the milk in m3. This law was put forth by Antoine Lavoisier
Since 1 L = 1000 cm3 in 1789. He performed careful experimental
be C

and 1m = 100 cm which gives studies for combustion reactions for reaching
to the above conclusion. This law formed the
1m 100cm basis for several later developments in
=1 =
N

100 cm 1m chemistry. Infact, this was the result of exact

measurement of masses of reactants and
To get m3 from the above unit factors, the
products, and carefully planned experiments
first unit factor is taken and it is cubed.
performed by Lavoisier.
©

3
 1m  1 m3
= (1) = 1
3
  ⇒ 6 3
1.5.2 Law of Definite Proportions
 100 cm  10 cm This law was given by, a
Now 2 L = 2×1000 cm3 French chemist, Joseph
The above is multiplied by the unit factor Proust. He stated that a
given compound always
1m 3 2 m3
2 × 1000 cm 3 × 6 3
= 3
= 2 ×10− 3 m 3 contains exactly the same
to

10 cm 10 proportion of elements by
Example weight.
How many seconds are there in 2 days? Proust worked with two Joseph Proust
(1754—1826)
Here, we know 1 day = 24 hours (h) samples of cupric carbonate
— one of which was of natural origin and the
t

1 day 24 h other was synthetic one. He found that the

no

or 24 h = 1 = 1d ay composition of elements present in it was same

for both the samples as shown below :
then 1h = 60 min
% of % of % of
1h 60 min copper oxygen carbon
or =1=
60 min 1h Natural Sample 51.35 9.74 38.91
so, for converting 2 days to seconds, Synthetic Sample 51.35 9.74 38.91
12 CHEMISTRY

Thus, irrespective of the source, a given Thus, 100 mL of hydrogen combine with
compound always contains same elements in 50 mL of oxygen to give 100 mL of water
the same proportion. The validity of this law vapour.
has been confirmed by various experiments. Hydrogen + Oxygen → Water
It is sometimes also referred to as Law of
100 mL 50 mL 100 mL
definite composition.
Thus, the volumes of hydrogen and oxygen
1.5.3 Law of Multiple Proportions which combine together (i.e. 100 mL and

ed
This law was proposed by Dalton in 1803. 50 mL) bear a simple ratio of 2:1.
According to this law, if two elements can Gay-Lussac’s discovery of integer ratio in
combine to form more than one compound, the volume relationship is actually the law of
masses of one element that combine with a definite proportions by volume. The law of
fixed mass of the other element, are in the definite proportions, stated earlier, was with

h
ratio of small whole numbers. respect to mass. The Gay-Lussac’s law was

pu T
For example, hydrogen combines with explained properly by the work of Avogadro

is
oxygen to form two compounds, namely, water in 1811.
and hydrogen peroxide.
re ER
Hydrogen + Oxygen → Water
1.5.5 Avogadro Law

bl
In 1811, Avogadro proposed
2g 16g 18g that equal volumes of gases
Hydrogen + Oxygen → Hydrogen Peroxide at the same temperature and
2g 32g 34g pressure should contain
be C

Here, the masses of oxygen (i.e. 16 g and 32 g) equal number of molecules.

Avogadro made a distinction
which combine with a fixed mass of hydrogen
between atoms and
(2g) bear a simple ratio, i.e. 16:32 or 1: 2.
molecules which is quite Lorenzo Romano
N

Amedeo Carlo
1.5.4 Gay Lussac’s Law of Gaseous understandable in the Avogadro di
Volumes present times. If we consider Quareqa edi

This law was given by Gay again the reaction of hydrogen Carreto
(1776-1856)
and oxygen to produce water,
©

Lussac in 1808. He observed

that when gases combine or we see that two volumes of hydrogen combine
are produced in a chemical with one volume of oxygen to give two volumes
reaction they do so in a of water without leaving any unreacted oxygen.
simple ratio by volume Note that in the Fig. 1.9, each box contains
provided all gases are at equal number of molecules. In fact, Avogadro
same temperature and Joseph Louis could explain the above result by considering
Gay Lussac the molecules to be polyatomic. If hydrogen
to

pressure.
t
no

Fig. 1.9 Two volumes of hydrogen react with One volume of oxygen to give Two volumes of water vapour
 SOME BASIC CONCEPTS OF CHEMISTRY 13

and oxygen were considered as diatomic as understand what we mean by atomic and
recognised now, then the above results are molecular masses.
easily understandable. However, Dalton and
1.7.1 Atomic Mass
others believed at that time that atoms of the
same kind cannot combine and molecules of The atomic mass or the mass of an atom is
oxygen or hydrogen containing two atoms did actually very-very small because atoms are
not exist. Avogadro’s proposal was published extremely small. Today, we have sophisticated
in the French Journal de Physidue. In spite techniques e.g., mass spectrometry for
determining the atomic masses fairly

ed
of being correct, it did not gain much support.
accurately. But, in the nineteenth century,
After about 50 years, in 1860, first scientists could determine mass of one atom
international conference on chemistry was held
relative to another by experimental means, as
in Karlsruhe, Germany to resolve various ideas.
has been mentioned earlier. Hydrogen, being
At the meeting, Stanislao Cannizaro presented

h
lightest atom was arbitrarily assigned a mass
a sketch of a course of chemical philosophy

pu T
of 1 (without any units) and other elements
which emphasised the importance of

is
were assigned masses relative to it. However,
Avogadro’s work.
the present system of atomic masses is based
re ER
1.6 DALTON’S ATOMIC THEORY on carbon - 12 as the standard and has been

bl
agreed upon in 1961. Here, Carbon - 12 is one
Although the origin of idea
of the isotopes of carbon and can be
that matter is composed of
represented as 12C. In this system, 12C is
small indivisible particles
assigned a mass of exactly 12 atomic mass unit
called ‘a-tomio’ (meaning —
(amu) and masses of all other atoms are given
be C

indivisible), dates back to the

relative to this standard. One atomic mass
time of Democritus, a Greek
unit is defined as a mass exactly equal to one-
Philosopher (460 — 370 BC),
twelfth the mass of one carbon - 12 atom.
N

it again started emerging as John Dalton

a result of several (1776—1884) And 1 amu = 1.66056×10–24 g
experimental studies which led to the Laws Mass of an atom of hydrogen
mentioned above. = 1.6736×10–24 g
©

In 1808, Dalton published ‘A New System Thus, in terms of amu, the mass
of Chemical Philosophy’ in which he proposed 1.6736 × 10–24 g
the following : of hydrogen atom =
1.66056 ×10– 24 g
1. Matter consists of indivisible atoms.
= 1.0078 amu
2. All the atoms of a given element have
= 1.0080 amu
identical properties including identical
mass. Atoms of different elements differ in Similarly, the mass of oxygen - 16 (16O)
to

mass. atom would be 15.995 amu.

Today, ‘amu’ has been replaced by ‘u’
3. Compounds are formed when atoms of which is known as unified mass.
different elements combine in a fixed ratio.
When we use atomic masses of elements in
4. Chemical reactions involve reorganisation calculations, we actually use average atomic
t

of atoms. These are neither created nor masses of elements which are explained
no

destroyed in a chemical reaction.

below.
Dalton’s theory could explain the laws of
chemical combination. 1.7.2 Average Atomic Mass
Many naturally occurring elements exist as
1.7 ATOMIC AND MOLECULAR MASSES more than one isotope. When we take into
After having some idea about the terms atoms account the existence of these isotopes and
and molecules, it is appropriate here to their relative abundance (per cent occurrence),
14 CHEMISTRY

the average atomic mass of that element can Solution

be computed. For example, carbon has the Molecular mass of glucose (C6H12O6)
following three isotopes with relative
= 6(12.011 u) + 12(1.008 u) +
abundances and masses as shown against 6(16.00 u)
each of them.
= (72.066 u) + (12.096 u) +
Isotope Relative Atomic (96.00 u)
Abundance Mass (amu)
= 180.162 u
(%)

ed
12
C 98.892 12 1.7.4 Formula Mass
13
C 1.108 13.00335
Some substances such as sodium chloride do
14
C 2 ×10–10 14.00317 not contain discrete molecules as their

h
From the above data, the average atomic constituent units. In such compounds, positive
(sodium) and negative (chloride) entities are

pu T
mass of carbon will come out to be :
arranged in a three-dimensional structure, as

is
(0.98892) (12 u) + ( 0.01108) (13.00335 u) + shown in Fig. 1.10.
(2 × 10–12) (14.00317 u)
re ER
= 12.011 u

bl
Similarly, average atomic masses for other
elements can be calculated. In the periodic
table of elements, the atomic masses mentioned
for different elements actually represented their
be C

average atomic masses.

1.7.3 Molecular Mass

N

Molecular mass is the sum of atomic masses

of the elements present in a molecule. It is
obtained by multiplying the atomic mass of
©

each element by the number of its atoms and

adding them together. For example, molecular
mass of methane which contains one carbon
atom and four hydrogen atoms can be obtained
as follows :
Molecular mass of methane,
(CH4) = (12.011 u) + 4 (1.008 u) Fig. 1.10 Packing of Na+ and Cl– ions in
to

sodium chloride
= 16.043 u
Similarly, molecular mass of water (H2O) It may be noted that in sodium chloride,
–
one Na+ is surrounded by six Cl and vice-versa.
= 2 × atomic mass of hydrogen + 1 × atomic
The formula such as NaCl is used to
t

mass of oxygen
calculate the formula mass instead of
no

= 2 (1.008 u) + 16.00 u molecular mass as in the solid state sodium

= 18.02 u chloride does not exist as a single entity.

Problem 1.1 Thus, formula mass of sodium chloride =

atomic mass of sodium + atomic mass of
Calculate molecular mass of glucose chlorine
(C6H 12O 6) molecule.
= 23.0 u + 35.5 u = 58.5 u
 SOME BASIC CONCEPTS OF CHEMISTRY 15

1.8 MOLE CONCEPT AND MOLAR

MASSES
Atoms and molecules are extremely small in
size and their numbers in even a small amount
of any substance is really very large. To handle
such large numbers, a unit of similar
magnitude is required.
Just as we denote one dozen for 12 items,

ed
score for 20 items, gross for 144 items, we
use the idea of mole to count entities at the
microscopic level (i.e. atoms/molecules/
particles, electrons, ions, etc). Fig. 1.11 One mole of various substances

h
In SI system, mole (symbol, mol) was
1 mol of sodium chloride = 6.022 × 1023

pu T
introduced as seventh base quantity for the
formula units of sodium chloride

is
amount of a substance.
One mole is the amount of a substance Having defined the mole, it is easier to know
re ER
that contains as many particles or entities mass of one mole of the substance or the

bl
as there are atoms in exactly 12 g (or 0.012 constituent entities. The mass of one mole
kg) of the 12C isotope. It may be emphasised of a substance in grams is called its
that the mole of a substance always contain molar mass. The molar mass in grams is
the same number of entities, no matter what numerically equal to atomic/molecular/
formula mass in u.
be C

the substance may be. In order to determine

this number precisely, the mass of a carbon– Molar mass of water = 18.02 g mol-1
12 atom was determined by a mass Molar mass of sodium chloride = 58.5 g mol-1
spectrometer and found to be equal to
N

1.992648 × 10–23 g. Knowing that one mole of 1.9 PERCENTAGE COMPOSITION

carbon weighs 12 g, the number of atoms in it So far, we were dealing with the number of
is equal to : entities present in a given sample. But many a
time, the information regarding the percentage
©

12 g/ mol 12C of a particular element present in a compound

1.992648 × 10−23 g/12 Catom is required. Suppose an unknown or new
compound is given to you, the first question
= 6.0221367 × 1023 atoms/mol you would ask is: what is its formula or what
This number of entities in 1 mol is so are its constituents and in what ratio are they
important that it is given a separate name and present in the given compound? For known
symbol. It is known as ‘Avogadro constant’, compounds also, such information provides a
to

denoted by NA in honour of Amedeo Avogadro. check whether the given sample contains the
To really appreciate largeness of this number, same percentage of elements as is present in a
let us write it with all the zeroes without using pure sample. In other words, one can check
any powers of ten. the purity of a given sample by analysing this
t

602213670000000000000000 data.
no

Hence, so many entities (atoms, molecules Let us understand it by taking the example
or any other particle) constitute one mole of a of water (H2O). Since water contains hydrogen
particular substance. and oxygen, the percentage composition of both
these elements can be calculated as follows :
We can, therefore, say that 1 mol of
hydrogen atoms = 6.022×1023 atoms Mass % of an element =
1 mol of water molecules = 6.022×1023 mass of that element in the compound × 100
water molecules molar mass of the compound
16 CHEMISTRY

Molar mass of water = 18.02 g Problem 1.2

2×1.008 A compound contains 4.07 % hydrogen,
Mass % of hydrogen = ×100
18.02 24.27 % carbon and 71.65 % chlorine.
= 11.18 Its molar mass is 98.96 g. What are its
empirical and molecular formulas ?
16.00
Mass % of oxygen = ×100 Solution
18.02
Step 1. Conversion of mass per cent
= 88.79

ed
to grams.
Let us take one more example. What is the
Since we are having mass per cent, it is
percentage of carbon, hydrogen and oxygen
convenient to use 100 g of the compound
in ethanol? as the starting material. Thus, in the
100 g sample of the above compound,

h
Molecular formula of ethanol is : C2H5OH
Molar mass of ethanol is : (2×12.01 + 6×1.008 4.07g hydrogen is present, 24.27g

pu T
+ 16.00) g carbon is present and 71.65 g chlorine is

is
present.
= 46.068 g
re ER
Mass per cent of carbon Step 2. Convert into number moles of
each element

bl
24.02g
= 46.068g ×100 = 52.14% Divide the masses obtained above by
respective atomic masses of various
Mass per cent of hydrogen elements.
be C

6.048 g 4.07 g
= ×100 = 13.13% Moles of hydrogen = = 4.04
46.068 g 1.008 g
N

Mass per cent of oxygen 24.27 g

16.00 g Moles of carbon = = 2.021
12.01g
= 46.068g ×100 = 34.73%
71.65g
After understanding the calculation of per
©

Moles of chlorine = 35.453 g = 2.021

cent of mass, let us now see what information
can be obtained from the per cent composition Step 3. Divide the mole value obtained
data. above by the smallest number
Since 2.021 is smallest value, division by
1.9.1 Empirical Formula for Molecular it gives a ratio of 2:1:1 for H:C:Cl .
Formula In case the ratios are not whole numbers,
to

An empirical formula represents the simplest then they may be converted into whole
whole number ratio of various atoms present number by multiplying by the suitable
in a compound whereas the molecular coefficient.
formula shows the exact number of different Step 4. Write empirical formula by
mentioning the numbers after writing
t

types of atoms present in a molecule of a

compound. the symbols of respective elements.
no

If the mass per cent of various elements CH2Cl is, thus, the empirical formula of
present in a compound is known, its the above compound.
empirical formula can be determined. Step 5. Writing molecular formula
Molecular formula can further be obtained if (a) Determine empirical formula mass
the molar mass is known. The following Add the atomic masses of various atoms
example illustrates this sequence. present in the empirical formula.
 SOME BASIC CONCEPTS OF CHEMISTRY 17

For CH2Cl, empirical formula mass is 1.10 STOICHIOMETRY AND

12.01 + 2 × 1.008 + 35.453 STOICHIOMETRIC CALCULATIONS
= 49.48 g The word ‘stoichiometry’ is derived from two
Greek words - stoicheion (meaning element)
(b) Divide Molar mass by empirical and metron (meaning measure).
formula mass Stoichiometry, thus, deals with the calculation
of masses (sometimes volumes also) of the
Molar mass 98.96 g reactants and the products involved in a
=

ed
Empirical formula mass 49.48 g chemical reaction. Before understanding how
to calculate the amounts of reactants required
= 2 = (n) or the products produced in a chemical
(c) Multiply empirical formula by n reaction, let us study what information is
obtained above to get the molecular available from the balanced chemical equation

h
formula of a given reaction. Let us consider the
combustion of methane. A balanced equation

pu T
Empirical formula = CH2Cl, n = 2. Hence
for this reaction is as given below :

is
molecular formula is C2H4Cl2.
CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g)
re ER
Balancing a chemical equation

bl
According to the law of conservation of mass, a balanced chemical equation has the same
number of atoms of each element on both sides of the equation. Many chemical equations can
be balanced by trial and err or. Let us take the reactions of a few metals and non-metals with
oxygen to give oxides
be C

4 Fe(s) + 3O2 (g) → 2Fe2 O3 (s) (a) balanced equation

2 Mg(s) + O 2(g) → 2MgO(s) (b) balanced equation
P4 (s) + O2 (g) → P4O 10(s) (c) unbalanced equation
N

Equations (a) and (b) are balanced since there are same number of metal and oxygen atoms on
each side of equations. However equation (c) is not balanced. In this equation, phosphorus
atoms are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on
the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the
right side of the equation.
©

P4 (s) + 5O2 (g) → P4O 10(s) balanced equation

Now let us take combustion of propane, C3 H8 . This equation can be balanced in steps.
Step 1 Write down the correct for mulas of reactants and products. Here propane and oxygen are
reactants, and carbon dioxide and water are products.
C3H 8(g) + O2 (g) → CO2 (g) +H2O(l) unbalanced equation
Step 2 Balance the number of C atoms: Since 3 carbon atoms are in the reactant, therefore,
three CO2 molecules are required on the right side.
C3H 8 (g) + O2 (g) → 3CO2 (g) + H2 O (l)
to

Step 3 Balance the number of H atoms : on the left there are 8 hydrogen atoms in the reactants
however, each molecule of water has two hydrogen atoms, so four molecules of water will be
required for eight hydrogen atoms on the right side.
C3H 8 (g) +O2 (g) → 3CO2 (g)+4H 2O (l)
Step 4 Balance the number of O atoms: There are ten oxygen atoms on the right side (3 × 2 = 6 in
t

CO2 and 4 × 1= 4 in water). Therefore, five O2 molecules are needed to supply the required ten
no

oxygen atoms.
C3H 8 (g) +5O2 (g) → 3CO2 (g) + 4H2 O (l)
Step 5 Verify that the number of atoms of each element is balanced in the final equation. The
equation shows three carbon atoms, eight hydrogen atoms, and ten oxygen atoms on each side.
All equations that have correct formulas for all reactants and products can be balanced. Always
remember that subscripts in formulas of reactants and products cannot be changed to balance
an equation.
18 CHEMISTRY

Here, methane and dioxygen are called CH 4 (g) gives 2 mol of H2O (g).
reactants and carbon dioxide and water are
2 mol of water (H2O) = 2 × (2+16)
called products. Note that all the reactants and
= 2 × 18 = 36 g
the products are gases in the above reaction
and this has been indicated by letter (g) in the 18 g H2O
brackets next to its formula. Similarly, in the 1 mol H2O = 18 g H2O ⇒ 1mol H O = 1
2
case of solids and liquids, (s) and (l) are written
respectively.
18 g H2O

ed
The coefficients 2 for O2 and H2O are called Hence 2 mol H2O ×
1 mol H2O
stoichiometric coefficients. Similarly the
coefficient for CH4 and CO2 is one in each case. = 2 × 18 g H 2O = 36 g H2O
They represent the number of molecules (and Problem 1.4
moles as well) taking part in the reaction or

h
formed in the reaction. How many moles of methane are required

pu T
to produce 22 g CO2 (g) after combustion?
Thus, according to the above chemical

is
reaction, Solution

•
re ER
One mole of CH4(g) reacts with two moles According to the chemical equation,

bl
of O2(g) to give one mole of CO2(g) and CH4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2H2O ( g )
two moles of H 2O(g)
44g CO 2 (g) is obtained from 16 g CH4 (g).
• One molecule of CH 4(g) reacts with
[ ∵ 1 mol CO2(g) is obtained from 1 mol of
2 molecules of O2(g) to give one molecule
be C

CH4(g)]
of CO2(g) and 2 molecules of H2O(g)
mole of CO 2 (g)
• 22.7 L of CH4(g) reacts with 45.4 L of O2 (g)
1 mol CO2 (g)
N

to give 22.7 L of CO2 (g) and 45.4 L of H2O(g)

= 22 g CO 2 (g) ×
• 16 g of CH4 (g) reacts with 2×32 g of O2 (g) 44 g CO2 (g)
to give 44 g of CO2 (g) and 2×18 g of = 0.5 mol CO 2 (g)
H 2O (g).
Hence, 0.5 mol CO2 (g) would be obtained
©

From these relationships, the given data from 0.5 mol CH4 (g) or 0.5 mol of CH4 (g)
can be interconverted as follows : would be required to produce 22 g
mass ⇌ moles ⇌ no. of molecules CO 2 (g).

Mass
= Density 1.10.1 Limiting Reagent
Volume
Many a time, the reactions are carried out
to

when the reactants are not present in the

Problem 1.3
amounts as required by a balanced chemical
Calculate the amount of water (g) reaction. In such situations, one reactant is in
produced by the combustion of 16 g of excess over the other. The reactant which is
methane. present in the lesser amount gets consumed
t

Solution after sometime and after that no further

no

reaction takes place whatever be the amount

The balanced equation for combustion of
methane is : of the other reactant present. Hence, the
reactant which gets consumed, limits the
CH4 ( g ) + 2O2 ( g ) → CO2 ( g ) + 2H2O ( g ) amount of product formed and is, therefore,
called the limiting reagent.
(i)16 g of CH4 corresponds to one mole.
In performing stoichiometric calculations,
(ii) From the above equation, 1 mol of this aspect is also to be kept in mind.
 SOME BASIC CONCEPTS OF CHEMISTRY 19

Problem 1.5 17.0 g NH3 (g)

3.30×103 mol NH3 (g) ×
50.0 kg of N2 (g) and 10.0 kg of H 2 (g) are 1 mol NH3 (g)
mixed to produce NH3 (g). Calculate the
NH3 (g) formed. Identify the limiting = 3.30×103×17 g NH 3 (g)
reagent in the production of NH3 in this = 56.1×103 g NH3
situation. = 56.1 kg NH3
Solution

ed
A balanced equation for the above reaction 1.10.2 Reactions in Solutions
is written as follows : A majority of reactions in the laboratories are
Calculation of moles : carried out in solutions. Therefore, it is
important to understand as how the amount
N 2 ( g ) + 3H2 ( g ) ⇌ 2 NH3 ( g )

h
of substance is expressed when it is present in
moles of N2 the form of a solution. The concentration of a

pu T
solution or the amount of substance present

is
1000 g N2 1 mol N2 in its given volume can be expressed in any of
= 50.0 kg N2 × ×
re ER 1 kg N2 28.0 g N2 the following ways.

bl
= 17.86×102 mol 1. Mass per cent or weight per cent (w/w %)
moles of H2 2. Mole fraction
1000 g H2 1mol H2 3. Molarity
= 10.00 kg H2 × ×
1 kg H2 2.016 g H2 4. Molality
be C

= 4.96×103 mol Let us now study each one of them in detail.

According to the above equation, 1 mol 1. Mass per cent
N

N2 (g) requires 3 mol H2 (g), for the reaction. It is obtained by using the following relation:
Hence, for 17.86×102 mol of N2, the moles
of H2 (g) required would be Massof solute
Massper cent = ×100
Massof solution
3 mol H2 (g)
©

17.86×102 mol N2 ×
1mol N2 (g)
Problem 1.6
= 5.36 ×103 mol H2
A solution is prepared by adding 2 g of a
But we have only 4.96×103 mol H2. Hence, substance A to 18 g of water. Calculate
dihydrogen is the limiting reagent in this the mass per cent of the solute.
case. So NH3(g) would be formed only from
that amount of available dihydrogen i.e.,
to

Solution
4.96 × 103 mol
Since 3 mol H2(g) gives 2 mol NH3(g) Massof A
Mass per centof A = ×100
Massof solution
2 mol NH3 (g)
4.96×103 mol H2 (g) ×
t

3mol H2 (g) 2g
= × 100
no

= 3.30×10 mol NH3 (g)

3
2 gof A +18 gof water
3.30×103 mol NH3 (g) is obtained.
2g
If they are to be converted to grams, it is = ×100
done as follows : 20 g
1 mol NH3 (g) = 17.0 g NH 3 (g) = 10 %
20 CHEMISTRY

2. Mole Fraction Thus, 200 mL of 1M NaOH are taken and

It is the ratio of number of moles of a particular enough water is added to dilute it to make it
component to the total number of moles of the 1 litre.
solution. If a substance ‘A’ dissolves in In fact for such calculations, a general
substance ‘B’ and their number of moles are formula, M1 × V1 = M 2 × V2 where M and V are
n A and nB respectively; then the mole fractions molarity and volume respectively can be used.
of A and B are given as In this case, M1 is equal to 0.2; V 1 = 1000 mL
and, M 2 = 1.0; V 2 is to be calculated.

ed
Mole fraction of A Substituting the values in the formula:
No.of moles of A 0.2 M × 1000 mL = 1.0 M × V2
=
No.of moles of solution 0.2M × 1000mL
nA ∴ V2 = = 200 mL

h
= 1.0M
n A + nB

pu T
Note that the number of moles of solute

is
(NaOH) was 0.2 in 200 mL and it has remained
Mole fraction of B the same, i.e., 0.2 even after dilution ( in 1000
re ER
No.of moles of B
mL) as we have changed just the amount of

bl
= solvent (i.e. water) and have not done anything
No.of moles of solution with respect to NaOH. But keep in mind the
nB concentration.
=
n A + nB Problem 1.7
be C

3. Molarity Calculate the molarity of NaOH in the

It is the most widely used unit and is denoted solution prepared by dissolving its 4 g in
enough water to form 250 mL of the
N

by M. It is defined as the number of moles of

the solute in 1 litre of the solution. Thus, solution.
Solution
No. of moles of solute
Molarity (M) = Since molarity (M)
Volume of solution in litres
©

Suppose we have 1 M solution of a No. of moles of solute

=
substance, say NaOH and we want to prepare Volume of solution in litres
a 0.2 M solution from it. Mass of NaOH/Molar mass of NaOH
1 M NaOH means 1 mol of NaOH present =
0.250 L
in 1 litre of the solution. For 0.2 M solution we
require 0.2 moles of NaOH in 1 litre solution. 4 g / 40 g 0.1 mol
= =
to

Hence, we have to take 0.2 moles of NaOH 0.250 L 0.250 L

and make the solution to 1 litre. = 0.4 mol L –1

Now how much volume of concentrated = 0.4 M

(1M) NaOH solution be taken which contains Note that molarity of a solution depends
t

0.2 moles of NaOH can be calculated as upon temperature because volume of a

no

follows: solution is temperature dependent.

If 1 mol is present in 1 L or 1000 mL
4. Molality
then 0.2 mol is present in
It is defined as the number of moles of solute
1000 mL present in 1 kg of solvent. It is denoted by m.
× 0.2 mol
1 mol No. of moles of solute
= 200 mL Thus, Molality (m) =
Mass of solvent in kg
 SOME BASIC CONCEPTS OF CHEMISTRY 21

Problem 1.8
The density of 3 M solution of NaCl is No. of moles of solute
Molality =
1.25 g mL–1. Calculate molality of the Mass of solvent in kg
solution.
3 mol
Solution =
1.0745 kg
M = 3 mol L–1
= 2.79 m
Mass of NaCl

ed
in 1 L solution = 3 × 58.5 = 175.5 g Often in a chemistry laboratory, a solution
of a desired concentration is prepared by
Mass of diluting a solution of known higher
1L solution = 1000 × 1.25 = 1250 g concentration. The solution of higher
–1
(since density = 1.25 g mL ) concentration is also known as stock

h
Mass of solution. Note that molality of a solution

pu T
water in solution = 1250 –175.5 does not change with temperature since

is
= 1074.5 mass remains unaffected with
g temperature.
re ER
bl SUMMARY
be C

The study of chemistry is very important as its domain encompasses every sphere of
life. Chemists study the properties and structure of substances and the changes
undergone by them. All substances contain matter which can exist in three states –
N

solid, liquid or gas. The constituent particles are held in different ways in these states of
matter and they exhibit their characteristic properties. Matter can also be classified into
elements, compounds or mixtures. An element contains particles of only one type which
may be atoms or molecules. The compounds are formed where atoms of two or more
©

elements combine in a fixed ratio to each other. Mixtures occur widely and many of the
substances present around us are mixtures.
When the properties of a substance are studied, measurement is inherent. The
quantification of properties requires a system of measurement and units in which the
quantities are to be expressed. Many systems of measurement exist out of which the
English and the Metric Systems are widely used. The scientific community, however, has
agreed to have a uniform and common system throughout the world which is abbreviated
as SI units (International System of Units).
to

Since measurements involve recording of data which are always associated with a
certain amount of uncertainty, the proper handling of data obtained by measuring the
quantities is very important. The measurements of quantities in chemistry are spread
over a wide range of 10–31 to 10+23. Hence, a convenient system of expressing the numbers
in scientific notation is used. The uncertainty is taken care of by specifying the number
t

of significant figures in which the observations are reported. The dimensional analysis
no

helps to express the measured quantities in different systems of units. Hence, it is possible
to interconvert the results from one system of units to another.
The combination of different atoms is governed by basic laws of chemical combination
– these being the Law of Conservation of Mass, Law of Definite Proportions, Law of
Multiple Proportions, Gay Lussac’s Law of Gaseous Volumes and Avogadro Law. All
these laws led to the Dalton’s atomic theory which states that atoms are building
blocks of matter. The atomic mass of an element is expressed relative to 12C isotope of
22 CHEMISTRY

carbon which has an exact value of 12u. Usually, the atomic mass used for an element is
the average atomic mass obtained by taking into account the natural abundance of
different isotopes of that element. The molecular mass of a molecule is obtained by
taking sum of the atomic masses of different atoms present in a molecule. The molecular
formula can be calculated by determining the mass per cent of different elements present
in a compound and its molecular mass.
The number of atoms, molecules or any other particles present in a given system are
expressed in the terms of Avogadro constant (6.022 × 10 23). This is known as 1 mol of

ed
the respective particles or entities.
Chemical reactions represent the chemical changes undergone by different elements
and compounds. A balanced chemical equation provides a lot of information. The
coefficients indicate the molar ratios and the respective number of particles taking part
in a particular reaction. The quantitative study of the reactants required or the products

h
formed is called stoichiometry. Using stoichiometric calculations, the amounts of one
or more reactant(s) required to produce a particular amount of product can be determined

pu T
and vice-versa. The amount of substance present in a given volume of a solution is

is
expressed in number of ways, e.g., mass per cent, mole fraction, molarity and molality.
re ER
bl
EXERCISES
be C

1.1 Calculate the molar mass of the following :

(i) H2O (ii) CO 2 (iii) CH4
1.2 Calculate the mass per cent of different elements present in sodium sulphate
N

(Na2 SO4).
1.3 Determine the empirical formula of an oxide of iron which has 69.9% iron and
30.1% dioxygen by mass.
1.4 Calculate the amount of carbon dioxide that could be produced when
©

(i) 1 mole of carbon is bur nt in air.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
1.5 Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of
0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1.
1.6 Calculate the concentration of nitric acid in moles per litre in a sample which
has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69 %.
to

1.7 How much copper can be obtained from 100 g of copper sulphate (CuSO4) ?
1.8 Determine the molecular formula of an oxide of iron in which the mass per cent
of iron and oxygen are 69.9 and 30.1 respectively.
1.9 Calculate the atomic mass (average) of chlorine using the following data :
t

% Natural Abundance Molar Mass

no

35
Cl 75.77 34.9689
37
Cl 24.23 36.9659
1.10 In three moles of ethane (C2H 6), calculate the following :
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
 SOME BASIC CONCEPTS OF CHEMISTRY 23

(iii) Number of molecules of ethane.

1.11 What is the concentration of sugar (C12H 22O 11) in mol L–1 if its 20 g are dissolved in
enough water to make a final volume up to 2L?
1.12 If the density of methanol is 0.793 kg L–1, what is its volume needed for making
2.5 L of its 0.25 M solution?
1.13 Pressure is determined as force per unit area of the surface. The SI unit of
pressure, pascal is as shown below :
1Pa = 1N m–2

ed
If mass of air at sea level is 1034 g cm–2 , calculate the pressure in pascal.
1.14 What is the SI unit of mass? How is it defined?
1.15 Match the following prefixes with their multiples:
Prefixes Multiples

h
(i) micro 106

pu T
(ii) deca 109

is
(iii) mega 10–6
re ER
(iv) giga
(v) femto 10
10–15

bl
1.16 What do you mean by significant figures ?
1.17 A sample of drinking water was found to be severely contaminated with chloroform,
CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15
ppm (by mass).
be C

(i) Express this in percent by mass.

(ii) Determine the molality of chloroform in the water sample.
1.18 Express the following in the scientific notation:
N

(i) 0.0048
(ii) 234,000
(iii) 8008
©

(iv) 500.0
(v) 6.0012
1.19 How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
to

(iv) 126,000
(v) 500.0
(vi) 2.0034
1.20 Round up the following upto three significant figures:
t

(i) 34.216
no

(ii) 10.4107
(iii) 0.04597
(iv) 2808
1.21 The following data are obtained when dinitrogen and dioxygen react together to
form different compounds :
Mass of dinitrogen Mass of dioxygen
(i) 14 g 16 g
24 CHEMISTRY

(ii) 14 g 32 g
(iii) 28 g 32 g
(iv) 28 g 80 g
(a) Which law of chemical combination is obeyed by the above experimental data?
Give its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = ...................... mm = ...................... pm

ed
(ii) 1 mg = ...................... kg = ...................... ng
(iii) 1 mL = ...................... L = ...................... dm3
1.22 If the speed of light is 3.0 × 108 m s–1 , calculate the distance covered by light in
2.00 ns.
1.23 In a reaction

h
A + B2 → AB2

pu T
Identify the limiting reagent, if any, in the following reaction mixtures.

is
(i) 300 atoms of A + 200 molecules of B
re ER
(ii) 2 mol A + 3 mol B

bl
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
1.24 Dinitrogen and dihydrogen react with each other to produce ammonia according
be C

to the following chemical equation:

N2 (g) + H 2 (g) → 2NH3 (g)
(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts
N

with 1.00 ×103 g of dihydrogen.

(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
1.25 How are 0.50 mol Na2CO3 and 0.50 M Na 2CO3 different?
©

1.26 If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how
many volumes of water vapour would be produced?
1.27 Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg
to

1.28 Which one of the following will have largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
t

(iv) 1 g of Cl2(g)
no

1.29 Calculate the molarity of a solution of ethanol in water in which the mole fraction
of ethanol is 0.040 (assume the density of water to be one).
1.30 What will be the mass of one 12C atom in g ?
1.31 How many significant figures should be present in the answer of the following
calculations?
0.02856 × 298.15 × 0.112
(i) (ii) 5 × 5.364
0.5785
 SOME BASIC CONCEPTS OF CHEMISTRY 25

(iii) 0.0125 + 0.7864 + 0.0215

1.32 Use the data given in the following table to calculate the molar mass of naturally
occuring argon isotopes:
Isotope Isotopic molar mass Abundance
36 –1
Ar 35.96755 g mol 0.33 7%
38
Ar 37.96272 g mol–1 0.06 3%
40 –1
Ar 39.9624 g mol 99.60 0%

ed
1.33 Calculate the number of atoms in each of the following (i) 52 moles of Ar
(ii) 52 u of He (iii) 52 g of He.
1.34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample
of it in oxygen gives 3.38 g carbon dioxide , 0.690 g of water and no other products.
A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g.

h
Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular

pu T
formula.

is
1.35 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to
the reaction, CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O(l)
re ER
What mass of CaCO 3 is required to react completely with 25 mL of 0.75 M HCl ?

bl
1.36 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with
aqueous hydrochloric acid according to the reaction
4 HCl (aq) + MnO2(s) → 2H2 O (l) + MnCl2 (aq) + Cl2 (g)
How many grams of HCl react with 5.0 g of manganese dioxide?
be C
N
t to ©
no
26 CHEMISTRY

UNIT 2

STRUCTURE OF ATOM

The rich diversity of chemical behaviour of different elements

can be traced to the differ ences in the internal structure of
atoms of these elements.

After studying this unit you will be

able to
• know about the discovery of
The existence of atoms has been proposed since the time
electron, proton and neutron and
of early Indian and Greek philosophers (400 B.C.) who
their characteristics;
were of the view that atoms are the fundamental building
• describe Thomson, Rutherford blocks of matter. According to them, the continued
and Bohr atomic models; subdivisions of matter would ultimately yield atoms which
• understand the important would not be further divisible. The word ‘atom’ has been
features of the quantum derived from the Greek word ‘a-tomio’ which means ‘uncut-
mechanical model of atom; able’ or ‘non-divisible’. These earlier ideas were mere
speculations and there was no way to test them
• understand nature of
experimentally. These ideas remained dormant for a very
electromagnetic radiation and
Planck’s quantum theory;
long time and were revived again by scientists in the
nineteenth century.
• explain the photoelectric effect
The atomic theory of matter was first proposed on a
and describe features of atomic
firm scientific basis by John Dalton, a British school
spectra;
teacher in 1808. His theory, called Dalton’s atomic
• state the de Broglie relation and theory, regarded the atom as the ultimate particle of
Heisenberg uncertainty principle; matter (Unit 1).
• define an atomic orbital in terms In this unit we start with the experimental
of quantum numbers; observations made by scientists towards the end of
• state aufbau principle, Pauli
nineteenth and beginning of twentieth century. These
exclusion principle and Hund’s established that atoms can be further divided into sub-
rule of maximum multiplicity; atomic particles, i.e., electrons, protons and neutrons—
a concept very different from that of Dalton. The major
• write the electronic configurations problems before the scientists at that time were:
of atoms.
• to account for the stability of atom after the discovery
of sub-atomic particles,
• to compare the behaviour of one element from other
in terms of both physical and chemical properties,

2015-16
STRUCTURE OF ATOM 27

• to explain the formation of different kinds

of molecules by the combination of
different atoms and,
• to understand the origin and nature of the
characteristics of electromagnetic
radiation absorbed or emitted by atoms.
2.1 SUB-ATOMIC PARTICLES
Dalton’s atomic theory was able to explain
the law of conservation of mass, law of Fig. 2.1(a) A cathode ray discharge tube
constant composition and law of multiple
proportion very successfully. However, it failed stream of particles moving in the tube from
to explain the results of many experiments, the negative electrode (cathode) to the positive
for example, it was known that substances electrode (anode). These were called cathode
like glass or ebonite when rubbed with silk or rays or cathode ray particles. The flow of
fur generate electricity. Many different kinds current from cathode to anode was further
of sub-atomic particles were discovered in the checked by making a hole in the anode and
twentieth century. However, in this section coating the tube behind anode with
we will talk about only two particles, namely phosphorescent material zinc sulphide. When
electron and proton. these rays, after passing through anode, strike
the zinc sulphide coating, a bright spot on
2.1.1 Discovery of Electron the coating is developed(same thing happens
In 1830, Michael Faraday showed that if in a television set) [Fig. 2.1(b)].
electricity is passed through a solution of an
electrolyte, chemical reactions occurred at the
electrodes, which resulted in the liberation
and deposition of matter at the electrodes. He
formulated certain laws which you will study
in class XII. These results suggested the
particulate nature of electricity.
An insight into the structure of atom was
obtained from the experiments on electrical
discharge through gases. Before we discuss
these results we need to keep in mind a basic Fig. 2.1(b) A cathode ray discharge tube with
rule regarding the behaviour of charged perforated anode
particles : “Like charges repel each other and The results of these experiments are
unlike charges attract each other”.
summarised below.
In mid 1850s many scientists mainly (i) The cathode rays start from cathode and
Faraday began to study electrical discharge
move towards the anode.
in partially evacuated tubes, known as
cathode ray discharge tubes. It is depicted (ii) These rays themselves are not visible but
in Fig. 2.1. A cathode ray tube is made of glass their behaviour can be observed with the
containing two thin pieces of metal, called help of certain kind of materials
electrodes, sealed in it. The electrical (fluorescent or phosphorescent) which
discharge through the gases could be glow when hit by them. Television
observed only at very low pressures and at picture tubes are cathode ray tubes and
very high voltages. The pressure of different television pictures result due to
gases could be adjusted by evacuation. When fluorescence on the television screen
sufficiently high voltage is applied across the coated with certain fluorescent or
electrodes, current starts flowing through a phosphorescent materials.

2015-16
28 CHEMISTRY

(iii) In the absence of electrical or magnetic (ii) the mass of the particle — lighter the
field, these rays travel in straight lines particle, greater the deflection.
(Fig. 2.2). (iii) the strength of the electrical or magnetic
(iv) In the presence of electrical or magnetic field — the deflection of electrons from
field, the behaviour of cathode rays are its original path increases with the
similar to that expected from negatively increase in the voltage across the
charged particles, suggesting that the electrodes, or the strength of the
cathode rays consist of negatively magnetic field.
charged particles, called electrons. When only electric field is applied, the
(v) The characteristics of cathode rays electrons deviate from their path and hit the
(electrons) do not depend upon the cathode ray tube at point A. Similarly when
material of electrodes and the nature of only magnetic field is applied, electron strikes
the gas present in the cathode ray tube. the cathode ray tube at point C. By carefully
Thus, we can conclude that electrons are balancing the electrical and magnetic field
basic constituent of all the atoms. strength, it is possible to bring back the
electron to the path followed as in the absence
2.1.2 Charge to Mass Ratio of Electron
of electric or magnetic field and they hit the
In 1897, British physicist J.J. Thomson screen at point B. By carrying out accurate
measured the ratio of electrical charge (e) to measurements on the amount of deflections
the mass of electron (me ) by using cathode observed by the electrons on the electric field
ray tube and applying electrical and magnetic strength or magnetic field strength, Thomson
field perpendicular to each other as well as to
was able to determine the value of e/me as:
the path of electrons (Fig. 2.2). Thomson
argued that the amount of deviation of the e
me = 1.758820 × 10 C kg (2.1)
11 –1
particles from their path in the presence of
electrical or magnetic field depends upon:
Where me is the mass of the electron in kg
(i) the magnitude of the negative charge on
and e is the magnitude of the charge on the
the particle, greater the magnitude of the
charge on the particle, greater is the electron in coulomb (C). Since electrons
interaction with the electric or magnetic are negatively charged, the charge on electron
field and thus greater is the deflection. is –e.

Fig. 2.2 The apparatus to deter mine the charge to the mass ratio of electron

C:\Chemistry XI\Unit-2\Unit-2(2)-Lay-3(reprint).pmd 27.7.6, 16.10.6 (Reprint)

2015-16
STRUCTURE OF ATOM 29

2.1.3 Charge on the Electron

Millikan’s Oil Drop Method
R.A. Millikan (1868-1953) devised a method
In this method, oil droplets in the form of
known as oil drop experiment (1906-14), to mist, pr oduced by the atomiser, were allowed
determine the charge on the electrons. He to enter thr ough a tiny hole in the upper plate
found that the charge on the electron to be of electrical condenser. The downward motion
– 1.6 × 10–19 C. The present accepted value of of these dr oplets was viewed through the
electrical charge is – 1.6022 × 10–19 C. The telescope, equipped with a micrometer eye
mass of the electron (me) was determined by piece. By measuring the rate of fall of these
combining these results with Thomson’s value droplets, Millikan was able to measure the
of e/me ratio. mass of oil dr oplets.The air inside the
chamber was ionized by passing a beam of
e 1.6022 × 10–19 C X-rays through it. The electrical charge on
me = = these oil dr oplets was acquired by collisions
e/ m e 1.758820 × 1011C kg –1
with gaseous ions. The fall of these charged
= 9.1094×10–31 kg (2.2) oil droplets can be retar ded, accelerated or
made stationary depending upon the charge
2.1.4 Discovery of Protons and Neutrons
on the droplets and the polarity and strength
Electrical discharge carried out in the of the voltage applied to the plate. By carefully
modified cathode ray tube led to the discovery measuring the ef fects of electrical field
of particles carrying positive charge, also strength on the motion of oil dr oplets,
known as canal rays. The characteristics of Millikan concluded that the magnitude of
these positively charged particles are listed electrical charge, q, on the dr oplets is always
an integral multiple of the electrical charge,
below.
e, that is, q = n e, where n = 1, 2, 3... .
(i) unlike cathode rays, the positively
charged particles depend upon the
nature of gas present in the cathode ray
tube. These are simply the positively
charged gaseous ions.
(ii) The charge to mass ratio of the particles
is found to depend on the gas from which
these originate.
(iii) Some of the positively charged particles
carry a multiple of the fundamental unit
of electrical charge.
(iv) The behaviour of these particles in the
magnetic or electrical field is opposite to
that observed for electron or cathode Fig. 2.3 The Millikan oil dr op apparatus for
measuring charge ‘e’. In chamber, the
rays.
forces acting on oil drop ar e :
The smallest and lightest positive ion was gravitational, electrostatic due to
obtained from hydrogen and was called electrical field and a viscous drag force
proton. This positively charged particle was when the oil drop is moving.
characterised in 1919. Later, a need was felt
properties of these fundamental particles are
for the presence of electrically neutral particle given in Table 2.1.
as one of the constituent of atom. These
particles were discovered by Chadwick (1932) 2.2 ATOMIC MODELS
by bombarding a thin sheet of beryllium by Observations obtained from the experiments
α-particles. When electrically neutral particles mentioned in the previous sections have
having a mass slightly greater than that of suggested that Dalton’s indivisible atom is
the protons was emitted. He named these composed of sub-atomic particles carrying
particles as neutr ons. The important positive and negative charges. Different

2015-16
30 CHEMISTRY

Table 2.1 Properties of Fundamental Particles

atomic models were proposed to explain the

distributions of these charged particles in an In the later half of the nineteenth century
atom. Although some of these models were different kinds of rays were discovered,
not able to explain the stability of atoms, two besides those mentioned earlier. Wilhalm
of these models, proposed by J. J. Thomson Röentgen (1845-1923) in 1895 showed
and Ernest Rutherford are discussed below. that when electrons strike a material in
2.2.1 Thomson Model of Atom the cathode ray tubes, produce rays
which can cause fluorescence in the
J. J. Thomson, in 1898, proposed that an fluorescent materials placed outside the
atom possesses a spherical shape (radius
cathode ray tubes. Since Röentgen did
approximately 10–10 m) in which the positive
not know the nature of the radiation, he
charge is uniformly distributed. The electrons
named them X-rays and the name is still
are embedded into it in such a manner as to
carried on. It was noticed that X-rays are
give the most stable electrostatic arrangement
(Fig. 2.4). Many different names are given to produced effectively when electrons
this model, for example, plum pudding, strike the dense metal anode, called
raisin pudding or watermelon. This model targets. These are not deflected by the
electric and magnetic fields and have a
very high penetrating power through the
matter and that is the reason that these
rays are used to study the interior of the
objects. These rays are of very short
wavelengths (∼0.1 nm) and possess
electro-magnetic character (Section
2.3.1).
Henri Becqueral (1852-1908)
Fig.2.4 Thomson model of atom observed that there are certain elements
can be visualised as a pudding or watermelon which emit radiation on their own and
of positive charge with plums or seeds named this phenomenon as
(electrons) embedded into it. An important radioactivity and the elements known
feature of this model is that the mass of the as radioactive elements. This field was
atom is assumed to be uniformly distributed developed by Marie Curie, Piere Curie,
over the atom. Although this model was able Rutherford and Fredrick Soddy. It was
to explain the overall neutrality of the atom, observed that three kinds of rays i.e., α,
but was not consistent with the results of later β- and γ-rays are emitted. Rutherford
experiments. Thomson was awarded Nobel found that α-rays consists of high energy
Prize for physics in 1906, for his theoretical particles carrying two units of positive
and experimental investigations on the charge and four unit of atomic mass. He
conduction of electricity by gases.

2015-16
STRUCTURE OF ATOM 31

concluded that α- particles are helium represented in Fig. 2.5. A stream of high
nuclei as when α- particles combined energy α–particles from a radioactive source
with two electrons yielded helium gas. was directed at a thin foil (thickness ∼ 100
β-rays are negatively charged particles nm) of gold metal. The thin gold foil had a
similar to electrons. The γ-rays are high circular fluorescent zinc sulphide screen
energy radiations like X-rays, are neutral around it. Whenever α–particles struck the
in nature and do not consist of particles. screen, a tiny flash of light was produced at
As regards penetrating power, α-particles that point.
are the least, followed by β-rays (100 The results of scattering experiment were
times that of α–particles) and γ-rays quite unexpected. According to Thomson
(1000 times of that α-particles). model of atom, the mass of each gold atom in
the foil should have been spread evenly over
2.2.2 Rutherford’s Nuclear Model of Atom the entire atom, and α– particles had enough
energy to pass directly through such a
Rutherford and his students (Hans Geiger and uniform distribution of mass. It was expected
Ernest Marsden) bombarded very thin gold that the particles would slow down and
foil with α–particles. Rutherford’s famous change directions only by a small angles as
α –particle scattering experiment is
they passed through the foil. It was observed
that :
(i) most of the α– particles passed through
the gold foil undeflected.
(ii) a small fraction of the α–particles was
deflected by small angles.
(iii) a very few α– particles (∼1 in 20,000)
bounced back, that is, were deflected by
nearly 180°.
A. Rutherford’s scattering experiment On the basis of the observations,
Rutherford drew the following conclusions
regarding the structure of atom :
(i) Most of the space in the atom is empty
as most of the α–particles passed
through the foil undeflected.
(ii) A few positively charged α– particles were
deflected. The deflection must be due to
enormous repulsive force showing that
the positive charge of the atom is not
spread throughout the atom as Thomson
had presumed. The positive charge has
to be concentrated in a very small volume
that repelled and deflected the positively
charged α– particles.
B. Schematic molecular view of the gold foil
(iii) Calculations by Rutherford showed that
Fig.2.5 Schematic view of Rutherford’s scattering the volume occupied by the nucleus is
experiment. When a beam of alpha (α) negligibly small as compared to the total
particles is “shot” at a thin gold foil, most volume of the atom. The radius of the
of them pass through without much effect. atom is about 10–10 m, while that of
Some, however, are deflected. nucleus is 10–15 m. One can appreciate

2015-16
32 CHEMISTRY

this difference in size by realising that if earlier protons and neutrons present in the
a cricket ball represents a nucleus, then nucleus are collectively known as nucleons.
the radius of atom would be about 5 km. The total number of nucleons is termed as
On the basis of above observations and mass number (A) of the atom.
conclusions, Rutherfor d proposed the mass number (A) = number of protons (Z)
nuclear model of atom (after the discovery of + number of
protons). According to this model : neutrons (n) (2.4)
(i) The positive charge and most of the mass 2.2.4 Isobars and Isotopes
of the atom was densely concentrated The composition of any atom can be
in extremely small region. This very small represented by using the normal element
portion of the atom was called nucleus symbol (X) with super-script on the left hand
by Rutherford. side as the atomic mass number (A) and
(ii) The nucleus is surrounded by electrons subscript (Z) on the left hand side as the
that move around the nucleus with a atomic number (i.e., AZ X).
very high speed in circular paths called Isobars are the atoms with same mass
orbits. Thus, Rutherford’s model of atom number but different atomic number for
14 14
resembles the solar system in which the example, 6 C and 7 N. On the other hand,
nucleus plays the role of sun and the atoms with identical atomic number but
electrons that of revolving planets. different atomic mass number are known as
Isotopes. In other words (according to
(iii) Electrons and the nucleus are held
equation 2.4), it is evident that difference
together by electrostatic forces of
between the isotopes is due to the presence
attraction.
of different number of neutrons present in
2.2.3 Atomic Number and Mass Number the nucleus. For example, considering of
The presence of positive charge on the hydrogen atom again, 99.985% of hydrogen
nucleus is due to the protons in the nucleus. atoms contain only one proton. This isotope
1
As established earlier, the charge on the is called protium( 1H). Rest of the percentage
proton is equal but opposite to that of of hydrogen atom contains two other isotopes,
electron. The number of protons present in the one containing 1 proton and 1 neutron
2
the nucleus is equal to atomic number (Z ). is called deuterium ( 1 D, 0.015%) and the
For example, the number of protons in the other one possessing 1 proton and 2 neutrons
3
hydrogen nucleus is 1, in sodium atom it is is called tritium ( 1 T ). The latter isotope is
11, therefore their atomic numbers are 1 and found in trace amounts on the earth. Other
11 respectively. In order to keep the electrical examples of commonly occuring isotopes are:
neutrality, the number of electrons in an carbon atoms containing 6, 7 and 8 neutrons
atom is equal to the number of protons besides 6 protons ( 12 13 14
6 C, 6 C, 6 C ); chlorine

(atomic number, Z ). For example, number of atoms containing 18 and 20 neutrons besides
electrons in hydrogen atom and sodium atom 17 protons ( 17
35 37
Cl, 17 Cl ).
are 1 and 11 respectively. Lastly an important point to mention
Atomic number (Z) = number of protons in regarding isotopes is that chemical properties
the nucleus of an atom of atoms are controlled by the number of
electrons, which are determined by the
= number of electrons number of protons in the nucleus. Number of
in a nuetral atom (2.3) neutrons present in the nucleus have very
While the positive charge of the nucleus little effect on the chemical properties of an
is due to protons, the mass of the nucleus, element. Therefore, all the isotopes of a given
due to protons and neutrons. As discussed element show same chemical behaviour.

2015-16
STRUCTURE OF ATOM 33

Problem 2.1 playing the role of the massive sun and the
electrons being similar to the lighter planets.
Calculate the number of protons, Further, the coulomb force (kq1q2/r2 where q1
neutrons and electrons in 80 35 Br
. and q2 are the charges, r is the distance of
Solution separation of the charges and k is the
In this case, 80
35 Br , Z = 35, A = 80, species proportionality constant) between electron and
is neutral the nucleus is mathematically similar to the
Number of protons = number of electrons  m1m 2 
= Z = 35 gravitational force  G. 2  where m1 and
 r 
Number of neutrons = 80 – 35 = 45, m 2 are the masses, r is the distance of
(equation 2.4) separation of the masses and G is the
gravitational constant. When classical
Problem 2.2
mechanics* is applied to the solar system,
The number of electrons, protons and it shows that the planets describe well-defined
neutrons in a species are equal to 18, orbits around the sun. The theory can also
16 and 16 respectively. Assign the proper calculate precisely the planetary orbits and
symbol to the species. these are in agreement with the experimental
Solution measurements. The similarity between the
The atomic number is equal to solar system and nuclear model suggests
number of protons = 16. The element is that electrons should move around the nucleus
sulphur (S). in well defined orbits. However, when a body
is moving in an orbit, it undergoes acceleration
Atomic mass number = number of
protons + number of neutrons (even if the body is moving with a constant
speed in an orbit, it must accelerate because
= 16 + 16 = 32 of changing direction). So an electron in the
Species is not neutral as the number of nuclear model describing planet like orbits is
protons is not equal to electrons. It is under acceleration. According to the
anion (negatively charged) with charge electromagnetic theory of Maxwell, charged
equal to excess electrons = 18 – 16 = 2. particles when accelerated should emit
32 2–
Symbol is 16 S . electromagnetic radiation (This feature does
Note : Before using the notation X , findA
Z
not exist for planets since they are uncharged).
out whether the species is a neutral Therefore, an electron in an orbit will emit
atom, a cation or an anion. If it is a radiation, the energy carried by radiation
neutral atom, equation (2.3) is valid, i.e., comes from electronic motion. The orbit will
number of protons = number of electrons thus continue to shrink. Calculations show
= atomic number. If the species is an ion, that it should take an electron only 10–8 s to
deter mine whether the number of spiral into the nucleus. But this does not
protons are larger (cation, positive ion) happen. Thus, the Rutherford model
or smaller (anion, negative ion) than the cannot explain the stability of an atom.
number of electrons. Number of neutrons If the motion of an electron is described on the
basis of the classical mechanics and
is always given by A–Z, whether the
electromagnetic theory, you may ask that
species is neutral or ion.
since the motion of electrons in orbits is
leading to the instability of the atom, then
2.2.5 Drawbacks of Rutherford Model
why not consider electrons as stationary
Rutherford nuclear model of an atom is like a around the nucleus. If the electrons were
small scale solar system with the nucleus stationary, electrostatic attraction between
* Classical mechanics is a theoretical science based on Newton’s laws of motion. It specifies the laws of motion of macroscopic
objects.

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34 CHEMISTRY

the dense nucleus and the electrons would 19th century when wave nature of light was
pull the electrons toward the nucleus to form established.
a miniature version of Thomson’s model of Maxwell was again the first to reveal that
atom. light waves are associated with oscillating
Another serious drawback of the electric and magnetic character (Fig. 2.6).
Rutherford model is that it says nothing Although electromagnetic wave motion is
about the electronic structure of atoms i.e., complex in nature, we will consider here only
how the electrons are distributed around the a few simple properties.
nucleus and what are the energies of these (i) The oscillating electric and magnetic
electrons. fields produced by oscillating charged
2.3 DEVELOPMENTS LEADING TO THE particles are perpendicular to each other
BOHR’S MODEL OF ATOM and both are perpendicular to the
direction of propagation of the wave.
Historically, results observed from the studies
Simplified picture of electromagnetic
of interactions of radiations with matter have
wave is shown in Fig. 2.6.
provided immense information regarding the
structure of atoms and molecules. Neils Bohr
utilised these results to improve upon the
model proposed by Rutherf o rd. Two
developments played a major role in the
formulation of Bohr’s model of atom. These
were:
(i) Dual character of the electromagnetic
radiation which means that radiations
possess both wave like and particle like
properties, and
(ii) Experimental results regarding atomic
spectra which can be explained only by Fig.2.6 The electric and magnetic field
assuming quantized (Section 2.4) components of an electromagnetic wave.
These components have the same
electronic energy levels in atoms.
wavelength, fr equency, speed and
2.3.1 Wave Nature of Electromagnetic amplitude, but they vibrate in two
Radiation mutually perpendicular planes.
James Maxwell (1870) was the first to give a (ii) Unlike sound waves or water waves,
comprehensive explanation about the electromagnetic waves do not require
interaction between the charged bodies and medium and can move in vacuum.
the behaviour of electrical and magnetic fields (iii) It is now well established that there are
on macroscopic level. He suggested that when many types of electromagnetic
electrically charged particle moves under radiations, which differ from one another
accelaration, alternating electrical and in wavelength (or frequency). These
magnetic fields are produced and constitute what is called
transmitted. These fields are transmitted in electromagnetic spectrum (Fig. 2.7).
the forms of waves called electromagnetic Different regions of the spectrum are
waves or electromagnetic radiation. identified by different names. Some
Light is the form of radiation known from examples are: radio frequency region
early days and speculation about its nature around 106 Hz, used for broadcasting;
dates back to remote ancient times. In earlier microwave region around 1010 Hz used
days (Newton) light was supposed to be made for radar; infrared region around 1013 Hz
of particles (corpuscules). It was only in the used for heating; ultraviolet region

2015-16
STRUCTURE OF ATOM 35

around 1016Hz a component of sun’s at the same speed, i.e., 3.0 × 10 8 m s–1
radiation. The small portion around 1015 (2.997925 × 108 m s –1, to be precise). This is
Hz, is what is ordinarily called visible called speed of light and is given the symbol
light. It is only this part which our eyes ‘c‘. The frequency (ν ), wavelength (λ) and velocity
can see (or detect). Special instruments of light (c) are related by the equation (2.5).
a re required to detect non-visible c=ν λ (2.5)
radiation.
The other commonly used quantity
(iv) Different kinds of units are used to specially in spectroscopy, is the wavenumber
represent electromagnetic radiation.
(ν ). It is defined as the number of wavelengths
These radiations are characterised by the
properties, namely, frequency ( ν ) and per unit length. Its units are reciprocal of
wavelength unit, i.e., m–1. However commonly
wavelength (λ).
used unit is cm–1 (not SI unit).
The SI unit for frequency (ν ) is hertz
(Hz, s–1), after Heinrich Hertz. It is defined as Problem 2.3
the number of waves that pass a given point The Vividh Bharati station of All India
in one second. Radio, Delhi, broadcasts on a frequency
Wavelength should have the units of of 1,368 kHz (kilo hertz). Calculate the
length and as you know that the SI units of wavelength of the electromagnetic
length is meter (m). Since electromagnetic radiation emitted by transmitter. Which
radiation consists of different kinds of waves part of the electromagnetic spectrum
of much smaller wavelengths, smaller units does it belong to?
are used. Fig.2.7 shows various types of
Solution
electro-magnetic radiations which differ from
The wavelength, λ, is equal to c/ν , where
one another in wavelengths and frequencies.
c is the speed of electromagnetic
In vaccum all types of electromagnetic radiation in vacuum and ν is the
radiations, regardless of wavelength, travel

ν
(a)

(b)

Fig. 2.7 (a) The spectrum of electromagnetic radiation. (b) Visible spectrum. The visible region is only
a small part of the entire spectrum .

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36 CHEMISTRY

frequency. Substituting the given values,

we have 1 1
ν= =
λ 5800×10–10 m
c
λ=
ν =1.724×106 m –1

3.00 × 10 8 m s –1 =1.724×104 cm –1
=
1368 kHz (b) Calculation of the frequency (ν )
3.00 × 10 m s
8 –1
c 3 ×108 m s–1
= ν= = = 5.172 ×1014 s– 1
1368 × 10 3 s –1 –10
λ 5800 ×10 m
= 219.3 m
This is a characteristic radiowave
wavelength. 2.3.2 Particle Nature of Electromagnetic
Radiation: Planck’s Quantum
Problem 2.4 Theory
The wavelength range of the visible Some of the experimental phenomenon such
spectrum extends from violet (400 nm) as diffraction* and interference** can be
to red (750 nm). Express these explained by the wave nature of the
wavelengths in frequencies (Hz). electromagnetic radiation. However, following
(1nm = 10–9 m) are some of the observations which could not
Solution be explained with the help of even the
Using equation 2.5, frequency of violet electromagentic theory of 19th century
light physics (known as classical physics):
c 3.00 × 10 8 m s –1 (i) the nature of emission of radiation from
ν = = hot bodies (black -body radiation)
λ 400 × 10– 9 m
(ii) ejection of electrons from metal surface
= 7.50 × 1014 Hz
when radiation strikes it (photoelectric
Frequency of red light effect)
c 3.00 × 108 ms–1 (iii) variation of heat capacity of solids as a
ν= = = 4.00 × 1014 Hz
λ 750 × 10–9m function of temperature
The range of visible spectrum is from (iv) line spectra of atoms with special
4.0 × 1014 to 7.5 × 1014 Hz in terms of reference to hydrogen.
frequency units. It is noteworthy that the first concrete
Problem 2.5 explanation for the phenomenon of the black
Calculate (a) wavenumber and (b) body radiation was given by Max Planck in
frequency of yellow radiation having 1900. This phenomenon is given below:
wavelength 5800 Å. When solids are heated they emit
radiation over a wide range of wavelengths.
Solution For example, when an iron rod is heated in a
(a) Calculation of wavenumber (ν ) furnace, it first turns to dull red and then
progressively becomes more and more red as
λ =5800Å =5800 × 10–8 cm the temperature increases. As this is heated
= 5800 × 10–10 m further, the radiation emitted becomes
white and then becomes blue as the
temperature becomes very high. In terms of
* Diffraction is the bending of wave around an obstacle.
** Interference is the combination of two waves of the same or differ ent frequencies to give a wave whose distribution at
each point in space is the algebraic or vector sum of disturbances at that point resulting from each interfering wave.

2015-16
STRUCTURE OF ATOM 37

frequency, it means that the frequency of to its frequency ( ν ) and is expressed by

emitted radiation goes from a lower frequency equation (2.6).
to a higher frequency as the temperature E = hν (2.6)
increases. The red colour lies in the lower
frequency region while blue colour belongs to The proportionality constant, ‘h’ is known
the higher frequency region of the as Planck’s constant and has the value
electromagnetic spectrum. The ideal body, 6.626×10–34 J s.
which emits and absorbs radiations of all With this theory, Planck was able to
frequencies, is called a black body and the explain the distribution of intensity in the
radiation emitted by such a body is called radiation from black body as a function of
black body radiation. The exact frequency frequency or wavelength at different
distribution of the emitted radiation (i.e., temperatures.
intensity versus frequency curve of the Photoelectric Effect
radiation) from a black body depends only on
In 1887, H. Hertz performed a very interesting
its temperature. At a given temperature,
experiment in which electrons (or electric
intensity of radiation emitted increases with
decrease of wavelength, reaches a maximum current) were ejected when certain metals (for
example potassium, rubidium, caesium etc.)
value at a given wavelength and then starts
were exposed to a beam of light as shown in
decreasing with further decrease of
Fig.2.9. The phenomenon is called
wavelength, as shown in Fig. 2.8.

Fig.2.9 Equipment for studying the photoelectric

effect. Light of a particular frequency strikes
a clean metal surface inside a vacuum
chamber. Electrons are ejected from the
metal and are counted by a detector that
measures their kinetic energy.

Fig. 2.8 Wavelength-intensity relationship Max Planck

(1858 – 1947)
The above experimental results cannot be Max Planck, a German physicist,
explained satisfactorily on the basis of the received his Ph.D in theoretical
wave theory of light. Planck suggested that physics from the University of
atoms and molecules could emit (or absorb) Munich in 1879. In 1888, he was
energy only in discrete quantities and not in
appointed Director of the Institute
a continuous manner, a belief popular at that
of Theoretical Physics at the
time. Planck gave the name quantum to the
University of Berlin. Planck was awarded the Nobel
smallest quantity of energy that can be
Prize in Physics in 1918 for his quantum theory.
emitted or absorbed in the form of
Planck also made significant contributions in
electromagnetic radiation. The energy (E ) of
thermodynamics and other areas of physics.
a quantum of radiation is proportional

2015-16
38 CHEMISTRY

Photoelectric effect. The results observed in (intensity) may shine on a piece of potassium
this experiment were: metal for hours but no photoelectrons are
(i) The electrons are ejected from the metal ejected. But, as soon as even a very weak
surface as soon as the beam of light yellow light (ν = 5.1–5.2 × 1014 Hz) shines on
strikes the surface, i.e., there is no time the potassium metal, the photoelectric effect
lag between the striking of light beam is observed. The threshold frequency (ν0) for
and the ejection of electrons from the potassium metal is 5.0×1014 Hz.
metal surface. Einstein (1905) was able to explain the
(ii) The number of electrons ejected is photoelectric effect using Planck’s quantum
proportional to the intensity or theory of electromagnetic radiation as a
brightness of light. starting point,
(iii) For each metal, there is a characteristic Shining a beam of light on to a metal
minimum frequency,ν0 (also known as surface can, therefore, be viewed as shooting
threshold frequency) below which a beam of particles, the photons. When a
photoelectric effect is not observed. At a photon of sufficient energy strikes an electron
frequency ν > ν 0, the ejected electrons in the atom of the metal, it transfers its energy
come out with certain kinetic energy. instantaneously to the electron during the
The kinetic energies of these electrons collision and the electron is ejected without
increase with the increase of frequency any time lag or delay. Greater the energy
of the light used. possessed by the photon, greater will be
transfer of energy to the electron and greater
All the above results could not be
the kinetic energy of the ejected electron. In
explained on the basis of laws of classical
other words, kinetic energy of the ejected
physics. According to latter, the energy
electron is proportional to the frequency of
content of the beam of light depends upon
the electromagnetic radiation. Since the
the brightness of the light. In other words,
striking photon has energy equal to h ν and
number of electrons ejected and kinetic
the minimum energy required to eject the
energy associated with them should depend
electron is h ν0 (also called work function, W0 ;
on the brightness of light. It has been
Table 2.2), then the difference in energy
observed that though the number of electrons
(h ν – h ν0 ) is transferred as the kinetic energy
ejected does depend upon the brightness of
of the photoelectron. Following the
light, the kinetic energy of the ejected
conservation of energy principle, the kinetic
electrons does not. For example, red light [ν
energy of the ejected electron is given by the
= (4.3 to 4.6) × 1014 Hz] of any brightness
equation 2.7.
Albert Einstein, a Ger m a n 1
bor n American physicist, is hν = hν0 + m e v2 (2.7)
2
regar ded by many as one of
the two great physicists the
where m e is the mass of the electron and v is
world has known (the other the velocity associated with the ejected
is Isaac Newton). His thr ee electron. Lastly, a more intense beam of light
resear ch papers (on special consists of larger number of photons,
relativity, Br ownian motion consequently the number of electrons ejected
and the photoelectric ef fect) Albert Einstein is also larger as compared to that in an
(1879 - 1955)
which he published in 1905, experiment in which a beam of weaker
while he was employed as a technical
intensity of light is employed.
assistant in a Swiss patent of fice in Ber ne
have profoundly influenced the development Dual Behaviour of Electromagnetic
of physics. He r eceived the Nobel Prize in Radiation
Physics in 192 1 for his explanation of the
The particle nature of light posed a
photoelectric effe ct.
dilemma for scientists. On the one hand, it

2015-16
 STRUCTURE OF ATOM 39

Table 2.2 Values of Work Function (W0 ) for a Few Metals

Metal Li Na K Mg Cu Ag

W0 /eV 2.42 2.3 2.25 3.7 4.8 4.3

could explain the black body radiation and the number of photons emitted per second
photoelectric effect satisfactorily but on the by the bulb.
other hand, it was not consistent with the
Solution
known wave behaviour of light which could
account for the phenomena of interference Power of the bulb = 100 watt
–1
and diffraction. The only way to resolve the = 100 J s
dilemma was to accept the idea that light Energy of one photon E = hν = hc/λ
possesses both particle and wave-like
properties, i.e., light has dual behaviour. 6.626 × 10−34 J s × 3 × 108 m s−1
Depending on the experiment, we find that =
400 × 10−9 m
light behaves either as a wave or as a stream
of particles. Whenever radiation interacts with = 4.969 ×10− 19 J
matter, it displays particle like properties in Number of photons emitted
contrast to the wavelike properties
(interference and diffraction), which it 100 J s−1
−19
= 2.012 ×1020 s−1
exhibits when it propagates. This concept was 4.969 ×10 J
totally alien to the way the scientists thought Problem 2.8
about matter and radiation and it took them
a long time to become convinced of its validity. When electromagnetic radiation of
It turns out, as you shall see later, that some wavelength 300 nm falls on the surface
microscopic particles like electrons also of sodium, electrons are emitted with a
exhibit this wave-particle duality. kinetic energy of 1.68 ×105 J mol–1. What
is the minimum energy needed to remove
Problem 2.6 an electron from sodium? What is the
maximum wavelength that will cause a
Calculate energy of one mole of photons photoelectron to be emitted ?
of radiation whose frequency is 5 ×1014
Hz. Solution
Solution The energy (E) of a 300 nm photon is
given by
Energy (E) of one photon is given by the
expression hν = hc / λ
E = hν 6.626 × 10 −34 J s × 3.0 × 108m s–1
=
h = 6.626 ×10–34 J s 300 × 10−9 m
ν = 5×10 14 s–1 (given) = 6.626 × 10-19 J
E = (6.626 ×10–34 J s) × (5 ×1014 s–1) The energy of one mole of photons
–19 23 –1
= 3.313 ×10–19 J = 6.626 ×10 J × 6.022 ×10 mol
Energy of one mole of photons = 3.99 × 105 J mol–1
–19 23 –1
= (3.313 ×10 J) × (6.022 × 10 mol ) The minimum energy needed to remove
= 199.51 kJ mol–1 one mole of electrons from sodium
Problem 2.7 = (3.99 –1.68) 105 J mol –1
5 –1
= 2.31 × 10 J mol
A 100 watt bulb emits monochromatic
light of wavelength 400 nm. Calculate The minimum energy for one electron

C:\Chemistry XI\Unit-2\Unit-2(2)-Lay-3(reprint).pmd 27.7.6, 16.10.6 (Reprint)

2015-16
40 CHEMISTRY

longest wavelength is deviated the least while

2.31 ×105 J mol –1 the violet light, which has shortest wavelength
=
6.022 × 1023 electrons mol –1 is deviated the most. The spectrum of white
= 3.84 ×10− 19 J light, that we can see, ranges from violet at
7.50 × 1014 Hz to red at 4×1014 Hz. Such a
This corresponds to the wavelength spectrum is called continuous spectrum.
hc Continuous because violet merges into blue,
∴λ = blue into green and so on. A similar spectrum
E
is produced when a rainbow forms in the sky.
6.626 × 10 −34 J s × 3.0 × 108m s− 1 Remember that visible light is just a small
=
3.84 ×10− 19 J portion of the electromagnetic radiation
(Fig.2.7). When electromagnetic radiation
= 517 nm
interacts with matter, atoms and molecules
(This corresponds to green light) may absorb energy and reach to a higher
Problem 2.9 energy state. With higher energy, these are in
The threshold frequency ν 0 for a metal an unstable state. For returning to their
is 7.0 ×1014 s–1. Calculate the kinetic normal (more stable, lower energy states)
energy of an electron emitted when energy state, the atoms and molecules emit
radiation of frequency ν =1.0 ×10 15 s –1 radiations in various regions of the
hits the metal. electromagnetic spectrum.
Solution Emission and Absorption Spectra
According to Einstein’s equation The spectrum of radiation emitted by a
2 substance that has absorbed energy is called
Kinetic energy = ½ mev =h(ν – ν0 )
–34 15 –1
an emission spectrum. Atoms, molecules or
= (6.626 ×10 J s) (1.0 × 10 s – 7.0 ions that have absorbed radiation are said to
×1014 s–1) be “excited”. To produce an emission
–34 14 –1
= (6.626 ×10 J s) (10.0 ×10 s – 7.0 spectrum, energy is supplied to a sample by
×1014 s–1) heating it or irradiating it and the wavelength
= (6.626 ×10–34 J s) × (3.0 ×1014 s–1) (or frequency) of the radiation emitted, as the
–19 sample gives up the absorbed energy, is
= 1.988 ×10 J
recorded.
2.3.3 Evidence for the quantized* An absorption spectrum is like the
Electronic Energy Levels: Atomic photographic negative of an emission
spectra spectrum. A continuum of radiation is passed
through a sample which absorbs radiation of
The speed of light depends upon the nature certain wavelengths. The missing wavelength
of the medium through which it passes. As a which corresponds to the radiation absorbed
result, the beam of light is deviated or
by the matter, leave dark spaces in the bright
refracted from its original path as it passes continuous spectrum.
from one medium to another. It is observed
that when a ray of white light is passed The study of emission or absorption
through a prism, the wave with shorter spectra is referred to as spectroscopy. The
wavelength bends more than the one with a spectrum of the visible light, as discussed
longer wavelength. Since ordinary white light above, was continuous as all wavelengths (red
consists of waves with all the wavelengths in to violet) of the visible light are represented
the visible range, a ray of white light is spread in the spectra. The emission spectra of atoms
out into a series of coloured bands called in the gas phase, on the other hand, do not
spectrum. The light of red colour which has show a continuous spread of wavelength from
* The restriction of any pr operty to discrete values is called quantization.

2015-16
STRUCTURE OF ATOM 41

red to violet, rather they emit light only at minerals were analysed by spectroscopic
specific wavelengths with dark spaces methods. The element helium (He) was
between them. Such spectra are called line discovered in the sun by spectroscopic
spectra or atomic spectra because the method.
emitted radiation is identified by the Line Spectrum of Hydrogen
appearance of bright lines in the spectra
When an electric discharge is passed through
(Fig, 2.10)
gaseous hydrogen, the H 2 molecules
Line emission spectra are of great dissociate and the energetically excited
interest in the study of electronic structure. hydrogen atoms produced emit
Each element has a unique line emission electromagnetic radiation of discrete
spectrum. The characteristic lines in atomic frequencies. The hydrogen spectrum consists
spectra can be used in chemical analysis to of several series of lines named after their
identify unknown atoms in the same way as discoverers. Balmer showed in 1885 on the
finger prints are used to identify people. The basis of experimental observations that if
exact matching of lines of the emission spectral lines are expressed in terms of
spectrum of the atoms of a known element wavenumber (ν ), then the visible lines of the
with the lines from an unknown sample hydrogen spectrum obey the following
quickly establishes the identity of the latter, formula :
German chemist, Robert Bunsen (1811-1899)
was one of the first investigators to use line  1 1  –1
spectra to identify elements. ν = 109,677  2 − 2  cm (2.8)
2 n 
Elements like rubidium (Rb), caesium (Cs)
thallium (Tl), indium (In), gallium (Ga) and where n is an integer equal to or greater than
scandium (Sc) were discovered when their 3 (i.e., n = 3,4,5,....)

(a)

(b)

Fig. 2.10 (a) Atomic emission. The light emitted by a sample of excited hydrogen atoms (or any other
element) can be passed through a prism and separated into certain discrete wavelengths. Thus an emission
spectrum, which is a photographic recording of the separated wavelengths is called as line spectrum. Any
sample of reasonable size contains an enormous number of atoms. Although a single atom can be in only
one excited state at a time, the collection of atoms contains all possible excited states. The light emitted as
these atoms fall to lower energy states is responsible for the spectrum. (b) Atomic absorption . When
white light is passed through unexcited atomic hydrogen and then through a slit and prism, the transmitted
light is lacking in intensity at the same wavelengths as are emitted in (a) The recorded absorption spectrum
is also a line spectrum and the photographic negative of the emission spectrum.

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42 CHEMISTRY

The series of lines described by this formula i) The electron in the hydrogen atom can
are called the Balmer series. The Balmer move around the nucleus in a circular
series of lines are the only lines in the hydrogen path of fixed radius and energy. These
spectrum which appear in the visible region of paths are called orbits, stationary states
the electromagnetic spectrum. The Swedish or allowed energy states. These orbits are
spectroscopist, Johannes Rydberg, noted that arranged concentrically around the
all series of lines in the hydrogen spectrum nucleus.
could be described by the following expression ii) The energy of an electron in the orbit does
: not change with time. However, the
 1 1  −1
ν = 109,677  2 − 2  cm (2.9) Table 2.3 The Spectral Lines for Atomic
n
 1 n 2  Hydrogen
where n 1=1,2........
n2 = n 1 + 1, n1 + 2......
The value 109,677 cm –1 is called the
Rydberg constant for hydrogen. The first five
series of lines that correspond to n 1 = 1, 2, 3,
4, 5 are known as Lyman, Balmer, Paschen,
Bracket and Pfund series, respectively,
Table 2.3 shows these series of transitions in
the hydrogen spectrum. Fig 2.11 shows the
L yman, Balmer and Paschen series of
transitions for hydrogen atom.
Of all the elements, hydrogen atom has
the simplest line spectrum. Line spectrum
becomes more and more complex for heavier
atom. There are however certain features
which are common to all line spectra, i.e.,
(i) line spectrum of element is unique and
(ii) there is regularity in the line spectrum of
each element. The questions which arise are
: What are the reasons for these similarities?
Is it something to do with the electronic
structure of atoms? These are the questions
need to be answered. We shall find later that
the answers to these questions provide the
key in understanding electronic structure of
these elements.
2.4 BOHR’S MODEL FOR HYDROGEN
ATOM
Neils Bohr (1913) was the first to explain
quantitatively the general features of
hydrogen atom structure and its spectrum.
Though the theory is not the modern
quantum mechanics, it can still be used to
rationalize many points in the atomic Fig. 2.11 T ransitions of the electron in the
structure and spectra. Bohr’s model for hydrogen atom (The diagram shows
hydrogen atom is based on the following the Lyman, Balmer and Paschen series
postulates: of transitions)

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STRUCTURE OF ATOM 43

electron will move from a lower stationary commonly known as Bohr’s frequency
state to a higher stationary state when rule.
required amount of energy is absorbed iv) The angular momentum of an electron
by the electron or energy is emitted when in a given stationary state can be
electron moves from higher stationary expressed as in equation (2.11)
state to lower stationary state (equation
h
2.16). The energy change does not take m e v r =n. n = 1,2,3..... (2.11)
place in a continuous manner. 2π
Thus an electron can move only in those
Angular Momentum orbits for which its angular momentum is
Just as linear momentum is the product integral multiple of h/2π that is why only
of mass (m) and linear velocity (v), angular certain fixed orbits are allowed.
momentum is the product of moment of The details regarding the derivation of
inertia (I ) and angular velocity (ω). For an energies of the stationary states used by Bohr,
electron of mass me, moving in a circular are quite complicated and will be discussed
path of radius r around the nucleus, in higher classes. However, according to
angular momentum = I × ω Bohr’s theory for hydrogen atom:
Since I = mer 2 , and ω = v/r where v is the a) The stationary states for electron are
linear velocity, numbered n = 1,2,3.......... These integral
∴angular momentum = mer2 × v/r = mevr numbers (Section 2.6.2) are known as
Principal quantum numbers.
iii) The frequency of radiation absorbed or b) The radii of the stationary states are
emitted when transition occurs between expressed as :
two stationary states that differ in energy rn = n 2 a0 (2.12)
by ∆E, is given by : where a 0 = 52,9 pm. Thus the radius of
the first stationary state, called the Bohr
∆E E2 − E1 orbit, is 52.9 pm. Normally the electron
ν = = (2.10)
h h in the hydrogen atom is found in this
Where E1 and E2 are the energies of the orbit (that is n=1). As n increases the
lower and higher allowed energy states value of r will increase. In other words
r espectively. This expression is the electron will be present away from
the nucleus.
c) The most important property associated
Niels Bohr
(1885–1962) with the electron, is the energy of its
stationary state. It is given by the
Niels B o hr, a Danish
expression.
physicist received his Ph.D.
f rom the University of  1 
En = − R H  2  n = 1,2,3.... (2.13)
Copenhagen in 1911. He n 
then spent a year with J.J. where RH is called Rydberg constant and its
Thomson and Er nest Rutherfor d in England. value is 2.18×10–18 J. The energy of the lowest
In 1913, he retur ned to Copenhagen wher e state, also called as the ground state, is
he remained for the rest of his life. In 1920
1
he was named Director of the Institute of E1 = –2.18×10 –18 ( 2 ) = –2.18×10
–18 J. The
1
theor etical Physics. After first World War,
Bohr worked energetically for peaceful uses energy of the stationary state for n = 2, will
of atomic energy. He received the first Atom s 1
be : E 2 = –2.18×10–18J ( 2 )= –0.545×10
–18 J.
for Peace award in 1957. Bohr was awar ded 2
the Nobel Prize in Physics in 1922. Fig. 2.11 depicts the energies of different

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44 CHEMISTRY

stationary states or energy levels of hydrogen where Z is the atomic number and has values
atom. This representation is called an energy 2, 3 for the helium and lithium atoms
level diagram. respectively. From the above equations, it is
evident that the value of energy becomes more
What does the negative electronic negative and that of radius becomes smaller
energy (E n) for hydrogen atom mean?
with increase of Z . This means that electron
The energy of the electron in a hydrogen will be tightly bound to the nucleus.
atom has a negative sign for all possible
orbits (eq. 2.13). What does this negative e) It is also possible to calculate the
sign convey? This negative sign means that velocities of electrons moving in these
the energy of the electron in the atom is orbits. Although the precise equation is
lower than the energy of a free electron at not given here, qualitatively the
rest. A free electron at rest is an electron magnitude of velocity of electro n
that is infinitely far away from the nucleus increases with increase of positive charge
and is assigned the energy value of zero. on the nucleus and decreases with
Mathematically, this corresponds to increase of principal quantum number.
setting n equal to infinity in the equation
2.4.1 Explanation of Line Spectrum of
(2.13) so that E∞=0. As the electron gets
Hydrogen
closer to the nucleus (as n decreases), En
becomes larger in absolute value and more Line spectrum observed in case of hydrogen
and more negative. The most negative atom, as mentioned in section 2.3.3, can be
energy value is given by n=1 which explained quantitatively using Bohr’s model.
corresponds to the most stable orbit. We According to assumption 2, radiation (energy)
call this the ground state. is absorbed if the electron moves from the
orbit of smaller Principal quantum number
When the electron is free from the influence to the orbit of higher Principal quantum
of nucleus, the energy is taken as zero. The number, whereas the radiation (energy) is
electron in this situation is associated with the emitted if the electron moves from higher orbit
stationary state of Principal Quantum number to lower orbit. The energy gap between the
= n = ∞ and is called as ionized hydrogen atom. two orbits is given by equation (2.16)
When the electron is attracted by the nucleus
∆E = Ef – Ei (2.16)
and is present in orbit n, the energy is emitted
and its energy is lowered. That is the reason Combining equations (2.13) and (2.16)
for the presence of negative sign in equation  R   R 
(2.13) and depicts its stability relative to the ∆ E =  − H2  −  − H2  (where n and n
reference state of zero energy and n = ∞ .  n f   ni  i f

d) Bohr’s theory can also be applied to the stand for initial orbit and final orbits)
ions containing only one electron, similar 1 1   1 1 
to that present in hydrogen atom. For ∆E = R H  2 − 2  = 2.18 ×10 −18 J  2 − 2 
n
 i n f  n
 i n f 
example, He+ Li2+ , Be3+ and so on. The
energies of the stationary states (2,17)
associated with these kinds of ions (also The frequency (ν ) associated with the
known as hydrogen like species) are given absorption and emission of the photon can
by the expression. be evaluated by using equation (2.18)
 Z2 
E n = − 2.18 ×10 −18  2  J (2.14) ∆ E RH  1 1 
n  ν = =  − 
and radii by the expression h h  n 2i n f2 

52.9 (n 2 ) 2.18 × 10 − 18 J  1 1 
rn = pm (2.15) =  −  (2.18)
Z 6.626 × 10 −34 J s  n i2 n f2 

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STRUCTURE OF ATOM 45

 1 1  It is an emission energy
= 3.29 ×1015  2 − 2  Hz (2.19) The frequency of the photon (taking
 ni n f  energy in terms of magnitude) is given
and in terms of wavenumbers (ν ) by

∆E
ν = RH  1 − 1  ν =
ν=
c hc  n 2 n 2 
i f
(2.20) h

4.58×10– 19 J
3.29 ×10 s  1
15 −1
1  =
= − 2 6.626×10–34 J s
8 −s  2
3 × 10 m s  n i n f 
= 6.91×1014 Hz
 1 1  c 3.0 × 108 m s −1
= 1.09677 × 107  2 − 2  m − 1 (2.21) λ= = = 434 nm
 ni nf  ν 6.91× 1014 Hz
In case of absorption spectrum, nf > ni and Problem 2.11
the term in the parenthesis is positive and
Calculate the energy associated with the
energy is absorbed. On the other hand in case first orbit of He+ . What is the radius of
of emission spectrum n i > nf , ∆ E is negative this orbit?
and energy is released.
The expression (2.17) is similar to that Solution
used by Rydberg (2.9) derived empirically (2.18 × 10 −18 J)Z 2
using the experimental data available at that En = − atom–1
n2
time. Further, each spectral line, whether in
absorption or emission spectrum, can be For He +, n = 1, Z = 2
associated to the particular transition in (2.18 ×10−18 J)(22 )
hydrogen atom. In case of large number of E1 = − 2 = −8.72 × 10−18 J
1
hydrogen atoms, different possible transitions
can be observed and thus leading to large The radius of the orbit is given by
number of spectral lines. The brightness or equation (2.15)
intensity of spectral lines depends upon the (0.0529 nm)n 2
number of photons of same wavelength or rn =
Z
frequency absorbed or emitted.
Since n = 1, and Z = 2
Problem 2.10 (0.0529 nm)12
rn = = 0.02645 nm
What are the frequency and wavelength 2
of a photon emitted during a transition
from n = 5 state to the n = 2 state in the 2.4.2 Limitations of Bohr’s Model
hydrogen atom? Bohr’s model of the hydrogen atom was no
Solution doubt an improvement over Rutherford’s
Since ni = 5 and nf = 2, this transition nuclear model, as it could account for the
gives rise to a spectral line in the visible stability and line spectra of hydrogen atom
region of the Balmer series. Fr o m and hydrogen like ions (for example, He+, Li2+ ,
equation (2.17) Be3+ , and so on). However, Bohr’s model was
too simple to account for the following points.
1 1
∆E = 2.18 × 10 −18 J  2 − 2  i) It fails to account for the finer details
5 2  (doublet, that is two closely spaced lines)
−19
= − 4.58 ×10 J of the hydrogen atom spectrum observed

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46 CHEMISTRY

by using sophisticated spectroscopic

Louis de Broglie (1892 – 1987)
techniques. This model is also unable to
explain the spectrum of atoms other than Louis de Broglie, a French
hydrogen, for example, helium atom which physicist, studied history as an
possesses only two electrons. Further, undergraduate in the early
Bohr’s theory was also unable to explain 1910’s. His interest turned to
the splitting of spectral lines in the science as a result of his
presence of magnetic field (Zeeman effect) assignment to radio
or an electric field (Stark effect). communications in World War I.
ii) It could not explain the ability of atoms to He received his Dr. Sc. from the University of
form molecules by chemical bonds. Paris in 1924. He was professor of theoretical
physics at the University of Paris from 1932 untill
In other words, taking into account the
his retirement in 1962. He was awarded the
points mentioned above, one needs a better
Nobel Prize in Physics in 1929.
theory which can explain the salient features
of the structure of complex atoms.
which is based on the wavelike behaviour of
2.5 TOWARDS QUANTUM MECHANICAL electrons just as an ordinary microscope
MODEL OF THE ATOM utilises the wave nature of light. An electron
In view of the shortcoming of the Bohr’s model, microscope is a powerful tool in modern
attempts were made to develop a more scientific research because it achieves a
suitable and general model for atoms. Two magnification of about 15 million times.
important developments which contributed It needs to be noted that according to de
significantly in the formulation of such a Broglie, every object in motion has a wave
model were : character. The wavelengths associated with
1. Dual behaviour of matter, ordinary objects are so short (because of their
large masses) that their wave properties
2. Heisenberg uncertainty principle. cannot be detected. The wavelengths
2.5.1 Dual Behaviour of Matter associated with electrons and other subatomic
particles (with very small mass) can however
The French physicist, de Broglie in 1924 be detected experimentally. Results obtained
proposed that matter, like radiation, should from the following problems prove these
also exhibit dual behaviour i.e., both particle points qualitatively.
and wavelike properties. This means that just
as the photon has momentum as well as Problem 2.12
wavelength, electrons should also have
What will be the wavelength of a ball of
momentum as well as wavelength, de Broglie, mass 0.1 kg moving with a velocity of 10
from this analogy, gave the following relation
m s–1 ?
between wavelength (λ) and momentum (p) of
a material particle. Solution
According to de Brogile equation (2.22)
h h
λ= = (2.22)
mv p h (6.626 × 10−34 Js)
λ= =
where m is the mass of the particle, v its mv (0.1 kg)(10 m s−1 )
velocity and p its momentum. de Broglie’s = 6.626×10–34 m (J = kg m2 s–2)
prediction was confirmed experimentally
Problem 2.13
when it was found that an electron beam
undergoes diffraction, a phenomenon The mass of an electron is 9.1×10–31 kg.
characteristic of waves. This fact has been put If its K.E. is 3.0×10–25 J, calculate its
to use in making an electron microscope, wavelength.

2015-16
STRUCTURE OF ATOM 47

Solution the other hand, if the velocity of the electron is

known precisely (∆(vx ) is small), then the
Since K. E. = ½ mv 2
position of the electron will be uncertain
1/2 1/ 2
 2 × 3.0 × 10 −25 kg m 2 s −2  (∆x will be large). Thus, if we carry out some
2K.E. 
v =   =  physical measurements on the electron’s
 m   9.1 × 10 −31 kg  position or velocity, the outcome will always
= 812 m s–1 depict a fuzzy or blur picture.
The uncertainty principle can be best
h 6.626 × 10−34 Js understood with the help of an example.
λ= =
m v (9.1× 10−31kg)(812 m s−1 ) Suppose you are asked to measure the
thickness of a sheet of paper with an
= 8967 × 10–10 m = 896.7 nm unmarked metrestick. Obviously, the results
Problem 2.14 obtained would be extremely inaccurate and
Calculate the mass of a photon with meaningless, In order to obtain any accuracy,
wavelength 3.6 Å. you should use an instrument graduated in
Solution units smaller than the thickness of a sheet of
the paper. Analogously, in order to determine
λ = 3.6 Å = 3.6 × 10− 10 m the position of an electron, we must use a
Velocity of photon = velocity of light meterstick calibrated in units of smaller than
the dimensions of electron (keep in mind that
h 6.626 × 10−34 Js an electron is considered as a point charge
m= =
λν (3.6 × 10–10 m)(3 × 108 m s−1) and is therefore, dimensionless). To observe
an electron, we can illuminate it with “light”
= 6.135 × 10–29 kg or electromagnetic radiation. The “light” used
must have a wavelength smaller than the
2.5.2 Heisenberg’s Uncertainty Principle dimensions of an electron. The high
Werner Heisenberg a German physicist in  h
1927, stated uncertainty principle which is momentum photons of such light  p = 
 λ
the consequence of dual behaviour of matter would change the energy of electrons by
and radiation. It states that it is impossible collisions. In this process we, no doubt, would
to determine simultaneously, the exact be able to calculate the position of the
position and exact momentum (or velocity) electron, but we would know very little about
of an electron. the velocity of the electron after the collision.
Mathematically, it can be given as in
equation (2.23). Significance of Uncertainty Principle

h One of the important implications of the

∆x × ∆p x ≥ (2.23) Heisenberg Uncertainty Principle is that it
4π
rules out existence of definite paths or
h trajectories of electrons and other similar
or ∆ x × ∆(m v x ) ≥
4π particles. The trajectory of an object is
determined by its location and velocity at
h
or ∆x × ∆ v x ≥ various moments. If we know where a body is
4 πm at a particular instant and if we also know its
where ∆x is the uncertainty in position and velocity and the forces acting on it at that
∆p x ( or ∆v x) is the uncertainty in momentum instant, we can tell where the body would be
(or velocity) of the particle. If the position of sometime later. We, therefore, conclude that
the electron is known with high degree of the position of an object and its velocity fix
accuracy (∆x is small), then the velocity of the its trajectory. Since for a sub-atomic object
electron will be uncertain [∆(vx ) is large]. On such as an electro n, it is not possible

2015-16
48 CHEMISTRY

simultaneously to determine the position and uncertainty of only 10–8 m, then the
velocity at any given instant to an arbitrary uncertainty ∆v in velocity would be
degree of precision, it is not possible to talk
10 –4 m 2s –1
of the trajectory of an electron. –8
≈ 10 4m s-1
The effect of Heisenberg Uncertainty 10 m
Principle is significant only for motion of which is so large that the classical picture of
microscopic objects and is negligible for electrons moving in Bohr’s orbits (fixed)
that of macroscopic objects. This can be cannot hold good. It, therefore, means that
seen from the following examples. the precise statements of the position and
If uncertainty principle is applied to an momentum of electrons have to be
object of mass, say about a milligram (10–6 replaced by the statements of probability,
kg), then that the electron has at a given position
and momentum. This is what happens in
h
∆v.∆x = the quantum mechanical model of atom.
4π.m
6.626×10–34 J s Problem 2.15
= ≈ 10 –28 m2 s–1 A microscope using suitable photons is
4×3.1416×10–6 kg
employed to locate an electron in an
The value of ∆v∆x obtained is extremely atom within a distance of 0.1 Å. What is
small and is insignificant. Therefore, one may the uncertainty involved in the
say that in dealing with milligram-sized or measurement of its velocity?
heavier objects, the associated Solution
uncertainties are hardly of any real
consequence. h h
∆x ∆p = or ∆x m ∆v =
In the case of a microscopic object like an 4π 4π
electron on the other hand. ∆v.∆x obtained is
much larger and such uncertainties are of h
∆v =
real consequence. For example, for an electron 4π∆x m
whose mass is 9.11×10–31 kg., according to
Heisenberg uncertainty principle 6.626×10 –34 Js
∆v =
h 4×3.14×0.1×10 –10 m × 9.11 × 10 –31 kg
∆v.∆x =
4π m = 0.579×107 m s–1 (1J = 1 kg m2 s–2)
= 5.79×106 m s –1
–34
6.626×10 Js
= Problem 2.16
4 × 3.1416×9.11×10–31 kg A golf ball has a mass of 40g, and a speed
= 10 –4 m 2 s–1 of 45 m/s. If the speed can be measured
within accuracy of 2%, calculate the
It, therefore, means that if one tries to find uncertainty in the position.
the exact location of the electron, say to an

Wer ner Heisenberg (1901 – 1976) Wer ner Heisenberg (1901 – 1976) received his Ph.D. in
physics from the University of Munich in 1923. He then spent a year working with Max
Born at Gottingen and three years with Niels Bohr in Copenhagen. He was professor of
physics at the University of Leipzig from 1927 to 1941. During World War II, Heisenberg
was in charge of German research on the atomic bomb. After the war he was named
director of Max Planck Institute for physics in Gottingen. He was also accomplished
mountain climber. Heisenberg was awarded the Nobel Prize in Physics in 1932.

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STRUCTURE OF ATOM 49

Solution Erwin Schrödinger, an

The uncertainty in the speed is 2%, i.e., Austrian physicist
received his Ph.D. in
2
45 × = 0.9 m s–1 . theoretical physics fr om
100 the University of Vienna
Using the equation (2.22) in 1910. In 1927
h Schrödinger succeeded
∆x = Max Planck at the
4π m ∆ v
University of Berlin at
6.626 × 10 –34 Js Planck’s request. In 1933, Erwin Schrödinger
= (1887-1961)
4×3.14×40g × 10–3 kg g –1 (0.9 m s –1 ) Schrödinger left Berlin
= 1.46×10 –33 m because of his opposition to Hitler and Nazi
policies and retur ned to Austria in 1936. After
This is nearly ~ 1018 times smaller than
the invasion of Austria by Ger many,
the diameter of a typical atomic nucleus.
Schrödinger was forcibly removed fr om his
As mentioned earlier for large particles,
the uncertainty principle sets no professorship. He then moved to Dublin, Ir eland
meaningful limit to the precision of where he remained for seventeen years.
measurements. Schrödinger shared the Nobel Prize for Physics
with P.A.M. Dirac in 1933.

Reasons for the Failure of the Bohr Model of all macroscopic objects such as a falling
One can now understand the reasons for the stone, orbiting planets etc., which have
failure of the Bohr model. In Bohr model, an essentially a particle-like behaviour as shown
electron is regarded as a charged particle in the previous section. However it fails when
moving in well defined circular orbits about applied to microscopic objects like electrons,
the nucleus. The wave character of the atoms, molecules etc. This is mainly because
electron is not considered in Bohr model. of the fact that classical mechanics ignores
Further, an orbit is a clearly defined path and the concept of dual behaviour of matter
this path can completely be defined only if especially for sub-atomic particles and the
both the position and the velocity of the uncertainty principle. The branch of science
electron are known exactly at the same time. that takes into account this dual behaviour
This is not possible according to the of matter is called quantum mechanics.
Heisenberg uncertainty principle. Bohr model
Quantum mechanics is a theoretical
of the hydrogen atom, therefore, not only
science that deals with the study of the
ignores dual behaviour of matter but also motions of the microscopic objects that have
contradicts Heisenberg uncertainty principle.
both observable wave like and particle like
In view of these inherent weaknesses in the
properties. It specifies the laws of motion that
Bohr model, there was no point in extending these objects obey. When quantum
Bohr model to other atoms. In fact an insight
mechanics is applied to macroscopic objects
into the structure of the atom was needed
(for which wave like properties are
which could account for wave-particle duality insignificant) the results are the same as
of matter and be consistent with Heisenberg
those from the classical mechanics.
uncertainty principle. This came with the
advent of quantum mechanics. Quantum mechanics was developed
independently in 1926 by Werner Heisenberg
2.6 QUANTUM MECHANICAL MODEL OF and Erwin Schrödinger. Here, however, we
ATOM shall be discussing the quantum mechanics
Classical mechanics, based on Newton’s laws which is based on the ideas of wave motion.
of motion, successfully describes the motion The fundamental equation of quantum

2015-16
50 CHEMISTRY

mechanics was developed by Schrödinger and probability of finding an electron at a point

it won him the Nobel Prize in Physics in 1933. within an atom is proportional to the |ψ |2 at
This equation which incorporates wave- that point. The quantum mechanical results
particle duality of matter as proposed by de of the hydrogen atom successfully predict all
Broglie is quite complex and knowledge of aspects of the hydrogen atom spectrum
higher mathematics is needed to solve it. You including some phenomena that could not
will learn its solutions for different systems in be explained by the Bohr model.
higher classes. Application of Schrödinger equation to
For a system (such as an atom or a multi-electron atoms presents a difficulty: the
molecule whose energy does not change with Schrödinger equation cannot be solved
time) the Schrödinger equation is written as exactly for a multi-electron atom. This
difficulty can be overcome by using
H ψ = E ψ where H is a mathematical approximate methods. Such calculations
operator called Hamiltonian. Schrödinger with the aid of modern computers show that
gave a recipe of constructing this operator orbitals in atoms other than hydrogen do not
from the expression for the total energy of differ in any radical way from the hydrogen
the system. The total energy of the system orbitals discussed above. The principal
takes into account the kinetic energies of all difference lies in the consequence of
the sub-atomic particles (electrons, nuclei), increased nuclear charge. Because of this all
attractive potential between the electrons and the orbitals are somewhat contracted.
nuclei and repulsive potential among the Further, as you shall see later (in subsections
electrons and nuclei individually. Solution of 2.6.3 and 2.6.4), unlike orbitals of hydrogen
this equation gives E and ψ. or hydrogen like species, whose energies
Hydrogen Atom and the Schrödinger depend only on the quantum number n, the
Equation energies of the orbitals in multi-electron
When Schrödinger equation is solved for atoms depend on quantum numbers n and l.
hydrogen atom, the solution gives the
possible energy levels the electron can occupy Important Features of the Quantum
and the corresponding wave function(s) (ψ) Mechanical Model of Atom
of the electron associated with each energy Quantum mechanical model of atom is the
level. These quantized energy states and picture of the structure of the atom, which
corresponding wave functions which are e m e rges fr om the application of the
Schrödinger equation to atoms. The following
characterized by a set of three quantum
are the important features of the quantum-
numbers (principal quantum number n,
mechanical model of atom:
azimuthal quantum number l and
magnetic quantum number m l ) arise as a 1. The ener gy of electrons in atoms is
natural consequence in the solution of the quantized (i.e., can only have certain
specific values), for example when
Schrödinger equation. When an electron is electrons are bound to the nucleus in
in any energy state, the wave function atoms.
corresponding to that energy state contains
2. The existence of quantized electronic
all information about the electron. The wave energy levels is a direct r esult of the wave
function is a mathematical function whose like properties of electrons and are allowed
value depends upon the coordinates of the solutions of Schrödinger wave equation.
electron in the atom and does not carry any 3. Both the exact position and exact velocity
physical meaning. Such wave functions of of an electron in an atom cannot be
hydrogen or hydrogen like species with one determined simultaneously (Heisenberg
electron are called atomic orbitals. Such uncertainty principle). The path of an
wave functions pertaining to one-electron electron in an atom therefore, can never
species are called one-electron systems. The be deter mined or known accurately. That

2015-16
STRUCTURE OF ATOM 51

is why, as you shall see later on, one The principal quantum number determines the
talks of only probability of finding the size and to large extent the energy of the
electron at different points in an atom. orbital. For hydrogen atom and hydrogen like
4. An atomic orbital is the wave function species (He+, Li 2+, .... etc.) energy and size of
ψ for an electron in an atom. Whenever the orbital depends only on ‘n’.
an electron is described by a wave The principal quantum number also
function, we say that the electron occupies identifies the shell. With the increase in the
that orbital. Since many such wave
value of ‘n’, the number of allowed orbital
functions ar e possible for an electron,
there are many atomic orbitals in an atom.
increases and are given by ‘n2 ’ All the
These “one electron orbital wave functions” orbitals of a given value of ‘n’ constitute a
or orbitals form the basis of the electronic single shell of atom and are represented by
structure of atoms. In each orbital, the the following letters
electron has a definite energy. An orbital n = 1 2 3 4 ............
cannot contain more than two electrons. Shell = K L M N ............
In a multi-electron atom, the electrons ar e
Size of an orbital increases with increase
filled in various orbitals in the order of
of principal quantum number ‘n’. In other
incr easing energy. For each electron of a
multi-electron atom, there shall, therefore,
words the electron will be located away from
be an orbital wave function characteristic the nucleus. Since energy is required in
of the orbital it occupies. All the shifting away the negatively charged electron
information about the electron in an atom from the positively charged nucleus, the
is stored in its orbital wave function ψ and energy of the orbital will increase with
quantum mechanics makes it possible to increase of n.
extract this infor mation out of ψ .
Azimuthal quantum number. ‘l’ is also
5. The probability of finding an electron at a known as orbital angular momentum or
point within an atom is proportional to the subsidiary quantum number. It defines the
square of the orbital wave function i.e.,
2 2 three dimensional shape of the orbital. For a
| ψ | at that point. |ψ | is known as
pr obability density and is always
given value of n, l can have n values ranging
positive. From the value of |ψ | at
2 from 0 to n – 1, that is, for a given value of n,
differ ent points within an atom, it is the possible value of l are : l = 0, 1, 2, ..........
possible to predict the region ar ound (n–1)
the nucleus where electron will most For example, when n = 1, value of l is only
probably be found. 0. For n = 2, the possible value of l can be 0
and 1. For n = 3, the possible l values are 0,
2.6.1 Orbitals and Quantum Numbers 1 and 2.
A large number of orbitals are possible in an Each shell consists of one or more sub-
atom. Qualitatively these orbitals can be shells or sub-levels. The number of sub-
distinguished by their size, shape and shells in a principal shell is equal to the value
orientation. An orbital of smaller size means of n. For example in the first shell (n = 1),
there is more chance of finding the electron there is only one sub-shell which corresponds
near the nucleus. Similarly shape and to l = 0. There are two sub-shells (l = 0, 1) in
orientation mean that there is more the second shell (n = 2), three (l = 0, 1, 2) in
probability of finding the electron along third shell (n = 3) and so on. Each sub-shell
certain directions than along others. Atomic is assigned an azimuthal quantum number
orbitals are precisely distinguished by what (l ). Sub-shells corresponding to different
are known as quantum numbers. Each values of l are represented by the following
orbital is designated by three quantum symbols.
numbers labelled as n, l and m l. Value for l : 0 1 2 3 4 5 ............
The principal quantum number ‘n’ is a notation for s p d f g h ............
positive integer with value of n = 1,2,3....... . sub-shell

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52 CHEMISTRY

Table 2.4 shows the permissible values of Thus for l = 0, the only permitted value of
‘l ’ for a given principal quantum number and ml = 0, [2(0)+1 = 1, one s orbital]. For l = 1, ml
the corresponding sub-shell notation. can be –1, 0 and +1 [2(1)+1 = 3, three p
Table 2.4 Subshell Notations orbitals]. For l = 2, ml = –2, –1, 0, +1 and +2,
[2(2)+1 = 5, five d orbitals]. It should be noted
that the values of m l are derived from l and
that the value of l are derived from n.
Each orbital in an atom, therefore, is
defined by a set of values for n, l and ml. An
orbital described by the quantum numbers
n = 2, l = 1, ml = 0 is an orbital in the p sub-
shell of the second shell. The following chart
gives the relation between the sub-shell and
the number of orbitals associated with it.

Value of l 0 1 2 3 4 5
Subshell notation s p d f g h
number of orbitals 1 3 5 7 9 11

Electron spin ‘s’ : The three quantum

numbers labelling an atomic orbital can be
used equally well to define its energy, shape
and orientation. But all these quantum
numbers are not enough to explain the line
Magnetic orbital quantum number. ‘ml ’ spectra observed in the case of multi-electron
gives information about the spatial atoms, that is, some of the lines actually occur
orientation of the orbital with respect to in doublets (two lines closely spaced), triplets
standard set of co-ordinate axis. For any (three lines, closely spaced) etc. This suggests
sub-shell (defined by ‘l’ value) 2l+1 values the presence of a few more energy levels than
of ml are possible and these values are given predicted by the three quantum numbers.
by :
In 1925, George Uhlenbeck and Samuel
ml = – l, – (l –1), – (l –2)... 0,1... (l –2), (l –1), l Goudsmit proposed the presence of the fourth

Orbit, orbital and its importance

Orbit and orbital are not synonymous. An orbit, as proposed by Bohr, is a circular path around
the nucleus in which an electron moves. A precise description of this path of the electron is
impossible according to Heisenberg uncertainty principle. Bohr orbits, therefore, have no real
meaning and their existence can never be demonstrated experimentally. An atomic orbital, on the
other hand, is a quantum mechanical concept and refers to the one electron wave function ψ in
an atom. It is characterized by three quantum numbers (n, l and m l ) and its value depends upon
the coordinates of the electron. ψ has, by itself, no physical meaning. It is the square of the wave
2 2
function i.e., |ψ| which has a physical meaning. |ψ| at any point in an atom gives the value of
probability density at that point. Probability density (| ψ| 2) is the probability per unit volume and
2
the product of |ψ| and a small volume (called a volume element) yields the probability of finding
2
the electron in that volume (the reason for specifying a small volume element is that |ψ| varies
from one r egion to another in space but its value can be assumed to be constant within a small
volume element). The total probability of finding the electron in a given volume can then be
calculated by the sum of all the products of |ψ| 2 and the corresponding volume elements. It is
thus possible to get the probable distribution of an electron in an orbital.

2015-16
STRUCTURE OF ATOM 53

quantum number known as the electron Solution

spin quantum number (m s ). An electron For n = 3, the possible values of l are 0, 1
spins around its own axis, much in a similar and 2. Thus there is one 3s orbital
way as earth spins around its own axis while (n = 3, l = 0 and ml = 0); there are three
revolving around the sun. In other words, an 3p orbitals (n = 3, l = 1 and ml = –1, 0,
electron has, besides charge and mass, +1); there are five 3d orbitals (n = 3, l = 2
intrinsic spin angular quantum number. Spin and m l = –2, –1, 0, +1+, +2).
angular momentum of the electron — a vector
Therefore, the total number of orbitals
quantity, can have two orientations relative
is 1+3+5 = 9
to the chosen axis. These two orientations are
distinguished by the spin quantum numbers The same value can also be obtained by
ms which can take the values of +½ or –½. using the relation; number of orbitals
These are called the two spin states of the = n2, i.e. 32 = 9.
electron and are normally represented by two Problem 2.18
arrows, ↑ (spin up) and ↓ (spin down). Two Using s, p, d, f notations, describe the
electrons that have different m s values (one orbital with the following quantum
+½ and the other –½) are said to have numbers
opposite spins. An orbital cannot hold more (a) n = 2, l = 1, (b) n = 4, l = 0, (c) n = 5,
than two electrons and these two electrons l = 3, (d) n = 3, l = 2
should have opposite spins.
To sum up, the four quantum numbers Solution
provide the following information : n l orbital
i) n defines the shell, determines the size a) 2 1 2p
of the orbital and also to a large extent b) 4 0 4s
the energy of the orbital. c) 5 3 5f
ii) There are n subshells in the n th shell. l d) 3 2 3d
identifies the subshell and determines the
shape of the orbital (see section 2.6.2). 2.6.2 Shapes of Atomic Orbitals
There are (2l+1) orbitals of each type in a The orbital wave function or ψ for an electron
subshell, that is, one s orbital (l = 0), three in an atom has no physical meaning. It is
p orbitals (l = 1) and five d orbitals (l = 2) simply a mathematical function of the
per subshell. To some extent l also coordinates of the electron. However, for
determines the energy of the orbital in a different orbitals the plots of corresponding
multi-electron atom. wave functions as a function of r (the distance
from the nucleus) are different. Fig. 2.12(a),
iii) m l designates the orientation of the
(page 54) gives such plots for 1s (n = 1, l = 0)
orbital. For a given value of l, ml has (2l+1)
and 2s (n = 2, l = 0) orbitals.
values, the same as the number of orbitals
per subshell. It means that the number According to the German physicist, Max
of orbitals is equal to the number of ways Born, the square of the wave function
in which they are oriented. (i.e.,ψ 2 ) at a point gives the probability
density of the electron at that point. The
iv) ms refers to orientation of the spin of the 2
variation of ψ as a function of r for 1s and
electron. 2s orbitals is given in Fig. 2.12(b), (page 54).
Problem 2.17 Here again, you may note that the curves for
1s and 2s orbitals are different.
What is the total number of orbitals
associated with the principal quantum It may be noted that for 1s orbital the
number n = 3 ? probability density is maximum at the
nucleus and it decreases sharply as we move

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54 CHEMISTRY

Fig. 2.13 (a) Probability density plots of 1s and

2s atomic orbitals. The density of the
dots represents the probability density
Fig. 2.12 The plots of (a) the orbital wave of finding the electron in that region.
function ψ (r ); (b) the variation of (b) Boundary surface diagram for 1s
probability density ψ (r) as a function
2
and 2s orbitals.
of distance r of the electron from the
nucleus for 1s and 2s orbitals.
probability density |ψ |2 is constant. In
principle many such boundary surfaces may
away from it. On the other hand, for 2s orbital be possible. However, for a given orbital, only
the probability density first decreases sharply that boundary surface diagram of constant
to zero and again starts increasing. After probability density* is taken to be good
reaching a small maxima it decreases again representation of the shape of the orbital
and approaches zero as the value of r which encloses a region or volume in which
increases further. The region where this the probability of finding the electron is very
probability density function reduces to zero high, say, 90%. The boundary surface
is called nodal surfaces or simply nodes. In diagram for 1s and 2s orbitals are given in
general, it has been found that ns-orbital has Fig. 2.13(b). One may ask a question : Why
(n – 1) nodes, that is, number of nodes do we not draw a boundary surface diagram,
increases with increase of principal quantum which bounds a region in which the
number n. In other words, number of nodes probability of finding the electron is, 100 %?
for 2s orbital is one, two for 3s and so on. The answer to this question is that the
These probability density variation can be probability density |ψ| 2 has always some
visualised in terms of charge cloud diagrams value, howsoever small it may be, at any finite
[Fig. 2.13(a)]. In these diagrams, the density distance from the nucleus. It is therefore, not
of the dots in a region represents electron possible to draw a boundary surface diagram
probability density in that region. of a rigid size in which the probability of
Boundary surface diagrams of constant finding the electron is 100%. Boundary
probability density for different orbitals give surface diagram for a s orbital is actually a
a fairly good representation of the shapes of sphere centred on the nucleus. In two
the orbitals. In this representation, a dimensions, this sphere looks like a circle. It
boundary surface or contour surface is drawn encloses a region in which probability of
in space for an orbital on which the value of finding the electron is about 90%.
2
* If pr obability density | ψ| is constant on a given surface, |ψ| is also constant over the surface. The boundary surface
for |ψ |2 and | ψ| are identical.

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STRUCTURE OF ATOM 55

Thus we see that 1s and 2s orbitals are because there are three possible values of m l,
spherical in shape. In reality all the s-orbitals there are, therefore, three p orbitals whose
are spherically symmetric, that is, the axes are mutually perpendicular. Like s
probability of finding the electron at a given orbitals, p orbitals increase in size and energy
distance is equal in all the directions. It is with increase in the principal quantum
also observed that the size of the s orbital number and hence the order of the energy
increases with increase in n, that is, 4s > 3s and size of various p orbitals is 4p > 3p > 2p.
> 2s > 1s and the electron is located further Further, like s orbitals, the probability density
away from the nucleus as the principal functions for p-orbital also pass through value
quantum number increases. zero, besides at zero and infinite distance, as
Boundary surface diagrams for three 2p the distance from the nucleus increases. The
orbitals (l = 1) are shown in Fig. 2.14. In these number of nodes are given by the n –2, that
diagrams, the nucleus is at the origin. Here, is number of radial node is 1 for 3p orbital,
unlike s-orbitals, the boundary surface two for 4p orbital and so on.
diagrams are not spherical. Instead each For l = 2, the orbital is known as d-orbital
p orbital consists of two sections called lobes and the minimum value of principal quantum
that are on either side of the plane that passes number (n) has to be 3. as the value of l cannot
through the nucleus. The probability density be greater than n–1. There are five ml values
function is zero on the plane where the two (–2, –1, 0, +1 and +2) for l = 2 and thus there
lobes touch each other. The size, shape and are five d orbitals. The boundary surface
energy of the three orbitals are identical. They diagram of d orbitals are shown in Fig. 2.15,
differ however, in the way the lobes are (page 56).
oriented. Since the lobes may be considered The five d-orbitals are designated as dxy,
to lie along the x, y or z axis, they are given dyz, dxz, dx 2–y 2 and dz2. The shapes of the first
the designations 2p x, 2py , and 2p z. It should four d-orbitals are similar to each other, where
be understood, however, that there is no as that of the fifth one, dz2, is different from
simple relation between the values of ml (–1, others, but all five 3d orbitals are equivalent
0 and +1) and the x, y and z directions. For in energy. The d orbitals for which n is greater
our purpose, it is sufficient to remember that, than 3 (4d, 5d...) also have shapes similar to
3d orbital, but differ in energy and size.
Besides the radial nodes (i.e., probability
density function is zero), the probability
density functions for the np and nd orbitals
are zero at the plane (s), passing through the
nucleus (origin). For example, in case of pz
orbital, xy-plane is a nodal plane, in case of
dxy orbital, there are two nodal planes passing
through the origin and bisecting the xy plane
containing z-axis. These are called angular
nodes and number of angular nodes are given
by ‘l’, i.e., one angular node for p orbitals, two
angular nodes for ‘d’ orbitals and so on. The
total number of nodes are given by (n–1),
i.e., sum of l angular nodes and (n – l – 1)
radial nodes.
2.6.3 Energies of Orbitals
Fig. 2.14 Boundary surface diagrams of the The energy of an electron in a hydrogen atom
three 2p orbitals. is determined solely by the principal quantum

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56 CHEMISTRY

Fig. 2.16 Energy level diagrams for the few

electronic shells of (a) hydrogen atom
and (b) multi-electronic atoms. Note that
orbitals for the same value of principal
quantum number, have the same
energies even for different azimuthal
quantum number for hydrogen atom.
In case of multi-electron atoms, orbitals
with same principal quantum number
possess different energies for different
azimuthal quantum numbers.
stable condition and is called the ground state
and an electron residing in this orbital is most
strongly held by the nucleus. An electron in
the 2s, 2p or higher orbitals in a hydrogen atom
Fig. 2.15 Boundary surface diagrams of the five is in excited state.
3d orbitals.
The energy of an electron in a multi-
electron atom, unlike that of the hydrogen
number. Thus the energy of the orbitals
atom, depends not only on its principal
increases as follows :
quantum number (shell), but also on its
1s < 2s = 2p < 3s = 3p = 3d <4s = 4p = 4d azimuthal quantum number (subshell). That
= 4f < (2.23) is, for a given principal quantum number, s,
and is depicted in Fig. 2.16. Although the p, d, f ... all have different energies. The main
shapes of 2s and 2p orbitals are different, an reason for having different energies of the
electron has the same energy when it is in the subshells is the mutual repulsion among the
2s orbital as when it is present in 2p orbital. electrons in a multi-electron atoms. The only
The orbitals having the same energy are called electrical interaction present in hydrogen
degenerate. The 1s orbital in a hydrogen atom is the attraction between the negatively
atom, as said earlier, corresponds to the most charged electron and the positively charged

2015-16
STRUCTURE OF ATOM 57

nucleus. In multi-electron atoms, besides the electron. In other words, for a given shell
presence of attraction between the electron and (principal quantum number), the Z eff
nucleus, there are repulsion terms between experienced by the electron decreases with
every electron and other electrons present in increase of azimuthal quantum number (l),
the atom. Thus the stability of an electron in that is, the s orbital electron will be more tightly
multi-electron atom is because total attractive bound to the nucleus than p orbital electron
interactions are more than the repulsive which in turn will be better tightly bound than
interactions. In general, the repulsive the d orbital electron. The energy of electrons
interaction of the electrons in the outer shell in s orbital will be lower (more negative) than
with the electrons in the inner shell are more that of p orbital electron which will have less
important. On the other hand, the attractive energy than that of d orbital electron and so
interactions of an electron increases with on. Since the extent of shielding from the
increase of positive charge (Ze) on the nucleus. nucleus is different for electrons in different
Due to the presence of electrons in the inner orbitals, it leads to the splitting of energy levels
shells, the electron in the outer shell will not within the same shell (or same principal
experience the full positive charge of the quantum number), that is, energy of electron
nucleus (Ze). The effect will be lowered due to in an orbital, as mentioned earlier, depends
the partial screening of positive charge on the upon the values of n and l. Mathematically,
the dependence of energies of the orbitals on n
nucleus by the inner shell electrons. This is
and l are quite complicated but one simple rule
known as the shielding of the outer shell
is that, the lower the value of (n + l) for an
electrons from the nucleus by the inner
orbital, the lower is its energy. If two
shell electrons, and the net positive charge
orbitals have the same value of (n + l), the
experienced by the outer electrons is known
orbital with lower value of n will have the
as effective nuclear charge (Zeff e). Despite lower energy. The Table 2.5 (page 58)
the shielding of the outer electrons from the illustrates the (n + l ) rule and Fig. 2.16 depicts
nucleus by the inner shell electrons, the the energy levels of multi-electrons atoms. It
attractive force experienced by the outer shell may be noted that different subshells of a
electrons increases with increase of nuclear particular shell have different energies in case
charge. In other words, the energy of of multi–electrons atoms. However, in
interaction between, the nucleus and electron hydrogen atom, these have the same energy.
(that is orbital energy) decreases (that is Lastly it may be mentioned here that energies
more negative) with the increase of atomic of the orbitals in the same subshell decrease
number (Z ). with increase in the atomic number (Zeff ).
Both the attractive and repulsive For example, energy of 2s orbital of hydrogen
interactions depend upon the shell and shape atom is greater than that of 2s orbital of lithium
of the orbital in which the electron is present. and that of lithium is greater than that of
For example electrons present in spherical sodium and so on, that is, E2s (H) > E2s(Li) >
shaped, s orbital shields the outer electrons E2s(Na) > E2s (K).
from the nucleus more effectively as compared 2.6.4 Filling of Orbitals in Atom
to electrons present in p orbital. Similarly
electrons present in p orbitals shield the outer The filling of electrons into the orbitals of
electrons from the nucleus more than the different atoms takes place according to the
electrons present in d orbitals, even though all aufbau principle which is based on the Pauli’s
these orbitals are present in the same shell. exclusion principle, the Hund’s rule of
Further within a shell, due to spherical shape maximum multiplicity and the relative
of s orbital, the s orbital electron spends more energies of the orbitals.
time close to the nucleus in comparison to p Aufbau Principle
orbital electron which spends more time in the The word ‘aufbau’ in German means ‘building
vicinity of nucleus in comparison to d orbital up’. The building up of orbitals means the

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58 CHEMISTRY

Table 2.5 Arrangement of Orbitals with

Increasing Ener gy on the Basis of
(n+l ) Rule

Fig.2.17 Order of filling of orbitals

Pauli Exclusion Principle

The number of electrons to be filled in various
filling up of orbitals with electrons. The orbitals is restricted by the exclusion
principle states : In the ground state of the principle, given by the Austrian scientist
atoms, the orbitals are filled in order of Wolfgang Pauli (1926). According to this
their increasing energies. In other words, principle : No two electrons in an atom can
electrons first occupy the lowest energy orbital have the same set of four quantum
available to them and enter into higher energy numbers. Pauli exclusion principle can also
orbitals only after the lower energy orbitals be stated as : “Only two electrons may exist
are filled. in the same orbital and these electrons
The order in which the energies of the must have opposite spin.” This means that
orbitals increase and hence the order in which the two electrons can have the same value of
the orbitals are filled is as follows : three quantum numbers n, l and ml, but must
have the opposite spin quantum number. The
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f,
restriction imposed by Pauli’s exclusion
5d, 6p, 7s...
principle on the number of electrons in an
The order may be remembered by using orbital helps in calculating the capacity of
the method given in Fig. 2.17. Starting from electrons to be present in any subshell. For
the top, the direction of the arrows gives the example, subshell 1s comprises of one orbital
order of filling of orbitals, that is starting from and thus the maximum number of electrons
right top to bottom left. present in 1s subshell can be two, in p and d

2015-16
STRUCTURE OF ATOM 59

subshells, the maximum number of electrons arrow (↑) a positive spin or an arrow (↓) a
can be 6 and 10 and so on. This can be negative spin. The advantage of second notation
summed up as : the maximum number of over the first is that it represents all the four
electrons in the shell with principal quantum numbers.
quantum number n is equal to 2n2. The hydrogen atom has only one electron
Hund’s Rule of Maximum Multiplicity which goes in the orbital with the lowest
This rule deals with the filling of electrons energy, namely 1s. The electronic
configuration of the hydrogen atom is 1s 1
into the orbitals belonging to the same
meaning that it has one electron in the 1s
subshell (that is, orbitals of equal energy,
orbital. The second electron in helium (He)
called degenerate orbitals). It states : pairing
can also occupy the 1s orbital. Its
of electrons in the orbitals belonging to
the same subshell (p, d or f) does not take configuration is, therefore, 1s2. As mentioned
above, the two electrons differ from each other
place until each orbital belonging to that
with opposite spin, as can be seen from the
subshell has got one electron each i.e., it
orbital diagram.
is singly occupied.
Since there are three p, five d and seven f
orbitals, therefore, the pairing of electrons will
start in the p, d and f orbitals with the entry The third electron of lithium (Li) is not
of 4th, 6th and 8th electron, respectively. It allowed in the 1s orbital because of Pauli
has been observed that half filled and fully exclusion principle. It, therefore, takes the
filled degenerate set of orbitals acquire extra next available choice, namely the 2s orbital.
stability due to their symmetry (see Section, The electronic configuration of Li is 1s22s1.
2.6.7). The 2s orbital can accommodate one more
electron. The configuration of beryllium (Be)
2.6.5 Electronic Configuration of Atoms atom is, therefore, 1s 2 2s2 (see Table 2.6, page
The distribution of electrons into orbitals of 62 for the electronic configurations of
an atom is called its electronic elements).
configuration. If one keeps in mind the basic In the next six elements-boro n
rules which govern the filling of different (B, 1s 22s 22p1), carbon (C, 1s 22s 22p2), nitrogen
atomic orbitals, the electronic configurations (N, 1s22s22p 3), oxygen (O, 1s 22s 22p 4), fluorine
of different atoms can be written very easily. (F, 1s 22s 22p5) and neon (Ne, 1s22s22p6), the
The electronic configuration of different 2p orbitals get progressively filled. This
atoms can be represented in two ways. For process is completed with the neon atom. The
example : orbital picture of these elements can be
(i) s a pbd c ...... notation represented as follows :
(ii) Orbital diagram

s p d
In the first notation, the subshell is
represented by the respective letter symbol
and the number of electrons present in the
subshell is depicted, as the super script, like
a, b, c, ... etc. The similar subshell represented
for different shells is differentiated by writing
the principal quantum number before the
respective subshell. In the second notation
each orbital of the subshell is represented by
a box and the electron is represented by an

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60 CHEMISTRY

The electronic configuration of the elements turn of the 6s orbital. In caesium (Cs) and the
sodium (Na, 1s 22s 22p 6 3s 1 ) to argon barium (Ba), this orbital contains one and two
(Ar,1s22s22p63s 23p6), follow exactly the same electrons, respectively. Then from lanthanum
pattern as the elements from lithium to neon (La) to mercury (Hg), the filling up of electrons
with the difference that the 3s and 3p orbitals takes place in 4f and 5d orbitals. After this,
are getting filled now. This process can be filling of 6p, then 7s and finally 5f and 6d
simplified if we represent the total number of orbitals takes place. The elements after
electrons in the first two shells by the name uranium (U) are all short-lived and all of them
of element neon (Ne). The electr onic are produced artificially. The electronic
configuration of the elements from sodium to configurations of the known elements (as
argon can be written as (Na, [Ne]3s 1) to (Ar, determined by spectroscopic methods) are
[Ne] 3s 23p6). The electrons in the completely tabulated in Table 2.6.
filled shells are known as core electrons and One may ask what is the utility of knowing
the electrons that are added to the electronic the electron configuration? The modern
shell with the highest principal quantum approach to the chemistry, infact, depends
number are called valence electrons. For
almost entirely on electronic distribution to
example, the electrons in Ne are the core understand and explain chemical behaviour.
electrons and the electrons from Na to Ar are
For example, questions like why two or more
the valence electrons. In potassium (K) and
atoms combine to form molecules, why some
calcium (Ca), the 4s orbital, being lower in
elements are metals while others are non-
energy than the 3d orbitals, is occupied by
metals, why elements like helium and argon
one and two electrons respectively.
are not reactive but elements like the halogens
A new pattern is followed beginning with are reactive, find simple explanation from the
scandium (Sc). The 3d orbital, being lower in electronic configuration. These questions have
energy than the 4p orbital, is filled first. no answer in the Daltonian model of atom. A
Consequently, in the next ten elements, detailed understanding of the electronic
scandium (Sc), titanium (Ti), vanadium (V), structure of atom is, therefore, very essential
chromium (Cr), manganese (Mn), iron (Fe), for getting an insight into the various aspects
cobalt (Co), nickel (Ni), copper (Cu) and zinc of modern chemical knowledge.
(Zn), the five 3d orbitals are progressively
occupied. We may be puzzled by the fact that 2.6.6 Stability of Completely Filled and
chromium and copper have five and ten Half Filled Subshells
electrons in 3d orbitals rather than four and The ground state electronic configuration of
nine as their position would have indicated the atom of an element always corresponds
with two-electrons in the 4s orbital. The to the state of the lowest total electronic
reason is that fully filled orbitals and half- energy. The electronic configurations of most
filled orbitals have extra stability (that is, of the atoms follow the basic rules given in
lower energy). Thus p3, p6, d5, d10,f 7, f 14 etc. Section 2.6.5. However, in certain elements
configurations, which are either half-filled or such as Cu, or Cr, where the two subshells
fully filled, are more stable. Chromium and (4s and 3d) differ slightly in their energies,
copper therefore adopt the d5 and d10 an electron shifts from a subshell of lower
configuration (Section 2.6.7)[caution: energy (4s) to a subshell of higher energy (3d),
exceptions do exist] provided such a shift results in all orbitals of
With the saturation of the 3d orbitals, the the subshell of higher energy getting either
filling of the 4p orbital starts at gallium (Ga) completely filled or half filled. The valence
and is complete at krypton (Kr). In the next electronic configurations of Cr and Cu,
eighteen elements from rubidium (Rb) to xenon therefore, are 3d5 4s1 and 3d10 4s 1 respectively
(Xe), the pattern of filling the 5s, 4d and 5p and not 3d4 4s2 and 3d9 4s 2. It has been found
orbitals are similar to that of 4s, 3d and 4p that there is extra stability associated with
orbitals as discussed above. Then comes the these electronic configurations.

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STRUCTURE OF ATOM 61

Causes of Stability of Completely Filled and Half Filled Sub-shells

The completely filled and completely half

filled sub-shells are stable due to the
following reasons:
1. Symmetrical distribution of
electrons: It is well known that symmetry
leads to stability. The completely filled or
half filled subshells have symmetrical
distribution of electrons in them and are
therefore more stable. Electrons in the
same subshell (her e 3d) have equal energy
but dif fer ent spatial distribution.
Consequently, their shielding of one-
another is relatively small and the
electrons are more strongly attracted by
the nucleus.
2. Exchange Energy : The stabilizing
e f fect arises whenever two or more
electrons with the same spin are present
in the degenerate orbitals of a subshell.
These electrons tend to exchange their
positions and the energy released due to
this exchange is called exchange energy.
The number of exchanges that can take
place is maximum when the subshell is
either half filled or completely filled (Fig.
2.18). As a result the exchange energy is
maximum and so is the stability.
You may note that the exchange
ener gy is at the basis of Hund’s rule that
electrons which enter orbitals of equal
energy have parallel spins as far as
possible. In other wor ds, the extra
stability of half-filled and completely filled
subshell is due to: (i) relatively small
shielding, (ii) smaller coulombic repulsion
ener gy, and (iii) larger exchange energy.
Details about the exchange energy will be Fig. 2.18 Possible exchange for a d5
dealt with in higher classes. configuration

2015-16
62 CHEMISTRY

Table 2.6 Electronic Configurations of the Elements

* Elements with exceptional electronic configurations

2015-16
STRUCTURE OF ATOM 63

** Elements with atomic number 112 and above have been reported but not yet fully authenticated and named.

2015-16
64 CHEMISTRY

SUMMARY

Atoms are the building blocks of elements. They are the smallest parts of an element that
chemically react. The first atomic theory, proposed by John Dalton in 1808, regarded atom
as the ultimate indivisible particle of matter. Towards the end of the nineteenth century, it
was proved experimentally that atoms are divisible and consist of three fundamental particles:
electrons , protons and neutrons. The discovery of sub-atomic particles led to the proposal
of various atomic models to explain the structure of atom.
Thomson in 1898 proposed that an atom consists of uniform sphere of positive electricity
with electrons embedded into it. This model in which mass of the atom is considered to be
evenly spread over the atom was pr oved wrong by Rutherford’s famous alpha-particle
scattering experiment in 1909. Rutherford concluded that atom is made of a tiny positively
charged nucleus, at its centre with electrons revolving around it in circular orbits .
Rutherford model, which resembles the solar system, was no doubt an improvement over
Thomson model but it could not account for the stability of the atom i.e., why the electron
does not fall into the nucleus. Further, it was also silent about the electronic structure of
atoms i.e., about the distribution and relative energies of electrons ar ound the nucleus.
The difficulties of the Rutherford model were over come by Niels Bohr in 1913 in his model
of the hydrogen atom. Bohr postulated that electron moves ar ound the nucleus in circular
orbits. Only certain orbits can exist and each orbit corresponds to a specific energy. Bohr
calculated the energy of electron in various orbits and for each orbit predicted the distance
between the electron and nucleus. Bohr model, though of fering a satisfactory model for
explaining the spectra of the hydrogen atom, could not explain the spectra of multi-electron
atoms. The reason for this was soon discovered. In Bohr model, an electron is regarded as
a charged particle moving in a well defined circular orbit about the nucleus. The wave
character of the electron is ignor ed in Bohr’s theory. An orbit is a clearly defined path and
this path can completely be defined only if both the exact position and the exact velocity of
the electron at the same time are known. This is not possible according to the Heisenberg
uncertainty principle. Bohr model of the hydrogen atom, therefore, not only ignores the
dual behaviour of electron but also contradicts Heisenberg uncertainty principle.
Erwin Schrödinger, in 1926, proposed an equation called Schrödinger equation to describe
the electron distributions in space and the allowed energy levels in atoms. This equation
incorporates de Broglie’s concept of wave-particle duality and is consistent with Heisenberg
uncertainty principle. When Schrödinger equation is solved for the electron in a hydrogen
atom, the solution gives the possible energy states the electron can occupy [and the
corresponding wave function(s) (ψ) (which in fact are the mathematical functions) of the
electron associated with each energy state]. These quantized energy states and corresponding
wave functions which are characterized by a set of three quantum numbers (principal
quantum number n, azimuthal quantum number l and magnetic quantum number ml )
arise as a natural consequence in the solution of the Schrödinger equation. The restrictions
on the values of these three quantum numbers also come naturally from this solution. The
quantum mechanical model of the hydrogen atom successfully predicts all aspects of the
hydrogen atom spectrum including some phenomena that could not be explained by the
Bohr model.
According to the quantum mechanical model of the atom, the electron distribution of an
atom containing a number of electrons is divided into shells. The shells, in tur n, are thought
to consist of one or more subshells and subshells are assumed to be composed of one or
more orbitals, which the electrons occupy. While for hydrogen and hydrogen like systems
+ 2+
(such as He , Li etc.) all the orbitals within a given shell have same energy, the energy of
the orbitals in a multi-electron atom depends upon the values of n and l: The lower the
value of (n + l ) for an orbital, the lower is its energy. If two orbitals have the same (n + l )
value, the orbital with lower value of n has the lower energy. In an atom many such orbitals
are possible and electrons are filled in those orbitals in order of increasing energy in

2015-16
STRUCTURE OF ATOM 65

accordance with Pauli exclusion principle (no two electrons in an atom can have the same
set of four quantum numbers) and Hund’s rule of maximum multiplicity (pairing of
electrons in the orbitals belonging to the same subshell does not take place until each
orbital belonging to that subshell has got one electron each, i.e., is singly occupied). This
forms the basis of the electronic structure of atoms.

EXERCISES

2.1 (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
2.2 (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C.
–27
(Assume that mass of a neutron = 1.675 × 10 kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at
STP.
Will the answer change if the temperature and pressure are changed ?
2.3 How many neutr ons and protons are ther e in the following nuclei ?
13 16 24 56 88
6 C, 8 O , 1 2 Mg, 26 Fe, 3 8 Sr

2.4 Write the complete symbol for the atom with the given atomic number (Z) and
atomic mass (A)
(i) Z = 17 , A = 35.
(ii) Z = 92 , A = 233.
(iii) Z = 4 , A = 9.
2.5 Yellow light emitted fr om a sodium lamp has a wavelength (λ) of 580 nm. Calculate
the fr equency (ν) and wavenumber ( ν ) of the yellow light.
2.6 Find energy of each of the photons which
(i) correspond to light of fr equency 3×1015 Hz.
(ii) have wavelength of 0.50 Å.
2.7 Calculate the wavelength, frequency and wavenumber of a light wave whose period
–10
is 2.0 × 10 s.
2.8 What is the number of photons of light with a wavelength of 4000 pm that provide
1J of energy?
–7
2.9 A photon of wavelength 4 × 10 m strikes on metal sur face, the work function of
the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic
ener gy of the emission, and (iii) the velocity of the photoelectron
–19
(1 eV= 1.6020 × 10 J).
2.10 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the
–1
sodium atom. Calculate the ionisation energy of sodium in kJ mol .
2.11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm.
Calculate the rate of emission of quanta per second.
2.12 Electrons are emitted with zer o velocity from a metal surface when it is exposed to
radiation of wavelength 6800 Å. Calculate threshold frequency (ν0 ) and work function
(W0 ) of the metal.
2.13 What is the wavelength of light emitted when the electron in a hydrogen atom
undergoes transition fr om an energy level with n = 4 to an energy level with n = 2?

2015-16
66 CHEMISTRY

2.14 How much energy is requir ed to ionise a H atom if the electron occupies n = 5
orbit? Compare your answer with the ionization enthalpy of H atom ( energy required
to remove the electr on from n =1 orbit).
2.15 What is the maximum number of emission lines when the excited electron of a H
atom in n = 6 drops to the ground state?
2.16 (i) The ener gy associated with the first orbit in the hydrogen atom is
–18 –1
–2.18 × 10 J atom . What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
2.17 Calculate the wavenumber for the longest wavelength transition in the Balmer
series of atomic hydrogen.
2.18 What is the energy in joules, requir ed to shift the electron of the hydrogen atom
from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the
light emitted when the electron retur ns to the ground state? The ground state
–11
electron energy is –2.18 × 10 ergs.
2.19 The electron energy in hydrogen atom is given by En = (–2.18 × 10–18 )/n2 J. Calculate
the energy r equired to remove an electron completely from the n = 2 orbit. What is
the longest wavelength of light in cm that can be used to cause this transition?
2.20 Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m s–1.
–31 –25
2.21 The mass of an electron is 9.1 × 10 kg. If its K.E. is 3.0 × 10 J, calculate its
wavelength.
2.22 Which of the following are isoelectronic species i.e., those having the same number
of electrons?
Na+ , K+, Mg2+, Ca2+, S2–, Ar.
– + 2–
2.23 (i) Write the electronic configurations of the following ions: (a) H (b) Na (c) O
(d) F–
(ii) What are the atomic numbers of elements whose outermost electrons are
1 3 5
represented by (a) 3 s (b) 2p and (c) 3p ?
(iii) Which atoms are indicated by the following configurations ?
1 2 3 2 1
(a) [He] 2s (b) [Ne] 3s 3p (c) [Ar] 4s 3d .
2.24 What is the lowest value of n that allows g orbitals to exist?
2.25 An electron is in one of the 3d orbitals. Give the possible values of n, l and m l for
this electron.
2.26 An atom of an element contains 29 electrons and 35 neutr ons. Deduce (i) the
number of protons and (ii) the electronic configuration of the element.
2.27 Give the number of electrons in the species H2 , H2 and O2
+ +

2.28 (i) An atomic orbital has n = 3. What are the possible values of l and ml ?
(ii) List the quantum numbers (m l and l ) of electrons for 3 d orbital.
(iii) Which of the following orbitals are possible?
1p, 2 s, 2p and 3f
2.29 Using s, p, d notations, describe the orbital with the following quantum numbers.
(a) n=1, l=0; (b) n = 3; l=1 (c) n = 4; l =2; (d) n=4; l=3.
2.30 Explain, giving reasons, which of the following sets of quantum numbers are not
possible.
(a) n = 0, l = 0, m l = 0, ms = + ½
(b) n = 1, l = 0, m l = 0, ms = – ½
(c) n = 1, l = 1, m l = 0, ms = + ½
(d) n = 2, l = 1, m l = 0, ms = – ½

2015-16
STRUCTURE OF ATOM 67

(e) n = 3, l = 3, ml = –3, ms = + ½
(f) n = 3, l = 1, ml = 0, ms = + ½
2.31 How many electrons in an atom may have the following quantum numbers?
(a) n = 4, ms = – ½ (b) n = 3, l = 0
2.32 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral
multiple of the de Broglie wavelength associated with the electron revolving around
the orbit.
2.33 What transition in the hydrogen spectrum would have the same wavelength as the
Balmer transition n = 4 to n = 2 of He+ spectrum ?
2.34 Calculate the energy required for the process
+ 2+ –
He (g) g He (g) + e
–18 –1
The ionization energy for the H atom in the gr ound state is 2.18 × 10 J atom
2.35 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms
which can be placed side by side in a straight line across length of scale of length
20 cm long.
8
2.36 2 ×10 atoms of carbon are arranged side by side. Calculate the radius of carbon
atom if the length of this arrangement is 2.4 cm.
2.37 The diameter of zinc atom is 2.6 Å.Calculate (a) radius of zinc atom in pm and (b)
number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side
by side lengthwise.
–16
2.38 A certain particle carries 2.5 × 10 C of static electric charge. Calculate the number
of electrons present in it.
2.39 In Milikan’s experiment, static electric charge on the oil drops has been obtained
by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10 –18C,
calculate the number of electrons present on it.
2.40 In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum
etc. have been used to be bombarded by the α-particles. If the thin foil of light
atoms like aluminium etc. is used, what difference would be observed from the
above results ?
79 79 35 35
2.41 Symbols 35 Br and Br can be written, whereas symbols 79 Br
and Br are not
acceptable. Answer briefly.
2.42 An element with mass number 81 contains 31.7% more neutrons as compar ed to
protons. Assign the atomic symbol.
2.43 An ion with mass number 37 possesses one unit of negative charge. If the ion
conatins 11.1% more neutrons than the electrons, find the symbol of the ion.
2.44 An ion with mass number 56 contains 3 units of positive charge and 30.4% mor e
neutrons than electr ons. Assign the symbol to this ion.
2.45 Arrange the following type of radiations in increasing order of frequency: (a) radiation
from microwave oven (b) amber light from traf fic signal (c) radiation from FM radio
(d) cosmic rays from outer space and (e) X-rays.
2.46 Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of
24
photons emitted is 5.6 × 10 , calculate the power of this laser.
2.47 Neon gas is generally used in the sign boards. If it emits str ongly at 616 nm,
calculate (a) the frequency of emission, (b) distance traveled by this radiation in
30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of
energy.
2.48 In astronomical observations, signals observed from the distant stars are generally

2015-16
68 CHEMISTRY

weak. If the photon detector receives a total of 3.15 × 10–18 J from the radiations of
600 nm, calculate the number of photons received by the detector.
2.49 Lifetimes of the molecules in the excited states are often measured by using
pulsed radiation source of duration nearly in the nano second range. If the
radiation source has the duration of 2 ns and the number of photons emitted
during the pulse source is 2.5 × 1015, calculate the energy of the source.
2.50 The longest wavelength doublet absorption transition is observed at 589 and
589.6 nm. Calcualte the frequency of each transition and energy difference
between two excited states.
2.51 The work function for caesium atom is 1.9 eV. Calculate (a) the thr eshold
wavelength and (b) the threshold frequency of the radiation. If the caesium
element is irradiated with a wavelength 500 nm, calculate the kinetic energy
and the velocity of the ejected photoelectron.
2.52 Following results are observed when sodium metal is irradiated with different
wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.
λ (nm) 500 450 400
–5 –1
v × 10 (cm s ) 2.55 4.35 5.35
2.53 The ejection of the photoelectron from the silver metal in the photoelectric effect
experiment can be stopped by applying the voltage of 0.35 V when the radiation
256.7 nm is used. Calculate the work function for silver metal.
2.54 If the photon of the wavelength 150 pm strikes an atom and one of tis inner bound
7 –1
electrons is ejected out with a velocity of 1.5 × 10 m s , calculate the energy with
which it is bound to the nucleus.
2.55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n
15 2 2
and can be represeted as v = 3.29 × 10 (Hz) [ 1/3 – 1/n ]
Calculate the value of n if the transition is observed at 1285 nm. Find the region of
the spectrum.
2.56 Calculate the wavelength for the emission transition if it starts from the orbit having
radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition
belongs and the region of the spectrum.
2.57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron
microscope often used for the highly magnified images of biological molecules and
6
other type of material. If the velocity of the electron in this micr oscope is 1.6 × 10
–1
ms , calculate de Br oglie wavelength associated with this electron.
2.58 Similar to electron diffraction, neutron diffraction micr oscope is also used for the
determination of the structure of molecules. If the wavelength used here is 800 pm,
calculate the characteristic velocity associated with the neutron.
6 –1
2.59 If the velocity of the electron in Bohr’s first orbit is 2.19 × 10 ms , calculate the
de Broglie wavelength associated with it.
2.60 The velocity associated with a proton moving in a potential differ ence of 1000 V is
4.37 × 105 ms–1 . If the hockey ball of mass 0.1 kg is moving with this velocity,
calcualte the wavelength associated with this velocity.
2.61 If the position of the electron is measured within an accuracy of + 0.002 nm, calculate
the uncertainty in the momentum of the electron. Suppose the momentum of the
electron is h/4πm × 0.05 nm, is ther e any problem in defining this value.
2.62 The quantum numbers of six electrons are given below. Arrange them in order of
increasing ener gies. If any of these combination(s) has/have the same energy lists:
1. n = 4, l = 2, ml = –2 , ms = –1/2
2. n = 3, l = 2, ml = 1 , ms = +1/2

2015-16
STRUCTURE OF ATOM 69

3. n = 4, l = 1, ml = 0 , ms = +1/2
4. n = 3, l = 2, ml = –2 , ms = –1/2
5. n = 3, l = 1, ml = –1 , ms = +1/2
6. n = 4, l = 1, ml = 0 , ms = +1/2
2.63 The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6
electr ons in 3p orbital and 5 electron in 4p orbital. Which of these electron
experiences the lowest effective nuclear char ge ?
2.64 Among the following pairs of orbitals which orbital will experience the larger ef fective
nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.
2.65 The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will
experience mor e effective nuclear charge fr om the nucleus ?
2.66 Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
2.67 (a) How many sub-shells are associated with n = 4 ? (b) How many electrons will be
present in the sub-shells having m s value of –1/2 for n = 4 ?

2015-16
70 CHEMISTRY

UNIT 3

CLASSIFICATION OF ELEMENTS AND

ed
PERIODICITY IN PROPERTIES

h
The Periodic Table is arguably the most important concept in

pu T
chemistry, both in principle and in practice. It is the everyday

is
support for students, it suggests new avenues of research to
re ER
After studying this Unit, you will be
able to
professionals, and it provides a succinct organization of the
whole of chemistry. It is a remarkable demonstration of the

bl
fact that the chemical elements are not a random cluster of
• app reciate how the concept of entities but instead display trends and lie together in families.
grouping elements in accordance to An awareness of the Periodic Table is essential to anyone who
their properties led to the
wishes to disentangle the world and see how it is built up
development of Periodic Table.
be C

from the fundamental building blocks of the chemistry, the

• understand the Periodic Law;
chemical elements.
• understand the significance of
atomic number and electronic Glenn T. Seaborg
N

configuration as the basis for

periodic classification;
• name the elements with In this Unit, we will study the historical development of the
Z >100 according to IUPAC Periodic Table as it stands today and the Modern Periodic
nomenclature; Law. We will also learn how the periodic classification
©

• classify elements into s, p, d, f follows as a logical consequence of the electronic

blocks and learn their main configuration of atoms. Finally, we shall examine some of
characteristics; the periodic trends in the physical and chemical properties
• recognise the periodic trends in of the elements.
physical and chemical properties of
elements; 3.1 WHY DO WE NEED TO CLASSIFY ELEMENTS ?
• compare the reactivity of elements
We know by now that the elements are the basic units of all
to

and correlate it with their

occurrence in nature; types of matter. In 1800, only 31 elements were known. By
• explain the relationship between 1865, the number of identified elements had more than
ionization enthalpy and metallic doubled to 63. At present 114 elements are known. Of
character; them, the recently discovered elements are man-made.
t

• use scientific vocabulary Efforts to synthesise new elements are continuing. With
appropriately to communicate ideas such a large number of elements it is very difficult to study
no

related to certain important individually the chemistry of all these elements and their
properties of atoms e.g., atomic/ innumerable compounds individually. To ease out this
ionic radii, ionization enthalpy,
problem, scientists searched for a systematic way to
electron gain enthalpy,
electronegativity, valence of
organise their knowledge by classifying the elements. Not
only that it would rationalize known chemical facts about
element s.
elements, but even predict new ones for undertaking further
study.
 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES 71

3.2 GENESIS OF PERIODIC chemist, John Alexander Newlands in 1865

CLASSIFICATION profounded the Law of Octaves. He arranged
the elements in increasing order of their atomic
Classification of elements into groups and weights and noted that every eighth element
development of Periodic Law and Periodic had properties similar to the first element
Table are the consequences of systematising (Table 3.2). The relationship was just like every
the knowledge gained by a number of scientists eighth note that resembles the first in octaves
through their observations and experiments. of music. Newlands’s Law of Octaves seemed

ed
The German chemist, Johann Dobereiner in to be true only for elements up to calcium.
early 1800’s was the first to consider the idea Although his idea was not widely accepted at
of trends among properties of elements. By that time, he, for his work, was later awarded
1829 he noted a similarity among the physical Davy Medal in 1887 by the Royal Society,
and chemical properties of several groups of London.

h
three elements (Triads). In each case, he

pu T
noticed that the middle element of each of the The Periodic Law, as we know it today owes

is
Triads had an atomic weight about half way its development to the Russian chemist, Dmitri
between the atomic weights of the other two Mendeleev (1834-1907) and the German
re ER
(Table 3.1). Also the properties of the middle chemist, Lothar Meyer (1830-1895). Working

bl
element were in between those of the other independently, both the chemists in 1869
Table 3.1 Dobereiner’s Triads

Element Atomic Element Atomic Element Atomic

weight weight weight
be C

Li 7 Ca 40 Cl 35.5
Na 23 Sr 88 Br 80
N

K 39 Ba 137 I 127

two members. Since Dobereiner’s relationship, proposed that on arranging elements in the
referred to as the Law of Triads, seemed to increasing order of their atomic weights,
©

work only for a few elements, it was dismissed similarities appear in physical and chemical
as coincidence. The next reported attempt to properties at regular intervals. Lothar Meyer
classify elements was made by a French plotted the physical properties such as atomic
geologist, A.E.B. de Chancourtois in 1862. He volume, melting point and boiling point
arranged the then known elements in order of against atomic weight and obtained a
increasing atomic weights and made a periodically repeated pattern. Unlike
cylindrical table of elements to display the Newlands, Lothar Meyer observed a change in
to

periodic recurrence of properties. This also did length of that repeating pattern. By 1868,
not attract much attention. The English Lothar Meyer had developed a table of the
Table 3.2 Newlands’ Octaves
t

Element Li Be B C N O F
no

At. wt. 7 9 11 12 14 16 19
Element Na Mg Al Si P S Cl
At. wt. 23 24 27 29 31 32 35.5
Element K Ca
At. wt. 39 40
72 CHEMISTRY

elements that closely resembles the Modern weights, thinking that the atomic
Periodic Table. However, his work was not measurements might be incorrect, and placed
published until after the work of Dmitri the elements with similar properties together.
Mendeleev, the scientist who is generally For example, iodine with lower atomic weight
credited with the development of the Modern than that of tellurium (Group VI) was placed
Periodic Table. in Group VII along with fluorine, chlorine,
bromine because of similarities in properties
While Dobereiner initiated the study of
(Fig. 3.1). At the same time, keeping his

ed
periodic relationship, it was Mendeleev who
was responsible for publishing the Periodic primary aim of arranging the elements of
Law for the first time. It states as follows : similar properties in the same group, he
proposed that some of the elements were still
The properties of the elements are a undiscovered and, therefore, left several gaps

h
periodic function of their atomic in the table. For example, both gallium and
weights. germanium were unknown at the time

pu T
Mendeleev published his Periodic Table. He left

is
Mendeleev arranged elements in horizontal
the gap under aluminium and a gap under
rows and vertical columns of a table in order
re ER
of their increasing atomic weights in such a
silicon, and called these elements Eka-

bl
Aluminium and Eka-Silicon. Mendeleev
way that the elements with similar properties
occupied the same vertical column or group. predicted not only the existence of gallium and
Mendeleev’s system of classifying elements was germanium, but also described some of their
more elaborate than that of Lothar Meyer’s. general physical properties. These elements
were discovered later. Some of the properties
be C

He fully recognized the significance of

periodicity and used broader range of physical predicted by Mendeleev for these elements and
and chemical properties to classify the those found experimentally are listed in
N

elements. In particular, Mendeleev relied on Table 3.3.

the similarities in the empirical formulas and
properties of the compounds formed by the The boldness of Mendeleev’s quantitative
elements. He realized that some of the elements predictions and their eventual success made
him and his Periodic Table famous.
©

did not fit in with his scheme of classification

if the order of atomic weight was strictly Mendeleev’s Periodic Table published in 1905
followed. He ignored the order of atomic is shown in Fig. 3.1.

Table 3.3 Mendeleev’s Predictions for the Elements Eka-aluminium (Gallium) and
Eka-silicon (Germanium)
to

Property Eka-aluminium Gallium Eka-silicon Germanium

(predicted) (found) (predicted) (found)

Atomic weight 68 70 72 72.6

t

Density / (g/cm3) 5.9 5.94 5.5 5.36

no

Melting point /K Low 302.93 High 1231

Formula of oxide E2O3 Ga2O 3 EO2 GeO2

Formula of chloride ECl3 GaCl 3 ECl4 GeCl4

 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES
PERIODIC SYSTEM OF THE ELEMENTS IN GROUPS AND SERIES

h ed
pu T
is
re ER
bl
be C
N
to ©

Fig. 3.1 Mendeleev’s Periodic Table published earlier

t

73
no
74 CHEMISTRY

ed
Dmitri Mendeleev was born in Tobalsk, Siberia in Russia. After his
father’s death, the family moved to St. Petersburg. He received his
Master’s degree in Chemistry in 1856 and the doctoral degree in
1865. He taught at the University of St.Petersburg where he was

h
appointed Professor of General Chemistry in 1867. Preliminary work
for his great textbook “Principles of Chemistry” led Mendeleev to

pu T
propose the Periodic Law and to construct his Periodic Table of

is
elements. At that time, the structure of atom was unknown and
re ER
Mendeleev’s idea to consider that the properties of the elements

bl
were in someway related to their atomic masses was a very Dmitri Ivanovich
imaginative one. To place certain elements into the correct group from Mendeleev
the point of view of their chemical properties, Mendeleev reversed the (1834-1907)
order of some pairs of elements and asserted that their atomic masses
were incorrect. Mendeleev also had the foresight to leave gaps in the Periodic Table for
be C

elements unknown at that time and predict their properties from the trends that he observed
among the properties of related elements. Mendeleev’s predictions were proved to be
astonishingly correct when these elements were discovered later.
N

Mendeleev’s Periodic Law spurred several areas of research during the subsequent
decades. The discovery of the first two noble gases helium and argon in 1890 suggested
the possibility that there must be other similar elements to fill an entire family. This idea
led Ramsay to his successful search for krypton and xenon. Work on the radioactive decay
©

series for uranium and thorium in the early years of twentieth century was also guided by
the Periodic Table.
Mendeleev was a versatile genius. He worked on many problems connected with
Russia’s natural resources. He invented an accurate barometer. In 1890, he resigned from
the Professorship. He was appointed as the Director of the Bureau of Weights and Measures.
He continued to carry out important research work in many areas until his death in 1907.
to

You will notice fr om the modern Period Table (Fig. 3.2) that Mendeleev’s name has
been immortalized by naming the element with atomic number 101, as Mendelevium. This
name was pr oposed by American scientist Glenn T. Seaborg, the discoverer of this element,
“in recognition of the pioneering role of the great Russian Chemist who was the first to use
the periodic system of elements to predict the chemical properties of undiscovered elements,
t

a principle which has been the key to the discovery of nearly all the transuranium elements”.
no
 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES 75

3.3 MODERN PERIODIC LAW AND THE Numerous forms of Periodic Table have
PRESENT FORM OF THE PERIODIC been devised from time to time. Some forms
TABLE emphasise chemical reactions and valence,
We must bear in mind that when Mendeleev whereas others stress the electronic
developed his Periodic Table, chemists knew configuration of elements. A modern version,
nothing about the internal structure of atom. the so-called “long form” of the Periodic Table
However, the beginning of the 20 th century of the elements (Fig. 3.2), is the most convenient
witnessed profound developments in theories and widely used. The horizontal rows (which

ed
about sub-atomic particles. In 1913, the Mendeleev called series) are called periods and
English physicist, Henry Moseley observed the vertical columns, groups. Elements having
regularities in the characteristic X-ray spectra similar outer electronic configurations in their
of the elements. A plot of ν (where ν is atoms are arranged in vertical columns,

h
frequency of X-rays emitted) against atomic referred to as groups or families. According
to the recommendation of International Union

pu T
number (Z ) gave a straight line and not the

is
of Pure and Applied Chemistry (IUPAC), the
plot of ν vs atomic mass. He thereby showed
groups are numbered from 1 to 18 replacing
re ER
that the atomic number is a more fundamental the older notation of groups IA … VIIA, VIII, IB

bl
property of an element than its atomic mass. … VIIB and 0.
Mendeleev’s Periodic Law was, therefore, There are altogether seven periods. The
accordingly modified. This is known as the period number corresponds to the highest
Modern Periodic Law and can be stated as : principal quantum number (n) of the elements
be C

The physical and chemical properties in the period. The first period contains 2
of the elements are periodic functions elements. The subsequent periods consists of
of their atomic numbers. 8, 8, 18, 18 and 32 elements, respectively. The
N

seventh period is incomplete and like the sixth

The Periodic Law revealed important
period would have a theoretical maximum (on
analogies among the 94 naturally occurring
elements (neptunium and plutonium like the basis of quantum numbers) of 32 elements.
actinium and protoactinium are also found in In this form of the Periodic Table, 14 elements
©

pitch blende – an ore of uranium). It stimulated of both sixth and seventh periods (lanthanoids
renewed interest in Inorganic Chemistry and and actinoids, respectively) are placed in
has carried into the present with the creation separate panels at the bottom* .
of artificially produced short-lived elements. 3.4 NOMENCLATURE OF ELEMENTS WITH
You may recall that the atomic number is ATOMIC NUMBERS > 100
equal to the nuclear charge (i.e., number of
protons) or the number of electrons in a neutral The naming of the new elements had been
to

atom. It is then easy to visualize the significance traditionally the privilege of the discoverer (or
of quantum numbers and electronic discoverers) and the suggested name was
configurations in periodicity of elements. In ratified by the IUPAC. In recent years this has
fact, it is now recognized that the Periodic Law led to some controversy. The new elements with
is essentially the consequence of the periodic very high atomic numbers are so unstable that
t

variation in electronic configurations, which only minute quantities, sometimes only a few
no

indeed determine the physical and chemical atoms of them are obtained. Their synthesis
properties of elements and their compounds. and characterisation, therefore, require highly

* Glenn T. Seaborg’s work in the middle of the 20t h century starting with the discovery of plutonium in 1940, followed by
those of all the transuranium elements from 94 to 102 led to reconfiguration of the periodic table placing the actinoids
below the lanthanoids. In 1951, Seaborg was awarded the Nobel Prize in chemistry for his work. Element 106 has been
named Seabor gium (Sg) in his honour.
 76
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CHEMISTRY
Fig. 3.2 Long form of the Periodic Table of the Elements with their atomic numbers and ground state outer
electronic configurations. The groups are numbered 1-18 in accordance with the 1984 IUPAC
t

recommendations. This notation replaces the old numbering scheme of IA–VIIA, VIII, IB–VIIB and 0 for
no

the elements.
 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES 77

sophisticated costly equipment and laboratory. which make up the atomic number and “ium”
Such work is carried out with competitive spirit is added at the end. The IUPAC names for
only in some laboratories in the world. elements with Z above 100 are shown in
Scientists, before collecting the reliable data on Table 3.5.
the new element, at times get tempted to claim
Table 3.4 Notation for IUPAC Nomenclature
for its discovery. For example, both American of Elements
and Soviet scientists claimed credit for
discovering element 104. The Americans Digit Name Abbreviation

ed
named it Rutherfordium whereas Soviets 0 nil n
named it Kurchatovium. To avoid such 1 un u
problems, the IUPAC has made 2 bi b
recommendation that until a new element’s 3 tri t

h
discovery is proved, and its name is officially 4 quad q

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recognized,,,,,,, a systematic nomenclature be 5 pent p

is
derived directly from the atomic number of the 6 hex h
element using the numerical roots for 0 and 7 sept s
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numbers 1-9. These are shown in Table 3.4. 8 oct o

bl
The roots are put together in order of digits 9 enn e

Table 3.5 Nomenclature of Elements with Atomic Number Above 100

be C

Atomic Name according to Symbol IUPAC IUPAC

Number IUP AC nomenclature Official Name Symbol
101 Unnilunium Unu Mendelevium Md
N

102 Unnilbium Unb Nobelium No

103 Unniltrium Unt Lawrencium Lr
104 Unnilquadium Unq Rutherfordium Rf
©

105 Unnilpentium Unp Dubnium Db

106 Unnilhexium Unh Seaborgium Sg
107 Unnilseptium Uns Bohrium Bh
108 Unniloctium Uno Hassium Hs
109 Unnilennium Une Meitnerium Mt
to

110 Ununnillium Uun Darmstadtium Ds

111 Unununnium Uuu Rontgenium Rg
112 Ununbium Uub Copernicium Cn
113 Ununtrium Uut * –
t

114 Ununquadium Uuq Flerovium Fl

no

115 Ununpentium Uup * –

116 Ununhexium Uuh Livermorium Lv
117 Ununseptium Uus * –
118 Ununoctium Uuo * –

* Official IUPAC name yet to be announced

78 CHEMISTRY

Thus, the new element first gets a be readily seen that the number of elements in
temporary name, with symbol consisting of each period is twice the number of atomic
three letters. Later permanent name and orbitals available in the energy level that is
symbol are given by a vote of IUPAC being filled. The first period (n = 1) starts with
representatives from each country. The the filling of the lowest level (1s) and therefore
permanent name might reflect the country (or has two elements — hydrogen (ls1) and helium
state of the country) in which the element was (ls2) when the first shell (K) is completed. The
discovered, or pay tribute to a notable scientist. second period (n = 2) starts with lithium and

ed
As of now, elements with atomic numbers up the third electron enters the 2s orbital. The next
to 118 have been discovered. Official names of element, beryllium has four electrons and has
2 2
elements with atomic numbers 113, 115, 117 the electronic configuration 1s 2s . Starting
and 118 are yet to be announced by IUPAC. from the next element boron, the 2p orbitals
are filled with electrons when the L shell is

h
Problem 3.1 completed at neon (2s 22p 6). Thus there are

pu T
What would be the IUPAC name and 8 elements in the second period. The third

is
symbol for the element with atomic period (n = 3) begins at sodium, and the added
number 120? electron enters a 3s orbital. Successive filling
re ER of 3s and 3p orbitals gives rise to the third
Solution

bl
period of 8 elements from sodium to argon. The
From Table 3.4, the roots for 1, 2 and 0 fourth period (n = 4) starts at potassium, and
are un, bi and nil, respectively. Hence, the the added electrons fill up the 4s orbital. Now
symbol and the name respectively are Ubn you may note that before the 4p orbital is filled,
and unbinilium. filling up of 3d orbitals becomes energetically
be C

favourable and we come across the so called

3d transition series of elements. This starts
3.5 ELECTRONIC CONFIGURATIONS OF from scandium (Z = 21) which has the electronic
N

ELEMENTS AND THE PERIODIC 1 2

configuration 3d 4s . The 3d orbitals are filled
TABLE at zinc (Z=30) with electronic configuration
In the preceding unit we have learnt that an 10 2
3d 4s . The fourth period ends at krypton
electron in an atom is characterised by a set of with the filling up of the 4p orbitals. Altogether
©

four quantum numbers, and the principal we have 18 elements in this fourth period. The
quantum number (n ) defines the main energy fifth period (n = 5) beginning with rubidium is
level known as shell. We have also studied similar to the fourth period and contains the
about the filling of electrons into different 4d transition series starting at yttrium
subshells, also referred to as orbitals (s, p, d, (Z = 39). This period ends at xenon with the
f ) in an atom. The distribution of electrons into filling up of the 5p orbitals. The sixth period
orbitals of an atom is called its electronic (n = 6) contains 32 elements and successive
to

configuration. An element’s location in the electrons enter 6s, 4f, 5d and 6p orbitals, in
Periodic Table reflects the quantum numbers the order — filling up of the 4f orbitals begins
of the last orbital filled. In this section we will with cerium (Z = 58) and ends at lutetium
observe a direct connection between the (Z = 71) to give the 4f-inner transition series
electronic configurations of the elements and which is called the lanthanoid series. The
t

the long form of the Periodic Table. seventh period (n = 7) is similar to the sixth
no

(a) Electronic Configurations in Periods period with the successive filling up of the 7s,
5f, 6d and 7p orbitals and includes most of
The period indicates the value of n for the the man-made radioactive elements. This
outermost or valence shell. In other words, period will end at the element with atomic
successive period in the Periodic Table is number 118 which would belong to the noble
associated with the filling of the next higher gas family. Filling up of the 5f orbitals after
principal energy level (n = 1, n = 2, etc.). It can actinium (Z = 89) gives the 5f-inner transition
 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES 79

series known as the actinoid series. The 4f- theoretical foundation for the periodic
and 5f-inner transition series of elements classification. The elements in a vertical column
are placed separately in the Periodic Table to of the Periodic Table constitute a group or
maintain its structure and to preserve the family and exhibit similar chemical behaviour.
principle of classification by keeping elements This similarity arises because these elements
with similar properties in a single column. have the same number and same distribution
of electrons in their outermost orbitals. We can
Problem 3.2 classify the elements into four blocks viz.,

ed
How would you justify the presence of 18 s-block, p-block, d-block and f-block
elements in the 5th period of the Periodic depending on the type of atomic orbitals that
Table? are being filled with electrons. This is illustrated
Solution in Fig. 3.3. We notice two exceptions to this

h
When n = 5, l = 0, 1, 2, 3. The order in categorisation. Strictly, helium belongs to the
s-block but its positioning in the p-block along

pu T
which the energy of the available orbitals
with other group 18 elements is justified

is
4d, 5s and 5p increases is 5s < 4d < 5p.
The total number of orbitals available are because it has a completely filled valence shell
re ER
9. The maximum number of electrons that (1s 2) and as a result, exhibits properties

bl
can be accommodated is 18; and therefore characteristic of other noble gases. The other
18 elements are there in the 5th period. exception is hydrogen. It has only one
s-electron and hence can be placed in group 1
(b) Groupwise Electronic Configurations (alkali metals). It can also gain an electron to
achieve a noble gas arrangement and hence it
be C

Elements in the same vertical column or group

have similar valence shell electronic can behave similar to a group 17 (halogen
configurations, the same number of electrons family) elements. Because it is a special case,
N

in the outer orbitals, and similar properties. we shall place hydrogen separately at the top
For example, the Group 1 elements (alkali of the Periodic Table as shown in Fig. 3.2 and
1
metals) all have ns valence shell electronic Fig. 3.3. We will briefly discuss the salient
configuration as shown below. features of the four types of elements marked in
©

Atomic number Symbol Electronic configuration

3 Li 1s2 2s1 (or) [He]2s1

11 Na 1s2 2s22p 63s1 (or) [Ne]3s1
19 K 1s2 2s22p 63s2 3p6 4s1 (or) [Ar]4 s1
37 Rb 1s2 2s22p 63s2 3p6 3d104s2 4p6 5s1 (or) [Kr]5s1
to

55 Cs 1s2 2s22p 63s23p 63d104s2 4p 64d105s2 5p 6 6s 1 (or) [Xe]6s1

87 Fr [Rn]7s 1

Thus it can be seen that the properties of the Periodic Table. More about these elements
t

an element have periodic dependence upon its will be discussed later. During the description
no

atomic number and not on relative atomic of their features certain terminology has been
mass.
used which has been classified in section 3.7.
3.6 ELECTRONIC CONFIGURATIONS
AND TYPES OF ELEMENTS: 3.6.1 The s-Block Elements
s-, p-, d-, f- BLOCKS
The aufbau (build up) principle and the The elements of Group 1 (alkali metals) and
electronic configuration of atoms provide a Group 2 (alkaline earth metals) which have ns1
 80
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CHEMISTRY
Fig. 3.3 The types of elements in the Periodic Table based on the orbitals that
are being filled. Also shown is the broad division of elements into METALS
t

( ) , NON-METALS ( ) and METALLOIDS ( ).

no
 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES 81

and ns2 outermost electronic configuration used as catalysts. However, Zn, Cd and Hg
belong to the s-Block Elements. They are all which have the electronic configuration,
reactive metals with low ionization enthalpies. (n-1) d10ns2 do not show most of the properties
They lose the outermost electron(s) readily to of transition elements. In a way, transition
form 1+ ion (in the case of alkali metals) or 2+ metals form a bridge between the chemically
ion (in the case of alkaline earth metals). The active metals of s-block elements and the less
metallic character and the reactivity increase active elements of Groups 13 and 14 and thus
as we go down the group. Because of high take their familiar name “ Transition

ed
reactivity they are never found pure in nature. Elements”.
The compounds of the s-block elements, with 3.6.4 The f-Block Elements
the exception of those of lithium and beryllium (Inner-Transition Elements)
are predominantly ionic.
The two rows of elements at the bottom of the

h
3.6.2 The p-Block Elements Periodic Table, called the Lanthanoids,
Ce(Z = 58) – Lu(Z = 71) and Actinoids,

pu T
The p -Block Elements comprise those
Th(Z = 90) – Lr (Z = 103) are characterised by

is
belonging to Group 13 to 18 and these
the outer electronic configuration (n-2)f 1-14
together with the s-Block Elements are called
re ER
the Representative Elements or Main Group
(n-1)d0–1 ns2. The last electron added to each
element is filled in f- orbital. These two series

bl
Elements. The outermost electronic
of elements are hence called the Inner-
configuration varies from ns 2np1 to ns2np6 in Transition Elements (f-Block Elements).
each period. At the end of each period is a noble They are all metals. Within each series, the
gas element with a closed valence shell ns2np6 properties of the elements are quite similar. The
be C

configuration. All the orbitals in the valence chemistry of the early actinoids is more
shell of the noble gases are completely filled complicated than the corresponding
by electrons and it is very difficult to alter this lanthanoids, due to the large number of
N

stable arrangement by the addition or removal oxidation states possible for these actinoid
of electrons. The noble gases thus exhibit very elements. Actinoid elements are radioactive.
low chemical reactivity. Preceding the noble gas Many of the actinoid elements have been made
family are two chemically important groups of only in nanogram quantities or even less by
non-metals. They are the halogens (Group 17)
©

nuclear reactions and their chemistry is not

and the chalcogens (Group 16). These two fully studied. The elements after uranium are
groups of elements have highly negative called Transuranium Elements.
electron gain enthalpies and readily add one
or two electrons respectively to attain the stable Problem 3.3
noble gas configuration. The non-metallic The elements Z = 117 and 120 have not
character increases as we move from left to right
yet been discovered. In which family /
across a period and metallic character increases
to

group would you place these elements

as we go down the group.
and also give the electronic configuration
3.6.3 The d-Block Elements (Transition in each case.
Elements) Solution
t

These are the elements of Group 3 to 12 in the We see from Fig. 3.2, that element with Z
no

centre of the Periodic Table. These are = 117, would belong to the halogen family
characterised by the filling of inner d orbitals (Group 17) and the electronic
by electrons and are therefore referred to as configuration would be [Rn]
d-Block Elements. These elements have the 5f 146d107s 27p5. The element with Z = 120,
general outer electronic configuration will be placed in Group 2 (alkaline earth
(n-1)d1-10ns0-2 . They are all metals. They mostly metals), and will have the electronic
form coloured ions, exhibit variable valence configuration [Uuo]8s2.
(oxidation states), paramagnetism and oftenly
82 CHEMISTRY

3.6.5 Metals, Non-metals and Metalloids from left to right. Hence the order of
increasing metallic character is: P < Si <
In addition to displaying the classification of
Be < Mg < Na.
elements into s-, p-, d-, and f-blocks, Fig. 3.3
shows another broad classification of elements 3.7 PERIODIC TRENDS IN PROPERTIES
based on their properties. The elements can OF ELEMENTS
be divided into Metals and Non-Metals. Metals
comprise more than 78% of all known elements There are many observable patterns in the
and appear on the left side of the Periodic physical and chemical properties of elements

ed
Table. Metals are usually solids at room as we descend in a group or move across a
temperature [mercury is an exception; gallium period in the Periodic Table. For example,
and caesium also have very low melting points within a period, chemical reactivity tends to be
(303K and 302K, respectively)]. Metals usually high in Group 1 metals, lower in elements

h
have high melting and boiling points. They are towards the middle of the table, and increases
good conductors of heat and electricity. They to a maximum in the Group 17 non-metals.

pu T
are malleable (can be flattened into thin sheets Likewise within a group of representative

is
by hammering) and ductile (can be drawn into metals (say alkali metals) reactivity increases
re ER
wires). In contrast, non-metals are located at
the top right hand side of the Periodic Table.
on moving down the group, whereas within a
group of non-metals (say halogens), reactivity

bl
In fact, in a horizontal row, the property of decreases down the group. But why do the
elements change from metallic on the left to properties of elements follow these trends? And
non-metallic on the right. Non-metals are how can we explain periodicity? To answer
usually solids or gases at room temperature these questions, we must look into the theories
be C

with low melting and boiling points (boron and of atomic structure and properties of the atom.
carbon are exceptions). They are poor In this section we shall discuss the periodic
conductors of heat and electricity. Most non- trends in certain physical and chemical
N

metallic solids are brittle and are neither properties and try to explain them in terms of
malleable nor ductile. The elements become number of electrons and energy levels.
more metallic as we go down a group; the non-
metallic character increases as one goes from 3.7.1 Trends in Physical Properties
©

left to right across the Periodic Table. The

There are numerous physical properties of
change from metallic to non-metallic character
elements such as melting and boiling points,
is not abrupt as shown by the thick zig-zag
heats of fusion and vaporization, energy of
line in Fig. 3.3. The elements (e.g., silicon,
atomization, etc. which show periodic
germanium, arsenic, antimony and tellurium)
variations. However, we shall discuss the
bordering this line and running diagonally
periodic trends with respect to atomic and ionic
across the Periodic Table show properties that
radii, ionization enthalpy, electron gain
to

are characteristic of both metals and non-

metals. These elements are called Semi-metals enthalpy and electronegativity.
or Metalloids. (a) Atomic Radius
Problem 3.4 You can very well imagine that finding the size
t

Considering the atomic number and of an atom is a lot more complicated than
no

position in the periodic table, arrange the measuring the radius of a ball. Do you know
following elements in the increasing order why? Firstly, because the size of an atom
of metallic character : Si, Be, Mg, Na, P. (~ 1.2 Å i.e., 1.2 × 10–10 m in radius) is very
small. Secondly, since the electron cloud
Solution surrounding the atom does not have a sharp
Metallic character increases down a group boundary, the determination of the atomic size
and decreases along a period as we move cannot be precise. In other words, there is no
 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES 83

practical way by which the size of an individual explain these trends in terms of nuclear charge
atom can be measured. However, an estimate and energy level. The atomic size generally
of the atomic size can be made by knowing the decreases across a period as illustrated in
distance between the atoms in the combined Fig. 3.4(a) for the elements of the second period.
state. One practical approach to estimate the It is because within the period the outer
size of an atom of a non-metallic element is to electrons are in the same valence shell and the
measure the distance between two atoms when effective nuclear charge increases as the atomic
they are bound together by a single bond in a number increases resulting in the increased

ed
covalent molecule and from this value, the attraction of electrons to the nucleus. Within a
“Covalent Radius” of the element can be family or vertical column of the periodic table,
calculated. For example, the bond distance in the atomic radius increases regularly with
the chlorine molecule (Cl2) is 198 pm and half atomic number as illustrated in Fig. 3.4(b). For
this distance (99 pm), is taken as the atomic

h
alkali metals and halogens, as we descend the
radius of chlorine. For metals, we define the groups, the principal quantum number (n)

pu T
term “Metallic Radius” which is taken as half
increases and the valence electrons are farther

is
the internuclear distance separating the metal
from the nucleus. This happens because the
cores in the metallic crystal. For example, the
re ER
distance between two adjacent copper atoms
inner energy levels are filled with electrons,
which serve to shield the outer electrons from

bl
in solid copper is 256 pm; hence the metallic
radius of copper is assigned a value of 128 pm. the pull of the nucleus. Consequently the size
For simplicity, in this book, we use the term of the atom increases as reflected in the atomic
Atomic Radius to refer to both covalent or radii.
Note that the atomic radii of noble gases
be C

metallic radius depending on whether the

element is a non-metal or a metal. Atomic radii are not considered here. Being monoatomic,
can be measured by X-ray or other their (non-bonded radii) values are very large.
N

spectroscopic methods. In fact radii of noble gases should be compared

The atomic radii of a few elements are listed not with the covalent radii but with the van der
in Table 3.6 . Two trends are obvious. We can Waals radii of other elements.

Table 3.6(a) Atomic Radii/pm Across the Periods

©

Atom (Period II) Li Be B C N O F

Atomic radius 152 111 88 77 74 66 64
Atom (Period III) Na Mg Al Si P S Cl
Atomic radius 186 160 143 117 110 104 99
to

Table 3.6(b) Atomic Radii/pm Down a Family

Atom Atomic Atom Atomic

t

(Group I) Radius (Group 17) Radius

no

Li 152 F 64
Na 186 Cl 99
K 231 Br 114
Rb 244 I 133

Cs 262 At 140
84 CHEMISTRY

h ed
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is
Fig. 3.4 (a) Variation of atomic radius with Fig. 3.4 (b) Variation of atomic radius with
re ER
atomic number across the second atomic number for alkali metals

bl
period and halogens

(b) Ionic Radius attraction of the electrons to the nucleus. Anion

with the greater negative charge will have the
The removal of an electron from an atom results
larger radius. In this case, the net repulsion of
in the formation of a cation, whereas gain of
be C

the electrons will outweigh the nuclear charge

an electron leads to an anion. The ionic radii
and the ion will expand in size.
can be estimated by measuring the distances
N

between cations and anions in ionic crystals.

Problem 3.5
In general, the ionic radii of elements exhibit
the same trend as the atomic radii. A cation is Which of the following species will have
smaller than its parent atom because it has the largest and the smallest size?
fewer electrons while its nuclear charge remains Mg, Mg2+, Al, Al3+ .
©

the same. The size of an anion will be larger

Solution
than that of the parent atom because the
addition of one or more electrons would result Atomic radii decrease across a period.
in increased repulsion among the electrons Cations are smaller than their parent
and a decrease in effective nuclear charge. For atoms. Among isoelectronic species, the
–
example, the ionic radius of fluoride ion (F ) is one with the larger positive nuclear charge
136 pm whereas the atomic radius of fluorine will have a smaller radius.
to

is only 64 pm. On the other hand, the atomic Hence the largest species is Mg; the
radius of sodium is 186 pm compared to the smallest one is Al3+.
+
ionic radius of 95 pm for Na .
When we find some atoms and ions which (c) Ionization Enthalpy
t

contain the same number of electrons, we call

no

them isoelectronic species*. For example, A quantitative measure of the tendency of an

O2–, F–, Na + and Mg2+ have the same number of element to lose electron is given by its
electrons (10). Their radii would be different Ionization Enthalpy. It represents the energy
because of their different nuclear charges. The required to remove an electron from an isolated
cation with the greater positive charge will have gaseous atom (X) in its ground state. In other
a smaller radius because of the greater words, the first ionization enthalpy for an
* Two or more species with same number of atoms, same number of valence electrons and same structure,
regardless of the nature of elements involved.
 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES 85

element X is the enthalpy change (∆i H) for the

reaction depicted in equation 3.1.

X(g) → X+(g) + e– (3.1)

The ionization enthalpy is expressed in

units of kJ mol–1. We can define the second
ionization enthalpy as the energy required to
remove the second most loosely bound

ed
electron; it is the energy required to carry out
the reaction shown in equation 3.2.
+ 2+ –
X (g) → X (g) + e (3.2)

h
Energy is always required to remove
Fig. 3.5 Variation of first ionization enthalpies

pu T
electrons from an atom and hence ionization
(∆i H) with atomic number for elements

is
enthalpies are always positive. The second
with Z = 1 to 60
ionization enthalpy will be higher than the first
re ER
ionization enthalpy because it is more difficult with their high reactivity. In addition, you will

bl
to remove an electron from a positively charged notice two trends the first ionization enthalpy
ion than from a neutral atom. In the same way generally increases as we go across a period
the third ionization enthalpy will be higher than and decreases as we descend in a group. These
the second and so on. The term “ionization trends are illustrated in Figs. 3.6(a) and 3.6(b)
enthalpy”, if not qualified, is taken as the first respectively for the elements of the second
be C

ionization enthalpy. period and the first group of the periodic table.
The first ionization enthalpies of elements You will appreciate that the ionization enthalpy
and atomic radius are closely related
N

having atomic numbers up to 60 are plotted

in Fig. 3.5. The periodicity of the graph is quite properties. To understand these trends, we
striking. You will find maxima at the noble gases have to consider two factors : (i) the attraction
which have closed electron shells and very of electrons towards the nucleus, and (ii) the
stable electron configurations. On the other repulsion of electrons from each other. The
©

hand, minima occur at the alkali metals and effective nuclear charge experienced by a
their low ionization enthalpies can be correlated valence electron in an atom will be less than
t to
no

3.6 (a) 3.6 (b)

Fig. 3.6(a) First ionization enthalpies (∆ i H) of elements of the second period as a
function of atomic number (Z) and Fig. 3.6(b) ∆i H of alkali metals as a function of Z.
86 CHEMISTRY

the actual charge on the nucleus because of to remove the 2p-electron from boron compared
“shielding” or “screening” of the valence to the removal of a 2s- electron from beryllium.
electron from the nucleus by the intervening Thus, boron has a smaller first ionization
core electrons. For example, the 2s electron in enthalpy than beryllium. Another “anomaly”
lithium is shielded from the nucleus by the is the smaller first ionization enthalpy of oxygen
inner core of 1s electrons. As a result, the compared to nitrogen. This arises because in
valence electron experiences a net positive the nitrogen atom, three 2p-electrons reside in
charge which is less than the actual charge of different atomic orbitals (Hund’s rule) whereas

ed
+3. In general, shielding is effective when the in the oxygen atom, two of the four 2p-electrons
orbitals in the inner shells are completely filled. must occupy the same 2p-orbital resulting in
This situation occurs in the case of alkali metals an increased electron-electron repulsion.
which have single outermost ns-electron Consequently, it is easier to remove the fourth
preceded by a noble gas electronic 2p-electron from oxygen than it is, to remove

h
configuration. one of the three 2p-electrons from nitrogen.

pu T
When we move from lithium to fluorine

is
Problem 3.6
across the second period, successive electrons
The first ionization enthalpy (∆i H ) values
re ER
are added to orbitals in the same principal
quantum level and the shielding of the nuclear of the third period elements, Na, Mg and

bl
Si are respectively 496, 737 and 786 kJ
charge by the inner core of electrons does not
mol–1. Predict whether the first ∆ i H value
increase very much to compensate for the
increased attraction of the electron to the for Al will be more close to 575 or 760 kJ
mol–1 ? Justify your answer.
nucleus. Thus, across a period, increasing
be C

nuclear charge outweighs the shielding. Solution

Consequently, the outermost electrons are held –1
It will be more close to 575 kJ mol . The
more and more tightly and the ionization
value for Al should be lower than that of
N

enthalpy increases across a period. As we go

Mg because of effective shielding of 3p
down a group, the outermost electron being electrons from the nucleus by
increasingly farther from the nucleus, there is
3s-electrons.
an increased shielding of the nuclear charge
©

by the electrons in the inner levels. In this case, (d) Electron Gain Enthalpy
increase in shielding outweighs the increasing When an electron is added to a neutral gaseous
nuclear charge and the removal of the atom (X) to convert it into a negative ion, the
outermost electron requires less energy down enthalpy change accompanying the process is
a group. defined as the Electron Gain Enthalpy (∆ egH).
From Fig. 3.6(a), you will also notice that Electron gain enthalpy provides a measure of
the first ionization enthalpy of boron (Z = 5) is the ease with which an atom adds an electron
to

slightly less than that of beryllium (Z = 4) even to form anion as represented by equation 3.3.
though the former has a greater nuclear charge.
When we consider the same principal quantum X(g) + e – → X –(g) (3.3)
level, an s-electron is attracted to the nucleus Depending on the element, the process of
more than a p-electron. In beryllium, the
t

adding an electron to the atom can be either

electron removed during the ionization is an endothermic or exothermic. For many elements
no

s-electron whereas the electron removed during energy is released when an electron is added
ionization of boron is a p-electron. The to the atom and the electron gain enthalpy is
penetration of a 2s-electron to the nucleus is negative. For example, group 17 elements (the
more than that of a 2p-electron; hence the 2p halogens) have very high negative electron gain
electron of boron is more shielded from the enthalpies because they can attain stable noble
nucleus by the inner core of electrons than the gas electronic configurations by picking up an
2s electrons of beryllium. Therefore, it is easier electron. On the other hand, noble gases have
 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES 87

Table 3.7 Electron Gain Enthalpies* / (kJ mol–1 ) of Some Main Group Elements

Group 1 ∆ eg H Group 16 ∆ eg H Group 17 ∆ eg H Group 0 ∆ eg H

H – 73 He + 48
Li – 60 O – 141 F – 328 Ne + 116
Na – 53 S – 200 Cl – 349 Ar + 96
K – 48 Se – 195 Br – 325 Kr + 96

ed
Rb – 47 Te – 190 I – 295 Xe + 77
Cs – 46 Po – 174 At – 270 Rn + 68

large positive electron gain enthalpies because

h
Problem 3.7
the electron has to enter the next higher

pu T
principal quantum level leading to a very Which of the following will have the most

is
unstable electronic configuration. It may be negative electron gain enthalpy and which
the least negative?
re ER
noted that electron gain enthalpies have large
negative values toward the upper right of the P, S, Cl, F.

bl
periodic table preceding the noble gases. Explain your answer.
The variation in electron gain enthalpies of
elements is less systematic than for ionization Solution
enthalpies. As a general rule, electron gain Electron gain enthalpy generally becomes
be C

enthalpy becomes more negative with increase more negative across a period as we move
in the atomic number across a period. The from left to right. Within a group, electron
effective nuclear charge increases from left to gain enthalpy becomes less negative down
N

right across a period and consequently it will a group. However, adding an electron to
be easier to add an electron to a smaller atom the 2p-orbital leads to greater repulsion
since the added electron on an average would than adding an electron to the larger
be closer to the positively charged nucleus. We 3p-orbital. Hence the element with most
should also expect electron gain enthalpy to
©

negative electron gain enthalpy is chlorine;

become less negative as we go down a group
the one with the least negative electron
because the size of the atom increases and the gain enthalpy is phosphorus.
added electron would be farther from the
nucleus. This is generally the case (Table 3.7).
(e) Electronegativity
However, electron gain enthalpy of O or F is
less negative than that of the succeeding A qualitative measure of the ability of an atom
element. This is because when an electron is in a chemical compound to attract shared
to

added to O or F, the added electron goes to the electrons to itself is called electronegativity.
smaller n = 2 quantum level and suffers Unlike ionization enthalpy and electron gain
significant repulsion from the other electrons enthalpy, it is not a measureable quantity.
present in this level. For the n = 3 quantum However, a number of numerical scales of
level (S or Cl), the added electron occupies a electronegativity of elements viz., Pauling scale,
t

larger region of space and the electron-electron Mulliken-Jaffe scale, Allred-Rochow scale have
no

repulsion is much less. been developed. The one which is the most

* In many books, the negative of the enthalpy change for the process depicted in equation 3.3 is defined as the
ELECTRON AFFINITY (Ae ) of the atom under consideration. If energy is released when an electron is added to an atom,
the electron affinity is taken as positive, contrary to thermodynamic convention. If energy has to be supplied to add an
electron to an atom, then the electron affinity of the atom is assigned a negative sign. However, electron affinity is
defined as absolute zero and, therefore at any other temperature (T) heat capacities of the reactants and the products
have to be taken into account in ∆egH = –Ae – 5/2 RT.
88 CHEMISTRY

widely used is the Pauling scale. Linus Pauling, electrons and the nucleus increases as the
an American scientist, in 1922 assigned atomic radius decreases in a period. The
arbitrarily a value of 4.0 to fluorine, the element electronegativity also increases. On the same
considered to have the greatest ability to attract account electronegativity values decrease with
electrons. Approximate values for the the increase in atomic radii down a group. The
electronegativity of a few elements are given in trend is similar to that of ionization enthalpy.
Table 3.8(a) Knowing the relationship between
The electronegativity of any given element electronegativity and atomic radius, can you

ed
is not constant; it varies depending on the now visualise the relationship between
element to which it is bound. Though it is not electronegativity and non-metallic properties?
a measurable quantity, it does provide a means
of predicting the nature of force
that holds a pair of atoms together

h
– a relationship that you will

pu T
explore later.

is
Electronegativity generally
re ER
increases across a period from left
to right (say from lithium to

bl
fluorine) and decrease down a group
(say from fluorine to astatine) in
the periodic table. How can these
trends be explained? Can the
be C

electronegativity be related to
atomic radii, which tend to
decrease across each period from
N

left to right, but increase down

each group ? The attraction
between the outer (or valence) Fig. 3.7 The periodic trends of elements in the periodic table
©

Table 3.8(a) Electronegativity Values (on Pauling scale) Across the Periods

Atom (Period II) Li Be B C N O F

Electronegativity 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Atom (Period III) Na Mg Al Si P S Cl

to

Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0

Table 3.8(b) Electronegativity Values (on Pauling scale) Down a Family

Atom Electronegativity Atom Electronegativity

t

(Group I) Value (Group 17) Value

no

Li 1.0 F 4.0
Na 0.9 Cl 3.0
K 0.8 Br 2.8
Rb 0.8 I 2.5
Cs 0.7 At 2.2
 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES 89

Non-metallic elements have strong tendency electronic configuration 2s 22p5, shares one
to gain electrons. Therefore, electronegativity electron with oxygen in the OF2 molecule. Being
is directly related to that non-metallic highest electronegative element, fluorine is
properties of elements. It can be further given oxidation state –1. Since there are two
extended to say that the electronegativity is fluorine atoms in this molecule, oxygen with
2
inversely related to the metallic properties of outer electronic configuration 2s 2p4 shares
elements. Thus, the increase in two electrons with fluorine atoms and thereby
electronegativities across a period is exhibits oxidation state +2. In Na2O, oxygen

ed
accompanied by an increase in non-metallic being more electronegative accepts two
properties (or decrease in metallic properties) electrons, one from each of the two sodium
of elements. Similarly, the decrease in atoms and, thus, shows oxidation state –2. On
electronegativity down a group is accompanied the other hand sodium with electronic
by a decrease in non-metallic properties (or configuration 3s 1 loses one electron to oxygen

h
increase in metallic properties) of elements. and is given oxidation state +1. Thus, the

pu T
All these periodic trends are summarised oxidation state of an element in a particular

is
in figure 3.7. compound can be defined as the charge
acquired by its atom on the basis of
re ER
3.7.2 Periodic Trends in Chemical electronegative consideration from other atoms

bl
Properties in the molecule.
Most of the trends in chemical properties of
elements, such as diagonal relationships, inert Problem 3.8
pair effect, effects of lanthanoid contraction etc. Using the Periodic Table, predict the
be C

will be dealt with along the discussion of each formulas of compounds which might be
group in later units. In this section we shall formed by the following pairs of elements;
study the periodicity of the valence state shown (a) silicon and bromine (b) aluminium and
N

by elements and the anomalous properties of sulphur.

the second period elements (from lithium to
fluorine). Solution
(a) Periodicity of Valence or Oxidation (a) Silicon is group 14 element with a
valence of 4; bromine belongs to the
©

States
halogen family with a valence of 1.
The valence is the most characteristic property
Hence the formula of the compound
of the elements and can be understood in terms
formed would be SiBr4.
of their electronic configurations. The valence
of representative elements is usually (though (b) Aluminium belongs to group 13 with
not necessarily) equal to the number of a valence of 3; sulphur belongs to
electrons in the outermost orbitals and / or group 16 elements with a valence of
to

equal to eight minus the number of outermost 2. Hence, the formula of the compound
electrons as shown below. formed would be Al2S3.
Nowadays the term oxidation state is
Some periodic trends observed in the
frequently used for valence. Consider the two valence of elements (hydrides and oxides) are
oxygen containing compounds: OF2 and Na2O.
t

shown in Table 3.9. Other such periodic trends

The order of electronegativity of the three
which occur in the chemical behaviour of the
no

elements involved in these compounds is F > elements are discussed elsewhere in this book.
O > Na. Each of the atoms of fluorine, with outer
Group 1 2 13 14 15 16 17 18
Number of valence 1 2 3 4 5 6 7 8
electron
Valence 1 2 3 4 3,5 2,6 1,7 0,8
90 CHEMISTRY

Table 3.9 Periodic Trends in Valence of Elements as shown by the Formulas

of Their Compounds

Group 1 2 13 14 15 16 17

Formula LiH B2 H6 CH4 NH3 H2O HF

of hydride NaH CaH2 AlH3 SiH4 PH3 H2 S HCl
KH GeH4 AsH3 H2 Se HBr

ed
SnH4 SbH3 H2 Te HI
Formula Li2O MgO B2 O3 CO2 N2O 3, N2 O5 –
of oxide Na2 O CaO Al2O3 SiO2 P4 O6, P4O 10 SO3 Cl2 O 7

h
K2O SrO Ga2 O3 GeO2 As2O 3, As2 O5 SeO3 –

pu T
BaO In2O3 SnO2 Sb2O 3, Sb2O 5 TeO3 –

is
PbO2 Bi2 O3 – –
re ER
bl
There are many elements which exhibit variable following group i.e., magnesium and
valence. This is particularly characteristic of aluminium, respectively. This sort of similarity
transition elements and actinoids, which we is commonly referred to as diagonal
shall study later. relationship in the periodic properties.
be C

(b) Anomalous Properties of Second Period What are the reasons for the different
Elements chemical behaviour of the first member of a
group of elements in the s- and p-blocks
The first element of each of the groups 1
N

compared to that of the subsequent members

(lithium) and 2 (beryllium) and groups 13-17 in the same group? The anomalous behaviour
(boron to fluorine) differs in many respects from is attributed to their small size, large charge/
the other members of their respective group. radius ratio and high electronegativity of the
For example, lithium unlike other alkali metals, elements. In addition, the first member of
©

and beryllium unlike other alkaline earth group has only four valence orbitals (2s and
metals, form compounds with pronounced 2p) available for bonding, whereas the second
covalent character; the other members of these member of the groups have nine valence
groups predominantly form ionic compounds. orbitals (3s, 3p, 3d). As a consequence of this,
In fact the behaviour of lithium and beryllium the maximum covalency of the first member of
is more similar with the second element of the each group is 4 (e.g., boron can only form
−
to

Property Element [BF4 ] , whereas the other members

of the groups can expand their
Metallic radius M/ pm Li Be B valence shell to accommodate more
152 111 88 than four pairs of electrons e.g.,
3−
aluminium forms [ AlF6 ] ).
t

Na Mg Al
Furthermore, the first member of
no

186 160 143

p-block elements displays greater
Li Be ability to form pπ – p π multiple bonds
+
to itself (e.g., C = C, C ≡ C, N = N,
Ionic radius M / pm 76 31 N ≡ Ν) and to other second period
Na Mg elements (e.g., C = O, C = N, C ≡ N,
N = O) compared to subsequent
102 72
members of the same group.
 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES 91

Problem 3.9 non-metallic character increases while moving

from left to right across the period. The
Are the oxidation state and covalency of chemical reactivity of an element can be best
2+
Al in [AlCl(H2 O)5] same ? shown by its reactions with oxygen and
Solution halogens. Here, we shall consider the reaction
of the elements with oxygen only. Elements on
No. The oxidation state of Al is +3 and the two extremes of a period easily combine with
covalency is 6. oxygen to form oxides. The normal oxide

ed
formed by the element on extreme left is the
3.7.3 Periodic Trends and Chemical most basic (e.g., Na2O), whereas that formed
Reactivity by the element on extreme right is the most
acidic (e.g., Cl2O 7). Oxides of elements in the
We have observed the periodic trends in certain
centre are amphoteric (e.g., Al2O 3, As2O3) or

h
fundamental properties such as atomic and
neutral (e.g., CO, NO, N2O). Amphoteric oxides
ionic radii, ionization enthalpy, electron gain

pu T
behave as acidic with bases and as basic with
enthalpy and valence. We know by now that

is
acids, whereas neutral oxides have no acidic
the periodicity is related to electronic
or basic properties.
re ER
configuration. That is, all chemical and
physical properties are a manifestation of the

bl
Problem 3.10
electronic configuration of elements. We shall
now try to explore relationships between these Show by a chemical reaction with water
fundamental properties of elements with their that Na2O is a basic oxide and Cl2O7 is an
chemical reactivity. acidic oxide.
be C

The atomic and ionic radii, as we know, Solution

generally decrease in a period from left to right. Na2O with water forms a strong base
As a consequence, the ionization enthalpies
N

whereas Cl2O7 forms strong acid.

generally increase (with some exceptions as
outlined in section 3.7.1(a)) and electron gain Na2O + H2O → 2NaOH
enthalpies become more negative across a
period. In other words, the ionization enthalpy Cl2O 7 + H2O → 2HClO4
©

of the extreme left element in a period is the

least and the electron gain enthalpy of the Their basic or acidic nature can be
element on the extreme right is the highest qualitatively tested with litmus paper.
negative (note : noble gases having completely
filled shells have rather positive electron gain Among transition metals (3d series), the change
enthalpy values). This results into high in atomic radii is much smaller as compared
chemical reactivity at the two extremes and the to those of representative elements across the
to

lowest in the centre. Thus, the maximum period. The change in atomic radii is still
chemical reactivity at the extreme left (among smaller among inner-transition metals
alkali metals) is exhibited by the loss of an (4f series). The ionization enthalpies are
electron leading to the formation of a cation intermediate between those of s- and p-blocks.
and at the extreme right (among halogens) As a consequence, they are less electropositive
t

shown by the gain of an electron forming an than group 1 and 2 metals.

no

anion. This property can be related with the In a group, the increase in atomic and ionic
reducing and oxidizing behaviour of the radii with increase in atomic number generally
elements which you will learn later. However, results in a gradual decrease in ionization
here it can be directly related to the metallic enthalpies and a regular decrease (with
and non-metallic character of elements. Thus, exception in some third period elements as
the metallic character of an element, which is shown in section 3.7.1(d)) in electron gain
highest at the extremely left decreases and the enthalpies in the case of main group elements.
92 CHEMISTRY

Thus, the metallic character increases down will learn later. In the case of transition
the group and non-metallic character elements, however, a reverse trend is observed.
decreases. This trend can be related with their This can be explained in terms of atomic size
reducing and oxidizing property which you and ionization enthalpy.

SUMMARY

ed
In this Unit, you have studied the development of the Periodic Law and the Periodic
Table. Mendeleev’s Periodic Table was based on atomic masses. Modern Periodic Table
arranges the elements in the order of their atomic numbers in seven horizontal rows
(periods) and eighteen vertical columns (groups or families). Atomic numbers in a period

h
are consecutive, whereas in a group they increase in a pattern. Elements of the same
group have similar valence shell electronic configuration and, therefore, exhibit similar

pu T
chemical properties. However, the elements of the same period have incrementally

is
increasing number of electrons from left to right, and, therefore, have different valencies.
re ER
Four types of elements can be recognized in the periodic table on the basis of their
electronic configurations. These are s-block, p-block, d-block and f -block elements.

bl
Hydrogen with one electron in the 1s orbital occupies a unique position in the periodic
table. Metals comprise more than seventy eight per cent of the known elements. Non-
metals, which are located at the top of the periodic table, are less than twenty in number.
Elements which lie at the border line between metals and non-metals (e.g., Si, Ge, As)
are called metalloids or semi-metals. Metallic character increases with increasing atomic
be C

number in a group whereas decreases from left to right in a period. The physical and
chemical properties of elements vary periodically with their atomic numbers.
Periodic trends are observed in atomic sizes, ionization enthalpies, electron
N

gain enthalpies, electronegativity and valence. The atomic radii decrease while going
from left to right in a period and increase with atomic number in a group. Ionization
enthalpies generally increase across a period and decrease down a group. Electronegativity
also shows a similar trend. Electron gain enthalpies, in general, become more negative
©

across a period and less negative down a group. There is some periodicity in valence, for
example, among representative elements, the valence is either equal to the number of
electrons in the outermost orbitals or eight minus this number. Chemical reactivity is
hightest at the two extremes of a period and is lowest in the centre. The reactivity on the
left extreme of a period is because of the ease of electron loss (or low ionization enthalpy).
Highly reactive elements do not occur in nature in free state; they usually occur in the
combined form. Oxides formed of the elements on the left are basic and of the elements
on the right are acidic in nature. Oxides of elements in the centre are amphoteric or
to

neutral.

EXERCISES
t
no

3.1 What is the basic theme of organisation in the periodic table?

3.2 Which important property did Mendeleev use to classify the elements in his periodic
table and did he stick to that?
3.3 What is the basic difference in approach between the Mendeleev’s Periodic Law
and the Modern Periodic Law?
3.4 On the basis of quantum numbers, justify that the sixth period of the periodic
table should have 32 elements.
 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES 93

3.5 In terms of period and group where would you locate the element with Z =114?
3.6 Write the atomic number of the element present in the third period and seventeenth
group of the periodic table.
3.7 Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group?
3.8 Why do elements in the same group have similar physical and chemical properties?
3.9 What does atomic radius and ionic radius really mean to you?

ed
3.10 How do atomic radius vary in a period and in a group? How do you explain the
variation?
3.11 What do you understand by isoelectronic species? Name a species that will be
isoelectronic with each of the following atoms or ions.
– +
(iii) Mg2+

h
(i) F (ii) Ar (iv) Rb
3.12 Consider the following species :

pu T
3– 2– – + 2+ 3+
N , O , F , Na , Mg and Al

is
(a) What is common in them?
re ER
(b) Arrange them in the order of increasing ionic radii.

bl
3.13 Explain why cation are smaller and anions larger in radii than their parent atoms?
3.14 What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’
while defining the ionization enthalpy and electron gain enthalpy?
Hint : Requirements for comparison purposes.
be C

3.15 Energy of an electron in the ground state of the hydrogen atom is

–2.18×10–18J. Calculate the ionization enthalpy of atomic hydrogen in terms of
J mol–1 .
Hint: Apply the idea of mole concept to derive the answer.
N

3.16 Among the second period elements the actual ionization enthalpies are in the
order Li < B < Be < C < O < N < F < Ne.
Explain why
(i) Be has higher ∆i H than B
©

(ii) O has lower ∆i H than N and F?

3.17 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?
3.18 What are the various factors due to which the ionization enthalpy of the main
group elements tends to decrease down a group?
3.19 The first ionization enthalpy values (in kJ mol–1) of group 13 elements are :
to

B Al Ga In Tl
801 577 579 558 589
How would you explain this deviation from the general trend ?
3.20 Which of the following pairs of elements would have a more negative electron gain
t

enthalpy?
no

(i) O or F (ii) F or Cl
3.21 Would you expect the second electron gain enthalpy of O as positive, more negative
or less negative than the first? Justify your answer.
3.22 What is the basic difference between the terms electron gain enthalpy and
electronegativity?
3.23 How would you react to the statement that the electronegativity of N on Pauling
scale is 3.0 in all the nitrogen compounds?
94 CHEMISTRY

3.24 Describe the theory associated with the radius of an atom as it

(a) gains an electron
(b) loses an electron
3.25 Would you expect the first ionization enthalpies for two isotopes of the same element
to be the same or differ ent? Justify your answer.
3.26 What are the major differences between metals and non-metals?
3.27 Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.

ed
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal, liquid as well as gas at the room
temperature.

h
3.28 The increasing order of reactivity among group 1 elements is Li < Na < K < Rb <Cs
whereas that among group 17 elements is F > CI > Br > I. Explain.

pu T
3.29 Write the general outer electronic configuration of s-, p-, d- and f- block elements.

is
3.30 Assign the position of the element having outer electronic configuration
re ER
(i) ns2 np4 for n=3 (ii) (n-1)d2 ns2 for n=4, and (iii) (n-2) f 7 (n-1)d1 ns2 for n=6, in the
periodic table.

bl
3.31 The first (∆i H 1) and the second (∆i H2) ionization enthalpies (in kJ mol –1) and the
(∆egH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:
Elements ∆H 1 ∆H 2 ∆ egH
I 520 7300 –60
be C

II 419 3051 –48

III 1681 3374 –328
IV 1008 1846 –295
N

V 2372 5251 +48

VI 738 1451 –40
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 2(X=halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula
MX (X=halogen)?
3.32 Predict the formulas of the stable binary compounds that would be formed by the
to

combination of the following pairs of elements.

(a) Lithium and oxygen (b) Magnesium and nitrogen
(c) Aluminium and iodine (d) Silicon and oxygen
(e) Phosphorus and fluorine (f) Element 71 and fluorine
t

3.33 In the modern periodic table, the period indicates the value of :
no

(a) atomic number

(b) atomic mass
(c) principal quantum number
(d) azimuthal quantum number.
3.34 Which of the following statements related to the modern periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all
the orbitals in a p-shell.
 CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPER TIES 95

(b) The d-block has 8 columns, because a maximum of 8 electrons

can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of
electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (l) for the
last subshell that received electrons in building up the electronic
configuration.
3.35 Anything that influences the valence electrons will affect the chemistry

ed
of the element. Which one of the following factors does not affect the
valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z )
(c) Nuclear mass

h
(d) Number of core electrons.

pu T
–
3.36 The size of isoelectronic species — F , Ne and Na+ is affected by

is
(a) nuclear charge ( Z )
(b) valence principal quantum number (n)
re ER(c) electron-electron interaction in the outer orbitals

bl
(d) none of the factors because their size is the same.
3.37 Which one of the following statements is incorrect in relation to
ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
be C

(b) The greatest increase in ionization enthalpy is experienced on

removal of electron from core noble gas configuration.
N

(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 having higher n value.
©

3.38 Considering the elements B, Al, Mg, and K, the correct order of their
metallic character is :
(a) B > Al > Mg > K (b) Al > Mg > B > K
(c) Mg > Al > K > B (d) K > Mg > Al > B
3.39 Considering the elements B, C, N, F, and Si, the correct or der of their
non-metallic character is :
to

(a) B > C > Si > N > F (b) Si > C > B > N > F
(c) F > N > C > B > Si (d) F > N > C > Si > B
3.40 Considering the elements F, Cl, O and N, the correct order of their
chemical reactivity in terms of oxidizing property is :
t

(a) F > Cl > O > N (b) F > O > Cl > N

no

(c) Cl > F > O > N (d) O > F > N > Cl

96 CHEMISTR Y

UNIT 4

CHEMICAL BONDING AND

ed
MOLECULAR STRUCTURE

h
Scientists ar e constantly discovering new compounds, orderly

pu T
arranging the facts about them, trying to explain with the

is
existing knowledge, or ganising to modify the earlier views or
re ER
After studying this Unit, you will be
able to
evolve theories for explaining the newly observed facts.

bl
• understand K Ö ssel-Lewis
appr oach to chemical bonding;

• explain the octet rule and its Matter is made up of one or different type of elements.
be C

limitations, draw Lewis Under normal conditions no other element exists as an

structur es of simple molecules;
independent atom in nature, except noble gases. However,
• explain the for mation of different a group of atoms is found to exist together as one species
N

types of bonds; having characteristic properties. Such a group of atoms is

• describe the VSEPR theory and called a molecule. Obviously there must be some force
predict the geometry of simple which holds these constituent atoms together in the
molecules; molecules. The attractive force which holds various
©

• explain the valence bond constituents (atoms, ions, etc.) together in different
appr oach for the for mation of chemical species is called a chemical bond. Since the
covalent bonds; formation of chemical compounds takes place as a result
of combination of atoms of various elements in different
• predict the directional properties
of covalent bonds; ways, it raises many questions. Why do atoms combine?
Why are only certain combinations possible? Why do some
• explain the dif fer ent types of atoms combine while certain others do not? Why do
to

hybridisation involving s, p and

d orbitals and draw shapes of molecules possess definite shapes? To answer such
simple covalent molecules; questions different theories and concepts have been put
forward from time to time. These are Kö ssel-Lewis
• describe the molecular orbital
theory of homonuclear diatomic approach, Valence Shell Electron Pair Repulsion (VSEPR)
t

molecules; Theory, Valence Bond (VB) Theory and Molecular Orbital

no

(MO) Theory. The evolution of various theories of valence

• explain the concept of hydrogen
bond. and the interpretation of the nature of chemical bonds have
closely been r elated to the developments in the
understanding of the structure of atom, the electronic
configuration of elements and the periodic table. Every
system tends to be more stable and bonding is nature’s
way of lowering the energy of the system to attain stability.
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 97

4.1 KÖSSEL-LEWIS APPROACH TO the number of valence electrons. This number

CHEMICAL BONDING of valence electrons helps to calculate the
In order to explain the formation of chemical common or group valence of the element. The
bond in terms of electrons, a number of group valence of the elements is generally
attempts were made, but it was only in 1916 either equal to the number of dots in Lewis
when Kössel and Lewis succeeded symbols or 8 minus the number of dots or
independently in giving a satisfactory valence electrons.
explanation. They were the first to provide Kössel, in relation to chemical bonding,

ed
some logical explanation of valence which was drew attention to the following facts:
based on the inertness of noble gases.
• In the periodic table, the highly
Lewis pictured the atom in terms of a electronegative halogens and the highly
positively charged ‘Kernel’ (the nucleus plus electropositive alkali metals are separated

h
the inner electrons) and the outer shell that by the noble gases;
could accommodate a maximum of eight

pu T
electrons. He, further assumed that these • The formation of a negative ion from a

is
halogen atom and a positive ion from an
eight electrons occupy the corners of a cube
alkali metal atom is associated with the
re ER
which surround the ‘Kernel’. Thus the single
outer shell electron of sodium would occupy
gain and loss of an electron by the

bl
respective atoms;
one corner of the cube, while in the case of a
noble gas all the eight corners would be • The negative and positive ions thus
occupied. This octet of electrons, represents formed attain stable noble gas electronic
a particularly stable electronic arrangement. configurations. The noble gases (with the
be C

Lewis postulated that atoms achieve the exception of helium which has a duplet
stable octet when they are linked by of electrons) have a particularly stable
chemical bonds. In the case of sodium and outer shell configuration of eight (octet)
N

chlorine, this can happen by the transfer of electrons, ns 2np6 .

an electron from sodium to chlorine thereby • The negative and positive ions are
–
giving the Na + and Cl ions. In the case of stabilized by electrostatic attraction.
other molecules like Cl2 , H2 , F2 , etc., the bond For example, the formation of NaCl from
©

is formed by the sharing of a pair of electrons sodium and chlorine, according to the above
between the atoms. In the process each atom scheme, can be explained as:
attains a stable outer octet of electrons. Na → Na+ + e–
Lewis Symbols: In the for mation of a [Ne] 3s1 [Ne]
molecule, only the outer shell electrons take
Cl + e– → Cl–
part in chemical combination and they are
2 5
known as valence electrons. The inner shell [Ne] 3s 3p [Ne] 3s2 3p6 or [Ar]
to

electrons are well protected and are generally Na+ + Cl– → NaCl or Na+Cl–
not involved in the combination process. Similarly the formation of CaF2 may be
G.N. Lewis, an American chemist introduced shown as:
simple notations to represent valence
Ca → Ca2 + + 2e–
electrons in an atom. These notations are
t

called Lewis symbols. For example, the Lewis [Ar]4s 2 [Ar]

no

symbols for the elements of second period are F + e– → F–

as under: [He] 2s 2p 2 5
[He] 2s2 2p6 or [Ne]
– –
Ca2+ + 2F → CaF2 or Ca2 +(F )2
The bond formed, as a result of the
Significance of Lewis Symbols : The electrostatic attraction between the
number of dots around the symbol represents positive and negative ions was termed as
98 CHEMISTR Y

the electrovalent bond. The electrovalence chlorine atoms attain the outer shell octet of
is thus equal to the number of unit the nearest noble gas (i.e., argon).
charge(s) on the ion. Thus, calcium is The dots represent electrons. Such
assigned a positive electrovalence of two, structures are referred to as Lewis dot
while chlorine a negative electrovalence of structures.
one.
The Lewis dot structures can be written
Kössel’s postulations provide the basis for for other molecules also, in which the
the modern concepts regarding ion-formation combining atoms may be identical or

ed
by electron transfer and the formation of ionic different. The important conditions being that:
crystalline compounds. His views have proved
• Each bond is formed as a result of sharing
to be of great value in the understanding and
of an electron pair between the atoms.
systematisation of the ionic compounds. At
the same time he did recognise the fact that • Each combining atom contributes at least

h
a large number of compounds did not fit into one electron to the shared pair.

pu T
these concepts. • The combining atoms attain the outer-

is
shell noble gas configurations as a result
4.1.1 Octet Rule
of the sharing of electrons.
re ER
Kössel and Lewis in 1916 developed an
• Thus in water and carbon tetrachloride

bl
important theory of chemical combination
molecules, formation of covalent bonds
between atoms known as electronic theory
can be represented as:
of chemical bonding. According to this,
atoms can combine either by transfer of
valence electrons from one atom to another
be C

(gaining or losing) or by sharing of valence

electrons in order to have an octet in their
valence shells. This is known as octet rule.
N

4.1.2 Covalent Bond

L a n g m u i r (1919) re f i n e d t h e L e w i s
postulations by abandoning the idea of the
©

stationary cubical arrangement of the octet, Thus, when two atoms share one
and by introducing the term covalent bond. electron pair they are said to be joined by
The Lewis-Langmuir theory can be a single covalent bond. In many compounds
understood by considering the formation of we have multiple bonds between atoms. The
the chlorine molecule,Cl2 . The Cl atom with for mation of multiple bonds envisages
electronic configuration, [Ne]3s2 3p5 , is one sharing of more than one electr on pair
electron short of the argon configuration. between two atoms. If two atoms share two
to

The formation of the Cl2 molecule can be pairs of electrons, the covalent bond
understood in terms of the sharing of a pair between them is called a double bond. For
of electrons between the two chlorine atoms, example, in the carbon dioxide molecule, we
each chlorine atom contributing one electron have two double bonds between the carbon
to the shared pair. In the process both and oxygen atoms. Similarly in ethene
t

molecule the two carbon atoms are joined by

no

a double bond.

or Cl – Cl
Covalent bond between two Cl atoms Double bonds in CO2 molecule
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 99

in subtraction of one electron from the

total number of valence electrons. For
example, for the CO32– ion, the two negative
charges indicate that there are two
additional electrons than those provided
by the neutral atoms. For NH +4 ion, one
positive charge indicates the loss of one
C2H 4 molecule electron from the group of neutral atoms.

ed
• Knowing the chemical symbols of the
When combining atoms share three
combining atoms and having knowledge
electron pairs as in the case of two
of the skeletal structure of the compound
nitrogen atoms in the N2 molecule and the
(known or guessed intelligently), it is easy
two carbon atoms in the ethyne molecule,
to distribute the total number of electrons

h
a triple bond is formed.
as bonding shared pairs between the

pu T
atoms in proportion to the total bonds.

is
• In general the least electronegative atom
occupies the central position in the
re ER molecule/ion. For example in the NF3 and

bl
2–
N2 molecule CO3 , nitrogen and carbon are the central
atoms whereas fluorine and oxygen
occupy the terminal positions.
• After accounting for the shared pairs of
be C

electrons for single bonds, the remaining

C2 H2 molecule
electron pairs are either utilized for
multiple bonding or remain as the lone
N

4.1.3 Lewis Representation of Simple pairs. The basic requirement being that
Molecules (the Lewis Structures) each bonded atom gets an octet of
The Lewis dot structures provide a picture electrons.
of bonding in molecules and ions in terms Lewis representations of a few molecules/
©

of the shared pairs of electrons and the ions are given in Table 4.1.
octet rule. While such a picture may not Table 4.1 The Lewis Representation of Some
explain the bonding and behaviour of a Molecules
molecule completely, it does help in
understanding the formation and properties
of a molecule to a large extent. Writing of
Lewis dot structures of molecules is,
to

therefore, very useful. The Lewis dot

structures can be written by adopting the
following steps:
• The total number of electrons required for
writing the structures are obtained by
t

adding the valence electrons of the

no

combining atoms. For example, in the CH4

molecule there are eight valence electrons
available for bonding (4 from carbon and
4 from the four hydrogen atoms).
• For anions, each negative charge would
mean addition of one electron. For * Each H atom attains the configuration of helium (a duplet
cations, each positive charge would result of electrons)
100 CHEMISTR Y

Problem 4.1 each of the oxygen atoms completing the

Write the Lewis dot structure of CO octets on oxygen atoms. This, however,
does not complete the octet on nitrogen
molecule.
if the remaining two electrons constitute
Solution lone pair on it.
Step 1. Count the total number of
valence electrons of carbon and oxygen
atoms. The outer (valence) shell

ed
configurations of carbon and oxygen Hence we have to resort to multiple
atoms are: 2s 2 2p 2 and 2 s 2 2 p 4 , bonding between nitrogen and one of the
respectively. The valence electrons oxygen atoms (in this case a double
available are 4 + 6 =10. bond). This leads to the following Lewis

h
Step 2. The skeletal structure of CO is dot structures.
written as: C O

pu T
Step 3. Draw a single bond (one shared

is
electron pair) between C and O and
re ER
complete the octet on O, the remaining
two electrons are the lone pair on C.

bl
This does not complete the octet on
be C

carbon and hence we have to resort to

multiple bonding (in this case a triple 4.1.4 Formal Charge
bond) between C and O atoms. This Lewis dot structures, in general, do not
N

satisfies the octet rule condition for both represent the actual shapes of the molecules.
atoms. In case of polyatomic ions, the net charge is
possessed by the ion as a whole and not by a
particular atom. It is, however, feasible to
©

assign a formal charge on each atom. The

formal charge of an atom in a polyatomic
molecule or ion may be defined as the
Problem 4.2
difference between the number of valence
Write the Lewis structure of the nitrite electrons of that atom in an isolated or free
ion, NO2– . state and the number of electrons assigned
to that atom in the Lewis structure. It is
to

Solution
expressed as :
Step 1. Count the total number of
valence electrons of the nitrogen atom, For mal charge (F.C.)
on an atom in a Lewis =
the oxygen atoms and the additional one structur e
negative charge (equal to one electron).
t

N(2s2 2p3 ), O (2s2 2p4 )

no

5 + (2 × 6) +1 = 18 electrons total number of valence total number of non

electr ons in the free — bonding (lone pair)
–
Step 2. The skeletal structure of NO2 is atom electrons
written as : O N O total number of
Step 3. Draw a single bond (one shared — (1/2) bonding(shared)
electrons
electron pair) between the nitrogen and
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 101

The counting is based on the assumption 4.1.5 Limitations of the Octet Rule
that the atom in the molecule owns one The octet rule, though useful, is not universal.
electron of each shared pair and both the It is quite useful for understanding the
electrons of a lone pair. structures of most of the organic compounds
Let us consider the ozone molecule (O3). and it applies mainly to the second period
The Lewis structure of O3 may be drawn as : elements of the periodic table. There are three
types of exceptions to the octet rule.
The incomplete octet of the central atom

ed
In some compounds, the number of electrons
surrounding the central atom is less than
eight. This is especially the case with elements
having less than four valence electrons.

h
The atoms have been numbered as 1, 2 Examples are LiCl, BeH2 and BCl3 .

pu T
and 3. The formal charge on:

is
• The central O atom marked 1
re ER
1 Li, Be and B have 1,2 and 3 valence electrons

bl
=6–2– (6) = +1
2 only. Some other such compounds are AlCl3
• The end O atom marked 2 and BF3 .
Odd-electron molecules
1
=6–4– (4) = 0 In molecules with an odd number of electrons
be C

2 like nitric oxide, NO and nitrogen dioxide,

• The end O atom marked 3 NO2, the octet rule is not satisfied for all the
atoms
N

1
=6–6– (2) = –1
2
Hence, we represent O 3 along with the The expanded octet
formal charges as follows:
©

Elements in and beyond the third period of

the periodic table have, apart from 3s and 3p
orbitals, 3d orbitals also available for bonding.
In a number of compounds of these elements
there are more than eight valence electrons
around the central atom. This is termed as
We must understand that formal charges the expanded octet. Obviously the octet rule
to

do not indicate real charge separation within does not apply in such cases.
the molecule. Indicating the charges on the Some of the examples of such compounds
atoms in the Lewis structure only helps in are: PF5 , SF 6 , H 2 SO4 and a number of
keeping track of the valence electrons in the coordination compounds.
molecule. Formal charges help in the
t

selection of the lowest energy structure from

no

a number of possible Lewis structures for a

given species. Generally the lowest energy
structure is the one with the smallest
formal charges on the atoms. The formal
charge is a factor based on a pure covalent
view of bonding in which electron pairs
are shared equally by neighbouring atoms.
102 CHEMISTR Y

Interestingly, sulphur also forms many Obviously ionic bonds will be formed
compounds in which the octet rule is obeyed. m o re easily between elements with
In sulphur dichloride, the S atom has an octet comparatively low ionization enthalpies
of electrons around it. and elements with comparatively high
negative value of electron gain enthalpy.
Most ionic compounds have cations
derived from metallic elements and anions
Other drawbacks of the octet theory
fr o m n o n - m e t a l l i c e l e m e n t s . T h e

ed
• It is clear that octet rule is based upon +
ammonium ion, NH4 (made up of two non-
the chemical inertness of noble gases. metallic elements) is an exception. It forms
However, some noble gases (for example the cation of a number of ionic compounds.
xenon and krypton) also combine with
Ionic compounds in the crystalline state
oxygen and fluorine to form a number of

h
c o n s i s t o f o r d e r l y t h re e - d i m e n s i o n a l
compounds like XeF2, KrF2, XeOF2 etc.,
arrangements of cations and anions held

pu T
• This theory does not account for the shape
together by coulombic interaction energies.

is
of molecules.
These compounds crystallise in different
• It does not explain the relative stability of
re ER
the molecules being totally silent about
crystal structures determined by the size
of the ions, their packing arrangements and

bl
the energy of a molecule.
other factors. The crystal structure of
4.2 IONIC OR ELECTROVALENT BOND sodium chloride, NaCl (rock salt), for
From the Kössel and Lewis treatment of the example is shown below.
formation of an ionic bond, it follows that the
be C

f o rmation of ionic compounds would

primarily depend upon:
• The ease of formation of the positive and
N

negative ions from the respective neutral

atoms;
• The arrangement of the positive and
negative ions in the solid, that is, the
©

lattice of the crystalline compound.

The formation of a positive ion involves
ionization, i.e., removal of electron(s) from
the neutral atom and that of the negative ion
involves the addition of electron(s) to the Rock salt structure
neutral atom. In ionic solids, the sum of the electron
→ M+ (g) + e– ;
to

M(g) gain enthalpy and the ionization enthalpy

Ionization enthalpy may be positive but still the crystal
– –
X(g) + e → X (g) ; structure gets stabilized due to the energy
Electron gain enthalpy r eleased in the formation of the crystal
+ –
M (g) + X (g) → MX(s) lattice. For example: the ionization
t

The electron gain enthalpy, ∆eg H, is the enthalpy for Na + (g) formation from Na(g)
no

enthalpy change (Unit 3), when a gas phase atom is 495.8 kJ mol–1 ; while the electron gain
in its ground state gains an electron. The enthalpy for the change Cl(g) + e – →
electron gain process may be exothermic or Cl – (g) is, – 348.7 kJ mol–1 only. The sum
endothermic. The ionization, on the other hand, of the two, 147.1 kJ mol -1 is more than
is always endothermic. Electron affinity, is the compensated for by the enthalpy of lattice
negative of the energy change accompanying f o rmation of NaCl(s) (–788 kJ mol –1).
electron gain. Therefore, the energy released in the
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 103

processes is more than the energy absorbed.

Thus a qualitative measure of the
stability of an ionic compound is
provided by its enthalpy of lattice
formation and not simply by achieving
octet of electrons around the ionic species
in gaseous state.
Since lattice enthalpy plays a key role

ed
in the formation of ionic compounds, it is
important that we learn more about it.
4.2.1 Lattice Enthalpy
The Lattice Enthalpy of an ionic solid is

h
defined as the energy required to

pu T
completely separate one mole of a solid

is
ionic compound into gaseous constituent
ions. For example, the lattice enthalpy of NaCl
re ER
is 788 kJ mol–1 . This means that 788 kJ of Fig. 4.1 The bond length in a covalent
molecule AB.

bl
energy is required to separate one mole of R = rA + rB (R is the bond length and rA and rB
solid NaCl into one mole of Na+ (g) and one are the covalent radii of atoms A and B
mole of Cl– (g) to an infinite distance. respectively)
This process involves both the attractive covalent bond in the same molecule. The van
be C

forces between ions of opposite charges and der Waals radius represents the overall size
the repulsive forces between ions of like of the atom which includes its valence shell
charge. The solid crystal being three- in a nonbonded situation. Further, the van
N

dimensional; it is not possible to calculate der Waals radius is half of the distance
lattice enthalpy directly from the interaction between two similar atoms in separate
of forces of attraction and repulsion only. molecules in a solid. Covalent and van der
Factors associated with the crystal geometry Waals radii of chlorine are depicted in Fig.4.2
have to be included.
©

rc = 99 pm 19
4.3 BOND PARAMETERS 8
pm
4.3.1 Bond Length
Bond length is defined as the equilibrium
distance between the nuclei of two bonded
atoms in a molecule. Bond lengths are
to

measured by spectroscopic, X-ray diffraction

and electron-diffraction techniques about
r vd

which you will learn in higher classes. Each

w
=
18

atom of the bonded pair contributes to the

0
pm

bond length (Fig. 4.1). In the case of a covalent

t

pm

bond, the contribution from each atom is

no

0
36

called the covalent radius of that atom.

The covalent radius is measure d
approximately as the radius of an atom’s Fig. 4.2 Covalent and van der Waals radii in a
core which is in contact with the core of chlorine molecule .The inner circles
an adjacent atom in a bonded situation. correspond to the size of the chlorine atom
The covalent radius is half of the distance (rvdw and r c ar e van der Waals and
covalent radii respectively).
between two similar atoms joined by a
104 CHEMISTR Y

Some typical average bond lengths for Table 4.2 Average Bond Lengths for Some
single, double and triple bonds are shown in Single, Double and Triple Bonds
Table 4.2. Bond lengths for some common
Bond Type Covalent Bond Length
molecules are given in Table 4.3.
(pm)
The covalent radii of some common
O–H 96
elements are listed in Table 4.4.
C–H 107
4.3.2 Bond Angle N–O 136
C–O 143

ed
It is defined as the angle between the orbitals
C–N 143
containing bonding electron pairs around the C–C 154
central atom in a molecule/complex ion. Bond C=O 121
angle is expressed in degree which can be N=O 122
experimentally determined by spectroscopic C=C 133

h
methods. It gives some idea regarding the C=N 138

pu T
distribution of orbitals around the central C≡N 116

is
atom in a molecule/complex ion and hence it C≡C 120
helps us in determining its shape. For
re ER
example H–O–H bond angle in water can be Table 4.3 Bond Lengths in Some Common
Molecules

bl
represented as under :
Molecule Bond Length
(pm)
H2 (H – H) 74
be C

F2 (F – F) 144
4.3.3 Bond Enthalpy Cl2 (Cl – Cl) 199
It is defined as the amount of energy required Br2 (Br – Br) 228
N

to break one mole of bonds of a particular I2 (I – I) 267

type between two atoms in a gaseous state. N2 (N ≡ N) 109
The unit of bond enthalpy is kJ mol–1. For O2 (O = O) 121
example, the H – H bond enthalpy in hydrogen HF (H – F) 92
©

molecule is 435.8 kJ mol–1. HCl (H – Cl) 127

V HBr (H – Br) 141
H2(g) → H(g) + H(g); ∆a H = 435.8 kJ mol–1
HI (H – I) 160
Similarly the bond enthalpy for molecules
containing multiple bonds, for example O2 and
Table 4.4 Covalent Radii, *rcov/(pm)
N2 will be as under :
O2 (O = O) (g) → O(g) + O(g);
V
∆a H = 498 kJ mol–1
to

N2 (N ≡ N) (g) → N(g) + N(g);

V
∆a H = 946.0 kJ mol–1
It is important that larger the bond
dissociation enthalpy, stronger will be the
t

bond in the molecule. For a heteronuclear

no

diatomic molecules like HCl, we have

V
HCl (g) → H(g) + Cl (g); ∆a H = 431.0 kJ mol –1
In case of polyatomic molecules, the
measurement of bond strength is more
complicated. For example in case of H2O * The values cited ar e for single bonds, except where
molecule, the enthalpy needed to break the otherwise indicated in parenthesis. (See also Unit 3 for
two O – H bonds is not the same. periodic trends).
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 105

V
H2O(g) → H(g) + OH(g); ∆ aH 1 = 502 kJ mol–1 shown below:
OH(g) → H(g) + O(g); ∆a HV2 = 427 kJ mol–1 In both structures we have a O–O single
V
The difference in the ∆aH value shows that
the second O – H bond undergoes some change
because of changed chemical environment.
This is the reason for some difference in energy
of the same O – H bond in different molecules
like C2H 5OH (ethanol) and water. Therefore in

ed
polyatomic molecules the term mean or
average bond enthalpy is used. It is obtained
by dividing total bond dissociation enthalpy
by the number of bonds broken as explained

h
below in case of water molecule,

pu T
502 + 427

is
Average bond enthalpy =
2 Fig. 4.3 Resonance in the O 3 molecule
re ER = 464.5 kJ mol –1
(structures I and II represent the two canonical

bl
forms while the structure III is the resonance
4.3.4 Bond Order
hybrid)
In the Lewis description of covalent bond,
the Bond Order is given by the number of bond and a O=O double bond. The normal
bonds between the two atoms in a O–O and O=O bond lengths are 148 pm and
be C

molecule. The bond order, for example in H2 121 pm respectively. Experimentally

(with a single shared electron pair), in O2 determined oxygen-oxygen bond lengths in
(with two shared electron pairs) and in N2 the O3 molecule are same (128 pm). Thus the
N

(with three shared electron pairs) is 1,2,3 oxygen-oxygen bonds in the O3 molecule are
respectively. Similarly in CO (three shared intermediate between a double and a single
electron pairs between C and O) the bond bond. Obviously, this cannot be represented
order is 3. For N2, bond order is 3 and its by either of the two Lewis structures shown
©

∆a H V is 946 kJ mol–1; being one of the above.

highest for a diatomic molecule. The concept of resonance was introduced
to deal with the type of difficulty experienced
Isoelectronic molecules and ions have
identical bond orders; for example, F 2 and in the depiction of accurate structures of
molecules like O3 . According to the concept
O22– have bond order 1. N2, CO and NO +
of resonance, whenever a single Lewis
have bond order 3.
structure cannot describe a molecule
to

A general correlation useful for accurately, a number of structures with

understanding the stablities of molecules similar energy, positions of nuclei, bonding
is that: with increase in bond order, bond and non-bonding pairs of electrons are taken
enthalpy increases and bond length as the canonical structures of the hybrid
decreases.
t

which describes the molecule accurately.

Thus for O3, the two structures shown above
no

4.3.5 Resonance Structures

constitute the canonical structure s o r
It is often observed that a single Lewis resonance structures and their hybrid i.e., the
structure is inadequate for the representation III structure represents the structure of O3
of a molecule in conformity with its more accurately. This is also called resonance
experimentally determined parameters. For hybrid. Resonance is represented by a double
example, the ozone, O 3 molecule can be headed arrow.
equally represented by the structures I and II
106 CHEMISTR Y

Some of the other examples of resonance

structures are provided by the carbonate ion
and the carbon dioxide molecule.
Fig. 4.5 Resonance in CO2 molecule, I, II
Problem 4.3
and III r epr esent the three
Explain the structure of CO32 – ion in canonical for ms.
terms of resonance.
Solution In general, it may be stated that

ed
The single Lewis structure based on the • Resonance stabilizes the molecule as the
presence of two single bonds and one energy of the resonance hybrid is less
double bond between carbon and oxygen than the energy of any single cannonical
atoms is inadequate to represent the structure; and,

h
molecule accurately as it represents • Resonance averages the bond
unequal bonds. According to the characteristics as a whole.

pu T
experimental findings, all carbon to Thus the energy of the O3 resonance

is
oxygen bonds in CO 2– 3
are equivalent. hybrid is lower than either of the two
re ER
Therefore the carbonate ion is best
described as a resonance hybrid of the
cannonical froms I and II (Fig 4.3).

bl
canonical forms I, II, and III shown below. Many misconceptions are associated
with resonance and the same need to be
dispelled. You should remember that :
• The cannonical forms have no real
be C

existence.
• The molecule does not exist for a
certain fraction of time in one
N

2– cannonical form and for other

Fig.4.4 Resonance in CO3 , I, II and
III r epresent the thr ee
fractions of time in other cannonical
canonical for ms. forms.
• There is no such equilibrium between
©

Problem 4.4
the cannonical forms as we have
Explain the structure of CO2 molecule.
between tautomeric forms (keto and
Solution enol) in tautomerism.
The experimentally determined carbon • The molecule as such has a single
to oxygen bond length in CO 2 i s structure which is the resonance
115 pm. The lengths of a normal hybrid of the cannonical forms and
carbon to oxygen double bond (C=O) which cannot as such be depicted by
to

and carbon to oxygen triple bond (C≡ O) a single Lewis structure.

are 121 pm and 110 pm respectively.
The carbon-oxygen bond lengths in 4.3.6 Polarity of Bonds
CO 2 (115 pm) lie between the values
The existence of a hundred percent ionic or
t

for C=O and C≡O. Obviously, a single

covalent bond represents an ideal situation.
no

Lewis structure cannot depict this

In reality no bond or a compound is either
position and it becomes necessary to
completely covalent or ionic. Even in case of
write more than one Lewis structures
covalent bond between two hydrogen atoms,
and to consider that the structure of
there is some ionic character.
CO 2 is best described as a hybrid of
the canonical or resonance forms I, II When covalent bond is formed between
and III. two similar atoms, for example in H2, O2, Cl2,
N2 or F2, the shared pair of electrons is equally
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 107

attracted by the two atoms. As a result electron In case of polyatomic molecules the dipole
pair is situated exactly between the two moment not only depend upon the individual
identical nuclei. The bond so formed is called dipole moments of bonds known as bond
nonpolar covalent bond. Contrary to this in dipoles but also on the spatial arrangement of
case of a heteronuclear molecule like HF, the various bonds in the molecule. In such case,
shared electron pair between the two atoms the dipole moment of a molecule is the vector
gets displaced more towards fluorine since the sum of the dipole moments of various bonds.
electronegativity of fluorine (Unit 3) is far For example in H2O molecule, which has a bent

ed
greater than that of hydrogen. The resultant structure, the two O–H bonds are oriented at
covalent bond is a polar covalent bond. an angle of 104.50. Net dipole moment of 6.17
As a result of polarisation, the molecule × 10–30 C m (1D = 3.33564 × 10–30 C m) is the
possesses the dipole moment (depicted resultant of the dipole moments of two O–H
below) which can be defined as the product bonds.

h
of the magnitude of the charge and the

pu T
distance between the centres of positive and

is
negative charge. It is usually designated by a
Greek letter ‘µ’. Mathematically, it is expressed
as follows :
re ER
bl
Dipole moment (µ) = charge (Q) × distance of
separation (r)
Dipole moment is usually expressed in Net Dipole moment, µ = 1.85 D
Debye units (D). The conversion factor is = 1.85 × 3.33564 × 10–30 C m = 6.17 ×10 –30 C m
be C

1 D = 3.33564 × 10–30 C m The dipole moment in case of BeF2 is zero.

where C is coulomb and m is meter. This is because the two equal bond dipoles
Further dipole moment is a vector quantity point in opposite directions and cancel the
N

and by convention it is depicted by a small effect of each other.

arrow with tail on the negative centre and head
pointing towards the positive centre. But in
chemistry presence of dipole moment is
©

represented by the crossed arrow ( ) put

on Lewis structure of the molecule. The cross In tetra-atomic molecule, for example in
is on positive end and arrow head is on negative BF3, the dipole moment is zero although the
end. For example the dipole moment of HF may B – F bonds are oriented at an angle of 120o to
be represented as : one another, the three bond moments give a
H F
net sum of zero as the resultant of any two is
equal and opposite to the third.
to

This arrow symbolises the direction of the

shift of electron density in the molecule. Note
that the direction of crossed arrow is opposite
to the conventional direction of dipole moment
t

vector.
no

Peter Debye, the Dutch chemist

received Nobel prize in 1936 for
his work on X-ray diffraction and Let us study an interesting case of NH3
dipole moments. The magnitude and NF3 molecule. Both the molecules have
of the dipole moment is given in pyramidal shape with a lone pair of electrons
Debye units in order to honour him. on nitrogen atom. Although fluorine is more
electronegative than nitrogen, the resultant
108 CHEMISTR Y

dipole moment of NH3 ( 4.90 × 10–30 C m) is in terms of the following rules:

greater than that of NF3 (0.8 × 10–30 C m). This • The smaller the size of the cation and the
is because, in case of NH3 the orbital dipole larger the size of the anion, the greater the
due to lone pair is in the same direction as the covalent character of an ionic bond.
resultant dipole moment of the N – H bonds, • The greater the charge on the cation, the
whereas in NF3 the orbital dipole is in the greater the covalent character of the ionic bond.
direction opposite to the resultant dipole • For cations of the same size and charge,
moment of the three N–F bonds. The orbital the one, with electronic configuration

ed
dipole because of lone pair decreases the effect (n-1)dnnso , typical of transition metals, is
of the resultant N – F bond moments, which more polarising than the one with a noble
results in the low dipole moment of NF3 as gas configuration, ns2 np6, typical of alkali
represented below : and alkaline earth metal cations.

h
The cation polarises the anion, pulling the
electronic charge toward itself and thereby

pu T
increasing the electronic charge between

is
the two. This is precisely what happens in
re ER a covalent bond, i.e., buildup of electron
charge density between the nuclei. The

bl
polarising power of the cation, the
polarisability of the anion and the extent
of distortion (polarisation) of anion are the
factors, which determine the per cent
be C

Dipole moments of some molecules are covalent character of the ionic bond.
shown in Table 4.5. 4.4 THE VALENCE SHELL ELECTRON
N

Just as all the covalent bonds have PAIR REPULSION (VSEPR) THEORY
some partial ionic character, the ionic As already explained, Lewis concept is unable
bonds also have partial covalent to explain the shapes of molecules. This theory
character. The partial covalent character provides a simple procedure to predict the
of ionic bonds was discussed by Fajans
©

shapes of covalent molecules. Sidgwick

Table 4.5 Dipole Moments of Selected Molecules

Type of Example Dipole Geometry

Molecule Moment, µ(D)

Molecule (AB) HF 1.78 linear

HCl 1.07 linear
to

HBr 0.79 linear

HI 0.38 linear
H2 0 linear
Molecule (AB2) H2O 1.85 bent
t

H 2S 0.95 bent
CO2 0 linear
no

Molecule (AB3) NH3 1.47 trigonal-pyramidal

NF3 0.23 trigonal-pyramidal
BF3 0 trigonal-planar
Molecule (AB4) CH4 0 tetrahedral
CHCl3 1.04 tetrahedral
CCl4 0 tetrahedral
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 109

and Powell in 1940, proposed a simple theory result in deviations from idealised shapes and
based on the repulsive interactions of the alterations in bond angles in molecules.
electron pairs in the valence shell of the atoms. For the prediction of geometrical shapes of
It was further developed and redefined by molecules with the help of VSEPR theory, it is
Nyholm and Gillespie (1957). convenient to divide molecules into two
The main postulates of VSEPR theory are categories as (i) molecules in which the
as follows: central atom has no lone pair and (ii)
• The shape of a molecule depends upon molecules in which the central atom has

ed
the number of valence shell electron pairs one or more lone pairs.
(bonded or nonbonded) around the central Table 4.6 (page110) shows the
atom. arrangement of electron pairs about a central
atom A (without any lone pairs) and
• Pairs of electrons in the valence shell repel

h
geometries of some molecules/ions of the type
one another since their electron clouds are
AB. Table 4.7 (page 111) shows shapes of

pu T
negatively charged. some simple molecules and ions in which the

is
• These pairs of electrons tend to occupy central atom has one or more lone pairs. Table
re ER
such positions in space that minimise
repulsion and thus maximise distance
4.8 (page 112) explains the reasons for the
distortions in the geometry of the molecule.

bl
between them. As depicted in Table 4.6, in the
• The valence shell is taken as a sphere with compounds of AB2, AB3, AB4, AB5 and AB6,
the electron pairs localising on the the arrangement of electron pairs and the B
spherical surface at maximum distance atoms around the central atom A are : linear,
be C

from one another. trigonal planar, tetrahedral, trigonal-

• A multiple bond is treated as if it is a single bipyramidal and octahedral, respectively.
electron pair and the two or three electron Such arrangement can be seen in the
N

pairs of a multiple bond are treated as a molecules like BF3 (AB3), CH4 (AB4) and PCl5
single super pair. (AB5) as depicted below by their ball and stick
models.
• Where two or more resonance structures
©

can represent a molecule, the VSEPR

model is applicable to any such structure.
The repulsive interaction of electron pairs
decrease in the order:
Lone pair (lp) – Lone pair (lp) > Lone pair (lp)
– Bond pair (bp) > Bond pair (bp) – Fig. 4.6 The shapes of molecules in which
Bond pair (bp) central atom has no lone pair
to

Nyholm and Gillespie (1957) refined the The VSEPR Theory is able to predict
VSEPR model by explaining the important geometry of a large number of molecules,
difference between the lone pairs and bonding especially the compounds of p-block elements
pairs of electrons. While the lone pairs are accurately. It is also quite successful in
t

localised on the central atom, each bonded pair determining the geometry quite-accurately
no

is shared between two atoms. As a result, the even when the energy difference between
lone pair electrons in a molecule occupy more possible structures is very small. The
space as compared to the bonding pairs of theoretical basis of the VSEPR theory
electrons. This results in greater repulsion regarding the effects of electron pair repulsions
between lone pairs of electrons as compared on molecular shapes is not clear and
to the lone pair - bond pair and bond pair - continues to be a subject of doubt and
bond pair repulsions. These repulsion effects discussion.
110 CHEMISTR Y

Table 4.6 Geometry of Molecules in which the Central Atom has No Lone Pair of Electrons

h ed
pu T
is
re ER
bl
be C
N
t to ©
no
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 111

Table 4.7 Shape (geometry) of Some Simple Molecules/Ions with Central Ions having One or
More Lone Pairs of Electrons(E).

h ed
pu T
is
re ER
bl
be C
N
t to ©
no
112 CHEMISTR Y

Table 4.8 Shapes of Molecules containing Bond Pair and Lone Pair
Molecule No. of No. of Arrangement Shape Reason for the
type bonding lone of electrons shape acquired
pairs pairs

AB2 E 4 1 Bent Theoretically the shape

should have been triangular
planar but actually it is found
to be bent or v-shaped. The

ed
reason being the lone pair -
bond pair repulsion is much
mor e as compared to the
bond pair -bond pair repul-
sion. So the angle is reduced

h
to 119.5° from 120°.

pu T
is
AB3E 3 1 T rigonal Had there been a bp in place
pyramidal of lp the shape would have
re ER been tetrahedral but one

bl
lone pair is present and due
to the r epulsion between
lp-bp (which is more than
bp-bp repulsion) the angle
between bond pairs is
be C

reduced to 107° from 109.5°.

N

Bent The shape should have been

AB2E2 2 2 tetrahedral if there were all bp
but two lp are present so the
shape is distorted tetrahedral
©

or angular. The r eason is

lp-lp repulsion is more than
lp-bp repulsion which is more
than bp-bp repulsion. Thus,
the angle is reduced to 104.5°
from 109.5°.
to

AB4E 4 1 See- In (a) the lp is present at axial

saw position so ther e are thr ee
t

lp—bp repulsions at 90°. In(b)

no

the lp is in an equatorial
position, and there are two
lp—bp repulsions. Hence,
arrangement (b) is mor e
stable. The shape shown in (b)
is described as a distorted
tetrahedron, a folded square or
(More stable) a see-saw.
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 113

Molecule No. of No. of Arrangement Shape Reason for the

type bonding lone of electrons shape acquired
pairs pairs

AB3E2 3 2 T -shape In (a) the lp are at

equatorial position so
there ar e less lp-bp
repulsions as

ed
compared to others in
which the lp are at
axial positions. So
structure (a) is most
stable. (T-shaped).

h
pu T
is
re ER
bl
be C

4.5 VALENCE BOND THEORY knowledge of atomic orbitals, electronic

N

As we know that Lewis approach helps in configurations of elements (Units 2), the
writing the structure of molecules but it fails overlap criteria of atomic orbitals, the
to explain the formation of chemical bond. It hybridization of atomic orbitals and the
also does not give any reason for the difference principles of variation and superposition. A
©

in bond dissociation enthalpies and bond rigorous treatment of the VB theory in terms
lengths in molecules like H2 (435.8 kJ mol-1, of these aspects is beyond the scope of this
74 pm) and F2 (155 kJ mol - 1, 144 pm), book. Therefore, for the sake of convenience,
although in both the cases a single covalent valence bond theory has been discussed in
bond is formed by the sharing of an electron terms of qualitative and non-mathematical
pair between the respective atoms. It also gives treatment only. To start with, let us consider
no idea about the shapes of polyatomic the formation of hydrogen molecule which is
to

molecules. the simplest of all molecules.

Similarly the VSEPR theory gives the Consider two hydrogen atoms A and B
geometry of simple molecules but approaching each other having nuclei NA and
theoretically, it does not explain them and also NB and electrons present in them are
it has limited applications. To overcome these
t

represented by eA and eB . When the two atoms

limitations the two important theories based
no

are at large distance from each other, there is

on quantum mechanical principles are
no interaction between them. As these two
introduced. These are valence bond (VB) theory
atoms approach each other, new attractive and
and molecular orbital (MO) theory.
repulsive forces begin to operate.
Valence bond theory was introduced by
Attractive forces arise between:
Heitler and London (1927) and developed
further by Pauling and others. A discussion (i) nucleus of one atom and its own electron
of the valence bond theory is based on the that is NA – eA and NB – eB .
114 CHEMISTR Y

(ii) nucleus of one atom and electron of other hydrogen atoms are said to be bonded together
atom i.e., NA – eB, NB – eA. to form a stable molecule having the bond
Similarly repulsive forces arise between length of 74 pm.
(i) electrons of two atoms like e A – eB , Since the energy gets released when the
(ii) nuclei of two atoms NA – NB . bond is formed between two hydrogen atoms,
Attractive forces tend to bring the two the hydrogen molecule is more stable than that
atoms close to each other whereas repulsive of isolated hydrogen atoms. The energy so
forces tend to push them apart (Fig. 4.7). released is called as bond enthalpy, which is

ed
corresponding to minimum in the curve
depicted in Fig. 4.8. Conversely, 435.8 kJ of
energy is required to dissociate one mole of
H 2 molecule.
H 2(g) + 435.8 kJ mol–1 → H(g) + H(g)

h
pu T
is
re ER
bl
be C
N

Fig. 4.8 The potential energy curve for the

formation of H2 molecule as a function of
©

internuclear distance of the H atoms. The

minimum in the curve corresponds to the
most stable state of H2.

4.5.1 Orbital Overlap Concept

In the formation of hydrogen molecule, there
is a minimum energy state when two hydrogen
to

atoms are so near that their atomic orbitals

undergo partial interpenetration. This partial
merging of atomic orbitals is called overlapping
Fig. 4.7 Forces of attraction and repulsion during
the for mation of H2 molecule. of atomic orbitals which results in the pairing
of electrons. The extent of overlap decides the
t

Experimentally it has been found that the strength of a covalent bond. In general, greater
no

magnitude of new attractive force is more than the overlap the stronger is the bond formed
the new repulsive forces. As a result, two between two atoms. Therefore, according to
atoms approach each other and potential orbital overlap concept, the formation of a
energy decreases. Ultimately a stage is
covalent bond between two atoms results by
reached where the net force of attraction
pairing of electrons present in the valence shell
balances the force of repulsion and system
having opposite spins.
acquires minimum energy. At this stage two
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 115

4.5.2 Directional Properties of Bonds

As we have already seen, the covalent bond is
formed by overlapping of atomic orbitals. The
molecule of hydrogen is formed due to the
overlap of 1s-orbitals of two H atoms.
In case of polyatomic molecules like CH4,
NH3 and H2O, the geometry of the molecules is
also important in addition to the bond

ed
formation. For example why is it so that CH4
molecule has tetrahedral shape and HCH bond
angles are 109.5°? Why is the shape of NH3
molecule pyramidal ?

h
The valence bond theory explains the

pu T
shape, the formation and directional properties

is
of bonds in polyatomic molecules like CH4, NH3
and H 2O, etc. in terms of overlap and
re ER
hybridisation of atomic orbitals.

bl
4.5.3 Overlapping of Atomic Orbitals
When orbitals of two atoms come close to form
bond, their overlap may be positive, negative
or zero depending upon the sign (phase) and
be C

direction of orientation of amplitude of orbital

wave function in space (Fig. 4.9). Positive and
negative sign on boundary surface diagrams
N

in the Fig. 4.9 show the sign (phase) of orbital

wave function and are not related to charge.
Fig.4.9 Positive, negative and zero overlaps of
Orbitals forming bond should have same sign
s and p atomic orbitals
(phase) and orientation in space. This is called
©

positive overlap. Various overlaps of s and p hydrogen.The four atomic orbitals of carbon,
orbitals are depicted in Fig. 4.9. each with an unpaired electron can overlap
The criterion of overlap, as the main factor with the 1s orbitals of the four H atoms which
for the formation of covalent bonds applies are also singly occupied. This will result in the
uniformly to the homonuclear/heteronuclear formation of four C-H bonds. It will, however,
diatomic molecules and polyatomic molecules. be observed that while the three p orbitals of
We know that the shapes of CH4 , NH3, and H2 O carbon are at 90° to one another, the HCH
to

molecules are tetrahedral, pyramidal and bent angle for these will also be 90°. That is three
respectively. It would be therefore interesting C-H bonds will be oriented at 90° to one
to use VB theory to find out if these geometrical another. The 2s orbital of carbon and the 1s
shapes can be explained in terms of the orbital orbital of H are spherically symmetrical and
t

overlaps. they can overlap in any direction. Therefore

no

Let us first consider the CH4 (methane) the direction of the fourth C-H bond cannot
molecule. The electronic configuration of be ascertained. This description does not fit
carbon in its ground state is [He]2s2 2p2 which in with the tetrahedral HCH angles of 109.5°.
in the excited state becomes [He] 2s 1 2p x1 2py1 Clearly, it follows that simple atomic orbital
2pz1 . The energy required for this excitation is overlap does not account for the directional
compensated by the release of energy due to characteristics of bonds in CH4 . Using similar
overlap between the orbitals of carbon and the procedure and arguments, it can be seen that in the
116 CHEMISTR Y

case of NH3 and H2 O molecules, the HNH and saucer type charged clouds above and
HOH angles should be 90 °. This is in below the plane of the participating atoms.
disagreement with the actual bond angles of
107 ° and 104.5 ° in the NH 3 and H 2 O
molecules respectively.
4.5.4 Types of Overlapping and Nature of
Covalent Bonds
The covalent bond may be classified into two

ed
types depending upon the types of
overlapping:
(i) Sigma(σ) bond, and (ii) pi(π) bond 4.5.5 Strength of Sigma and pi Bonds
(i) Sigma(σ σ) bond : This type of covalent bond Basically the strength of a bond depends upon

h
is formed by the end to end (head-on) the extent of overlapping. In case of sigma bond,

pu T
overlap of bonding orbitals along the the overlapping of orbitals takes place to a

is
internuclear axis. This is called as head larger extent. Hence, it is stronger as compared
on overlap or axial overlap. This can be
re ER
formed by any one of the following types
to the pi bond where the extent of overlapping
occurs to a smaller extent. Further, it is

bl
of combinations of atomic orbitals. important to note that in the formation of
• s-s overlapping : In this case, there is multiple bonds between two atoms of a
overlap of two half filled s-orbitals along molecule, pi bond(s) is formed in addition to a
the internuclear axis as shown below : sigma bond.
be C

4.6 HYBRIDISATION
In order to explain the characteristic
N

geometrical shapes of polyatomic molecules

like CH4, NH3 and H2O etc., Pauling introduced
• s-p overlapping: This type of overlap the concept of hybridisation. According to him
occurs between half filled s-orbitals of one the atomic orbitals combine to form new set of
©

atom and half filled p-orbitals of another equivalent orbitals known as hybrid orbitals.
atom. Unlike pure orbitals, the hybrid orbitals are
used in bond formation. The phenomenon is
known as hybridisation which can be defined
as the process of intermixing of the orbitals of
slightly different energies so as to redistribute
their energies, resulting in the formation of new
to

• p–p overlapping : This type of overlap set of orbitals of equivalent energies and shape.
takes place between half filled p-orbitals
For example when one 2s and three 2p-orbitals
of the two approaching atoms.
of carbon hybridise, there is the formation of
four new sp 3 hybrid orbitals.
t

Salient features of hybridisation: The main

no

features of hybridisation are as under :

(ii) pi(π ) bond : In the formation of π bond 1. The number of hybrid orbitals is equal to
the atomic orbitals overlap in such a way the number of the atomic orbitals that get
that their axes remain parallel to each hybridised.
other and perpendicular to the
internuclear axis. The orbitals formed due 2. The hybridised orbitals are always
to sidewise overlapping consists of two equivalent in energy and shape.
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 117

3. The hybrid orbitals are more effective in vacant 2p orbital to account for its bivalency.
forming stable bonds than the pure atomic One 2s and one 2p-orbital gets hybridised to
orbitals. form two sp hybridised orbitals. These two
4. These hybrid orbitals are directed in space sp hybrid orbitals are oriented in opposite
in some preferred direction to have direction forming an angle of 180°. Each of
minimum repulsion between electron the sp hybridised orbital overlaps with the
pairs and thus a stable arrangement. 2p-orbital of chlorine axially and form two Be-
Therefore, the type of hybridisation Cl sigma bonds. This is shown in Fig. 4.10.

ed
indicates the geometry of the molecules.
Important conditions for hybridisation
(i) The orbitals present in the valence shell
of the atom are hybridised.

h
Be
(ii) The orbitals undergoing hybridisation

pu T
should have almost equal energy.

is
(iii) Promotion of electron is not essential
condition prior to hybridisation.
re ER
(iv) It is not necessary that only half filled

bl
orbitals participate in hybridisation. In
some cases, even filled orbitals of valence
shell take part in hybridisation.
4.6.1 Types of Hybridisation Fig.4.10 (a) Formation of sp hybrids from s and
be C

p orbitals; (b) Formation of the linear

There are various types of hybridisation BeCl2 molecule
involving s, p and d orbitals. The different
(II) sp2 hybridisation : In this hybridisation
N

types of hybridisation are as under:

there is involvement of one s and two
(I) sp hybridisation: This type of p-orbitals in order to form three equivalent sp2
hybridisation involves the mixing of one s and hybridised orbitals. For example, in BCl3
one p orbital resulting in the formation of two molecule, the ground state electronic
©

equivalent sp hybrid orbitals. The suitable configuration of central boron atom is

orbitals for sp hybridisation are s and pz, if 1s 22s 22p1. In the excited state, one of the 2s
the hybrid orbitals are to lie along the z-axis. electrons is promoted to vacant 2p orbital as
Each sp hybrid orbitals has 50% s-character
and 50% p-character. Such a molecule in
which the central atom is sp-hybridised and
linked directly to two other central atoms
to

possesses linear geometry. This type of

hybridisation is also known as diagonal
hybridisation.
The two sp hybrids point in the opposite
direction along the z-axis with projecting
t

positive lobes and very small negative lobes,

no

which provides more effective overlapping

resulting in the formation of stronger bonds.
Example of molecule having s p
hybridisation
BeCl2 : The ground state electronic
configuration of Be is 1s22s 2. In the exited state Fig.4.11 Formation of sp2 hybrids and the BCl3
one of the 2s -electrons is promoted to molecule
118 CHEMISTR Y

1 1
a result boron has three unpaired electrons. ground state is 2s 22 px 2 p y 2 p 1z having three
These three orbitals (one 2s and two 2p) unpaired electrons in the sp3 hybrid orbitals
hybridise to form three sp2 hybrid orbitals. The and a lone pair of electrons is present in the
three hybrid orbitals so formed are oriented in fourth one. These three hybrid orbitals overlap
a trigonal planar arrangement and overlap with with 1s orbitals of hydrogen atoms to form
2p orbitals of chlorine to form three B-Cl three N–H sigma bonds. We know that the
bonds. Therefore, in BCl 3 (Fig. 4.11), the force of repulsion between a lone pair and a
geometry is trigonal planar with ClBCl bond bond pair is more than the force of repulsion

ed
angle of 120°. between two bond pairs of electrons. The
(III) sp 3 hybridisation: This type of molecule thus gets distorted and the bond
hybridisation can be explained by taking the angle is reduced to 107° from 109.5°. The
example of CH4 molecule in which there is geometry of such a molecule will be pyramidal
mixing of one s-orbital and three p-orbitals of as shown in Fig. 4.13.

h
the valence shell to form four sp3 hybrid orbital

pu T
of equivalent energies and shape. There is 25%

is
s-character and 75% p-character in each sp3
hybrid orbital. The four sp3 hybrid orbitals so
re ER
formed are directed towards the four corners

bl
of the tetrahedron. The angle between sp3
hybrid orbital is 109.5° as shown in Fig. 4.12.
The structure of NH3 and H 2O molecules
be C

Fig.4.13 Formation of NH3 molecule

N

In case of H 2O molecule, the four oxygen

orbitals (one 2s and three 2p) undergo sp3
hybridisation forming four sp3 hybrid orbitals
out of which two contain one electron each and
©

the other two contain a pair of electrons. These

four sp3 hybrid orbitals acquire a tetrahedral
geometry, with two corners occupied by
σ
hydrogen atoms while the other two by the lone
pairs. The bond angle in this case is reduced
σ σ to 104.5° from 109.5°
(Fig. 4.14) and the molecule thus acquires a
to

V-shape or angular geometry.

σ
t

Fig.4.12 For mation of sp 3 hybrids by the

no

combination of s , px , py and pz atomic

orbitals of carbon and the formation of
CH4 molecule

can also be explained with the help of sp3

hybridisation. In NH3, the valence shell (outer)
electronic configuration of nitrogen in the
Fig.4.14 For mation of H2O molecule
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 119

4.6.2 Other Examples of sp3, sp2 and sp sp2 hybrid orbitals of each carbon atom are
Hybridisation used for making sp2–s sigma bond with two
hydrogen atoms. The unhybridised orbital (2px
sp3 Hybridisation in C2 H6 molecule: In
or 2py ) of one carbon atom overlaps sidewise
ethane molecule both the carbon atoms with the similar orbital of the other carbon
assume sp3 hybrid state. One of the four sp3 atom to form weak π bond, which consists of
hybrid orbitals of carbon atom overlaps axially two equal electron clouds distributed above
with similar orbitals of other atom to form and below the plane of carbon and hydrogen
sp3-sp3 sigma bond while the other three atoms.

ed
hybrid orbitals of each carbon atom are used
in forming sp3–s sigma bonds with hydrogen Thus, in ethene molecule, the carbon-
carbon bond consists of one sp2–sp2 sigma
atoms as discussed in section 4.6.1(iii).
bond and one pi (π ) bond between p orbitals
Therefore in ethane C–C bond length is 154
which are not used in the hybridisation and

h
pm and each C–H bond length is 109 pm.
are perpendicular to the plane of molecule;

pu T
sp2 Hybridisation in C2 H4: In the formation the bond length 134 pm. The C–H bond is

is
of ethene molecule, one of the sp2 hybrid sp2–s sigma with bond length 108 pm. The H–
orbitals of carbon atom overlaps axially with C–H bond angle is 117.6° while the H–C–C
re ER
sp2 hybridised orbital of another carbon atom angle is 121°. The formation of sigma and pi

bl
to form C–C sigma bond. While the other two bonds in ethene is shown in Fig. 4.15.
be C
N
t to ©
no

Fig. 4.15 Formation of sigma and pi bonds in ethene

120 CHEMISTR Y

sp Hybridisation in C2H2 : In the formation 4.6.3 Hybridisation of Elements involving

of ethyne molecule, both the carbon atoms d Orbitals
undergo sp-hybridisation having two The elements present in the third period
unhybridised orbital i.e., 2py and 2p x. contain d orbitals in addition to s and p
One sp hybrid orbital of one carbon atom orbitals. The energy of the 3d orbitals are
overlaps axially with sp hybrid orbital of the comparable to the energy of the 3s and 3p
orbitals. The energy of 3d orbitals are also
other carbon atom to form C–C sigma bond,
comparable to those of 4s and 4p orbitals. As
while the other hybridised orbital of each

ed
a consequence the hybridisation involving
carbon atom overlaps axially with the half either 3s, 3p and 3d or 3d, 4s and 4p is
filled s orbital of hydrogen atoms forming σ possible. However, since the difference in
bonds. Each of the two unhybridised p orbitals energies of 3p and 4s orbitals is significant, no
of both the carbon atoms overlaps sidewise to hybridisation involving 3p, 3d and 4s orbitals

h
form two π bonds between the carbon atoms. is possible.

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So the triple bond between the two carbon The important hybridisation schemes

is
atoms is made up of one sigma and two pi involving s, p and d orbitals are summarised
bonds as shown in Fig. 4.16.
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bl
Shape of Hybridisation Atomic Examples
molecules/ type orbitals
ions
Square dsp2 d+s+p(2) [Ni(CN)4 ]2–,
be C

planar [Pt(Cl)4]2–

Trigonal sp3d s+p(3)+ d PF5, PCl5

N

bipyramidal

Square sp3d2 s+p(3)+ d(2) BrF5

pyramidal
©

Octahedral sp3d2 s+p(3)+ d(2) SF6, [CrF 6]3–

d2sp3 d(2)+s+p(3) [Co(NH3 )6]3+

(i) Formation of PCl5 (sp3d hybridisation):

The ground state and the excited state outer
electronic configurations of phosphorus (Z=15)
are represented below.
t to
no

Fig.4.16 For mation of sigma and pi bonds in sp3 d hybrid orbitals filled by electron pairs
ethyne donated by five Cl atoms.
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 121

Now the five orbitals (i.e., one s, three p and hybrid orbitals overlap with singly occupied
one d orbitals) are available for hybridisation orbitals of fluorine atoms to form six S–F sigma
to yield a set of five sp3d hybrid orbitals which bonds. Thus SF6 molecule has a regular
are directed towards the five corners of a octahedral geometry as shown in Fig. 4.18.
trigonal bipyramidal as depicted in the Fig.
4.17.

h ed
pu T
is
re ER sp3 d2 hybridisation

bl
Fig. 4.17 Trigonal bipyramidal geometry of PCl5
molecule
be C

It should be noted that all the bond angles

in trigonal bipyramidal geometry are not
equivalent. In PCl5 the five sp3d orbitals of
N

phosphorus overlap with the singly occupied

p orbitals of chlorine atoms to form five P–Cl
sigma bonds. Three P–Cl bond lie in one plane
and make an angle of 120° with each other;
these bonds are termed as equatorial bonds.
©

Fig. 4.18 Octahedral geometry of SF6 molecule

The remaining two P–Cl bonds–one lying
above and the other lying below the equatorial
plane, make an angle of 90° with the plane. 4.7 MOLECULAR ORBITAL THEORY
These bonds are called axial bonds. As the axial Molecular orbital (MO) theory was developed
bond pairs suffer more repulsive interaction by F. Hund and R.S. Mulliken in 1932. The
from the equatorial bond pairs, therefore axial salient features of this theory are :
bonds have been found to be slightly longer
to

and hence slightly weaker than the equatorial (i) The electrons in a molecule are present
in the various molecular orbitals as the
bonds; which makes PCl5 molecule more
reactive. electrons of atoms are present in the
various atomic orbitals.
(ii) Formation of SF6 (sp3d2 hybridisation):
t

In SF 6 the central sulphur atom has the (ii) The atomic orbitals of comparable
no

ground state outer electronic configuration energies and proper symmetry combine
to form molecular orbitals.
3s23p 4. In the exited state the available six
orbitals i.e., one s, three p and two d are singly (iii) While an electron in an atomic orbital is
occupied by electrons. These orbitals hybridise influenced by one nucleus, in a molecular
to form six new sp3d2 hybrid orbitals, which orbital it is influenced by two or more
are projected towards the six corners of a nuclei depending upon the number of
regular octahedron in SF6. These six sp3d2 atoms in the molecule. Thus, an atomic
122 CHEMISTR Y

orbital is monocentric while a molecular Mathematically, the formation of molecular

orbital is polycentric. orbitals may be described by the linear
(iv) The number of molecular orbital formed combination of atomic orbitals that can take
is equal to the number of combining place by addition and by subtraction of wave
atomic orbitals. When two atomic functions of individual atomic orbitals as
orbitals combine, two molecular orbitals shown below :
are formed. One is known as bonding ψ MO = ψA + ψB
molecular orbital while the other is

ed
called antibonding molecular orbital. Therefore, the two molecular orbitals
σ and σ* are formed as :
(v) The bonding molecular orbital has lower
energy and hence greater stability than σ = ψA + ψB
the corresponding antibonding σ* = ψA – ψB

h
molecular orbital. The molecular orbital σ formed by the

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(vi) Just as the electron probability addition of atomic orbitals is called the

is
distribution around a nucleus in an bonding molecular orbital while the
atom is given by an atomic orbital, the molecular orbital σ* formed by the subtraction
re ER
electron probability distribution around of atomic orbital is called antibonding

bl
a group of nuclei in a molecule is given molecular orbital as depicted in Fig. 4.19.
by a molecular orbital.
(vii) The molecular orbitals like atomic
orbitals are filled in accordance with the
be C

aufbau principle obeying the Pauli’s

exclusion principle and the Hund’s rule.
4.7.1 Formation of Molecular Orbitals
N

Linear Combination of Atomic

σ* = ψ A – ψ B
Orbitals (LCAO)
According to wave mechanics, the atomic
orbitals can be expressed by wave functions
©

ψA ψB
(ψ ’s) which represent the amplitude of the
electron waves. These are obtained from the σ = ψA + ψB
solution of Schrödinger wave equation.
However, since it cannot be solved for any
system containing more than one electron,
molecular orbitals which are one electron wave
to

functions for molecules are difficult to obtain Fig.4.19 For mation of bonding ( σ) and
directly from the solution of Schrödinger wave antibonding (σ*) molecular orbitals by the
equation. To overcome this problem, an linear combination of atomic orbitals ψ A
approximate method known as linear and ψB centered on two atoms A and B
combination of atomic orbitals (LCAO) has respectively.
t

been adopted.
Qualitatively, the formation of molecular
no

Let us apply this method to the orbitals can be understood in terms of the
homonuclear diatomic hydrogen molecule. constructive or destructive interference of the
Consider the hydrogen molecule consisting electron waves of the combining atoms. In the
of two atoms A and B. Each hydrogen atom in formation of bonding molecular orbital, the two
the ground state has one electron in 1s orbital. electron waves of the bonding atoms reinforce
The atomic orbitals of these atoms may be each other due to constructive interference
represented by the wave functions ψA and ψ B. while in the formation of antibonding
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 123

molecular orbital, the electron waves cancel taken as the molecular axis. It is important
each other due to destructive interference. As to note that atomic orbitals having same
a result, the electron density in a bonding or nearly the same energy will not combine
molecular orbital is located between the nuclei if they do not have the same symmetry.
of the bonded atoms because of which the For example, 2p z orbital of one atom can
repulsion between the nuclei is very less while combine with 2p z orbital of the other atom
in case of an antibonding molecular orbital, but not with the 2p x or 2py orbitals because
most of the electron density is located away of their dif ferent symmetries.

ed
from the space between the nuclei. Infact, there 3.The combining atomic orbitals must
is a nodal plane (on which the electron density overlap to the maximum extent. Greater
is zero) between the nuclei and hence the the extent of overlap, the greater will be the
repulsion between the nuclei is high. Electrons electron-density between the nuclei of a
placed in a bonding molecular orbital tend to

h
molecular orbital.
hold the nuclei together and stabilise a

pu T
4.7.3 Types of Molecular Orbitals
molecule. Therefore, a bonding molecular

is
orbital always possesses lower energy than Molecular orbitals of diatomic molecules are
designated as σ (sigma), π (pi), δ (delta), etc.
re ER
either of the atomic orbitals that have combined
to form it. In contrast, the electrons placed in In this nomenclature, the sigma ( σ )

bl
the antibonding molecular orbital destabilise molecular orbitals are symmetrical around
the molecule. This is because the mutual the bond-axis while pi (π) molecular orbitals
repulsion of the electrons in this orbital is more are not symmetrical. For example, the linear
than the attraction between the electrons and combination of 1s orbitals centered on two
be C

the nuclei, which causes a net increase in nuclei produces two molecular orbitals which
energy. are symmetrical around the bond-axis. Such
It may be noted that the energy of the molecular orbitals are of the σ type and are
N

antibonding orbital is raised above the energy designated as σ1s and σ*1s [Fig. 4.20(a),page
of the parent atomic orbitals that have 124]. If internuclear axis is taken to be in
combined and the energy of the bonding the z-direction, it can be seen that a linear
orbital has been lowered than the parent combination of 2pz - orbitals of two atoms
©

orbitals. The total energy of two molecular also produces two sigma molecular orbitals
orbitals, however, remains the same as that designated as σ2pz and σ *2pz. [Fig. 4.20(b)]
of two original atomic orbitals. Molecular orbitals obtained from 2px and
2py orbitals are not symmetrical around the
4.7.2 Conditions for the Combination of bond axis because of the presence of positive
Atomic Orbitals lobes above and negative lobes below the
The linear combination of atomic orbitals to molecular plane. Such molecular orbitals, are
to

form molecular orbitals takes place only if the labelled as π and π * [Fig. 4.20(c)]. A π bonding
following conditions are satisfied: MO has larger electron density above and
1.The combining atomic orbitals must below the inter -nuclear axis. The π*
have the same or nearly the same energy. antibonding MO has a node between the nuclei.
t

This means that 1s orbital can combine with 4.7.4 Energy Level Diagram for Molecular
no

another 1s orbital but not with 2s orbital Orbitals

because the energy of 2s orbital is appreciably We have seen that 1s atomic orbitals on two
higher than that of 1s orbital. This is not true atoms form two molecular orbitals designated
if the atoms are very different. as σ1s and σ*1s. In the same manner, the 2s
2.The combining atomic orbitals must and 2p atomic orbitals (eight atomic orbitals
have the same symmetry about the on two atoms) give rise to the following eight
molecular axis. By convention z-axis is molecular orbitals:
124 CHEMISTR Y

h ed
pu T
is
re ER
bl
be C
N
t to ©

Fig. 4.20 Contours and ener gies of bonding and antibonding molecular orbitals formed through
no

combinations of (a) 1s atomic orbitals; (b) 2p z atomic orbitals and (c) 2px atomic orbitals.

Antibonding MOs σ*2s σ*2pz π*2px π*2py from spectroscopic data for homonuclear
Bonding MOs σ2s σ2pz π2px π2py diatomic molecules of second row elements of
the periodic table. The increasing order of
The energy levels of these molecular
energies of various molecular orbitals for O2
orbitals have been determined experimentally and F2 is given below :
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 125

σ1s < σ*1s < σ2s < σ*2s <σ2pz<(π 2px = π 2py) terms of bond order as follows: A positive bond
< (π *2px = π *2py)<σ*2pz order (i.e., Nb > Na ) means a stable molecule
However, this sequence of energy levels of while a negative (i.e., Nb<Na ) or zero (i.e.,
molecular orbitals is not correct for the Nb = Na ) bond order means an unstable
remaining molecules Li2, Be 2, B2, C2, N2. For molecule.
instance, it has been observed experimentally Nature of the bond
that for molecules such as B2, C2, N2 etc. the
Integral bond order values of 1, 2 or 3
increasing order of energies of various

ed
correspond to single, double or triple bonds
molecular orbitals is
respectively as studied in the classical
σ1s < σ*1s < σ2s < σ*2s < (π 2px = π 2py) <σ2pz concept.
< (π *2px= π *2py) < σ*2pz
Bond-length
The important characteristic feature of

h
The bond order between two atoms in a
this order is that the energy of σ 2p z
molecule may be taken as an approximate

pu T
molecular orbital is higher than that of
measure of the bond length. The bond length

is
π 2px and π 2py molecular orbitals.
decreases as bond order increases.
re ER
4.7.5 Electronic Configuration and Magnetic nature

bl
Molecular Behaviour If all the molecular orbitals in a molecule are
The distribution of electrons among various doubly occupied, the substance is
molecular orbitals is called the electronic diamagnetic (repelled by magnetic field).
configuration of the molecule. From the However if one or more molecular orbitals are
be C

electronic configuration of the molecule, it is singly occupied it is paramagnetic (attracted

possible to get important information about by magnetic field), e.g., O2 molecule.
the molecule as discussed below.
N

4.8 BONDING IN SOME HOMONUCLEAR

Stability of Molecules: If Nb is the number DIATOMIC MOLECULES
of electrons occupying bonding orbitals and
In this section we shall discuss bonding in
Na the number occupying the antibonding
some homonuclear diatomic molecules.
orbitals, then
©

1. Hydrogen molecule (H2 ): It is formed by

(i) the molecule is stable if Nb is greater than
the combination of two hydrogen atoms. Each
Na, and
hydrogen atom has one electron in 1s orbital.
(ii) the molecule is unstable if N b is less Therefore, in all there are two electrons in
than Na . hydrogen molecule which are present in σ1s
In (i) more bonding orbitals are occupied molecular orbital. So electronic configuration
and so the bonding influence is stronger and a of hydrogen molecule is
to

stable molecule results. In (ii) the antibonding H2 : (σ1s)2

influence is stronger and therefore the molecule
The bond order of H2 molecule can be
is unstable.
calculated as given below:
Bond order
Nb − Na 2 − 0
t

Bond order (b.o.) is defined as one half the Bond order = = =1

no

difference between the number of electrons 2 2

present in the bonding and the antibonding This means that the two hydrogen atoms
orbitals i.e., are bonded together by a single covalent bond.
Bond order (b.o.) = ½ (Nb–Na ) The bond dissociation energy of hydrogen
molecule has been found to be 438 kJ mol–1
The rules discussed above regarding the and bond length equal to 74 pm. Since no
stability of the molecule can be restated in unpaired electron is present in hydrogen
126 CHEMISTR Y

molecule, therefore, it is diamagnetic. molecules a double bond is made up of a sigma

2. Helium molecule (He2 ): The electronic bond and a pi bond. In a similar fashion the
configuration of helium atom is 1s 2. Each bonding in N2 molecule can be discussed.
helium atom contains 2 electrons, therefore, 5. Oxygen molecule (O 2 ): The electronic
in He2 molecule there would be 4 electrons. configuration of oxygen atom is 1s2 2s2 2p4.
These electrons will be accommodated in σ1s Each oxygen atom has 8 electrons, hence, in
and σ*1s molecular orbitals leading to
O 2 molecule there are 16 electrons. The
electronic configuration:
electronic configuration of O2 molecule,

ed
He2 : (σ1s)2 (σ*1s)2 therefore, is
Bond order of He2 is ½(2 – 2) = 0
He2 molecule is therefore unstable and does O 2: ( σ1s) 2( σ*1s)2( σ2s)2 ( σ*2s)2( σ2p z )2
not exist.
(π 2 p x 2 ≡ π 2 p y2 ) (π * 2p x1 ≡ π * 2 p y1 )

h
or
Similarly, it can be shown that Be2 molecule

pu T
(σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 also does not exist.
O2 :  KK (σ2 s )2 ( σ * 2s )2 (σ2 p z )2 

is
3. Lithium molecule (Li2 ): The electronic  
 (π 2 p x ≡ π 2 p y ) , ( π * 2 px ≡ π * 2 p y ) 
2 2 1 1
re ER
configuration of lithium is 1s 2, 2s 1 . There are
six electrons in Li 2. The electronic

bl
From the electronic configuration of O2
configuration of Li2 molecule, therefore, is
molecule it is clear that ten electrons are
Li 2 : (σ1s) 2 (σ*1s)2 (σ2s)2 present in bonding molecular orbitals and six
The above configuration is also written as electr ons are present in antibonding
KK(σ2s)2 where KK represents the closed K
be C

molecular orbitals. Its bond order, therefore,

shell structure (σ1s)2 (σ*1s)2. is
From the electronic configuration of Li2 1 1
Bond order = [ N b − N a ] = [10 − 6] = 2
N

molecule it is clear that there are four 2 2

electrons present in bonding molecular So in oxygen molecule, atoms are held by
orbitals and two electrons present in a double bond. Moreover, it may be noted that
antibonding molecular orbitals. Its bond order, it contains two unpaired electrons in π *2px
therefore, is ½ (4 – 2) = 1. It means that Li2 and π *2py molecular orbitals, therefore, O 2
©

molecule is stable and since it has no unpaired molecule should be paramagnetic, a

electrons it should be diamagnetic. Indeed prediction that corresponds to
diamagnetic Li2 molecules are known to exist experimental observation. In this way, the
in the vapour phase. theory successfully explains the paramagnetic
4. Carbon molecule (C2 ): The electronic nature of oxygen.
configuration of carbon is 1s2 2s2 2p2. There Similarly, the electronic configurations of
to

are twelve electrons in C2. The electronic

other homonuclear diatomic molecules of the
configuration of C2 molecule, therefore, is
second row of the periodic table can be written.
2 2 2 2 2 2
C2 : (σ1s ) (σ *1s ) (σ2s ) (σ * 2s ) (π 2p x = π 2 py ) In Fig.4.21 are given the molecular orbital
occupancy and molecular properties for B2
KK ( σ2s ) 2 ( σ * 2s ) 2 ( π 2p x2 = π 2p y2 )
t

or through Ne2. The sequence of MOs and their

no

The bond order of C2 is ½ (8 – 4) = 2 and C 2 electron population are shown. The bond
should be diamagnetic. Diamagnetic C 2 energy, bond length, bond order, magnetic
molecules have indeed been detected in properties and valence electron configuration
vapour phase. It is important to note that appear below the orbital diagrams.
double bond in C2 consists of both pi bonds
4.9 HYDROGEN BONDING
because of the presence of four electrons in two
pi molecular orbitals. In most of the other Nitrogen, oxygen and fluorine are the higly
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 127

h ed
pu T
is
re ER
bl
be C
N

Fig. 4.21 MO occupancy and molecular properties for B2 through Ne2.

©

electronegative elements. When they are bond. Thus, hydrogen bond can be defined
attached to a hydrogen atom to form covalent as the attractive force which binds
bond, the electrons of the covalent bond are hydrogen atom of one molecule with the
shifted towards the more electronegative atom. electronegative atom (F, O or N) of another
This partially positively charged hydrogen molecule.
atom forms a bond with the other more
4.9.1 Cause of Formation of Hydrogen
to

electronegative atom. This bond is known as

hydrogen bond and is weaker than the Bond
covalent bond. For example, in HF molecule, When hydrogen is bonded to strongly
the hydrogen bond exists between hydrogen electronegative element ‘X’, the electron pair
atom of one molecule and fluorine atom of shared between the two atoms moves far away
t

another molecule as depicted below : from hydrogen atom. As a result the hydrogen
no

atom becomes highly electropositive with

− − −H δ + – F δ − − − − H δ+ – F δ− − − − H δ+ – F δ− respect to the other atom ‘X’. Since there is
Here, hydrogen bond acts as a bridge between displacement of electrons towards X, the
two atoms which holds one atom by covalent hydrogen acquires fractional positive charge
bond and the other by hydrogen bond. (δ + ) while ‘X’ attain fractional negative charge
Hydrogen bond is represented by a dotted line (δ–). This results in the formation of a polar
(– – –) while a solid line represents the covalent molecule having electrostatic force of attraction
128 CHEMISTR Y

which can be represented as : (2) Intramolecular hydrogen bond : It is

formed when hydrogen atom is in between the
H δ+ − X δ− − − − H δ+ − X δ− − − − H δ+ − X δ− two highly electronegative (F, O, N) atoms
The magnitude of H-bonding depends on present within the same molecule. For example,
the physical state of the compound. It is in o-nitrophenol the hydrogen is in between
maximum in the solid state and minimum in the two oxygen atoms.
the gaseous state. Thus, the hydrogen bonds EXERCISES
have strong influence on the structure and

ed
properties of the compounds.
4.9.2 Types of H-Bonds
There are two types of H-bonds
(i) Intermolecular hydrogen bond

h
(ii) Intramolecular hydrogen bond

pu T
(1) Intermolecular hydrogen bond : It is

is
formed between two different molecules of the
re ER
same or different compounds. For example, H-
bond in case of HF molecule, alcohol or water Fig. 4.22 Intramolecular hydrogen bonding in

bl
molecules, etc. o-nitrophenol molecule
be C

SUMMARY

Kössel’s first insight into the mechanism of formation of electropositive and electronegative
N

ions related the process to the attainment of noble gas configurations by the respective
ions. Electrostatic attraction between ions is the cause for their stability. This gives the
concept of electrovalency.
The first description of covalent bonding was provided by Lewis in terms of the sharing
of electron pairs between atoms and he related the pr ocess to the attainment of noble gas
©

configurations by reacting atoms as a result of sharing of electrons. The Lewis dot symbols
show the number of valence electrons of the atoms of a given element and Lewis dot
structures show pictorial repr esentations of bonding in molecules.
An ionic compound is pictured as a three-dimensional aggr egation of positive and
negative ions in an order ed arrangement called the crystal lattice. In a crystalline solid
there is a charge balance between the positive and negative ions. The crystal lattice is
stabilized by the enthalpy of lattice formation.
to

While a single covalent bond is formed by sharing of an electron pair between two
atoms, multiple bonds result fr om the sharing of two or three electron pairs. Some bonded
atoms have additional pairs of electrons not involved in bonding. These are called lone-
pairs of electrons. A Lewis dot structur e shows the arrangement of bonded pairs and lone
pairs around each atom in a molecule. Important parameters, associated with chemical
t

bonds, like: bond length, bond angle, bond enthalpy, bond order and bond polarity
no

have significant effect on the properties of compounds.

A number of molecules and polyatomic ions cannot be described accurately by a single
Lewis structur e and a number of descriptions (representations) based on the same skeletal
structure are written and these taken together represent the molecule or ion. This is a very
important and extremely useful concept called resonance. The contributing structures or
canonical for ms taken together constitute the resonance hybrid which repr esents the
molecule or ion.
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 129

The VSEPR model used for pr edicting the geometrical shapes of molecules is based on
the assumption that electr on pairs repel each other and, therefore, tend to remain as far
apart as possible. According to this model, molecular geometry is determined by repulsions
between lone pairs and lone pairs ; lone pairs and bonding pairs and bonding pairs and
bonding pairs . The order of these repulsions being : lp-lp > lp-bp > bp-bp
The valence bond (VB) approach to covalent bonding is basically concerned with the
energetics of covalent bond formation about which the Lewis and VSEPR models are silent.
Basically the VB theory discusses bond formation in terms of overlap of orbitals. For

ed
example the formation of the H2 molecule from two hydrogen atoms involves the overlap of
the 1s orbitals of the two H atoms which are singly occupied. It is seen that the potential
energy of the system gets lowered as the two H atoms come near to each other. At the
equilibrium inter-nuclear distance (bond distance) the energy touches a minimum. Any
attempt to bring the nuclei still closer results in a sudden increase in energy and consequent
destabilization of the molecule. Because of orbital overlap the electron density between the

h
nuclei increases which helps in bringing them closer. It is however seen that the actual

pu T
bond enthalpy and bond length values are not obtained by overlap alone and other variables

is
have to be taken into account.
re ER
For explaining the characteristic shapes of polyatomic molecules Pauling introduced
the concept of hybridisation of atomic orbitals. sp,sp2, sp3 hybridizations of atomic orbitals

bl
of Be, B,C, N and O are used to explain the formation and geometrical shapes of molecules
like BeCl2, BCl 3, CH4, NH3 and H2O. They also explain the formation of multiple bonds in
molecules like C2H 2 and C2H 4.
The molecular orbital (MO) theory describes bonding in terms of the combination
and arrangment of atomic orbitals to form molecular orbitals that ar e associated with the
be C

molecule as a whole. The number of molecular orbitals are always equal to the number of
atomic orbitals from which they are formed. Bonding molecular orbitals increase electron
density between the nuclei and are lower in energy than the individual atomic orbitals.
N

Antibonding molecular orbitals have a region of zer o electron density between the nuclei
and have more energy than the individual atomic orbitals.
The electr onic configuration of the molecules is written by filling electrons in the
molecular orbitals in the order of incr easing energy levels. As in the case of atoms, the
©

Pauli exclusion principle and Hund’s rule are applicable for the filling of molecular orbitals.
Molecules are said to be stable if the number of elctrons in bonding molecular orbitals is
gr eater than that in antibonding molecular orbitals.
Hydrogen bond is for med when a hydrogen atom finds itself between two highly
electronegative atoms such as F, O and N. It may be intermolecular (existing between two
or more molecules of the same or different substances) or intramolecular (present within
the same molecule). Hydr ogen bonds have a powerful effect on the structure and properties
of many compounds.
to

4.1 Explain the formation of a chemical bond.

4.2 Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br.
t

4.3 Write Lewis symbols for the following atoms and ions:
no

–
S and S2–; Al and Al3+; H and H
4.4 Draw the Lewis structures for the following molecules and ions :

H2 S, SiCl4, BeF 2, CO 23− , HCOOH

4.5 Define octet rule. Write its significance and limitations.
4.6 Write the favourable factors for the formation of ionic bond.
130 CHEMISTR Y

4.7 Discuss the shape of the following molecules using the VSEPR model:
BeCl2, BCl 3, SiCl4 , AsF5, H2 S, PH3
4.8 Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond
angle in water is less than that of ammonia. Discuss.
4.9 How do you express the bond strength in ter ms of bond order ?
4.10 Define the bond length.

4.11 Explain the important aspects of resonance with reference to the CO23 − ion.

ed
4.12 H 3PO3 can be represented by structures 1 and 2 shown below. Can these two
structures be taken as the canonical forms of the resonance hybrid representing
H 3PO3 ? If not, give reasons for the same.

h
pu T
is
re ER
bl
4.13 Write the resonance structures for SO 3, NO2 and NO −3 .
4.14 Use Lewis symbols to show electron transfer between the following atoms to form
cations and anions : (a) K and S (b) Ca and O (c) Al and N.
4.15 Although both CO2 and H 2O are triatomic molecules, the shape of H 2O molecule is
be C

bent while that of CO2 is linear. Explain this on the basis of dipole moment.
4.16 Write the significance/applications of dipole moment.
4.17 Define electronegativity. How does it differ from electron gain enthalpy ?
N

4.18 Explain with the help of suitable example polar covalent bond.
4.19 Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O,
N2, SO2 and ClF3.
4.20 The skeletal structure of CH3COOH as shown below is correct, but some of the
©

bonds are shown incorrectly. Write the corr ect Lewis structure for acetic acid.

4.21 Apart from tetrahedral geometry, another possible geometry for CH4 is square planar
to

with the four H atoms at the corners of the square and the C atom at its centre.
Explain why CH4 is not square planar ?
4.22 Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are
polar.
t

4.23 Which out of NH3 and NF 3 has higher dipole moment and why ?
no

4.24 What is meant by hybridisation of atomic orbitals? Describe the shapes of sp,
sp2, sp3 hybrid orbitals.
4.25 Describe the change in hybridisation (if any) of the Al atom in the following
reaction.
AlCl3 + Cl− → AlCl4−
4.26 Is there any change in the hybridisation of B and N atoms as a result of the following
 CHEMICAL BONDING AND MOLECULAR STRUCTURE 131

reaction ?
BF3 + NH 3 → F3 B.NH 3
4.27 Draw diagrams showing the formation of a double bond and a triple bond between
carbon atoms in C2 H4 and C2H 2 molecules.
4.28 What is the total number of sigma and pi bonds in the following molecules ?
(a) C2H 2 (b) C2 H4
4.29 Considering x-axis as the internuclear axis which out of the following will not form

ed
a sigma bond and why? (a) 1s and 1s (b) 1 s and 2 p x ; (c) 2p y and 2p y
(d) 1s and 2s.
4.30 Which hybrid orbitals are used by carbon atoms in the following molecules ?
CH3 –CH3; (b) CH 3–CH=CH2; (c) CH 3-CH 2-OH; (d) CH 3-CHO (e) CH3 COOH

h
4.31 What do you understand by bond pairs and lone pairs of electrons ? Illustrate by
giving one exmaple of each type.

pu T
4.32 Distinguish between a sigma and a pi bond.

is
4.33 Explain the formation of H2 molecule on the basis of valence bond theory.
4.34
re ER
Write the important conditions required for the linear combination of atomic orbitals

bl
to form molecular orbitals.
4.35 Use molecular orbital theory to explain why the Be2 molecule does not exist.
4.36 Compare the relative stability of the following species and indicate their magnetic
properties;
be C

O2 , O 2+, O 2− (superoxide), O22 − (peroxide)

4.37 Write the significance of a plus and a minus sign shown in representing the orbitals.
4.38 Describe the hybridisation in case of PCl5 . Why are the axial bonds longer as
N

compared to equatorial bonds ?

4.39 Define hydrogen bond. Is it weaker or stronger than the van der Waals for ces?
+
4.40 What is meant by the term bond order ? Calculate the bond order of : N2, O 2, O2
–
and O2 .
t to ©
no
 132 CHEMISTRY

UNIT 5

STATES OF MATTER

d
he
The snowflake falls, yet lays not long
Its feath’ry grasp on Mother Earth

is
Ere Sun returns it to the vapors Whence it came,
Or to waters tumbling down the rocky slope.
After studying this unit you will be

bl
able to Rod O’ Connor
• explain the existence of different
states of matter in terms of
balance between intermolecular
pu
forces and thermal energy of
particles; INTRODUCTION
In previous units we have learnt about the properties
be T

• explain the laws governing

behaviour of ideal gases; related to single particle of matter, such as atomic size,
re
• apply gas laws in various real life ionization enthalpy, electronic charge density, molecular
o R

situations; shape and polarity, etc. Most of the observable

characteristics of chemical systems with which we are
• explain the behaviour of real
gases; familiar represent bulk properties of matter, i.e., the
tt E

properties associated with a collection of a large number

• describe the conditions required of atoms, ions or molecules. For example, an individual
for liquifaction of gases;
molecule of a liquid does not boil but the bulk boils.
C

• realise that there is continuity in Collection of water molecules have wetting properties;
gaseous and liquid state;
individual molecules do not wet. Water can exist as ice,
• differentiate between gaseous
no N

which is a solid; it can exist as liquid; or it can exist in

state and vapours; the gaseous state as water vapour or steam. Physical
• explain properties of liquids in properties of ice, water and steam are very different. In
terms of intermolecular all the three states of water chemical composition of water
attractions.
©

remains the same i.e., H2O. Characteristics of the three

states of water depend on the energies of molecules and
on the manner in which water molecules aggregate. Same
is true for other substances also.
Chemical properties of a substance do not change with
the change of its physical state; but rate of chemical
reactions do depend upon the physical state. Many times
in calculations while dealing with data of experiments we
require knowledge of the state of matter. Therefore, it
becomes necessary for a chemist to know the physical

132 C:\ChemistryXI\Unit-5\Unit-5(4)-Lay-2.pmd 14.1.6 (Final), 17.1.6, 24.1.6

 STATES OF MATTER 133

laws which govern the behaviour of matter in so happen that momentarily electronic charge
different states. In this unit, we will learn distribution in one of the atoms, say ‘A’,
more about these three physical states of becomes unsymmetrical i.e., the charge cloud
matter particularly liquid and gaseous states. is more on one side than the other (Fig. 5.1 b
To begin with, it is necessary to understand and c). This results in the development of
the nature of intermolecular forces, molecular instantaneous dipole on the atom ‘A’ for a very
interactions and effect of thermal energy on short time. This instantaneous or transient
the motion of particles because a balance dipole distorts the electron density of the
between these determines the state of a other atom ‘B’, which is close to it and as a

d
substance. consequence a dipole is induced in the
5.1 INTERMOLECULAR FORCES atom ‘B’.

he
Intermolecular forces are the forces of The temporary dipoles of atom ‘A’ and ‘B’
attraction and repulsion between interacting attract each other. Similarly temporary dipoles
particles (atoms and molecules). This term are induced in molecules also. This force of
does not include the electrostatic forces that attraction was first proposed by the German

is
exist between the two oppositely charged ions physicist Fritz London, and for this reason
and the forces that hold atoms of a molecule force of attraction between two temporary
together i.e., covalent bonds.

bl
Attractive intermolecular forces are known
as van der Waals forces, in honour of Dutch
scientist Johannes van der Waals (1837-
pu
1923), who explained the deviation of real
gases from the ideal behaviour through these
forces. We will learn about this later in this
be T

unit. van der Waals forces vary considerably

re
in magnitude and include dispersion forces
o R

or London forces, dipole-dipole forces, and

dipole-induced dipole forces. A particularly
strong type of dipole-dipole interaction is
tt E

hydrogen bonding. Only a few elements can

participate in hydrogen bond formation,
therefore it is treated as a separate
C

category. We have already learnt about this

interaction in Unit 4.
At this point, it is important to note that
no N

attractive forces between an ion and a dipole

are known as ion-dipole forces and these are
not van der Waals forces. We will now learn
about different types of van der Waals forces.
©

5.1.1 Dispersion Forces or London Forces

Atoms and nonpolar molecules are electrically
symmetrical and have no dipole moment
because their electronic charge cloud is
symmetrically distributed. But a dipole may
develop momentarily even in such atoms and
molecules. This can be understood as follows.
Suppose we have two atoms ‘A’ and ‘B’ in the Fig. 5.1 Dispersion forces or London forces
close vicinity of each other (Fig. 5.1a). It may between atoms.

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 134 CHEMISTRY

dipoles is known as London force. Another proportional to 1/r 6, where r is the distance
name for this force is dispersion force. These between polar molecules. Besides dipole-
forces are always attractive and interaction dipole interaction, polar molecules can
energy is inversely proportional to the sixth interact by London forces also. Thus
power of the distance between two interacting cumulative effect is that the total of
particles (i.e., 1/r 6 where r is the distance intermolecular forces in polar molecules
between two particles). These forces are increase.
important only at short distances (~500 pm)
5.1.3 Dipole–Induced Dipole Forces
and their magnitude depends on the

d
polarisability of the particle. This type of attractive forces operate between
the polar molecules having permanent dipole

he
5.1.2 Dipole - Dipole Forces and the molecules lacking permanent dipole.
Dipole-dipole forces act between the molecules Permanent dipole of the polar molecule
possessing permanent dipole. Ends of the induces dipole on the electrically neutral
dipoles possess “partial charges” and these molecule by deforming its electronic cloud
charges are shown by Greek letter delta (δ). (Fig. 5.3). Thus an induced dipole is developed

is
Partial charges are always less than the unit in the other molecule. In this case also
electronic charge (1.610 –19 C). The polar interaction energy is proportional to 1/r 6
molecules interact with neighbouring where r is the distance between two

bl
molecules. Fig 5.2 (a) shows electron cloud molecules. Induced dipole moment depends
distribution in the dipole of hydrogen chloride upon the dipole moment present in the
and Fig. 5.2 (b) shows dipole-dipole interaction
pu permanent dipole and the polarisability of the
between two HCl molecules. This interaction electrically neutral molecule. We have already
is stronger than the London forces but is learnt in Unit 4 that molecules of larger size
weaker than ion-ion interaction because only can be easily polarized. High polarisability
be T

partial charges are involved. The attractive increases the strength of attractive
re
force decreases with the increase of distance interactions.
o R

between the dipoles. As in the above case here

also, the interaction energy is inversely
proportional to distance between polar
tt E

molecules. Dipole-dipole interaction energy

between stationary polar molecules (as in
solids) is proportional to 1/r 3 and that
C

between rotating polar molecules is

no N
©

Fig. 5.3 Dipole - induced dipole interaction

between permanent dipole and induced
dipole
In this case also cumulative effect of
dispersion forces and dipole-induced dipole
interactions exists.
5.1.4 Hydrogen bond
Fig. 5.2 (a) Distribution of electron cloud in HCl – As already mentioned in section (5.1); this is
a polar molecule, (b) Dipole-dipole special case of dipole-dipole interaction. We
interaction between two HCl molecules have already learnt about this in Unit 4. This

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 STATES OF MATTER 135

is found in the molecules in which highly polar thermal energy of the molecules tends to keep
N–H, O–H or H–F bonds are present. Although them apart. Three states of matter are the result
hydrogen bonding is regarded as being limited of balance between intermolecular forces and
to N, O and F; but species such as Cl may the thermal energy of the molecules.
also participate in hydrogen bonding. Energy When molecular interactions are very
of hydrogen bond varies between 10 to 100 weak, molecules do not cling together to make
kJ mol–1. This is quite a significant amount of liquid or solid unless thermal energy is
energy; therefore, hydrogen bonds are reduced by lowering the temperature. Gases
powerful force in determining the structure and do not liquify on compression only, although

d
properties of many compounds, for example molecules come very close to each other and
proteins and nucleic acids. Strength of the intermolecular forces operate to the maximum.

he
hydrogen bond is determined by the coulombic However, when thermal energy of molecules
interaction between the lone-pair electrons of is reduced by lowering the temperature; the
the electronegative atom of one molecule and gases can be very easily liquified.
the hydrogen atom of other molecule. Predominance of thermal energy and the
Following diagram shows the formation of

is
molecular interaction energy of a substance
hydrogen bond. in three states is depicted as follows :
δ+ δ− δ+ δ−

bl
H− F ⋅⋅ ⋅ H− F

Intermolecular forces discussed so far are

pu
all attractive. Molecules also exert repulsive
forces on one another. When two molecules
are brought into close contact with each other,
be T

the repulsion between the electron clouds and

that between the nuclei of two molecules comes
re
into play. Magnitude of the repulsion rises very We have already learnt the cause for the
o R

rapidly as the distance separating the existence of the three states of matter. Now
molecules decreases. This is the reason that we will learn more about gaseous and liquid
states and the laws which govern the
tt E

liquids and solids are hard to compress. In

these states molecules are already in close behaviour of matter in these states. We shall
contact; therefore they resist further deal with the solid state in class XII.
C

compression; as that would result in the

5.4 THE GASEOUS STATE
increase of repulsive interactions.
This is the simplest state of matter.
no N

5.2 THERMAL ENERGY Throughout our life we remain immersed in

Thermal energy is the energy of a body arising the ocean of air which is a mixture of gases.
from motion of its atoms or molecules. It is We spend our life in the lowermost layer of
directly proportional to the temperature of the the atmosphere called troposphere, which is
©

substance. It is the measure of average kinetic held to the surface of the earth by gravitational
energy of the particles of the matter and is force. The thin layer of atmosphere is vital to
thus responsible for movement of particles. our life. It shields us from harmful radiations
This movement of particles is called thermal and contains substances like dioxygen,
motion. dinitrogen, carbon dioxide, water vapour, etc.
Let us now focus our attention on the
5.3 INTERMOLECULAR FORCES vs behaviour of substances which exist in the
THERMAL INTERACTIONS gaseous state under normal conditions of
We have already learnt that intermolecular temperature and pressure. A look at the
forces tend to keep the molecules together but periodic table shows that only eleven elements

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 136 CHEMISTRY

centuries on the physical properties of gases.

The first reliable measurement on properties
of gases was made by Anglo-Irish scientist
Robert Boyle in 1662. The law which he
formulated is known as Boyle’s Law. Later
on attempts to fly in air with the help of hot
air balloons motivated Jaccques Charles and
Joseph Lewis Gay Lussac to discover
additional gas laws. Contribution from

d
Avogadro and others provided lot of
information about gaseous state.

he
5.5.1 Boyle’s Law (Pressure - Volume
Relationship)
Fig. 5.4 Eleven elements that exist as gases On the basis of his experiments, Robert Boyle

is
reached to the conclusion that at constant
exist as gases under normal conditions
temperature, the pressure of a fixed
(Fig 5.4).
amount (i.e., number of moles n) of gas

bl
The gaseous state is characterized by the varies inversely with its volume. This is
following physical properties. known as Boyle’s law. Mathematically, it can
• Gases are highly compressible.
pu be written as
• Gases exert pressure equally in all
1
directions. p ∝ ( at constant T and n) (5.1)
V
• Gases have much lower density than the
be T

solids and liquids. 1

⇒ p = k1
re
• The volume and the shape of gases are (5.2)
V
o R

not fixed. These assume volume and shape

of the container. where k1 is the proportionality constant. The
• Gases mix evenly and completely in all value of constant k 1 depends upon the
tt E

proportions without any mechanical aid. amount of the gas, temperature of the gas
and the units in which p and V are expressed.
Simplicity of gases is due to the fact that
C

the forces of interaction between their On rearranging equation (5.2) we obtain

molecules are negligible. Their behaviour is
pV = k1 (5.3)
governed by same general laws, which were
no N

discovered as a result of their experimental It means that at constant temperature,

studies. These laws are relationships between product of pressure and volume of a fixed
measurable properties of gases. Some of these amount of gas is constant.
properties like pressure, volume, temperature If a fixed amount of gas at constant
©

and mass are very important because temperature T occupying volume V 1 at

relationships between these variables describe pressure p1 undergoes expansion, so that
state of the gas. Interdependence of these volume becomes V2 and pressure becomes p2,
variables leads to the formulation of gas laws.
then according to Boyle’s law :
In the next section we will learn about gas
laws. p V = p V = constant (5.4)
1 1 2 2

5.5 THE GAS LAWS

The gas laws which we will study now are the p1 V2
⇒ = (5.5)
result of research carried on for several p2 V1

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 STATES OF MATTER 137

Figure 5.5 shows two conventional ways

of graphically presenting Boyle’s law.
Fig. 5.5 (a) is the graph of equation (5.3) at
different temperatures. The value of k1 for
each curve is different because for a given
mass of gas, it varies only with temperature.
Each curve corresponds to a different
constant temperature and is known as an
isotherm (constant temperature plot). Higher

d
curves correspond to higher temperature. It
should be noted that volume of the gas

he
doubles if pressure is halved. Table 5.1 gives
effect of pressure on volume of 0.09 mol of
CO2 at 300 K.

Fig 5.5 (b) represents the graph between

is
1
Fig. 5.5(a) Graph of pressure, p vs. Volume, V of p and . It is a straight line passing through
a gas at different temperatures. V

bl
origin. However at high pressures, gases
deviate from Boyle’s law and under such
conditions a straight line is not obtained in
the graph.
pu Experiments of Boyle, in a quantitative
manner prove that gases are highly
be T

compressible because when a given mass of

a gas is compressed, the same number of
re
molecules occupy a smaller space. This means
o R

that gases become denser at high pressure.

A relationship can be obtained between
density and pressure of a gas by using Boyle’s
tt E

law :
By definition, density ‘d’ is related to the
C

mass ‘m’ and the volume ‘V’ by the relation

m
d= . If we put value of V in this equation
no N

1 V
Fig. 5.5 (b) Graph of pressure of a gas, p vs.
V
Table 5.1 Effect of Pressure on the Volume of 0.09 mol CO2 Gas at 300 K.
Pressure/104 Pa Volume/10–3 m3 (1/V )/m–3 pV/102 Pa m3
©

2.0 112.0 8.90 22.40

2.5 89.2 11.2 22.30
3.5 64.2 15.6 22.47
4.0 56.3 17.7 22.50
6.0 37.4 26.7 22.44
8.0 28.1 35.6 22.48
10.0 22.4 44.6 22.40

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 138 CHEMISTRY

from Boyle’s law equation, we obtain the

⎛ 273.15 + t ⎞
relationship. ⇒ V t = V0 ⎜ ⎟ (5.6)
⎝ 273.15 ⎠
⎛m ⎞
d = ⎜ ⎟ p = k′ p At this stage, we define a new scale of
⎝ k1 ⎠ temperature such that t °C on new scale is
This shows that at a constant
given by T = 273.15 + t and 0 °C will be given
temperature, pressure is directly proportional
by T0 = 273.15. This new temperature scale
to the density of a fixed mass of the gas.
is called the Kelvin temperature scale or
Absolute temperature scale.

d
Problem 5.1
A balloon is filled with hydrogen at room Thus 0°C on the celsius scale is equal to
273.15 K at the absolute scale. Note that

he
temperature. It will burst if pressure
exceeds 0.2 bar. If at 1 bar pressure the degree sign is not used while writing the
gas occupies 2.27 L volume, upto what temperature in absolute temperature scale,
volume can the balloon be expanded ? i.e., Kelvin scale. Kelvin scale of temperature
is also called Thermodynamic scale of
Solution

is
temperature and is used in all scientific
According to Boyle’s Law p1V1 = p2V2 works.
If p1 is 1 bar, V1 will be 2.27 L Thus we add 273 (more precisely 273.15)

bl
to the celsius temperature to obtain
p1V1 temperature at Kelvin scale.
If p2 = 0.2 bar, then V2 =
pu p2 If we write Tt = 273.15 + t and T0 = 273.15
1bar × 2.27 L in the equation (5.6) we obtain the
⇒ V2 = =11.35 L relationship
0.2 bar
be T

Since balloon bursts at 0.2 bar pressure, ⎛T ⎞

Vt = V0 ⎜ t ⎟
re
the volume of balloon should be less than ⎝ T0 ⎠
o R

11.35 L.
Vt T
⇒ = t (5.7)
5.5.2 Charles’ Law (Temperature - Volume V0 T0
tt E

Relationship) Thus we can write a general equation as

Charles and Gay Lussac performed several follows.
C

experiments on gases independently to V2 T

improve upon hot air balloon technology. = 2 (5.8)
Their investigations showed that for a fixed V1 T1
no N

mass of a gas at constant pressure, volume

V1 V2
of a gas increases on increasing temperature ⇒ =
and decreases on cooling. They found that T1 T2
for each degree rise in temperature, volume
©

V
1 ⇒ = constant = k 2 (5.9)
of a gas increases by of the original T
273.15
Thus V = k2 T (5.10)
volume of the gas at 0 °C. Thus if volumes of
the gas at 0 °C and at t °C are V0 and Vt The value of constant k2 is determined by
respectively, then the pressure of the gas, its amount and the
t units in which volume V is expressed.
Vt = V0 + V0
273.15 Equation (5.10) is the mathematical
⎛ t ⎞ expression for Charles’ law, which states that
⇒ Vt = V0 ⎜1 + ⎟ pressure remaining constant, the volume
⎝ 273.15 ⎠

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 STATES OF MATTER 139

of a fixed mass of a gas is directly with 2 L air. What will be the volume of
proportional to its absolute temperature. the balloon when the ship reaches Indian
Charles found that for all gases, at any given ocean, where temperature is 26.1°C ?
pressure, graph of volume vs temperature (in
celsius) is a straight line and on extending to Solution
zero volume, each line intercepts the V1 = 2 L T2 = 26.1 + 273
temperature axis at – 273.15 °C. Slopes of
T1 = (23.4 + 273) K = 299.1 K
lines obtained at different pressure are
different but at zero volume all the lines meet = 296.4 K

d
the temperature axis at – 273.15 °C (Fig. 5.6).
From Charles law

he
V1 V2
=
T1 T2

V1T2
⇒ V2 =

is
T1

2 L × 299.1K
⇒ V2 =

bl
296.4 K
pu = 2 L × 1.009
= 2.018 L
be T

5.5.3 Gay Lussac’s Law (Pressure-

Temperature Relationship)
re
o R

Pressure in well inflated tyres of automobiles

is almost constant, but on a hot summer day
this increases considerably and tyre may
tt E

Fig. 5.6 Volume vs Temperature ( °C) graph burst if pressure is not adjusted properly.
Each line of the volume vs temperature During winters, on a cold morning one may
graph is called isobar. find the pressure in the tyres of a vehicle
C

Observations of Charles can be interpreted decreased considerably. The mathematical

if we put the value of t in equation (5.6) as relationship between pressure and
– 273.15 °C. We can see that the volume of temperature was given by Joseph Gay Lussac
no N

the gas at – 273.15 °C will be zero. This means and is known as Gay Lussac’s law. It states
that gas will not exist. In fact all the gases get that at constant volume, pressure of a fixed
liquified before this temperature is reached. amount of a gas varies directly with the
The lowest hypothetical or imaginary temperature. Mathematically,
©

temperature at which gases are supposed to

p∝T
occupy zero volume is called Absolute zero.
p
All gases obey Charles’ law at very low ⇒ = constant = k 3
pressures and high temperatures. T
This relationship can be derived from
Problem 5.2 Boyle’s law and Charles’ Law. Pressure vs
On a ship sailing in pacific ocean where temperature (Kelvin) graph at constant molar
temperature is 23.4 °C , a balloon is filled volume is shown in Fig. 5.7. Each line of this
graph is called isochore.

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 140 CHEMISTRY

will find that this is the same number which

we came across while discussing definition
of a ‘mole’ (Unit 1).
Since volume of a gas is directly
proportional to the number of moles; one mole
of each gas at standard temperature and
pressure (STP)* will have same volume.
Standard temperature and pressure means
273.15 K (0°C) temperature and 1 bar (i.e.,

d
exactly 10 5 pascal) pressure. These
values approximate freezing temperature

he
of water and atmospheric pressure at sea
level. At STP molar volume of an ideal gas
or a combination of ideal gases is
22.71098 L mol–1.

is
Molar volume of some gases is given in
(Table 5.2).
Table 5.2 Molar volume in litres per mole of

bl
Fig. 5.7 Pressure vs temperature (K) graph some gases at 273.15 K and 1 bar
(Isochores) of a gas. (STP).

Argon 22.37
pu
5.5.4 Avogadro Law (Volume - Amount
Relationship) Carbon dioxide 22.54
In 1811 Italian scientist Amedeo Avogadro Dinitrogen 22.69
be T

tried to combine conclusions of Dalton’s Dioxygen 22.69

re
atomic theory and Gay Lussac’s law of Dihydrogen 22.72
o R

combining volumes (Unit 1) which is now

Ideal gas 22.71
known as Avogadro law. It states that equal
volumes of all gases under the same Number of moles of a gas can be calculated
tt E

conditions of temperature and pressure as follows

contain equal number of molecules. This
m
means that as long as the temperature and n= (5.12)
C

pressure remain constant, the volume M

depends upon number of molecules of the gas Where m = mass of the gas under
or in other words amount of the gas. investigation and M = molar mass
no N

Mathematically we can write Thus,

V ∝ n where n is the number of m
moles of the gas. V = k4 (5.13)
M
©

⇒ V = k4 n (5.11) Equation (5.13) can be rearranged as

follows :
The number of molecules in one mole of a
gas has been determined to be 6.022 1023 m
M = k4 = k4d (5.14)
and is known as Avogadro constant. You V
* The previous standard is still often used, and applies to all chemistry data more than decade old. In this definition STP
denotes the same temperature of 0°C (273.15 K), but a slightly higher pressure of 1 atm (101.325 kPa). One mole of any gas
of a combination of gases occupies 22.413996 L of volume at STP.
Standard ambient temperature and pressure (SATP), conditions are also used in some scientific works. SATP conditions
means 298.15 K and 1 bar (i.e., exactly 105 Pa). At SATP (1 bar and 298.15 K), the molar volume of an ideal gas is
24.789 L mol–1.

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 STATES OF MATTER 141

Here ‘d’ is the density of the gas. We can this equation we can see that at constant
conclude from equation (5.14) that the density temperature and pressure n moles of any gas
of a gas is directly proportional to its molar
mass. will have the same volume because V =
nRT
A gas that follows Boyle’s law, Charles’ p
law and Avogadro law strictly is called an and n,R,T and p are constant. This equation
will be applicable to any gas, under those
ideal gas. Such a gas is hypothetical. It is
conditions when behaviour of the gas
assumed that intermolecular forces are not approaches ideal behaviour. Volume of

d
present between the molecules of an ideal gas. one mole of an ideal gas under STP
Real gases follow these laws only under conditions (273.15 K and 1 bar pressure) is

he
certain specific conditions when forces of 22.710981 L mol–1. Value of R for one mole of
interaction are practically negligible. In all an ideal gas can be calculated under these
other situations these deviate from ideal conditions as follows :
behaviour. You will learn about the deviations
(10 5
Pa )( 22.71 ×10 –3 m 3 )

is
later in this unit. R=
(1mol )( 273.15 K )
5.6 IDEAL GAS EQUATION
= 8.314 Pa m3 K –1 mol–1

bl
The three laws which we have learnt till now
can be combined together in a single equation = 8.314 10–2 bar L K –1 mol–1
which is known as ideal gas equation.
= 8.314 J K –1 mol–1
pu
At constant T and n; V ∝
1
Boyle’s Law AtSTP conditions used earlier
p (0 °C
and 1 atm pressure), value of R is
be T

At constant p and n; V ∝ T Charles’ Law 8.20578 10–2 L atm K–1 mol–1.

re
At constant p and T ; V ∝ n Avogadro Law Ideal gas equation is a relation between
o R

Thus, four variables and it describes the state of

nT any gas, therefore, it is also called equation
V ∝
tt E

(5.15) of state.
p
nT Let us now go back to the ideal gas
⇒ V =R (5.16)
C

p equation. This is the relationship for the

simultaneous variation of the variables. If
where R is proportionality constant. On
temperature, volume and pressure of a fixed
no N

rearranging the equation (5.16) we obtain

amount of gas vary from T1, V1 and p1 to T2,
pV = n RT (5.17)
V2 and p2 then we can write
pV
⇒ R= (5.18) p1V1 p 2V2
©

nT = nR and = nR
T1 T2
R is called gas constant. It is same for all
gases. Therefore it is also called Universal p1V1 p V
⇒ = 2 2 (5.19)
Gas Constant. Equation (5.17) is called ideal T1 T2
gas equation.
Equation (5.19) is a very useful equation.
Equation (5.18) shows that the value of R If out of six, values of five variables are known,
depends upon units in which p, V and T are the value of unknown variable can be
measured. If three variables in this equation calculated from the equation (5.19). This
are known, fourth can be calculated. From equation is also known as Combined gas law.

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 142 CHEMISTRY

5.6.2 Dalton’s Law of Partial Pressures

Problem 5.3
At 25°C and 760 mm of Hg pressure a The law was formulated by John Dalton in
gas occupies 600 mL volume. What will 1801. It states that the total pressure
be its pressure at a height where exerted by the mixture of non-reactive
temperature is 10°C and volume of the gases is equal to the sum of the partial
gas is 640 mL. pressures of individual gases i.e., the
Solution pressures which these gases would exert if
they were enclosed separately in the same

d
p1 = 760 mm Hg, V1= 600 mL
volume and under the same conditions of
T1 = 25 + 273 = 298 K
temperature. In a mixture of gases, the

he
V2 = 640 mL and T2 = 10 + 273 = 283
pressure exerted by the individual gas is
K
called partial pressure. Mathematically,
According to Combined gas law
pTotal = p1+p2+p3+......(at constant T, V) (5.23)
p1V1 p2V2
=

is
T1 T2 where pTotal is the total pressure exerted
by the mixture of gases and p1, p2 , p3 etc. are
p1V1T2 partial pressures of gases.

bl
⇒ p2 =
T1V2 Gases are generally collected over water
and therefore are moist. Pressure of dry gas
pu ( 760 mm Hg ) × ( 600 mL ) × ( 283 K ) can be calculated by subtracting vapour
⇒ p2 =
( 640 mL ) × ( 298 K ) pressure of water from the total pressure of
= 676.6 mm Hg the moist gas which contains water vapours
be T

also. Pressure exerted by saturated water

vapour is called aqueous tension. Aqueous
re
5.6.1 Density and Molar Mass of a
o R

tension of water at different temperatures is

Gaseous Substance
given in Table 5.3.
Ideal gas equation can be rearranged as
follows: pDry gas = pTotal – Aqueous tension (5.24)
tt E

n p Table 5.3 Aqueous Tension of Water (Vapour

= Pressure) as a Function of
C

V RT Temperature
m
Replacing n by , we get
no N

m p
= (5.20)
MV RT
©

d p
=
M R T (where d is the density) (5.21)
On rearranging equation (5.21) we get the Partial pressure in terms of mole fraction
relationship for calculating molar mass of a Suppose at the temperature T, three gases,
gas. enclosed in the volume V, exert partial
pressure p1, p2 and p3 respectively. then,
d RT
M= (5.22) n1RT
p p1 = (5.25)
V

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 STATES OF MATTER 143

n 2 RT
p2 = (5.26) 70.6 g
V =
32 g mol −1
n 3 RT = 2.21 mol
p3 = (5.27)
V Number of moles of neon
where n1 n2 and n3 are number of moles of 167.5 g
these gases. Thus, expression for total =
20 g mol −1
pressure will be

d
= 8.375 mol
pTotal = p1 + p2 + p3
Mole fraction of dioxygen

he
RT RT RT 2.21
= n1 + n2 + n3 =
V V V 2.21 + 8.375
RT 2.21
= (n1 + n2 + n3) (5.28) =

is
V 10.585
On dividing p1 by ptotal we get = 0.21

bl
p1 ⎛ n1 ⎞ RTV 8.375
=⎜ ⎟ Mole fraction of neon =
p total ⎝ n1 +n 2 +n 3 ⎠ RTV 2.21 + 8.375
pu=
n1 n
= 1 = x1
n1 +n 2 +n 3 n
Alternatively,
= 0.79

mole fraction of neon = 1– 0.21 = 0.79

be T

where n = n1+n2+n3 Partial pressure = mole fraction

re
x1 is called mole fraction of first gas. of a gas total pressure
o R

Thus, p1 = x1 ptotal
⇒ Partial pressure = 0.21 (25 bar)
Similarly for other two gases we can write
of oxygen = 5.25 bar
tt E

p2 = x2 ptotal and p3 = x3 ptotal

Partial pressure = 0.79 (25 bar)
Thus a general equation can be written as
C

of neon = 19.75 bar

pi = xi ptotal (5.29)
where pi and xi are partial pressure and mole
no N

fraction of ith gas respectively. If total pressure 5.7 KINETIC MOLECULAR THEORY OF
of a mixture of gases is known, the equation GASES
(5.29) can be used to find out pressure exerted So far we have learnt the laws (e.g., Boyle’s
by individual gases. law, Charles’ law etc.) which are concise
©

statements of experimental facts observed in

Problem 5.4 the laboratory by the scientists. Conducting
A neon-dioxygen mixture contains careful experiments is an important aspect
70.6 g dioxygen and 167.5 g neon. If of scientific method and it tells us how the
pressure of the mixture of gases in the particular system is behaving under different
cylinder is 25 bar. What is the partial conditions. However, once the experimental
pressure of dioxygen and neon in the facts are established, a scientist is curious to
mixture ? know why the system is behaving in that way.
Number of moles of dioxygen For example, gas laws help us to predict that
pressure increases when we compress gases

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 144 CHEMISTRY

but we would like to know what happens at motion of molecules will stop and gases
molecular level when a gas is compressed ? A will settle down. This is contrary to what
theory is constructed to answer such is actually observed.
questions. A theory is a model (i.e., a mental • At any particular time, different particles
picture) that enables us to better understand in the gas have different speeds and hence
our observations. The theory that attempts different kinetic energies. This
to elucidate the behaviour of gases is known assumption is reasonable because as the
as kinetic molecular theory. particles collide, we expect their speed to
change. Even if initial speed of all the

d
Assumptions or postulates of the kinetic-
molecular theory of gases are given below. particles was same, the molecular
These postulates are related to atoms and collisions will disrupt this uniformity.

he
molecules which cannot be seen, hence it is Consequently the particles must have
said to provide a microscopic model of gases. different speeds, which go on changing
constantly. It is possible to show that
• Gases consist of large number of identical
though the individual speeds are
particles (atoms or molecules) that are so

is
changing, the distribution of speeds
small and so far apart on the average that
remains constant at a particular
the actual volume of the molecules is
temperature.

bl
negligible in comparison to the empty
space between them. They are considered • If a molecule has variable speed, then it
as point masses. This assumption must have a variable kinetic energy.
explains the great compressibility of gases.
pu Under these circumstances, we can talk
only about average kinetic energy. In
• There is no force of attraction between the
kinetic theory it is assumed that average
particles of a gas at ordinary temperature
kinetic energy of the gas molecules is
be T

and pressure. The support for this

directly proportional to the absolute
assumption comes from the fact that
re
temperature. It is seen that on heating a
o R

gases expand and occupy all the space

gas at constant volume, the pressure
available to them.
increases. On heating the gas, kinetic
• Particles of a gas are always in constant energy of the particles increases and these
tt E

and random motion. If the particles were strike the walls of the container more
at rest and occupied fixed positions, then frequently thus exerting more pressure.
a gas would have had a fixed shape which
C

Kinetic theory of gases allows us to derive

is not observed.
theoretically, all the gas laws studied in the
• Particles of a gas move in all possible previous sections. Calculations and
no N

directions in straight lines. During their predictions based on kinetic theory of gases
random motion, they collide with each agree very well with the experimental
other and with the walls of the container. observations and thus establish the
Pressure is exerted by the gas as a result correctness of this model.
©

of collision of the particles with the walls

of the container. 5.8 BEHAVIOUR OF REAL GASES:
DEVIATION FROM IDEAL GAS
• Collisions of gas molecules are perfectly
BEHAVIOUR
elastic. This means that total energy of
molecules before and after the collision Our theoritical model of gases corresponds
remains same. There may be exchange of very well with the experimental observations.
energy between colliding molecules, their Difficulty arises when we try to test how far
individual energies may change, but the the relation pV = nRT reproduce actual
sum of their energies remains constant. pressure-volume-temperature relationship of
If there were loss of kinetic energy, the gases. To test this point we plot pV vs p plot

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 STATES OF MATTER 145

of gases because at constant temperature, pV theoretically calculated from Boyle’s law (ideal
will be constant (Boyle’s law) and pV vs p gas) should coincide. Fig 5.9 shows these
graph at all pressures will be a straight line plots. It is apparent that at very high pressure
parallel to x-axis. Fig. 5.8 shows such a plot the measured volume is more than the
constructed from actual data for several gases calculated volume. At low pressures,
at 273 K. measured and calculated volumes approach
each other.

d
he
is
bl
pu
be T

Fig. 5.8 Plot of pV vs p for real gas and Fig. 5.9 Plot of pressure vs volume for real gas
re
ideal gas and ideal gas
o R

It can be seen easily that at constant It is found that real gases do not follow,
temperature pV vs p plot for real gases is not Boyle’s law, Charles law and Avogadro law
tt E

a straight line. There is a significant deviation perfectly under all conditions. Now two
from ideal behaviour. Two types of curves are questions arise.
seen.In the curves for dihydrogen and helium, (i) Why do gases deviate from the ideal
C

as the pressure increases the value of pV also behaviour?

increases. The second type of plot is seen in
(ii) What are the conditions under which
the case of other gases like carbon monoxide
gases deviate from ideality?
no N

and methane. In these plots first there is a

negative deviation from ideal behaviour, the We get the answer of the first question if
pV value decreases with increase in pressure we look into postulates of kinetic theory once
and reaches to a minimum value again. We find that two assumptions of the
©

characteristic of a gas. After that pV value kinetic theory do not hold good. These are
starts increasing. The curve then crosses the (a) There is no force of attraction between the
line for ideal gas and after that shows positive molecules of a gas.
deviation continuously. It is thus, found that (b) Volume of the molecules of a gas is
real gases do not follow ideal gas equation negligibly small in comparison to the space
perfectly under all conditions. occupied by the gas.
Deviation from ideal behaviour also If assumption (a) is correct, the gas will
becomes apparent when pressure vs volume never liquify. However, we know that gases
plot is drawn. The pressure vs volume plot of do liquify when cooled and compressed. Also,
experimental data (real gas) and that liquids formed are very difficult to compress.

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 146 CHEMISTRY

This means that forces of repulsion are van der Waals constants and their value
powerful enough and prevent squashing of depends on the characteristic of a gas. Value
molecules in tiny volume. If assumption (b) of ‘a’ is measure of magnitude of
is correct, the pressure vs volume graph of intermolecular attractive forces within the gas
experimental data (real gas) and that and is independent of temperature and
theoritically calculated from Boyles law (ideal pressure.
gas) should coincide. Also, at very low temperature,
Real gases show deviations from ideal gas intermolecular forces become significant. As
law because molecules interact with each the molecules travel with low average speed,

d
other. At high pressures molecules of gases these can be captured by one another due to
are very close to each other. Molecular attractive forces. Real gases show ideal

he
interactions start operating. At high pressure, behaviour when conditions of temperature
molecules do not strike the walls of the and pressure are such that the intermolecular
container with full impact because these are forces are practically negligible. The real gases
dragged back by other molecules due to show ideal behaviour when pressure

is
molecular attractive forces. This affects the approaches zero.
pressure exerted by the molecules on the walls The deviation from ideal behaviour can
of the container. Thus, the pressure exerted be measured in terms of compressibility

bl
by the gas is lower than the pressure exerted factor Z, which is the ratio of product pV and
by the ideal gas. nRT. Mathematically
an 2 pV
pu
pideal = preal +
V2
observed correction
(5.30) Z =
n RT (5.32)

For ideal gas Z = 1 at all temperatures

be T

pressure term and pressures because pV = n RT. The graph

Here, a is a constant. of Z vs p will be a straight line parallel to
re
o R

Repulsive forces also become significant. pressure axis (Fig. 5.10). For gases which
Repulsive interactions are short-range deviate from ideality, value of Z deviates from
interactions and are significant when unity. At very low pressures all gases shown
tt E

molecules are almost in contact. This is the

situation at high pressure. The repulsive
forces cause the molecules to behave as small
C

but impenetrable spheres. The volume

occupied by the molecules also becomes
significant because instead of moving in
no N

volume V, these are now restricted to volume

(V–nb) where nb is approximately the total
volume occupied by the molecules
themselves. Here, b is a constant. Having
©

taken into account the corrections for

pressure and volume, we can rewrite equation
(5.17) as

⎛ an 2 ⎞
⎜ p + ⎟ (V − nb ) = nRT (5.31)
⎝ V2 ⎠
Equation (5.31) is known as van der Waals
equation. In this equation n is number of Fig. 5.10 Variation of compressibility factor for
moles of the gas. Constants a and b are called some gases

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 STATES OF MATTER 147

have Z ≈1 and behave as ideal gas. At high gaseous state and liquid state and that liquids
pressure all the gases have Z > 1. These are may be considered as continuation of gas
more difficult to compress. At intermediate phase into a region of small volumes and very
pressures, most gases have Z < 1. Thus gases high molecular attraction. We will also see
show ideal behaviour when the volume how we can use isotherms of gases for
occupied is large so that the volume of the predicting the conditions for liquifaction of
molecules can be neglected in comparison gases.
to it. In other words, the behaviour of the
5.9 LIQUIFACTION OF GASES
gas becomes more ideal when pressure is very

d
low. Upto what pressure a gas will follow the First complete data on pressure - volume -
ideal gas law, depends upon nature of the temperature relations of a substance in both

he
gas and its temperature. The temperature at gaseous and liquid state was obtained by
which a real gas obeys ideal gas law over an Thomas Andrews on carbon dioxide. He
appreciable range of pressure is called Boyle plotted isotherms of carbon dioxide at various
temperature or Boyle point. Boyle point of temperatures (Fig. 5.11). Later on it was found
a gas depends upon its nature. Above their that real gases behave in the same manner

is
Boyle point, real gases show positive as carbon dioxide. Andrews noticed that at
deviations from ideality and Z values are high temperatures isotherms look like that
of an ideal gas and the gas cannot be liquified

bl
greater than one. The forces of attraction
between the molecules are very feeble. Below even at very high pressure. As the
Boyle temperature real gases first show temperature is lowered, shape of the curve
decrease in Z value with increasing pressure, changes and data shows considerable
pu
which reaches a minimum value. On further
increase in pressure, the value of Z increases
deviation from ideal behaviour. At 30.98 °C
be T

continuously. Above explanation shows that

at low pressure and high temperature gases
re
show ideal behaviour. These conditions are
o R

different for different gases.

More insight is obtained in the
significance of Z if we note the following
tt E

derivation
pVreal
C

Z = (5.33)
n RT
If the gas shows ideal behaviour then
no N

n RT nRT
Videal =
p . On putting this value of p
in equation (5.33) we have
©

Vreal
Z = (5.34)
Videal
From equation (5.34) we can see that
compressibility factor is the ratio of actual
molar volume of a gas to the molar volume of
it, if it were an ideal gas at that temperature
and pressure.
In the following sections we will see that Fig. 5.11 Isotherms of carbon dioxide at various
it is not possible to distinguish between temperatures

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 148 CHEMISTRY

carbon dioxide remains gas upto 73 cooled below their critical temperature for
atmospheric pressure. (Point E in Fig. 5.11). liquification. Critical temperature of a gas is
At 73 atmospheric pressure, liquid carbon highest temperature at which liquifaction of
dioxide appears for the first time. The the gas first occurs. Liquifaction of so called
temperature 30.98 ° C is called critical permanent gases (i.e., gases which show
temperature (TC) of carbon dioxide. This is continuous positive deviation in Z value)
the highest temperature at which liquid requires cooling as well as considerable
carbon dioxide is observed. Above this compression. Compression brings the
temperature it is gas. Volume of one mole of molecules in close vicinity and cooling slows

d
the gas at critical temperature is called down the movement of molecules therefore,
critical volume (VC) and pressure at this intermolecular interactions may hold the

he
temperature is called critical pressure (pC). closely and slowly moving molecules together
The critical temperature, pressure and volume and the gas liquifies.
are called critical constants. Further increase
It is possible to change a gas into liquid
in pressure simply compresses the liquid
or a liquid into gas by a process in which

is
carbon dioxide and the curve represents the
always a single phase is present. For example
compressibility of the liquid. The steep line
in Fig. 5.11 we can move from point A to F
represents the isotherm of liquid. Even a
vertically by increasing the temperature, then

bl
slight compression results in steep rise in
we can reach the point G by compressing the
pressure indicating very low compressibility
gas at the constant temperature along this
of the liquid. Below 30.98 °C, the behaviour
pu isotherm (isotherm at 31.1°C). The pressure
of the gas on compression is quite different.
will increase. Now we can move vertically
At 21.5 °C, carbon dioxide remains as a gas
down towards D by lowering the temperature.
only upto point B. At point B, liquid of a
As soon as we cross the point H on the critical
be T

particular volume appears. Further

isotherm we get liquid. We end up with liquid
compression does not change the pressure.
re
but in this series of changes we do not pass
Liquid and gaseous carbon dioxide coexist
o R

through two-phase region. If process is carried

and further application of pressure results
out at the critical temperature, substance
in the condensation of more gas until the
always remains in one phase.
tt E

point C is reached. At point C, all the gas has

been condensed and further application of Thus there is continuity between the
pressure merely compresses the liquid as gaseous and liquid state. The term fluid is
C

shown by steep line. A slight compression used for either a liquid or a gas to recognise
from volume V2 to V3 results in steep rise in this continuity. Thus a liquid can be viewed
pressure from p2 to p3 (Fig. 5.11). Below 30.98 as a very dense gas. Liquid and gas can be
no N

°C (critical temperature) each curve shows the distinguished only when the fluid is below its
similar trend. Only length of the horizontal critical temperature and its pressure and
line increases at lower temperatures. At volume lie under the dome, since in that
critical point horizontal portion of the situation liquid and gas are in equilibrium
©

isotherm merges into one point. Thus we see and a surface separating the two phases is
that a point like A in the Fig. 5.11 represents visible. In the absence of this surface there is
gaseous state. A point like D represents liquid no fundamental way of distinguishing
state and a point under the dome shaped area between two states. At critical temperature,
represents existence of liquid and gaseous liquid passes into gaseous state imperceptibly
carbon dioxide in equilibrium. All the gases and continuously; the surface separating two
upon compression at constant temperature phases disappears (Section 5.10.1). A gas
(isothermal compression) show the same below the critical temperature can be liquified
behaviour as shown by carbon dioxide. Also by applying pressure, and is called vapour of
above discussion shows that gases should be the substance. Carbon dioxide gas below its

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 STATES OF MATTER 149

critical temperature is called carbon dioxide sections we will look into some of the physical
vapour. Critical constants for some common properties of the liquids such as vapour
substances are given in Table 5.4. pressure, surface tension and viscosity.
Table 5.4 Critical Constants for Some 5.10.1 Vapour Pressure
Substances
If an evacuated container is partially filled with
a liquid, a portion of liquid evaporates to fill
the remaining volume of the container with
vapour. Initially the liquid evaporates and

d
pressure exerted by vapours on the walls of
the container (vapour pressure) increases.

he
After some time it becomes constant, an
equilibrium is established between liquid
phase and vapour phase. Vapour pressure at
this stage is known as equilibrium vapour
pressure or saturated vapour pressure..

is
Since process of vapourisation is temperature
dependent; the temperature must be
mentioned while reporting the vapour

bl
Problem 5.5
Gases possess characteristic critical pressure of a liquid.
temperature which depends upon the When a liquid is heated in an open vessel,
magnitude of intermolecular forces
pu
between the gas particles. Critical
temperatures of ammonia and carbon
the liquid vapourises from the surface. At the
temperature at which vapour pressure of the
liquid becomes equal to the external pressure,
be T

dioxide are 405.5 K and 304.10 K vapourisation can occur throughout the bulk
respectively. Which of these gases will of the liquid and vapours expand freely into
re
liquify first when you start cooling from the surroundings. The condition of free
o R

500 K to their critical temperature ? vapourisation throughout the liquid is called

Solution boiling. The temperature at which vapour
pressure of liquid is equal to the external
tt E

Ammonia will liquify first because its

critical temperature will be reached first. pressure is called boiling temperature at that
Liquifaction of CO2 will require more pressure. Vapour pressure of some common
C

cooling. liquids at various temperatures is given in

(Fig. 5.12). At 1 atm pressure boiling
temperature is called normal boiling point.
5.10 LIQUID STATE
no N

If pressure is 1 bar then the boiling point is

Intermolecular forces are stronger in liquid called standard boiling point of the liquid.
state than in gaseous state. Molecules in Standard boiling point of the liquid is slightly
liquids are so close that there is very little lower than the normal boiling point because
©

empty space between them and under normal 1 bar pressure is slightly less than 1 atm
conditions liquids are denser than gases. pressure. The normal boiling point of water
Molecules of liquids are held together by is 100 °C (373 K), its standard boiling point
attractive intermolecular forces. Liquids have is 99.6 °C (372.6 K).
definite volume because molecules do not At high altitudes atmospheric pressure is
separate from each other. However, molecules low. Therefore liquids at high altitudes boil
of liquids can move past one another freely, at lower temperatures in comparison to that
therefore, liquids can flow, can be poured and at sea level. Since water boils at low
can assume the shape of the container in temperature on hills, the pressure cooker is
which these are stored. In the following used for cooking food. In hospitals surgical

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 150 CHEMISTRY

5.10.2 Surface Tension

It is well known fact that liquids assume the
shape of the container. Why is it then small
drops of mercury form spherical bead instead
of spreading on the surface. Why do particles
of soil at the bottom of river remain separated
but they stick together when taken out ? Why
does a liquid rise (or fall) in a thin capillary
as soon as the capillary touches the surface

d
of the liquid ? All these phenomena are caused
due to the characteristic property of liquids,

he
called surface tension. A molecule in the
bulk of liquid experiences equal
intermolecular forces from all sides. The
molecule, therefore does not experience any

is
net force. But for the molecule on the surface
of liquid, net attractive force is towards the
interior of the liquid (Fig. 5.13), due to the

bl
molecules below it. Since there are no
molecules above it.
pu Liquids tend to minimize their surface
area. The molecules on the surface experience
a net downward force and have more energy
than the molecules in the bulk, which do not
be T

experience any net force. Therefore, liquids

tend to have minimum number of molecules
re
o R

Fig. 5.12 Vapour pressure vs temperature curve

at their surface. If surface of the liquid is
of some common liquids. increased by pulling a molecule from the bulk,
attractive forces will have to be overcome. This
instruments are sterilized in autoclaves in
tt E

will require expenditure of energy. The energy

which boiling point of water is increased by required to increase the surface area of the
increasing the pressure above the liquid by one unit is defined as surface energy.
C

atmospheric pressure by using a weight

covering the vent.
Boiling does not occur when liquid is
no N

heated in a closed vessel. On heating

continuously vapour pressure increases. At
first a clear boundary is visible between liquid
and vapour phase because liquid is more
©

dense than vapour. As the temperature

increases more and more molecules go to
vapour phase and density of vapours rises.
At the same time liquid becomes less dense.
It expands because molecules move apart.
When density of liquid and vapours becomes
the same; the clear boundary between liquid
and vapours disappears. This temperature is Fig. 5.13 Forces acting on a molecule on liquid
called critical temperature about which we surface and on a molecule inside the
have already discussed in section 5.9. liquid

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 STATES OF MATTER 151

Its dimensions are J m–2. Surface tension is of upper layers increases as the distance of
defined as the force acting per unit length layers from the fixed layer increases. This
perpendicular to the line drawn on the surface type of flow in which there is a regular
of liquid. It is denoted by Greek letter γ gradation of velocity in passing from one layer
(Gamma). It has dimensions of kg s–2 and in to the next is called laminar flow. If we
SI unit it is expressed as N m–1. The lowest choose any layer in the flowing liquid
energy state of the liquid will be when surface (Fig.5.14), the layer above it accelerates its
area is minimum. Spherical shape satisfies flow and the layer below this retards its flow.
this condition, that is why mercury drops are

d
spherical in shape. This is the reason that
sharp glass edges are heated for making them

he
smooth. On heating, the glass melts and the
surface of the liquid tends to take the rounded
shape at the edges, which makes the edges
smooth. This is called fire polishing of glass.

is
Liquid tends to rise (or fall) in the capillary
because of surface tension. Liquids wet the
things because they spread across their Fig. 5.14 Gradation of velocity in the laminar flow

bl
surfaces as thin film. Moist soil grains are
pulled together because surface area of thin If the velocity of the layer at a distance dz
film of water is reduced. It is surface tension
pu is changed by a value du then velocity
which gives stretching property to the surface
du
of a liquid. On flat surface, droplets are gradient is given by the amount . A force
slightly flattened by the effect of gravity; but dz
be T

in the gravity free environments drops are is required to maintain the flow of layers. This
force is proportional to the area of contact of
re
perfectly spherical.
o R

layers and velocity gradient i.e.

The magnitude of surface tension of a
liquid depends on the attractive forces F ∝ A (A is the area of contact)
between the molecules. When the attractive
tt E

du du
forces are large, the surface tension is large. F ∝ (where, is velocity gradient;
Increase in temperature increases the kinetic dz dz
the change in velocity with distance)
C

energy of the molecules and effectiveness of

intermolecular attraction decreases, so
du
surface tension decreases as the temperature F ∝ A.
no N

is raised. dz
du
5.10.3 Viscosity ⇒ F = ηA
dz
It is one of the characteristic properties of
‘ η ’ is proportionality constant and is
©

liquids. Viscosity is a measure of resistance

to flow which arises due to the internal called coefficient of viscosity. Viscosity
friction between layers of fluid as they slip coefficient is the force when velocity gradient
past one another while liquid flows. Strong is unity and the area of contact is unit area.
intermolecular forces between molecules hold Thus ‘ η ’ is measure of viscosity. SI unit of
them together and resist movement of layers viscosity coefficient is 1 newton second per
past one another. square metre (N s m –2) = pascal second
When a liquid flows over a fixed surface, (Pa s = 1kg m–1s–1). In cgs system the unit of
the layer of molecules in the immediate coefficient of viscosity is poise (named after
contact of surface is stationary. The velocity great scientist Jean Louise Poiseuille).

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 152 CHEMISTRY

1 poise = 1 g cm–1s–1 = 10–1kg m–1s–1 These become thicker at the bottom than at
Greater the viscosity, the more slowly the the top.
liquid flows. Hydrogen bonding and van der Viscosity of liquids decreases as the
Waals forces are strong enough to cause high temperature rises because at high
viscosity. Glass is an extremely viscous liquid. temperature molecules have high kinetic
It is so viscous that many of its properties
energy and can overcome the intermolecular
resemble solids. However, property of flow of
forces to slip past one another between the
glass can be experienced by measuring the
layers.
thickness of windowpanes of old buildings.

d
he
SUMMARY

Intermolecular forces operate between the particles of matter. These forces differ from pure
electrostatic forces that exist between two oppositely charged ions. Also, these do not include
forces that hold atoms of a covalent molecule together through covalent bond. Competition

is
between thermal energy and intermolecular interactions determines the state of matter.
“Bulk” properties of matter such as behaviour of gases, characteristics of solids and liquids
and change of state depend upon energy of constituent particles and the type of interaction

bl
between them. Chemical properties of a substance do not change with change of state, but
the reactivity depends upon the physical state.
pu Forces of interaction between gas molecules are negligible and are almost independent
of their chemical nature. Interdependence of some observable properties namely pressure,
volume, temperature and mass leads to different gas laws obtained from experimental
studies on gases. Boyle’s law states that under isothermal condition, pressure of a fixed
be T

amount of a gas is inversely proportional to its volume. Charles’ law is a relationship

between volume and absolute temperature under isobaric condition. It states that volume
re
of a fixed amount of gas is directly proportional to its absolute temperature (V ∝ T ) . If
o R

state of a gas is represented by p1, V1 and T1 and it changes to state at p2, V2 and T2, then
relationship between these two states is given by combined gas law according to which
tt E

p1V1 p 2V2
= . Any one of the variables of this gas can be found out if other five variables
T1 T2
C

are known. Avogadro law states that equal volumes of all gases under same conditions of
temperature and pressure contain equal number of molecules. Dalton’s law of partial
pressure states that total pressure exerted by a mixture of non-reacting gases is equal to
no N

the sum of partial pressures exerted by them. Thus p = p1+p2+p3+ ... . Relationship between
pressure, volume, temperature and number of moles of a gas describes its state and is
called equation of state of the gas. Equation of state for ideal gas is pV=nRT, where R is a
gas constant and its value depends upon units chosen for pressure, volume and temperature.
©

At high pressure and low temperature intermolecular forces start operating strongly
between the molecules of gases because they come close to each other. Under suitable
temperature and pressure conditions gases can be liquified. Liquids may be considered as
continuation of gas phase into a region of small volume and very strong molecular attractions.
Some properties of liquids e.g., surface tension and viscosity are due to strong intermolecular
attractive forces.

EXERCISES

5.1 What will be the minimum pressure required to compress 500 dm3 of air at 1 bar
to 200 dm3 at 30°C?

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 STATES OF MATTER 153

5.2 A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2
bar pressure. The gas is transferred to another vessel of volume 180 mL at 35
°C. What would be its pressure?
5.3 Using the equation of state pV=nRT; show that at a given temperature density
of a gas is proportional to gas pressure p.
5.4 At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen
at 5 bar. What is the molecular mass of the oxide?
5.5 Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another
ideal gas B is introduced in the same flask at same temperature the pressure

d
becomes 3 bar. Find a relationship between their molecular masses.
5.6 The drain cleaner, Drainex contains small bits of aluminum which react with caustic

he
soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will
be released when 0.15g of aluminum reacts?
5.7 What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of
carbon dioxide contained in a 9 dm3 flask at 27 °C ?

is
5.8 What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and
2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?
5.9 Density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be

bl
its density at STP?
5.10 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure.
pu What is the molar mass of phosphorus?
5.11 A student forgot to add the reaction mixture to the round bottomed flask at 27 °C
but instead he/she placed the flask on the flame. After a lapse of time, he realized
his mistake, and using a pyrometer he found the temperature of the flask was 477
be T

°C. What fraction of air would have been expelled out?

5.12 Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar.
re
(R = 0.083 bar dm3 K–1 mol–1).
o R

5.13 Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
5.14 How much time would it take to distribute one Avogadro number of wheat grains,
if 1010 grains are distributed each second ?
tt E

5.15 Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen

confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K–1 mol–1.
C

5.16 Pay load is defined as the difference between the mass of displaced air and the
mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass
100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m–3 and
no N

R = 0.083 bar dm3 K–1 mol–1).

5.17 Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure.
R = 0.083 bar L K–1 mol–1.
5.18 2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C,
©

at the same pressure. What is the molar mass of the gas?

5.19 A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight
of dihydrogen. Calculate the partial pressure of dihydrogen.
5.20 What would be the SI unit for the quantity pV 2T 2/n ?
5.21 In terms of Charles’ law explain why –273 °C is the lowest possible temperature.
5.22 Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C
respectively. Which of these has stronger intermolecular forces and why?
5.23 Explain the physical significance of van der Waals parameters.

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 154 CHEMISTRY

UNIT 6

THERMODYNAMICS

h ed
It is the only physical theory of universal content concerning

pu T
which I am convinced that, within the framework of the

is
applicability of its basic concepts, it will never be overthrown.

be able to
re ER
After studying this Unit, you will
Albert Einstein

bl
• explain the terms : system and
surroundings;
• discriminate between close,
open and isolated systems;
be C

• explain internal energy, work Chemical energy stored by molecules can be released as heat
and heat; during chemical reactions when a fuel like methane, cooking
• state first law of gas or coal burns in air. The chemical energy may also be
thermodynamics and express
N

it mathematically;
used to do mechanical work when a fuel burns in an engine
• calculate energy changes as
or to provide electrical energy through a galvanic cell like
work and heat contributions dry cell. Thus, various forms of energy are interrelated and
in chemical systems; under certain conditions, these may be transformed from
©

• explain state functions: U, H. one form into another. The study of these energy
• correlate ∆U and ∆H; transformations forms the subject matter of thermodynamics.
• measure experimentally ∆U The laws of thermodynamics deal with energy changes of
and ∆H; macroscopic systems involving a large number of molecules
• define standard states for ∆H; rather than microscopic systems containing a few molecules.
• calculate enthalpy changes for Thermodynamics is not concerned about how and at what
various types of reactions; rate these energy transformations are carried out, but is
•
to

state and apply Hess’s law of based on initial and final states of a system undergoing the
constant heat summation;
change. Laws of thermodynamics apply only when a system
• differentiate between extensive
and intensive properties; is in equilibrium or moves from one equilibrium state to
• define spontaneous and non- another equilibrium state. Macroscopic properties like
spontaneous processes; pressure and temperature do not change with time for a
t

• explain entropy as a system in equilibrium state. In this unit, we would like to

no

thermodynamic state function answer some of the important questions through

and apply it for spontaneity; thermodynamics, like:
• explain Gibbs energy change
(∆G); How do we determine the energy changes involved in a
• establish relationship between chemical reaction/process? Will it occur or not?
∆G and spontaneity, ∆G and
What drives a chemical reaction/process?
equilibrium constant.
To what extent do the chemical reactions proceed?
 THERMODYNAMICS 155

6.1 THERMODYNAMIC TERMS be real or imaginary. The wall that separates

the system from the surroundings is called
We are interested in chemical reactions and the
boundary. This is designed to allow us to
energy changes accompanying them. For this
control and keep track of all movements of
we need to know certain thermodynamic
matter and energy in or out of the system.
terms. These are discussed below.
6.1.2 Types of the System
6.1.1 The System and the Surroundings
We, further classify the systems according to
A system in thermodynamics refers to that

ed
the movements of matter and energy in or out
part of universe in which observations are
of the system.
made and remaining universe constitutes the
surroundings. The surroundings include 1. Open System
everything other than the system. System and In an open system, there is exchange of energy
and matter between system and surroundings

h
the surroundings together constitute the
universe . [Fig. 6.2 (a)]. The presence of reactants in an

pu T
open beaker is an example of an open system*.

is
The universe = The system + The surroundings
However, the entire universe other than Here the boundary is an imaginary surface
re ER
the system is not affected by the changes enclosing the beaker and reactants.
2. Closed System
taking place in the system. Therefore, for

bl
all practical purposes, the surroundings In a closed system, there is no exchange of
are that portion of the remaining universe matter, but exchange of energy is possible
which can interact with the system. between system and the surroundings
Usually, the region of space in the [Fig. 6.2 (b)]. The presence of reactants in a
be C

neighbourhood of the system constitutes closed vessel made of conducting material e.g.,
its surroundings. copper or steel is an example of a closed
For example, if we are studying the system.
N

reaction between two substances A and B

kept in a beaker, the beaker containing the
reaction mixture is the system and the room
where the beaker is kept is the surroundings
©

(Fig. 6.1).
to

Fig. 6.1 System and the surroundings

t

Note that the system may be defined by

physical boundaries, like beaker or test tube,
no

or the system may simply be defined by a set

of Cartesian coordinates specifying a
particular volume in space. It is necessary to
think of the system as separated from the
surroundings by some sort of wall which may Fig. 6.2 Open, closed and isolated systems.

* We could have chosen only the reactants as system then walls of the beakers will act as boundary.
 156 CHEMISTRY

3. Isolated System a quantity which represents the total energy

In an isolated system, there is no exchange of of the system. It may be chemical, electrical,
energy or matter between the system and the mechanical or any other type of energy you
surroundings [Fig. 6.2 (c)]. The presence of may think of, the sum of all these is the energy
reactants in a thermos flask or any other closed of the system. In thermodynamics, we call it
insulated vessel is an example of an isolated the internal energy, U of the system, which may
system. change, when
6.1.3 The State of the System • heat passes into or out of the system,

ed
The system must be described in order to make • work is done on or by the system,
any useful calculations by specifying • matter enters or leaves the system.
quantitatively each of the properties such as These systems are classified accordingly as
its pressure (p), volume (V), and temperature you have already studied in section 6.1.2.

h
(T ) as well as the composition of the system.
We need to describe the system by specifying (a) Work

pu T
it before and after the change. You would recall

is
from your Physics course that the state of a Let us first examine a change in internal
energy by doing work. We take a system
re ER
system in mechanics is completely specified at
a given instant of time, by the position and containing some quantity of water in a

bl
velocity of each mass point of the system. In thermos flask or in an insulated beaker. This
thermodynamics, a different and much simpler would not allow exchange of heat between the
concept of the state of a system is introduced. system and surroundings through its
It does not need detailed knowledge of motion boundary and we call this type of system as
be C

of each particle because, we deal with average adiabatic. The manner in which the state of
measurable properties of the system. We specify such a system may be changed will be called
the state of the system by state functions or adiabatic process. Adiabatic process is a
N

state variables. process in which there is no transfer of heat

The state of a thermodynamic system is between the system and surroundings. Here,
described by its measurable or macroscopic the wall separating the system and the
(bulk) properties. We can describe the state of surroundings is called the adiabatic wall
(Fig 6.3).
©

a gas by quoting its pressure (p), volume (V),

temperature (T ), amount (n) etc. Variables like
p, V, T are called state variables or state
functions because their values depend only
on the state of the system and not on how it is
reached. In order to completely define the state
of a system it is not necessary to define all the
to

properties of the system; as only a certain

number of properties can be varied
independently. This number depends on the
nature of the system. Once these minimum
number of macroscopic properties are fixed,
t

others automatically have definite values.

no

The state of the surroundings can never Fig. 6.3 An adiabatic system which does not
be completely specified; fortunately it is not permit the transfer of heat through its
necessary to do so. boundary.
6.1.4 The Internal Energy as a State Let us bring the change in the internal
Function energy of the system by doing some work on
When we talk about our chemical system it. Let us call the initial state of the system as
losing or gaining energy, we need to introduce state A and its temperature as T A. Let the
 THERMODYNAMICS 157

internal energy of the system in state A be the change in temperature is independent of

called UA. We can change the state of the system the route taken. Volume of water in a pond,
in two different ways. for example, is a state function, because
One way: We do some mechanical work, say change in volume of its water is independent
1 kJ, by rotating a set of small paddles and of the route by which water is filled in the
thereby churning water. Let the new state be pond, either by rain or by tubewell or by both,
called B state and its temperature, as T B. It is (b) Heat
found that T B > T A and the change in We can also change the internal energy of a

ed
temperature, ∆T = T B–TA . Let the internal system by transfer of heat from the
energy of the system in state B be UB and the surroundings to the system or vice-versa
change in internal energy, ∆U =U B – UA. without expenditure of work. This exchange
Second way: We now do an equal amount of energy, which is a result of temperature

h
(i.e., 1kJ) electrical work with the help of an difference is called heat, q. Let us consider
bringing about the same change in temperature

pu T
immersion rod and note down the temperature
(the same initial and final states as before in

is
change. We find that the change in
temperature is same as in the earlier case, say, section 6.1.4 (a) by transfer of heat through
TB – TA.
re ER thermally conducting walls instead of
adiabatic walls (Fig. 6.4).

bl
In fact, the experiments in the above
manner were done by J. P. Joule between
1840–50 and he was able to show that a given
amount of work done on the system, no matter
be C

how it was done (irrespective of path) produced

the same change of state, as measured by the
change in the temperature of the system.
N

So, it seems appropriate to define a

quantity, the internal energy U, whose value
is characteristic of the state of a system,
whereby the adiabatic work, wad required to
©

bring about a change of state is equal to the Fig. 6.4 A system which allows heat transfer
difference between the value of U in one state through its boundary.
and that in another state, ∆U i.e.,
We take water at temperature, TA in a
∆U = U 2 − U1 = w ad container having thermally conducting walls,
Therefore, internal energy, U, of the system say made up of copper and enclose it in a huge
is a state function. heat reservoir at temperature, TB . The heat
to

The positive sign expresses that wad is absorbed by the system (water), q can be
positive when work is done on the system. measured in terms of temperature difference ,
Similarly, if the work is done by the system,wad TB – TA. In this case change in internal energy,
will be negative. ∆U= q, when no work is done at constant
volume.
t

Can you name some other familiar state

functions? Some of other familiar state The q is positive, when heat is transferred
no

functions are V, p, and T. For example, if we from the surroundings to the system and q is
bring a change in temperature of the system negative when heat is transferred from
from 25°C to 35°C, the change in temperature system to the surroundings.
is 35°C–25°C = +10°C, whether we go straight
up to 35°C or we cool the system for a few (c) The general case
degrees, then take the system to the final Let us consider the general case in which a
temperature. Thus, T is a state function and change of state is brought about both by
 158 CHEMISTRY

doing work and by transfer of heat. We write Solution

change in internal energy for this case as: (i) ∆ U = w , wall is adiabatic
ad
∆U = q + w (6.1) (ii) ∆ U = – q, thermally conducting walls
For a given change in state, q and w can (iii) ∆ U = q – w, closed system.
vary depending on how the change is carried
out. However, q +w = ∆U will depend only on 6.2 APPLICATIONS
initial and final state. It will be independent of Many chemical reactions involve the generation

ed
the way the change is carried out. If there is of gases capable of doing mechanical work or
no transfer of energy as heat or as work the generation of heat. It is important for us to
(isolated system) i.e., if w = 0 and q = 0, then quantify these changes and relate them to the
∆ U = 0. changes in the internal energy. Let us see how!

h
The equation 6.1 i.e., ∆U = q + w is 6.2.1 Work

pu T
mathematical statement of the first law of First of all, let us concentrate on the nature of

is
thermodynamics, which states that work a system can do. We will consider only
mechanical work i.e., pressure-volume work.
constant.
re ER
The energy of an isolated system is
For understanding pressure-volume

bl
It is commonly stated as the law of work, let us consider a cylinder which
conservation of energy i.e., energy can neither contains one mole of an ideal gas fitted with a
be created nor be destroyed. frictionless piston. Total volume of the gas is
Vi and pressure of the gas inside is p. If
be C

Note: There is considerable difference between external pressure is p ex which is greater than
the character of the thermodynamic property p, piston is moved inward till the pressure
energy and that of a mechanical property such inside becomes equal to p ex. Let this change
N

as volume. We can specify an unambiguous

(absolute) value for volume of a system in a
particular state, but not the absolute value of
the internal energy. However, we can measure
©

only the changes in the internal energy, ∆U of

the system.
Problem 6.1
Express the change in internal energy of
a system when
(i) No heat is absorbed by the system
to

from the surroundings, but work (w)

is done on the system. What type of
wall does the system have ?
(ii) No work is done on the system, but
q amount of heat is taken out from
t

the system and given to the

no

surroundings. What type of wall does

the system have?
(iii) w amount of work is done by the Fig. 6.5(a) Work done on an ideal gas in a
system and q amount of heat is cylinder when it is compressed by a
supplied to the system. What type of constant external pressure, p ex
system would it be? (in single step) is equal to the shaded
area.
 THERMODYNAMICS 159

be achieved in a single step and the final If the pressure is not constant but changes
volume be V f . During this compression, during the process such that it is always
suppose piston moves a distance, l and is infinitesimally greater than the pressure of the
cross-sectional area of the piston is A gas, then, at each stage of compression, the
[Fig. 6.5(a)]. volume decreases by an infinitesimal amount,
then, volume change = l × A = ∆V = (Vf – Vi ) dV. In such a case we can calculate the work
done on the gas by the relation
force
We also know, pressure = Vf

ed
area w= − ∫p ex dV ( 6.3)
Therefore, force on the piston = pex . A Vi

If w is the work done on the system by Here, p ex at each stage is equal to (p in + dp) in
movement of the piston then case of compression [Fig. 6.5(c)]. In an

h
w = force × distance = pex . A .l expansion process under similar conditions,
the external pressure is always less than the

pu T
= pex . (–∆V) = – pex ∆V = – pex (V f – V i ) (6.2)
pressure of the system i.e., pex = (pin– dp). In

is
The negative sign of this expression is general case we can write, pex = (pin + dp). Such
required to obtain conventional sign for w,
re ER
which will be positive. It indicates that in case
processes are called reversible processes.
A process or change is said to be

bl
of compression work is done on the system. reversible, if a change is brought out in
Here (V f – V i ) will be negative and negative such a way that the process could, at any
multiplied by negative will be positive. Hence moment, be reversed by an infinitesimal
the sign obtained for the work will be positive. change. A reversible process proceeds
be C

If the pressure is not constant at every infinitely slowly by a series of equilibrium

stage of compression, but changes in number states such that system and the
of finite steps, work done on the gas will be surroundings are always in near
N

summed over all the steps and will be equal equilibrium with each other. Processes
to − ∑ p ∆V [Fig. 6.5 (b)]
t to ©
no

Fig. 6.5 (c) pV-plot when pressure is not constant

Fig. 6.5 (b) pV-plot when pressure is not constant and changes in infinite steps
and changes in finite steps during (reversible conditions) during
compression from initial volume, Vi to compression from initial volume, Vi to
final volume, Vf . Work done on the gas final volume, Vf . Work done on the gas
is represented by the shaded area. is represented by the shaded area.
 160 CHEMISTRY

other than reversible processes are known Isothermal and free expansion of an
as irreversible processes. ideal gas
In chemistry, we face problems that can For isothermal (T = constant) expansion of an
be solved if we relate the work term to the ideal gas into vacuum ; w = 0 since p ex = 0.
internal pressure of the system. We can Also, Joule determined experimentally that
relate work to internal pressure of the system q = 0; therefore, ∆U = 0
under reversible conditions by writing Equation 6.1, ∆ U = q + w can be
equation 6.3 as follows: expressed for isothermal irreversible and

ed
Vf Vf reversible changes as follows:
wrev = − ∫p ex dV =− ∫ (p in ± dp )dV 1. For isothermal irreversible change
Vi Vi q = – w = pex (Vf – Vi )
2. For isothermal reversible change

h
Since dp × dV is very small we can write

pu T
Vf Vf
w rev = − ∫ p in dV

is
(6.4) q = – w = nRT ln V
i
Vi
re ER
Now, the pressure of the gas (pin which we = 2.303 nRT log
Vf

bl
Vi
can write as p now) can be expressed in terms
of its volume through gas equation. For n mol 3. For adiabatic change, q = 0,
of an ideal gas i.e., pV =nRT ∆U = wad
nRT
be C

⇒p = Problem 6.2
V
Therefore, at constant temperature (isothermal Two litres of an ideal gas at a pressure of
N

process), 10 atm expands isothermally into a

Vf
vacuum until its total volume is 10 litres.
dV Vf How much heat is absorbed and how
w rev = − ∫ nRT = −nRT ln
Vi V Vi much work is done in the expansion ?
©

Solution
Vf
= – 2.303 nRT log (6.5) We have q = – w = p ex (10 – 2) = 0(8) = 0
Vi No work is done; no heat is absorbed.
Free expansion: Expansion of a gas in
vacuum (pex = 0) is called free expansion. No Problem 6.3
work is done during free expansion of an ideal Consider the same expansion, but this
gas whether the process is reversible or time against a constant external pressure
to

irreversible (equation 6.2 and 6.3). of 1 atm.

Now, we can write equation 6.1 in number Solution
of ways depending on the type of processes. We have q = – w = pex (8) = 8 litre-atm
Let us substitute w = – p ex∆V (eq. 6.2) in Problem 6.4
t

equation 6.1, and we get Consider the same expansion, to a final

no

∆U = q − p ex ∆V volume of 10 litres conducted reversibly.

Solution
If a process is carried out at constant volume
(∆V = 0), then 10
We have q = – w = 2.303 × 10 log
∆U = qV 2
the subscript V in qV denotes that heat is = 16.1 litre-atm
supplied at constant volume.
 THERMODYNAMICS 161

6.2.2 Enthalpy, H ∆H is positive for endothermic reactions

(a) A useful new state function which absorb heat from the surroundings.
We know that the heat absorbed at constant At constant volume (∆V = 0), ∆U = qV,
volume is equal to change in the internal therefore equation 6.8 becomes
energy i.e., ∆U = qV. But most of chemical ∆H = ∆U = q
V
reactions are carried out not at constant
volume, but in flasks or test tubes under The difference between ∆H and ∆U is not
constant atmospheric pressure. We need to usually significant for systems consisting of

ed
define another state function which may be only solids and / or liquids. Solids and liquids
suitable under these conditions. do not suffer any significant volume changes
upon heating. The difference, however,
We may write equation (6.1) as becomes significant when gases are involved.
∆U = q p − p ∆V at constant pressure, where Let us consider a reaction involving gases. If

h
qp is heat absorbed by the system and –p∆V VA is the total volume of the gaseous reactants,

pu T
represent expansion work done by the system. VB is the total volume of the gaseous products,

is
Let us represent the initial state by nA is the number of moles of gaseous reactants
subscript 1 and final state by 2 and nB is the number of moles of gaseous
re ER
We can rewrite the above equation as products, all at constant pressure and

bl
temperature, then using the ideal gas law, we
U2–U 1 = qp – p (V2 – V1) write,
On rearranging, we get
pVA = n ART
qp = (U 2 + pV2) – (U 1 + pV1) (6.6)
be C

Now we can define another thermodynamic and pVB = n BRT

function, the enthalpy H [Greek word Thus, pVB – pVA = nB RT – nART = (n B–nA)RT
enthalpien, to warm or heat content] as :
N

H = U + pV (6.7) or p (VB – VA) = (nB – nA) RT

so, equation (6.6) becomes or p ∆V = ∆n gRT (6.9)
qp= H 2 – H1 = ∆H Here, ∆ng refers to the number of moles of
©

Although q is a path dependent function, gaseous products minus the number of moles
H is a state function because it depends on U, of gaseous reactants.
p and V, all of which are state functions. Substituting the value of p∆V from
Therefore, ∆H is independent of path. Hence,
equation 6.9 in equation 6.8, we get
qp is also independent of path.
For finite changes at constant pressure, we ∆H = ∆U + ∆n gRT (6.10)
can write equation 6.7 as The equation 6.10 is useful for calculating
to

∆H = ∆U + ∆pV ∆H from ∆U and vice versa.

Since p is constant, we can write Problem 6.5
∆H = ∆U + p∆V (6.8) If water vapour is assumed to be a perfect
t

It is important to note that when heat is gas, molar enthalpy change for
absorbed by the system at constant pressure,
no

vapourisation of 1 mol of water at 1bar

we are actually measuring changes in the and 100°C is 41kJ mol–1. Calculate the
enthalpy. internal energy change, when
Remember ∆H = qp , heat absorbed by the (i) 1 mol of water is vaporised at 1 bar
system at constant pressure. pressure and 100°C.
∆H is negative for exothermic reactions (ii) 1 mol of water is converted into ice.
which evolve heat during the reaction and
 162 CHEMISTRY

Solution halved, each part [Fig. 6.6 (b)] now has one
V
half of the original volume, , but the
(i) The change H2 O (l ) → H 2O ( g ) 2
temperature will still remain the same i.e., T.
∆H = ∆U + ∆ n g RT It is clear that volume is an extensive property
and temperature is an intensive property.
or ∆U = ∆H – ∆n g R T , substituting the
values, we get

ed
∆U = 41.00 kJ mol −1 − 1
× 8.3 J mol −1 K −1 × 373 K

= 41.00 kJ mol −1 − 3.096 kJ mol −1

h
= 37.904 kJ mol–1 Fig. 6.6(a) A gas at volume V and temperature T
(ii) The change H2O ( l ) → H2O ( s)

pu T
is
There is negligible change in volume,
re ER
So, we can put p ∆V = ∆n g R T ≈ 0 in this

bl
case,
∆H ≅ ∆U

so, ∆U = 41.00 kJ mol − 1 Fig. 6.6 (b) Partition, each part having half the
volume of the gas
be C

(b) Extensive and Intensive Properties (c) Heat Capacity

In thermodynamics, a distinction is made In this sub-section, let us see how to measure
N

between extensive properties and intensive heat transferred to a system. This heat appears
properties. An extensive property is a as a rise in temperature of the system in case
property whose value depends on the quantity of heat absorbed by the system.
or size of matter present in the system. For The increase of temperature is proportional
©

example, mass, volume, internal energy, to the heat transferred

enthalpy, heat capacity, etc. are extensive
properties. q = coeff × ∆T
Those properties which do not depend on The magnitude of the coefficient depends
the quantity or size of matter present are on the size, composition and nature of the
known as intensive properties. For example system. We can also write it as q = C ∆T
temperature, density, pressure etc. are The coefficient, C is called the heat capacity.
to

intensive properties. A molar property, χm, is Thus, we can measure the heat supplied
the value of an extensive property χ of the by monitoring the temperature rise, provided
system for 1 mol of the substance. If n is the we know the heat capacity.
χ When C is large, a given amount of heat
t

amount of matter, χm = is independent of

n results in only a small temperature rise. Water
no

the amount of matter. Other examples are has a large heat capacity i.e., a lot of energy is
molar volume, Vm and molar heat capacity, Cm. needed to raise its temperature.
Let us understand the distinction between C is directly proportional to amount of
extensive and intensive properties by substance. The molar heat capacity of a
considering a gas enclosed in a container of
volume V and at temperature T [Fig. 6.6(a)]. C 
substance, C m =   , is the heat capacity for
Let us make a partition such that volume is n 
 THERMODYNAMICS 163

one mole of the substance and is the quantity heat capacity of the liquid in which calorimeter
of heat needed to raise the temperature of one is immersed and the heat capacity of
mole by one degree celsius (or one kelvin). calorimeter, it is possible to determine the heat
Specific heat, also called specific heat capacity evolved in the process by measuring
is the quantity of heat required to raise the temperature changes. Measurements are
temperature of one unit mass of a substance made under two different conditions:
by one degree celsius (or one kelvin). For i) at constant volume, qV
finding out the heat, q, required to raise the
ii) at constant pressure, qp

ed
temperatures of a sample, we multiply the
specific heat of the substance, c, by the mass (a) ∆U measurements
m, and temperatures change, ∆T as For chemical reactions, heat absorbed at
constant volume, is measured in a bomb
q = c × m × ∆T = C ∆T (6.11)
calorimeter (Fig. 6.7). Here, a steel vessel (the

h
(d) The relationship between Cp and CV for bomb) is immersed in a water bath. The whole

pu T
an ideal gas device is called calorimeter. The steel vessel is

is
At constant volume, the heat capacity, C is immersed in water bath to ensure that no heat
denoted by CV and at constant pressure, this is lost to the surroundings. A combustible
re ER
is denoted by C p . Let us find the relationship substance is burnt in pure dioxygen supplied

bl
between the two. in the steel bomb. Heat evolved during the
We can write equation for heat, q reaction is transferred to the water around the
bomb and its temperature is monitored. Since
at constant volume as qV = CV ∆T = ∆U the bomb calorimeter is sealed, its volume does
be C

not change i.e., the energy changes associated

at constant pressure as qp = C p ∆T = ∆H with reactions are measured at constant
The difference between Cp and CV can be volume. Under these conditions, no work is
N

derived for an ideal gas as:

For a mole of an ideal gas, ∆H = ∆U + ∆(pV )
= ∆U + ∆(RT )
= ∆U + R∆T
©

∴ ∆H = ∆U + R ∆T (6.12)
On putting the values of ∆H and ∆U,
we have
C p∆T = CV ∆T + R∆T
to

C p = CV + R

C p − CV = R (6.13)
t

6.3 MEASUREMENT OF ∆ U AND ∆ H:

no

CALORIMETRY
We can measure energy changes associated
with chemical or physical processes by an
experimental technique called calorimetry. In
calorimetry, the process is carried out in a
vessel called calorimeter, which is immersed
in a known volume of a liquid. Knowing the Fig. 6.7 Bomb calorimeter
 164 CHEMISTRY

done as the reaction is carried out at constant Problem 6.6

volume in the bomb calorimeter. Even for
reactions involving gases, there is no work 1g of graphite is burnt in a bomb
done as ∆V = 0. Temperature change of the calorimeter in excess of oxygen at 298 K
calorimeter produced by the completed and 1 atmospheric pressure according to
reaction is then converted to qV , by using the the equation
known heat capacity of the calorimeter with C (graphite) + O2 (g) → CO2 (g)
the help of equation 6.11. During the reaction, temperature rises

ed
(b) ∆ H measurements from 298 K to 299 K. If the heat capacity
Measurement of heat change at constant of the bomb calorimeter is 20.7kJ/K,
pressure (generally under atmospheric what is the enthalpy change for the above
pressure) can be done in a calorimeter shown reaction at 298 K and 1 atm?
in Fig. 6.8. We know that ∆H = q p (at Solution

h
constant p) and, therefore, heat absorbed or Suppose q is the quantity of heat from

pu T
evolved, qp at constant pressure is also called the reaction mixture and CV is the heat

is
the heat of reaction or enthalpy of reaction, ∆ rH. capacity of the calorimeter, then the
re ER
In an exothermic reaction, heat is evolved, quantity of heat absorbed by the
calorimeter.

bl
and system loses heat to the surroundings.
Therefore, qp will be negative and ∆r H will also q = C V × ∆T
be negative. Similarly in an endothermic Quantity of heat from the reaction will
reaction, heat is absorbed, qp is positive and have the same magnitude but opposite
∆ rH will be positive. sign because the heat lost by the system
be C

(reaction mixture) is equal to the heat

gained by the calorimeter.
N

q = − C V × ∆T = − 20.7 kJ/K × (299 − 298)K

= − 20.7 kJ
(Here, negative sign indicates the
exothermic nature of the reaction)
©

Thus, ∆U for the combustion of the 1g of

graphite = – 20.7 kJK –1
For combustion of 1 mol of graphite,
12.0 g mol − 1 × ( −20.7 kJ )
=
1g
to

= – 2.48 ×102 kJ mol–1 , Since ∆ n g = 0,

∆ H = ∆ U = – 2.48 ×10 kJ mol–1
2

6.4 ENTHALPY CHANGE, ∆ r H OF A

t

REACTION – REACTION ENTHALPY

no

In a chemical reaction, reactants are converted

into products and is represented by,
Reactants → Products
The enthalpy change accompanying a
Fig. 6.8 Calorimeter for measuring heat changes reaction is called the reaction enthalpy. The
at constant pressure (atmospheric enthalpy change of a chemical reaction, is
pressure). given by the symbol ∆rH
 THERMODYNAMICS 165

∆ rH = (sum of enthalpies of products) – (sum melting. Normally this melting takes place at
of enthalpies of reactants) constant pressure (atmospheric pressure) and
during phase change, temperature remains
= ∑ a i H products − ∑ bi Hreactants (6.14) constant (at 273 K).
i i

H2O(s) → H2O(l ); ∆ fus H V = 6.00 kJ mol −1

(Here symbol ∑ (sigma) is used for
summation and ai and bi are the stoichiometric Here ∆fus H 0 is enthalpy of fusion in standard
coefficients of the products and reactants state. If water freezes, then process is reversed

ed
respectively in the balanced chemical and equal amount of heat is given off to the
equation. For example, for the reaction surroundings.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) The enthalpy change that accompanies
melting of one mole of a solid substance
∆r H = ∑ a i H products − ∑ bi Hreactants

h
in standard state is called standard

pu T
i i
enthalpy of fusion or molar enthalpy of
0
fusion, ∆fusH .

is
= [H m (CO2 ,g) + 2Hm (H2O, l)]– [H m (CH4 , g)
+ 2Hm (O2, g)] Melting of a solid is endothermic, so all
re ER
where Hm is the molar enthalpy. enthalpies of fusion are positive. Water requires

bl
Enthalpy change is a very useful quantity. heat for evaporation. At constant temperature
Knowledge of this quantity is required when of its boiling point Tb and at constant pressure:
one needs to plan the heating or cooling H2O(l ) → H2 O(g); ∆vap H V = + 40.79 kJ mol − 1
required to maintain an industrial chemical
be C

reaction at constant temperature. It is also ∆vapH 0 is the standard enthalpy of vaporization.

required to calculate temperature dependence Amount of heat required to vaporize
of equilibrium constant.
one mole of a liquid at constant
N

(a) Standard enthalpy of reactions temperature and under standard pressure

Enthalpy of a reaction depends on the (1bar) is called its standard enthalpy of
conditions under which a reaction is carried vaporization or molar enthalpy of
out. It is, therefore, necessary that we must vaporization, ∆ vapH 0.
©

specify some standard conditions. The

Sublimation is direct conversion of a solid
standard enthalpy of reaction is the
enthalpy change for a reaction when all into its vapour. Solid CO2 or ‘dry ice’ sublimes
the participating substances are in their at 195K with ∆ s u bH 0=25.2 kJ mol –1 ;
standard states. naphthalene sublimes slowly and for this
The standard state of a substance at a ∆sub H V = 73.0 kJ mol −1 .
specified temperature is its pure form at
to

Standard enthalpy of sublimation,

1 bar. For example, the standard state of liquid ∆subH 0 is the change in enthalpy when one
ethanol at 298 K is pure liquid ethanol at mole of a solid substance sublimes at a
1 bar; standard state of solid iron at 500 K is constant temperature and under standard
pure iron at 1 bar. Usually data are taken at pressure (1bar).
t

298 K.
no

The magnitude of the enthalpy change

Standard conditions are denoted by adding
depends on the strength of the intermolecular
the superscript V to the symbol ∆H, e.g., ∆H V interactions in the substance undergoing the
(b) Enthalpy changes during phase phase transfomations. For example, the strong
transformations hydrogen bonds between water molecules hold
Phase transformations also involve energy them tightly in liquid phase. For an organic
changes. Ice, for example, requires heat for liquid, such as acetone, the intermolecular
 166 CHEMISTRY

Table 6.1 Standard Enthalpy Changes of Fusion and Vaporisation

h ed
pu T
is
re ER (T f and Tb are melting and boiling points, respectively)

bl
dipole-dipole interactions are significantly
weaker. Thus, it requires less heat to vaporise ∆ vapH V − ∆ng RT = 40.66 kJ mol −1
1 mol of acetone than it does to vaporize 1 mol − (1)(8.314 JK −1mol −1 )(373K )(10 −3 kJ J −1 )
of water. Table 6.1 gives values of standard
be C

enthalpy changes of fusion and vaporisation ∆vapU V = 40.66 kJ mol −1 − 3.10 kJ mol −1
for some substances. = 37.56 kJ mol −1
N

Problem 6.7
(c) Standard enthalpy of formation
A swimmer coming out from a pool is
The standard enthalpy change for the
covered with a film of water weighing
about 18g. How much heat must be formation of one mole of a compound from
©

supplied to evaporate this water at its elements in their most stable states of
298 K ? Calculate the internal energy of aggregation (also known as reference
vaporisation at 100°C. states) is called Standard Molar Enthalpy
of Formation. Its symbol is ∆ f H 0, where
∆vapHV for water the subscript ‘ f ’ indicates that one mole of
the compound in question has been formed in
at 373K = 40.66 kJ mol–1
its standard state from its elements in their
Solution
to

most stable states of aggregation. The reference

We can r epresent the process of state of an element is its most stable state of
evaporation as aggregation at 25°C and 1 bar pressure.
For example, the reference state of dihydrogen
18 gH2O(l) →
vaporisation
18 g H2O(g)
is H2 gas and those of dioxygen, carbon and
t

No. of moles in 18 g H2O(l) is sulphur are O 2 gas, Cgraphite and Srhombic

no

respectively. Some reactions with standard

18g
= = 1 mol molar enthalpies of formation are given below.
18 g mol − 1
H2 (g) + ½O2 (g) → H2 O(1);
∆vap U = ∆vap H V − p ∆V = ∆ vap H V − ∆ n gRT
∆ f H V = −285.8 kJ mol −1
(assuming steam behaving as an ideal
gas). C (graphite, s) +2H2 (g) → CH4 (g);
 THERMODYNAMICS 167

∆ f H y) at 298K of a
Table 6.2 Standard Molar Enthalpies of Formation (∆
Few Selected Substances

h ed
pu T
is
re ER
bl
be C
N
©

∆ f H V = −74.81 kJ mol−1 is not an enthalpy of formation of calcium

carbonate, since calcium carbonate has been
2C( graphite,s) + 3H2 (g ) + ½O2 (g) → C2 H5OH(1);
to

formed from other compounds, and not from

∆ f H V = −277.7kJ mol −1 its constituent elements. Also, for the reaction
given below, enthalpy change is not standard
0
It is important to understand that a enthalpy of formation, ∆ fH for HBr(g).
standard molar enthalpy of formation, ∆f H 0,
H2 (g) + Br2 (l ) → 2HBr(g);
t

is just a special case of ∆r H 0, where one mole

no

of a compound is formed from its constituent ∆r H V = −72.8 kJ mol −1

elements, as in the above three equations, Here two moles, instead of one mole of the
where 1 mol of each, water, methane and
product is formed from the elements, i.e.,
ethanol is formed. In contrast, the enthalpy
change for an exothermic reaction: ∆ r H V = 2∆ f H V .
CaO(s) + CO2 (g) → CaCO3 (s); Therefore, by dividing all coefficients in the
−1
∆r H V
= −178.3kJ mol balanced equation by 2, expression for
 168 CHEMISTRY

enthalpy of formation of HBr (g) is written as (alongwith allotropic state) of the substance in
an equation. For example:
½H 2 (g) + ½Br2 (1) → HBr(g );
∆ f H V = − 36.4 kJ m ol− 1 C2H 5 OH(l ) + 3O 2 (g ) → 2CO 2 (g ) + 3H 2 O(l ) :
Standard enthalpies of formation of some ∆ r H V = −1367 kJ mol −1
common substances are given in Table 6.2. The abov e equation describes the
By convention, standard enthalpy for combustion of liquid ethanol at constant
formation, ∆f H 0, of an element in reference

ed
temperature and pressure. The negative sign
state, i.e., its most stable state of aggregation of enthalpy change indicates that this is an
is taken as zero. exothermic reaction.
Suppose, you are a chemical engineer and It would be necessary to remember the
want to know how much heat is required to following conventions regarding thermo-

h
decompose calcium carbonate to lime and chemical equations.

pu T
carbon dioxide, with all the substances in their
1. The coefficients in a balanced thermo-

is
standard state.
chemical equation refer to the number of
re ER
CaCO3 (s) → CaO(s) + CO2 (g); ∆r H V = ? moles (never molecules) of reactants and

bl
Here, we can make use of standard enthalpy products involved in the reaction.
0
of formation and calculate the enthalpy 2. The numerical value of ∆r H refers to the
change for the reaction. The following general number of moles of substances specified
equation can be used for the enthalpy change by an equation. Standard enthalpy change
be C

0
calculation. ∆ rH will have units as kJ mol–1.
∆ r H V = ∑ ai ∆ f H V (products) − ∑ bi ∆ f H V ( reactants) To illustrate the concept, let us consider
i i the calculation of heat of reaction for the
N

(6.15) following reaction :

where a and b represent the coefficients of the
products and reactants in the balanced Fe2O3 ( s) + 3H2 ( g ) → 2Fe ( s ) + 3H2O ( l ) ,
equation. Let us apply the above equation for From the Table (6.2) of standard enthalpy of
©

decomposition of calcium carbonate. Here, formation (∆ f H 0), we find :

coefficients ‘a’ and ‘b’ are 1 each.
Therefore, ∆ f H V ( H2O, l ) = –285.83 kJ mol–1;

∆r H V = ∆ f H V [CaO(s)] + ∆ f H V [CO2 (g)] ∆ f H V ( Fe 2O 3 ,s ) = – 824.2 kJ mol–1;

− ∆ f H V [CaCO3 (s)] A lso ∆ f H V(Fe, s) = 0 and

to

∆ f H V (H 2 ,g ) = 0 a s per convention
=1( − 635.1 kJ mol− 1 ) + 1( − 393.5 kJ mol − 1)
− 1( −1206.9 kJ mol− 1 ) Then,
V
∆r H 1 = 3(–285.83 kJ mol–1)
= 178.3 kJ mol–1
t

Thus, the decomposition of CaCO3 (s) is an – 1(– 824.2 kJ mol–1)

no

endothermic process and you have to heat it = (–857.5 + 824.2) kJ mol –1

for getting the desired products.
= –33.3 kJ mol–1
(d) Thermochemical equations Note that the coefficients used in these
A balanced chemical equation together with calculations are pure numbers, which are
the value of its ∆ rH is called a thermochemical equal to the respective stoichiometric
equation. We specify the physical state coefficients. The unit for ∆ r H 0 is
 THERMODYNAMICS 169

kJ mol–1, which means per mole of reaction. Consider the enthalpy change for the
Once we balance the chemical equation in a reaction
particular way, as above, this defines the mole 1
of reaction. If we had balanced the equation C ( graphite,s) + O2 ( g ) → CO ( g ) ; ∆r H V = ?
2
differently, for example,
1 3 3 Although CO(g) is the major product, some
Fe2O3 ( s) + H2 ( g ) → Fe ( s ) + H2O ( l ) CO2 gas is always produced in this reaction.
2 2 2
Therefore, we cannot measure enthalpy change

ed
then this amount of reaction would be one
0 for the above reaction directly. However, if we
mole of reaction and ∆ rH would be
can find some other reactions involving related
3 species, it is possible to calculate the enthalpy
V
∆ r H2 =
2
(− 285.83 kJ mol −1 ) change for the above reaction.

h
Let us consider the following reactions:
1
− ( − 824.2 kJ mol− 1 )

pu T
C ( graphite,s) + O2 ( g ) → CO2 ( g ) ;

is
2
(i)
= (– 428.7 + 412.1) kJ mol –1
∆r H V = − 393.5 kJ mol − 1
re ER V

bl
= –16.6 kJ mol–1 = ½ ∆r H 1 1
CO ( g ) + O2 ( g ) → CO2 ( g ) ;
It shows that enthalpy is an extensive quantity. 2 (ii)
3. When a chemical equation is reversed, the ∆r H V = −283.0 kJ mol −1
value of ∆ r H 0 is reversed in sign. For
be C

example We can combine the above two reactions

in such a way so as to obtain the desired
N 2(g) + 3H2 ( g ) → 2NH3 (g); reaction. To get one mole of CO(g) on the right,
N

∆r H V = −91.8 kJ mol − 1 we reverse equation (ii). In this, heat is

absorbed instead of being released, so we
2NH3 (g) → N 2(g) + 3H2 (g); change sign of ∆rH 0 value
∆r H V = + 91.8 kJ mol −1
©

1
CO2 ( g) → CO ( g ) + O2 ( g ) ;
(e) Hess’s Law of Constant Heat 2
Summation ∆r H V = + 283.0 kJ mol − 1 (iii)
We know that enthalpy is a state function,
therefore the change in enthalpy is Adding equation (i) and (iii), we get the
independent of the path between initial state desired equation,
(reactants) and final state (products). In other
to

1
words, enthalpy change for a reaction is the C ( graphite,s) + O2 ( g ) → CO ( g ) ;
same whether it occurs in one step or in a 2
series of steps. This may be stated as follows
in the form of Hess’s Law. for which ∆r H V = ( −393.5 + 283.0)
t

If a reaction takes place in several steps = – 110.5 kJ mol–1

no

then its standard reaction enthalpy is the In general, if enthalpy of an overall reaction
sum of the standard enthalpies of the A→B along one route is ∆ rH and ∆r H1, ∆rH 2,
intermediate reactions into which the ∆rH 3..... representing enthalpies of reactions
overall reaction may be divided at the same
leading to same product, B along another
temperature.
route,then we have
Let us understand the importance of this
law with the help of an example. ∆rH = ∆ rH 1 + ∆ rH 2 + ∆r H3 ... (6.16)
 170 CHEMISTRY

It can be represented as: combustion, CO2(g) and H2O (1) are

∆rH produced and 3267.0 kJ of heat is
A B liberated. Calculate the standard enthalpy
of formation, ∆ f H 0 of benzene. Standard
∆rH1 ∆rH3 enthalpies of formation of CO 2(g) and
∆rH2 H 2O(l) are –393.5 kJ mol–1 and – 285.83
C D
kJ mol –1 respectively.

ed
6.5 ENTHALPIES FOR DIFFERENT TYPES Solution
OF REACTIONS The formation reaction of benezene is
It is convenient to give name to enthalpies given by :
specifying the types of reactions.
6C ( graphite ) + 3H2 ( g ) → C6 H6 ( l ) ;

h
(a) Standard enthalpy of combustion
(symbol : ∆ cH 0 ) ∆ f H V = ?... ( i )

pu T
is
Combustion reactions are exothermic in The enthalpy of combustion of 1 mol of
nature. These are important in industry, benzene is :
re ER
rocketry, and other walks of life. Standard

bl
15
enthalpy of combustion is defined as the C6H 6 ( l ) + O2 → 6CO2 ( g ) + 3H2 O ( l ) ;
enthalpy change per mole (or per unit amount) 2
of a substance, when it undergoes combustion ∆c H V = −3267 kJ mol -1... ( ii )
and all the reactants and products being in
The enthalpy of formation of 1 mol of
be C

their standard states at the specified

temperature. CO 2(g) :

Cooking gas in cylinders contains mostly C ( graphite ) + O2 ( g ) → CO2 ( g ) ;

N

butane (C4H10). During complete combustion ∆ f H V = −393.5 kJ mol-1... ( iii )

of one mole of butane, 2658 kJ of heat is
released. We can write the thermochemical The enthalpy of formation of 1 mol of
reactions for this as: H 2O(l) is :
©

13 1
C4 H10 (g) + O2 (g) → 4CO2 (g) + 5H2 O(1); H2 ( g ) + O 2 ( g ) → H 2O ( l ) ;
2 2
∆c H V = −2658.0 kJ mol −1 ∆ f H V = − 285.83 kJ mol -1... ( iv )
Similarly, combustion of glucose gives out multiplying eqn. (iii) by 6 and eqn. (iv)
2802.0 kJ/mol of heat, for which the overall by 3 we get:
equation is : 6C ( graphite ) + 6O 2 ( g ) → 6CO 2 ( g) ;
to

C6H12O6 (g) + 6O2 (g) → 6CO2 (g) + 6H2O(1); ∆ f H V = −2361 kJ mol -1

∆c H V = −2802.0 kJ mol −1
3
Our body also generates energy from food 3H 2 ( g ) + O 2 ( g) → 3H 2O ( l ) ;
t

by the same overall process as combustion, 2

no

although the final products are produced after ∆ f H V = − 857.49 kJ mol –1

a series of complex bio-chemical reactions
involving enzymes. Summing up the above two equations :
15
6C ( graphite ) + 3H2 ( g ) + O2 ( g ) → 6CO2 ( g )
Problem 6.8 2
The combustion of one mole of benzene + 3H2 O ( l) ;
takes place at 298 K and 1 atm. After
 THERMODYNAMICS 171

(i) Bond dissociation enthalpy

∆ f H V = −3218.49 kJ mol -1 ... (v )
(ii) Mean bond enthalpy
Reversing equation (ii); Let us discuss these terms with reference
15 to diatomic and polyatomic molecules.
6CO2 ( g ) + 3H 2O ( l ) → C 6H6 ( l ) +
O2 ; Diatomic Molecules: Consider the following
2
∆ f H V = 3267.0 kJ mol- 1 ... ( vi ) process in which the bonds in one mole of
dihydrogen gas (H2) are broken:

ed
Adding equations (v) and (vi), we get H2(g) → 2H(g) ; ∆H–HH 0 = 435.0 kJ mol–1
6C ( graphite ) + 3H2 ( g ) → C6 H6 ( l ) ; The enthalpy change involved in this process
is the bond dissociation enthalpy of H–H bond.
∆ f H V = 48.51 kJ mol -1 The bond dissociation enthalpy is the change
in enthalpy when one mole of covalent bonds

h
(b) Enthalpy of atomization of a gaseous covalent compound is broken to

pu T
(symbol: ∆aH 0 ) form products in the gas phase.

is
Consider the following example of atomization Note that it is the same as the enthalpy of
re ER
of dihydrogen atomization of dihydrogen. This is true for all
diatomic molecules. For example:
H2(g) → 2H(g); ∆aH 0 = 435.0 kJ mol–1

bl
You can see that H atoms are formed by Cl2(g) → 2Cl(g) ; ∆Cl–ClH 0 = 242 kJ mol–1
breaking H–H bonds in dihydrogen. The 0 –1
enthalpy change in this process is known as O2(g) → 2O(g) ; ∆O=OH = 428 kJ mol
enthalpy of atomization, ∆aH 0 . It is the In the case of polyatomic molecules, bond
be C

enthalpy change on breaking one mole of dissociation enthalpy is different for different
bonds completely to obtain atoms in the gas bonds within the same molecule.
phase. Polyatomic Molecules: Let us now consider
N

In case of diatomic molecules, like a polyatomic molecule like methane, CH4. The
dihydrogen (given above), the enthalpy of overall thermochemical equation for its
atomization is also the bond dissociation atomization reaction is given below:
enthalpy. The other examples of enthalpy of
©

CH4 (g) → C(g) + 4H(g);

atomization can be
CH4(g) → C(g) + 4H(g); ∆ aH 0 = 1665 kJ mol–1 ∆a H V = 1665 kJ mol −1
Note that the products are only atoms of C In methane, all the four C – H bonds are
and H in gaseous phase. Now see the following identical in bond length and energy. However,
reaction: the energies required to break the individual
Na(s) → Na(g) ; ∆aH 0 = 108.4 kJ mol–1 C – H bonds in each successive step differ :
to

In this case, the enthalpy of atomization is CH4(g) → CH3(g) + H(g); ∆bond HV = +427 kJ mol−1
same as the enthalpy of sublimation.
CH3(g) → CH2(g) + H(g); ∆bond H V = +439kJ mol− 1
(c) Bond Enthalpy (symbol: ∆ bondH 0)
CH2 (g) → CH(g) + H(g); ∆bo nd H V = +452 kJ mol −1
t

Chemical reactions involve the breaking and

making of chemical bonds. Energy is required
no

to break a bond and energy is released when CH(g) → C(g) + H(g);∆bond H V = +347 kJ mol −1
a bond is formed. It is possible to relate heat
Therefore,
of reaction to changes in energy associated
with breaking and making of chemical bonds. CH4(g) → C(g) + 4H(g); ∆a H V = 1665 kJ mol −1
With reference to the enthalpy changes
associated with chemical bonds, two different In such cases we use mean bond enthalpy
terms are used in thermodynamics. of C – H bond.
 172 CHEMISTRY

For example in CH4, ∆C–HH 0 is calculated as: products in gas phase reactions as:

∆C− H H V = ¼( ∆a H V ) = ¼ (1665 kJ mol −1 ) ∆r H V = ∑ bond enthalpiesreact ants

= 416 kJ mol–1 − ∑ bond enthalpies product s
We find that mean C–H bond enthalpy in (6.17)**
methane is 416 kJ/mol. It has been found that This relationship is particularly more
mean C–H bond enthalpies differ slightly from useful when the required values of ∆ f H0 are
compound to compound, as in not available. The net enthalpy change of a

ed
CH 3CH 2 Cl, CH 3 NO 2 , etc, but it does not differ reaction is the amount of energy required to
in a great deal*. Using Hess’s law, bond break all the bonds in the reactant molecules
enthalpies can be calculated. Bond enthalpy minus the amount of energy required to break
values of some single and multiple bonds are all the bonds in the product molecules.

h
given in Table 6.3. The reaction enthalpies are Remember that this relationship is
approximate and is valid when all substances

pu T
very important quantities as these arise from

is
the changes that accompany the breaking of (reactants and products) in the reaction are in
old bonds and formation of the new bonds. We gaseous state.
re ER
can predict enthalpy of a reaction in gas phase,
(d) Enthalpy of Solution (symbol : ∆solH 0 )

bl
if we know different bond enthalpies. The
standard enthalpy of reaction, ∆r H0 is related Enthalpy of solution of a substance is the
to bond enthalpies of the reactants and enthalpy change when one mole of it dissolves

–1
be C

Table 6.3(a) Some Mean Single Bond Enthalpies in kJ mol at 298 K

H C N O F Si P S Cl Br I
N

435.8 414 389 464 569 293 318 339 431 368 297 H
347 293 351 439 289 264 259 330 276 238 C
159 201 272 - 209 - 201 243 - N
138 184 368 351 - 205 - 201 O
©

155 540 490 327 255 197 - F

176 213 226 360 289 213 Si
213 230 331 272 213 P
213 251 213 - S
243 218 209 Cl
192 180 Br
151 I
to

–1
Table 6.3(b) Some Mean Multiple Bond Enthalpies in kJ mol at 298 K

N=N 418 C=C 611 O=O 498

N ≡N 946 C ≡ C 837
t

C=N 615 C=O 741

no

C ≡N 891 C ≡O 1070

* Note that symbol used for bond dissociation enthalpy and mean bond enthalpy is the same.
** If we use enthalpy of bond formation, (∆f H bond
0
), which is the enthalpy change when one mole of a particular type of
bond is formed from gaseous atom, then ∆ H = ∑ ∆ H − ∑∆ H
V V V
r f bonds of products f bonds of reactants
 THERMODYNAMICS 173

in a specified amount of solvent. The enthalpy

Na +Cl− ( s ) → Na + (g) + Cl − ( g ) ;
of solution at infinite dilution is the enthalpy
change observed on dissolving the substance ∆lat tice H V = +788 kJ mol −1
in an infinite amount of solvent when the
Since it is impossible to determine lattice
interactions between the ions (or solute
enthalpies directly by experiment, we use an
molecules) are negligible.
indirect method where we construct an
When an ionic compound dissolves in a enthalpy diagram called a Born-Haber Cycle
solvent, the ions leave their ordered positions (Fig. 6.9).

ed
on the crystal lattice. These are now more free in Let us now calculate the lattice enthalpy
solution. But solvation of these ions (hydration
of Na+ Cl–(s) by following steps given below :
in case solvent is water) also occurs at the same
time. This is shown diagrammatically, for an 1. Na(s) → Na(g) , sublimation of sodium

h
ionic compound, AB (s) metal, ∆sub H V = 108.4 kJ mol −1

pu T
2. Na(g) → Na +(g) + e −1 (g) , the ionization of

is
sodium atoms, ionization enthalpy
re ER ∆ iH 0 = 496 kJ mol–1

bl
1
3. Cl 2 (g) → Cl(g) , the dissociation of
2
chlorine, the reaction enthalpy is half the
be C

The enthalpy of solution of AB(s), ∆ solH 0, in

N

water is, therefore, determined by the selective

values of the lattice enthalpy,∆latticeH 0 and
enthalpy of hydration of ions, ∆hyd H0 as
©

∆sol HV = ∆lattice HV + ∆hyd H V

For most of the ionic compounds, ∆sol H0 is

positive and the dissociation process is
endothermic. Therefore the solubility of most
salts in water increases with rise of
temperature. If the lattice enthalpy is very
to

high, the dissolution of the compound may not

take place at all. Why do many fluorides tend
to be less soluble than the corresponding
chlorides? Estimates of the magnitudes of
enthalpy changes may be made by using tables
t

of bond energies (enthalpies) and lattice

no

energies (enthalpies).
Lattice Enthalpy
The lattice enthalpy of an ionic compound is
the enthalpy change which occurs when one
mole of an ionic compound dissociates into its Fig. 6.9 Enthalpy diagram for lattice enthalpy
ions in gaseous state. of NaCl
 174 CHEMISTRY

bond dissociation enthalpy. Internal energy is smaller by 2RT ( because

∆n g = 2) and is equal to + 783 kJ mol–1.
1
∆bon d H = 121kJ mol−1 .
V
Now we use the value of lattice enthalpy to
2
calculate enthalpy of solution from the
4. Cl(g) + e−1(g) → Cl(g) electron gained by expression:
chlorine atoms. The electron gain enthalpy,
∆sol H V = ∆la ttic eH V + ∆hyd H V
∆ egH 0 = –348.6 kJ mol–1 .
You have learnt about ionization enthalpy For one mole of NaCl(s),

ed
and electron gain enthalpy in Unit 3. In lattice enthalpy = + 788 kJ mol–1
0
fact, these terms have been taken from and ∆ hydH = – 784 kJ mol –1 ( from the
thermodynamics. Earlier terms, ionization literature)
energy and electron affinity were in practice ∆sol H 0 = + 788 kJ mol –1 – 784 kJ mol –1

h
in place of the above terms (see the box for = + 4 kJ mol–1
justification). The dissolution of NaCl(s) is accompanied

pu T
is
by very little heat change.
Ionization Energy and Electron Affinity
6.6 SPONTANEITY
re ER
Ionization energy and electron affinity are
The first law of thermodynamics tells us about

bl
defined at absolute zero. At any other
temperature, heat capacities for the the relationship between the heat absorbed
reactants and the products have to be and the work performed on or by a system. It
taken into account. Enthalpies of reactions puts no restrictions on the direction of heat
for flow. However, the flow of heat is unidirectional
be C

+ –
M(g) → M (g) + e (for ionization) from higher temperature to lower temperature.
– –
M(g) + e → M (g) (for electron gain) In fact, all naturally occurring processes
at temperature, T is whether chemical or physical will tend to
N

T
proceed spontaneously in one direction only.
For example, a gas expanding to fill the
∫∆C
V
0
∆r H (T ) = ∆rH (0) +
0
r p dT
0
available volume, burning carbon in dioxygen
The value of Cp for each species in the giving carbon dioxide.
©

above reaction is 5/2 R (CV = 3/2R) But heat will not flow from colder body to
0
So, ∆rCp = + 5/2 R (for ionization) warmer body on its own, the gas in a container
0
∆rCp = – 5/2 R (for electron gain) will not spontaneously contract into one corner
Therefore, or carbon dioxide will not form carbon and
0
∆r H (ionization enthalpy) dioxygen spontaneously. These and many
= E0 (ionization energy) + 5/2 RT other spontaneously occurring changes show
0
∆r H (electron gain enthalpy) unidirectional change. We may ask ‘what is the
to

= – A( electr on affinity) – 5/2 RT

driving force of spontaneously occurring
changes ? What determines the direction of a
5. Na + (g ) + Cl − (g) → Na + Cl − (s) spontaneous change ? In this section, we shall
The sequence of steps is shown in Fig. 6.9, establish some criterion for these processes
and is known as a Born-Haber cycle. The whether these will take place or not.
t

importance of the cycle is that, the sum of Let us first understand what do we mean
no

the enthalpy changes round a cycle is zero. by spontaneous reaction or change ? You may
Applying Hess’s law, we get, think by your common observation that
spontaneous reaction is one which occurs
∆lattice H V = 411.2 +108.4 + 121 + 496 − 348.6
immediately when contact is made between the
∆lattic e H V = +788 kJ reactants. Take the case of combination of
hydrogen and oxygen. These gases may be
for NaCl(s) → Na + (g) + Cl −(g) mixed at room temperature and left for many
 THERMODYNAMICS 175

years without observing any perceptible

change. Although the reaction is taking place
between them, it is at an extremely slow rate.
It is still called spontaneous reaction. So
spontaneity means ‘having the potential to
proceed without the assistance of external
agency’. However, it does not tell about the
rate of the reaction or process. Another aspect

ed
of spontaneous reaction or process, as we see
is that these cannot reverse their direction on
their own. We may summarise it as follows:
A spontaneous process is an
irreversible process and may only be

h
reversed by some external agency. Fig. 6.10 (a) Enthalpy diagram for exothermic

pu T
reactions
(a) Is decrease in enthalpy a criterion for

is
spontaneity ? C(graphite, s) + 2 S(l) → CS2(l);
re ER
If we examine the phenomenon like flow of
0
∆r H = +128.5 kJ mol–1

bl
water down hill or fall of a stone on to the These reactions though endothermic, are
ground, we find that there is a net decrease in spontaneous. The increase in enthalpy may be
potential energy in the direction of change. By represented on an enthalpy diagram as shown
analogy, we may be tempted to state that a in Fig. 6.10(b).
be C

chemical reaction is spontaneous in a given

direction, because decrease in energy has
taken place, as in the case of exothermic
reactions. For example:
N

1 3
N (g) + H2(g) = NH3(g) ;
2 2 2
∆r H0 = – 46.1 kJ mol–1
©

1 1
H2(g) + Cl2(g) = HCl (g) ;
2 2
∆r H 0 = – 92.32 kJ mol–1
1
H2(g) + O (g) → H 2O(l) ;
2 2
Fig. 6.10 (b) Enthalpy diagram for endothermic
∆r H0 = –285.8 kJ mol–1
to

reactions
The decrease in enthalpy in passing from
reactants to products may be shown for any Therefore, it becomes obvious that while
exothermic reaction on an enthalpy diagram decrease in enthalpy may be a contributory
as shown in Fig. 6.10(a). factor for spontaneity, but it is not true for all
t

Thus, the postulate that driving force for a cases.

no

chemical reaction may be due to decrease in (b) Entropy and spontaneity

energy sounds ‘reasonable’ as the basis of Then, what drives the spontaneous process in
evidence so far ! a given direction ? Let us examine such a case
Now let us examine the following reactions: in which ∆H = 0 i.e., there is no change in
1 enthalpy, but still the process is spontaneous.
N (g) + O2(g) → NO2(g); Let us consider diffusion of two gases into
2 2
∆r H 0 = +33.2 kJ mol–1 each other in a closed container which is
 176 CHEMISTRY

isolated from the surroundings as shown in At this point, we introduce another

Fig. 6.11. thermodynamic function, entropy denoted as
S . The above mentioned disorder is the
manifestation of entropy. To form a mental
picture, one can think of entropy as a measure
of the degree of randomness or disorder in the
system. The greater the disorder in an isolated
system, the higher is the entropy. As far as a

ed
chemical reaction is concerned, this entropy
change can be attributed to rearrangement of
atoms or ions from one pattern in the reactants
to another (in the products). If the structure
of the products is very much disordered than

h
that of the reactants, there will be a resultant

pu T
increase in entropy. The change in entropy

is
accompanying a chemical reaction may be
estimated qualitatively by a consideration of
re ER the structures of the species taking part in the

bl
reaction. Decrease of regularity in structure
would mean increase in entropy. For a given
substance, the crystalline solid state is the
state of lowest entropy (most ordered), The
be C

gaseous state is state of highest entropy.

Fig. 6.11 Diffusion of two gases Now let us try to quantify entropy. One way
to calculate the degree of disorder or chaotic
N

The two gases, say, gas A and gas B are distribution of energy among molecules would
represented by black dots and white dots be through statistical method which is beyond
respectively and separated by a movable the scope of this treatment. Other way would
partition [Fig. 6.11 (a)]. When the partition is be to relate this process to the heat involved in
withdrawn [Fig.6.11( b)], the gases begin to
©

a process which would make entropy a

diffuse into each other and after a period of thermodynamic concept. Entropy, like any
time, diffusion will be complete. other thermodynamic property such as
Let us examine the process. Before internal energy U and enthalpy H is a state
partition, if we were to pick up the gas function and ∆S is independent of path.
molecules from left container, we would be Whenever heat is added to the system, it
sure that these will be molecules of gas A and increases molecular motions causing
to

similarly if we were to pick up the gas increased randomness in the system. Thus
molecules from right container, we would be heat (q) has randomising influence on the
sure that these will be molecules of gas B. But, system. Can we then equate ∆S with q ? Wait !
if we were to pick up molecules from container Experience suggests us that the distribution
when partition is removed, we are not sure of heat also depends on the temperature at
t

whether the molecules picked are of gas A or which heat is added to the system. A system
no

gas B. We say that the system has become less at higher temperature has greater randomness
predictable or more chaotic. in it than one at lower temperature. Thus,
We may now formulate another postulate: temperature is the measure of average
in an isolated system, there is always a chaotic motion of particles in the system.
tendency for the systems’ energy to become Heat added to a system at lower temperature
more disordered or chaotic and this could be causes greater randomness than when the
a criterion for spontaneous change ! same quantity of heat is added to it at higher
 THERMODYNAMICS 177

temperature. This suggests that the entropy (ii) At 0 K, the contituent particles are
change is inversely proportional to the static and entropy is minimum. If
temperature. ∆S is related with q and T for a temperature is raised to 115 K, these
reversible reaction as : begin to move and oscillate about
qrev their equilibrium positions in the
∆S = (6.18) lattice and system becomes more
T
The total entropy change ( ∆Stotal) for the disordered, therefore entropy
system and surroundings of a spontaneous increases.

ed
process is given by (iii) Reactant, NaHCO3 is a solid and it
has low entropy. Among products
∆Stot al = ∆S syst em + ∆S surr > 0 (6.19) there are one solid and two gases.
When a system is in equilibrium, the Therefore, the products represent a

h
entropy is maximum, and the change in condition of higher entropy.
entropy, ∆S = 0. (iv) Here one molecule gives two atoms

pu T
is
We can say that entropy for a spontaneous i.e., number of particles increases
process increases till it reaches maximum and leading to more disordered state.
re ER
at equilibrium the change in entropy is zero. Two moles of H atoms have higher

bl
Since entropy is a state property, we can entropy than one mole of dihydrogen
calculate the change in entropy of a reversible molecule.
process by Problem 6.10
q sys ,rev For oxidation of iron,
∆Ssys =
be C

T 4Fe ( s ) + 3O2 ( g ) → 2Fe2O3 ( s )

We find that both for reversible and entropy change is – 549.4 JK–1mol–1at
N

irreversible expansion for an ideal gas, under 298 K. Inspite of negative entropy change
isothermal conditions, ∆U = 0, but ∆Stotal i.e., of this reaction, why is the reaction
∆Ssys + ∆Ssurr is not zero for irreversible spontaneous?
process. Thus, ∆U does not discriminate (∆ r H 0for this reaction is
©

between reversible and irreversible process, –1648 × 103 J mol –1)

whereas ∆S does. Solution
One decides the spontaneity of a reaction
Problem 6.9
by considering
Predict in which of the following, entropy
increases/decreases : ∆St otal ( ∆S sys + ∆S surr ) . For calculating
(i) A liquid crystallizes into a solid. ∆S surr, we have to consider the heat
to

(ii) Temperature of a crystalline solid is absorbed by the surroundings which is

0
raised from 0 K to 115 K. equal to – ∆ rH . At temperature T, entropy
change of the surroundings is
(iii ) 2NaHCO3 ( s ) → Na 2CO3 ( s ) +
∆r H V
(at constant pressure)
t

CO2 ( g ) + H2 O ( g ) ∆Ssurr = −
T
no

(iv) H2 ( g ) → 2H ( g ) =−
( −1648×10 3
J mol −1 )

Solution 298 K
(i) After freezing, the molecules attain an = 5530 JK −1mol −1
ordered state and therefore, entropy Thus, total entropy change for this
decreases. reaction
 178 CHEMISTRY

∆r S tota l = 5530 JK −1mol−1 + ( − 549.4 JK−1 mol−1 ) Now let us consider how ∆G is related to
reaction spontaneity.
= 4980.6 JK −1mol −1 We know,
This shows that the above reaction is ∆Stotal = ∆Ssys + ∆Ssurr
spontaneous.
If the system is in thermal equilibrium with
the surrounding, then the temperature of the
(c) Gibbs energy and spontaneity surrounding is same as that of the system.

ed
We have seen that for a system, it is the total Also, increase in enthalpy of the surrounding
entropy change, ∆S total which decides the is equal to decrease in the enthalpy of the
spontaneity of the process. But most of the system.
chemical reactions fall into the category of Therefore, entropy change of
either closed systems or open systems. surroundings,

h
Therefore, for most of the chemical reactions
∆Hsurr ∆H sys

pu T
there are changes in both enthalpy and
∆Ssurr = =−

is
entropy. It is clear from the discussion in T T
previous sections that neither decrease in
re ER
enthalpy nor increase in entropy alone can  ∆Hsys 
∆Stot al = ∆Ssys +  − 

bl
determine the direction of spontaneous change  T 
for these systems. Rearranging the above equation:
For this purpose, we define a new T∆Stotal = T∆Ssys – ∆Hsys
thermodynamic function the Gibbs energy or
For spontaneous process, ∆ S tot al > 0 , so
be C

Gibbs function, G, as
G = H – TS (6.20) T∆Ssys – ∆Hsys > 0
Gibbs function, G is an extensive property
⇒ − ( ∆H sys − T ∆Ssys ) > 0
N

and a state function.

The change in Gibbs energy for the system, Using equation 6.21, the above equation can
∆Gsys can be written as be written as
−∆G > 0
©

∆G sys = ∆H sy s − T ∆Ssys − S sy s∆T

At constant temperature, ∆T = 0 ∆G = ∆H − T ∆S < 0 (6.22)

∆Hsys is the enthalpy change of a reaction,
∴ ∆G sys = ∆Hsys − T ∆S sys
T∆Ssys is the energy which is not available to
Usually the subscript ‘system’ is dropped do useful work. So ∆G is the net energy
and we simply write this equation as available to do useful work and is thus a
measure of the ‘free energy’. For this reason, it
to

∆G = ∆H − T ∆S (6.21)
is also known as the free energy of the reaction.
Thus, Gibbs energy change = enthalpy
∆G gives a criteria for spontaneity at
change – temperature × entropy change, and
constant pressure and temperature.
is referred to as the Gibbs equation, one of the
t

most important equations in chemistry. Here, (i) If ∆G is negative (< 0), the process is
no

we have considered both terms together for spontaneous.

spontaneity: energy (in terms of ∆H) and (ii) If ∆G is positive (> 0), the process is non
entropy (∆S , a measure of disorder) as spontaneous.
indicated earlier. Dimensionally if we analyse, Note : If a reaction has a positive enthalpy
we find that ∆G has units of energy because, change and positive entropy change, it can be
both ∆H and the T ∆S are energy terms, since spontaneous when T∆S is large enough to
T∆S = (K) (J/K) = J. outweigh ∆H. This can happen in two ways;
 THERMODYNAMICS 179

(a) The positive entropy change of the system at equilibrium the free energy of the system is
can be ‘small’ in which case T must be large. minimum. If it is not, the system would
(b) The positive entropy change of the system spontaneously change to configuration of
can be ’large’, in which case T may be small. lower free energy.
The former is one of the reasons why reactions So, the criterion for equilibrium
are often carried out at high temperature.
A + B ⇌ C + D ; is
Table 6.4 summarises the effect of temperature
on spontaneity of reactions. ∆r G = 0

ed
Gibbs energy for a reaction in which all
6.7 GIBBS ENERGY CHANGE AND reactants and products are in standard state,
EQUILIBRIUM 0
∆r G is related to the equilibrium constant of
We have seen how a knowledge of the sign and the reaction as follows:
magnitude of the free energy change of a 0 = ∆r G0 + RT ln K

h
chemical reaction allows: 0
or ∆r G = – RT ln K

pu T
(i) Prediction of the spontaneity of the
∆r G0 = – 2.303 RT log K

is
chemical reaction. or (6.23)
We also know that
re ER
(ii) Prediction of the useful work that could be
extracted from it. ∆r G V = ∆r H V − T ∆r S V = − RT ln K (6.24)

bl
So far we have considered free energy For strongly endothermic reactions, the
changes in irreversible reactions. Let us now value of ∆r H0 may be large and positive. In
examine the free energy changes in reversible such a case, value of K will be much smaller
reactions. than 1 and the reaction is unlikely to form
be C

‘Reversible’ under strict thermodynamic much product. In case of exothermic reactions,

sense is a special way of carrying out a process ∆r H0 is large and negative, and ∆r G0 is likely to
such that system is at all times in perfect be large and negative too. In such cases, K will
N

equilibrium with its surroundings. When be much larger than 1. We may expect strongly
applied to a chemical reaction, the term exothermic reactions to have a large K, and
0
‘reversible’ indicates that a given reaction can hence can go to near completion. ∆ rG also
proceed in either direction simultaneously, so depends upon ∆r S , if the changes in the
0
©

that a dynamic equilibrium is set up. This entropy of reaction is also taken into account,
means that the reactions in both the directions the value of K or extent of chemical reaction
should proceed with a decrease in free energy, will also be affected, depending upon whether
0
which seems impossible. It is possible only if ∆r S is positive or negative.

Table 6.4 Effect of Temperature on Spontaneity of Reactions

to

0 0 0
∆r H ∆ rS ∆rG Description*

– + – Reaction spontaneous at all temperature

– – – (at low T ) Reaction spontaneous at low temperature
t

– – + (at high T ) Reaction nonspontaneous at high temperature

no

+ + + (at low T ) Reaction nonspontaneous at low temperature

+ + – (at high T ) Reaction spontaneous at high temperature
+ – + (at all T ) Reaction nonspontaneous at all temperatures

* The term low temperature and high temperature are relative. For a particular reaction, high temperature could even
mean room temperature.
 180 CHEMISTRY

Using equation (6.24),

(i) It is possible to obtain an estimate of ∆GV ( –13.6 × 10 3
J mol –1 )
=
from the measurement of ∆H V and ∆SV, 2.303 ( 8.314 JK (298 K )
–1
mol –1 )
and then calculate K at any temperature = 2.38
for economic yields of the products. Hence K = antilog 2.38 = 2.4 × 102.
(ii) If K is measured directly in the laboratory, Problem 6.13
value of ∆G0 at any other temperature can
be calculated. At 60°C, dinitrogen tetroxide is fifty

ed
percent dissociated. Calculate the
Problem 6.11 standard free energy change at this
0
Calculate ∆ rG for conversion of oxygen temperature and at one atmosphere.
to ozone, 3/2 O2(g) → O3(g) at 298 K. if Kp Solution

h
for this conversion is 2.47 × 10 –29 . N2O 4(g) 
↽⇀ 2NO2(g)
Solution

pu T
If N2O 4 is 50% dissociated, the mole

is
We know ∆ rG 0 = – 2.303 RT log Kp and fraction of both the substances is given
R = 8.314 JK –1 mol–1 by
re ER
Therefore, ∆ rG 0 = 1 − 0.5 2 × 0.5
xN O =

bl
– 2.303 (8.314 J K–1 mol–1) ; x NO2 =
2 4
1 + 0.5 1 + 0.5
× (298 K) (log 2.47 × 10 –29)
= 163000 J mol–1 pN 0.5
2 O4
=× 1 atm, p NO2 =
= 163 kJ mol–1. 1.5
be C

Problem 6.12 1
× 1 atm.
Find out the value of equilibrium constant 1.5
The equilibrium constant K p is given by
N

for the following reaction at 298 K.

(p )
2
2NH3 ( g ) + CO2 ( g ) ⇌ NH2CONH2 ( aq ) . NO2 1.5
Kp = =
+ H2 O ( l ) p N2O 4 (1.5)2 (0.5)
©

Standard Gibbs energy change, ∆rG 0 at = 1.33 atm.

the given temperature is –13.6 kJ mol –1. Since
∆r G0 = –RT ln Kp
Solution
∆r G0 = (– 8.314 JK–1 mol–1) × (333 K)
– ∆ rG V
We know, log K = × (2.303) × (0.1239)
2.303 RT = – 763.8 kJ mol–1
t to
no
 THERMODYNAMICS 181

SUMMARY

Thermodynamics deals with energy changes in chemical or physical processes and

enables us to study these changes quantitatively and to make useful predictions. For
these purposes, we divide the universe into the system and the surroundings. Chemical
or physical processes lead to evolution or absorption of heat (q), part of which may be
converted into work (w). These quantities are related through the first law of
thermodynamics via ∆U = q + w. ∆U, change in internal energy, depends on initial and

ed
final states only and is a state function, whereas q and w depend on the path and are
not the state functions. We follow sign conventions of q and w by giving the positive sign
to these quantities when these are added to the system. We can measur e the transfer of
heat from one system to another which causes the change in temperature. The magnitude
of rise in temperature depends on the heat capacity (C) of a substance. Therefore, heat

h
absorbed or evolved is q = C∆T. Work can be measured by w = –p ex∆V, in case of expansion
of gases. Under reversible process, we can put pex = p for infinitesimal changes in the

pu T
volume making wrev = – p dV. In this condition, we can use gas equation, pV = nRT.

is
At constant volume, w = 0, then ∆U = q V , heat transfer at constant volume. But in
re ER
study of chemical reactions, we usually have constant pressure. We define another
state function enthalpy. Enthalpy change, ∆H = ∆U + ∆ngRT, can be found directly from

bl
the heat changes at constant pressure, ∆H = q p.
There are varieties of enthalpy changes. Changes of phase such as melting,
vaporization and sublimation usually occur at constant temperature and can be
characterized by enthalpy changes which are always positive. Enthalpy of formation,
be C

combustion and other enthalpy changes can be calculated using Hess’s law. Enthalpy
change for chemical reactions can be determined by

∆r H = ∑ (a i ∆ f H pr oducts ) − ∑ (b i ∆ f H reactio ns )
N

f i

and in gaseous state by

∆rH
0
= Σ bond enthalpies of the reactants – Σ bond enthalpies of the products
First law of thermodynamics does not guide us about the direction of chemical
©

reactions i.e., what is the driving force of a chemical reaction. For isolated systems,
∆U = 0. We define another state function, S, entropy for this purpose. Entr opy is a
measure of disorder or randomness. For a spontaneous change, total entropy change is
positive. Therefore, for an isolated system, ∆U = 0, ∆S > 0, so entropy change distinguishes
a spontaneous change, while energy change does not. Entropy changes can be measured
q rev q rev
by the equation ∆S = for a reversible process. is independent of path.
T T
to

Chemical reactions are generally carried at constant pressure, so we define another

state function Gibbs energy , G, which is related to entropy and enthalpy changes of
the system by the equation:
∆rG = ∆rH – T ∆rS
t

For a spontaneous change, ∆Gsys < 0 and at equilibrium, ∆Gsys = 0.

no

Standard Gibbs energy change is related to equilibrium constant by

∆rG 0 = – RT ln K.
0
K can be calculated from this equation, if we know ∆r G which can be found from
∆rG V = ∆r H V − T ∆r S V . Temperature is an important factor in the equation. Many reactions
which are non-spontaneous at low temperature, are made spontaneous at high
temperature for systems having positive entropy of reaction.
 182 CHEMISTRY

EXERCISES

6.1 Choose the corr ect answer. A thermodynamic state function is a

quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work

ed
(iv) whose value depends on temperature only.
6.2 For the process to occur under adiabatic conditions, the correct
condition is:
(i) ∆T = 0

h
(ii) ∆p = 0

pu T
(iii) q = 0

is
(iv) w = 0
re ER
6.3 The enthalpies of all elements in their standard states are:
(i) unity

bl
(ii) zero
(iii) < 0
(iv) different for each element
0 0
6.4 ∆U of combustion of methane is – X kJ mol–1. The value of ∆H is
be C

0
(i) = ∆U
(ii) > ∆U 0
N

0
(iii) < ∆U
(iv) = 0
6.5 The enthalpy of combustion of methane, graphite and dihydrogen
at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol –1, and –285.8 kJ mol–1
©

respectively. Enthalpy of formation of CH 4(g) will be

(i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1
(iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1.
6.6 A reaction, A + B → C + D + q is found to have a positive entropy
change. The reaction will be
(i) possible at high temperature
to

(ii) possible only at low temperature

(iii) not possible at any temperature
(v) possible at any temperature
6.7 In a process, 701 J of heat is absorbed by a system and 394 J of
t

work is done by the system. What is the change in internal energy

no

for the process?

6.8 The reaction of cyanamide, NH 2CN (s), with dioxygen was carried
out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol–1
at 298 K. Calculate enthalpy change for the reaction at 298 K.

3
NH2 CN(g) + O (g) → N2 (g) + CO2 (g) + H2O(l)
2 2
 THERMODYNAMICS 183

6.9 Calculate the number of kJ of heat necessary to raise the temperature

of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al
is 24 J mol–1 K–1.
6.10 Calculate the enthalpy change on freezing of 1.0 mol of water
at10.0°C to ice at –10.0°C. ∆fus H = 6.03 kJ mol–1 at 0°C.
Cp [H2O(l)] = 75.3 J mol–1 K–1
Cp [H2O(s)] = 36.8 J mol–1 K–1
6.11 Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate

ed
the heat released upon formation of 35.2 g of CO2 from carbon and
dioxygen gas.
6.12 Enthalpies of formation of CO(g), CO2 (g), N2 O(g) and N2O 4(g) are –110,
– 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ∆r H for the

h
reaction:
N2 O4 (g) + 3CO(g) → N2 O(g) + 3CO2(g)

pu T
6.13 Given

is
N2 (g) + 3H2(g) → 2NH3(g) ; ∆rH 0 = –92.4 kJ mol–1
re ER What is the standard enthalpy of formation of NH3 gas?

bl
6.14 Calculate the standard enthalpy of formation of CH3 OH(l) from the
following data:

3 0
CH3OH (l) + O (g) → CO2 (g) + 2H2O(l) ; ∆r H = –726 kJ mol–1
2 2
be C

0
C(graphite) + O2 (g) → CO2(g) ; ∆c H = –393 kJ mol–1

1 0
O (g) → H2 O(l) ; ∆f H = –286 kJ mol–1.
N

H 2(g) +
2 2
6.15 Calculate the enthalpy change for the process
CCl4 (g) → C(g) + 4 Cl(g)
and calculate bond enthalpy of C – Cl in CCl4(g).
©

0
∆vapH (CCl4 ) = 30.5 kJ mol–1.
∆fH 0 (CCl4 ) = –135.5 kJ mol–1.
0 0
∆aH (C) = 715.0 kJ mol–1 , where ∆aH is enthalpy of atomisation
0
∆aH (Cl2 ) = 242 kJ mol–1
6.16 For an isolated system, ∆U = 0, what will be ∆S ?
to

6.17 For the reaction at 298 K,

2A + B → C
∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1
At what temperature will the reaction become spontaneous
t

considering ∆H and ∆S to be constant over the temperature range.

no

6.18 For the reaction,

2 Cl(g) → Cl2(g), what are the signs of ∆H and ∆S ?
6.19 For the reaction
2 A(g) + B(g) → 2D(g)
0 0
∆U = –10.5 kJ and ∆S = –44.1 JK–1.
0
Calculate ∆G for the reaction, and predict whether the reaction
may occur spontaneously.
 184 CHEMISTRY

6.20 The equilibrium constant for a reaction is 10. What will be the value
0
of ∆G ? R = 8.314 JK–1 mol–1, T = 300 K.
6.21 Comment on the thermodynamic stability of NO(g), given

1 1 0
N 2(g) + O (g) → NO(g) ; ∆rH = 90 kJ mol–1
2 2 2

1 0
NO(g) + O (g) → NO2 (g) : ∆rH = –74 kJ mol–1

ed
2 2
6.22 Calculate the entropy change in surroundings when 1.00 mol of
0
H 2O(l) is formed under standard conditions. ∆f H = –286 kJ mol–1.

h
pu T
is
re ER
bl
be C
N
t to ©
no
 EQUILIBRIUM 185

UNIT 7

EQUILIBRIUM

h ed
Chemical equilibria are important in numerous biological

pu T
and environmental processes. For example, equilibria

is
involving O2 molecules and the protein hemoglobin play a

able to
re ER
After studying this unit you will be crucial role in the transport and delivery of O2 from our
lungs to our muscles. Similar equilibria involving CO

bl
• ide ntify dynamic nature of molecules and hemoglobin account for the toxicity of CO.
equilibrium involved in physical
and chemical processes; When a liquid evaporates in a closed container,
• state the law of equilibrium; molecules with relatively higher kinetic energy escape the
• explain characteristics of
be C

liquid surface into the vapour phase and number of liquid

equilibria involved in physical
molecules from the vapour phase strike the liquid surface
and chemical processes;
• write expressions for and are retained in the liquid phase. It gives rise to a constant
N

equilibrium constants; vapour pressure because of an equilibrium in which the

• establish a relationship between number of molecules leaving the liquid equals the number
Kp and K c; returning to liquid from the vapour. We say that the system
• explain various factors that
affect the equilibrium state of a has reached equilibrium state at this stage. However, this
©

reaction; is not static equilibrium and there is a lot of activity at the

• classify substances as acids or boundary between the liquid and the vapour. Thus, at
bases according to Arrhenius, equilibrium, the rate of evaporation is equal to the rate of
Bronsted-Lowry and Lewis
concepts;
condensation. It may be represented by
• classify acids and bases as weak H2O (l) ⇌ H2O (vap)
or strong in terms of their
ionization constants;
The double half arrows indicate that the processes in
• explain the dependence of both the directions are going on simultaneously. The mixture
to

degree of ionization on of reactants and products in the equilibrium state is called

concentration of the electrolyte an equilibrium mixture.
and that of the common ion;
• describe pH scale for Equilibrium can be established for both physical
representing hydrogen ion processes and chemical reactions. The reaction may be fast
t

concentration; or slow depending on the experimental conditions and the

• explain ionisation of water and
no

nature of the reactants. When the reactants in a closed vessel

its duel role as acid and base;
• describe ionic product (Kw ) and at a particular temperature react to give products, the
pK w for water; concentrations of the reactants keep on decreasing, while
• appreciate use of buffer those of products keep on increasing for some time after
solutions; which there is no change in the concentrations of either of
• calculate solubility product
constant.
the reactants or products. This stage of the system is the
dynamic equilibrium and the rates of the forward and
186 CHEMISTRY

reverse reactions become equal. It is due to and the atmospheric pressure are in
this dynamic equilibrium stage that there is equilibrium state and the system shows
no change in the concentrations of various interesting characteristic features. We observe
species in the reaction mixture. Based on the that the mass of ice and water do not change
extent to which the reactions proceed to reach with time and the temperature remains
the state of chemical equilibrium, these may constant. However, the equilibrium is not
be classified in three groups. static. The intense activity can be noticed at
(i) The reactions that proceed nearly to the boundary between ice and water.

ed
completion and only negligible Molecules from the liquid water collide against
concentrations of the reactants are left. In ice and adhere to it and some molecules of ice
some cases, it may not be even possible to escape into liquid phase. There is no change
detect these experimentally. of mass of ice and water, as the rates of transfer
of molecules from ice into water and of reverse

h
(ii) The reactions in which only small amounts
transfer from water into ice are equal at
of products are formed and most of the

pu T
atmospheric pressure and 273 K.
reactants remain unchanged at

is
equilibrium stage. It is obvious that ice and water are in
re ER
(iii) The reactions in which the concentrations
equilibrium only at particular temperature
and pressure. For any pure substance at

bl
of the reactants and products are
comparable, when the system is in atmospheric pressure, the temperature at
which the solid and liquid phases are at
equilibrium.
equilibrium is called the normal melting point
The extent of a reaction in equilibrium or normal freezing point of the substance.
be C

varies with the experimental conditions such The system here is in dynamic equilibrium and
as concentrations of reactants, temperature, we can infer the following:
etc. Optimisation of the operational conditions
(i) Both the opposing processes occur
N

is very important in industry and laboratory

so that equilibrium is favorable in the simultaneously.
direction of the desired product. Some (ii) Both the processes occur at the same rate
important aspects of equilibrium involving so that the amount of ice and water
physical and chemical processes are dealt in remains constant.
©

this unit along with the equilibrium involving 7.1.2 Liquid-Vapour Equilibrium
ions in aqueous solutions which is called as
ionic equilibrium. This equilibrium can be better understood if
we consider the example of a transparent box
7.1 EQUILIBRIUM IN PHYSICAL carrying a U-tube with mercury (manometer).
PROCESSES Drying agent like anhydrous calcium chloride
The characteristics of system at equilibrium (or phosphorus penta-oxide) is placed for a
to

are better understood if we examine some few hours in the box. After removing the
physical processes. The most familiar drying agent by tilting the box on one side, a
examples are phase transformation watch glass (or petri dish) containing water is
processes, e.g., quickly placed inside the box. It will be
observed that the mercury level in the right
t

solid ⇌ liquid limb of the manometer slowly increases and

no

liquid ⇌ gas finally attains a constant value, that is, the

solid ⇌ gas pressure inside the box increases and reaches
a constant value. Also the volume of water in
7.1.1 Solid-Liquid Equilibrium the watch glass decreases (Fig. 7.1). Initially
Ice and water kept in a perfectly insulated there was no water vapour (or very less) inside
thermos flask (no exchange of heat between the box. As water evaporated the pressure in
its contents and the surroundings) at 273K the box increased due to addition of water
 EQUILIBRIUM 187

ed
Fig.7.1 Measuring equilibrium vapour pressure of water at a constant temperature

h
molecules into the gaseous phase inside the dispersed into large volume of the room. As a
box. The rate of evaporation is constant.

pu T
consequence the rate of condensation from

is
However, the rate of increase in pressure vapour to liquid state is much less than the
decreases with time due to condensation of rate of evaporation. These are open systems
re ER
vapour into water. Finally it leads to an and it is not possible to reach equilibrium in

bl
equilibrium condition when there is no net an open system.
evaporation. This implies that the number of Water and water vapour are in equilibrium
water molecules from the gaseous state into position at atmospheric pressure (1.013 bar)
the liquid state also increases till the and at 100°C in a closed vessel. The boiling
equilibrium is attained i.e.,
be C

point of water is 100°C at 1.013 bar pressure.

rate of evaporation= rate of condensation For any pure liquid at one atmospheric
H2O(l) ⇌ H 2O (vap) pressure (1.013 bar), the temperature at
N

which the liquid and vapours are at

At equilibrium the pressure exerted by the
equilibrium is called normal boiling point of
water molecules at a given temperature
the liquid. Boiling point of the liquid depends
remains constant and is called the equilibrium
on the atmospheric pressure. It depends on
vapour pressure of water (or just vapour
©

pressure of water); vapour pressure of water the altitude of the place; at high altitude the
boiling point decreases.
increases with temperature. If the above
experiment is repeated with methyl alcohol, 7.1.3 Solid – Vapour Equilibrium
acetone and ether, it is observed that different Let us now consider the systems where solids
liquids have different equilibrium vapour sublime to vapour phase. If we place solid iodine
pressures at the same temperature, and the in a closed vessel, after sometime the vessel gets
liquid which has a higher vapour pressure is filled up with violet vapour and the intensity of
to

more volatile and has a lower boiling point. colour increases with time. After certain time the
If we expose three watch glasses intensity of colour becomes constant and at this
containing separately 1mL each of acetone, stage equilibrium is attained. Hence solid iodine
ethyl alcohol, and water to atmosphere and sublimes to give iodine vapour and the iodine
t

repeat the experiment with different volumes vapour condenses to give solid iodine. The
of the liquids in a warmer room, it is observed
no

equilibrium can be represented as,

that in all such cases the liquid eventually
I2(solid) ⇌ I2 (vapour)
disappears and the time taken for complete
evaporation depends on (i) the nature of the Other examples showing this kind of
liquid, (ii) the amount of the liquid and (iii) the equilibrium are,
temperature. When the watch glass is open to Camphor (solid) ⇌ Camphor (vapour)
the atmosphere, the rate of evaporation
remains constant but the molecules are NH4Cl (solid) ⇌ NH4Cl (vapour)
188 CHEMISTRY

7.1.4 Equilibrium Involving Dissolution pressure of the gas above the solvent. This
of Solid or Gases in Liquids amount decreases with increase of
Solids in liquids temperature. The soda water bottle is sealed
under pressure of gas when its solubility in
We know from our experience that we can
water is high. As soon as the bottle is opened,
dissolve only a limited amount of salt or sugar
some of the dissolved carbon dioxide gas
in a given amount of water at room
escapes to reach a new equilibrium condition
temperature. If we make a thick sugar syrup
required for the lower pressure, namely its
solution by dissolving sugar at a higher

ed
partial pressure in the atmosphere. This is how
temperature, sugar crystals separate out if we
the soda water in bottle when left open to the
cool the syrup to the room temperature. We
air for some time, turns ‘flat’. It can be
call it a saturated solution when no more of
generalised that:
solute can be dissolved in it at a given
(i) For solid ⇌ liquid equilibrium, there is

h
temperature. The concentration of the solute
in a saturated solution depends upon the only one temperature (melting point) at

pu T
temperature. In a saturated solution, a 1 atm (1.013 bar) at which the two phases

is
dynamic equilibrium exits between the solute can coexist. If there is no exchange of heat
re ER
molecules in the solid state and in the solution: with the surroundings, the mass of the two
phases remains constant.
Sugar (solution) ⇌ Sugar (solid), and

bl
the rate of dissolution of sugar = rate of (ii) For liquid ⇌ vapour equilibrium, the
crystallisation of sugar. vapour pressure is constant at a given
Equality of the two rates and dynamic temperature.
(iii) For dissolution of solids in liquids, the
be C

nature of equilibrium has been confirmed with

the help of radioactive sugar. If we drop some solubility is constant at a given
radioactive sugar into saturated solution of temperature.
(iv) For dissolution of gases in liquids, the
N

non-radioactive sugar, then after some time

radioactivity is observed both in the solution concentration of a gas in liquid is
and in the solid sugar. Initially there were no proportional to the pressure
radioactive sugar molecules in the solution (concentration) of the gas over the liquid.
but due to dynamic nature of equilibrium, These observations are summarised in
©

there is exchange between the radioactive and Table 7.1

non-radioactive sugar molecules between the Table 7.1 Some Features of Physical
two phases. The ratio of the radioactive to non- Equilibria
radioactive molecules in the solution increases
Process Conclusion
till it attains a constant value.
Gases in liquids Liquid ⇌ Vapour p H2 O constant at given
to

When a soda water bottle is opened, some of H 2O (l) ⇌ H 2O (g) temperature

the carbon dioxide gas dissolved in it fizzes Solid ⇌ Liquid Melting point is fixed at
out rapidly. The phenomenon arises due to H 2O (s) ⇌ H 2O (l) constant pressure
difference in solubility of carbon dioxide at Solute(s) ⇌ Solute Concentration of solute
different pressures. There is equilibrium (solution) in solution is constant
t

between the molecules in the gaseous state Sugar(s) ⇌ Sugar at a given temperature
no

and the molecules dissolved in the liquid (solution)

under pressure i.e.,
Gas(g) ⇌ Gas (aq) [gas(aq)]/[gas(g)] is
CO2(gas) ⇌ CO2(in solution) constant at a given
This equilibrium is governed by Henry’s temperature
law, which states that the mass of a gas CO2(g) ⇌ CO2 (aq) [CO 2 (aq)]/[CO 2 (g)] is
dissolved in a given mass of a solvent at constant at a given
temperature
any temperature is proportional to the
 EQUILIBRIUM 189

7.1.5 General Characteristics of Equilibria

Involving Physical Processes
For the physical processes discussed above,
following characteristics are common to the
system at equilibrium:
(i) Equilibrium is possible only in a closed
system at a given temperature.

ed
(ii) Both the opposing processes occur at the
same rate and there is a dynamic but
stable condition.
(iii) All measurable properties of the system
remain constant.

h
(iv) When equilibrium is attained for a physical

pu T
process, it is characterised by constant

is
Fig. 7.2 Attainment of chemical equilibrium.
value of one of its parameters at a given
re ER
temperature. Table 7.1 lists such
quantities. same rate and the system reaches a state of

bl
(v) The magnitude of such quantities at any equilibrium.
stage indicates the extent to which the Similarly, the reaction can reach the state
physical process has proceeded before of equilibrium even if we start with only C and
reaching equilibrium. D; that is, no A and B being present initially,
be C

7.2 EQUILIBRIUM IN CHEMICAL as the equilibrium can be reached from either

PROCESSES – DYNAMIC direction.
EQUILIBRIUM The dynamic nature of chemical
N

Analogous to the physical systems chemical equilibrium can be demonstrated in the

reactions also attain a state of equilibrium. synthesis of ammonia by Haber’s process. In
These reactions can occur both in forward and a series of experiments, Haber started with
known amounts of dinitrogen and dihydrogen
backward directions. When the rates of the
©

maintained at high temperature and pressure

forward and reverse reactions become equal,
and at regular intervals determined the
the concentrations of the reactants and the
amount of ammonia present. He was
products remain constant. This is the stage of
successful in determining also the
chemical equilibrium. This equilibrium is concentration of unreacted dihydrogen and
dynamic in nature as it consists of a forward dinitrogen. Fig. 7.4 (page 191) shows that after
reaction in which the reactants give product(s) a certain time the composition of the mixture
to

and reverse reaction in which product(s) gives remains the same even though some of the
the original reactants. reactants are still present. This constancy in
For a better comprehension, let us composition indicates that the reaction has
consider a general case of a reversible reaction, reached equilibrium. In order to understand
the dynamic nature of the reaction, synthesis
t

A+B ⇌ C+D
of ammonia is carried out with exactly the
no

With passage of time, there is

same starting conditions (of partial pressure
accumulation of the products C and D and and temperature) but using D2 (deuterium)
depletion of the reactants A and B (Fig. 7.2). in place of H2 . The reaction mixtures starting
This leads to a decrease in the rate of forward either with H2 or D2 reach equilibrium with
reaction and an increase in he rate of the the same composition, except that D2 and ND3
reverse reaction, are present instead of H2 and NH3. After
Eventually, the two reactions occur at the equilibrium is attained, these two mixtures
190 CHEMISTRY

Dynamic Equilibrium – A Student’s Activity

Equilibrium whether in a physical or in a chemical system, is always of dynamic
nature. This can be demonstrated by the use of radioactive isotopes. This is not feasible
in a school laboratory. However this concept can be easily comprehended by performing
the following activity. The activity can be performed in a group of 5 or 6 students.
Take two 100mL measuring cylinders (marked as 1 and 2) and two glass tubes
each of 30 cm length. Diameter of the tubes may be same or different in the range of

ed
3-5mm. Fill nearly half of the measuring cylinder-1 with colour ed water (for this
purpose add a crystal of potassium permanganate to water) and keep second cylinder
(number 2) empty.
Put one tube in cylinder 1 and second in cylinder 2. Immerse one tube in cylinder

h
1, close its upper tip with a finger and transfer the coloured water contained in its

pu T
lower portion to cylinder 2. Using second tube, kept in 2 nd cylinder , transfer the coloured

is
water in a similar manner from cylinder 2 to cylinder 1. In this way keep on transferring
coloured water using the two glass tubes from cylinder 1 to 2 and from 2 to 1 till you
re ER
notice that the level of coloured water in both the cylinders becomes constant.

bl
If you continue intertransferring coloured solution between the cylinders, there will
not be any further change in the levels of coloured water in two cylinders. If we take
analogy of ‘level’ of coloured water with ‘concentration’ of reactants and products in the
two cylinders, we can say the process of transfer, which continues even after the constancy
of level, is indicative of dynamic nature of the process. If we repeat the experiment taking
be C

two tubes of different diameters we find that at equilibrium the level of coloured water in
two cylinders is different. How far diameters are responsible for change in levels in two
cylinders? Empty cylinder (2) is an indicator of no product in it at the beginning.
N
t to ©
no

Fig.7.3 Demonstrating dynamic nature of equilibrium. (a) initial stage (b) final stage after the
equilibrium is attained.
 EQUILIBRIUM 191

2NH3 (g) ⇌ N2(g) + 3H2(g)

Similarly let us consider the reaction,
H2(g) + I 2(g) ⇌ 2HI(g). If we start with equal
initial concentration of H2 and I2, the reaction
proceeds in the forward direction and the
concentration of H2 and I2 decreases while that
of HI increases, until all of these become

ed
constant at equilibrium (Fig. 7.5). We can also
start with HI alone and make the reaction to
proceed in the reverse direction; the
concentration of HI will decrease and
concentration of H2 and I2 will increase until

h
they all become constant when equilibrium is

pu T
reached (Fig.7.5). If total number of H and I

is
atoms are same in a given volume, the same
equilibrium mixture is obtained whether we
re ER
Fig 7.4 Depiction of equilibrium for the reaction start it from pure reactants or pure product.
N 2 ( g ) + 3H2 ( g ) ⇌ 2 NH3 ( g )

bl
(H2 , N2, NH 3 and D 2, N2 , ND3 ) are mixed
together and left for a while. Later, when this
mixture is analysed, it is found that the
be C

concentration of ammonia is just the same as

before. However, when this mixture is
analysed by a mass spectrometer, it is found
N

that ammonia and all deuterium containing

forms of ammonia (NH3, NH2D, NHD2 and ND3)
and dihydrogen and its deutrated forms
(H2, HD and D2) are present. Thus one can
©

conclude that scrambling of H and D atoms

in the molecules must result from a
continuation of the forward and reverse
reactions in the mixture. If the reaction had
simply stopped when they reached
equilibrium, then there would have been no Fig.7.5 Chemical equilibrium in the reaction
mixing of isotopes in this way. H2(g) + I2(g) ⇌ 2HI(g) can be attained
to

Use of isotope (deuterium) in the formation from either direction

of ammonia clearly indicates that chemical 7.3 LAW OF CHEMICAL EQUILIBRIUM
reactions reach a state of dynamic AND EQUILIBRIUM CONSTANT
equilibrium in which the rates of forward
A mixture of reactants and products in the
t

and reverse reactions are equal and there

equilibrium state is called an equilibrium
no

is no net change in composition.

mixture. In this section we shall address a
Equilibrium can be attained from both number of important questions about the
sides, whether we start reaction by taking, composition of equilibrium mixtures: What is
H2 (g) and N2(g) and get NH3(g) or by taking the relationship between the concentrations of
NH 3(g) and decomposing it into N2(g) and reactants and products in an equilibrium
H2 (g). mixture? How can we determine equilibrium
N2(g) + 3H2 (g) ⇌ 2NH3(g) concentrations from initial concentrations?
192 CHEMISTRY

What factors can be exploited to alter the Six sets of experiments with varying initial
composition of an equilibrium mixture? The conditions were performed, starting with only
last question in particular is important when gaseous H 2 and I 2 in a sealed reaction vessel
choosing conditions for synthesis of industrial in first four experiments (1, 2, 3 and 4) and
chemicals such as H2, NH3, CaO etc. only HI in other two experiments (5 and 6).
To answer these questions, let us consider Experiment 1, 2, 3 and 4 were performed
a general reversible reaction: taking different concentrations of H2 and / or
A+B ⇌ C+D I 2, and with time it was observed that intensity

ed
where A and B are the reactants, C and D are of the purple colour remained constant and
the products in the balanced chemical equilibrium was attained. Similarly, for
equation. On the basis of experimental studies experiments 5 and 6, the equilibrium was
of many reversible reactions, the Norwegian attained from the opposite direction.

h
chemists Cato Maximillian Guldberg and Peter Data obtained from all six sets of

pu T
Waage pr oposed in 1864 that the experiments are given in Table 7.2.

is
concentrations in an equilibrium mixture are It is evident from the experiments 1, 2, 3
related by the following equilibrium and 4 that number of moles of dihydrogen
equation,
re ER reacted = number of moles of iodine reacted =
[C ][D]

bl
½ (number of moles of HI formed). Also,
Kc =
[ A ][B] (7.1) experiments 5 and 6 indicate that,
where K c is the equilibrium constant and the [H2(g)]eq = [I2(g)] eq
expression on the right side is called the
be C

equilibrium constant expression. Knowing the above facts, in order to

The equilibrium equation is also known as establish a relationship between
the law of mass action because in the early concentrations of the reactants and products,
N

days of chemistry, concentration was called several combinations can be tried. Let us
“active mass”. In order to appreciate their work consider the simple expression,
better, let us consider reaction between
[HI(g)] eq / [H2(g)] eq [I 2(g)] eq
gaseous H2 and I 2 carried out in a sealed vessel
©

at 731K. It can be seen from Table 7.3 that if we

H2(g) + I2(g) ⇌ 2HI(g) put the equilibrium concentrations of the
1 mol 1 mol 2 mol reactants and products, the above expression

Table 7.2 Initial and Equilibrium Concentrations of H2 , I2 and HI

t to
no
 EQUILIBRIUM 193

Table 7.3 Expression Involving the The equilibrium constant for a general
Equilibrium Concentration of reaction,
Reactants
H2(g) + I2 (g) 2HI(g) aA + bB ⇌ cC + dD
is expressed as,
Kc = [C] c[D]d / [A]a[B]b (7.4)
where [A], [B], [C] and [D] are the equilibrium
concentrations of the reactants and products.

ed
Equilibrium constant for the reaction,
4NH3(g) + 5O 2(g) ⇌ 4NO(g) + 6H2O(g) is
written as

h
4 6 4 5
Kc = [NO] [H2O] / [NH3] [O2]

pu T
Molar concentration of different species is

is
indicated by enclosing these in square bracket
is far from constant. However, if we consider and, as mentioned above, it is implied that
re ER
the expression, these are equilibrium concentrations. While

bl
[HI(g)]2eq / [H2(g)]eq [I2(g)]eq writing expression for equilibrium constant,
we find that this expression gives constant symbol for phases (s, l, g) are generally
value (as shown in Table 7.3) in all the six ignored.
cases. It can be seen that in this expression Let us write equilibrium constant for the
be C

the power of the concentration for reactants reaction, H2(g) + I2(g) ⇌ 2HI(g) (7.5)
and products are actually the stoichiometric as, Kc = [HI]2 / [H 2] [I2] = x (7.6)
coefficients in the equation for the chemical
The equilibrium constant for the reverse
N

reaction. Thus, for the reaction H2(g) + I2(g) ⇌

2HI(g), following equation 7.1, the equilibrium reaction, 2HI(g) ⇌ H 2(g) + I2(g), at the same
constant K c is written as, temperature is,
2
K c = [HI(g)]eq / [H2(g)]eq [I2(g)]eq (7.2) K′c = [H 2] [I2] / [HI] 2 = 1/ x = 1 / K c (7.7)
©

Generally the subscript ‘eq’ (used for Thus, K′c = 1 / Kc (7.8)

equilibrium) is omitted from the concentration Equilibrium constant for the reverse
terms. It is taken for granted that the reaction is the inverse of the equilibrium
concentrations in the expression for Kc are constant for the reaction in the forward
equilibrium values. We, therefore, write, direction.
K c = [HI(g)]2 / [H2(g)] [I2(g)] (7.3) If we change the stoichiometric coefficients
to

in a chemical equation by multiplying

The subscript ‘c’ indicates that K c is throughout by a factor then we must make
expressed in concentrations of mol L–1. sure that the expression for equilibrium
At a given temperature, the product of constant also reflects that change. For
concentrations of the reaction products example, if the reaction (7.5) is written as,
t

raised to the respective stoichiometric

no

coefficient in the balanced chemical ½ H2(g) + ½ I2(g) ⇌ HI(g) (7.9)

equation divided by the product of the equilibrium constant for the above reaction
concentrations of the reactants raised to is given by
their individual stoichiometric
K″c = [HI] / [H2]
1/2 1/2 2 1/2
[I2] = {[HI] / [H2][I2]}
coefficients has a constant value. This is
known as the Equilibrium Law or Law of = x1/2 = K c1/2 (7.10)
Chemical Equilibrium. On multiplying the equation (7.5) by n, we get
194 CHEMISTRY

nH 2(g) + nI2(g) ⇌ 2nHI(g) (7.11) 800K. What will be Kc for the reaction
Therefore, equilibrium constant for the N2(g) + O2(g) ⇌ 2NO(g)
reaction is equal to K cn. These findings are
Solution
summarised in Table 7.4. It should be noted
that because the equilibrium constants K c and For the reaction equilibrium constant,
K ′c have different numerical values, it is K c can be written as,
important to specify the form of the balanced
chemical equation when quoting the value of [NO ]2

ed
Kc =
an equilibrium constant. [N2 ][O 2 ]
Table 7.4 Relations between Equilibrium
(2.8 × 10 M )
2
Constants for a General Reaction -3

and its Multiples. =

(3.0 × 10 M )(4.2 × 10
−3 −3
M)

h
Chemical equation Equilibrium

pu T
constant = 0.622

is
a A + b B ⇌ c C + dD K
cC+dD ⇌ aA+bB
re ER K′c =(1/Kc )
7.4 HOMOGENEOUS EQUILIBRIA

bl
In a homogeneous system, all the reactants
K′″c = ( Kc ) and products are in the same phase. For
n
na A + nb B ⇌ nc C + nd D
example, in the gaseous reaction,
N2(g) + 3H2(g) ⇌ 2NH3(g), reactants and
Problem 7.1
be C

products are in the homogeneous phase.

The following concentrations were Similarly, for the reactions,
obtained for the formation of NH3 from N2 CH3COOC2H5 (aq) + H2O (l) ⇌ CH3COOH (aq)
N

and H2 at equilibrium at 500K. + C2H 5OH (aq)

[N2] = 1.5 × 10–2M. [H2] = 3.0 ×10–2 M and
[NH3] = 1.2 ×10–2M. Calculate equilibrium and, Fe3+ (aq) + SCN–(aq) ⇌ Fe(SCN)2+ (aq)
constant. all the reactants and products are in
©

Solution homogeneous solution phase. We shall now

The equilibrium constant for the reaction, consider equilibrium constant for some
homogeneous reactions.
N2(g) + 3H2(g) ⇌ 2NH3(g) can be written
as, 7.4.1 Equilibrium Constant in Gaseous
Systems
 NH3 ( g ) 
2
So far we have expressed equilibrium constant
Kc =
to

of the reactions in terms of molar

 N 2 ( g )   H2 ( g ) 
3

concentration of the reactants and products,

and used symbol, Kc for it. For reactions

=
(1.2 ×10 ) −2 2 involving gases, however, it is usually more
convenient to express the equilibrium
(1.5 × 10 )(3.0 ×10 ) −2 3
t

−2
constant in terms of partial pressure.
no

The ideal gas equation is written as,

= 0.106 × 104 = 1.06 × 103
pV = nRT
Problem 7.2
n
At equilibrium, the concentrations of ⇒ p= RT
N2=3.0 × 10 –3M, O2 = 4.2 × 10–3M and V
NO= 2.8 × 10–3M in a sealed vessel at Here, p is the pressure in Pa, n is the number
of moles of the gas, V is the volume in m 3 and
 EQUILIBRIUM 195

T is the temperature in Kelvin

NH3 ( g )  [ RT ]
2 −2

=  = K c ( RT )
Therefore, −2

 N 2 ( g )   H 2 ( g ) 
3
n/V is concentration expressed in mol/m3
If concentration c, is in mol/L or mol/dm3,
or K p = K c ( RT )
−2
and p is in bar then (7.14)
p = cRT, Similarly, for a general reaction
We can also write p = [gas]RT. aA + bB ⇌ cC + dD

ed
Here, R= 0.0831 bar litre/mol K
At constant temperature, the pressure of
Kp =
( p )( p ) [ C] [ D] ( RT ) (
c
C
=
d
D
c d c +d )

the gas is proportional to its concentration i.e.,

( p )( p ) [ A ] [B ] ( RT )(
a
A
b
B
a b a +b )

p ∝ [gas]

h
[C] [D] RT ( c +d ) − (a +b )
c d
For reaction in equilibrium

pu T
= ( )

is
H2(g) + I2(g) ⇌ 2HI(g)
re ER [ A ]a [B]b
We can write either
[C] [D] RT ∆n
c d

bl
∆n
HI ( g )  ( ) = Kc ( RT )
2
= (7.15)
Kc =
 H2 ( g )   I2 ( g ) 
[ A ]a [B]b
where ∆n = (number of moles of gaseous
or K c =
( pH I )2 products) – (number of moles of gaseous
be C

( p )( p )
H2 I2
(7.12)
reactants) in the balanced chemical equation.
It is necessary that while calculating the value
Further, since p HI =  HI ( g )  RT of K p, pressure should be expressed in bar
N

because standard state for pressure is 1 bar.

p I2 =  I2 ( g)  RT
We know from Unit 1 that :
pH 2 =  H2 ( g )  RT 1pascal, Pa=1Nm–2, and 1bar = 105 Pa
©

Therefore, Kp values for a few selected reactions at

different temperatures are given in Table 7.5
( p HI ) HI ( g )  [ RT ]
2 2 2

Kp = = Table 7.5 Equilibrium Constants, Kp for a

( p )( p )
H2 I2 H 2 ( g )  RT .  I 2 ( g )  RT Few Selected Reactions

 HI ( g ) 
2

= = Kc (7.13)
H2 ( g )  I2 ( g ) 
to

In this example, K p = K c i.e., both

equilibrium constants are equal. However, this
is not always the case. For example in reaction
t

N2(g) + 3H2(g) ⇌ 2NH3(g)

no

(p )
2
NH3
Kp =
( p )( p )
3
N2 H2 Problem 7.3
PCl5, PCl3 and Cl2 are at equilibrium at
 NH3 ( g )  [ RT ]
2 2
500 K and having concentration 1.59M
= PCl3, 1.59M Cl2 and 1.41 M PCl5.
 N 2 ( g )  RT .  H2 ( g )  ( RT )
3 3
196 CHEMISTRY

Calculate Kc for the reaction, the value 0.194 should be neglected

PCl5 ⇌ PCl3 + Cl2 because it will give concentration of the
Solution reactant which is more than initial
The equilibrium constant K c for the above concentration.
reaction can be written as, Hence the equilibrium concentrations are,
[CO 2] = [H2- ] = x = 0.067 M
Kc =
[PCl 3 ][Cl 2 ] = (1.59)2 = 1.79 [CO] = [H2O] = 0.1 – 0.067 = 0.033 M
[PCl5 ] (1.41)

ed
Problem 7.5
Problem 7.4
For the equilibrium,
The value of Kc = 4.24 at 800K for the reaction,
CO (g) + H2O (g) ⇌ CO 2 (g) + H2 (g) 2NOCl(g) ⇌ 2NO(g) + Cl2(g)

h
Calculate equilibrium concentrations of the value of the equilibrium constant, K c

pu T
is 3.75 × 10 –6 at 1069 K. Calculate the K p
CO2, H2, CO and H2O at 800 K, if only CO

is
for the reaction at this temperature?
and H 2O are present initially at
re ER
concentrations of 0.10M each. Solution
Solution We know that,

bl
For the reaction, K p = Kc(RT)∆ n
CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g) For the above reaction,
Initial concentration: ∆n = (2+1) – 2 = 1
be C

0.1M 0.1M 0 0 K p = 3.75 ×10–6 (0.0831 × 1069)

Let x mole per litre of each of the product K p = 0.033
be formed.
N

At equilibrium:
7.5 HETEROGENEOUS EQUILIBRIA
(0.1-x) M (0.1-x) M xM xM
Equilibrium in a system having more than one
where x is the amount of CO2 and H2 at phase is called heterogeneous equilibrium.
©

equilibrium. The equilibrium between water vapour and

Hence, equilibrium constant can be liquid water in a closed container is an
written as, example of heterogeneous equilibrium.
K c = x2/(0.1-x)2 = 4.24 H 2O(l) ⇌ H2O(g)
x2 = 4.24(0.01 + x2-0.2x)
In this example, there is a gas phase and a
x2 = 0.0424 + 4.24x2-0.848x liquid phase. In the same way, equilibrium
to

3.24x2 – 0.848x + 0.0424 = 0 between a solid and its saturated solution,

a = 3.24, b = – 0.848, c = 0.0424 Ca(OH) 2 (s) + (aq) ⇌ Ca2+ (aq) + 2OH–(aq)
(for quadratic equation ax2 + bx + c = 0,
is a heterogeneous equilibrium.

(− b ± )
t

b2 − 4ac Heterogeneous equilibria often involve pure

solids or liquids. We can simplify equilibrium
no

x=
2a expressions for the heterogeneous equilibria
x = 0.848±√(0.848)2– 4(3.24)(0.0424)/ involving a pure liquid or a pure solid, as the
(3.24×2) molar concentration of a pure solid or liquid
x = (0.848 ± 0.4118)/ 6.48 is constant (i.e., independent of the amount
x1 = (0.848 – 0.4118)/6.48 = 0.067 present). In other words if a substance ‘X’ is
x2 = (0.848 + 0.4118)/6.48 = 0.194 involved, then [X(s)] and [X(l)] are constant,
whatever the amount of ‘X’ is taken. Contrary
 EQUILIBRIUM 197

to this, [X(g)] and [X(aq)] will vary as the This shows that at a particular
amount of X in a given volume varies. Let us temperature, there is a constant concentration
take thermal dissociation of calcium carbonate or pressure of CO2 in equilibrium with CaO(s)
which is an interesting and important example and CaCO3(s). Experimentally it has been
of heterogeneous chemical equilibrium. found that at 1100 K, the pressure of CO2 in
CaCO3 (s) ∆⇀
 CaO (s) + CO2 (g) (7.16) equilibrium with CaCO3(s) and CaO(s), is
↽ 

2.0 ×105 Pa. Therefore, equilibrium constant
On the basis of the stoichiometric equation, at 1100K for the above reaction is:

ed
we can write, K p = p CO2 = 2 ×105 Pa /105 Pa = 2.00
CaO ( s )  CO 2 ( g)  Similarly, in the equilibrium between
Kc =  nickel, carbon monoxide and nickel carbonyl
CaCO 3 ( s ) 
(used in the purification of nickel),

h
Since [CaCO3(s)] and [CaO(s)] are both Ni (s) + 4 CO (g) ⇌ Ni(CO)4 (g),

pu T
constant, therefore modified equilibrium
the equilibrium constant is written as

is
constant for the thermal decomposition of
calcium carbonate will be  Ni ( CO) 4 
re ER Kc = 
K´c = [CO2(g)] (7.17) [CO]4

bl
or K p = p CO2 (7.18) It must be remembered that for the
existence of heterogeneous equilibrium pure
Units of Equilibrium Constant solids or liquids must also be present
be C

The value of equilibrium constant Kc can (however small the amount may be) at
be calculated by substituting the equilibrium, but their concentrations or
concentration terms in mol/L and for Kp partial pressures do not appear in the
N

partial pressure is substituted in Pa, kPa, expression of the equilibrium constant. In the
bar or atm. This results in units of reaction,
equilibrium constant based on molarity or Ag2O(s) + 2HNO3(aq) ⇌ 2AgNO3(aq) +H2O(l)
pressure, unless the exponents of both the
numerator and denominator are same.
[AgNO3 ]
2
©

For the reactions, Kc =

H 2(g) + I2 (g) ⇌ 2HI, Kc and Kp have no unit. [HNO3 ]2
N2 O4 (g) ⇌ 2NO 2 (g), Kc has unit mol/L
and Kp has unit bar Problem 7.6
Equilibrium constants can also be The value of K p for the reaction,
expressed as dimensionless quantities if CO2 (g) + C (s) ⇌ 2CO (g)
to

the standard state of reactants and is 3.0 at 1000 K. If initially

products are specified. For a pure gas, the
standard state is 1bar. Therefore a pressure p CO2 = 0.48 bar and p CO = 0 bar and pure
of 4 bar in standard state can be expressed graphite is present, calculate the equilibrium
as 4 bar/1 bar = 4, which is a
partial pressures of CO and CO2.
t

dimensionless number. Standard state (c0 )

for a solute is 1 molar solution and all Solution
no

concentrations can be measured with For the reaction,

respect to it. The numerical value of
equilibrium constant depends on the
let ‘x’ be the decrease in pressure of CO2,
standard state chosen. Thus, in this
then
system both Kp and Kc are dimensionless CO2(g) + C(s) ⇌ 2CO(g)
quantities but have different numerical Initial
values due to different standard states. pressure: 0.48 bar 0
198 CHEMISTRY

At equilibrium: 5. The equilibrium constant K for a reaction

(0.48 – x)bar 2x bar is related to the equilibrium constant of the
corresponding reaction, whose equation is
2
p obtained by multiplying or dividing the
Kp = CO

p CO2 equation for the original reaction by a small

integer.
K p = (2x)2/(0.48 – x) = 3 Let us consider applications of equilibrium
4x 2 = 3(0.48 – x) constant to:

ed
4x 2 = 1.44 – x • predict the extent of a reaction on the basis
4x 2 + 3x – 1.44 = 0 of its magnitude,
a = 4, b = 3, c = –1.44 • predict the direction of the reaction, and
•
(− b ± )
calculate equilibrium concentrations.

h
b2 − 4ac
x= 7.6.1 Predicting the Extent of a Reaction

pu T
2a

is
The numerical value of the equilibrium
= [–3 ± √(3)2– 4(4)(–1.44)]/2 × 4 constant for a reaction indicates the extent of
re ER
= (–3 ± 5.66)/8 the reaction. But it is important to note that

bl
= (–3 + 5.66)/ 8 (as value of x cannot be an equilibrium constant does not give any
negative hence we neglect that value) information about the rate at which the
x = 2.66/8 = 0.33 equilibrium is reached. The magnitude of Kc
or K p is directly proportional to the
The equilibrium partial pressures are, concentrations of products (as these appear
be C

p CO = 2x = 2 × 0.33 = 0.66 bar in the numerator of equilibrium constant

expression) and inversely proportional to the
p CO = 0.48 – x = 0.48 – 0.33 = 0.15 bar concentrations of the reactants (these appear
N

2
in the denominator). This implies that a high
value of K is suggestive of a high concentration
7.6 APPLICATIONS OF EQUILIBRIUM of products and vice-versa.
CONSTANTS
We can make the following generalisations
©

Before considering the applications of concerning the composition of

equilibrium constants, let us summarise the equilibrium mixtures:
important features of equilibrium constants as
follows: • If Kc > 103, products predominate over
reactants, i.e., if Kc is very large, the reaction
1. Expression for equilibrium constant is proceeds nearly to completion. Consider
applicable only when concentrations of the the following examples:
reactants and products have attained
to

constant value at equilibrium state. (a) The reaction of H2 with O2 at 500 K has a
2. The value of equilibrium constant is very large equilibrium c o n s t a n t ,
independent of initial concentrations of the K c = 2.4 × 1047.
reactants and products. (b) H 2(g) + Cl2(g) ⇌ 2HCl(g) at 300K has
t

3. Equilibrium constant is temperature K c = 4.0 × 1031.

no

dependent having one unique value for a

particular reaction represented by a (c) H 2(g) + Br2(g) ⇌ 2HBr (g) at 300 K,
balanced equation at a given temperature. K c = 5.4 × 10 18
4. The equilibrium constant for the reverse • If K c < 10–3, reactants predominate over
reaction is equal to the inverse of the products, i.e., if K c is very small, the reaction
equilibrium constant for the forward proceeds rarely. Consider the following
reaction. examples:
 EQUILIBRIUM 199

(a) The decomposition of H2O into H2 and O2 If Qc = Kc , the reaction mixture is already
at 500 K has a very small equilibrium at equilibrium.
constant, K c = 4.1 × 10– 48 Consider the gaseous reaction of H2
(b) N2(g) + O2(g) ⇌ 2NO(g), with I2,
at 298 K has K c = 4.8 ×10 – 31. H2(g) + I2(g) ⇌ 2HI(g); Kc = 57.0 at 700 K.
Suppose we have molar concentrations
• If K c is in the range of 10 – 3 to 10 3,
appreciable concentrations of both [H2]t=0.10M, [I 2] t = 0.20 M and [HI]t = 0.40 M.
reactants and products are present. (the subscript t on the concentration symbols

ed
Consider the following examples: means that the concentrations were measured
(a) For reaction of H2 with I 2 to give HI, at some arbitrary time t, not necessarily at
equilibrium).
K c = 57.0 at 700K. Thus, the reaction quotient, Qc at this stage

h
(b) Also, gas phase decomposition of N2O 4 to of the reaction is given by,
NO2 is another reaction with a value

pu T
Qc = [HI]t2 / [H2] t [I2] t = (0.40)2/ (0.10)×(0.20)
of Kc = 4.64 × 10 –3 at 25°C which is neither

is
too small nor too large. Hence, = 8.0
Now, in this case, Q c (8.0) does not equal
re ER
equilibrium mixtures contain appreciable
concentrations of both N2O4 and NO2. Kc (57.0), so the mixture of H2(g), I2(g) and HI(g)

bl
is not at equilibrium; that is, more H2(g) and
These generarlisations are illustrated in
I2(g) will react to form more HI(g) and their
Fig. 7.6
concentrations will decrease till Qc = K c.
The reaction quotient, Qc is useful in
be C

predicting the direction of reaction by

comparing the values of Qc and K c.
Thus, we can make the following
N

generalisations concerning the direction of the

Fig.7.6 Dependence of extent of reaction on Kc reaction (Fig. 7.7) :

7.6.2 Predicting the Direction of the

©

Reaction
The equilibrium constant helps in predicting
the direction in which a given reaction will
proceed at any stage. For this purpose, we
calculate the reaction quotient Q. The
reaction quotient, Q (Q c with molar
to

concentrations and QP with partial pressures) Fig. 7.7 Predicting the direction of the reaction
is defined in the same way as the equilibrium
constant K c except that the concentrations in • If Qc < Kc, net reaction goes from left to right
Qc are not necessarily equilibrium values. • If Q c > K c, net reaction goes from right to
For a general reaction: left.
t

aA+bB ⇌ cC+dD (7.19) • If Qc = K c, no net reaction occurs.

no

Qc = [C] c[D]d / [A]a[B]b (7.20) Problem 7.7

Then, The value of Kc for the reaction
If Qc > Kc , the reaction will proceed in the 2A ⇌ B + C is 2 × 10 –3. At a given time,
direction of reactants (reverse reaction). the composition of reaction mixture is
If Qc < Kc , the reaction will proceed in the [A] = [B] = [C] = 3 × 10 –4 M. In which
direction of the products (forward reaction). direction the reaction will proceed?
200 CHEMISTRY

Solution The total pressure at equilbrium was

For the reaction the reaction quotient Qc found to be 9.15 bar. Calculate K c, K p and
is given by, partial pressure at equilibrium.
Q c = [B][C]/ [A]2 Solution
as [A] = [B] = [C] = 3 × 10 – 4M We know pV = nRT
Q c = (3 ×10 –4)(3 × 10 –4) / (3 ×10 –4)2 = 1 Total volume (V ) = 1 L
as Qc > K c so the reaction will proceed in Molecular mass of N2O 4 = 92 g

ed
the reverse direction. Number of moles = 13.8g/92 g = 0.15
7.6.3 Calculating Equilibrium of the gas (n)
Concentrations Gas constant (R) = 0.083 bar L mol–1K –1

h
In case of a problem in which we know the Temperature (T ) = 400 K
initial concentrations but do not know any of

pu T
pV = nRT

is
the equilibrium concentrations, the following p × 1L = 0.15 mol × 0.083 bar L mol–1K –1
three steps shall be followed: × 400 K
re ER
Step 1. Write the balanced equation for the p = 4.98 bar

bl
reaction. N2O 4 ⇌ 2NO2
Step 2. Under the balanced equation, make a Initial pressure: 4.98 bar 0
table that lists for each substance involved in
the reaction: At equilibrium: (4.98 – x) bar 2x bar
Hence,
be C

(a) the initial concentration,

p total at equilibrium = p N2 O4 + p NO2
(b) the change in concentration on going to
equilibrium, and 9.15 = (4.98 – x) + 2x
N

(c) the equilibrium concentration. 9.15 = 4.98 + x

In constructing the table, define x as the x = 9.15 – 4.98 = 4.17 bar
concentration (mol/L) of one of the substances
Partial pressures at equilibrium are,
that reacts on going to equilibrium, then use
©

the stoichiometry of the reaction to determine p N 2O4 = 4.98 – 4.17 = 0.81bar

the concentrations of the other substances in
p NO2 = 2x = 2 × 4.17 = 8.34 bar
terms of x.
K p = ( p NO2 ) / p N 2O4
2
Step 3. Substitute the equilibrium
concentrations into the equilibrium equation
for the reaction and solve for x. If you are to = (8.34)2/0.81 = 85.87
∆n
solve a quadratic equation choose the K p = K c(RT)
to

mathematical solution that makes chemical 85.87 = K c(0.083 × 400) 1

sense. K c = 2.586 = 2.6
Step 4. Calculate the equilibrium
concentrations from the calculated value of x. Problem 7.9
t

Step 5. Check your results by substituting 3.00 mol of PCl 5 kept in 1L closed reaction
no

them into the equilibrium equation. vessel was allowed to attain equilibrium
at 380K. Calculate composition of the
Problem 7.8 mixture at equilibrium. K c= 1.80
13.8g of N 2O 4 was placed in a 1L reaction Solution
vessel at 400K and allowed to attain PCl5 ⇌ PCl 3 + Cl 2
equilibrium Initial
N 2O4 (g) ⇌ 2NO2 (g) concentration: 3.0 0 0
 EQUILIBRIUM 201

Let x mol per litre of PCl5 be dissociated, K = e – ∆G

V
/ RT (7.23)
At equilibrium:
(3-x) x x Hence, using the equation (7.23), the
reaction spontaneity can be interpreted in
K c = [PCl3][Cl2]/[PCl5]
terms of the value of ∆G0.
1.8 = x2/ (3 – x) • If ∆G0 < 0, then –∆G0/RT is positive, and
x2 + 1.8x – 5.4 = 0 V
e– ∆G /RT >1, making K >1, which implies
x = [–1.8 ± √(1.8)2 – 4(–5.4)]/2 a spontaneous reaction or the reaction

ed
x = [–1.8 ± √ 3.24 + 21.6]/2 which proceeds in the forward direction to
such an extent that the products are
x = [–1.8 ± 4.98]/2 present predominantly.
x = [–1.8 + 4.98]/2 = 1.59 • If ∆G0 > 0, then –∆G0/RT is negative, and
V
e– ∆G /RT < 1, that is , K < 1, which implies

h
[PCl5] = 3.0 – x = 3 –1.59 = 1.41 M
[PCl3] = [Cl2] = x = 1.59 M a non-spontaneous reaction or a reaction

pu T
which proceeds in the forward direction to

is
7.7 RELATIONSHIP BETWEEN such a small degree that only a very minute
re ER
EQUILIBRIUM CONSTANT K, quantity of product is formed.

bl
REACTION QUOTIENT Q AND Problem 7.10
GIBBS ENERGY G
The value of ∆G0 for the phosphorylation
The value of K c for a reaction does not depend
of glucose in glycolysis is 13.8 kJ/mol.
on the rate of the reaction. However, as you Find the value of K c at 298 K.
be C

have studied in Unit 6, it is directly related

to the thermodynamics of the reaction and Solution
in particular, to the change in Gibbs energy, ∆G0 = 13.8 kJ/mol = 13.8 × 103J/mol
∆G. If,
N

Also, ∆G0 = – RT lnK c

• ∆G is negative, then the reaction is Hence, ln K c = –13.8 × 103J/mol
spontaneous and proceeds in the forward (8.314 J mol –1K –1 × 298 K)
direction.
ln Kc = – 5.569
©

• ∆G is positive, then reaction is considered

Kc = e–5.569
non-spontaneous. Instead, as reverse
reaction would have a negative ∆G, the Kc = 3.81 × 10 –3
products of the forward reaction shall be Problem 7.11
converted to the reactants.
Hydrolysis of sucrose gives,
• ∆G is 0, reaction has achieved equilibrium;
at this point, there is no longer any free Sucrose + H2O ⇌ Glucose + Fructose
to

energy left to drive the reaction. Equilibrium constant K c for the reaction
A mathematical expression of this is 2 ×1013 at 300K. Calculate ∆G0 at
thermodynamic view of equilibrium can be 300K.
described by the following equation: Solution
0
∆G = ∆G + RT lnQ (7.21) 0
t

∆G = – RT lnK c
where, G0 is standard Gibbs energy.
no

∆G0 = – 8.314J mol–1K –1×

At equilibrium, when ∆G = 0 and Q = K c, 300K × ln(2×1013)
the equation (7.21) becomes, ∆G = – 7.64 ×10 J mol–1
0 4

∆G = ∆G0 + RT ln K = 0
0
∆G = – RT lnK (7.22) 7.8 FACTORS AFFECTING EQUILIBRIA
lnK = – ∆G / RT
0
One of the principal goals of chemical synthesis
Taking antilog of both sides, we get, is to maximise the conversion of the reactants
202 CHEMISTRY

to products while minimizing the expenditure “When the concentration of any of the
of energy. This implies maximum yield of reactants or products in a reaction at
products at mild temperature and pressure equilibrium is changed, the composition
conditions. If it does not happen, then the of the equilibrium mixture changes so as
experimental conditions need to be adjusted. to minimize the effect of concentration
For example, in the Haber process for the changes”.
synthesis of ammonia from N2 and H2, the Let us take the reaction,
choice of experimental conditions is of real
H 2(g) + I2(g) ⇌ 2HI(g)

ed
economic importance. Annual world
production of ammonia is about hundred If H 2 is added to the reaction mixture at
million tones, primarily for use as fertilizers. equilibrium, then the equilibrium of the
Equilibrium constant, Kc is independent of reaction is disturbed. In order to restore it, the
initial concentrations. But if a system at reaction proceeds in a direction wherein H2 is

h
equilibrium is subjected to a change in the consumed, i.e., more of H2 and I2 react to form

pu T
concentration of one or more of the reacting HI and finally the equilibrium shifts in right

is
(forward) direction (Fig.7.8). This is in
substances, then the system is no longer at
accordance with the Le Chatelier’s principle
re ER
equilibrium; and net reaction takes place in
some direction until the system returns to
which implies that in case of addition of a

bl
reactant/product, a new equilibrium will be
equilibrium once again. Similarly, a change in
temperature or pressure of the system may set up in which the concentration of the
reactant/product should be less than what it
also alter the equilibrium. In order to decide
was after the addition but more than what it
what course the reaction adopts and make a
was in the original mixture.
be C

qualitative prediction about the effect of a

change in conditions on equilibrium we use
Le Chatelier’s principle. It states that a
N

change in any of the factors that

determine the equilibrium conditions of a
system will cause the system to change
in such a manner so as to reduce or to
©

counteract the effect of the change. This

is applicable to all physical and chemical
equilibria.
We shall now be discussing factors which
can influence the equilibrium.
7.8.1 Effect of Concentration Change
to

In general, when equilibrium is disturbed by

the addition/removal of any reactant/
products, Le Chatelier’s principle predicts that:
• The concentration stress of an added
t

reactant/product is relieved by net reaction Fig. 7.8 Effect of addition of H2 on change of

no

in the direction that consumes the added concentration for the reactants and
substance. products in the reaction,
• The concentration stress of a removed H 2(g) + I2 (g) ⇌ 2HI(g)
reactant/product is relieved by net reaction
in the direction that replenishes the The same point can be explained in terms
removed substance. of the reaction quotient, Qc,
2
or in other words, Q c = [HI] / [H2][I2]
 EQUILIBRIUM 203

Addition of hydrogen at equilibrium results replenish the Fe 3+ ions. Because the

in value of Qc being less than K c . Thus, in order concentration of [Fe(SCN)]2+ decreases, the
to attain equilibrium again reaction moves in intensity of red colour decreases.
the forward direction. Similarly, we can say Addition of aq. HgCl2 also decreases red
that removal of a product also boosts the colour because Hg2+ reacts with SCN– ions to
forward reaction and increases the form stable complex ion [Hg(SCN)4]2–. Removal
concentration of the products and this has –
of free SCN (aq) shifts the equilibrium in
great commercial application in cases of equation (7.24) from right to left to replenish

ed
reactions, where the product is a gas or a SCN – ions. Addition of potassium thiocyanate
volatile substance. In case of manufacture of on the other hand increases the colour
ammonia, ammonia is liquified and removed intensity of the solution as it shift the
from the reaction mixture so that reaction equilibrium to right.
keeps moving in forward direction. Similarly,

h
in the large scale production of CaO (used as 7.8.2 Effect of Pressure Change

pu T
important building material) from CaCO3, A pressure change obtained by changing the

is
constant removal of CO2 from the kiln drives volume can affect the yield of products in case
the reaction to completion. It should be of a gaseous reaction where the total number
re ER
remembered that continuous removal of a of moles of gaseous reactants and total

bl
product maintains Qc at a value less than Kc number of moles of gaseous products are
and reaction continues to move in the forward different. In applying Le Chatelier’s principle
direction. to a heterogeneous equilibrium the effect of
Effect of Concentration – An experiment pressure changes on solids and liquids can
be C

be ignored because the volume (and

This can be demonstrated by the following
concentration) of a solution/liquid is nearly
reaction:
independent of pressure.
Fe3+(aq)+ SCN –(aq) ⇌ [Fe(SCN)]2+(aq) (7.24)
N

Consider the reaction,

yellow colourless deep red CO(g) + 3H2(g) ⇌ CH4(g) + H2O(g)
 Fe ( SCN ) 2+ ( aq )  Here, 4 mol of gaseous reactants (CO + 3H2)
Kc =   become 2 mol of gaseous products (CH4 +
©

(7.25)
 Fe3+ ( aq )   SCN – ( aq )  H2O). Suppose equilibrium mixture (for above
reaction) kept in a cylinder fitted with a piston
A reddish colour appears on adding two at constant temperature is compressed to one
drops of 0.002 M potassium thiocynate half of its original volume. Then, total pressure
solution to 1 mL of 0.2 M iron(III) nitrate will be doubled (according to
solution due to the formation of [Fe(SCN)]2+. pV = constant). The partial pressure and
The intensity of the red colour becomes therefore, concentration of reactants and
to

constant on attaining equilibrium. This products have changed and the mixture is no
equilibrium can be shifted in either forward longer at equilibrium. The direction in which
or reverse directions depending on our choice the reaction goes to re-establish equilibrium
of adding a reactant or a product. The can be predicted by applying the Le Chatelier’s
equilibrium can be shifted in the opposite principle. Since pressure has doubled, the
t

direction by adding reagents that remove Fe3+ equilibrium now shifts in the forward
no

or SCN – ions. For example, oxalic acid direction, a direction in which the number of
(H2C2O 4), reacts with Fe3+ ions to form the moles of the gas or pressure decreases (we
stable complex ion [Fe(C 2O 4) 3] 3 – , thus know pressure is proportional to moles of the
decreasing the concentration of free Fe3+ (aq). gas). This can also be understood by using
In accordance with the Le Chatelier’s principle, reaction quotient, Qc. Let [CO], [H2], [CH4] and
the concentration stress of removed Fe3+ is [H 2O] be the molar concentrations at
relieved by dissociation of [Fe(SCN)] 2+ to equilibrium for methanation reaction. When
204 CHEMISTRY

volume of the reaction mixture is halved, the Production of ammonia according to the
partial pressure and the concentration are reaction,
doubled. We obtain the reaction quotient by N2(g) + 3H2(g) ⇌ 2NH3(g) ;
replacing each equilibrium concentration by
double its value. ∆H= – 92.38 kJ mol –1
is an exothermic process. According to
 CH4 ( g )  H2O ( g )  Le Chatelier’s principle, raising the
Qc =  temperature shifts the equilibrium to left and
CO ( g )   H2 ( g ) 
3

ed
decreases the equilibrium concentration of
As Q c < Kc , the reaction proceeds in the ammonia. In other words, low temperature is
forward direction. favourable for high yield of ammonia, but
In reaction C(s) + CO2(g) ⇌ 2CO(g), when practically very low temperatures slow down
the reaction and thus a catalyst is used.

h
pressure is increased, the reaction goes in the
reverse direction because the number of moles Effect of Temperature – An experiment

pu T
of gas increases in the forward direction. Effect of temperature on equilibrium can be

is
demonstrated by taking NO 2 gas (brown in
7.8.3 Effect of Inert Gas Addition
re ER
If the volume is kept constant and an inert gas
colour) which dimerises into N 2O 4 gas
(colourless).

bl
such as argon is added which does not take
2NO2(g) ⇌ N2O4(g); ∆H = –57.2 kJ mol–1
part in the reaction, the equilibrium remains
undisturbed. It is because the addition of an NO 2 gas prepared by addition of Cu
inert gas at constant volume does not change turnings to conc. HNO 3 is collected in two
be C

the partial pressures or the molar 5 mL test tubes (ensuring same intensity of
concentrations of the substance involved in the colour of gas in each tube) and stopper sealed
reaction. The reaction quotient changes only with araldite. Three 250 mL beakers 1, 2 and
N

if the added gas is a reactant or product 3 containing freezing mixture, water at room
involved in the reaction. temperature and hot water (36 3K ),
respectively, are taken (Fig. 7.9). Both the test
7.8.4 Effect of Temperature Change tubes are placed in beaker 2 for 8-10 minutes.
©

Whenever an equilibrium is disturbed by a After this one is placed in beaker 1 and the
change in the concentration, pressure or other in beaker 3. The effect of temperature
volume, the composition of the equilibrium on direction of reaction is depicted very well
mixture changes because the reaction in this experiment. At low temperatures in
quotient, Qc no longer equals the equilibrium beaker 1, the forward reaction of formation of
constant, K c . However, when a change in N2O4 is preferred, as reaction is exothermic, and
temperature occurs, the value of equilibrium thus, intensity of brown colour due to NO2
to

constant, Kc is changed. decreases. While in beaker 3, high

temperature favours the reverse reaction of
In general, the temperature dependence of
the equilibrium constant depends on the sign
of ∆H for the reaction.
t

• The equilibrium constant for an exothermic

no

reaction (negative ∆H) decreases as the

temperature increases.
• The equilibrium constant for an
endothermic reaction (positive ∆H)
increases as the temperature increases.
Temperature changes affect the Fig. 7.9 Effect of temperature on equilibrium for
equilibrium constant and rates of reactions. the reaction, 2NO2 (g) ⇌ N2 O4 (g)
 EQUILIBRIUM 205

formation of NO2 and thus, the brown colour Similarly, in manufacture of sulphuric
intensifies. acid by contact process,
Effect of temperature can also be seen in 2SO2(g) + O2(g) ⇌ 2SO3(g); Kc = 1.7 × 1026
an endothermic reaction, though the value of K is suggestive of reaction
[Co(H2O) 6]3+ (aq) + 4Cl– (aq) ⇌ [CoCl4]2–(aq) + going to completion, but practically the oxidation
6H2O(l) of SO2 to SO3 is very slow. Thus, platinum or
pink colourless blue divanadium penta-oxide (V2O5) is used as
At room temperature, the equilibrium catalyst to increase the rate of the reaction.

ed
mixture is blue due to [CoCl 4] 2–. When cooled Note: If a reaction has an exceedingly small
in a freezing mixture, the colour of the mixture K, a catalyst would be of little help.
turns pink due to [Co(H2O)6]3+ .
7.9 IONIC EQUILIBRIUM IN SOLUTION
7.8.5 Effect of a Catalyst

h
Under the effect of change of concentration on
A catalyst increases the rate of the chemical

pu T
the direction of equilibrium, you have
reaction by making available a new low energy

is
incidently come across with the following
pathway for the conversion of reactants to
equilibrium which involves ions:
re ER
products. It increases the rate of forward and
reverse reactions that pass through the same Fe3+(aq) + SCN–(aq) ⇌ [Fe(SCN)]2+(aq)

bl
transition state and does not affect There are numerous equilibria that involve
equilibrium. Catalyst lowers the activation ions only. In the following sections we will
energy for the forward and reverse reactions study the equilibria involving ions. It is well
by exactly the same amount. Catalyst does not known that the aqueous solution of sugar
be C

affect the equilibrium composition of a does not conduct electricity. However, when
reaction mixture. It does not appear in the common salt (sodium chloride) is added to
balanced chemical equation or in the water it conducts electricity. Also, the
N

equilibrium constant expression. conductance of electricity increases with an

Let us consider the formation of NH3 from increase in concentration of common salt.
dinitrogen and dihydrogen which is highly Michael Faraday classified the substances into
exothermic reaction and proceeds with two categories based on their ability to conduct
©

decrease in total number of moles formed as electricity. One category of substances

compared to the reactants. Equilibrium conduct electricity in their aqueous solutions
and are called electrolytes while the other do
constant decreases with increase in
temperature. At low temperature rate not and are thus, referred to as non-
electrolytes. Faraday further classified
decreases and it takes long time to reach at
electrolytes into strong and weak electrolytes.
equilibrium, whereas high temperatures give
Strong electrolytes on dissolution in water are
satisfactory rates but poor yields.
to

ionized almost completely, while the weak

German chemist, Fritz Haber discovered electrolytes are only partially dissociated.
that a catalyst consisting of iron catalyse the For example, an aqueous solution of
reaction to occur at a satisfactory rate at sodium chloride is comprised entirely of
temperatures, where the equilibrium sodium ions and chloride ions, while that
t

concentration of NH3 is reasonably favourable. of acetic acid mainly contains unionized

no

Since the number of moles formed in the acetic acid molecules and only some acetate
reaction is less than those of reactants, the ions and hydronium ions. This is because
yield of NH3 can be improved by increasing there is almost 100% ionization in case of
the pressure. sodium chloride as compared to less
Optimum conditions of temperature and than 5% ionization of acetic acid which is
pressure for the synthesis of NH 3 using a weak electrolyte. It should be noted
catalyst are around 500 °C and 200 atm. that in weak electrolytes, equilibrium is
206 CHEMISTRY

established between ions and the unionized exists in solid state as a cluster of positively
molecules. This type of equilibrium involving charged sodium ions and negatively charged
ions in aqueous solution is called ionic chloride ions which are held together due to
equilibrium. Acids, bases and salts come electrostatic interactions between oppositely
under the category of electrolytes and may act charged species (Fig.7.10). The electrostatic
as either strong or weak electrolytes. forces between two charges are inversely
proportional to dielectric constant of the
7.10 ACIDS, BASES AND SALTS medium. Water, a universal solvent, possesses

ed
Acids, bases and salts find widespread a very high dielectric constant of 80. Thus,
occurrence in nature. Hydrochloric acid when sodium chloride is dissolved in water,
present in the gastric juice is secreted by the the electrostatic interactions are reduced by a
lining of our stomach in a significant amount factor of 80 and this facilitates the ions to move
of 1.2-1.5 L/day and is essential for digestive freely in the solution. Also, they are well-

h
processes. Acetic acid is known to be the main separated due to hydration with water

pu T
constituent of vinegar. Lemon and orange molecules.

is
juices contain citric and ascorbic acids, and
tartaric acid is found in tamarind paste. As
re ER
most of the acids taste sour, the word “acid”

bl
has been derived from a latin word “acidus”
meaning sour. Acids are known to turn blue
litmus paper into red and liberate dihydrogen
on reacting with some metals. Similarly, bases
be C

are known to turn red litmus paper blue, taste

bitter and feel soapy. A common example of a
base is washing soda used for washing
N

purposes. When acids and bases are mixed in

the right proportion they react with each other
to give salts. Some commonly known Fig.7.10 Dissolution of sodium chloride in water.
+ –
examples of salts are sodium chloride, barium Na and Cl ions are stablised by their
©

sulphate, sodium nitrate. Sodium chloride hydration with polar water molecules.
(common salt ) is an important component of Comparing, the ionization of hydrochloric
our diet and is formed by reaction between acid with that of acetic acid in water we find
hydrochloric acid and sodium hydroxide. It that though both of them are polar covalent

Faraday w as born near London into a family of very limited means. At the age of 14 he
was an apprentice to a kind bookbinder who allowed Faraday to read the books he
to

was binding. Through a fortunate chance he became laboratory assistant to Davy, and
during 1813-4, Faraday accompanied him to the Continent. During this trip he gained
much from the experience of coming into contact with many of the leading scientists of
the time. In 1825, he succeeded Davy as Director of the Royal Institution laboratories,
t

and in 1833 he also became the first Fullerian Professor of Chemistry. Faraday’s first Michael Faraday
important work was on analytical chemistry. After 1821 much of his work was on (1791–1867)
no

electricity and magnetism and different electromagnetic phenomena. His ideas have led to the establishment
of modern field theory. He discovered his two laws of electrolysis in 1834. Faraday was a very modest
and kind hearted person. He declined all honours and avoided scientific controversies. He preferred to
work alone and never had any assistant. He disseminated science in a variety of ways including his
Friday evening discourses, which he founded at the Royal Institution. He has been very famous for his
Christmas lecture on the ‘Chemical History of a Candle’. He published nearly 450 scientific papers.
 EQUILIBRIUM 207

molecules, former is completely ionized into

its constituent ions, while the latter is only Hydronium and Hydroxyl Ions
partially ionized (< 5%). The extent to which Hydrogen ion by itself is a bare proton with
ionization occurs depends upon the strength very small size (~10–15 m radius) and
of the bond and the extent of solvation of ions intense electric field, binds itself with the
produced. The terms dissociation and water molecule at one of the two available
ionization have earlier been used with different lone pairs on it giving H3 O +. This species
has been detected in many compounds
meaning. Dissociation refers to the process of
(e.g., H3O +Cl –) in the solid state. In aqueous

ed
separation of ions in water already existing as
solution the hydronium ion is further
such in the solid state of the solute, as in hydrated to give species like H5O 2+ , H7O 3+ and
sodium chloride. On the other hand, ionization H9 O4+. Similarly the hydroxyl ion is hydrated
corresponds to a process in which a neutral to give several ionic species like H3O 2–, H5O 3–
molecule splits into charged ions in the

h
and H7 O4– etc.
solution. Here, we shall not distinguish

pu T
between the two and use the two terms

is
interchangeably.
re ER
7.10.1 Arrhenius Concept of Acids and

bl
Bases
+
According to Arrhenius theory, acids are H9O 4
substances that dissociates in water to give
+
hydrogen ions H (aq) and bases are 7.10.2 The Brönsted-Lowry Acids and
be C

substances that produce hydroxyl ions

Bases
OH –(aq). The ionization of an acid HX (aq) can
be represented by the following equations: The Danish chemist, Johannes Brönsted and
the English chemist, Thomas M. Lowry gave a
N

HX (aq) → H+(aq) + X – (aq) more general definition of acids and bases.

or According to Brönsted-Lowry theory, acid is
HX(aq) + H2O(l) → H3O+(aq) + X –(aq) a substance that is capable of donating a
A bare proton, H + is very reactive and hydrogen ion H+ and bases are substances
©

cannot exist freely in aqueous solutions. Thus, capable of accepting a hydrogen ion, H+. In
it bonds to the oxygen atom of a solvent water short, acids are proton donors and bases are
molecule to give trigonal pyramidal proton acceptors.
+ +
hydronium ion, H3O {[H (H2O)] } (see box). Consider the example of dissolution of NH3
In this chapter we shall use H (aq) and H3O+(aq)
+
in H2O represented by the following equation:
interchangeably to mean the same i.e., a
to

hydrated proton.
Similarly, a base molecule like MOH
ionizes in aqueous solution according to the
equation:
t

MOH(aq) → M+(aq) + OH– (aq)

no

The hydroxyl ion also exists in the hydrated

form in the aqueous solution. Arrhenius
concept of acid and base, however, suffers The basic solution is formed due to the
from the limitation of being applicable only to presence of hydroxyl ions. In this reaction,
aqueous solutions and also, does not account water molecule acts as proton donor and
for the basicity of substances like, ammonia ammonia molecule acts as proton acceptor
which do not possess a hydroxyl group. and are thus, called Lowry-Brönsted acid and
208 CHEMISTRY

Arrhenius was born near Uppsala, Sweden. He presented his thesis, on the conductivities
of electrolyte solutions, to the University of Uppsala in 1884. For the next five years he
travelled extensively and visited a number of research centers in Europe. In 1895 he was
appointed professor of physics at the newly formed University of Stockholm, serving its
rector from 1897 to 1902. From 1905 until his death he was Director of physical chemistry
at the Nobel Institute in Stockholm. He continued to work for many years on electrolytic
solutions. In 1899 he discussed the temperature dependence of reaction rates on the
basis of an equation, now usually known as Arrhenius equation.
He worked in a variety of fields, and made important contributions to

ed
immunochemistry, cosmology, the origin of life, and the causes of ice age. He was the
Svante Arrhenius
first to discuss the ‘green house effect’ calling by that name. He received Nobel Prize in
(1859-1927)
Chemistry in 1903 for his theory of electrolytic dissociation and its use in the development
of chemistry.

h
base, respectively. In the reverse reaction, H+ ammonia it acts as an acid by donating a
+ –
is transferred from NH4 to OH . In this case, proton.

pu T
NH4+ acts as a Bronsted acid while OH – acted

is
Problem 7.12
as a Brönsted base. The acid-base pair that
re ER
differs only by one proton is called a conjugate
–
What will be the conjugate bases for the
following Brönsted acids: HF, H2SO4 and

bl
acid-base pair. Therefore, OH is called the
conjugate base of an acid H2O and NH4 is
+ HCO3– ?
called conjugate acid of the base NH3. If Solution
Brönsted acid is a strong acid then its The conjugate bases should have one
conjugate base is a weak base and vice- proton less in each case and therefore the
be C

versa. It may be noted that conjugate acid has corresponding conjugate bases are: F –,
one extra proton and each conjugate base has HSO4– and CO32– respectively.
one less proton.
N

Problem 7.13
Consider the example of ionization of Write the conjugate acids for the following
hydrochloric acid in water. HCl(aq) acts as an –
Brönsted bases: NH2 , NH3 and HCOO–.
acid by donating a proton to H2O molecule
which acts as a base. Solution
©

The conjugate acid should have one extra

proton in each case and therefore the
corresponding conjugate acids are: NH3,
NH 4+ and HCOOH respectively.
Problem 7.14
– –
The species: H 2O, HCO3 , HSO4 and NH3
to

can act both as Bronsted acids and bases.

For each case give the corresponding
It can be seen in the above equation, that conjugate acid and conjugate base.
water acts as a base because it accepts the Solution
proton. The species H3O + is produced when
t

The answer is given in the following Table:

water accepts a proton from HCl. Therefore,
no

– Species Conjugate Conjugate

Cl is a conjugate base of HCl and HCl is the
conjugate acid of base Cl –. Similarly, H 2O is a acid base
+ + –
conjugate base of an acid H3O and H3O is a H 2O H 3O+ OH
conjugate acid of base H2O. HCO3
–
H2CO3 CO2–
3
It is interesting to observe the dual role of HSO4– H2SO4 SO42–
water as an acid and a base. In case of reaction –
NH3 NH4+ NH2
with HCl water acts as a base while in case of
 EQUILIBRIUM 209

7.10.3 Lewis Acids and Bases hydrochloric acid (HCl), hydrobromic acid
G.N. Lewis in 1923 defined an acid as a (HBr), hyrdoiodic acid (HI), nitric acid (HNO3)
species which accepts electron pair and base and sulphuric acid (H2SO 4) are termed strong
which donates an electron pair. As far as because they are almost completely
bases are concerned, there is not much dissociated into their constituent ions in an
difference between Brönsted-Lowry and Lewis aqueous medium, thereby acting as proton
concepts, as the base provides a lone pair in (H+ ) donors. Similarly, strong bases like
both the cases. However, in Lewis concept lithium hydroxide (LiOH), sodium hydroxide

ed
many acids do not have proton. A typical (NaOH), potassium hydroxide (KOH), caesium
example is reaction of electron deficient species hydroxide (CsOH) and barium hydroxide
BF3 with NH3. Ba(OH)2 are almost completely dissociated into
BF3 does not have a proton but still acts ions in an aqueous medium giving hydroxyl
ions, OH – . According to Arrhenius concept

h
as an acid and reacts with NH3 by accepting
its lone pair of electrons. The reaction can be they are strong acids and bases as they are

pu T
+
represented by, able to completely dissociate and produce H3O

is
–
and OH ions respectively in the medium.
BF3 + :NH3 → BF 3:NH3 Alternatively, the strength of an acid or base
re ER
Electron deficient species like AlCl3, Co 3+, may also be gauged in terms of Brönsted-

bl
2+
Mg , etc. can act as Lewis acids while species Lowry concept of acids and bases, wherein a
–
like H2O, NH3, OH etc. which can donate a pair strong acid means a good proton donor and a
of electrons, can act as Lewis bases. strong base implies a good proton acceptor.
Consider, the acid-base dissociation
Problem 7.15
be C

equilibrium of a weak acid HA,

Classify the following species into Lewis HA(aq) + H2O(l) –
⇌ H3O+ (aq) + A (aq)
acids and Lewis bases and show how
conjugate conjugate
N

these act as such:

acid base acid base
(a) HO – (b)F – (c) H + (d) BCl3
In section 7.10.2 we saw that acid (or base)
Solution
dissociation equilibrium is dynamic involving
(a) Hydroxyl ion is a Lewis base as it can a transfer of proton in forward and reverse
©

–
donate an electron lone pair (:OH ). directions. Now, the question arises that if the
(b) Flouride ion acts as a Lewis base as equilibrium is dynamic then with passage of
it can donate any one of its four time which direction is favoured? What is the
electron lone pairs. driving force behind it? In order to answer
(c) A proton is a Lewis acid as it can these questions we shall deal into the issue of
accept a lone pair of electrons from comparing the strengths of the two acids (or
bases like hydroxyl ion and fluoride bases) involved in the dissociation equilibrium.
to

ion. Consider the two acids HA and H3O + present

(d) BCl3 acts as a Lewis acid as it can in the above mentioned acid-dissociation
accept a lone pair of electrons from equilibrium. We have to see which amongst
species like ammonia or amine them is a stronger proton donor. Whichever
molecules.
t

exceeds in its tendency of donating a proton

over the other shall be termed as the stronger
no

7.11 IONIZATION OF ACIDS AND BASES acid and the equilibrium will shift in the
direction of weaker acid. Say, if HA is a
Arrhenius concept of acids and bases becomes stronger acid than H3O + , then HA will donate
useful in case of ionization of acids and bases protons and not H3O+ , and the solution will
as mostly ionizations in chemical and –
mainly contain A and H 3O + ions. The
biological systems occur in aqueous medium. equilibrium moves in the direction of
Strong acids like perchloric acid (HClO4), formation of weaker acid and weaker base
210 CHEMISTRY

because the stronger acid donates a proton H 2O(l) + H2O(l) ⇌ H3O+ (aq) + OH–(aq)
to the stronger base.
acid base conjugate conjugate
It follows that as a strong acid dissociates acid base
completely in water, the resulting base formed
would be very weak i.e., strong acids have The dissociation constant is represented by,
–
very weak conjugate bases. Strong acids like K = [H3O+ ] [OH ] / [H2O] (7.26)
perchloric acid (HClO 4), hydrochloric acid The concentration of water is omitted from
(HCl), hydrobromic acid (HBr), hydroiodic acid the denominator as water is a pure liquid and

ed
(HI), nitric acid (HNO3) and sulphuric acid its concentration remains constant. [H2O] is
(H2SO 4) will give conjugate base ions ClO4– , Cl, incorporated within the equilibrium constant
Br–, I– , NO3– and HSO4– , which are much weaker to give a new constant, Kw, which is called the
bases than H2O. Similarly a very strong base ionic product of water.
would give a very weak conjugate acid. On the

h
K w = [H+][OH– ] (7.27)
other hand, a weak acid say HA is only partially

pu T
+
dissociated in aqueous medium and thus, the The concentration of H has been found

is
solution mainly contains undissociated HA out experimentally as 1.0 × 10 –7 M at 298 K.
And, as dissociation of water produces equal
re ER
molecules. Typical weak acids are nitrous acid
(HNO2), hydrofluoric acid (HF) and acetic acid number of H + and OH – ions, the concentration

bl
(CH3COOH). It should be noted that the weak of hydroxyl ions, [OH – ] = [H+] = 1.0 × 10 –7 M.
acids have very strong conjugate bases. For Thus, the value of K w at 298K,
2– + –
example, NH 2– , O and H – are very good proton K w = [H3O ][OH ] = (1 × 10–7)2 = 1 × 10–14 M 2
acceptors and thus, much stronger bases than (7.28)
be C

H2O.
The value of Kw is temperature dependent
Certain water soluble organic compounds as it is an equilibrium constant.
like phenolphthalein and bromothymol blue
N

The density of pure water is 1000 g / L

behave as weak acids and exhibit different
and its molar mass is 18.0 g /mol. From this
colours in their acid (HIn) and conjugate base
–
(In ) forms. the molarity of pure water can be given as,
+
HIn(aq) + H2O(l) ⇌ H3O (aq) + In (aq)
– [H2O] = (1000 g /L)(1 mol/18.0 g) = 55.55 M.
©

acid conjugate conjugate Therefore, the ratio of dissociated water to that

indicator acid base of undissociated water can be given as:
colour A colourB 10 –7 / (55.55) = 1.8 × 10–9 or ~ 2 in 10–9 (thus,
Such compounds are useful as indicators equilibrium lies mainly towards undissociated
+
in acid-base titrations, and finding out H ion water)
concentration. We can distinguish acidic, neutral and
to

7.11.1 The Ionization Constant of Water basic aqueous solutions by the relative values
and its Ionic Product of the H3O + and OH– concentrations:
–
Some substances like water are unique in their Acidic: [H3O +] > [OH ]
–
ability of acting both as an acid and a base. Neutral: [H3O +] = [OH ]
t

We have seen this in case of water in section –

Basic : [H3O +] < [OH ]
7.10.2. In presence of an acid, HA it accepts a
no

proton and acts as the base while in the 7.11.2 The pH Scale
–
presence of a base, B it acts as an acid by Hydronium ion concentration in molarity is
donating a proton. In pure water, one H2O more conveniently expressed on a logarithmic
molecule donates proton and acts as an acid
scale known as the pH scale. The pH of a
and another water molecules accepts a proton solution is defined as the negative logarithm
and acts as a base at the same time. The
following equilibrium exists: ( )
to base 10 of the activity a H+ of hydrogen
 EQUILIBRIUM 211

ion. In dilute solutions (< 0.01 M), activity of change in pH by just one unit also means
+
hydrogen ion (H ) is equal in magnitude to change in [H+ ] by a factor of 10. Similarly, when
+ the hydrogen ion concentration, [H+] changes
molarity represented by [H ]. It should
be noted that activity has no units and is by a factor of 100, the value of pH changes by
defined as: 2 units. Now you can realise why the change
in pH with temperature is often ignored.
a H+ = [H+ ] / mol L–1
Measurement of pH of a solution is very
From the definition of pH, the following essential as its value should be known when

ed
can be written, dealing with biological and cosmetic
pH = – log aH+ = – log {[H+] / mol L–1} applications. The pH of a solution can be found
roughly with the help of pH paper that has
Thus, an acidic solution of HCl (10–2 M)
different colour in solutions of different pH.
will have a pH = 2. Similarly, a basic solution

h
– –4 + Now-a-days pH paper is available with four
of NaOH having [OH ] =10 M and [H3O ] = strips on it. The different strips have different

pu T
–10
10 M will have a pH = 10. At 25 °C, pure colours (Fig. 7.11) at the same pH. The pH in

is
water has a concentration of hydrogen ions, the range of 1-14 can be determined with an

given as:
re ER
[H +] = 10–7 M. Hence, the pH of pure water is accuracy of ~0.5 using pH paper.

bl
pH = –log(10–7) = 7
Acidic solutions possess a concentration
of hydrogen ions, [H +] > 10 –7 M, while basic
be C

solutions possess a concentration of hydrogen

ions, [H+] < 10 –7 M. thus, we can summarise
Fig.7.11 pH-paper with four strips that may
that have different colours at the same pH
N

Acidic solution has pH < 7

Basic solution has pH > 7 For greater accuracy pH meters are used.
Neutral solution has pH = 7 pH meter is a device that measures the
pH-dependent electrical potential of the test
Now again, consider the equation (7.28) at
©

solution within 0.001 precision. pH meters of

298 K
– the size of a writing pen are now available in
K w = [H3O +] [OH ] = 10–14 the market. The pH of some very common
Taking negative logarithm on both sides substances are given in Table 7.5 (page 212).
of equation, we obtain
– Problem 7.16
–log K w = – log {[H3O+ ] [OH ]}
– The concentration of hydrogen ion in a
to

= – log [H3O +] – log [OH ] sample of soft drink is 3.8 × 10–3M. what
= – log 10 –14 is its pH ?
pKw = pH + pOH = 14 (7.29) Solution
Note that although K w may change with pH = – log[3.8 × 10 –3 ]
t

temperature the variations in pH with = – {log[3.8] + log[10 –3]}

no

temperature are so small that we often = – {(0.58) + (– 3.0)} = – { – 2.42} = 2.42

ignore it.
Therefore, the pH of the soft drink is 2.42
pK w is a very important quantity for
and it can be inferred that it is acidic.
aqueous solutions and controls the relative
concentrations of hydrogen and hydroxyl ions Problem 7.17
–8
as their product is a constant. It should be Calculate pH of a 1.0 × 10 M solution
noted that as the pH scale is logarithmic, a of HCl.
212 CHEMISTRY

Table 7.5 The pH of Some Common Substances

Name of the Fluid pH Name of the Fluid pH

Saturated solution of NaOH ~15 Black Coffee 5.0

0.1 M NaOH solution 13 Tomato juice ~4.2
Lime water 10.5 Soft drinks and vinegar ~3.0
Milk of magnesia 10 Lemon juice ~2.2
Egg white, sea water 7.8 Gastric juice ~1.2

ed
Human blood 7.4 1M HCl solution ~0
Milk 6.8 Concentrated HCl ~–1.0
Human Saliva 6.4

Solution constant for the above discussed acid-

h
– dissociation equilibrium:
2H2O (l) ⇌ H3O + (aq) + OH (aq)
K a = c2α 2 / c(1-α) = cα2 / 1-α

pu T
–
K w = [OH ][H3O+ ]

is
K a is called the dissociation or ionization
= 10–14 constant of acid HX. It can be represented
re ER
–
Let, x = [OH ] = [H3O +] from H2O. The H3O+ alternatively in terms of molar concentration

bl
concentration is generated (i) from as follows,
the ionization of HCl dissolved i.e., + –
+ –
K a = [H ][X ] / [HX] (7.30)
HCl(aq) + H2 O(l) ⇌ H3O (aq) + Cl (aq), At a given temperature T, K a is a
and (ii) from ionization of H2O. In these measure of the strength of the acid HX
be C

very dilute solutions, both sources of i.e., larger the value of K a, the stronger is
H3O + must be considered: the acid. Ka is a dimensionless quantity
[H3O + ] = 10 –8 + x with the understanding that the standard
N

–8 –14 state concentration of all species is 1M.

K w = (10 + x)(x) = 10
The values of the ionization constants of
or x2 + 10 –8 x – 10–14 = 0
some selected weak acids are given in
[OH – ] = x = 9.5 × 10–8 Table 7.6.
©

So, pOH = 7.02 and pH = 6.98 Table 7.6 The Ionization Constants of Some
Selected Weak Acids (at 298K)
7.11.3 Ionization Constants of Weak Acids
Acid Ionization Constant,
Consider a weak acid HX that is partially
Ka
ionized in the aqueous solution. The
equilibrium can be expressed by: Hydrofluoric Acid (HF) 3.5 × 10–4
to

–4
HX(aq) + H2O(l) ⇌ H3O+(aq) + X –(aq) Nitrous Acid (HNO2) 4.5 × 10
–4
Initial Formic Acid (HCOOH) 1.8 × 10
concentration (M) Niacin (C5H 4NCOOH) 1.5 × 10
–5

c 0 0 Acetic Acid (CH3COOH) 1.74 × 10

–5

Let α be the extent of ionization

t

–5
Benzoic Acid (C6H5COOH) 6.5 × 10
Change (M)
no

–8
-cα +cα +cα Hypochlorous Acid (HCIO) 3.0 × 10
Equilibrium concentration (M) Hydrocyanic Acid (HCN) 4.9 × 10–10
c-cα cα cα Phenol (C6H5OH) 1.3 × 10
–10

Here, c = initial concentration of the

undissociated acid, HX at time, t = 0. α = extent The pH scale for the hydrogen ion
up to which HX is ionized into ions. Using concentration has been so useful that besides
these notations, we can derive the equilibrium pKw , it has been extended to other species and
 EQUILIBRIUM 213

quantities. Thus, we have: Solution

pKa = –log (Ka ) (7.31) The following proton transfer reactions are
Knowing the ionization constant, Ka of an possible:
acid and its initial concentration, c, it is 1) HF + H2O ⇌ H 3O+ + F–
possible to calculate the equilibrium Ka = 3.2 × 10–4
concentration of all species and also the degree
2) H2O + H 2O ⇌ H3O + OH –
+
of ionization of the acid and the pH of the
solution. Kw = 1.0 × 10–14

ed
As Ka >> Kw, [1] is the principle reaction.
A general step-wise approach can be
HF + H2O ⇌ H3O+ + F–
adopted to evaluate the pH of the weak
Initial
electrolyte as follows:
concentration (M)
Step 1. The species present before dissociation

h
0.02 0 0
are identified as Brönsted-Lowry Change (M)

pu T
acids / bases. –0.02α +0.02α +0.02α

is
Step 2. Balanced equations for all possible Equilibrium
re ER
reactions i.e., with a species acting both as
acid as well as base are written.
concentration (M)
0.02 – 0.02 α 0.02 α 0.02α

bl
Step 3. The reaction with the higher Ka is Substituting equilibrium concentrations
identified as the primary reaction whilst the in the equilibrium reaction for principal
other is a subsidiary reaction. reaction gives:
Ka = (0.02α) 2 / (0.02 – 0.02α)
be C

Step 4. Enlist in a tabular form the following

values for each of the species in the primary = 0.02 α 2 / (1 –α) = 3.2 × 10–4
reaction We obtain the following quadratic
N

(a) Initial concentration, c. equation:

(b) Change in concentration on proceeding to α2 + 1.6 × 10–2α – 1.6 × 10–2 = 0
equilibrium in terms of α, degree of The quadratic equation in α can be solved
ionization. and the two values of the roots are:
©

(c) Equilibrium concentration. α = + 0.12 and – 0.12

Step 5. Substitute equilibrium concentrations The negative root is not acceptable and
into equilibrium constant equation for hence,
principal reaction and solve for α. α = 0.12
Step 6. Calculate the concentration of species This means that the degree of ionization,
in principal reaction. α = 0.12, then equilibrium concentrations
to

of other species viz., HF, F – and H3O+ are

Step 7. Calculate pH = – log[H3O +]
given by:
The above mentioned methodology has –
[H3O+ ] = [F ] = cα = 0.02 × 0.12
been elucidated in the following examples.
= 2.4 × 10–3 M
t

Problem 7.18 [HF] = c(1 – α) = 0.02 (1 – 0.12)

no

The ionization constant of HF is = 17.6 × 10-3 M

3.2 × 10 –4. Calculate the degree of pH = – log[H+] = –log(2.4 × 10–3) = 2.62
dissociation of HF in its 0.02 M solution.
Calculate the concentration of all species Problem 7.19
present (H3O +, F – and HF) in the solution The pH of 0.1M monobasic acid is 4.50.
+
and its pH. Calculate the concentration of species H ,
214 CHEMISTRY

A– and HA at equilibrium. Also, determine Percent dissociation

the value of Ka and pK a of the monobasic = {[HOCl]dissociated / [HOCl]initial }× 100
acid.
= 1.41 × 10 –3 × 102/ 0.08 = 1.76 %.
Solution
+ pH = –log(1.41 × 10–3) = 2.85.
pH = – log [H ]
Therefore, [H+ ] = 10 –pH = 10 –4.50 7.11.4 Ionization of Weak Bases
= 3.16 × 10 –5 The ionization of base MOH can be represented

ed
+ –
[H ] = [A ] = 3.16 × 10 –5 by equation:

Thus, Ka = [H+ ][A- ] / [HA] MOH(aq) ⇌ M + (aq) + OH– (aq)

In a weak base there is partial ionization
[HA]eqlbm = 0.1 – (3.16 × 10-5) ∫ 0.1
of MOH into M + and OH– , the case is similar to

h
K a = (3.16 × 10–5)2 / 0.1 = 1.0 × 10–8 that of acid-dissociation equilibrium. The

pu T
pKa = – log(10–8) = 8 equilibrium constant for base ionization is

is
called base ionization constant and is
Alternatively, “Percent dissociation” is
re ER
another useful method for measure of
represented by K b. It can be expressed in terms
of concentration in molarity of various species

bl
strength of a weak acid and is given as:
in equilibrium by the following equation:
Percent dissociation K b = [M+ ][OH–] / [MOH] (7.33)
= [HA]dissociated/[HA]initial × 100% (7.32) Alternatively, if c = initial concentration of
base and α = degree of ionization of base i.e.
be C

Problem 7.20
the extent to which the base ionizes. When
Calculate the pH of 0.08M solution of equilibrium is reached, the equilibrium
hypochlorous acid, HOCl. The ionization constant can be written as:
N

constant of the acid is 2.5 × 10 –5 .

K b = (cα)2 / c (1-α) = cα2 / (1-α)
Determine the percent dissociation of
HOCl. The values of the ionization constants of
some selected weak bases, K b are given in
Solution
©

Table 7.7.
HOCl(aq) + H2O (l) ⇌ H 3O +(aq) + ClO–(aq)
Table 7.7 The Values of the Ionization
Initial concentration (M) Constant of Some Weak Bases at
0.08 0 0 298 K
Change to reach Base Kb
equilibrium concentration
–4
(M) Dimethylamine, (CH3 )2NH 5.4 × 10
to

–5
–x +x +x Triethylamine, (C2H 5)3N 6.45 × 10
–5
equilibrium concentartion (M) Ammonia, NH3 or NH4OH 1.77 × 10
–6
Quinine, (A plant product) 1.10 × 10
0.08 – x x x
Pyridine, C 5H5N 1.77 × 10–9
K a = {[H3O+ ][ClO– ] / [HOCl]}
t

–10
= x2 / (0.08 –x) Aniline, C6H 5NH2 4.27 × 10
no

–14
As x << 0.08, therefore 0.08 – x ∫ 0.08 Urea, CO (NH 2)2 1.3 × 10

x2 / 0.08 = 2.5 × 10–5

Many organic compounds like amines are
x2 = 2.0 × 10–6, thus, x = 1.41 × 10–3 weak bases. Amines are derivatives of
[H+] = 1.41 × 10–3 M. ammonia in which one or more hydrogen
Therefore, atoms are replaced by another group. For
example, methylamine, codeine, quinine and
 EQUILIBRIUM 215

nicotine all behave as very weak bases due to Kb = 10–4.75 = 1.77 × 10–5 M
their very small Kb. Ammonia produces OH– + –
in aqueous solution: NH3 + H 2O ⇌ NH4 + OH

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH– (aq) Initial concentration (M)

0.10 0.20 0
The pH scale for the hydrogen ion
concentration has been extended to get: Change to reach
pKb = –log (Kb ) (7.34) equilibrium (M)

ed
–x +x +x
Problem 7.21 At equilibrium (M)
The pH of 0.004M hydrazine solution is 0.10 – x 0.20 + x x
9.7. Calculate its ionization constant K b + –
and pKb. Kb = [NH4 ][OH ] / [NH3]

h
= (0.20 + x)(x) / (0.1 – x) = 1.77 × 10–5
Solution

pu T
As Kb is small, we can neglect x in

is
–
NH2NH2 + H2O ⇌ NH2NH3+ + OH comparison to 0.1M and 0.2M. Thus,
re ER
From the pH we can calculate the
hydrogen ion concentration. Knowing
–
[OH ] = x = 0.88 × 10–5

bl
hydrogen ion concentration and the ionic Therefore, [H+ ] = 1.12 × 10–9
product of water we can calculate the pH = – log[H+ ] = 8.95.
concentration of hydroxyl ions. Thus we
have: 7.11.5 Relation between K a and K b
be C

[H+ ] = antilog (–pH) As seen earlier in this chapter, K a and K b

= antilog (–9.7) = 1.67 ×10–10 represent the strength of an acid and a base,
respectively. In case of a conjugate acid-base
N

[OH–] = K w / [H+ ] = 1 × 10–14 / 1.67 × 10–10

pair, they are related in a simple manner so
= 5.98 × 10–5
that if one is known, the other can be deduced.
The concentration of the corresponding Considering the example of NH4+ and NH3
hydrazinium ion is also the same as that we see,
©

of hydroxyl ion. The concentration of both NH4+(aq) + H2O(l) ⇌ H3O +(aq) + NH3(aq)
these ions is very small so the
Ka = [H3O+ ][ NH3] / [NH4+ ] = 5.6 × 10–10
concentration of the undissociated base
can be taken equal to 0.004M. NH3(aq) + H 2O(l) ⇌ NH4+ (aq) + OH –(aq)
Thus, Kb =[ NH4+ ][ OH – ] / NH3 = 1.8 × 10–5
K b = [NH 2NH3+][OH–] / [NH 2NH2] Net: 2 H2O(l) ⇌ H3O+ (aq) + OH – (aq)
to

= (5.98 × 10–5)2 / 0.004 = 8.96 × 10–7 Kw = [H3O +][ OH – ] = 1.0 × 10–14 M

+
pKb = –logK b = –log(8.96 × 10–7) = 6.04. Where, K a represents the strength of NH4 as
an acid and K b represents the strength of NH3
Problem 7.22 as a base.
t

Calculate the pH of the solution in which It can be seen from the net reaction that
no

0.2M NH4Cl and 0.1M NH3 are present. The the equilibrium constant is equal to the
pK b of ammonia solution is 4.75. product of equilibrium constants K a and K b
Solution for the reactions added. Thus,
NH 3 + H2O ⇌ NH4+ + OH – K a × K b = {[H3O +][ NH3] / [NH4+ ]} × {[NH4+ ]
–
The ionization constant of NH3, [ OH ] / [NH3]}

K b = antilog (–pKb) i.e. = [H3O+ ][ OH–] = Kw

= (5.6x10–10) × (1.8 × 10–5) = 1.0 × 10–14 M
216 CHEMISTRY

This can be extended to make a NH 3 + H2O ⇌ NH4 + OH

+ –

generalisation. The equilibrium constant for

We use equation (7.33) to calculate
a net reaction obtained after adding two
(or more) reactions equals the product of hydroxyl ion concentration,
the equilibrium constants for individual [OH – ] = c α = 0.05 α
reactions: K b = 0.05 α 2 / (1 – α)
K NET = K1 × K2 × …… (3.35) The value of α is small, therefore the
Similarly, in case of a conjugate acid-base

ed
quadratic equation can be simplified by
pair, neglecting α in comparison to 1 in the
Ka × Kb = Kw (7.36) denominator on right hand side of the
Knowing one, the other can be obtained. It equation,
should be noted that a strong acid will have

h
Thus,
a weak conjugate base and vice-versa.
K b = c α 2 or α = √ (1.77 × 10–5 / 0.05)

pu T
Alternatively, the above expression

is
= 0.018.
K w = K a × K b, can also be obtained by
re ER
considering the base-dissociation equilibrium
–
[OH ] = c α = 0.05 × 0.018 = 9.4 × 10–4M.
reaction: [H+ ] = Kw / [OH– ] = 10–14 / (9.4 × 10–4)

bl
B(aq) + H2O(l) ⇌ BH+ (aq) + OH – (aq) = 1.06 × 10 –11
+ –
K b = [BH ][OH ] / [B] pH = –log(1.06 × 10–11) = 10.97.
As the concentration of water remains Now, using the relation for conjugate
be C

constant it has been omitted from the

acid-base pair,
denominator and incorporated within the
dissociation constant. Then multiplying and K a × K b = Kw
N

dividing the above expression by [H+], we get: using the value of K b of NH 3 from
+ – + + Table 7.7.
Kb = [BH ][OH ][H ] / [B][H ]
– + + +
={[ OH ][H ]}{[BH ] / [B][H ]} We can determine the concentration of
conjugate acid NH4+
©

= Kw / Ka
or K a × K b = K w K a = K w / Kb = 10–14 / 1.77 × 10–5
It may be noted that if we take negative = 5.64 × 10–10.
logarithm of both sides of the equation, then
pK values of the conjugate acid and base are 7.11.6 Di- and Polybasic Acids and Di- and
related to each other by the equation: Polyacidic Bases
pK a + pK b = pK w = 14 (at 298K) Some of the acids like oxalic acid, sulphuric
to

acid and phosphoric acids have more than one

Problem 7.23 ionizable proton per molecule of the acid.
Such acids are known as polybasic or
Determine the degree of ionization and pH
polyprotic acids.
of a 0.05M of ammonia solution. The The ionization reactions for example for a
t

ionization constant of ammonia can be dibasic acid H 2X are represented by the

no

taken from Table 7.7. Also, calculate the equations:

+ –
ionization constant of the conjugate acid H 2X(aq) ⇌ H (aq) + HX (aq)
of ammonia. – + 2–
HX (aq) ⇌ H (aq) + X (aq)
Solution And the corresponding equilibrium
The ionization of NH 3 in water is constants are given below:
represented by equation: K a1 = {[H+ ][HX–]} / [H2X] and
 EQUILIBRIUM 217

K a 2 = {[H+][X2- ]} / [HX-] In general, when strength of H-A bond

decreases, that is, the energy required to break
Here, K a1 and K a2 are called the first and second the bond decreases, HA becomes a stronger
ionization constants respectively of the acid H2
acid. Also, when the H-A bond becomes more
X. Similarly, for tribasic acids like H3PO4 we polar i.e., the electronegativity difference
have three ionization constants. The values of between the atoms H and A increases and
the ionization constants for some common
there is marked charge separation, cleavage
polyprotic acids are given in Table 7.8. of the bond becomes easier thereby increasing

ed
Table 7.8 The Ionization Constants of Some the acidity.
Common Polyprotic Acids (298K)
But it should be noted that while
comparing elements in the same group of the
periodic table, H-A bond strength is a more
important factor in determining acidity than

h
its polar nature. As the size of A increases

pu T
down the group, H-A bond strength decreases

is
and so the acid strength increases. For
re ER example,
Size increases

bl
It can be seen that higher order ionization HF << HCl << HBr << HI
( )
constants K a2 , K a3 are smaller than the
Acid strength increases
lower order ionization constant ( K a1 ) of a Similarly, H2S is stronger acid than H2O.
be C

polyprotic acid. The reason for this is that it is But, when we discuss elements in the same
more difficult to remove a positively charged row of the periodic table, H-A bond polarity
proton from a negative ion due to electrostatic becomes the deciding factor for determining
N

forces. This can be seen in the case of removing the acid strength. As the electronegativity of A
a proton from the uncharged H2CO 3 as
increases, the strength of the acid also
compared from a negatively charged HCO3–.
Similarly, it is more difficult to remove a proton increases. For example,
2– Electronegativity of A increases
from a doubly charged HPO4 anion as
©

–
compared to H2PO4 . CH4 < NH3 < H2O < HF
Polyprotic acid solutions contain a mixture
2–
of acids like H2A, HA– and A in case of a Acid strength increases
diprotic acid. H2A being a strong acid, the 7.11.8 Common Ion Effect in the
primary reaction involves the dissociation of Ionization of Acids and Bases
H2 A, and H3O + in the solution comes mainly Consider an example of acetic acid dissociation
from the first dissociation step.
to

equilibrium represented as:

7.11.7 Factors Affecting Acid Strength CH3COOH(aq) ⇌ H+ (aq) + CH3COO– (aq)
Having discussed quantitatively the strengths or HAc(aq) ⇌ H + (aq) + Ac– (aq)
of acids and bases, we come to a stage where
we can calculate the pH of a given acid Ka = [H+ ][Ac– ] / [HAc]
t

solution. But, the curiosity rises about why Addition of acetate ions to an acetic acid
no

should some acids be stronger than others? solution results in decreasing the
What factors are responsible for making them concentration of hydrogen ions, [H+ ]. Also, if
stronger? The answer lies in its being a H+ ions are added from an external source then
complex phenomenon. But, broadly speaking the equilibrium moves in the direction of
we can say that the extent of dissociation of undissociated acetic acid i.e., in a direction of
an acid depends on the strength and polarity reducing the concentration of hydrogen ions,
of the H-A bond. [H+]. This phenomenon is an example of
218 CHEMISTRY

common ion effect. It can be defined as a

Thus, x = 1.33 × 10–3 = [OH–]
shift in equilibrium on adding a substance
that provides more of an ionic species already Therefore,[H +] = K w / [OH–] = 10–14 /
present in the dissociation equilibrium. Thus, (1.33 × 10–3) = 7.51 × 10–12
we can say that common ion effect is a pH = –log(7.5 × 10–12) = 11.12
phenomenon based on the Le Chatelier’s On addition of 25 mL of 0.1M HCl
principle discussed in section 7.8. solution (i.e., 2.5 mmol of HCl) to 50 mL
In order to evaluate the pH of the solution of 0.1M ammonia solution (i.e., 5 mmol

ed
resulting on addition of 0.05M acetate ion to of NH3), 2.5 mmol of ammonia molecules
0.05M acetic acid solution, we shall consider are neutralized. The resulting 75 mL
the acetic acid dissociation equilibrium once solution contains the remaining
again, unneutralized 2.5 mmol of NH3 molecules
and 2.5 mmol of NH4+.

h
–
HAc(aq) ⇌ H+ (aq) + Ac (aq)
→ NH4+ + Cl–

pu T
Initial concentration (M) NH 3 + HCl

is
0.05
re ER 0 0.05 2.5 2.5 0 0
Let x be the extent of ionization of acetic At equilibrium
acid. 0 0 2.5 2.5

bl
Change in concentration (M) The resulting 75 mL of solution contains
+
–x +x +x 2.5 mmol of NH 4 ions (i.e., 0.033 M) and
Equilibrium concentration (M) 2.5 mmol (i.e., 0.033 M ) of uneutralised
NH 3 molecules. This NH3 exists in the
be C

0.05-x x 0.05+x following equilibrium:

Therefore,
–
NH 4OH ⇌ NH4+ + OH –
K a= [H ][Ac ]/[H Ac] = {(0.05+x)(x)}/(0.05-x)
+
N

0.033M – y y y
As K a is small for a very weak acid, x<<0.05. – +
where, y = [OH ] = [NH 4 ]
Hence, (0.05 + x) ≈ (0.05 – x) ≈ 0.05
The final 75 mL solution after
Thus, neutralisation already contains
©

1.8 × 10–5 = (x) (0.05 + x) / (0.05 – x) 2.5 m mol NH 4+ ions (i.e. 0.033M), thus
= x(0.05) / (0.05) = x = [H+ ] = 1.8 × 10–5M total concentration of NH4+ ions is given as:
pH = – log(1.8 × 10–5) = 4.74 [NH 4+] = 0.033 + y
As y is small, [NH 4OH] ∫ 0.033 M and
Problem 7.24 [NH 4+] ∫ 0.033M.
Calculate the pH of a 0.10M ammonia We know,
to

solution. Calculate the pH after 50.0 mL

K b = [NH4+][OH– ] / [NH4OH]
of this solution is treated with 25.0 mL of
0.10M HCl. The dissociation constant of = y(0.033)/(0.033) = 1.77 × 10 –5 M
–
ammonia, K b = 1.77 × 10–5 Thus, y = 1.77 × 10–5 = [OH ]
[H+ ] = 10–14 / 1.77 × 10–5 = 0.56 × 10–9
t

Solution
→
no

NH 3 + H 2O NH4+ + OH– Hence, pH = 9.24

+ – –5
K b = [NH4 ][OH ] / [NH3] = 1.77 × 10
Before neutralization, 7.11.9 Hydrolysis of Salts and the pH of
their Solutions
[NH4+ ] = [OH–] = x
Salts formed by the reactions between acids
[NH3] = 0.10 – x ∫ 0.10
and bases in definite proportions, undergo
x2 / 0.10 = 1.77 × 10–5 ionization in water. The cations/anions formed
 EQUILIBRIUM 219

on ionization of salts either exist as hydrated increased of H+ ion concentration in solution

ions in aqueous solutions or interact with making the solution acidic. Thus, the pH of
water to reform corresponding acids/bases NH4Cl solution in water is less than 7.
depending upon the nature of salts. The later Consider the hydrolysis of CH3COONH4 salt
process of interaction between water and formed from weak acid and weak base. The
cations/anions or both of salts is called ions formed undergo hydrolysis as follow:
hydrolysis. The pH of the solution gets affected CH3COO– + NH4+ + H2O CH3COOH +
+ +
by this interaction. The cations (e.g., Na , K , NH4OH

ed
2+ 2+
Ca , Ba , etc.) of strong bases and anions
– – – –
(e.g., Cl , Br , NO3, ClO4 etc.) of strong acids CH3COOH and NH4OH, also remain into
simply get hydrated but do not hydrolyse, and partially dissociated form:
therefore the solutions of salts formed from CH3COOH CH3COO – + H+
strong acids and bases are neutral i.e., their NH4OH NH4+ + OH –

h
pH is 7. However, the other category of salts H2O H+ + OH –

pu T
do undergo hydrolysis.

is
Without going into detailed calculation, it
We now consider the hydrolysis of the salts can be said that degree of hydrolysis is
re ER
of the following types : independent of concentration of solution, and

bl
(i) salts of weak acid and strong base e.g., pH of such solutions is determined by their pK
CH3COONa. values:
(ii) salts of strong acid and weak base e.g., pH = 7 + ½ (pK a – pKb) (7.38)
NH4Cl, and The pH of solution can be greater than 7,
be C

(iii) salts of weak acid and weak base, e.g., if the difference is positive and it will be less
CH3COONH4. than 7, if the difference is negative.
In the first case, CH3COONa being a salt of
N

Problem 7.25
weak acid, CH3COOH and strong base, NaOH
gets completely ionised in aqueous solution. The pK a of acetic acid and pK b of
CH3COONa(aq) → CH 3COO (aq)+ Na (aq)
– + ammonium hydroxide are 4.76 and 4.75
respectively. Calculate the pH of
Acetate ion thus formed undergoes
©

ammonium acetate solution.

hydrolysis in water to give acetic acid and OH–
ions Solution

CH3COO –(aq)+H2O(l) ⇌ CH3COOH(aq)+OH– (aq) pH = 7 + ½ [pK a – pK b]

Acetic acid being a weak acid = 7 + ½ [4.76 – 4.75]
(K a = 1.8 × 10–5) remains mainly unionised in = 7 + ½ [0.01] = 7 + 0.005 = 7.005
–
solution. This results in increase of OH ion
to

concentration in solution making it alkaline. 7.12 BUFFER SOLUTIONS

The pH of such a solution is more than 7. Many body fluids e.g., blood or urine have
Similarly, NH4Cl formed from weak base, definite pH and any deviation in their pH
NH4 OH and strong acid, HCl, in water indicates malfunctioning of the body. The
t

dissociates completely. control of pH is also very important in many

no

NH4Cl(aq) → NH+4 (aq) +Cl– (aq) chemical and biochemical processes. Many
Ammonium ions undergo hydrolysis with medical and cosmetic formulations require
water to form NH4OH and H+ ions that these be kept and administered at a
NH4 (aq ) + H2 O ( l ) ⇌ NH4 OH ( aq ) + H ( aq )
+ +
particular pH. The solutions which resist
Ammonium hydroxide is a weak base change in pH on dilution or with the
(Kb = 1.77 × 10–5) and therefore remains almost addition of small amounts of acid or alkali
unionised in solution. This results in are called Buffer Solutions. Buffer solutions
220 CHEMISTRY

of known pH can be prepared from the We shall now consider the equilibrium
knowledge of pK a of the acid or pK b of base between the sparingly soluble ionic salt and
and by controlling the ratio of the salt and acid its saturated aqueous solution.
or salt and base. A mixture of acetic acid and
7.13.1 Solubility Product Constant
sodium acetate acts as buffer solution around
pH 4.75 and a mixture of ammonium chloride Let us now have a solid like barium sulphate
and ammonium hydroxide acts as a buffer in contact with its saturated aqueous solution.
around pH 9.25. You will learn more about The equilibrium between the undisolved solid
and the ions in a saturated solution can be

ed
buffer solutions in higher classes.
represented by the equation:
7.13 SOLUBILITY EQUILIBRIA OF
SPARINGLY SOLUBLE SALTS BaSO 4(s) Ba2+ (aq) + SO42–(aq),
We have already known that the solubility of

h
The equilibrium constant is given by the
ionic solids in water varies a great deal. Some equation:

pu T
of these (like calcium chloride) are so soluble
K = {[Ba2+ ][SO2–]} / [BaSO4]

is
that they are hygroscopic in nature and even 4

absorb water vapour from atmosphere. Others For a pure solid substance the
re ER
(such as lithium fluoride) have so little concentration remains constant and we can

bl
solubility that they are commonly termed as write
insoluble. The solubility depends on a number K sp = K[BaSO4] = [Ba2+][SO42–] (7.39)
of factors important amongst which are the We call Ksp the solubility product constant
lattice enthalpy of the salt and the solvation or simply solubility product. The experimental
be C

enthalpy of the ions in a solution. For a salt to value of K sp in above equation at 298K is
dissolve in a solvent the strong forces of 1.1 × 10 –10. This means that for solid barium
attraction between its ions (lattice enthalpy) sulphate in equilibrium with its saturated
N

must be overcome by the ion-solvent solution, the product of the concentrations of

interactions. The solvation enthalpy of ions is barium and sulphate ions is equal to its
referred to in terms of solvation which is always solubility product constant. The
negative i.e. energy is released in the process concentrations of the two ions will be equal to
of solvation. The amount of solvation enthalpy
©

the molar solubility of the barium sulphate. If

depends on the nature of the solvent. In case molar solubility is S, then
of a non-polar (covalent) solvent, solvation
1.1 × 10 –10 = (S)(S) = S2
enthalpy is small and hence, not sufficient to
overcome lattice enthalpy of the salt. or S = 1.05 × 10–5.
Consequently, the salt does not dissolve in Thus, molar solubility of barium sulphate
non-polar solvent. As a general rule , for a salt will be equal to 1.05 × 10 –5 mol L–1.
to be able to dissolve in a particular solvent
to

A salt may give on dissociation two or more

its solvation enthalpy must be greater than its than two anions and cations carrying different
lattice enthalpy so that the latter may be charges. For example, consider a salt like
overcome by former. Each salt has its zirconium phosphate of molecular formula
characteristic solubility which depends on 4+ 3–
(Zr ) 3(PO4 )4. It dissociates into 3 zirconium
t

temperature. We classify salts on the basis of cations of charge +4 and 4 phosphate anions
no

their solubility in the following three categories. of charge –3. If the molar solubility of
Category I Soluble Solubility > 0.1M zirconium phosphate is S, then it can be seen
from the stoichiometry of the compound that
Category II Slightly 0.01M<Solubility< 0.1M
Soluble [Zr4+ ] = 3S and [PO43–] = 4S

Category III Sparingly Solubility < 0.01M and K sp = (3S)3 (4S)4 = 6912 (S)7
Soluble or S = {Ksp / (33 × 44)}1/7 = (Ksp / 6912)1/7
 EQUILIBRIUM 221

A solid salt of the general formula Table 7.9 The Solubility Product Constants,
p+ q−
M X y with molar solubility S in equilibrium Ksp of Some Common Ionic Salts at
x
298K.
with its saturated solution may be represented
by the equation:
M xX y (s) ⇌ xM p+(aq) + yXq– (aq)
(where x × p+ = y × q–)
And its solubility product constant is given

ed
by:
K sp = [M p+] x[X q– ] y = (xS)x(yS) y
(7.40) = x x . yy . S(x + y)
S(x + y) = K sp / xx . yy

h
S = (Ksp / xx . y y)1 / x + y (7.41)

pu T
The term K sp in equation is given by Q sp

is
(section 7.6.2) when the concentration of one
or more species is not the concentration under
re ER
equilibrium. Obviously under equilibrium

bl
conditions K sp = Qsp but otherwise it gives the
direction of the processes of precipitation or
dissolution. The solubility product constants
of a number of common salts at 298K are given
be C

in Table 7.9.

Problem 7.26
N

Calculate the solubility of A2X 3 in pure

water, assuming that neither kind of ion
reacts with water. The solubility product
of A 2X 3, K sp = 1.1 × 10–23.
©

Solution
A2X 3 → 2A3+ + 3X2–
K sp = [A3+ ]2 [X2–]3 = 1.1 × 10 –23
If S = solubility of A2X 3, then
[A3+ ] = 2S; [X2–] = 3S
therefore, Ksp = (2S) 2(3S)3 = 108S5
to

= 1.1 × 10–23
thus, S5 = 1 × 10–25
S = 1.0 × 10–5 mol/L.
Problem 7.27
t
no

The values of K sp of two sparingly soluble

salts Ni(OH)2 and AgCN are 2.0 × 10–15
and 6 × 0–17 respectively. Which salt is
more soluble? Explain.
Solution
AgCN ⇌ Ag+ + CN–
222 CHEMISTRY

K sp = [Ag+][CN–] = 6 × 10–17 S mol/L of Ni2+ and 2S mol/L of OH – , but

the total concentration of OH – = (0.10 +
Ni(OH)2 ⇌ Ni2+ + 2OH–
2S) mol/L because the solution already
K sp = [Ni2+ ][OH–]2 = 2 × 10–15 contains 0.10 mol/L of OH – from NaOH.
Let [Ag+ ] = S1, then [CN-] = S1 K sp = 2.0 × 10–15 = [Ni2+ ] [OH–]2
Let [Ni2+ ] = S2, then [OH–] = 2S2 = (S) (0.10 + 2S)2
S12 = 6 × 10–17 , S1 = 7.8 × 10–9 As Ksp is small, 2S << 0.10,

ed
(S2)(2S2)2 = 2 × 10–15, S 2 = 0.58 × 10 –4 thus, (0.10 + 2S) ≈ 0.10
Ni(OH)2 is more soluble than AgCN. Hence,
2.0 × 10–15 = S (0.10)2
7.13.2 Common Ion Effect on Solubility
of Ionic Salts S = 2.0 × 10–13 M = [Ni2+]

h
It is expected from Le Chatelier’s principle that

pu T
The solubility of salts of weak acids like
if we increase the concentration of any one of

is
phosphates increases at lower pH. This is
the ions, it should combine with the ion of its
because at lower pH the concentration of the
re ER
opposite charge and some of the salt will be
precipitated till once again K sp = Qsp. Similarly,
anion decreases due to its protonation. This

bl
in turn increase the solubility of the salt so
if the concentration of one of the ions is
that Ksp = Qsp. We have to satisfy two equilibria
decreased, more salt will dissolve to increase
simultaneously i.e.,
the concentration of both the ions till once
again Ksp = Qsp. This is applicable even to K sp = [M+ ] [X–],
be C

soluble salts like sodium chloride except that

HX ( aq ) ⇌ H+ ( aq ) + X − ( aq ) ;
due to higher concentrations of the ions, we
use their activities instead of their molarities  H+ ( aq )   X − ( aq ) 
N

in the expression for Qsp. Thus if we take a Ka =

saturated solution of sodium chloride and  HX ( aq ) 
pass HCl gas through it, then sodium chloride [X – ] / [HX] = Ka / [H+ ]
is precipitated due to increased concentration
Taking inverse of both side and adding 1
©

(activity) of chloride ion available from the

dissociation of HCl. Sodium chloride thus we get
obtained is of very high purity and we can get [HX ]  H + 
rid of impurities like sodium and magnesium +1 = +1
 X −  Ka
sulphates. The common ion effect is also used
for almost complete precipitation of a particular
ion as its sparingly soluble salt, with very low [HX ] + H −   H+  + K a
=
to

value of solubility product for gravimetric  X −  Ka

estimation. Thus we can precipitate silver ion
as silver chloride, ferric ion as its hydroxide Now, again taking inverse, we get
(or hydrated ferric oxide) and barium ion as – –
its sulphate for quantitative estimations. [X ] / {[X ] + [HX]} = f = Ka / (Ka + [H+ ]) and it
t

can be seen that ‘f ’ decreases as pH decreases.

Problem 7.28
no

If S is the solubility of the salt at a given pH

Calculate the molar solubility of Ni(OH)2 then
in 0.10 M NaOH. The ionic product of K sp = [S] [f S] = S2 {Ka / (Ka + [H+ ])} and
Ni(OH)2 is 2.0 × 10–15. S = {K sp ([H+ ] + K a ) / K a }1/2
Solution0 (7.42)
Let the solubility of Ni(OH)2 be equal to S. Thus solubility S increases with increase
Dissolution of S mol/L of Ni(OH)2 provides
in [H + ] or decrease in pH.
 EQUILIBRIUM 223

SUMMARY

When the number of molecules leaving the liquid to vapour equals the number of
molecules returning to the liquid fr om vapour, equilibrium is said to be attained and is
dynamic in nature. Equilibrium can be established for both physical and chemical
processes and at this stage rate of forward and reverse reactions are equal. Equilibrium
constant, K c is expressed as the concentration of products divided by reactants, each
term raised to the stoichiometric coefficient.

ed
For reaction, a A + b B ⇌ c C +d D
Kc = [C] c[D]d/[A]a [B]b
Equilibrium constant has constant value at a fixed temperature and at this stage
all the macroscopic properties such as concentration, pressure, etc. become constant.
For a gaseous reaction equilibrium constant is expressed as Kp and is written by replacing

h
concentration terms by partial pressures in Kc expression. The direction of reaction can

pu T
be predicted by reaction quotient Q c which is equal to Kc at equilibrium. Le Chatelier’s

is
principle states that the change in any factor such as temperature, pressure,
concentration, etc. will cause the equilibrium to shift in such a direction so as to reduce
re ER
or counteract the effect of the change. It can be used to study the effect of various

bl
factors such as temperature, concentration, pressure, catalyst and inert gases on the
direction of equilibrium and to control the yield of products by controlling these factors.
Catalyst does not effect the equilibrium composition of a reaction mixture but increases
the rate of chemical reaction by making available a new lower energy pathway for
conversion of reactants to products and vice-versa.
be C

All substances that conduct electricity in aqueous solutions are called electrolytes.
Acids, bases and salts are electrolytes and the conduction of electricity by their aqueous
solutions is due to anions and cations produced by the dissociation or ionization of
electrolytes in aqueous solution. The strong electrolytes are completely dissociated. In
N

weak electrolytes there is equilibrium between the ions and the unionized electrolyte
molecules. According to Arrhenius, acids give hydrogen ions while bases produce
hydroxyl ions in their aqueous solutions. Brönsted-Lowry on the other hand, defined
an acid as a proton donor and a base as a proton acceptor. When a Brönsted-Lowry
©

acid reacts with a base, it produces its conjugate base and a conjugate acid corresponding
to the base with which it reacts. Thus a conjugate pair of acid-base differs only by one
proton. Lewis further generalised the definition of an acid as an electron pair acceptor
and a base as an electron pair donor. The expr essions for ionization (equilibrium)
constants of weak acids (Ka ) and weak bases (Kb ) are developed using Arrhenius definition.
The degree of ionization and its dependence on concentration and common ion are
discussed. The pH scale (pH = -log[H+]) for the hydrogen ion concentration (activity) has
–
been introduced and extended to other quantities (pOH = – log[OH ]) ; p Ka = –log[Ka] ;
to

pKb = –log[ Kb ]; and pKw = –log[ Kw] etc.). The ionization of water has been considered and
we note that the equation: pH + pOH = pKw is always satisfied. The salts of strong acid
and weak base, weak acid and strong base, and weak acid and weak base undergo
hydrolysis in aqueous solution.The definition of buffer solutions, and their importance
are discussed briefly. The solubility equilibrium of sparingly soluble salts is discussed
t

and the equilibrium constant is introduced as solubility product constant (Ksp). Its
no

relationship with solubility of the salt is established. The conditions of precipitation of

the salt from their solutions or their dissolution in water are worked out. The role of
common ion and the solubility of sparingly soluble salts is also discussed.
224 CHEMISTRY

SUGGESTED ACTIVITIES FOR STUDENTS REGARDING THIS UNIT

(a) The student may use pH paper in determining the pH of fresh juices of various
vegetables and fruits, soft drinks, body fluids and also that of water samples available.
(b) The pH paper may also be used to determine the pH of different salt solutions and
from that he/she may determine if these are formed from strong/weak acids and
bases.
(c) They may prepare some buffer solutions by mixing the solutions of sodium acetate
and acetic acid and deter mine their pH using pH paper.

ed
(d) They may be provided with different indicators to observe their colours in solutions
of varying pH.
(e) They may perform some acid-base titrations using indicators.
(f) They may observe common ion effect on the solubility of sparingly soluble salts.
(g) If pH meter is available in their school, they may measure the pH with it and compare

h
the r esults obtained with that of the pH paper.

pu T
is
EXERCISES
re ER
bl
7.1 A liquid is in equilibrium with its vapour in a sealed container at a fixed
temperature. The volume of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
be C

c) What happens when equilibrium is restored finally and what will be the final
vapour pressure?
7.2 What is Kc for the following equilibrium when the equilibrium concentration of
N

each substance is: [SO 2]= 0.60M, [O2] = 0.82M and [SO 3] = 1.90M ?
2SO2(g) + O 2(g) ⇌ 2SO3 (g)
7.3 At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40%
©

by volume of I atoms
I2 (g) ⇌ 2I (g)
Calculate Kp for the equilibrium.
7.4 Write the expr ession for the equilibrium constant, Kc for each of the following
reactions:
(i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g)
to

(ii) 2Cu(NO 3)2 (s) ⇌ 2CuO (s) + 4NO2 (g) + O2 (g)

(iii) CH3 COOC2 H5 (aq) + H2O(l) ⇌ CH3 COOH (aq) + C2H 5OH (aq)
(iv) Fe 3+ (aq) + 3OH – (aq) ⇌ Fe(OH)3 (s)

(v) I 2 (s) + 5F2 ⇌ 2IF5

t
no

7.5 Find out the value of Kc for each of the following equilibria from the value of Kp:
(i) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 at 500 K
(ii) CaCO 3 (s) ⇌ CaO(s) + CO2(g); Kp= 167 at 1073 K
7.6 For the following equilibrium, Kc = 6.3 × 1014 at 1000 K
NO (g) + O 3 (g) ⇌ NO 2 (g) + O2 (g)
 EQUILIBRIUM 225

Both the forward and reverse reactions in the equilibrium are elementary
bimolecular reactions. What is Kc, for the reverse reaction?
7.7 Explain why pure liquids and solids can be ignored while writing the equilibrium
constant expression?
7.8 Reaction between N2 and O2– takes place as follows:
2N2 (g) + O2 (g) ⇌ 2N 2O (g)
If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction

ed
vessel and allowed to form N2 O at a temperature for which Kc= 2.0 × 10–37,
determine the composition of equilibrium mixture.
7.9 Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given
below:
2NO (g) + Br2 (g) ⇌ 2NOBr (g)

h
When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at

pu T
constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate

is
equilibrium amount of NO and Br2 .

7.10
re ER
At 450K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium.

bl
2SO2 (g) + O2(g) ⇌ 2SO3 (g)
What is Kc at this temperature ?
7.11 A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the
be C

partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium ?
2HI (g) ⇌ H2 (g) + I 2 (g)
7.12 A mixture of 1.57 mol of N2 , 1.92 mol of H2 and 8.13 mol of NH3 is introduced into
N

a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant,

Kc for the reaction N2 (g) + 3H2 (g) ⇌ 2NH3 (g) is 1.7 × 10 2. Is the reaction mixture
at equilibrium? If not, what is the direction of the net reaction?
7.13 The equilibrium constant expression for a gas reaction is,
©

Kc =
[ NH3 ]4 [O 2 ]5
[ NO ]4 [ H2 O ]6
Write the balanced chemical equation corresponding to this expression.
7.14 One mole of H2O and one mole of CO are taken in 10 L vessel and heated to
725 K. At equilibrium 40% of water (by mass) reacts with CO according to the
to

equation,
H 2O (g) + CO (g) ⇌ H2 (g) + CO2 (g)
Calculate the equilibrium constant for the reaction.
7.15 At 700 K, equilibrium constant for the reaction:
t

H2 (g) + I2 (g) ⇌ 2HI (g)

no

–1
is 54.8. If 0.5 mol L of HI(g) is present at equilibrium at 700 K, what are the
concentration of H2 (g) and I2 (g) assuming that we initially started with HI(g) and
allowed it to reach equilibrium at 700 K?
7.16 What is the equilibrium concentration of each of the substances in the
equilibrium when the initial concentration of ICl was 0.78 M ?
2ICl (g) ⇌ I2 (g) + Cl2 (g); Kc = 0.14
226 CHEMISTRY

7.17 Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium
concentration of C2H 6 when it is placed in a flask at 4.0 atm pressure and allowed
to come to equilibrium?
C2 H6 (g) ⇌ C2 H4 (g) + H2 (g)
7.18 Ethyl acetate is formed by the reaction between ethanol and acetic acid and the
equilibrium is represented as:
CH3COOH (l) + C2H 5 OH (l) ⇌ CH3COOC2H 5 (l) + H 2O (l)

ed
(i) Write the concentration ratio (reaction quotient), Qc , for this reaction (note:
water is not in excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol,
there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate
the equilibrium constant.

h
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining

pu T
it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium

is
been reached?
7.19 A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After
re ER
equilibrium was attained, concentration of PCl 5 was found to be

bl
0.5 × 10–1 mol L –1. If value of Kc is 8.3 × 10 –3, what are the concentrations of PCl3
and Cl2 at equilibrium?
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)
7.20 One of the reaction that takes place in producing steel from iron ore is the
reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2 .
be C

FeO (s) + CO (g) ⇌ Fe (s) + CO2 (g); Kp = 0.265 atm at 1050K

What are the equilibrium partial pressures of CO and CO2 at 1050 K if the
N

initial partial pressures are: p CO= 1.4 atm and p CO 2 = 0.80 atm?
7.21 Equilibrium constant, Kc for the reaction
N2 (g) + 3H 2 (g) ⇌ 2NH3 (g) at 500 K is 0.061
©

At a particular time, the analysis shows that composition of the reaction mixture
is 3.0 mol L–1 N2 , 2.0 mol L–1 H2 and 0.5 mol L–1 NH3 . Is the reaction at equilibrium?
If not in which direction does the reaction tend to proceed to reach equilibrium?
7.22 Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches
the equilibrium:
2BrCl (g) ⇌ Br2 (g) + Cl2 (g)
for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of
to

3.3 × 10–3 mol L–1, what is its molar concentration in the mixture at equilibrium?
7.23 At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium
with soild carbon has 90.55% CO by mass
C (s) + CO2 (g) ⇌ 2CO (g)
t

Calculate Kc for this reaction at the above temperature.

no

0
7.24 Calculate a) ∆G and b) the equilibrium constant for the formation of NO2 from
NO and O2 at 298K
NO (g) + ½ O2 (g) ⇌ NO2 (g)
where
∆fG 0 (NO2 ) = 52.0 kJ/mol
 EQUILIBRIUM 227

0
∆fG (NO) = 87.0 kJ/mol
0
∆fG (O2 ) = 0 kJ/mol
7.25 Does the number of moles of reaction products increase, decrease or remain
same when each of the following equilibria is subjected to a decrease in pressure
by increasing the volume?
(a) PCl5 (g) ⇌ PCl3 (g) + Cl2 (g)

(b) CaO (s) + CO2 (g) ⇌ CaCO3 (s)

ed
(c) 3Fe (s) + 4H2 O (g) ⇌ Fe3 O 4 (s) + 4H2 (g)
7.26 Which of the following reactions will get affected by increasing the pressure?
Also, mention whether change will cause the reaction to go into forward or
backward direction.

h
(i) COCl2 (g) ⇌ CO (g) + Cl2 (g)

pu T
(ii) CH4 (g) + 2S2 (g) ⇌ CS 2 (g) + 2H 2S (g)

is
(iii) CO2 (g) + C (s) ⇌ 2CO (g)
(iv)
re ER
2H2 (g) + CO (g) ⇌ CH3 OH (g)

bl
(v) CaCO3 (s) ⇌ CaO (s) + CO2 (g)

(vi) 4 NH3 (g) + 5O2 (g) ⇌ 4NO (g) + 6H2O(g)

7.27 The equilibrium constant for the following reaction is 1.6 ×105 at 102 4K
be C

H2 (g) + Br2(g) ⇌ 2HBr(g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a
sealed container at 1024 K.
N

7.28 Dihydrogen gas is obtained from natural gas by partial oxidation with steam as
per following endothermic reaction:
CH4 (g) + H2O (g) ⇌ CO (g) + 3H2 (g)
©

(a) Write as expression for Kp for the above reaction.

(b) How will the values of Kp and composition of equilibrium mixture be affected
by
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst ?
7.29 Describe the effect of :
to

a) addition of H2
b) addition of CH3OH
c) removal of CO
d) removal of CH3 OH
t

on the equilibrium of the reaction:

no

2H2 (g) + CO (g) ⇌ CH3OH (g)

7.30 At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride,
PCl5 is 8.3 ×10-3. If decomposition is depicted as,
0
PCl5 (g) ⇌ PCl3 (g) + Cl2 (g) ∆r H = 124.0 kJ mol–1
a) write an expression for Kc for the reaction.
b) what is the value of Kc for the reverse reaction at the same temperature ?
228 CHEMISTRY

c) what would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased
(iii) the temperature is increased ?
7.31 Dihydrogen gas used in Haber’s process is produced by reacting methane from
natural gas with high temperature steam. The first stage of two stage reaction
involves the formation of CO and H2. In second stage, CO formed in first stage is
reacted with more steam in water gas shift reaction,
CO (g) + H2O (g) ⇌ CO2 (g) + H2 (g)
If a reaction vessel at 400 °C is charged with an equimolar mixture of CO and

ed
steam such that p CO = p H O = 4.0 bar, what will be the partial pressure of H2 at
2

equilibrium? Kp = 10.1 at 400°C

7.32 Predict which of the following reaction will have appreciable concentration of
reactants and products:

h
a) Cl2 (g) ⇌ 2Cl (g) Kc = 5 ×10–39

pu T
b) Cl2 (g) + 2NO (g) ⇌ 2NOCl (g) Kc = 3.7 × 108

is
c) Cl2 (g) + 2NO2 (g) ⇌ 2NO2 Cl (g) Kc = 1.8

7.33
re ER
The value of Kc for the reaction 3O2 (g) ⇌ 2O3 (g) is 2.0 ×10–50 at 25°C. If the

bl
equilibrium concentration of O 2 in air at 25°C is 1.6 ×10–2, what is the
concentration of O3?
7.34 The reaction, CO(g) + 3H2(g) ⇌ CH4 (g) + H2O(g)
is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol
be C

of H2 and 0.02 mol of H 2O and an unknown amount of CH4 in the flask. Determine
the concentration of CH4 in the mixture. The equilibrium constant, Kc for the
reaction at the given temperature is 3.90.
N

7.35 What is meant by the conjugate acid-base pair? Find the conjugate acid/base
for the following species:
HNO2 , CN–, HClO 4, F –, OH–, CO3 2–, and S2–
+ +
7.36 Which of the followings are Lewis acids? H2O, BF3, H , and NH4
©

–
7.37 What will be the conjugate bases for the Brönsted acids: HF , H2SO4 and HCO 3 ?
– –
7.38 Write the conjugate acids for the following Brönsted bases: NH2 , NH3 and HCOO .
– –
7.39 The species: H2 O, HCO3 , HSO4 and NH 3 can act both as Brönsted acids and
bases. For each case give the corresponding conjugate acid and base.
7.40 Classify the following species into Lewis acids and Lewis bases and show how
– –
these act as Lewis acid/base: (a) OH (b) F (c) H+ (d) BCl3 .
to

7.41 The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. what
is its pH?
7.42 The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen
ion in it.
7.43 The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4 ,
t

1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the
no

corresponding conjugate base.

7.44 The ionization constant of phenol is 1.0 × 10–10. What is the concentration of
phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization
if the solution is also 0.01M in sodium phenolate?
7.45 The first ionization constant of H2 S is 9.1 × 10–8. Calculate the concentration of
–
HS ion in its 0.1M solution. How will this concentration be affected if the solution
 EQUILIBRIUM 229

is 0.1M in HCl also ? If the second dissociation constant of H2 S is

1.2 × 10–13, calculate the concentration of S 2– under both conditions.
7.46 The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of
dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of
acetate ion in the solution and its pH.
7.47 It has been found that the pH of a 0.01M solution of an organic acid is 4.15.
Calculate the concentration of the anion, the ionization constant of the acid
and its pKa.

ed
7.48 Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH
7.49 Calculate the pH of the following solutions:
a) 2 g of TlOH dissolved in water to give 2 litre of solution.

h
b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution.

pu T
c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.

is
d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.
7.50
re ER
The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate
the pH of the solution and the pKa of bromoacetic acid.

bl
7.51 The pH of 0.005M codeine (C18H 21NO3 ) solution is 9.95. Calculate its ionization
constant and pKb.
7.52 What is the pH of 0.001M aniline solution ? The ionization constant of aniline
can be taken from Table 7.7. Calculate the degree of ionization of aniline in the
be C

solution. Also calculate the ionization constant of the conjugate acid of aniline.
7.53 Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74.
How is the degree of dissociation affected when its solution also contains
N

(a) 0.01M (b) 0.1M in HCl ?

7.54 The ionization constant of dimethylamine is 5.4 × 10–4 . Calculate its degree of
ionization in its 0.02M solution. What percentage of dimethylamine is ionized if
the solution is also 0.1M in NaOH?
©

7.55 Calculate the hydrogen ion concentration in the following biological fluids whose
pH are given below:
(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2
(c) Human blood, 7.38 (d) Human saliva, 6.4.
7.56 The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8,
5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion
concentration in each.
to

7.57 If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K.

Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What
is its pH?
7.58 The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the
t

concentrations of strontium and hydroxyl ions and the pH of the solution.

The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of
no

7.59
ionization of the acid in its 0.05M solution and also its pH. What will be its
degree of ionization if the solution is 0.01M in HCl also?
7.60 The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization
constant of the acid and its degree of ionization in the solution.
7.61 The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M
sodium nitrite solution and also its degree of hydrolysis.
230 CHEMISTRY

7.62 A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the

ionization constant of pyridine.
7.63 Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH4 NO3, NaNO2 and KF
7.64 The ionization constant of chloroacetic acid is 1.35 × 10–3 . What will be the pH of
0.1M acid and its 0.1M sodium salt solution?
7.65 Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at
this temperature?

ed
7.66 Calculate the pH of the resultant mixtures:
a) 10 mL of 0.2M Ca(OH) 2 + 25 mL of 0.1M HCl
b) 10 mL of 0.01M H2 SO4 + 10 mL of 0.01M Ca(OH)2
c) 10 mL of 0.1M H 2SO4 + 10 mL of 0.1M KOH

h
7.67 Determine the solubilities of silver chromate, barium chromate, ferric hydroxide,

pu T
lead chloride and mercurous iodide at 298K from their solubility product

is
constants given in Table 7.9. Determine also the molarities of individual ions.
7.68 The solubility product constant of Ag2 CrO4 and AgBr are 1.1 × 10 –12 and
re ER
5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated

bl
solutions.
7.69 Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are
mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate
Ksp = 7.4 × 10–8 ).
The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate
be C

7.70
is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH
3.19 compared to its solubility in pure water?
7.71 What is the maximum concentration of equimolar solutions of ferrous sulphate
N

and sodium sulphide so that when mixed in equal volumes, there is no

precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10–18).
7.72 What is the minimum volume of water required to dissolve 1g of calcium sulphate
at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6).
©

7.73 The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen
sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of
the following: FeSO4 , MnCl2 , ZnCl2 and CdCl2 . in which of these solutions
precipitation will take place?
t to
no
 REDOX RE ACTIONS 25 5

UNIT 8

REDOX REACTIONS

d
he
Wher e ther e is oxidation, ther e is alw ays re duct ion –
Chemist ry is essent ially a study of redox s ystems.

is
After studying this unit you will be
able to Chemistry deals with varieties of matter and change of one

bl
· identify redox reactions as a class
kind of matter into the other. Transformation of matter from
of reactions in which oxidation one kind into another occurs through the various types of
and reduction reactions occur reactions. One important category of such reactions is
pu
simultaneously; Redox Reactions. A number of phenomena, both physical
· de fi ne the terms ox id atio n, as well as biological, are concerned with redox reactions.
red uction , oxid ant ( oxidis ing These reactions find extensive use in pharmaceutical,
be T

agent) and reductant (reducing biological, industrial, metallurgical and agricultural areas.
agent); The importance of these reactions is apparent from the fact
re
o R

· exp lain mec hani sm o f re dox that burning of different types of fuels for obtaining energy
reactions by electron transfer for domestic, transport and other commercial purposes,
process;
electrochemical processes for extraction of highly reactive
tt E

· use the concept of oxidation metals and non-metals, manufacturing of chemical

number to identify oxidant and compounds like caustic soda, operation of dry and wet
reductant in a reaction;
batteries and corrosion of metals fall within the purview of
C

· cl as sif y red ox r eac ti on in to redox processes. Of late, environmental issues like

c omb i nati on ( sy n the s i s) ,
deco mposition, di splace ment
Hydrogen Economy (use of liquid hydrogen as fuel) and
an d d i sp r o po r tio n ati o n development of ‘Ozone Hole’ have started figuring under
no N

reactions; redox phenomenon.

· sug gest a c ompar ativ e or der
among various reductants and
8.1 CLASSICAL IDEA OF REDO X REACTIONS –
oxidants; OXIDATION AND REDUCTION REACTIONS
©

· balance c hemic al equatio ns Originally, the term oxidation was used to describe the
usi ng ( i) o xi dati on n umbe r addition of oxygen to an element or a compound. Because
(ii) half reaction method; of the presence of dioxygen in the atmosphere (~2 0%),
· l earn the co nc ep t o f re do x many elements combine with it and this is the principal
reactions in terms of electrode reason why they commonly occur on the earth in the
processes. form of their oxides. The following reactions represent
oxidation processes according to the limited definition of
oxidation:
2 Mg (s) + O2 (g) ® 2 MgO (s) (8.1)
S (s) + O2 (g) ® SO2 (g) (8.2)
 25 6 CHE MIST RY

In reactions (8.1) and (8.2), the elements been broadened these days to include removal
magnesium and sulphur are oxidised on of oxygen/electronegative element from a
acc ount of addition of oxygen to them. s ub stance or additio n of hy drogen/
Similarly, methane is oxidised owing to the electropositive element to a substance.
addition of oxygen to it. According to the definition given above, the
CH4 (g) + 2O2 (g) ® CO2 (g) + 2H2O (l) (8.3) following are the examp les of reduc tion
processes:
A careful examination of reaction (8.3) in
which hydrogen has been replaced by oxygen 2 HgO (s) 2 Hg (l) + O2 (g) (8.8)

d
prompted chemists to reinterpret oxidation in (removal of oxygen from mercuric oxide )
terms of removal of hydrogen from it and,
2 FeCl3 (aq) + H2 (g) ® 2 FeCl2 (aq) + 2 HCl(aq)

he
therefore, the scope of term oxidation was
(8.9)
broadened to include the removal of hydrogen
from a substance. The following illustration is (removal of electronegative element, chlorine
another reaction where removal of hydrogen from ferric chloride)
can also be cited as an oxidation reaction. CH2 = CH2 (g) + H2 (g) ® H3C – CH3 (g) (8.10)

is
2 H2S(g) + O2 (g) ® 2 S (s) + 2 H2O (l) (8.4) (addition of hydrogen)
As knowledge of chemists grew, it was 2HgCl2 (aq) + SnCl2 (aq) ® Hg2Cl2 (s)+SnCl4 (aq)

bl
natural to extend the term oxidation for (8.11)
reactions similar to (8.1 to 8.4), which do not (addition of mercury to mercuric chloride)
involve oxygen but other electronegative In reaction (8.11) simultaneous oxidation
pu
elements. The oxidation of magnesium with of stannous chloride to stannic chloride is also
fluorine, chlorine and sulphur etc. occurs occ urr ing be c aus e of the ad dition of
according to the following reactions : electronegative element chlorine to it. It was
be T

soon realised that oxidation and reduction

Mg (s) + F2 (g) ® MgF2 (s) (8.5) always occ ur simultaneously (as will be
re

apparent by re-examining all the equations

o R

Mg (s) + Cl2 (g) ® MgCl2 (s) (8.6)

given above), hence, the word “redox” was
Mg (s) + S (s) ® MgS (s) (8.7) coined for this class of chemical reactions.
tt E

Incorporating the reactions (8.5 to 8.7) Problem 8.1

w ithin the f old of oxid ation r e ac tions In the reactions given below, identify the
encouraged chemists to consider not only the s pe cies unde r going oxidation and
C

removal of hydrogen as oxidation, but also the reduction:

re moval of e le ctr op os itive elements as
oxidation. Thus the reaction : (i) H2S (g) + Cl2 (g) ® 2 HCl (g) + S (s)
no N

(ii) 3Fe3O4 (s) + 8 Al (s) ® 9 Fe (s)

2K4 [Fe(CN)6](aq) + H2O2 (aq) ®2K3[Fe(CN)6](aq)
+ 2 KOH (aq) + 4Al2O3 (s)
is interpreted as oxidation due to the removal (iii) 2 Na (s) + H2 (g) ® 2 NaH (s)
©

of electropositive element potassium from Solution

potassium ferrocyanide before it changes to
potassium ferricyanide. To summarise, the (i) H2S is oxid ise d be cause a more
electronegative element, chlorine is added
term “oxidation” is defined as the addition
of oxygen/electronegative element to a to hydrogen (or a more electropositive
element, hydrogen has been removed
s ub stance o r remo val o f hydro gen/
electropositive element from a substance. from S). Chlorine is reduced due to
addition of hydrogen to it.
I n the b e ginning, r e d uc tion w as
(ii) Aluminium is oxidised because
considered as remov al of oxygen from a
compound. However, the term reduction has oxygen is added to it. Ferrous ferric oxide
 REDOX RE ACTIONS 25 7

(Fe3O4) is reduced because oxygen has For convenience, each of the above
been removed from it. processes can be considered as two separate
(iii) With the careful application of the steps, one involving the loss of electrons and
concept of electronegativity only we may the othe r the gain of ele ctr ons. As an
infe r that s od ium is oxidise d and illustration, we may further elaborate one of
these, say, the formation of sodium chloride.
hydrogen is reduced.
Reaction (iii) chosen here prompts us to 2 Na(s) ® 2 Na+(g) + 2e–
think in terms of another way to define Cl2(g) + 2e– ® 2 Cl– (g)

d
redox reactions.
Each of the above steps is called a half
8.2 REDOX REACTIONS IN TERMS OF reaction, which explicitly shows involvement

he
of electrons. Sum of the half reactions gives
ELECTRON TRANSFER REACTIONS
the overall reaction :
We have already learnt that the reactions
2Na(s) + Cl2(g) ® 2NaCl (s) (8.12) 2 Na(s) + Cl2 (g) ® 2 Na+ Cl– (s) or 2 NaCl (s)
4Na(s) + O2(g) ® 2Na2O(s) Reactions 8. 12 to 8.14 suggest that half

is
(8.13)
reactions that involve loss of electrons are
2Na(s) + S(s) ® Na2S(s) (8.14) called oxidation reactions. Similarly, the
are redox reactions because in each of these half reactions that involve gain of electrons

bl
reactions sodium is oxid ised due to the are called reduction reactions. It may not
ad d ition of e ithe r oxyge n or mor e be out of context to mention here that the new
e le c tr one gativ e e le me nt to s od ium. way of defining oxidation and reduction has
pu
Simultaneously, chlorine, oxygen and sulphur b ee n ac hie ve d only by es tablishing a
are reduced because to each of these, the correlation between the behaviour of species
electropositive element sodium has been as per the classical idea and their interplay in
be T

added. From our knowledge of chemical electron-transfer change. In reactions (8.12 to

re
bonding we also know that sodium chloride, 8.14) sodium, which is oxidise d, acts as
o R

sodium oxide and sodium sulphide are ionic a reducing agent because it donates electron
compounds and perhaps better written as to each of the elements interacting with it and
Na+Cl – (s ), (Na+)2O 2– (s), and (Na+)2 S 2– (s ). thus helps in reducing them. Chlorine, oxygen
tt E

Deve lopment of c harges on the spec ies and sulphur are reduced and act as oxidising
produced suggests us to rewrite the reactions agents because these accept electrons from
(8.12 to 8.14) in the following manner : sodium. To summarise, we may mention that
C

Oxidation: Loss of electron(s) by any species.

Reduction: Gain of electron(s) by any species.
no N

Oxidising agent : Acceptor of electron(s).

Reducing agent : Donor of electron(s).
Problem 8.2 Justify that the reaction :
©

2 Na(s) + H2(g) ® 2 NaH (s) is a redox

change.
Solution
Since in the above reaction the compound
formed is an ionic compound, which may
also be repre sented as Na+H– (s), this
suggests that one half reaction in this
process is :
2 Na (s) ® 2 Na+(g) + 2e–
 25 8 CHE MIST RY

and the other half reaction is: At this stage we may investigate the state
H2 (g) + 2e– ® 2 H – (g) of equilibrium for the reaction represented by
equation (8.15). For this purpose, let us place
This splitting of the reaction under a strip of metallic copper in a zinc sulphate
examination into two half re actions solution. No visible reaction is noticed and
automatically reveals that here sodium is
attempt to detect the presence of Cu2+ ions by
oxidis ed and hydr oge n is re duce d, passing H2S gas through the solution to
therefore, the complete reaction is a redox produce the black colour of cupric sulphide,
change. CuS, does not succeed. Cupric sulphide has

d
such a low solubility that this is an extremely
8.2.1 Competitive Electron T rans fer
sensitive test; yet the amount of Cu2+ formed
Reactions

he
cannot be detected. We thus conclude that the
Place a str ip of metallic zinc in an aqueous
state of equilibrium for the reac tion (8.15)
solution of copper nitrate as shown in Fig. 8.1, greatly favours the products over the reactants.
for about one hour. You may notice that the
strip becomes coated with reddish metallic Let us extend electron transfer reaction now

is
copper and the blue colour of the solution to copper metal and silver nitrate solution in
disappears. Formation of Zn2+ ions among the water and arrange a set-up as shown in
products can easily be judged when the blue Fig. 8.2. The solution develops blue colour due
to the formation of Cu2+ ions on account of the

bl
colour of the s olution due to Cu2+ has
disappeare d. If hydrogen sulphide gas is reaction:
passed through the colourless solution
pu
containing Zn2+ ions, appearance of white zinc
sulphide, ZnS can be seen on making the
solution alkaline with ammonia.
be T

The reaction between metallic zinc and the (8.16)

re
aqueous solution of copper nitrate is : 2+
He re, Cu(s ) is oxidised to Cu (aq) and
o R

Zn(s) + Cu2+ (aq) ® Zn2+ (aq) + Cu(s) (8.15) Ag+(aq) is reduced to Ag(s). Equilibrium greatly
In reaction (8.15), zinc has lost electrons favours the products Cu2+ (aq) and Ag(s).
to form Zn2+ and, therefore, zinc is oxidised. By way of contrast, let us also compare the
tt E

Evidently, now if zinc is oxidised, releasing reaction of metallic cobalt place d in nickel
ele ctrons, s omething must be r educ ed, sulphate solution. The reaction that occurs
C

accepting the electrons lost by zinc. Copper here is :

ion is reduced by gaining electrons from the zinc.
Reaction (8.15) may be rewritten as :
no N

(8.17)
©

Fig. 8.1 Redox reaction between zinc and aqueous solution of copper nitrate occurring in a beaker.

C:\Chemistry XI\Unit-8\Unit-8(5)(reprint).pmd 27.7.6, 16.10.6 (reprint)

 REDOX RE ACTIONS 25 9

Fig. 8.2 Redox reaction between copper and aqueous solution of silver nitrate occurring in a beaker.

d
At equilibrium, chemical tests reveal that both However, as we shall see later, the charge
Ni2+(aq) and Co2+(aq) are present at moderate transfer is only partial and is perhaps better

he
concentrations. In this case, neither the described as an electron shift rather than a
reactants [Co(s) and Ni2+(aq)] nor the products complete loss of electron by H and gain by O.
[Co2+(aq) and Ni (s)] are greatly favoured. What has b een said here with respect to
This competition for release of electrons equation (8.18) may be true for a good number
of othe r r e actions inv olv ing c ov alent

is
incidently reminds us of the competition for
release of protons among acids. The similarity compounds. Two such examples of this class
suggests that we might develop a table in of the reactions are:

bl
which metals and their ions are listed on the H2(s) + Cl2(g) ® 2HCl(g) (8.19)
basis of their tendency to release electrons just and,
as we do in the case of acids to indicate the CH 4(g) + 4Cl2(g) ® CCl4(l) + 4HCl(g) (8.20)
pu
strength of the acids. As a matter of fact we
have already made certain comparisons. By In order to keep track of electron shifts in
comparison we have come to know that zinc chemical re actions involving formation of
be T

releases e lectrons to copper and copper covalent compounds, a more practical method
releases electrons to silver and, therefore, the of using o xidation numb er has b ee n
re

electron releasing tendency of the metals is in dev eloped . In this method , it is alw ays
o R

the order: Zn>Cu>Ag. We would love to make assumed that there is a complete transfer of
our list more vast and design a metal activity electron from a less electronegative atom to a
series o r electro chemical series. The more electonegative atom. For example, we
tt E

competition for electrons betwe en various rewrite equations (8.18 to 8.2 0) to show
charge on each of the atoms forming part of
metals help s us to design a clas s of cells,
the reaction :
C

named as Galvanic cells in which the chemical

0 0 +1 –2
reactions b ecome the source of electrical
energy. We would study more about these cells 2H2(g) + O2(g) ® 2H2O (l) (8.21)
no N

in Class XII. 0 0 +1 –1
H2 (s) + Cl2(g) ® 2HCl(g) (8.22)
8.3 OXIDATION NUMBER
–4 + 1 0 +4 –1 +1 –1
A less obvious example of electron transfer is
CH4(g) + 4Cl2(g) ® CCl4(l) +4HCl(g) (8.23)
©

realised when hydrogen combines with oxygen

to form water by the reaction: It may be emphasised that the assumption
2H2(g) + O2 (g) ® 2H2O (l) (8.18) of electron transfer is made for book-keeping
Though not simple in its approach, yet we purpose only and it will become obvious at a
can visualise the H atom as going from a later stage in this unit that it leads to the simple
neutral (zero) state in H2 to a positive state in description of redox reactions.
H2O, the O atom goes from a zero state in O2 O x idatio n numb er deno tes the
to a dinegative state in H2O. It is assumed that o xidatio n s tate of an element in a
there is an electron transfer from H to O and compound ascertained according to a set
consequently H2 is oxidised and O2 is reduced. of rules formulated on the basis that

C:\Chemistry XI\Unit-8\Unit-8(5)(reprint).pmd 27.7.6, 16.10.6 (reprint)

 26 0 CHE MIST RY

electron pair in a covalent bo nd belongs of oxygen but this number would now be
entirely to more electronegative element. a positive figure only.
It is not always possible to remember or 4. The oxidation number of hydrogen is +1,
make out e asily in a compound/ ion, which except when it is bonded to metals in binary
element is more electronegative than the other. compounds (that is compounds containing
Therefore, a set of rules has been formulated two elements). For example, in LiH, NaH,
to determine the oxidation number of an and CaH2, its oxidation number is –1.
element in a compound /ion. If two or more 5. In all its compounds, fluorine has an
than two atoms of an element are present in oxidation number of –1. Other halogens (Cl,

d
2–
the molecule/ion such as Na2S2O3/Cr2O7 , the Br, and I) also have an oxidation number
oxidation number of the atom of that element of –1, when they occ ur as halide ions in

he
will then b e the average of the oxidation their compounds. Chlorine, bromine and
number of all the atoms of that element. We iodine whe n combined with oxygen, for
may at this stage, state the rules for the example in oxoacids and oxoanions, have
calculation of oxidation number. These rules are: positive oxidation numbers.
6. The algebraic sum of the oxidation number

is
1. In elements, in the free or the uncombined
of all the atoms in a compound must be
state, each atom bears an oxidation
zero. In polyatomic ion, the algebraic sum
number of zero. Evidently each atom in H2,
of all the oxidation numbers of atoms of

bl
O2, Cl2, O3, P4, S8, Na, Mg, Al has the
the ion must equal the charge on the ion.
oxidation number zero.
Thus, the sum of oxidation number of three
2. For ions composed of only one atom, the oxygen atoms and one carbon atom in the
pu
oxidation number is equal to the charge carbonate ion, (CO3)2– must equal –2.
on the ion. Thus Na+ ion has an oxidation By the application of above rules, we can
number of +1, Mg2+ ion, +2, Fe3+ ion, +3, find out the oxidation number of the desired
be T

Cl– ion, –1, O2– ion, –2; and so on. In their element in a molecule or in an ion. It is clear
c omp ound s all alk ali me tals hav e
re
that the metallic elements have positive
o R

oxidation number of +1, and all alkaline oxidation number and nonmetallic elements
earth metals have an oxidation number of have positive or negative oxidation number.
+2. Aluminium is regarded to have an The atoms of transition elements usually
tt E

oxid ation numb e r of +3 in all its display several positive oxidation states. The
compounds. highest oxidation number of a representative
3. The oxidation number of oxygen in most element is the group number for the first two
C

compounds is –2. However, we come across groups and the group number minus 10
two kinds of exceptions here. One arises (following the long form of periodic table) for
in the case of peroxides and superoxides, the other groups. Thus, it implies that the
no N

the compounds of oxygen in which oxygen highest value of oxidation number exhibited
atoms are directly linked to e ach other. by an atom of an element generally increases
While in peroxides (e.g., H2O2, Na2O2), each across the period in the periodic table. In the
oxygen atom is assigned an oxidation third period, the highest value of oxidation
©

number of –1, in superoxides (e.g., KO2, number changes from 1 to 7 as indicated below
RbO2) each oxygen atom is assigned an in the compounds of the elements.
oxidation number of –(½). T he second A term that is often used interchangeably
exception appears rarely, i.e. when oxygen with the oxidation number is the oxidation
is bonded to fluorine. In such compounds state. Thus in CO2, the oxidation state of
e.g., oxygen difluoride (OF2) and dioxygen carbon is +4, that is also its oxidation number
difluor ide (O2F2), the oxygen is assigned and similar ly the oxidation state as well as
an oxidation numb er of + 2 and + 1, oxidation number of oxygen is – 2. This implies
respectiv ely. The number as signed to that the oxidation number de note s the
oxygen will depend upon the bonding state oxidation state of an element in a compound.
 REDOX RE ACTIONS 26 1

Group 1 2 13 14 15 16 17
Element Na Mg Al Si P S Cl
Compound Na Cl MgSO4 AlF3 SiCl4 P4O10 SF6 HClO4
Highest oxidation +1 +2 +3 +4 +5 +6 +7
number state of
the group element

The oxidation number/state of a metal in a The idea of oxidation number has been

d
compound is sometimes presented according invariab ly app lied to d ef ine oxid ation,
to the notation given by German chemist, reduction, oxidising agent (oxidant), reducing
Alfred Stock. It is popularly known as Stock

he
agent (reductant) and the redox reaction. To
notation. According to this, the oxidation summarise, we may say that:
number is expressed by putting a Roman Oxidation: An increas e in the oxidation
numeral representing the oxidation number
number of the element in the given substance.
in parenthesis after the symbol of the metal in
Reduction: A dec reas e in the oxidation

is
the molecular formula. Thus aurous chloride
and auric chloride are written as Au(I)Cl and number of the element in the given substance.
Au(III)Cl3. Similarly, stannous chloride and Oxidising agent : A re age nt w hic h c an

bl
stannic chloride are written as Sn(II)Cl2 and increase the oxidation number of an element
Sn(IV)Cl4. This change in oxidation number in a given substance. These reagents are called
implies change in oxidation state, which in as oxidants also.
pu
turn helps to identify whether the species is Reducing agent: A reagent which lowers the
present in oxidised form or red uced form. oxidation number of an element in a given
Thus, Hg2(I)Cl2 is the reduced form of Hg(II) Cl2. substance. These reagents are also called as
be T

Problem 8.3 reductants.

re

Using Stock notation, represe nt the Redox reactions: Reactions which involve
o R

following compounds :HAuCl4, Tl2O, FeO, change in oxidation number of the interacting
Fe2O3, CuI, CuO, MnO and MnO2. species.
Solution
tt E

Problem 8.4
By applying various rules of calculating Justify that the reaction:
the oxidation numb er of the desired
C

element in a compound, the oxidation 2Cu2O(s) + Cu2S(s) ® 6Cu(s) + SO2(g)

number of each metallic element in its is a redox reaction. Identify the species
compound is as follows: oxidised/ reduced, which acts as an
no N

HAuCl4 ® Au has 3 oxidant and which acts as a reductant.

Tl2O ® Tl has 1 Solution
FeO ® Fe has 2 Let us assign oxidation number to each
©

Fe2O3 ® Fe has 3 of the sp ecies in the reaction under

CuI ® Cu has 1 examination. This results into:
CuO ® Cu has 2 +1 –2 +1 –2 0 +4 –2
MnO ® Mn has 2 2Cu2O(s) + Cu2S(s) ® 6Cu(s) + SO2
MnO2 ® Mn has 4
We therefor e, conc lude that in this
Therefore, these compounds may be reaction copper is reduced from +1 state
represented as:
to zero oxidation state and sulphur is
HAu(III)Cl4, Tl2(I)O, Fe(II)O, Fe2(III)O3, oxidised from –2 state to +4 state. The
Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O2. above reaction is thus a redox reaction.
 26 2 CHE MIST RY

Further, Cu2O helps sulphur in Cu2S to that all decomposition reactions are not redox
increase its oxidation number, therefore, reactions. For example, decomp osition of
Cu(I) is an oxidant; and sulphur of Cu2S calcium carbonate is not a redox reaction.
helps copper both in Cu2S itself and Cu2O +2 + 4 –2 +2 –2 +4 –2
to d ec re as e its oxidation numbe r; CaCO3 (s) CaO(s) + CO2(g)
therefore, sulphur of Cu2S is reductant.
3. Displacement reactions
8.3.1 Types of Redox Reactions In a displacement reaction, an ion (or an atom)
in a compound is replaced by an ion (or an

d
1. Combination reactions atom) of another element. It may be denoted
A combination reaction may be denoted in the as:

he
manner: X + YZ ® XZ + Y
A+ B ® C
Displacement reactions fit into two categories:
Either A and B or both A and B must be in the me tal d is p lac e me nt and non-me tal
elemental form for such a reaction to be a redox displacement.

is
reaction. All combustion reactions, which
make use of elemental dioxygen, as well as (a) Metal displacement : A metal in a
other reactions involving elements other than compound can be displaced by another metal

bl
dioxygen, are redox reactions. Some important in the unc ombined state. We have already
examples of this category are: discussed about this class of the reactions
0 0 +4 –2
under section 8.2.1. Metal displacement
pu
C(s) + O2 (g) CO2(g) (8.24) r e ac tions f ind many ap p lic ations in
metallurgical processes in which pure metals
0 0 + 2 –3 are obtained from their compounds in ores. A
be T

3Mg(s) + N2(g) Mg3N2(s) (8.25) few such examples are:

re
+2 +6 – 2 0 0 +2 +6 –2
–4 +1 0 +4 –2 +1 –2
CuSO4(aq) + Zn (s) ® Cu(s) + ZnSO4 (aq)
o R

CH4(g) + 2O2(g) CO2(g) + 2H2O (l)

(8.29)
2. Decomposition reactions +5 –2 0 0 +2 –2
tt E

Decomposition reactions are the opposite of V2O5 (s) + 5Ca (s) 2V (s) + 5CaO (s)
c omb ination r e ac tions . P r e c is e ly, a (8.30)
decomposition reaction leads to the breakdown
C

+4 –1 0 0 +2 –1
of a compound into two or more components
at least one of which must be in the elemental TiCl4 (l) + 2Mg (s) Ti (s) + 2 MgCl2 (s)
no N

state. Examples of this class of reactions are: (8.31)

+1 –2 0 0 +3 –2 0 +3 – 2 0
2H2O (l) 2H2 (g) + O2(g) (8.26) Cr2O3 (s) + 2 Al (s) Al2O3 (s) + 2Cr(s)
(8.32)
©

+1 –1 0 0

2NaH (s) 2Na (s) + H2(g) (8.27) In each case, the reducing metal is a better
reducing agent than the one that is being
+1 +5 – 2 +1 –1 0 reduced which evidently shows more capability
2KClO3 (s) 2KCl (s) + 3O2(g) (8.28) to lose electrons as compared to the one that
It may care fully be noted that there is no is reduced.
change in the oxidation number of hydrogen (b) Non-metal displacement: The non-metal
in methane under combination reactions and d is place me nt re d ox r e ac tions includ e
that of potassium in potassium chlorate in hydrogen displacement and a rarely occurring
reaction (8.28). This may also be noted here reaction involving oxygen displacement.
 REDOX RE ACTIONS 26 3

All alkali metals and some alkaline earth order Zn> Cu>Ag. Like metals, activity series
metals (Ca, Sr, and Ba) which are very good also exists for the halogens. The power of these
reductants, will displace hydrogen from cold elements as oxidising agents decreases as we
water. move down from fluorine to iodine in group
0 +1 –2 +1 –2 +1 0 17 of the p eriodic table. This implies that
2Na(s) + 2H2O(l) ® 2NaOH(aq) + H2(g) fluorine is so reactive that it can replace
(8.33) chloride, bromide and iodide ions in solution.
0 +1 –2 +2 – 2 +1 0 In fact, fluorine is so reactive that it attacks
Ca(s) + 2H2O(l) ® Ca(OH)2 (aq) + H2(g) water and displaces the oxygen of water :

d
(8.34) +1 –2 0 +1 –1 0
Less active metals such as magnesium and 2H2O (l) + 2F2 (g) ® 4HF(aq) + O2(g) (8.40)

he
iron react with steam to produce dihydrogen gas: It is for this reason that the displacement
0 +1 –2 +2 –2 +1 0 reactions of chlorine, bromine and iodine
Mg(s) + 2H2O(l) Mg(OH)2(s) + H2(g) using fluorine are not generally carried out in
(8.35) aqueous solution. On the other hand, chlorine

is
can displac e bromide and iodide ions in an
0 + 1 –2 +3 –2 0
aqueous solution as shown below:
2Fe(s) + 3H2O(l) Fe2O3(s) + 3H2(g) (8.36)
0 + 1 –1 +1 –1 0

bl
Many metals, including those which do not Cl2 (g) + 2KBr (aq) ® 2 KCl (aq) + Br2 (l)
react with cold water, are capable of displacing (8.41)
hydrogen from acids. Dihydrogen from acids 0 +1–1 +1 –1 0
pu
may even be produced by such metals which Cl2 (g) + 2KI (aq) ® 2 KCl (aq) + I2 (s)
do not react with steam. Cadmium and tin are (8.42)
the examples of such metals. A few examples As Br2 and I2 are coloured and dissolve in CCl4,
be T

for the displacement of hydrogen from acids can easily be identified from the colour of the
solution. The above reactions can be written
re
are:
o R

0 +1 –1 +2 –1 0 in ionic form as:

Zn(s) + 2HCl(aq) ® ZnCl2(aq) + H2 (g) 0 –1 –1 0
(8.37) Cl2 (g) + 2Br– (aq) ® 2Cl– (aq) + Br2(l) (8.41a)
tt E

0 +1 –1 +2 –1 0 0 –1 –1 0
Mg (s) + 2HCl (aq) ® MgCl2 (aq) + H2 (g) Cl2 (g) + 2I – (aq) ® 2Cl– (aq) + I2 (s) (8.42b)
(8.38)
C

Reactions (8.41) and (8.42) form the basis

0 + 1 –1 +2 –1 0
of identifying Br– and I – in the laboratory
Fe(s) + 2HCl(aq) ® FeCl2(aq) + H2(g)
through the test popularly known as ‘Layer
(8.39)
no N

Test’. It may not be out of place to mention

Reactions (8.37 to 8 .39) are used to here that bromine likewise can displace iodide
prepare dihydrogen gas in the laboratory. ion in solution:
Here, the reactivity of metals is reflected in the 0 –1 –1 0
©

rate of hydrogen gas evolution, which is the Br2 (l) + 2I – (aq) ® 2Br– (aq) + I2 (s) (8.43)
slowest for the least active metal Fe, and the
fastest for the most reactive metal, Mg. Very The halogen displacement reactions have
less active metals, which may occur in the a direct industrial application. The recovery
native state such as silver (Ag), and gold (Au) of halogens from their halides requires an
do not react even with hydrochloric acid. oxidation process, which is represented by:
I n se ction (8 .2 .1 ) we hav e alre ad y 2X– ® X2 + 2e– (8.44)
discussed that the metals – zinc (Zn), copper here X denotes a halogen element. Whereas
(Cu) and silver (Ag) through tendency to lose chemical means are available to oxidise Cl –,
electrons show their reducing activity in the Br– and I –, as fluorine is the strongest oxidising
 26 4 CHE MIST RY

–
agent; there is no way to convert F ions to F2 fluorine shows deviation from this behaviour
by chemical means. The only way to achieve when it reacts with alkali. The reaction that
F2 from F– is to oxidise electrolytic ally, the takes place in the case of fluorine is as follows:
details of which you will study at a later stage. 2 F2(g) + 2OH– (aq) ® 2 F – (aq) + OF2(g) + H2O(l)
4. Disproportionation reactions (8.49)
Disproportionation reactions are a special type (It is to be noted with care that fluorine in
of redox re actions. In a disprop ortionation reaction (8.49) will undoubtedly attack water
reaction an element in one oxidation state is to produce some oxygen also). This departure

d
simultaneously oxidised and reduced. One of shown by fluorine is not surprising for us as
the r e ac ting s ub s tanc e s in a we know the limitation of fluorine that, being
disproportionation reaction always contains the most electronegative element, it cannot

he
an element that can exist in at least three exhibit any positive oxidation state. This
oxidation states. The element in the form of means that among halogens, fluorine does not
reacting substance is in the intermediate show a disproportionation tendency.
oxidation s tate; and both highe r and lower
Problem 8.5

is
oxidation states of that element are formed in
the reaction. The decomposition of hydrogen Which of the following species, do not
peroxide is a familiar example of the reaction, show disp roportionation reaction and

bl
where oxygen experiences disproportionation. why ?
– – – –
+1 –1 +1 –2 0 ClO , ClO2 , ClO3 and ClO4
2H2O2 (aq) ® 2H2O(l) + O2(g) (8.45) Also write reaction for each of the species
pu
Here the oxygen of peroxide, which is present that disproportionates.
in –1 state, is converted to zero oxidation state Solution
be T

in O2 and decreases to –2 oxidation state in Among the oxoanions of chlorine listed

H2O. above, ClO4– does not dis proportionate
re

Phos phorous, s ulp hur and c hlorine because in this oxoanion chlorine is
o R

undergo dis proportionation in the alkaline present in its highest oxidation state that
medium as shown below : is, +7. The disproportionation reactions
for the other three oxoanions of chlorine
tt E

0 –3 +1
P4(s) + 3OH– (aq)+ 3H2O(l) ® PH3(g) + 3H2PO2– are as follows:
(aq) +1 –1 +5
C

– – –
(8.46) 3ClO ® 2Cl + ClO3
0 –2 +2 +3 +5 –1
S8(s) + 12 OH– (aq) ® 4S2– (aq) + 2S2O32– (aq) 6 ClO2
–
4ClO3 + 2Cl
– –
no N

+ 6H2O(l)
+5 –1 +7
(8.47) – – –
4ClO3 ® Cl + 3 ClO4
0 +1 –1
Cl2 (g) + 2 OH– (aq) ® ClO– (aq) + Cl– (aq) + Problem 8.6
©

H2O (l) Suggest a scheme of classification of the

(8.48) following redox reactions
The reaction (8.48) describes the formation (a) N2 (g) + O2 (g) ® 2 NO (g)
of hous e hold b le ac hing age nts. T he
hypochlorite ion (ClO– ) formed in the reaction (b) 2Pb(NO3)2(s) ® 2PbO(s) + 2 NO2 (g) +
oxidises the colour-bearing stains of the ½ O2 (g)
substances to colourless compounds. (c) NaH(s) + H2O(l) ® NaOH(aq) + H2 (g)
– –
It is of interest to mention here that whereas (d) 2NO2(g) + 2OH (aq) ® NO2(aq) +
–
bromine and iodine follow the same trend as NO3 (aq)+H2O(l)
exhib ited by chlorine in reac tion (8. 48),
 REDOX RE ACTIONS 26 5

Solution (c), hydrogen of water has been displaced

In reaction (a), the compound nitric oxide by hydrid e ion into dihydr ogen gas.
is formed by the c ombination of the T he r ef ore , this may b e calle d as
ele mental subs tances , nitrogen and displacement redox reaction. The reaction
oxygen; therefore, this is an example of (d) involv es disproportionation of NO2
c ombination r ed ox r e ac tions . The (+4 state) into NO2– (+3 state ) and NO3–
reaction (b) involves the breaking down (+5 state). Therefore reaction (d) is an
of lead nitrate into three c omponents;
ther efore, this is categorised under examp le of disproportionation redox

d
decomposition redox reaction. In reaction reaction.

he
The Paradox of Fractional Oxidation Number
Sometimes, we come across with certain compounds in which the oxidation nu mber of a
particular element in th e compound is in fraction. Examples are:
C3O2 [where oxidation number of carbon is (4/3)],

is
Br3O8 [where o xidation number of bromine is (16/3)]
and Na2S 4O6 (where oxidation number of sulphur is 2.5).
We know t hat the idea of fractional oxidat ion number is un convincing to us, because

bl
electron s are never shared/transferred in fraction. Act ually this fract ional oxidation state is
the avera ge oxidation state of the element un der examination an d the structural parameters
reveal t hat the element for whom fractio nal oxidation state is realised is present in different
pu 2–
oxidatio n states. Structure of the species C3O2, Br3O8 and S 4O6 reveal the fo llowing bonding
situations:
+2 0 +2
be T

O = C = C*= C = O
Structure of C3O2
re
(carbon suboxide)
o R
tt E

2–
Structu re of Br3O8 (tribromo octaoxide) Structu re of S 4O6 (tetrathionate ion)
The element marked with asterisk in each species is exhibiting the different oxidation
C

state (oxidation number) fro m rest of the atoms of the same element in each of th e species.
This reveals that in C3O2, two carbon atoms are present in +2 oxidation state each , whereas
the third one is present in zero oxidation state and th e average is 4/3. However, the realistic
picture is +2 for two terminal carbons and zero for the middle carbon. Likewise in Br3O8, each
no N

of the tw o terminal bromine atoms are present in +6 oxidation state and the middle bromine
is present in +4 oxidation state. Once again t he average, that is different from reality, is
2–
16/3. In the same fash ion, in the species S 4O6 , each o f the two extreme sulphurs exhibits
oxidation state of +5 and t he two middle sulphurs as zero. The average of four oxidation
©

2–
numbers of sulphurs of the S 4O6 is 2.5, wh ereas the reality being + 5,0,0 and +5 oxidation
number respectively for each sulphur.
We may th us, in general, conclude that the idea of fractional oxidation state should be
taken with care and the reality is revealed by the structures only. Fu rther, whenever we come
across with fractional oxidation state of an y particular element in any species, we must
understa nd that this is the average oxidation number o nly. In reality (revealed by st ructures
only), the element in that particular species is present in more than one whole number oxidation
states. Fe3O4, Mn 3O4, Pb3O4 are some of the other examples of the compounds, which are
mixed oxides, where we come across with fractional oxidation states of the metal atom. However,
+ –
the oxidation states may be in fraction as in O2 and O2 where it is +½ and –½ respectively.
 26 6 CHE MIST RY

Problem 8.7 (a) Oxidation Number Method: In writing

equations for oxidation-reduction reactions,
Why do the following reactions proceed just as for other reactions, the compositions
differently ? and f or mulas must b e know n for the
Pb3O4 + 8HCl ® 3PbCl2 + Cl2 + 4H2O substances that react and for the products that
and are formed. The oxidation number method is
Pb3O4 + 4HNO3 ® 2Pb(NO3)2 + PbO2 + now best illustrated in the following steps:
2H2O Step 1: Write the correct formula for each
Solution reactant and product.

d
Pb3 O4 is ac tually a s toic hiometric Step 2: Identify atoms which undergo change
mixture of 2 mol of PbO and 1 mol of in oxid ation number in the reac tion by

he
PbO2. In PbO2, lead is pr esent in +4 assigning the oxidation number to all elements
oxidation s tate , whe reas the stable in the reaction.
oxidation state of lead in PbO is +2. PbO2 Step 3: Calculate the increase or decrease in
thus can act as an oxidant (oxidising the oxidation number per atom and for the

is
agent) and, therefore, can oxidise Cl – ion entire molecule/ion in which it occurs. If these
of HCl into chlorine. We may also keep in are not equal the n multiply by suitable
mind that PbO is a basic oxide. Therefore, number so that these become equal. (If you

bl
the reaction realise that two substances are reduced and
Pb3O4 + 8HCl ® 3PbCl2 + Cl2 + 4H2O nothing is oxidised or vice-versa, something
can be splitted into two reactions namely: is wrong. Either the formulas of reactants or
pu
products are wrong or the oxidation numbers
2PbO + 4HCl ® 2PbCl2 + 2H2O
have not been assigned properly).
(acid-base reaction)
Step 4: Ascertain the involvement of ions if
be T

+4 –1 +2 0
the reaction is taking place in water, add H+ or
PbO2 + 4HCl ® PbCl2 + Cl2 +2H2O
re
OH – ions to the expression on the appropriate
(redox reaction)
o R

side so that the total ionic charges of reactants

Since HNO3 itself is an oxidising agent and products are equal. If the reaction is
therefore, it is unlikely that the reaction carried out in acidic solution, use H+ ions in
tt E

may occur b etween PbO2 and HNO3. the equation; if in basic solution, use OH– ions.
However, the acid-base reaction occurs
Step 5 : Make the numbers of hydrogen atoms
between PbO and HNO3 as:
C

in the expression on the two sides equal by

2PbO + 4HNO3 ® 2Pb(NO3)2 + 2H2O adding water (H2O) molecules to the reactants
It is the passive nature of PbO2 against or products. Now, also check the number of
no N

HNO3 that makes the reaction different oxygen atoms. If there are the same number
from the one that follows with HCl. of oxyge n atoms in the reactants and
products, the equation then represents the
8.3.2 Balancing of Redox Reactions balanced redox reaction.
©

Two methods are used to balance chemical Let us now explain the steps involved in
equations for redox processes. One of these the method with the help of a few problems
methods is bas ed on the change in the given below:
oxidation number of reducing agent and the
oxidising agent and the other method is based Problem 8.8
on splitting the redox reaction into two half Write the net ionic equation f or the
reactions — one involving oxidation and the reaction of potassium dichromate(VI),
other involving reduction. Both these methods K2Cr2O7 with sodium sulphite, Na2SO3,
are in use and the choice of their use rests with in an acid solution to give chromium(III)
the individual using them. ion and the sulphate ion.
 REDOX RE ACTIONS 26 7

Solution the oxidant and bromide ion is the

Step 1: The skeletal ionic equation is: reductant.
2– 2– 3+
Cr2O7 (aq) + SO3 (aq) ® Cr (aq) Step 3: Calc ulate the inc re ase and
2– decrease of oxidation number, and make
+ SO4 (aq) the increase equal to the decrease.
Step 2: Assign oxidation numbers for +7 –1 +4 +5
Cr and S – – –
2MnO4(aq)+Br (aq) ® 2MnO2(s)+BrO3(aq)
+6 –2 +4 –2 +3 +6 –2
2– 2–
Cr2O7 (aq) + SO3 (aq) ® Cr(aq)+SO4 (aq)
2– Step 4: As the reaction occurs in the basic

d
medium, and the ionic charges are not
This indicates that the dichromate ion is equal on both sides, add 2 OH– ions on
the oxidant and the sulphite ion is the

he
the right to make ionic charges equal.
reductant. – –
2MnO4 (aq) + Br (aq) ® 2MnO2(s) +
Step 3: Calc ulate the inc re ase and – –
decrease of oxidation number, and make BrO3 (aq) + 2OH (aq)
them equal: Step 5: Finally, count the hydrogen

is
+6 –2 +4 –2 +3
atoms and add appr opriate number of
Cr2O72– (aq) + 3SO32– (aq) ® 2Cr3+ (aq) + water molecules (i.e. one H2O molecule)
+6 –2 on the left side to achieve balanced redox

bl
3SO42– (aq) change.
– –
Step 4: As the reaction occurs in the 2MnO4 (aq) + Br (aq) + H2O(l) ® 2MnO2(s)
– –
acidic me dium, and further the ionic + BrO3 (aq) + 2OH (aq)
pu
charges are not equal on both the sides, (b) Half Reaction Method: In this method,
add 8H+ on the left to make ionic charges the two half equations are balanced separately
be T

equal and then added together to giv e balanced

2– 2– + 3+
Cr2O7 (aq) + 3SO3 (aq)+ 8H ® 2Cr (aq) equation.
re
2–
+ 3SO4 (aq)
o R

Suppose we are to b alance the equation

Step 5: Finally, count the hydrogen showing the oxidation of Fe2+ ions to Fe3+ions
atoms, and add appropriate number of by dichromate ions (Cr2O7)2– in acidic medium,
tt E

water molecules (i.e., 4H2O) on the right wherein, Cr2O72– ions are reduced to Cr3+ ions.
to achieve balanced redox change. The following steps are involved in this task.
2– 2– +
Cr2O7 (aq) + 3SO3 (aq)+ 8H (aq) ® Step 1: Produce unbalanced equation for the
C

3+ 2–
2Cr (aq) + 3SO4 (aq) +4H2O (l) reaction in ionic form :
2+ 2– 3+ 3+
Problem 8.9 Fe (aq) + Cr2O7 (aq) ® Fe (aq) + Cr (aq)
no N

(8.50)
Permanganate ion reacts with bromide ion Step 2: Se parate the equation into half-
in basic medium to give manganese reactions:
dioxide and br omate ion. Write the
+2 +3
balanced ionic equation for the reaction.
©

2+
Oxidation half : Fe (aq) ® Fe3+(aq) (8.51)
Solution +6 –2 +3
Step 1 : The skeletal ionic equation is : 2–
Reduction half : Cr2O7 (aq) ® Cr (aq)
3+
– – –
MnO4 (aq) + Br (aq) ® MnO2(s) + BrO3 (aq) (8.52)
Step 2 : Assign oxidation numbers for Step 3: Balance the atoms other than O and
Mn and Br H in each half reaction individually. Here the
oxidation half reaction is already balanced with
+7 –1 +4 +5
– – – respect to Fe atoms. For the red uction half
MnO4 (aq) + Br (aq) ®MnO2 (s) + BrO3 (aq) reaction, we multiply the Cr3+ by 2 to balance
this indicates that permanganate ion is Cr atoms.
 26 8 CHE MIST RY

2– 3+
Cr2O7 (aq) ® 2 Cr (aq) (8.53) Problem 8.10
Step 4: For re actions occ urring in acidic –
Permanganate(V II) ion, MnO4 in basic
medium, add H2O to balance O atoms and H+ solution oxidises iodide ion, I– to produce
to balance H atoms. molecular iodine (I2) and manganese (IV)
Thus, we get : oxide (MnO2). Write a balanced ionic
2– + 3+
Cr2O7 (aq) + 14H (aq) ® 2 Cr (aq) + 7H2O (l) equation to represent this redox reaction.
(8.54) Solution
Step 1: First we write the skeletal ionic

d
Step 5: Add electrons to one side of the half
reaction to balance the charges. If need be, equation, which is
–
make the number of electrons equal in the two MnO4 (aq) + I – (aq) ® MnO2(s) + I2(s)

he
half reactions by multiplying one or both half Step 2: The two half-reactions are:
reactions by appropriate number. –1 0
The oxidation half reaction is thus rewritten Oxidation half : I – (aq) ® I2 (s)
to balance the charge:

is
+7 +4
Fe2+ (aq) ® Fe3+ (aq) + e– (8.55) –
Reduction half: MnO4 (aq) ® MnO2(s)
Now in the reduction half reaction there are Step 3: To balance the I atoms in the

bl
net twelve positive charges on the left hand side oxidation half reaction, we rewrite it as:
and only six positive charges on the right hand 2I – (aq) ® I2 (s)
side. Therefore, we add six electrons on the left
pu
side. Step 4: To balance the O atoms in the
2– + – 3+ reduction half reaction, we add two water
Cr2O7 (aq) + 14H (aq) + 6e ® 2Cr (aq) + molecules on the right:
be T

7H2O (l) (8.56) MnO4– (aq) ® MnO2 (s) + 2 H2O (l)

To equalise the number of electrons in both To balance the H atoms, we add four H+
re
o R

the half reactions, we multiply the oxidation ions on the left:

half reaction by 6 and write as : –
MnO4 (aq) + 4 H+ (aq) ® MnO2(s) + 2H2O (l)
6Fe2+ (aq) ® 6Fe3+(aq) + 6e– (8.57)
As the reaction takes place in a basic
tt E

Step 6: We add the two half reactions to solution, therefore, for four H+ ions, we
achieve the overall reaction and cancel the add four OH – ions to both sides of the
C

electrons on each side. This gives the net ionic equation:

equation as : –
MnO4 (aq) + 4H+ (aq) + 4OH– (aq) ®
2+ 2– + 3+
6Fe (aq) + Cr2O7 (aq) + 14H (aq) ® 6 Fe (aq) + MnO2 (s) + 2 H2O(l) + 4OH – (aq)
no N

3+
2Cr (aq) + 7H2O(l) (8.58) Replacing the H+ and OH– ions with water,
Step 7: Verify that the equation contains the the resultant equation is:
same type and number of atoms and the same MnO4– (aq) + 2H2O (l) ® MnO2 (s) + 4 OH– (aq)
charges on both sides of the equation. This last
©

Step 5 : In this step we balance the

che ck rev eals that the equation is fully charges of the two half-reactions in the
balanced with respect to number of atoms and manner depicted as:
the charges. –
2I (aq) ® I2 (s) + 2e–
For the reaction in a basic medium, first –
MnO4 (aq) + 2H2O(l) + 3e– ® MnO2(s)
balance the atoms as is done in acidic medium. –
Then for each H+ ion, add an equal number of + 4OH (aq)
–
OH ions to both sides of the equation. Where Now to equalise the number of electrons,
–
H+ and OH appear on the s ame side of the we multiply the oxidation half-reaction by
equation, combine these to give H2O. 3 and the reduction half-reaction by 2.
 REDOX RE ACTIONS 26 9

– –
6I (aq) ® 3I2 (s) + 6e (iii) There is yet another method which is
– – interesting and quite common. Its use is
2 MnO4 (aq) + 4H2O (l) +6e ® 2MnO2(s)
– restricted to– those reagents which are able
+ 8OH (aq)
to oxidise I ions, say, for example, Cu(II):
Step 6: Add two half-reactions to –
2Cu2+(aq) + 4I (aq) ® Cu2I2(s) + I2(aq) (8.59)
obtain the net reactions after cancelling
electrons on both sides. This method relies on the facts that iodine
– – itself gives an intense blue colour with starch
6I (aq) + 2MnO4(aq) + 4H2O(l) ® 3I2(s) +
– and has a ve ry s pec if ic r eaction with
2MnO2(s) +8 OH (aq)
thiosulphate ions (S2O32– ), which too is a redox

d
Step 7: A final verification shows that reaction:
the equation is balanced in respect of the 2–
I2(aq) + 2 S2O3 (aq)®2I–(aq) + S4O62– (aq) (8.60)

he
number of atoms and charges on both
sides. I2, though insoluble in water, remains in
solution containing KI as KI3.
8.3.3 Redox Reactions as the Basis for On addition of starch after the liberation of
Titrations iodine from the reaction of Cu2+ ions on iodide

is
In acid-base systems we come across with a ions, an intense blue colour ap pears. This
titration method for finding out the strength colour disappears as soon as the iodine is

bl
of one solution against the other using a pH consumed by the thiosulphate ions. Thus, the
sensitive indicator. Similarly, in redox systems, end-point can easily be tracked and the rest
the titration method can b e adopted to is the stoichiometric calculation only.
pu
determine the strength of a reductant/oxidant 8.3.4 Limitations of Concept of Oxidation
using a redox sensitive indicator. The usage Number
of indicator s in redox titration is illustrated
As you have observed in the above discussion,
be T

below:
the concep t of redox processe s has been
(i) In one situation, the reagent itself is
re
evolving with time. This process of evolution
o R

intensely coloured, e.g., permanganate ion, is continuing. In fact, in rece nt past the
MnO–4. Here MnO4– acts as the self indicator. oxidation process is visualised as a decrease
The v isible end point in this case is in electron density and reduction process as
achieved after the last of the reductant (Fe2+
tt E

an increase in electron density around the

or C2O42– ) is oxidised and the first lasting atom(s) involved in the reaction.
tinge of pink colour appear s at MnO4–
C

concentration as low as 10–6 mol dm–3 8.4 REDOX REACTIONS AND ELECTRODE
(10 –6 mol L –1). This ensur es a minimal PROCESSES
‘ov e r s hoot’ in c olour b e yond the The experiment corresponding to reaction
no N

equivalenc e point, the point where the (8.15), can also be observed if zinc rod is
reductant and the oxidant ar e equal in dipped in copper sulphate solution. The redox
terms of their mole stoichiometry. reaction takes place and during the reaction,
(ii) If there is no dramatic auto-colour change zinc is oxidised to zinc ions and copper ions
©

(as with MnO –4 titr ation), ther e ar e are reduced to metallic copper due to direct
indicators which are oxidised immediately transfer of electrons from zinc to copper ion.
af ter the las t b it of the r eac tant is During this reaction heat is also evolved. Now
consumed, producing a dramatic colour we modify the experiment in such a manner
change. The best example is afforded by that for the same redox reaction transfer of
–
Cr2O27 , which is not a self-indicator, but e le ctrons take s p lace indire ctly. T his
oxid is e s the indic ator s ub s tanc e necessitates the separation of zinc metal from
diphenylamine just after the equivalence copper sulp hate solution. We take copper
point to produce an intense blue colour, sulphate solution in a beaker and put a copper
thus signalling the end point. strip or rod in it. We also take zinc sulphate
 27 0 CHE MIST RY

solution in another beaker and put a zinc rod jelly like substance). This provides an electric
or strip in it. Now reaction takes place in either contact betwe en the two solutions without
of the beakers and at the interface of the metal allowing them to mix w ith each other. The
and its salt solution in each beaker both the zinc and copper rods are connected by a metallic
reduced and oxidized forms of the same wire with a provision for an ammeter and a
species are present. These rep resent the switch. The set-up as shown in Fig.8.3 is known
species in the reduction and oxidation half as Daniell cell. When the switch is in the off
reactions. A redox couple is defined as having position, no reaction takes place in either of
together the oxidised and reduced forms of a the beakers and no current flows through the

d
substance taking part in an oxidation or metallic wire. As soon as the switch is in the
reduction half reaction. on p os ition, w e mak e the f ollow ing

he
This is re presented by separ ating the observations:
oxidised f orm fr om the reduced form by a 1. The transfer of electrons now does not take
ver tical line or a slash re prese nting an place directly from Zn to Cu2+ but through
interface (e.g. solid/solution). For example the metallic wire connecting the two rods
as is apparent from the arrow which

is
in this experiment the two redox couples are
represented as Zn2+/Zn and Cu2+/Cu. In both indicates the flow of current.
cases, oxidised form is put before the reduced 2. The electricity from solution in one beaker

bl
form. Now we put the b eaker containing to solution in the other beaker flows by the
copp er sulphate s olution and the be aker migration of ions through the salt bridge.
containing zinc sulphate solution side by side We know that the flow of current is possible
(Fig. 8.3). We connect s olutions in two only if there is a potential difference
pu
beakers by a salt bridge (a U-tube containing between the copper and zinc rods known
a s olution of p otas s ium c hlorid e or as electrodes here.
be T

ammonium nitrate usually solidifie d b y The potential associated with each

boiling with agar agar and later cooling to a electrode is known as electrode potential. If
re

the concentration of each species taking part

o R

in the electrode reaction is unity (if any gas

appears in the electrode reaction, it is confined
to 1 atmospheric pres sure) and further the
tt E

reaction is car ried out at 298K, then the

potential of each electrode is said to be the
C

Standard El ectr o de Po tential . By

convention, the standard electrode potential
0
(E ) of hydrogen electrode is 0.00 volts. The
no N

electrode potential value for each electrode

process is a measure of the relative tendency
of the active species in the process to remain
0
in the oxidised/reduced form. A negative E
means that the redox couple is a stronger
©

reducing agent than the H+/H2 couple. A

0
positive E means that the redox couple is a
Fig.8.3 The set-up for Daniell cell . El ectrons weaker reducing agent than the H+/H2 couple.
produced at the anode due to oxidation
The standar d electrode potentials are very
of Zn travel through the external circuit
to the cathode where these reduce the important and we can get a lot of other useful
copp er ions. The circuit is completed information from them. The values of standard
inside the cell by the migrati on of ions electrode potentials for some selected electrode
through the salt bridge. It may be noted processes (reduction reactions) are given in
that the di rection of current i s opposite Table 8.1. You will learn more about electrode
to the di rection of electron flow. reactions and cells in Class XII.

C:\Chemistry XI\Unit-8\Unit-8(5)(reprint).pmd 27.7.6, 16.10.6 (reprint)

 REDOX RE ACTIONS 27 1

Table 8.1 The Standard Ele ctrode Potentials at 298 K

Ions are present as aqueous species and H2O as liquid; gases and
solids are shown by g and s respectively.
0
® Reduced f orm)
–
Reaction (Oxidised form + ne E /V

F2(g) + 2e– ® 2F – 2.87

Co 3+ + e– ® Co 2+ 1.81
H2O2 + 2H+ + 2e– ® 2H2O 1.78
® Mn 2+ + 4H2O

d
MnO4– + 8H+ + 5e– 1.51
Au3+ + 3e– ® Au(s) 1.40

he
Cl2(g) + 2e– ® 2Cl– 1.36
Cr2O72– + 14H+ + 6e– ® 2Cr3+ + 7H2O 1.33
O2(g) + 4H+ + 4e– ® 2H2O 1.23
MnO2(s) + 4H+ + 2e– ® Mn 2+ + 2H2O 1.23

is
Br2 + 2e– ® 2Br– 1.09

Increasing strength of reducing agent

Increasing strength of oxidising agent

NO3– + 4H+ + 3e– ® NO(g) + 2H2O 0.97

2Hg2+ + 2e– ® Hg22+ 0.92

bl
Ag+ + e– ® Ag(s) 0.80
Fe3+ + e– ® Fe2+ 0.77
® H2O2
pu
O2(g) + 2H+ + 2e– 0.68
I2(s) + 2e– ® 2I– 0.54
Cu+ + e– ® Cu(s) 0.52
be T

Cu2+ + 2e– ® Cu(s) 0.34

® Ag(s) + Cl –
re
AgCl(s) + e– 0.22
o R

AgBr(s) + e– ® Ag(s) + Br – 0.10

2H+ + 2e– ® H2(g) 0.00
Pb2+ + 2e– ® Pb(s) –0.13
tt E

Sn 2+ + 2e– ® Sn(s) –0.14

Ni2+ + 2e– ® Ni(s) –0.25
C

Fe2+ + 2e– ® Fe(s) –0.44

Cr3+ + 3e– ® Cr(s) –0.74
Zn 2+ + 2e– ® Zn(s) –0.76
no N

–
2H2O + 2e– ® H2(g) + 2OH –0.83
Al3+ + 3e– ® Al(s) –1.66
Mg2+ + 2e– ® Mg(s) –2.36
® Na(s)
©

Na+ + e– –2.71
Ca2+ + 2e– ® Ca(s) –2.87
K+ + e– ® K(s) –2.93
Li+ + e– ® Li(s) –3.05
+
1. A negative E 0 means that the redox couple is a stronger reducing agent than the +H /H2 couple.
2. A positive E 0 means that the redox couple is a weaker reducing agent than the H /H2 couple.
 27 2 CHE MIST RY

SUMMARY

Redox reactions form an important class of reactions in which oxidation and reduction
occur simultaneously. Three tier conceptualisation viz, classical, electronic and oxidation
number, which is usually available in the texts, has been presented in detail. Oxidation,
reductio n, oxidising a gent (oxidant) and reducing agent (reduc tant) have been viewed
according to each conceptualisation. Oxidation numbers are assigned in accordance
with a co nsistent set of rules. Oxidation number and ion-electron method both are

d
useful means in writing equations for the redox reactions. Redox reactions are classified
into four categories: combination, decomposition displacement a nd disproportionation
reactio ns. The concept of redox c ouple and electrode processes is introduced here.

he
The redox reactions find wide applications in the study of electrode processes a nd cells.

is
EXERCISES

8.1 Assign oxidation number to t he underlined elements in each of the following

bl
species:
(a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4
(e) CaO2 (f) NaBH4 (g) H2S 2O7 (h) KAl(SO4)2.12 H2O
pu
8.2 What are the oxidation number of the underlined elements in each of the
following and how do you rationalise your results ?
be T

(a) KI3 (b) H2S 4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH
8.3 Justify th at the following reactions are redox reactions:
re

(a) CuO(s) + H2(g) ® Cu(s) + H2O(g)

o R

(b) Fe2O3(s) + 3CO(g) ® 2Fe(s) + 3CO2(g)

(c) 4BCl3(g) + 3LiAlH4(s) ® 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)
tt E

+ –
(d) 2K(s) + F2(g) ® 2K F (s)
(e) 4 NH3(g) + 5 O2(g) ® 4NO(g) + 6H2O(g)
C

8.4 Fluorine reacts with ice a nd results in th e change:

H2O(s) + F2(g) ® HF(g) + HOF(g)
no N

Justify that this reaction is a redox reaction.

8.5 Calculate the oxidation nu mber of sulphur, chromium and nitro gen in H2SO5,
2– –
Cr2O7 and NO3. Suggest structure of these compounds. Count for the fallacy.
8.6 Write formulas for the following compounds:
©

(a) Mercury(II) chloride (b) Nickel(II) sulphate

(c) Tin(IV) oxide (d) Thallium(I) sulphate
(e) Iron(III) sulphate (f) Chromiu m(III) oxide
8.7 Suggest a list of th e substances where carbon can exhibit o xidation states from
–4 to +4 and nitro gen from –3 to +5.
8.8 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as
reducing agents in their reactions, ozon e and nitric acid act only as o xidants.
Why ?
8.9 Consider the reactions:
(a) 6 CO2(g) + 6H2O(l) ® C6 H12 O6(aq) + 6O2(g)
 REDOX RE ACTIONS 27 3

(b) O3(g) + H2O2(l) ® H2O(l) + 2O2(g)

Why it is more appropriate t o write these reactions as :
(a) 6CO2(g) + 12H2O(l) ® C6 H12 O6(aq) + 6H2O(l) + 6O2(g)
(b) O3(g) + H2O2 (l) ® H2O(l) + O2(g) + O2(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox
react ions.
8.10 The compo und AgF2 is unstable compound. However, if formed, the compound
acts as a very strong oxidising agen t. Why ?

d
8.11 Whenever a reaction between an oxidisin g agent and a reducing agent is carried
out, a compound of lower oxidation stat e is formed if the reducing agent is in

he
excess and a compound of higher oxidation state is formed if the oxidising agent
is in ex cess. Justify th is statement giving three illust rations.
8.12 How do you co unt for the following observations ?
(a) Tho u gh a lka line pot a ssiu m perma n ga n a te a nd a cidic po ta ssiu m

is
permangana te both are used a s oxidants, yet in the manufacture o f benzoic
acid from toluene we use alcoholic potassium permanganate as an oxidant.
Why ? Writ e a balanced redox equation for the reaction.

bl
(b) Wh en con cen tra ted su lph uric a cid is added to an ino rga nic mixtu re
containing chloride, we get colourless pungent smelling gas HCl, but if the
mixture contains bromide then we get red vapour of bromin e. Why ?
8.13 Identify the substance oxidised reduced, oxidising agent and reducing agent for
pu
each of the following reactions:
(a) 2AgBr (s) + C6H6O2(aq) ® 2Ag(s) + 2HBr (aq) + C6H4O2(aq)
be T

+ – –
(b) HCHO(l) + 2[Ag (NH3)2] (aq) + 3OH (aq) ® 2Ag(s) + HCOO (aq) + 4NH3(aq)
re
+ 2H2O(l)
o R

2+ – –
(c) HCHO (l) + 2 Cu (aq) + 5 OH (aq) ® Cu2O(s) + HCOO (aq) + 3H2O(l)
(d) N2H4(l) + 2H2O2(l) ® N2(g) + 4H2O(l)
(e) Pb(s) + PbO2(s) + 2H2SO4(aq) ® 2PbSO4(s) + 2H2O(l)
tt E

8.14 Consider the reactions :

2– 2– –
2 S 2O3 (aq) + I2(s) ® S 4 O6 (aq) + 2I (aq)
C

2– 2– – +
S 2O3 (aq) + 2Br2(l) + 5 H2O(l) ® 2SO4 (aq) + 4Br (aq) + 10H (aq)
Why does t he same reductant, thiosulphate react differently with iodine and
no N

bromine ?
8.15 Justify giving reactions that among halogens, fluorine is the best oxidant and
among hydrohalic compounds, hydroiodic acid is the best reductant.
8.16 Why does t he following reaction occur ?
©

4– – +
XeO6 (aq) + 2F (aq) + 6H (aq) ® XeO3(g)+ F2(g) + 3H2O(l)
4–
What conclusion about the compound Na4XeO6 (of which XeO6 is a part) can be
drawn from t he reaction.
8.17 Consider the reactions:
(a) H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) ® H3PO4(aq) + 4Ag(s) + 4HNO3(aq)
(b) H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) ® H3PO4(aq) + 2Cu(s) + H2SO4(aq)
+ – –
(c) C6H5CHO(l) + 2[Ag (NH3)2] (aq) + 3OH (aq) ® C6H5COO (aq) + 2Ag(s) +
4NH3 (aq) + 2 H2O(l)
2+ –
(d) C6H5CHO(l) + 2Cu (aq) + 5OH (aq) ® No change observed.
 27 4 CHE MIST RY

+ 2+
What inference do you draw about the behavio ur of Ag and Cu from these
reactions ?
8.18 Balance the following redox reactions by ion – electron method :
– –
(a) MnO4 (aq) + I (aq) ® MnO2 (s) + I2(s) (in basic medium)
– 2+ –
(b) MnO4 (aq) + SO2 (g) ® Mn (aq) + HSO4 (aq) (in acidic solution)
2+ 3+
(c) H2O2 (aq) + Fe (aq) ® Fe (aq) + H2O (l) (in acidic solution)
2– 3+ 2–
(d) Cr2O7 + SO2(g) ® Cr (aq) + SO4 (aq) (in acidic solution)

d
8.19 Balance th e following equations in basic mediu m by ion-electron method and
oxidation number methods and identify the oxidising agent and th e reducing

he
agent.
– –
(a) P4(s) + OH (aq) ® PH3(g) + HPO2 (aq)
– –
(b) N2H4(l) + ClO3(aq) ® NO(g) + Cl (g)

is
– +
(c) Cl2O7 (g) + H2O2(aq) ® ClO2(aq) + O2(g) + H
8.20 What sort s of informations can you draw fro m the following reaction ?
– – –
(CN)2(g) + 2OH (aq) ® CN (aq) + CNO (aq) + H2O(l)

bl
3+
8.21 The Mn ion is u nstable in solution and undergoes disproportionatio n to give
2+ +
Mn , MnO2, and H ion. Write a balanced ionic equation for the reaction.
pu
8.22 Consider the elements :
Cs, Ne, I and F
be T

(a) Identify the element that exhibits only n egative oxidatio n state.
(b) Identify the element tha t exhibits only postive oxidatio n state.
re

(c) Identify the element that exhibits both positive and negative oxidation states.
o R

(d) Identify the element which exhibits neither the negative nor does the positive
oxidation state.
8.23 Chlorine is used to purify drinking water. Excess of chlorine is harmful. The
tt E

excess of chlorine is removed by treating with sulphur dioxide. Present a balanced

equation for this redox cha nge taking place in water.
C

8.24 Refer to t he periodic table given in your book and now answer the following
questions:
(a) Select the possible non meta ls that can show disproportionation reaction.
no N

(b) Select t hree metals tha t can show disproportionation reaction.

8.25 In Ostwa ld’s process for the manufacture of nitric acid, the first step involves
the oxidation of ammonia gas by oxygen gas to give nit ric oxide gas an d steam.
What is the maximum weight of nitric ox ide that can be obtained start ing only
©

with 10.00 g. of ammonia and 20.00 g of oxygen ?

8.26 Using th e standard elect rode potentials given in the Table 8.1, predict if the
reaction between the fo llowing is feasible:
3+ –
(a) Fe (aq) a nd I (aq)
+
(b) Ag (aq) an d Cu(s)
3+
(c) Fe (aq) an d Cu(s)
3+
(d) Ag(s) a nd Fe (aq)
2+
(e) Br2(aq) a nd Fe (aq).
 REDOX RE ACTIONS 27 5

8.27 Predict the products of electrolysis in each of the fo llowing:

(i) An aqueous solution o f AgNO3 with silver electrodes
(ii) An aqueous so lution AgNO3 with platinu m electrodes
(iii) A dilute solution of H2SO4 with platinum electrodes
(iv) An aqueous solution o f CuCl2 with platinu m electrodes.
8.28 Arrange th e following metals in the order in wh ich they displace each other
from the solution of their salts.
Al, Cu, Fe, Mg and Zn.

d
8.29 Given the standard electrode potentials,
+ +
K /K = –2.93V, Ag /Ag = 0.80V,

he
2+
Hg /Hg = 0.79V
2+ 3+
Mg /Mg = –2.37V. Cr /Cr = –0.74V
arrange t hese metals in t heir increasing order of reducin g power.
+ 2+
8.30 Depict the galvanic cell in which the reaction Zn(s) + 2Ag (aq) ® Zn (aq) +2Ag(s)

is
takes place, Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the cu rrent in the cell, and

bl
(iii) individual reaction at each electrode.
pu
be T
re
o R
tt E
C
no N
©
 276 CHEMISTRY

UNIT 9

HYDROGEN

d
he
Hydrogen, the most abundant element in the universe and the
third most abundant on the surface of the globe, is being
visualised as the major future source of energy.

is
After studying this unit, you will be
able to
• present informed opinions on the

bl
position of hydrogen in the
periodic table; Hydrogen has the simplest atomic structure among all the
• identify the modes of occurrence
pu elements around us in Nature. In atomic form it consists
and preparation of dihydrogen on of only one proton and one electron. However, in elemental
a small and commercial scale;
describe isotopes of hydrogen;
form it exists as a diatomic (H2) molecule and is called
• explain how different elements dihydrogen. It forms more compounds than any other
be T

combine with hydrogen to form element. Do you know that the global concern related to
ionic, molecular and non- energy can be overcome to a great extent by the use of
re
o R

stoichiometric compounds; hydrogen as a source of energy? In fact, hydrogen is of

• describe how an understanding of great industrial importance as you will learn in this unit.
its properties can lead to the
production of useful substances, 9.1 POSITION OF HYDROGEN IN THE PERIODIC
tt E

and new technologies;

• understand the structure of water
TABLE
and use the knowledge for Hydrogen is the first element in the periodic table.
C

explaining physical and chemical However, its placement in the periodic table has been a
properties; subject of discussion in the past. As you know by now
• explain how environmental water
that the elements in the periodic table are arranged
no N

quality depends on a variety of

dissolved substances; difference according to their electronic configurations.
1
between 'hard' and 'soft' water and Hydrogen has electronic configuration 1s . On one
learn about water softening; hand, its electronic configuration is similar to the outer
• acquire the knowledge about 1
electronic configuration (ns ) of alkali metals , which belong
©

heavy water and its importance;

• understand the structure of to the first group of the periodic table. On the other hand,
2 5
hydrogen peroxide, learn its like halogens (with ns np configuration belonging to the
preparatory methods and seventeenth group of the periodic table), it is short by one
properties leading to the electron to the corresponding noble gas configuration,
2
manufacture of useful chemicals helium (1s ). Hydrogen, therefore, has resemblance to
and cleaning of environment;
alkali metals, which lose one electron to form unipositive
• understand and use certain terms
e.g., electron-deficient, electron- ions, as well as with halogens, which gain one electron to
precise, electron-rich, hydrogen form uninegative ion. Like alkali metals, hydrogen forms
economy, hydrogenation etc. oxides, halides and sulphides. However, unlike alkali
metals, it has a very high ionization enthalpy and does not

276 C:\ChemistryXI\Unit-9\Unit-9(6).pmd 10.2.6, 6.3.6, 14.3.6

 HYDROGEN 277

possess metallic characteristics under normal solar atmosphere. The giant planets Jupiter
conditions. In fact, in terms of ionization and Saturn consist mostly of hydrogen.
enthalpy, hydrogen resembles more However, due to its light nature, it is much less
–1
with halogens, Δi H of Li is 520 kJ mol , F is abundant (0.15% by mass) in the earth’s
–1
1680 kJ mol–1 and that of H is 1312 kJ mol . atmosphere. Of course, in the combined form
Like halogens, it forms a diatomic molecule, it constitutes 15.4% of the earth's crust and
combines with elements to form hydrides and the oceans. In the combined form besides in
a large number of covalent compounds. water, it occurs in plant and animal tissues,
However, in terms of reactivity, it is very low as carbohydrates, proteins, hydrides including

d
compared to halogens. hydrocarbons and many other compounds.
Inspite of the fact that hydrogen, to a 9.2.2 Isotopes of Hydrogen

he
certain extent resembles both with alkali Hydrogen has three isotopes: protium, 1 H,
1

metals and halogens, it differs from them as 2 3

deuterium, 1H or D and tritium, 1H or T. Can
well. Now the pertinent question arises as you guess how these isotopes differ from each
where should it be placed in the periodic table? other ? These isotopes differ from one another

is
Loss of the electron from hydrogen atom in respect of the presence of neutrons. Ordinary
+
results in nucleus (H ) of ~1.510–3 pm size. hydrogen, protium, has no neutrons,
This is extremely small as compared to normal deuterium (also known as heavy hydrogen) has

bl
atomic and ionic sizes of 50 to 200pm. As a one and tritium has two neutrons in the
+
consequence, H does not exist freely and is nucleus. In the year 1934, an American
always associated with other atoms or scientist, Harold C. Urey, got Nobel Prize for
molecules. Thus, it is unique in behaviour and
pu
is, therefore, best placed separately in the
periodic table (Unit 3).
separating hydrogen isotope of mass number
2 by physical methods.
The predominant form is protium.
be T

9.2 DIHYDROGEN, H2 Terrestrial hydrogen contains 0.0156% of

deuterium mostly in the form of HD. The
re
9.2.1 Occurrence tritium concentration is about one atom per
o R

18
Dihydrogen is the most abundant element in 10 atoms of protium. Of these isotopes, only
the universe (70% of the total mass of the tritium is radioactive and emits low energy
–
universe) and is the principal element in the β particles (t , 12.33 years).
tt E

Table 9.1 Atomic and Physical Properties of Hydrogen

Property Hydrogen Deuterium Tritium
C

–15
Relative abundance (%) 99.985 0.0156 10
–1
Relative atomic mass (g mol ) 1.008 2.014 3.016
no N

Melting point / K 13.96 18.73 20.62

Boiling point/ K 20.39 23.67 25.0
–1
Density / gL 0.09 0.18 0.27
©

–1
Enthalpy of fusion/kJ mol 0.117 0.197 -
–1
Enthalpy of vaporization/kJ mol 0.904 1.226 -
Enthalpy of bond
–1
dissociation/kJ mol at 298.2K 435.88 443.35 -
Internuclear distance/pm 74.14 74.14 -
–1
Ionization enthalpy/kJ mol 1312 - -
–1
Electron gain enthalpy/kJ mol –73 - -
Covalent radius/pm 37 - -
–
Ionic radius(H )/pm 208

277 C:\ChemistryXI\Unit-9\Unit-9(6).pmd 10.2.6, 6.3.6, 14.3.6

 278 CHEMISTRY

Since the isotopes have the same electronic 1270K

C n H2n nH2O nCO (2n 1)H2
configuration, they have almost the same 2 Ni

chemical properties. The only difference is in e.g.,

their rates of reactions, mainly due to their CH4 ( g ) + H2 O ( g ) ⎯⎯⎯⎯
1270K
Ni
→ CO ( g ) + 3H2 ( g )
different enthalpy of bond dissociation (Table
9.1). However, in physical properties these The mixture of CO and H2 is called water
isotopes differ considerably due to their large gas. As this mixture of CO and H2 is used for
mass differences. the synthesis of methanol and a number of
hydrocarbons, it is also called synthesis gas

d
9.3 PREPARATION OF DIHYDROGEN, H2 or 'syngas'. Nowadays 'syngas' is produced
There are a number of methods for preparing from sewage, saw-dust, scrap wood,

he
dihydrogen from metals and metal hydrides. newspapers etc. The process of producing
'syngas' from coal is called 'coal gasification'.
9.3.1 Laboratory Preparation of
Dihydrogen C ( s ) + H2O ( g ) ⎯⎯⎯⎯
1270K
→ CO ( g ) + H2 ( g )
(i) It is usually prepared by the reaction of The production of dihydrogen can be

is
granulated zinc with dilute hydrochloric increased by reacting carbon monoxide of
acid. syngas mixtures with steam in the presence of
+ 2+
Zn + 2H → Zn + H2 iron chromate as catalyst.

bl
(ii) It can also be prepared by the reaction of
zinc with aqueous alkali. CO ( g ) + H2O ( g ) ⎯⎯⎯⎯
673 K
catalyst
→ CO2 ( g ) + H2 ( g )
Zn + 2NaOH → Na2ZnO2 + H2
pu Sodium zincate
This is called water-gas shift reaction.
Carbon dioxide is removed by scrubbing with
sodium arsenite solution.
9.3.2 Commercial Production of
be T

Dihydrogen Presently ~77% of the industrial

The commonly used processes are outlined dihydrogen is produced from petro-chemicals,
re
o R

below: 18% from coal, 4% from electrolysis of aqueous

(i) Electrolysis of acidified water using solutions and 1% from other sources.
platinum electrodes gives hydrogen. 9.4 PROPERTIES OF DIHYDROGEN
tt E

2H2O ( l ) ⎯⎯⎯⎯⎯⎯⎯→
Electrolysis
Traces of acid / base
2H2 ( g ) + O2 ( g ) 9.4.1 Physical Properties
(ii) High purity (>99.95%) dihydrogen is Dihydrogen is a colourless, odourless,
C

obtained by electrolysing warm aqueous tasteless, combustible gas. It is lighter than

barium hydroxide solution between nickel air and insoluble in water. Its other physical
electrodes. properties alongwith those of deuterium are
no N

(iii) It is obtained as a byproduct in the given in Table 9.1.

manufacture of sodium hydroxide and
chlorine by the electrolysis of brine 9.4.2 Chemical Properties
solution. During electrolysis, the reactions The chemical behaviour of dihydrogen (and for
©

that take place are: that matter any molecule) is determined, to a

– –
at anode: 2Cl (aq) → Cl2(g) + 2e large extent, by bond dissociation enthalpy.
– –
at cathode: 2H2O (l) + 2e → H2(g) + 2OH (aq) The H–H bond dissociation enthalpy is the
The overall reaction is highest for a single bond between two atoms
+ –
2Na (aq) + 2Cl (aq) + 2H2O(l) of any element. What inferences would you
↓ draw from this fact ? It is because of this factor
+ –
Cl2(g) + H2(g) + 2Na (aq) + 2OH (aq) that the dissociation of dihydrogen into its
(iv) Reaction of steam on hydrocarbons or coke atoms is only ~0.081% around 2000K which
at high temperatures in the presence of increases to 95.5% at 5000K. Also, it is
catalyst yields hydrogen. relatively inert at room temperature due to the

278 C:\ChemistryXI\Unit-9\Unit-9(6).pmd 10.2.6, 6.3.6, 14.3.6

 HYDROGEN 279

high H–H bond enthalpy. Thus, the atomic (i) Hydrogenation of vegetable oils using
hydrogen is produced at a high temperature nickel as catalyst gives edible fats
in an electric arc or under ultraviolet (margarine and vanaspati ghee)
radiations. Since its orbital is incomplete with (ii) Hydroformylation of olefins yields
1
1s electronic configuration, it does combine aldehydes which further undergo
with almost all the elements. It accomplishes reduction to give alcohols.
reactions by (i) loss of the only electron to
+
give H , (ii) gain of an electron to form H , and
–
H 2 + CO + RCH = CH 2 → RCH 2 CH 2 CHO
(iii) sharing electrons to form a single covalent bond.

d
H2 + RCH2 CH2 CHO → RCH2 CH 2 CH2 OH
The chemistry of dihydrogen can be
illustrated by the following reactions: Problem 9.1

he
Reaction with halogens: It reacts with Comment on the reactions of dihydrogen
halogens, X2 to give hydrogen halides, HX, with (i) chlorine, (ii) sodium, and (iii)
H2 ( g ) + X 2 ( g ) → 2HX ( g ) (X = F,Cl, Br,I) copper(II) oxide
Solution

is
While the reaction with fluorine occurs even in
the dark, with iodine it requires a catalyst. (i) Dihydrogen reduces chlorine into
–
chloride (Cl ) ion and itself gets oxidised
Reaction with dioxygen: It reacts with +

bl
to H ion by chlorine to form hydrogen
dioxygen to form water. The reaction is highly
chloride. An electron pair is shared
exothermic.
between H and Cl leading to the formation
of a covalent molecule.
pu
2H2(g) + O2 (g) 2H2O(l);
ΔH V = –285.9 kJ mol
–1
(ii) Dihydrogen is reduced by sodium to
Reaction with dinitrogen: With dinitrogen form NaH. An electron is transferred from
be T

it forms ammonia. Na to H leading to the formation of an ionic

+ –
compound, Na H .
re
3H2 ( g ) + N 2 ( g ) ⎯⎯⎯⎯⎯⎯
673K,200atm
→ 2NH3 ( g ) ;
o R

Fe (iii) Dihydrogen reduces copper(II) oxide

ΔH V = −92.6 kJ mol −1 to copper in zero oxidation state and itself
gets oxidised to H2O, which is a covalent
This is the method for the manufacture of
tt E

molecule.
ammonia by the Haber process.
Reactions with metals: With many metals it 9.4.3 Uses of Dihydrogen
C

combines at a high temperature to yield the • The largest single use of dihydrogen is in
corresponding hydrides (section 9.5) the synthesis of ammonia which is used in
H2(g) +2M(g) → 2MH(s); the manufacture of nitric acid and
no N

where M is an alkali metal nitrogenous fertilizers.

Reactions with metal ions and metal • Dihydrogen is used in the manufacture of
oxides: It reduces some metal ions in aqueous vanaspati fat by the hydrogenation of
©

solution and oxides of metals (less active than polyunsaturated vegetable oils like
iron) into corresponding metals. soyabean, cotton seeds etc.
• It is used in the manufacture of bulk
H2 ( g ) + Pd2 + ( aq ) → Pd ( s ) + 2H + ( aq )
organic chemicals, particularly methanol.
yH2 ( g ) + M x O y ( s ) → xM ( s ) + yH 2O ( l )
CO ( g ) + 2H2 ( g ) ⎯⎯⎯⎯
cobalt
catalyst
→ CH3OH ( l )
Reactions with organic compounds: It
• It is widely used for the manufacture of
reacts with many organic compounds in the
metal hydrides (Section 9.5)
presence of catalysts to give useful
hydrogenated products of commercial • It is used for the preparation of hydrogen
importance. For example : chloride, a highly useful chemical.

279 C:\ChemistryXI\Unit-9\Unit-9(6).pmd 10.2.6, 6.3.6, 14.3.6

 280 CHEMISTRY

• In metallurgical processes, it is used to Lithium hydride is rather unreactive at

reduce heavy metal oxides to metals. moderate temperatures with O2 or Cl2. It is,
• Atomic hydrogen and oxy-hydrogen therefore, used in the synthesis of other useful
torches find use for cutting and welding hydrides, e.g.,
purposes. Atomic hydrogen atoms 8LiH + Al2Cl6 → 2LiAlH4 + 6LiCl
(produced by dissociation of dihydrogen 2LiH + B2H6 → 2LiBH4
with the help of an electric arc) are allowed
to recombine on the surface to be welded 9.5.2 Covalent or Molecular Hydride
to generate the temperature of 4000 K. Dihydrogen forms molecular compounds with

d
• It is used as a rocket fuel in space research. most of the p-block elements. Most familiar
examples are CH4, NH3, H2O and HF. For

he
• Dihydrogen is used in fuel cells for convenience hydrogen compounds of non-
generating electrical energy. It has many metals have also been considered as hydrides.
advantages over the conventional fossil Being covalent, they are volatile compounds.
fuels and electric power. It does not produce
any pollution and releases greater energy Molecular hydrides are further classified

is
according to the relative numbers of electrons
per unit mass of fuel in comparison to
and bonds in their Lewis structure into :
gasoline and other fuels.
(i) electron-deficient, (ii) electron-precise,

bl
9.5 HYDRIDES and (iii) electron-rich hydrides.
Dihydrogen, under certain reaction conditions, An electron-deficient hydride, as the name
combines with almost all elements, except
pu suggests, has too few electrons for writing its
noble gases, to form binary compounds, called conventional Lewis structure. Diborane (B2H6)
hydrides. If ‘E’ is the symbol of an element then is an example. In fact all elements of group 13
hydride can be expressed as EHx (e.g., MgH2) will form electron-deficient compounds. What
be T

or EmHn (e.g., B2H6). do you expect from their behaviour? They act
as Lewis acids i.e., electron acceptors.
re
The hydrides are classified into three
o R

categories : Electron-precise compounds have the

required number of electrons to write their
(i) Ionic or saline or saltlike hydrides conventional Lewis structures. All elements of
(ii) Covalent or molecular hydrides
tt E

group 14 form such compounds (e.g., CH4)

(iii) Metallic or non-stoichiometric hydrides which are tetrahedral in geometry.
9.5.1 Ionic or Saline Hydrides Electron-rich hydrides have excess
C

These are stoichiometric compounds of electrons which are present as lone pairs.
dihydrogen formed with most of the s-block Elements of group 15-17 form such
elements which are highly electropositive in compounds. (NH3 has 1- lone pair, H2O – 2
no N

character. However, significant covalent and HF –3 lone pairs). What do you expect from
character is found in the lighter metal hydrides the behaviour of such compounds ? They will
such as LiH, BeH2 and MgH2. In fact BeH2 and behave as Lewis bases i.e., electron donors. The
MgH2 are polymeric in structure. The ionic presence of lone pairs on highly electronegative
©

hydrides are crystalline, non-volatile and non- atoms like N, O and F in hydrides results in
conducting in solid state. However, their melts hydrogen bond formation between the
conduct electricity and on electrolysis liberate molecules. This leads to the association of
dihydrogen gas at anode, which confirms the molecules.
–
existence of H ion.
Problem 9.2
2H – ( melt ) ⎯⎯⎯⎯
anode
→ H2 ( g ) + 2e − Would you expect the hydrides of N, O
Saline hydrides react violently with water and F to have lower boiling points than
producing dihydrogen gas. the hydrides of their subsequent group
NaH ( s ) + H2 O ( aq ) → NaOH ( aq ) + H2 ( g ) members ? Give reasons.

280 C:\ChemistryXI\Unit-9\Unit-9(6).pmd 10.2.6, 6.3.6, 14.3.6

 HYDROGEN 281

Solution Solution
On the basis of molecular masses of NH3, Although phosphorus exhibits +3 and +5
H2O and HF, their boiling points are oxidation states, it cannot form PH5.
expected to be lower than those of the Besides some other considerations, high
subsequent group member hydrides. ΔaH value of dihydrogen and ΔegH value
However, due to higher electronegativity of hydrogen do not favour to exhibit the
of N, O and F, the magnitude of hydrogen highest oxidation state of P, and
bonding in their hydrides will be quite consequently the formation of PH5.
appreciable. Hence, the boiling points

d
NH3, H2O and HF will be higher than the
hydrides of their subsequent group 9.6 WATER

he
members. A major part of all living organisms is made
up of water. Human body has about 65% and
9.5.3 Metallic or Non-stoichiometric some plants have as much as 95% water. It is
(or Interstitial ) Hydrides a crucial compound for the survival of all life

is
These are formed by many d-block and f-block forms. It is a solvent of great importance. The
elements. However, the metals of group 7, 8 distribution of water over the earth’s surface
and 9 do not form hydride. Even from group is not uniform. The estimated world water

bl
6, only chromium forms CrH. These hydrides supply is given in Table 9.2
conduct heat and electricity though not as Table 9.2 Estimated World Water Supply
efficiently as their parent metals do. Unlike
pu
saline hydrides, they are almost always non- Source % of Total
stoichiometric, being deficient in hydrogen. For
Oceans 97.33
example, LaH2.87, YbH2.55, TiH1.5–1.8, ZrH1.3–1.75,
be T

Saline lakes and inland seas 0.008

VH 0.56 , NiH 0.6–0.7 , PdH 0.6–0.8 etc. In such
hydrides, the law of constant composition does Polar ice and glaciers 2.04
re
o R

not hold good. Ground water 0.61

Earlier it was thought that in these Lakes 0.009
hydrides, hydrogen occupies interstices in the Soil moisture 0.005
tt E

metal lattice producing distortion without any Atmospheric water vapour 0.001
change in its type. Consequently, they were Rivers 0.0001
termed as interstitial hydrides. However, recent
C

studies have shown that except for hydrides 9.6.1 Physical Properties of Water
of Ni, Pd, Ce and Ac, other hydrides of this class It is a colourless and tasteless liquid. Its
no N

have lattice different from that of the parent physical properties are given in Table 9.3 along
metal. The property of absorption of hydrogen with the physical properties of heavy water.
on transition metals is widely used in catalytic The unusual properties of water in the
reduction / hydrogenation reactions for the
condensed phase (liquid and solid states) are
©

preparation of large number of compounds.

due to the presence of extensive hydrogen
Some of the metals (e.g., Pd, Pt) can
bonding between water molecules. This leads
accommodate a very large volume of hydrogen
to high freezing point, high boiling point, high
and, therefore, can be used as its storage
heat of vaporisation and high heat of fusion in
media. This property has high potential for
comparison to H2S and H2Se. In comparison
hydrogen storage and as a source of energy.
to other liquids, water has a higher specific
Problem 9.3 heat, thermal conductivity, surface tension,
dipole moment and dielectric constant, etc.
Can phosphorus with outer electronic These properties allow water to play a key role
2 3
configuration 3s 3p form PH5 ?
in the biosphere.

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 282 CHEMISTRY

Table 9.3 Physical Properties of H2O and D2O

Property H2 O D2O
–1
Molecular mass (g mol ) 18.0151 20.0276
Melting point/K 273.0 276.8
Boiling point/K 373.0 374.4
–1
Enthalpy of formation/kJ mol –285.9 –294.6
–1
Enthalpy of vaporisation (373K)/kJ mol 40.66 41.61

d
–1
Enthalpy of fusion/kJ mol 6.01 -
Temp of max. density/K 276.98 284.2

he
–3
Density (298K)/g cm 1.0000 1.1059
Viscosity/centipoise 0.8903 1.107
2 2
Dielectric constant/C /N.m 78.39 78.06
–1 –1 –8

is
Electrical conductivity (293K/ohm cm ) 5.7 10 -

The high heat of vaporisation and heat polar molecule, (Fig 9.1(b)). Its orbital overlap

bl
capacity are responsible for moderation of the picture is shown in Fig. 9.1(c). In the liquid
climate and body temperature of living beings. phase water molecules are associated together
It is an excellent solvent for transportation of by hydrogen bonds.
ions and molecules required for plant and
pu
animal metabolism. Due to hydrogen bonding
with polar molecules, even covalent
The crystalline form of water is ice. At
atmospheric pressure ice crystallises in the
hexagonal form, but at very low temperatures
be T

compounds like alcohol and carbohydrates it condenses to cubic form. Density of ice is
dissolve in water. less than that of water. Therefore, an ice cube
re
floats on water. In winter season ice formed
o R

9.6.2 Structure of Water

on the surface of a lake provides thermal
In the gas phase water is a bent molecule with
insulation which ensures the survival of the
a bond angle of 104.5°, and O–H bond length
aquatic life. This fact is of great ecological
tt E

of 95.7 pm as shown in Fig 9.1(a). It is a highly

significance.
9.6.3 Structure of Ice
C

Ice has a highly ordered three dimensional

hydrogen bonded structure as shown in
no N

Fig. 9.2. Examination of ice crystals with

©

Fig. 9.1 (a) The bent structure of water; (b) the

water molecule as a dipole and
(c) the orbital overlap picture in water
molecule. Fig. 9.2 The structure of ice

282 C:\ChemistryXI\Unit-9\Unit-9(6).pmd 10.2.6, 6.3.6, 14.3.6

 HYDROGEN 283

X-rays shows that each oxygen atom is N 3 − ( s ) + 3H2O ( l ) → NH3 ( g ) + 3OH − ( aq )

surrounded tetrahedrally by four other oxygen
atoms at a distance of 276 pm. (4) Hydrates Formation: From aqueous
solutions many salts can be crystallised as
Hydrogen bonding gives ice a rather open
hydrated salts. Such an association of water
type structure with wide holes. These holes can
is of different types viz.,
hold some other molecules of appropriate size
interstitially. (i) coordinated water e.g.,
3+

9.6.4 Chemical Properties of Water

⎡⎣Cr ( H2O )6 ⎤⎦ 3Cl –

d
Water reacts with a large number of (ii) interstitial water e.g., BaCl 2 .2H2 O
substances. Some of the important reactions (iii) hydrogen-bonded water e.g.,

he
are given below. 2+
⎡⎣Cu ( H2O )4 ⎤⎦ SO2–
4 .H2 O in CuSO 4 .5H 2 O,
(1) Amphoteric Nature: It has the ability to
act as an acid as well as a base i.e., it behaves Problem 9.4
as an amphoteric substance. In the Brönsted
How many hydrogen-bonded water

is
sense it acts as an acid with NH3 and a base
with H2S. molecule(s) are associated in
CuSO4.5H2O?
H2 O ( l ) + NH3 ( aq ) ( aq ) + NH 4 ( aq )
– +

bl
OH Solution
H2 O ( l ) + H2 S ( aq ) H3O + ( aq ) + HS – ( aq ) Only one water molecule, which is outside
the brackets (coordination sphere), is
The auto-protolysis (self-ionization) of water
pu
takes place as follows :
hydrogen-bonded. The other four
molecules of water are coordinated.
H2 O ( l ) + H2 O ( l )
H3O + ( aq ) + OH – ( aq )
be T

9.6.5 Hard and Soft Water

acid-1 base-2 acid-2 base-1
re
(acid) (base) (conjugate (conjugate Rain water is almost pure (may contain some
o R

acid) base) dissolved gases from the atmosphere). Being a

(2) Redox Reactions Involving Water: Water good solvent, when it flows on the surface of
can be easily reduced to dihydrogen by highly the earth, it dissolves many salts. Presence of
tt E

electropositive metals. calcium and magnesium salts in the form of

hydrogencarbonate, chloride and sulphate in
2H2O ( l ) + 2Na ( s ) → 2NaOH ( aq ) + H2 ( g ) water makes water ‘hard’. Hard water does
C

Thus, it is a great source of dihydrogen. not give lather with soap. Water free from
Water is oxidised to O2 during photosynthesis. soluble salts of calcium and magnesium is
no N

called Soft water. It gives lather with soap

6CO2(g) + 12H2O(l) → C6H12O6(aq) + 6H2O(l)
easily.
+ 6O2(g)
Hard water forms scum/precipitate with
With fluorine also it is oxidised to O2. soap. Soap containing sodium stearate
©

+ –
2F2(g) + 2H2O(l) → 4H (aq) + 4F (aq) + O2(g) (C17H35COONa) reacts with hard water to
(3) Hydrolysis Reaction: Due to high precipitate out Ca/Mg stearate.
dielectric constant, it has a very strong 2C17 H35 COONa ( aq ) + M2 + ( aq ) →
hydrating tendency. It dissolves many ionic
compounds. However, certain covalent and ( C17 H35COO )2 M ↓ +2Na + ( aq ) ; M is Ca / Mg
some ionic compounds are hydrolysed in water.
It is, therefore, unsuitable for laundry. It is
P4 O10 ( s ) + 6H2 O ( l ) → 4H3 PO4 ( aq ) harmful for boilers as well, because of
deposition of salts in the form of scale. This
SiCl 4 ( l ) + 2H2 O ( l ) → SiO2 ( s ) + 4HCl ( aq ) reduces the efficiency of the boiler. The

283 C:\ChemistryXI\Unit-9\Unit-9(6).pmd 10.2.6, 6.3.6, 14.3.6

 284 CHEMISTRY

hardness of water is of two types: (i) temporary Na 6 P6 O18 → 2Na + + Na 4 P6 O18

2–

hardness, and (ii) permanent hardness. (M = Mg, Ca)

9.6.6 Temporary Hardness
→ [ Na 2 MP6 O18 ]
2−
M2 + + Na 4 P6 O18
2−
+ 2Na +
Temporary hardness is due to the presence of 2+ 2+
magnesium and calcium hydrogen- The complex anion keeps the Mg and Ca
carbonates. It can be removed by : ions in solution.
(i) Boiling: During boiling, the soluble (iii) Ion-exchange method: This method is
Mg(HCO3)2 is converted into insoluble Mg(OH)2 also called zeolite/permutit process. Hydrated

d
and Ca(HCO3)2 is changed to insoluble CaCO3. sodium aluminium silicate is zeolite/permutit.
It is because of high solubility product of For the sake of simplicity, sodium aluminium

he
Mg(OH)2 as compared to that of MgCO3, that silicate (NaAlSiO4) can be written as NaZ. When
Mg(OH)2 is precipitated. These precipitates can this is added in hard water, exchange reactions
be removed by filtration. Filtrate thus obtained take place.
will be soft water. 2NaZ ( s ) + M2 + ( aq ) → MZ 2 ( s ) + 2Na + ( aq )

is
Mg ( HCO3 )2 ⎯⎯⎯⎯
Heating
→ Mg ( OH )2 ↓ + 2CO2 ↑ (M = Mg, Ca)
Ca ( HCO3 )2 ⎯⎯⎯⎯
Heating
→ CaCO3 ↓ +H2O + CO2 ↑ Permutit/zeolite is said to be exhausted

bl
when all the sodium in it is used up. It is
(ii) Clark’s method: In this method calculated regenerated for further use by treating with an
amount of lime is added to hard water. It aqueous sodium chloride solution.
precipitates out calcium carbonate and
pu
magnesium hydroxide which can be filtered off. MZ 2 ( s ) + 2NaCl ( aq ) → 2NaZ ( s ) + MCl 2 ( aq )
Ca ( HCO 3 )2 + Ca ( OH )2 → 2CaCO 3 ↓ +2H 2 O (iv) Synthetic resins method: Nowadays
be T

hard water is softened by using synthetic

Mg ( HCO3 )2 + 2Ca ( OH)2 → 2CaCO3 ↓ cation exchangers. This method is more efficient
re
than zeolite process. Cation exchange resins
o R

+ Mg ( OH )2 ↓ +2H2O contain large organic molecule with - SO3H

9.6.7 Permanent Hardness group and are water insoluble. Ion exchange
tt E

resin (RSO3H) is changed to RNa by treating it

It is due to the presence of soluble salts of +
with NaCl. The resin exchanges Na ions with
magnesium and calcium in the form of 2+ 2+
Ca and Mg ions present in hard water to
chlorides and sulphates in water. Permanent
C

make the water soft. Here R is resin anion.

hardness is not removed by boiling. It can be
removed by the following methods: 2RNa ( s ) + M2+ ( aq ) → R 2 M ( s ) + 2Na + ( aq )
no N

(i) Treatment with washing soda (sodium The resin can be regenerated by adding
carbonate): Washing soda reacts with soluble aqueous NaCl solution.
calcium and magnesium chlorides and
Pure de-mineralised (de-ionized) water free
sulphates in hard water to form insoluble
from all soluble mineral salts is obtained by
©

carbonates. passing water successively through a cation

+
MCl 2 + Na 2 CO3 → MCO3 ↓ + 2NaCl exchange (in the H form) and an anion-
–
exchange (in the OH form) resins:
(M = Mg, Ca)
MSO4 + Na 2 CO3 → MCO3 ↓ + Na 2 SO 4 2RH ( s ) + M2+ ( aq ) MR 2 ( s ) + 2H + ( aq )
+
In this cation exchange process, H exchanges
(ii) Calgon’s method: Sodium hexameta- + 2+ 2+
for Na , Ca , Mg and other cations present
phosphate (Na6P6O18), commercially called in water. This process results in proton release
‘calgon’, when added to hard water, the and thus makes the water acidic. In the anion
following reactions take place. exchange process:

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 HYDROGEN 285

RNH 2 ( s ) + H 2 O ( l ) RNH 3+ .OH − ( s ) ⎯⎯⎯⎯

2 − ethylanthraquinol ←⎯⎯⎯
O2 ( air )
→ H2 O 2 +
⎯ H2 / Pd

RNH .OH +
3
–
( s ) + X ( aq )
−
RNH .X +
3 (s)
−
( oxidised product )
+ OH − ( aq ) In this case 1% H 2O2 is formed. It is
– – – 2– extracted with water and concentrated to ~30%
OH exchanges for anions like Cl , HCO3, SO4 (by mass) by distillation under reduced
–
etc. present in water. OH ions, thus, liberated pressure. It can be further concentrated to
+
neutralise the H ions set free in the cation ~85% by careful distillation under low

d
exchange. pressure. The remaining water can be frozen
H + ( aq ) + OH − ( aq ) → H2O ( l ) out to obtain pure H2O2.

he
The exhausted cation and anion exchange 9.7.2 Physical Properties
resin beds are regenerated by treatment with In the pure state H2O2 is an almost colourless
(very pale blue) liquid. Its important physical
dilute acid and alkali solutions respectively.
properties are given in Table 9.4.
9.7 HYDROGEN PEROXIDE (H2O2) H 2 O 2 is miscible with water in all

is
proportions and forms a hydrate H2O2.H2O
Hydrogen peroxide is an important chemical
(mp 221K). A 30% solution of H2O2 is marketed
used in pollution control treatment of domestic
as ‘100 volume’ hydrogen peroxide. It means

bl
and industrial effluents.
that one millilitre of 30% H2O2 solution will give
9.7.1 Preparation 100 mL of oxygen at STP. Commercially
It can be prepared by the following methods.
pu marketed sample is 10 V, which means that
the sample contains 3% H2O2.
(i) Acidifying barium peroxide and removing
excess water by evaporation under reduced Problem 9.5
be T

pressure gives hydrogen peroxide. Calculate the strength of 10 volume

solution of hydrogen peroxide.
BaO2 .8H2O ( s ) + H2 SO4 ( aq ) → BaSO4 ( s ) +
re
Solution
o R

H2O2 ( aq ) + 8H2O ( l ) 10 volume solution of H2O2 means that

1L of this H2O2 solution will give 10 L of
(ii) Peroxodisulphate, obtained by electrolytic
oxygen at STP
tt E

oxidation of acidified sulphate solutions at

high current density, on hydrolysis yields 2H2O2 ( l ) → O2 ( g ) + H2O ( l )
hydrogen peroxide. 234 g 22.7 L at STP
C

68 g
2HSO4− ( aq ) ⎯⎯⎯⎯⎯
Electrolysis
→ HO3 SOOSO3H ( aq )
On the basis of above equation 22.7 L of
⎯⎯⎯⎯⎯
Hydrolysis
→ 2HSO4− ( aq ) + 2H+ ( aq ) + H2O2 ( aq )
no N

O2 is produced from 68 g H2O2 at STP

This method is now used for the laboratory 10 L of O 2 at STP is produced from
preparation of D2O2. 68 10
g = 29.9 g 30 g H2O2
K2S2O8 ( s) + 2D2O ( l ) → 2KDSO4 ( aq ) + D2O2 ( l ) 22.7
©

Therefore, strength of H2O2 in 10 volume

(iii) Industrially it is prepared by the auto- H 2 O 2 solution = 30 g/L = 3% H 2 O 2
oxidation of 2-alklylanthraquinols. solution
Table 9.4 Physical Properties of Hydrogen Peroxide
–3
Melting point/K 272.4 Density (liquid at 298 K)/g cm 1.44
Boiling point(exrapolated)/K 423 Viscosity (290K)/centipoise 1.25
2 2
Vapour pressure(298K)/mmHg 1.9 Dielectric constant (298K)/C /N m 70.7
–3 –1 –1 –8
Density (solid at 268.5K)/g cm 1.64 Electrical conductivity (298K)/Ω cm 5.1 10

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 286 CHEMISTRY

9.7.3 Structure reaction is catalysed. It is, therefore, stored in

Hydrogen peroxide has a non-planar wax-lined glass or plastic vessels in dark. Urea
structure. The molecular dimensions in the gas can be added as a stabiliser. It is kept away
phase and solid phase are shown in Fig 9.3 from dust because dust can induce explosive
decomposition of the compound.
9.7.6 Uses
Its wide scale use has led to tremendous
increase in the industrial production of H2O2.

d
Some of the uses are listed below:
(i) In daily life it is used as a hair bleach and

he
as a mild disinfectant. As an antiseptic it is
sold in the market as perhydrol.
Fig. 9.3 (a) H2O2 structure in gas phase, dihedral
angle is 111.5°. (b) H2O2 structure in solid (ii) It is used to manufacture chemicals like
phase at 110K, dihedral angle is 90.2°. sodium perborate and per-carbonate,

is
which are used in high quality detergents.
9.7.4 Chemical Properties (iii) It is used in the synthesis of hydroquinone,
It acts as an oxidising as well as reducing agent tartaric acid and certain food products and

bl
in both acidic and alkaline media. Simple pharmaceuticals (cephalosporin) etc.
reactions are described below. (iv) It is employed in the industries as a
(i) Oxidising action in acidic medium
pu bleaching agent for textiles, paper pulp,
leather, oils, fats, etc.
2Fe2+ ( aq ) + 2H+ ( aq ) + H2O2 ( aq ) → (v) Nowadays it is also used in Environmental
2Fe3+ ( aq ) + 2H2O ( l ) (Green) Chemistry. For example, in
be T

pollution control treatment of domestic and

PbS ( s ) + 4H2O2 ( aq ) → PbSO4 ( s ) + 4H2O ( l )
re
industrial effluents, oxidation of cyanides,
o R

restoration of aerobic conditions to sewage

(ii) Reducing action in acidic medium
wastes, etc.
2MnO4– + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2
tt E

9.8 HEAVY WATER, D2O

HOCl + H2O2 → H3O+ + Cl − + O2 It is extensively used as a moderator in nuclear
reactors and in exchange reactions for the
C

(iii) Oxidising action in basic medium

study of reaction mechanisms. It can be
2Fe2 + + H2 O2 → 2Fe3 + + 2OH− prepared by exhaustive electrolysis of water or
as a by-product in some fertilizer industries.
no N

Mn2+ + H2O2 → Mn 4 + + 2OH− Its physical properties are given in Table 9.3.
(iv) Reducing action in basic medium It is used for the preparation of other deuterium
compounds, for example:
I2 + H2O2 + 2OH − → 2I− + 2H2 O + O2
©

2MnO4– + 3H2O2 → 2MnO2 + 3O2 + CaC2 + 2D2O → C2 D2 + Ca ( OD )2

2H2O + 2OH – SO3 + D2O → D2 SO4

Al 4 C3 + 12D2 O → 3CD4 + 4Al ( OD )3
9.7.5 Storage
H2O2 decomposes slowly on exposure to light. 9.9 DIHYDROGEN AS A FUEL
2H2 O2 ( l ) → 2H2 O ( l ) + O2 ( g ) Dihydrogen releases large quantities of heat on
combustion. The data on energy released by
In the presence of metal surfaces or traces of combustion of fuels like dihydrogen, methane,
alkali (present in glass containers), the above LPG etc. are compared in terms of the same

286 C:\ChemistryXI\Unit-9\Unit-9(6).pmd 10.2.6, 6.3.6, 14.3.6

 HYDROGEN 287

amounts in mole, mass and volume, are shown limitations have prompted researchers to
in Table 9.5. search for alternative techniques to use
From this table it is clear that on a mass dihydrogen in an efficient way.
for mass basis dihydrogen can release more In this view Hydrogen Economy is an
energy than petrol (about three times). alternative. The basic principle of hydrogen
Moreover, pollutants in combustion of economy is the transportation and storage of
dihydrogen will be less than petrol. The only energy in the form of liquid or gaseous
pollutants will be the oxides of dinitrogen (due dihydrogen. Advantage of hydrogen economy
to the presence of dinitrogen as impurity with

d
is that energy is transmitted in the form of
dihydrogen). This, of course, can be minimised dihydrogen and not as electric power. It is for
by injecting a small amount of water into the the first time in the history of India that a pilot

he
cylinder to lower the temperature so that the project using dihydrogen as fuel was launched
reaction between dinitrogen and dioxygen may in October 2005 for running automobiles.
not take place. However, the mass of the Initially 5% dihydrogen has been mixed in
containers in which dihydrogen will be kept CNG for use in four-wheeler vehicles. The

is
must be taken into consideration. A cylinder percentage of dihydrogen would be gradually
of compressed dihydrogen weighs about 30 increased to reach the optimum level.
times as much as a tank of petrol containing
Nowadays, it is also used in fuel cells for

bl
the same amount of energy. Also, dihydrogen
gas is converted into liquid state by cooling to generation of electric power. It is expected that
20K. This would require expensive insulated economically viable and safe sources of
dihydrogen will be identified in the years to
pu
tanks. Tanks of metal alloy like NaNi5, Ti–TiH2,
Mg–MgH 2 etc. are in use for storage of
dihydrogen in small quantities. These
come, for its usage as a common source of
energy.
be T

Table 9.5 The Energy Released by Combustion of Various Fuels in Moles, Mass and Volume
re
o R

Energy released on Dihydrogen Dihydrogen LPG CH4 gas Octane

combustion in kJ (in gaseous (in liquid) (in liquid
state) state) state)
tt E

per mole 286 285 2220 880 5511

per gram 143 142 50 53 47
C

per litre 12 9968 25590 35 34005

no N

SUMMARY
©

Hydrogen is the lightest atom with only one electron. Loss of this electron results in an
elementary particle, the proton. Thus, it is unique in character. It has three isotopes,
1 2 3
namely : protium (1H), deuterium (D or 1H) and tritium (T or 1H). Amongst these three,
only tritium is radioactive. Inspite of its resemblance both with alkali metals and halogens,
it occupies a separate position in the periodic table because of its unique properties.
Hydrogen is the most abundant element in the universe. In the free state it is almost
not found in the earth’s atmosphere. However, in the combined state, it is the third most
abundant element on the earth’s surface.
Dihydrogen on the industrial scale is prepared by the water-gas shift reaction from
petrochemicals. It is obtained as a byproduct by the electrolysis of brine.

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 288 CHEMISTRY

–1
The H–H bond dissociation enthalpy of dihydrogen (435.88 kJ mol ) is the highest
for a single bond between two atoms of any elements. This property is made use of in the
atomic hydrogen torch which generates a temperature of ~4000K and is ideal for welding
of high melting metals.
Though dihydrogen is rather inactive at room temperature because of very high
negative dissociation enthalpy, it combines with almost all the elements under appropriate
conditions to form hydrides. All the type of hydrides can be classified into three categories:
ionic or saline hydrides, covalent or molecular hydrides and metallic or non-stoichiometric
hydrides. Alkali metal hydrides are good reagents for preparing other hydride compounds.

d
Molecular hydrides (e.g., B2H6, CH4, NH3, H2O) are of great importance in day-to-day life.
Metallic hydrides are useful for ultrapurification of dihydrogen and as dihydrogen storage

he
media.
Among the other chemical reactions of dihydrogen, reducing reactions leading to
the formation hydrogen halides, water, ammonia, methanol, vanaspati ghee, etc. are of
great importance. In metallurgical process, it is used to reduce metal oxides. In space
programmes, it is used as a rocket fuel. In fact, it has promising potential for use as a

is
non-polluting fuel of the near future (Hydrogen Economy).
Water is the most common and abundantly available substance. It is of a great
chemical and biological significance. The ease with which water is transformed from

bl
liquid to solid and to gaseous state allows it to play a vital role in the biosphere. The
water molecule is highly polar in nature due to its bent structure. This property leads to
hydrogen bonding which is the maximum in ice and least in water vapour. The polar
nature of water makes it: (a) a very good solvent for ionic and partially ionic compounds;
pu
(b) to act as an amphoteric (acid as well as base) substance; and (c) to form hydrates of
different types. Its property to dissolve many salts, particularly in large quantity, makes
it hard and hazardous for industrial use. Both temporary and permanent hardness can
be T

be removed by the use of zeolites, and synthetic ion-exchangers.

re
Heavy water, D2O is another important compound which is manufactured by the
o R

electrolytic enrichment of normal water. It is essentially used as a moderator in nuclear

reactors.
Hydrogen peroxide, H2O2 has an interesting non-polar structure and is widely used
tt E

as an industrial bleach and in pharmaceutical and pollution control treatment of

industrial and domestic effluents.
C

EXERCISES
no N

9.1 Justify the position of hydrogen in the periodic table on the basis of its electronic
configuration.
9.2 Write the names of isotopes of hydrogen. What is the mass ratio of these isotopes?
9.3 Why does hydrogen occur in a diatomic form rather than in a monoatomic form
©

under normal conditions?

9.4 How can the production of dihydrogen, obtained from ‘coal gasification’, be
increased?
9.5 Describe the bulk preparation of dihydrogen by electrolytic method. What is the
role of an electrolyte in this process ?
9.6 Complete the following reactions:
Δ
(i) H 2 ( g ) + M m O o ( s ) ⎯⎯⎯→

Δ
(ii) CO ( g ) + H 2 ( g ) ⎯ ⎯ ⎯ ⎯→
catalyst

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 HYDROGEN 289

Δ
(iii) C3H8 ( g ) + 3H2O ( g ) ⎯⎯⎯⎯
catalyst
→

(iv) Zn ( s ) + NaOH ( aq ) ⎯⎯⎯⎯

heat
→
9.7 Discuss the consequences of high enthalpy of H–H bond in terms of chemical
reactivity of dihydrogen.
9.8 What do you understand by (i) electron-deficient, (ii) electron-precise, and (iii)
electron-rich compounds of hydrogen? Provide justification with suitable examples.
9.9 What characteristics do you expect from an electron-deficient hydride with respect
to its structure and chemical reactions?

d
9.10 Do you expect the carbon hydrides of the type (CnH2n + 2) to act as ‘Lewis’ acid or
base? Justify your answer.

he
9.11 What do you understand by the term “non-stoichiometric hydrides”? Do you
expect this type of the hydrides to be formed by alkali metals? Justify your answer.
9.12 How do you expect the metallic hydrides to be useful for hydrogen storage?
Explain.

is
9.13 How does the atomic hydrogen or oxy-hydrogen torch function for cutting and
welding purposes ? Explain.
9.14 Among NH3, H2O and HF, which would you expect to have highest magnitude of

bl
hydrogen bonding and why?
9.15 Saline hydrides are known to react with water violently producing fire. Can CO2,
pu a well known fire extinguisher, be used in this case? Explain.
9.16 Arrange the following
(i) CaH2, BeH2 and TiH2 in order of increasing electrical conductance.
be T

(ii) LiH, NaH and CsH in order of increasing ionic character.

(iii) H–H, D–D and F–F in order of increasing bond dissociation enthalpy.
re
(iv) NaH, MgH2 and H2O in order of increasing reducing property.
o R

9.17 Compare the structures of H2O and H2O2.

9.18 What do you understand by the term ’auto-protolysis’ of water? What is its
tt E

significance?
9.19 Consider the reaction of water with F2 and suggest, in terms of oxidation and
reduction, which species are oxidised/reduced.
C

9.20 Complete the following chemical reactions.

(i) Pb S ( s ) + H 2O2 ( aq ) →
no N

(ii) MnO4– ( aq ) + H2O2 ( aq ) →

(iii) CaO ( s ) + H2O ( g ) →

(v) AlCl 3 ( g ) + H2O ( l ) →
©

(vi) Ca 3 N 2 ( s ) + H2O ( l ) →
Classify the above into (a) hydrolysis, (b) redox and (c) hydration reactions.
9.21 Describe the structure of the common form of ice.
9.22 What causes the temporary and permanent hardness of water ?
9.23 Discuss the principle and method of softening of hard water by synthetic ion-
exchange resins.
9.24 Write chemical reactions to show the amphoteric nature of water.
9.25 Write chemical reactions to justify that hydrogen peroxide can function as an
oxidising as well as reducing agent.

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 290 CHEMISTRY

9.26 What is meant by ‘demineralised’ water and how can it be obtained ?

9.27 Is demineralised or distilled water useful for drinking purposes? If not, how can
it be made useful?
9.28 Describe the usefulness of water in biosphere and biological systems.
9.29 What properties of water make it useful as a solvent? What types of compound
can it (i) dissolve, and (ii) hydrolyse ?
9.30 Knowing the properties of H2O and D2O, do you think that D2O can be used for
drinking purposes?

d
9.31 What is the difference between the terms ‘hydrolysis’ and ‘hydration’ ?
9.32 How can saline hydrides remove traces of water from organic compounds?

he
9.33 What do you expect the nature of hydrides is, if formed by elements of atomic
numbers 15, 19, 23 and 44 with dihydrogen? Compare their behaviour towards
water.
9.34 Do you expect different products in solution when aluminium(III) chloride and
potassium chloride treated separately with (i) normal water (ii) acidified water,

is
and (iii) alkaline water? Write equations wherever necessary.
9.35 How does H2O2 behave as a bleaching agent?

bl
9.36 What do you understand by the terms:
(i) hydrogen economy (ii) hydrogenation (iii) ‘syngas’ (iv) water-gas shift reaction
(v) fuel-cell ?
pu
be T
re
o R
tt E
C
no N
©

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 THE s-BLOCK ELEMENTS 291

UNIT 10

THE s -BLOCK ELEMENTS

d
he
The first element of alkali and alkaline earth metals differs
in many respects from the other members of the group

is
After studying this unit, you will be
able to

bl
The s-block elements of the Periodic Table are those in
• describe the general charact- which the last electron enters the outermost s-orbital. As
eristics of the alkali metals and
pu the s-orbital can accommodate only two electrons, two
their compounds;
groups (1 & 2) belong to the s-block of the Periodic Table.
• explain the general characteristics Group 1 of the Periodic Table consists of the elements:
of the alkaline earth metals and lithium, sodium, potassium, rubidium, caesium and
be T

their compounds; francium. They are collectively known as the alkali metals.
These are so called because they form hydroxides on
re
• describe the manufacture,
o R

properties and uses of industrially reaction with water which are strongly alkaline in nature.
important sodium and calcium The elements of Group 2 include beryllium, magnesium,
compounds including Portland calcium, strontium, barium and radium. These elements
tt E

cement; with the exception of beryllium are commonly known as

• appreciate the biological the alkaline earth metals. These are so called because their
significance of sodium, oxides and hydroxides are alkaline in nature and these
C

potassium, magnesium and metal oxides are found in the earth’s crust*.
calcium. Among the alkali metals sodium and potassium are
abundant and lithium, rubidium and caesium have much
no N

lower abundances (Table 10.1). Francium is highly

223
radioactive; its longest-lived isotope Fr has a half-life of
only 21 minutes. Of the alkaline earth metals calcium and
magnesium rank fifth and sixth in abundance respectively
©

in the earth’s crust. Strontium and barium have much

lower abundances. Beryllium is rare and radium is the
–10
rarest of all comprising only 10 per cent of igneous
rocks† (Table 10.2, page 299).
The general electronic configuration of s-block elements
1 2
is [noble gas]ns for alkali metals and [noble gas] ns for
alkaline earth metals.
† A type of rock formed
* The thin, rocky outer layer of the Earth is crust.
from magma (molten rock) that has cooled and hardened.

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 292 CHEMISTRY

Lithium and beryllium, the first elements increase in atomic number, the atom becomes
+
of Group 1 and Group 2 respectively exhibit larger. The monovalent ions (M ) are smaller
some properties which are different from those than the parent atom. The atomic and ionic
of the other members of the respective group. radii of alkali metals increase on moving down
In these anomalous properties they resemble the group i.e., they increase in size while going
the second element of the following group. from Li to Cs.
Thus, lithium shows similarities to magnesium 10.1.3 Ionization Enthalpy
and beryllium to aluminium in many of their
The ionization enthalpies of the alkali metals
properties. This type of diagonal similarity is

d
are considerably low and decrease down the
commonly referred to as diagonal relationship
group from Li to Cs. This is because the effect
in the periodic table. The diagonal relationship
of increasing size outweighs the increasing

he
is due to the similarity in ionic sizes and /or
nuclear charge, and the outermost electron is
charge/radius ratio of the elements.
very well screened from the nuclear charge.
Monovalent sodium and potassium ions and
divalent magnesium and calcium ions are 10.1.4 Hydration Enthalpy
found in large proportions in biological fluids.

is
The hydration enthalpies of alkali metal ions
These ions perform important biological decrease with increase in ionic sizes.
functions such as maintenance of ion balance + + + + +
Li > Na > K > Rb > Cs
and nerve impulse conduction.

bl
+
Li has maximum degree of hydration and
10.1 GROUP 1 ELEMENTS: ALKALI for this reason lithium salts are mostly
METALS
pu hydrated, e.g., LiCl· 2H2O
The alkali metals show regular trends in their 10.1.5 Physical Properties
physical and chemical properties with the
All the alkali metals are silvery white, soft and
increasing atomic number. The atomic,
be T

light metals. Because of the large size, these

physical and chemical properties of alkali
elements have low density which increases
re
metals are discussed below.
down the group from Li to Cs. However,
o R

10.1.1 Electronic Configuration potassium is lighter than sodium. The melting

All the alkali metals have one valence electron, and boiling points of the alkali metals are low
1 indicating weak metallic bonding due to the
tt E

ns (Table 10.1) outside the noble gas core.

The loosely held s-electron in the outermost presence of only a single valence electron in
valence shell of these elements makes them the them. The alkali metals and their salts impart
C

most electropositive metals. They readily lose characteristic colour to an oxidizing flame. This
+
electron to give monovalent M ions. Hence they is because the heat from the flame excites the
are never found in free state in nature. outermost orbital electron to a higher energy
no N

level. When the excited electron comes back to

Element Symbol Electronic configuration the ground state, there is emission of radiation
in the visible region as given below:
Lithium Li 1s22s1
Metal Li Na K Rb Cs
Sodium Na 1s22s22p63s1
©

Potassium K 1s22s22p63s23p64s1 Colour Crimson Yellow Violet Red Blue

Rubidium Rb 1s 2s 2p 3s 3p 3d 4s 4p 5s
2 2 6 2 6 10 2 6 1 red violet
Caesium Cs 1s22s22p63s23p63d104s2 λ/nm 670.8 589.2 766.5 780.0 455.5
4p64d105s25p66s1 or [Xe] 6s1 Alkali metals can therefore, be detected by
Francium Fr [Rn]7s1 the respective flame tests and can be
determined by flame photometry or atomic
10.1.2 Atomic and Ionic Radii absorption spectroscopy. These elements when
The alkali metal atoms have the largest sizes irradiated with light, the light energy absorbed
in a particular period of the periodic table. With may be sufficient to make an atom lose electron.

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 THE s-BLOCK ELEMENTS 293

Table 10.1 Atomic and Physical Properties of the Alkali Metals

Property Lithium Sodium Potassium Rubidium Caesium Francium

Li Na K Rb Cs Fr
Atomic number 3 11 19 37 55 87
–1
Atomic mass (g mol ) 6.94 22.99 39.10 85.47 132.91 (223)
1 1 1 1 1 1
Electronic [He] 2s [Ne] 3s [Ar] 4s [Kr] 5s [Xe] 6s [Rn] 7s
configuration

d
Ionization 520 496 419 403 376 ~375
–1
enthalpy / kJ mol

he
Hydration –506 –406 –330 –310 –276 –
–1
enthalpy/kJ mol
Metallic 152 186 227 248 265 –
radius / pm

is
Ionic radius 76 102 138 152 167 (180)
+
M / pm

bl
m.p. / K 454 371 336 312 302 –
b.p / K 1615 1156 1032 961 944 –
pu
Density / g cm
–3

Standard potentials
0 +
0.53
–3.04
0.97
–2.714
0.86
–2.925
1.53
–2.930
1.90
–2.927
–
–
E / V for (M / M)
be T

Occurrence in 18* 2.27** 1.84** 78-12* 2-6* ~ 10

–18 *
re
lithosphere†
o R

*ppm (part per million), ** percentage by weight; † Lithosphere: The Earth’s outer layer: its crust
and part of the upper mantle
tt E

This property makes caesium and potassium 2 Na + O2 → Na 2 O 2 (peroxide)

useful as electrodes in photoelectric cells.
M + O 2 → MO 2 (superoxide)
C

10.1.6 Chemical Properties

(M = K, Rb, Cs)
The alkali metals are highly reactive due to
In all these oxides the oxidation state of the
their large size and low ionization enthalpy. The
no N

alkali metal is +1. Lithium shows exceptional

reactivity of these metals increases down the
behaviour in reacting directly with nitrogen of
group.
air to form the nitride, Li3N as well. Because of
(i) Reactivity towards air: The alkali metals their high reactivity towards air and water,
tarnish in dry air due to the formation of
©

alkali metals are normally kept in kerosene oil.

their oxides which in turn react with
moisture to form hydroxides. They burn Problem 10.1
vigorously in oxygen forming oxides. What is the oxidation state of K in KO2?
Lithium forms monoxide, sodium forms
peroxide, the other metals form Solution
–
superoxides. The superoxide O2 ion is The superoxide species is represented as
–
stable only in the presence of large cations O 2 ; since the compound is neutral,
such as K, Rb, Cs. therefore, the oxidation state of potassium
is +1.
4 Li + O 2 → 2 Li 2 O (oxide)

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 294 CHEMISTRY

(ii) Reactivity towards water: The alkali the highest hydration enthalpy which
0
metals react with water to form hydroxide accounts for its high negative E value and
and dihydrogen. its high reducing power.
2 M + 2H2O → 2 M+ + 2 OH− + H2
(M = an alkali metal) Problem 10.2
0 – –
It may be noted that although lithium has The E for Cl 2 /Cl is +1.36, for I2/I is
0 + +
most negative E value (Table 10.1), its + 0.53, for Ag /Ag is +0.79, Na /Na is
+
reaction with water is less vigorous than –2.71 and for Li /Li is – 3.04. Arrange
that of sodium which has the least negative

d
0
the following ionic species in decreasing
E value among the alkali metals. This order of reducing strength:
behaviour of lithium is attributed to its – –

he
I , Ag, Cl , Li, Na
small size and very high hydration energy.
Other metals of the group react explosively Solution
– –
with water. The order is Li > Na > I > Ag > Cl
They also react with proton donors such

is
as alcohol, gaseous ammonia and alkynes. (vi) Solutions in liquid ammonia: The alkali
(iii) Reactivity towards dihydrogen: The metals dissolve in liquid ammonia giving
alkali metals react with dihydrogen at deep blue solutions which are conducting

bl
about 673K (lithium at 1073K) to form in nature.
hydrides. All the alkali metal hydrides are M + (x + y)NH3 →[M(NH3 )x ]+ + [e(NH3 )y ]−
ionic solids with high melting points.
pu The blue colour of the solution is due to
the ammoniated electron which absorbs
2 M + H2 → 2 M + H −
energy in the visible region of light and thus
(iv) Reactivity towards halogens : The alkali imparts blue colour to the solution. The
be T

metals readily react vigorously with solutions are paramagnetic and on

+ –
re
halogens to form ionic halides, M X . standing slowly liberate hydrogen resulting
o R

However, lithium halides are somewhat in the formation of amide.

covalent. It is because of the high
polarisation capability of lithium ion (The M + (am) + e − + NH3 (1) → MNH2(am) + ½H2 (g)
tt E

distortion of electron cloud of the anion by (where ‘am’ denotes solution in ammonia.)
+
the cation is called polarisation). The Li ion In concentrated solution, the blue colour
is very small in size and has high tendency changes to bronze colour and becomes
C

to distort electron cloud around the diamagnetic.

negative halide ion. Since anion with large
size can be easily distorted, among halides, 10.1.7 Uses
no N

lithium iodide is the most covalent in Lithium metal is used to make useful alloys,
nature. for example with lead to make ‘white metal’
(v) Reducing nature: The alkali metals are bearings for motor engines, with aluminium
strong reducing agents, lithium being the
©

to make aircraft parts, and with magnesium

most and sodium the least powerful to make armour plates. It is used in
(Table 10.1). The standard electrode thermonuclear reactions. Lithium is also used
0
potential (E ) which measures the reducing to make electrochemical cells. Sodium is used
power represents the overall change : to make a Na/Pb alloy needed to make PbEt4
M(s) → M(g) sublimationenthalpy and PbMe4. These organolead compounds were
+
M(g) → M (g) + e −
ionizationenthalpy earlier used as anti-knock additives to petrol,
but nowadays vehicles use lead-free petrol.
M+ (g) + H2O → M+ (aq) hydrationenthalpy Liquid sodium metal is used as a coolant in
With the small size of its ion, lithium has fast breeder nuclear reactors. Potassium has

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 THE s-BLOCK ELEMENTS 295

a vital role in biological systems. Potassium The hydroxides which are obtained by the
chloride is used as a fertilizer. Potassium reaction of the oxides with water are all white
hydroxide is used in the manufacture of soft crystalline solids. The alkali metal hydroxides
soap. It is also used as an excellent absorbent are the strongest of all bases and dissolve freely
of carbon dioxide. Caesium is used in devising in water with evolution of much heat on
photoelectric cells. account of intense hydration.

10.2 GENERAL CHARACTERISTICS OF 10.2.2 Halides

THE COMPOUNDS OF THE ALKALI The alkali metal halides, MX, (X=F,Cl,Br,I) are

d
METALS all high melting, colourless crystalline solids.
All the common compounds of the alkali metals They can be prepared by the reaction of the

he
are generally ionic in nature. General appropriate oxide, hydroxide or carbonate with
characteristics of some of their compounds are aqueous hydrohalic acid (HX). All of these
discussed here. halides have high negative enthalpies of
0
10.2.1 Oxides and Hydroxides formation; the Δf H values for fluorides

is
become less negative as we go down the group,
On combustion in excess of air, lithium forms 0
whilst the reverse is true for Δf H for chlorides,
mainly the oxide, Li2O (plus some peroxide
bromides and iodides. For a given metal
Li2O2), sodium forms the peroxide, Na2O2 (and

bl
0
Δf H always becomes less negative from
some superoxide NaO2) whilst potassium,
fluoride to iodide.
rubidium and caesium form the superoxides,
MO 2. Under appropriate conditions pure
pu The melting and boiling points always
compounds M2O, M 2O 2 and MO 2 may be follow the trend: fluoride > chloride > bromide
prepared. The increasing stability of the > iodide. All these halides are soluble in water.
The low solubility of LiF in water is due to its
be T

peroxide or superoxide, as the size of the metal

ion increases, is due to the stabilisation of large high lattice enthalpy whereas the low solubility
re
anions by larger cations through lattice energy of CsI is due to smaller hydration enthalpy of
o R

effects. These oxides are easily hydrolysed by its two ions. Other halides of lithium are soluble
water to form the hydroxides according to the in ethanol, acetone and ethylacetate; LiCl is
following reactions : soluble in pyridine also.
tt E

M2O + H2O → 2M+ + 2 OH – 10.2.3 Salts of Oxo-Acids

Oxo-acids are those in which the acidic proton
C

M2O2 + 2H2O → 2M + + 2 OH – + H2O2 is on a hydroxyl group with an oxo group

attached to the same atom e.g., carbonic acid,
2 MO2 + 2 H2O → 2M+ + 2 OH – + H2O2 + O2
no N

H 2 CO 3 (OC(OH) 2 ; sulphuric acid, H 2 SO 4

The oxides and the peroxides are colourless (O2S(OH)2). The alkali metals form salts with
when pure, but the superoxides are yellow or all the oxo-acids. They are generally soluble
orange in colour. The superoxides are also in water and thermally stable. Their
©

paramagnetic. Sodium peroxide is widely used carbonates (M2CO3) and in most cases the
as an oxidising agent in inorganic chemistry. hydrogencarbonates (MHCO3) also are highly
stable to heat. As the electropositive character
Problem 10.3 increases down the group, the stability of the
Why is KO2 paramagnetic ? carbonates and hydorgencarbonates increases.
Solution Lithium carbonate is not so stable to heat;
– lithium being very small in size polarises a
The superoxide O 2 is paramagnetic 2–
large CO3 ion leading to the formation of more
because of one unpaired electron in π*2p
stable Li2O and CO2. Its hydrogencarbonate
molecular orbital.
does not exist as a solid.

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 296 CHEMISTRY

10.3 ANOMALOUS PROPERTIES OF and lighter than other elements in the

LITHIUM respective groups.
The anomalous behaviour of lithium is due to (ii) Lithium and magnesium react slowly with
the : (i) exceptionally small size of its atom and water. Their oxides and hydroxides are
ion, and (ii) high polarising power (i.e., charge/ much less soluble and their hydroxides
radius ratio). As a result, there is increased decompose on heating. Both form a nitride,
covalent character of lithium compounds which Li3N and Mg3N2, by direct combination
is responsible for their solubility in organic with nitrogen.
solvents. Further, lithium shows diagonal (iii) The oxides, Li2O and MgO do not combine

d
relationship to magnesium which has been with excess oxygen to give any superoxide.
discussed subsequently. (iv) The carbonates of lithium and magnesium

he
10.3.1 Points of Difference between decompose easily on heating to
Lithium and other Alkali Metals form the oxides and CO 2 . Solid
hydrogencarbonates are not formed by
(i) Lithium is much harder. Its m.p. and b.p.
lithium and magnesium.
are higher than the other alkali metals.

is
(v) Both LiCl and MgCl2 are soluble in ethanol.
(ii) Lithium is least reactive but the strongest
reducing agent among all the alkali metals. (vi) Both LiCl and MgCl2 are deliquescent and
crystallise from aqueous solution as

bl
On combustion in air it forms mainly
monoxide, Li2O and the nitride, Li3N unlike hydrates, LiCl·2H2O and MgCl2·8H2O.
other alkali metals.
pu 10.4 SOME IMPORTANT COMPOUNDS OF
(iii) LiCl is deliquescent and crystallises as a SODIUM
hydrate, LiCl.2H2O whereas other alkali Industrially important compounds of sodium
metal chlorides do not form hydrates. include sodium carbonate, sodium hydroxide,
be T

(iv) Lithium hydrogencarbonate is not sodium chloride and sodium bicarbonate. The
obtained in the solid form while all other large scale production of these compounds
re
o R

elements form solid hydrogencarbonates. and their uses are described below:
(v) Lithium unlike other alkali metals forms Sodium Carbonate (Washing Soda),
no ethynide on reaction with ethyne. Na2CO3·10H2O
tt E

(vi) Lithium nitrate when heated gives lithium Sodium carbonate is generally prepared by
oxide, Li2O, whereas other alkali metal Solvay Process. In this process, advantage is
C

nitrates decompose to give the taken of the low solubility of sodium

corresponding nitrite. hydrogencarbonate whereby it gets
precipitated in the reaction of sodium chloride
4LiNO 3 → 2 Li 2 O + 4 NO 2 + O 2
no N

with ammonium hydrogencarbonate. The

2 NaNO 3 → 2 NaNO 2 + O 2 latter is prepared by passing CO 2 to a
concentrated solution of sodium chloride
(vii) LiF and Li2O are comparatively much less saturated with ammonia, where ammonium
soluble in water than the corresponding
©

carbonate followed by ammonium

compounds of other alkali metals. hydrogencarbonate are formed. The equations
10.3.2 Points of Similarities between for the complete process may be written as :
Lithium and Magnesium 2 NH 3 + H 2 O + CO 2 → ( NH 4 )2 CO 3
The similarity between lithium and magnesium
( NH 4 )2 CO 3 + H 2O + CO2 → 2 NH 4 HCO 3
is particularly striking and arises because of
their similar sizes : atomic radii, Li = 152 pm, NH 4 HCO 3 + NaCl → NH 4 Cl + NaHCO 3
+
Mg = 160 pm; ionic radii : Li = 76 pm, Sodium hydrogencarbonate crystal
2+
Mg = 72 pm. The main points of similarity are: separates. These are heated to give sodium
(i) Both lithium and magnesium are harder carbonate.

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 THE s-BLOCK ELEMENTS 297

2 NaHCO 3 → Na 2 CO 3 + CO 2 + H 2 O of brine solution, contains sodium sulphate,

calcium sulphate, calcium chloride and
In this process NH3 is recovered when the magnesium chloride as impurities. Calcium
solution containing NH4Cl is treated with chloride, CaCl2, and magnesium chloride,
Ca(OH)2. Calcium chloride is obtained as a
MgCl 2 are impurities because they are
by-product. deliquescent (absorb moisture easily from the
2 NH 4 Cl + Ca ( OH )2 → 2 NH3 + CaCl 2 + H2 O atmosphere). To obtain pure sodium chloride,
It may be mentioned here that Solvay the crude salt is dissolved in minimum amount
of water and filtered to remove insoluble

d
process cannot be extended to the
manufacture of potassium carbonate because impurities. The solution is then saturated with
potassium hydrogencarbonate is too soluble hydrogen chloride gas. Crystals of pure

he
to be precipitated by the addition of sodium chloride separate out. Calcium and
ammonium hydrogencarbonate to a saturated magnesium chloride, being more soluble than
solution of potassium chloride. sodium chloride, remain in solution.
Properties : Sodium carbonate is a white Sodium chloride melts at 1081K. It has a

is
crystalline solid which exists as a decahydrate, solubility of 36.0 g in 100 g of water at 273 K.
Na2CO3·10H2O. This is also called washing The solubility does not increase appreciably
soda. It is readily soluble in water. On heating, with increase in temperature.

bl
the decahydrate loses its water of crystallisation
to form monohydrate. Above 373K, the Uses :
monohydrate becomes completely anhydrous
pu (i) It is used as a common salt or table salt for
and changes to a white powder called soda ash. domestic purpose.
Na2CO3 10H2O ⎯⎯⎯⎯
375 K
→ Na2CO3 H2O + 9H2O (ii) It is used for the preparation of Na2O2,
NaOH and Na2CO3.
be T

>373K
Na 2CO3 H2O ⎯⎯⎯⎯ → Na 2CO3 + H2O
Sodium Hydroxide (Caustic Soda), NaOH
re
Carbonate part of sodium carbonate gets
o R

hydrolysed by water to form an alkaline Sodium hydroxide is generally prepared

solution. commercially by the electrolysis of sodium
chloride in Castner-Kellner cell. A brine
3 + H2 O → HCO3 + OH
CO2– – –
tt E

solution is electrolysed using a mercury

Uses:
cathode and a carbon anode. Sodium metal
(i) It is used in water softening, laundering discharged at the cathode combines with
C

and cleaning. mercury to form sodium amalgam. Chlorine

(ii) It is used in the manufacture of glass, gas is evolved at the anode.
soap, borax and caustic soda.
no N

(iii) It is used in paper, paints and textile Cathode : Na + + e − ⎯⎯⎯Hg

→ Na – amalgam
industries. 1
Anode : Cl – → Cl 2 + e –
(iv) It is an important laboratory reagent both 2
©

in qualitative and quantitative analysis. The amalgam is treated with water to give
sodium hydroxide and hydrogen gas.
Sodium Chloride, NaCl
2Na-amalgam + 2H2OÆ2NaOH+ 2Hg +H2
The most abundant source of sodium chloride
is sea water which contains 2.7 to 2.9% by Sodium hydroxide is a white, translucent
mass of the salt. In tropical countries like India, solid. It melts at 591 K. It is readily soluble in
common salt is generally obtained by water to give a strong alkaline solution.
evaporation of sea water. Approximately 50 Crystals of sodium hydroxide are deliquescent.
lakh tons of salt are produced annually in The sodium hydroxide solution at the surface
India by solar evaporation. Crude sodium reacts with the CO2 in the atmosphere to form
chloride, generally obtained by crystallisation Na2CO3.

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 298 CHEMISTRY

Uses: It is used in (i) the manufacture of soap, found on the opposite sides of cell membranes.
paper, artificial silk and a number of chemicals, As a typical example, in blood plasma, sodium
–1
(ii) in petroleum refining, (iii) in the purification is present to the extent of 143 mmolL ,
of bauxite, (iv) in the textile industries for whereas the potassium level is only
–1
mercerising cotton fabrics, (v) for the 5 mmolL within the red blood cells. These
–1 +
preparation of pure fats and oils, and (vi) as a concentrations change to 10 mmolL (Na ) and
–1 +
laboratory reagent. 105 mmolL (K ). These ionic gradients
Sodium Hydrogencarbonate (Baking demonstrate that a discriminatory mechanism,
Soda), NaHCO3 called the sodium-potassium pump, operates

d
across the cell membranes which consumes
Sodium hydrogencarbonate is known as
more than one-third of the ATP used by a
baking soda because it decomposes on heating

he
resting animal and about 15 kg per 24 h in a
to generate bubbles of carbon dioxide (leaving
resting human.
holes in cakes or pastries and making them
light and fluffy). 10.6 GROUP 2 ELEMENTS : ALKALINE
Sodium hydrogencarbonate is made by EARTH METALS

is
saturating a solution of sodium carbonate with The group 2 elements comprise beryllium,
carbon dioxide. The white crystalline powder magnesium, calcium, strontium, barium and
of sodium hydrogencarbonate, being less radium. They follow alkali metals in the

bl
soluble, gets separated out. periodic table. These (except beryllium) are
known as alkaline earth metals. The first
Na 2 CO3 H2O CO 2 2 NaHCO 3
pu element beryllium differs from the rest of the
Sodium hydrogencarbonate is a mild members and shows diagonal relationship to
antiseptic for skin infections. It is used in fire aluminium. The atomic and physical
extinguishers. properties of the alkaline earth metals are
be T

shown in Table 10.2.

10.5 BIOLOGICAL IMPORTANCE OF
re
SODIUM AND POTASSIUM 10.6.1 Electronic Configuration
o R

A typical 70 kg man contains about 90 g of Na These elements have two electrons in the
and 170 g of K compared with only 5 g of iron s -orbital of the valence shell (Table 10.2). Their
tt E

and 0.06 g of copper. general electronic configuration may be

2
Sodium ions are found primarily on the represented as [noble gas] ns . Like alkali
outside of cells, being located in blood plasma metals, the compounds of these elements are
C

and in the interstitial fluid which surrounds also predominantly ionic.

the cells. These ions participate in the
Element Symbol Electronic
transmission of nerve signals, in regulating the configuration
no N

flow of water across cell membranes and in the

transport of sugars and amino acids into cells. Beryllium Be 1s22s2
Sodium and potassium, although so similar Magnesium Mg 1s22s22p63s2
chemically, differ quantitatively in their ability Calcium Ca 1s22s22p63s23p64s2
©

to penetrate cell membranes, in their transport Strontium Sr 1s22s22p63s23p63d10

mechanisms and in their efficiency to activate 4s24p65s2
enzymes. Thus, potassium ions are the most Barium Ba 1s22s22p63s23p63d104s2
abundant cations within cell fluids, where they 4p64d105s25p66s2 or
activate many enzymes, participate in the [Xe]6s2
oxidation of glucose to produce ATP and, with Radium Ra [Rn]7s2
sodium, are responsible for the transmission
of nerve signals. 10.6.2 Atomic and Ionic Radii
There is a very considerable variation in the The atomic and ionic radii of the alkaline earth
concentration of sodium and potassium ions metals are smaller than those of the

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 THE s-BLOCK ELEMENTS 299

Table 10.2 Atomic and Physical Properties of the Alkaline Earth Metals

Property Beryllium Magnesium Calcium Strontium Barium Radium

Be Mg Ca Sr Ba Ra
Atomic number 4 12 20 38 56 88
–1
Atomic mass (g mol ) 9.01 24.31 40.08 87.62 137.33 226.03
Electronic [He] 2s2 [Ne] 3s2 [Ar] 4s2 [Kr] 5s2 [Xe] 6s2 [Rn] 7s2
configuration

d
Ionization 899 737 590 549 503 509
enthalpy (I) / kJ mol–1
Ionization 1757 1450 1145 1064 965 979

he
enthalpy (II) /kJ mol–1
Hydration enthalpy – 2494 – 1921 –1577 – 1443 – 1305 –
(kJ/mol)
Metallic 111 160 197 215 222 –

is
radius / pm
Ionic radius 31 72 100 118 135 148
M2+ / pm

bl
m.p. / K 1560 924 1124 1062 1002 973
b.p / K 2745 1363 1767 1655 2078 (1973)
Density / g cm–3
pu 1.84 1.74 1.55 2.63 3.59 (5.5)
Standard potential –1.97 –2.36 –2.84 –2.89 – 2.92 –2.92
E0 / V for (M2+/ M)
be T

Occurrence in 2* 2.76** 4.6** 384* 390 * 10–6*

lithosphere
re
*ppm (part per million); ** percentage by weight
o R

corresponding alkali metals in the same increase in ionic size down the group.
periods. This is due to the increased nuclear 2+ 2+ 2+
Be > Mg > Ca > Sr > Ba
2+ 2+
tt E

charge in these elements. Within the group, the

The hydration enthalpies of alkaline earth
atomic and ionic radii increase with increase
metal ions are larger than those of alkali metal
in atomic number.
C

ions. Thus, compounds of alkaline earth metals

10.6.3 Ionization Enthalpies are more extensively hydrated than those of
The alkaline earth metals have low ionization alkali metals, e.g., MgCl2 and CaCl2 exist as
no N

enthalpies due to fairly large size of the atoms. MgCl2.6H2O and CaCl2· 6H2O while NaCl and
Since the atomic size increases down the KCl do not form such hydrates.
group, their ionization enthalpy decreases 10.6.5 Physical Properties
(Table 10.2). The first ionisation enthalpies of
The alkaline earth metals, in general, are silvery
©

the alkaline earth metals are higher than those

white, lustrous and relatively soft but harder
of the corresponding Group 1 metals. This is
than the alkali metals. Beryllium and
due to their small size as compared to the
magnesium appear to be somewhat greyish.
corresponding alkali metals. It is interesting
The melting and boiling points of these metals
to note that the second ionisation enthalpies
are higher than the corresponding alkali metals
of the alkaline earth metals are smaller than
due to smaller sizes. The trend is, however, not
those of the corresponding alkali metals.
systematic. Because of the low ionisation
10.6.4 Hydration Enthalpies enthalpies, they are strongly electropositive in
Like alkali metal ions, the hydration enthalpies nature. The electropositive character increases
of alkaline earth metal ions decrease with down the group from Be to Ba. Calcium,

299
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27.7.6(reprint)
 300 CHEMISTRY

strontium and barium impart characteristic (iv) Reactivity towards acids: The alkaline
brick red, crimson and apple green colours earth metals readily react with acids liberating
respectively to the flame. In flame the electrons dihydrogen.
are excited to higher energy levels and when M + 2HCl → MCl2 + H2
they drop back to the ground state, energy is
(v) Reducing nature: Like alkali metals, the
emitted in the form of visible light. The
alkaline earth metals are strong reducing
electrons in beryllium and magnesium are too
strongly bound to get excited by flame. Hence, agents. This is indicated by large negative
these elements do not impart any colour to the values of their reduction potentials

d
flame. The flame test for Ca, Sr and Ba is (Table 10.2). However their reducing power is
helpful in their detection in qualitative analysis less than those of their corresponding alkali
metals. Beryllium has less negative value

he
and estimation by flame photometry. The
alkaline earth metals like those of alkali metals compared to other alkaline earth metals.
have high electrical and thermal conductivities However, its reducing nature is due to large
which are typical characteristics of metals. hydration energy associated with the small
2+
size of Be ion and relatively large value of the
10.6.6 Chemical Properties

is
atomization enthalpy of the metal.
The alkaline earth metals are less reactive than
(vi) Solutions in liquid ammonia: Like
the alkali metals. The reactivity of these
alkali metals, the alkaline earth metals dissolve

bl
elements increases on going down the group.
in liquid ammonia to give deep blue black
(i) Reactivity towards air and water: solutions forming ammoniated ions.
Beryllium and magnesium are kinetically inert
pu 2+
M + ( x + y ) NH3 → ⎡⎣M ( NH3 ) X ⎤⎦ + 2 ⎡⎣e ( NH3 ) Y ⎤⎦
–
to oxygen and water because of the formation
of an oxide film on their surface. However,
powdered beryllium burns brilliantly on From these solutions, the ammoniates,
be T

2+
ignition in air to give BeO and Be 3 N 2 . [M(NH3)6] can be recovered.
Magnesium is more electropositive and burns
re
with dazzling brilliance in air to give MgO and 10.6.7 Uses
o R

Mg3N2. Calcium, strontium and barium are Beryllium is used in the manufacture of alloys.
readily attacked by air to form the oxide and Copper-beryllium alloys are used in the
nitride. They also react with water with
tt E

preparation of high strength springs. Metallic

increasing vigour even in cold to form beryllium is used for making windows of
hydroxides. X-ray tubes. Magnesium forms alloys with
C

(ii) Reactivity towards the halogens: All aluminium, zinc, manganese and tin.
the alkaline earth metals combine with halogen Magnesium-aluminium alloys being light in
at elevated temperatures forming their halides. mass are used in air-craft construction.
no N

M + X 2 → MX 2 ( X = F, Cl, Br, l ) Magnesium (powder and ribbon) is used in

Thermal decomposition of (NH4)2BeF4 is the flash powders and bulbs, incendiary bombs
best route for the preparation of BeF2, and and signals. A suspension of magnesium
BeCl2 is conveniently made from the oxide. hydroxide in water (called milk of magnesia)
©

600 − 800K is used as antacid in medicine. Magnesium

BeO + C + Cl 2 BeCl 2 + CO carbonate is an ingredient of toothpaste.
(iii) Reactivity towards hydrogen: All the Calcium is used in the extraction of metals from
elements except beryllium combine with oxides which are difficult to reduce with
hydrogen upon heating to form their hydrides, carbon. Calcium and barium metals, owing
MH2. to their reactivity with oxygen and nitrogen at
BeH2, however, can be prepared by the reaction elevated temperatures, have often been used
of BeCl2 with LiAlH4. to remove air from vacuum tubes. Radium
salts are used in radiotherapy, for example, in
2BeCl 2 + LiAlH 4 → 2BeH 2 + LiCl + AlCl 3
the treatment of cancer.

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 THE s-BLOCK ELEMENTS 301

10.7 GENERAL CHARACTERISTICS OF In the vapour phase BeCl2 tends to form a

COMPOUNDS OF THE ALKALINE chloro-bridged dimer which dissociates into the
EARTH METALS linear monomer at high temperatures of the
2+
The dipositive oxidation state (M ) is the order of 1200 K. The tendency to form halide
predominant valence of Group 2 elements. The hydrates gradually decreases (for example,
alkaline earth metals form compounds which MgCl2·8H2O, CaCl2·6H2O, SrCl2·6H2O and
are predominantly ionic but less ionic than the BaCl2·2H2O) down the group. The dehydration
corresponding compounds of alkali metals. of hydrated chlorides, bromides and iodides
This is due to increased nuclear charge and of Ca, Sr and Ba can be achieved on heating;

d
smaller size. The oxides and other compounds however, the corresponding hydrated halides
of beryllium and magnesium are more covalent of Be and Mg on heating suffer hydrolysis. The

he
than those formed by the heavier and large fluorides are relatively less soluble than the
sized members (Ca, Sr, Ba). The general chlorides owing to their high lattice energies.
characteristics of some of the compounds of (iii) Salts of Oxoacids: The alkaline earth
alkali earth metals are described below. metals also form salts of oxoacids. Some of

is
(i) Oxides and Hydroxides: The alkaline these are :
earth metals burn in oxygen to form the Carbonates: Carbonates of alkaline earth
monoxide, MO which, except for BeO, have metals are insoluble in water and can be

bl
rock-salt structure. The BeO is essentially precipitated by addition of a sodium or
covalent in nature. The enthalpies of formation
ammonium carbonate solution to a solution
of these oxides are quite high and consequently
of a soluble salt of these metals. The solubility
pu
they are very stable to heat. BeO is amphoteric
while oxides of other elements are ionic in
of carbonates in water decreases as the atomic
number of the metal ion increases. All the
nature. All these oxides except BeO are basic
be T

in nature and react with water to form sparingly carbonates decompose on heating to give
soluble hydroxides. carbon dioxide and the oxide. Beryllium
re
carbonate is unstable and can be kept only in
o R

MO + H2O → M(OH)2
the atmosphere of CO2. The thermal stability
The solubility, thermal stability and the increases with increasing cationic size.
basic character of these hydroxides increase
tt E

Sulphates: The sulphates of the alkaline earth

with increasing atomic number from Mg(OH)2
to Ba(OH) 2 . The alkaline earth metal metals are all white solids and stable to heat.
hydroxides are, however, less basic and less BeSO4, and MgSO4 are readily soluble in water;
C

stable than alkali metal hydroxides. Beryllium the solubility decreases from CaSO4 to BaSO4.
2+
hydroxide is amphoteric in nature as it reacts The greater hydration enthalpies of Be and
2+
with acid and alkali both. Mg ions overcome the lattice enthalpy factor
no N

– 2– and therefore their sulphates are soluble in

Be(OH)2 + 2OH → [Be(OH)4]
Beryllate ion water.
Be(OH)2 + 2HCl + 2H2O → [Be(OH)4]Cl2 Nitrates: The nitrates are made by dissolution
©

of the carbonates in dilute nitric acid.

(ii) Halides: Except for beryllium halides, all Magnesium nitrate crystallises with six
other halides of alkaline earth metals are ionic molecules of water, whereas barium nitrate
in nature. Beryllium halides are essentially crystallises as the anhydrous salt. This again
covalent and soluble in organic solvents. shows a decreasing tendency to form hydrates
Beryllium chloride has a chain structure in the with increasing size and decreasing hydration
solid state as shown below:
enthalpy. All of them decompose on heating to
give the oxide like lithium nitrate.
2M ( NO 3 )2 → 2MO + 4NO2 + O2
(M = Be, Mg, Ca, Sr, Ba)

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 302 CHEMISTRY

Problem 10.4 (iii) The oxide and hydroxide of beryllium,

unlike the hydroxides of other elements in
Why does the solubility of alkaline earth the group, are amphoteric in nature.
metal hydroxides in water increase down
the group? 10.8.1 Diagonal Relationship between
Beryllium and Aluminium
Solution 2+
Among alkaline earth metal hydroxides, The ionic radius of Be is estimated to be
the anion being common the cationic 31 pm; the charge/radius ratio is nearly the
3+
radius will influence the lattice enthalpy. same as that of the Al ion. Hence beryllium

d
Since lattice enthalpy decreases much resembles aluminium in some ways. Some of
more than the hydration enthalpy with the similarities are:

he
increasing ionic size, the solubility (i) Like aluminium, beryllium is not readily
increases as we go down the group. attacked by acids because of the presence
of an oxide film on the surface of the metal.
Problem 10.5
(ii) Beryllium hydroxide dissolves in excess of
Why does the solubility of alkaline earth 2–

is
alkali to give a beryllate ion, [Be(OH)4] just
metal carbonates and sulphates in water as aluminium hydroxide gives aluminate
decrease down the group? ion, [Al(OH)4] .
–

bl
Solution (iii) The chlorides of both beryllium and
–
The size of anions being much larger aluminium have Cl bridged chloride
compared to cations, the lattice enthalpy
pu structure in vapour phase. Both the
will remain almost constant within a chlorides are soluble in organic solvents
particular group. Since the hydration and are strong Lewis acids. They are used
enthalpies decrease down the group, as Friedel Craft catalysts.
be T

solubility will decrease as found for (iv) Beryllium and aluminium ions have strong
alkaline earth metal carbonates and 2–
tendency to form complexes, BeF4 , AlF6 .
3–
re
o R

sulphates.
10.9 SOME IMPORTANT COMPOUNDS OF
CALCIUM
10.8 ANOMALOUS BEHAVIOUR OF
tt E

Important compounds of calcium are calcium

BERYLLIUM
oxide, calcium hydroxide, calcium sulphate,
Beryllium, the first member of the Group 2 calcium carbonate and cement. These are
C

metals, shows anomalous behaviour as industrially important compounds. The large

compared to magnesium and rest of the scale preparation of these compounds and
members. Further, it shows diagonal their uses are described below.
no N

relationship to aluminium which is discussed

Calcium Oxide or Quick Lime, CaO
subsequently.
It is prepared on a commercial scale by
(i) Beryllium has exceptionally small atomic
heating limestone (CaCO3) in a rotary kiln at
and ionic sizes and thus does not compare
©

well with other members of the group. 1070-1270 K.

Because of high ionisation enthalpy and CaCO3
heat
CaO + CO2
small size it forms compounds which are
largely covalent and get easily hydrolysed. The carbon dioxide is removed as soon as
(ii) Beryllium does not exhibit coordination it is produced to enable the reaction to proceed
number more than four as in its valence to completion.
shell there are only four orbitals. The Calcium oxide is a white amorphous solid.
remaining members of the group can have It has a melting point of 2870 K. On exposure
a coordination number of six by making to atmosphere, it absorbs moisture and carbon
use of d-orbitals. dioxide.

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 THE s-BLOCK ELEMENTS 303

CaO + H2 O → Ca ( OH )2 (ii) It is used in white wash due to its

disinfectant nature.
CaO + CO 2 → CaCO 3
(iii) It is used in glass making, in tanning
The addition of limited amount of water industry, for the preparation of bleaching
breaks the lump of lime. This process is called powder and for purification of sugar.
slaking of lime. Quick lime slaked with soda Calcium Carbonate, CaCO3
gives solid sodalime. Being a basic oxide, it Calcium carbonate occurs in nature in several
combines with acidic oxides at high forms like limestone, chalk, marble etc. It can
temperature.

d
be prepared by passing carbon dioxide
CaO + SiO 2 → CaSiO 3 through slaked lime or by the addition of
sodium carbonate to calcium chloride.

he
6CaO + P4 O10 → 2Ca 3 ( PO 4 )2
Ca ( OH )2 + CO2 → CaCO3 + H2 O
Uses:
(i) It is an important primary material for CaCl 2 + Na 2 CO 3 → CaCO 3 + 2NaCl
manufacturing cement and is the cheapest Excess of carbon dioxide should be

is
form of alkali. avoided since this leads to the formation of
(ii) It is used in the manufacture of sodium water soluble calcium hydrogencarbonate.
Calcium carbonate is a white fluffy powder.

bl
carbonate from caustic soda.
It is almost insoluble in water. When heated
(iii) It is employed in the purification of sugar to 1200 K, it decomposes to evolve carbon
and in the manufacture of dye stuffs.
pu dioxide.
Calcium Hydroxide (Slaked lime), Ca(OH)2 CaCO3 ⎯⎯⎯⎯
1200 K
→ CaO + CO2
Calcium hydroxide is prepared by adding It reacts with dilute acid to liberate carbon
be T

water to quick lime, CaO. dioxide.

It is a white amorphous powder. It is
re
CaCO 3 + 2HCl → CaCl 2 + H 2 O + CO 2
o R

sparingly soluble in water. The aqueous

solution is known as lime water and a CaCO 3 + H 2 SO 4 → CaSO 4 + H 2 O + CO 2
suspension of slaked lime in water is known Uses:
tt E

as milk of lime. It is used as a building material in the form of

When carbon dioxide is passed through marble and in the manufacture of quick lime.
lime water it turns milky due to the formation Calcium carbonate along with magnesium
C

of calcium carbonate. carbonate is used as a flux in the extraction of

metals such as iron. Specially precipitated
Ca ( OH )2 + CO2 → CaCO 3 + H2 O CaCO3 is extensively used in the manufacture
no N

On passing excess of carbon dioxide, the of high quality paper. It is also used as an
precipitate dissolves to form calcium antacid, mild abrasive in tooth paste, a
hydrogencarbonate. constituent of chewing gum, and a filler in
cosmetics.
©

CaCO3 + CO2 + H2 O → Ca ( HCO3 )2 Calcium Sulphate (Plaster of Paris),

Milk of lime reacts with chlorine to form CaSO4· H2O
hypochlorite, a constituent of bleaching It is a hemihydrate of calcium sulphate. It is
powder. obtained when gypsum, CaSO 4·2H 2 O, is
heated to 393 K.
2Ca ( OH )2 + 2Cl2 → CaCl 2 + Ca OCl ( ) 2
+ 2H2O
2 ( CaSO4 .2H2O ) → 2 ( CaSO4 ) .H2O + 3H2 O
Bleaching powder
Uses: Above 393 K, no water of crystallisation is left
(i) It is used in the preparation of mortar, a and anhydrous calcium sulphate, CaSO4 is
building material. formed. This is known as ‘dead burnt plaster’.

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 304 CHEMISTRY

It has a remarkable property of setting with silicate (Ca 3 SiO 5 ) 51% and tricalcium
water. On mixing with an adequate quantity aluminate (Ca3Al2O6) 11%.
of water it forms a plastic mass that gets into a Setting of Cement: When mixed with water,
hard solid in 5 to 15 minutes. the setting of cement takes place to give a hard
Uses: mass. This is due to the hydration of the
The largest use of Plaster of Paris is in the molecules of the constituents and their
building industry as well as plasters. It is used rearrangement. The purpose of adding
for immoblising the affected part of organ where gypsum is only to slow down the process of
there is a bone fracture or sprain. It is also setting of the cement so that it gets sufficiently

d
employed in dentistry, in ornamental work and hardened.
for making casts of statues and busts. Uses: Cement has become a commodity of

he
Cement: Cement is an important building national necessity for any country next to iron
material. It was first introduced in England in and steel. It is used in concrete and reinforced
1824 by Joseph Aspdin. It is also called concrete, in plastering and in the construction
Portland cement because it resembles with the of bridges, dams and buildings.

is
natural limestone quarried in the Isle of
10.10 BIOLOGICAL IMPORTANCE OF
Portland, England. MAGNESIUM AND CALCIUM
Cement is a product obtained by

bl
An adult body contains about 25 g of Mg and
combining a material rich in lime, CaO with 1200 g of Ca compared with only 5 g of iron
other material such as clay which contains and 0.06 g of copper. The daily requirement
silica, SiO 2 along with the oxides of
pu in the human body has been estimated to be
aluminium, iron and magnesium. The average 200 – 300 mg.
composition of Portland cement is : CaO, 50-
All enzymes that utilise ATP in phosphate
60%; SiO2, 20-25%; Al2O3, 5-10%; MgO, 2-
be T

transfer require magnesium as the cofactor.

3%; Fe2O3, 1-2% and SO3, 1-2%. For a good
The main pigment for the absorption of light
re
quality cement, the ratio of silica (SiO2) to
in plants is chlorophyll which contains
o R

alumina (Al2O3) should be between 2.5 and 4

magnesium. About 99 % of body calcium is
and the ratio of lime (CaO) to the total of the
present in bones and teeth. It also plays
oxides of silicon (SiO2) aluminium (Al2O3)
important roles in neuromuscular function,
tt E

and iron (Fe2O3) should be as close as possible

interneuronal transmission, cell membrane
to 2.
integrity and blood coagulation. The calcium
The raw materials for the manufacture of
C

concentration in plasma is regulated at about

cement are limestone and clay. When clay and 100 mgL–1. It is maintained by two hormones:
lime are strongly heated together they fuse and calcitonin and parathyroid hormone. Do you
no N

react to form ‘cement clinker’. This clinker is know that bone is not an inert and unchanging
mixed with 2-3% by weight of gypsum substance but is continuously being
(CaSO4·2H2O) to form cement. Thus important solubilised and redeposited to the extent of
ingredients present in Portland cement are 400 mg per day in man? All this calcium
dicalcium silicate (Ca2SiO4) 26%, tricalcium
©

passes through the plasma.

SUMMARY

The s-Block of the periodic table constitutes Group1 (alkali metals) and Group 2
(alkaline earth metals). They are so called because their oxides and hydroxides are alkaline
in nature. The alkali metals are characterised by one s-electron and the alkaline earth
metals by two s-electrons in the+ valence shell of their atoms. These are highly reactive
2+
metals forming monopositive (M ) and dipositve (M ) ions respectively.

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 THE s-BLOCK ELEMENTS 305

There is a regular trend in the physical and chemical properties of the alkali metal
with increasing atomic numbers. The atomic and ionic sizes increase and the ionization
enthalpies decrease systematically down the group. Somewhat similar trends are
observed among the properties of the alkaline earth metals.
The first element in each of these groups, lithium in Group 1 and beryllium in
Group 2 shows similarities in properties to the second member of the next group. Such
similarities are termed as the ‘diagonal relationship’ in the periodic table. As such
these elements are anomalous as far as their group characteristics are concerned.
The alkali metals are silvery white, soft and low melting. They are highly reactive.

d
The compounds of alkali metals are predominantly ionic. Their oxides and hydroxides
are soluble in water forming strong alkalies. Important compounds of sodium includes
sodium carbonate, sodium chloride, sodium hydroxide and sodium hydrogencarbonate.

he
Sodium hydroxide is manufactured by Castner-Kellner process and sodium carbonate
by Solvay process.
The chemistry of alkaline earth metals is very much like that of the alkali metals.
However, some differences arise because of reduced atomic and ionic sizes and increased
cationic charges in case of alkaline earth metals. Their oxides and hydroxides are less

is
basic than the alkali metal oxides and hydroxides. Industrially important compounds of
calcium include calcium oxide (lime), calcium hydroxide (slaked lime), calcium sulphate
(Plaster of Paris), calcium carbonate (limestone) and cement. Portland cement is an

bl
important constructional material. It is manufactured by heating a pulverised mixture
of limestone and clay in a rotary kiln. The clinker thus obtained is mixed with some
gypsum (2-3%) to give a fine powder of cement. All these substances find variety of uses
pu
in different areas.
Monovalent sodium and potassium ions and divalent magnesium and calcium ions
are found in large proportions in biological fluids. These ions perform important
be T

biological functions such as maintenance of ion balance and nerve impulse conduction.
re
o R

EXERCISES
tt E

10.1 What are the common physical and chemical features of alkali metals ?
10.2 Discuss the general characteristics and gradation in properties of alkaline earth
metals.
C

10.3 Why are alkali metals not found in nature ?

10.4 Find out the oxidation state of sodium in Na2O2.
10.5 Explain why is sodium less reactive than potassium.
no N

10.6 Compare the alkali metals and alkaline earth metals with respect to (i) ionisation
enthalpy (ii) basicity of oxides and (iii) solubility of hydroxides.
10.7 In what ways lithium shows similarities to magnesium in its chemical behaviour?
10.8 Explain why can alkali and alkaline earth metals not be obtained by chemical
©

reduction methods?
10.9 Why are potassium and caesium, rather than lithium used in photoelectric cells?
10.10 When an alkali metal dissolves in liquid ammonia the solution can acquire
different colours. Explain the reasons for this type of colour change.
10.11 Beryllium and magnesium do not give colour to flame whereas other alkaline
earth metals do so. Why ?
10.12 Discuss the various reactions that occur in the Solvay process.
10.13 Potassium carbonate cannot be prepared by Solvay process. Why ?
10.14 Why is Li2CO3 decomposed at a lower temperature whereas Na2CO3 at higher
temperature?

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 306 CHEMISTRY

10.15 Compare the solubility and thermal stability of the following compounds of the
alkali metals with those of the alkaline earth metals. (a) Nitrates (b) Carbonates
(c) Sulphates.
10.16 Starting with sodium chloride how would you proceed to prepare (i) sodium metal
(ii) sodium hydroxide (iii) sodium peroxide (iv) sodium carbonate ?
10.17 What happens when (i) magnesium is burnt in air (ii) quick lime is heated with
silica (iii) chlorine reacts with slaked lime (iv) calcium nitrate is heated ?
10.18 Describe two important uses of each of the following : (i) caustic soda (ii) sodium
carbonate (iii) quicklime.

d
10.19 Draw the structure of (i) BeCl2 (vapour) (ii) BeCl2 (solid).
10.20 The hydroxides and carbonates of sodium and potassium are easily soluble in

he
water while the corresponding salts of magnesium and calcium are sparingly
soluble in water. Explain.
10.21 Describe the importance of the following : (i) limestone (ii) cement (iii) plaster of
paris.
10.22 Why are lithium salts commonly hydrated and those of the other alkali ions

is
usually anhydrous?
10.23 Why is LiF almost insoluble in water whereas LiCl soluble not only in water but
also in acetone ?

bl
10.24 Explain the significance of sodium, potassium, magnesium and calcium in
biological fluids.
10.25
pu What happens when
(i) sodium metal is dropped in water ?
(ii) sodium metal is heated in free supply of air ?
(iii) sodium peroxide dissolves in water ?
be T

10.26 Comment on each of the following observations:

re
+ + +
(a) The mobilities of the alkali metal ions in aqueous solution are Li < Na < K
o R

+ +
< Rb < Cs
(b) Lithium is the only alkali metal to form a nitride directly.
0 2+ –
(c) E for M (aq) + 2e → M(s) (where M = Ca, Sr or Ba) is nearly constant.
tt E

10.27 State as to why

(a) a solution of Na2CO3 is alkaline ?
C

(b) alkali metals are prepared by electrolysis of their fused chlorides ?

(c) sodium is found to be more useful than potassium ?
10.28 Write balanced equations for reactions between
no N

(a) Na2O2 and water

(b) KO2 and water
(c) Na2O and CO2.
10.29 How would you explain the following observations?
©

(i) BeO is almost insoluble but BeSO4 is soluble in water,

(ii) BaO is soluble but BaSO4 is insoluble in water,
(iii) LiI is more soluble than KI in ethanol.
10.30 Which of the alkali metal is having least melting point ?
(a) Na (b) K (c) Rb (d) Cs
10.31 Which one of the following alkali metals gives hydrated salts ?
(a) Li (b) Na (c) K (d) Cs
10.32 Which one of the alkaline earth metal carbonates is thermally the most stable ?
(a) MgCO3 (b) CaCO3 (c) SrCO3 (d) BaCO3

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 THE p-BLOCK ELEMENTS 307

UNIT 11

THE p -BLOCK ELEMENTS

d
he
The variation in properties of the p-block elements due to the
influence of d and f electrons in the inner core of the heavier

is
elements makes their chemistry interesting

After studying this unit, you will be

bl
able to
• appreciate the general trends in the In p-block elements the last electron enters the outermost
pu chemistry of p-block elements; p orbital. As we know that the number of p orbitals is three
• describe the trends in physical and and, therefore, the maximum number of electrons that can
chemical properties of group 13 and be accommodated in a set of p orbitals is six. Consequently
14 elements; there are six groups of p–block elements in the periodic
be T

• explain anomalous behaviour of table numbering from 13 to 18. Boron, carbon, nitrogen,
oxygen, fluorine and helium head the groups. Their valence
re
boron and carbon;
o R

2 1-6
shell electronic configuration is ns np (except for He).
• describe allotropic forms of carbon;
The inner core of the electronic configuration may,
• know the chemistry of some however, differ. The difference in inner core of elements
tt E

important compounds of boron, greatly influences their physical properties (such as atomic
carbon and silicon; and ionic radii, ionisation enthalpy, etc.) as well as chemical
• list the important uses of group 13 properties. Consequently, a lot of variation in properties of
C

and 14 elements and their elements in a group of p-block is observed. The maximum
compounds. oxidation state shown by a p-block element is equal to the
total number of valence electrons (i.e., the sum of the s-
no N

and p-electrons). Clearly, the number of possible oxidation

states increases towards the right of the periodic table. In
addition to this so called group oxidation state, p-block
elements may show other oxidation states which normally,
©

but not necessarily, differ from the total number of valence

electrons by unit of two. The important oxidation states
exhibited by p-block elements are shown in Table 11.1. In
boron, carbon and nitrogen families the group oxidation
state is the most stable state for the lighter elements in the
group. However, the oxidation state two unit less than the
group oxidation state becomes progressively more stable
for the heavier elements in each group. The occurrence of
oxidation states two unit less than the group oxidation
states are sometime attributed to the ‘inert pair effect’.

307
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 308 CHEMISTRY

Table 11.1 General Electronic Configuration and Oxidation States of p-Block Elements

Group 13 14 15 16 17 18
General
electronic ns2np1 ns2np2 ns2np3 ns2np4 ns2np5 ns2np6
configuration (1s2 for He)

First member
of the B C N O F He

d
group

Group

he
oxidation +3 +4 +5 +6 +7 +8
state
Other
oxidation +1 +2, – 4 +3, – 3 +4, +2, –2 +5, + 3, +1, –1 +6, +4, +2

is
states

bl
The relative stabilities of these two oxidation The first member of p-block differs from the
states – group oxidation state and two unit less remaining members of their corresponding
than the group oxidation state – may vary from
pu group in two major respects. First is the size
group to group and will be discussed at and all other properties which depend on size.
appropriate places. Thus, the lightest p-block elements show the
It is interesting to note that the non-metals same kind of differences as the lightest s-block
be T

and metalloids exist only in the p-block of the elements, lithium and beryllium. The second
important difference, which applies only to the
re
periodic table. The non-metallic character of
o R

elements decreases down the group. In fact the p-block elements, arises from the effect of d-
heaviest element in each p-block group is the orbitals in the valence shell of heavier elements
most metallic in nature. This change from non- (starting from the third period onwards) and
tt E

metallic to metallic character brings diversity their lack in second period elements. The
in the chemistry of these elements depending second period elements of p-groups starting
on the group to which they belong. from boron are restricted to a maximum
C

covalence of four (using 2s and three 2p

In general, non-metals have higher ionisation orbitals). In contrast, the third period elements
enthalpies and higher electronegativities than of p-groups with the electronic configuration
no N

the metals. Hence, in contrast to metals which n

3s23p have the vacant 3d orbitals lying
readily form cations, non-metals readily form between the 3p and the 4s levels of energy.
anions. The compounds formed by highly Using these d-orbitals the third period
reactive non-metals with highly reactive metals elements can expand their covalence above
©

are generally ionic because of large differences four. For example, while boron forms only
in their electronegativities. On the other hand, – 3–
[BF 4] , aluminium gives [AlF 6] ion. The
compounds formed between non-metals presence of these d-orbitals influences the
themselves are largely covalent in character chemistry of the heavier elements in a number
because of small differences in their of other ways. The combined effect of size and
electronegativities. The change of non-metallic availability of d orbitals considerably
to metallic character can be best illustrated by influences the ability of these elements to form
the nature of oxides they form. The non-metal π bonds. The first member of a group differs
oxides are acidic or neutral whereas metal from the heavier members in its ability to form
oxides are basic in nature. pπ - pπ multiple bonds to itself ( e.g., C=C, C≡C,

308
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 THE p-BLOCK ELEMENTS 309

N≡N) and to other second row elements (e.g., 11.1.1 Electronic Configuration
C=O, C=N, C≡N, N=O). This type of π - bonding The outer electronic configuration of these
2 1
is not particularly strong for the heavier elements is ns np . A close look at the
p-block elements. The heavier elements do form electronic configuration suggests that while
π bonds but this involves d orbitals (dπ – pπ boron and aluminium have noble gas
or dπ –dπ ). As the d orbitals are of higher core, gallium and indium have noble gas plus
energy than the p orbitals, they contribute less 10 d-electrons, and thallium has noble gas
to the overall stability of molecules than does plus 14 f- electrons plus 10 d-electron cores.
pπ - pπ bonding of the second row elements. Thus, the electronic structures of these

d
However, the coordination number in species elements are more complex than for the first
of heavier elements may be higher than for two groups of elements discussed in unit 10.

he
the first element in the same oxidation state. This difference in electronic structures affects
For example, in +5 oxidation state both N and the other properties and consequently the
–
P form oxoanions : NO3 (three-coordination chemistry of all the elements of this group.
with π – bond involving one nitrogen p-orbital) 11.1.2 Atomic Radii

is
and PO34− (four-coordination involving s, p and On moving down the group, for each successive
d orbitals contributing to the π – bond). In member one extra shell of electrons is added
this unit we will study the chemistry of group and, therefore, atomic radius is expected to

bl
13 and 14 elements of the periodic table. increase. However, a deviation can be seen.
Atomic radius of Ga is less than that of Al. This
11.1 GROUP 13 ELEMENTS: THE BORON
can be understood from the variation in the
pu
FAMILY
This group elements show a wide variation in
inner core of the electronic configuration. The
presence of additional 10 d-electrons offer
properties. Boron is a typical non-metal, only poor screening effect (Unit 2) for the outer
be T

aluminium is a metal but shows many electrons from the increased nuclear charge in
re
chemical similarities to boron, and gallium, gallium. Consequently, the atomic radius of
o R

indium and thallium are almost exclusively gallium (135 pm) is less than that of
metallic in character. aluminium (143 pm).
Boron is a fairly rare element, mainly
tt E

11.1.3 Ionization Enthalpy

occurs as orthoboric acid, (H3BO3), borax,
The ionisation enthalpy values as expected
Na2B4O7·10H2O, and kernite, Na2B4O7·4H2O.
from the general trends do not decrease
C

In India borax occurs in Puga Valley (Ladakh)

smoothly down the group. The decrease from
and Sambhar Lake (Rajasthan). The
B to Al is associated with increase in size. The
abundance of boron in earth crust is less than
observed discontinuity in the ionisation
no N

0.0001% by mass. There are two isotopic

10 11 enthalpy values between Al and Ga, and
forms of boron B (19%) and B (81%).
between In and Tl are due to inability of d- and
Aluminium is the most abundant metal and
f-electrons ,which have low screening effect, to
the third most abundant element in the earth’s compensate the increase in nuclear charge.
©

crust (8.3% by mass) after oxygen (45.5%) and

Si (27.7%). Bauxite, Al2O3. 2H2O and cryolite, The order of ionisation enthalpies, as
Na 3 AlF 6 are the important minerals of expected, is Δi H1<Δi H2<Δi H3. The sum of the
aluminium. In India it is found as mica in first three ionisation enthalpies for each of the
Madhya Pradesh, Karnataka, Orissa and elements is very high. Effect of this will be
Jammu. Gallium, indium and thallium are less apparent when you study their chemical
abundant elements in nature. properties.
The atomic, physical and chemical 11.1.4 Electronegativity
properties of these elements are discussed Down the group, electronegativity first
below. decreases from B to Al and then increases

309
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 310 CHEMISTRY

marginally (Table 11.2). This is because of the only covalent compounds. But as we move from
discrepancies in atomic size of the elements. B to Al, the sum of the first three ionisation
enthalpies of Al considerably decreases, and
11.1.5 Physical Properties 3+
is therefore able to form Al ions. In fact,
Boron is non-metallic in nature. It is extremely aluminium is a highly electropositive metal.
hard and black coloured solid. It exists in many However, down the group, due to poor
allotropic forms. Due to very strong crystalline shielding effect of intervening d and f orbitals,
lattice, boron has unusually high melting point. the increased effective nuclear charge holds ns
Rest of the members are soft metals with low electrons tightly (responsible for inert pair

d
melting point and high electrical conductivity. effect) and thereby, restricting their
It is worthwhile to note that gallium with participation in bonding. As a result of this,

he
unusually low melting point (303 K), could only p-orbital electron may be involved in
exist in liquid state during summer. Its high bonding. In fact in Ga, In and Tl, both +1 and
boiling point (2676 K) makes it a useful +3 oxidation states are observed. The relative
material for measuring high temperatures. stability of +1 oxidation state progressively
Density of the elements increases down the increases for heavier elements: Al<Ga<In<Tl. In

is
group from boron to thallium. thallium +1 oxidation state is predominant
whereas the +3 oxidation state is highly
11.1.6 Chemical Properties oxidising in character. The compounds in

bl
Oxidation state and trends in chemical +1 oxidation state, as expected from energy
reactivity considerations, are more ionic than those in
Due to small size of boron, the sum of its first
pu +3 oxidation state.
three ionization enthalpies is very high. This In trivalent state, the number of electrons
prevents it to form +3 ions and forces it to form around the central atom in a molecule
Table 11.2 Atomic and Physical Properties of Group 13 Elements
be T

Element
re
Property Boron Aluminium Gallium Indium Thallium
o R

B Al Ga In Tl
Atomic number 5 13 31 49 81
tt E

–1
Atomic mass(g mol ) 10.81 26.98 69.72 114.82 204.38
Electronic Configuration [He]2s22p1 [Ne]3s23p1 [Ar]3d104s24p1 [Kr]4d105s25p1 [Xe]4f145d106s26p1
Atomic radius/pm a
(88) 143 135 167 170
C

Ionic radius (27) 53.5 62.0 80.0 88.5

M3+/pmb
no N

Ionic radius - - 120 140 150

M+/pm
Ionization Δi H 1 801 577 579 558 589
enthalpy Δi H 2 2427 1816 1979 1820 1971
(kJ mol–1) Δi H 3 3659 2744 2962 2704 2877
©

Electronegativityc 2.0 1.5 1.6 1.7 1.8

Density /g cm–3 2.35 2.70 5.90 7.31 11.85
at 298 K
Melting point / K 2453 933 303 430 576
Boiling point / K 3923 2740 2676 2353 1730
V 3+
E / V for (M /M) - –1.66 –0.56 –0.34 +1.26
V +
E / V for (M /M) - +0.55 -0.79(acid) –0.18 –0.34
–1.39(alkali)
a b c
Metallic radius, 6-coordination, Pauling scale,

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 THE p-BLOCK ELEMENTS 311

of the compounds of these elements solution but is a powerful oxidising agent

+
(e.g., boron in BF3) will be only six. Such also. Thus Tl is more stable in solution
3+
electron deficient molecules have tendency than Tl . Aluminium being able to form
to accept a pair of electrons to achieve stable +3 ions easily, is more electropositive than
electronic configuration and thus, behave as thallium.
Lewis acids. The tendency to behave as Lewis
acid decreases with the increase in the size (i) Reactivity towards air
down the group. BCl3 easily accepts a lone pair Boron is unreactive in crystalline form.
of electrons from ammonia to form BCl3⋅NH3. Aluminium forms a very thin oxide layer on

d
the surface which protects the metal from
further attack. Amorphous boron and

he
aluminium metal on heating in air form B2O3
and Al2O3 respectively. With dinitrogen at high
temperature they form nitrides.

2E ( s ) + 3O2 ( g ) ⎯⎯→
Δ
2E 2O3 ( s )

is
AlCl3 achieves stability by forming a dimer
2E ( s ) + N 2 ( g ) ⎯⎯→
Δ
2EN ( s )
(E = element)

bl
The nature of these oxides varies down the
group. Boron trioxide is acidic and reacts with
pu basic (metallic) oxides forming metal borates.
Aluminium and gallium oxides are amphoteric
and those of indium and thallium are basic in
their properties.
be T

In trivalent state most of the compounds

(ii) Reactivity towards acids and alkalies
being covalent are hydrolysed in water. For
re
Boron does not react with acids and alkalies
o R

example, the trichlorides on hyrolysis in water

− even at moderate temperature; but aluminium
form tetrahedral ⎡⎣ M ( OH )4 ⎤⎦ species; the dissolves in mineral acids and aqueous alkalies
3
hybridisation state of element M is sp . and thus shows amphoteric character.
tt E

Aluminium chloride in acidified aqueous Aluminium dissolves in dilute HCl and

3+
solution forms octahedral ⎡⎣ Al ( H2 O )6 ⎤⎦ ion. liberates dihydrogen.
C

3+ –
In this complex ion, the 3d orbitals of Al are 2Al(s) + 6HCl (aq) → 2Al (aq) + 6Cl (aq)
involved and the hybridisation state of Al is + 3H2 (g)
sp3d2. However, concentrated nitric acid renders
no N

aluminium passive by forming a protective

Problem 11.1 oxide layer on the surface.
V
Standard electrode potential values, E Aluminium also reacts with aqueous alkali
3+ 3+
for Al /Al is –1.66 V and that of Tl /Tl and liberates dihydrogen.
©

is +1.26 V. Predict about the formation of

3+
M ion in solution and compare the 2Al (s) + 2NaOH(aq) + 6H2O(l)
electropositive character of the two ↓
+ –
metals. 2 Na [Al(OH)4] (aq) + 3H2(g)
Sodium
Solution tetrahydroxoaluminate(III)
Standard electrode potential values for two (iii) Reactivity towards halogens
half cell reactions suggest that aluminium These elements react with halogens to form
3+
has high tendency to make Al (aq) ions, trihalides (except Tl I3).
3+
whereas Tl is not only unstable in
2E(s) + 3 X2 (g) → 2EX3 (s) (X = F, Cl, Br, I)

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 312 CHEMISTRY

Problem 11.2 11.3 SOME IMPORTANT COMPOUNDS OF

BORON
White fumes appear around the bottle of
anhydrous aluminium chloride. Give Some useful compounds of boron are borax,
reason. orthoboric acid and diborane. We will briefly
study their chemistry.
Solution
Anhydrous aluminium chloride is 11.3.1 Borax
partially hydrolysed with atmospheric It is the most important compound of boron.
moisture to liberate HCl gas. Moist HCl

d
It is a white crystalline solid of formula
appears white in colour. Na 2 B 4 O 7⋅ 10H 2 O. In fact it contains the
2−
tetranuclear units ⎡⎣ B4 O5 ( OH )4 ⎤⎦ and correct

he
11.2 IMPORTANT TRENDS AND
ANOMALOUS PROPERTIES OF formula; therefore, is Na2[B4O5 (OH)4].8H2O.
BORON Borax dissolves in water to give an alkaline
solution.
Certain important trends can be observed

is
in the chemical behaviour of group Na2B4O7 + 7H2O → 2NaOH + 4H3BO3
13 elements. The tri-chlorides, bromides Orthoboric acid
and iodides of all these elements being

bl
On heating, borax first loses water
covalent in nature are hydrolysed in water.
– molecules and swells up. On further heating it
Species like tetrahedral [M(OH) 4] and
3+
octahedral [M(H2O)6] , except in boron, exist
pu turns into a transparent liquid, which solidifies
in aqueous medium. into glass like material known as borax
bead.
The monomeric trihalides, being electron
Δ Δ
be T

deficient, are strong Lewis acids. Boron Na2B4O7.10H2O ⎯⎯→ Na2B4O7 ⎯⎯→ 2NaBO2
trifluoride easily reacts with Lewis bases such
Sodium + B2O3
re
as NH 3 to complete octet around
o R

metaborate Boric
boron.
anhydride
F3 B + :NH3 → F3 B ← NH3 The metaborates of many transition metals
tt E

It is due to the absence of d orbitals that have characteristic colours and, therefore,
the maximum covalence of B is 4. Since the borax bead test can be used to identify them
d orbitals are available with Al and other in the laboratory. For example, when borax is
C

elements, the maximum covalence can be heated in a Bunsen burner flame with CoO on
expected beyond 4. Most of the other metal a loop of platinum wire, a blue coloured
halides (e.g., AlCl3) are dimerised through
no N

Co(BO2)2 bead is formed.

halogen bridging (e.g., Al2Cl6). The metal
species completes its octet by accepting 11.3.2 Orthoboric acid
electrons from halogen in these halogen Orthoboric acid, H3BO3 is a white crystalline
bridged molecules. solid, with soapy touch. It is sparingly soluble
©

in water but highly soluble in hot water. It can

Problem 11.3 be prepared by acidifying an aqueous solution
3–
Boron is unable to form BF6 ion. Explain. of borax.
Solution Na2B4O7 + 2HCl + 5H2O → 2NaCl + 4B(OH)3
Due to non-availability of d orbitals, boron It is also formed by the hydrolysis (reaction
is unable to expand its octet. Therefore, with water or dilute acid) of most boron
the maximum covalence of boron cannot
compounds (halides, hydrides, etc.). It has a
exceed 4.
layer structure in which planar BO3 units are

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 THE p-BLOCK ELEMENTS 313

joined by hydrogen bonds as shown in 2NaBH4 + I2 → B2H6 + 2NaI + H2

Fig. 11.1. Diborane is produced on an industrial scale
Boric acid is a weak monobasic acid. It is by the reaction of BF3 with sodium hydride.
not a protonic acid but acts as a Lewis acid 450K
2BF3 + 6NaH ⎯⎯⎯ →B2 H6 + 6NaF
by accepting electrons from a hydroxyl
Diborane is a colourless, highly toxic gas
ion:
– + with a b.p. of 180 K. Diborane catches fire
B(OH)3 + 2HOH → [B(OH)4] + H3O
spontaneously upon exposure to air. It burns
On heating, orthoboric acid above 370K in oxygen releasing an enormous amount of

d
forms metaboric acid, HBO2 which on further energy.
heating yields boric oxide, B2O3.
B2 H6 +3O2 → B2 O3 + 3H2 O;

he
H3BO3 ⎯Δ→ HBO2 ⎯Δ→ B2O3 V
Δc H = −1976 kJ mol−1
Most of the higher boranes are also
spontaneously flammable in air. Boranes are
readily hydrolysed by water to give boric acid.

is
B2H6(g) + 6H2O(l) → 2B(OH)3(aq) + 6H2(g)
Diborane undergoes cleavage reactions

bl
with Lewis bases(L) to give borane adducts,
BH3⋅L
pu B2H6 + 2 NMe3 → 2BH3⋅NMe3
B2H6 + 2 CO → 2BH3⋅CO
Reaction of ammonia with diborane gives
be T

initially B2H6.2NH3 which is formulated as

+ –
[BH2(NH3)2] [BH4] ; further heating gives
re
borazine, B 3 N 3 H 6 known as “inorganic
o R

Fig. 11. 1 Structure of boric acid; the dotted lines benzene” in view of its ring structure with
represent hydrogen bonds alternate BH and NH groups.
tt E

–
3B2 H6 +6NH3 → 3[BH2 (NH3 )2 ]+ [BH4 ]
Problem 11.4 Heat
⎯⎯⎯ →2B3 N3 H6 +12H2
Why is boric acid considered as a weak
C

acid? The structure of diborane is shown in

Fig.11.2(a). The four terminal hydrogen atoms
Solution
and the two boron atoms lie in one plane.
no N

+
Because it is not able to release H ions Above and below this plane, there are two
–
on its own. It receives OH ions from water bridging hydrogen atoms. The four terminal
molecule to complete its octet and in turn B-H bonds are regular two centre-two electron
+
releases H ions. bonds while the two bridge (B-H-B) bonds are
©

different and can be described in terms of three

11.3.3 Diborane, B2H6
The simplest boron hydride known, is
diborane. It is prepared by treating boron
trifluoride with LiAlH4 in diethyl ether.
4BF3 + 3 LiAlH4 → 2B2H6 + 3LiF + 3AlF3
A convenient laboratory method for the
preparation of diborane involves the oxidation
of sodium borohydride with iodine. Fig.11.2(a) The structure of diborane, B2H6

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 314 CHEMISTRY

centre–two electron bonds shown in orthoboric acid is generally used as a mild

Fig.11.2 (b). antiseptic.
Boron also forms a series of hydridoborates; Aluminium is a bright silvery-white metal,
–
the most important one is the tetrahedral [BH4] with high tensile strength. It has a high
ion. Tetrahydridoborates of several metals are electrical and thermal conductivity. On a
known. Lithium and sodium tetra- weight-to-weight basis, the electrical
hydridoborates, also known as borohydrides, conductivity of aluminium is twice that of
are prepared by the reaction of metal hydrides copper. Aluminium is used extensively in
with B2H6 in diethyl ether. industry and every day life. It forms alloys

d
+ –
2MH + B2H6 → 2 M [BH4] (M = Li or Na) with Cu, Mn, Mg, Si and Zn. Aluminium and
its alloys can be given shapes of pipe, tubes,

he
rods, wires, plates or foils and, therefore, find
uses in packing, utensil making,
construction, aeroplane and transportation
industry. The use of aluminium and its

is
compounds for domestic purposes is now
reduced considerably because of their toxic
Fig.11.2(b) Bonding in diborane. Each B atom nature.

bl
uses sp3 hybrids for bonding. Out
of the four sp3 hybrids on each B 11.5 GROUP 14 ELEMENTS: THE CARBON
atom, one is without an electron FAMILY
shown in broken lines. The terminal
pu B-H bonds are normal 2-centre-2-
electron bonds but the two bridge
Carbon (C), silicon (Si), germanium (Ge), tin (Sn)
and lead (Pb) are the members of group 14.
Carbon is the seventeenth most abundant
bonds are 3-centre-2-electron bonds.
be T

The 3-centre-2-electron bridge bonds

element by mass in the earth’s crust. It is
are also referred to as banana bonds. widely distributed in nature in free as well as
re
in the combined state. In elemental state it is
o R

Both LiBH 4 and NaBH 4 are used as available as coal, graphite and diamond;
reducing agents in organic synthesis. They are however, in combined state it is present as
useful starting materials for preparing other metal carbonates, hydrocarbons and carbon
tt E

metal borohydrides. dioxide gas (0.03%) in air. One can

emphatically say that carbon is the most
11.4 USES OF BORON AND ALUMINIUM
C

versatile element in the world. Its combination

AND THEIR COMPOUNDS
Boron being extremely hard refractory solid of with other elements such as dihydrogen,
high melting point, low density and very low dioxygen, chlorine and sulphur provides an
no N

electrical conductivity, finds many astonishing array of materials ranging from

applications. Boron fibres are used in making living tissues to drugs and plastics. Organic
bullet-proof vest and light composite material chemistry is devoted to carbon containing
10
for aircraft. The boron-10 ( B) isotope has high compounds. It is an essential constituent of
©

ability to absorb neutrons and, therefore, all living organisms. Naturally occurring
12
metal borides are used in nuclear industry as carbon contains two stable isotopes: C and
13 14
protective shields and control rods. The main C. In addition to these, third isotope, C is
industrial application of borax and boric acid also present. It is a radioactive isotope with half-
is in the manufacture of heat resistant glasses life 5770 years and used for radiocarbon
(e.g., Pyrex), glass-wool and fibreglass. Borax dating. Silicon is the second (27.7 % by mass)
is also used as a flux for soldering metals, for most abundant element on the earth’s crust
heat, scratch and stain resistant glazed coating and is present in nature in the form of silica
to earthenwares and as constituent of and silicates. Silicon is a very important
medicinal soaps. An aqueous solution of component of ceramics, glass and cement.

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 THE p-BLOCK ELEMENTS 315

Germanium exists only in traces. Tin occurs due to the presence of completely filled d and f
mainly as cassiterite, SnO 2 and lead as orbitals in heavier members.
galena, PbS.
11.5.3 Ionization Enthalpy
Ultrapure form of germanium and silicon
The first ionization enthalpy of group 14
are used to make transistors and
members is higher than the corresponding
semiconductor devices.
members of group 13. The influence of inner
The important atomic and physical core electrons is visible here also. In general the
properties of the group 14 elements along ionisation enthalpy decreases down the group.

d
with their electronic configuration are given Small decrease in ΔiH from Si to Ge to Sn and
in Table 11.3 Some of the atomic, physical slight increase in Δi H from Sn to Pb is the
and chemical properties are discussed

he
consequence of poor shielding effect of
below: intervening d and f orbitals and increase in size
11.5.1 Electronic Configuration of the atom.
The valence shell electronic configuration of 11.5.4 Electronegativity
2 2

is
these elements is ns np . The inner core of the Due to small size, the elements of this group
electronic configuration of elements in this are slightly more electronegative than group
group also differs. 13 elements. The electronegativity values for

bl
11.5.2 Covalent Radius elements from Si to Pb are almost the same.
There is a considerable increase in covalent 11.5.5 Physical Properties
radius from C to Si, thereafter from Si to Pb a
pu All group 14 members are solids. Carbon and
small increase in radius is observed. This is silicon are non-metals, germanium is a metalloid,
Table 11.3 Atomic and Physical Properties of Group 14 Elements
be T

Element
re
Carbon Silicon Germanium Tin Lead
o R

Property
C Si Ge Sn Pb
Atomic Number 6 14 32 50 82
tt E

–1
Atomic mass (g mol ) 12.01 28.09 72.60 118.71 207.2
2 2 2 2 10 2 2 10 2 2 14 2 2
Electronic [He]2s 2p [Ne]3s 3p [Ar]3d 4s 4p [Kr]4d 5s 5p [Xe]4f 5d6s 6p
configuration
C

a
Covalent radius/pm 77 118 122 140 146
4+ b
Ionic radius M /pm – 40 53 69 78
no N

2+ b
Ionic radius M /pm – – 73 118 119
Ionization Δ iH 1 1086 786 761 708 715
enthalpy/ Δ iH 2 2352 1577 1537 1411 1450
kJ mol–1 Δ iH 3 4620 3228 3300 2942 3081
©

Δ iH 4 6220 4354 4409 3929 4082

c
Electronegativity 2.5 1.8 1.8 1.8 1.9
d –3 e f
Density /g cm 3.51 2.34 5.32 7.26 11.34
Melting point/K 4373 1693 1218 505 600
Boiling point/K – 3550 3123 2896 2024
14 16 –5 –5
Electrical resistivity/ 10 –10 50 50 10 2 10
ohm cm (293 K)
a IV b c d e
for M oxidation state; 6–coordination; Pauling scale; 293 K; for diamond; for graphite, density is
f
2.22; β-form (stable at room temperature)

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 316 CHEMISTRY

whereas tin and lead are soft metals with low those in lower oxidation states. The dioxides
melting points. Melting points and boiling points — CO2, SiO2 and GeO2 are acidic, whereas
of group 14 elements are much higher than those SnO2 and PbO2 are amphoteric in nature.
of corresponding elements of group 13. Among monoxides, CO is neutral, GeO is
distinctly acidic whereas SnO and PbO are
11.5.6 Chemical Properties
amphoteric.
Oxidation states and trends in chemical
reactivity Problem 11.5
The group 14 elements have four electrons in

d
Select the member(s) of group 14 that
outermost shell. The common oxidation states (i) forms the most acidic dioxide, (ii) is
exhibited by these elements are +4 and +2. commonly found in +2 oxidation state,

he
Carbon also exhibits negative oxidation states. (iii) used as semiconductor.
Since the sum of the first four ionization
enthalpies is very high, compounds in +4 Solution
oxidation state are generally covalent in nature. (i) carbon (ii) lead

is
In heavier members the tendency to show +2 (iii) silicon and germanium
oxidation state increases in the sequence
2
Ge<Sn<Pb. It is due to the inability of ns (ii) Reactivity towards water

bl
electrons of valence shell to participate in
bonding. The relative stabilities of these two Carbon, silicon and germanium are not
oxidation states vary down the group. Carbon affected by water. Tin decomposes steam to
pu
and silicon mostly show +4 oxidation state. form dioxide and dihydrogen gas.
Germanium forms stable compounds in +4 Δ
Sn + 2H2 O ⎯
→ SnO2 + 2H2
state and only few compounds in +2 state. Tin
Lead is unaffected by water, probably
be T

forms compounds in both oxidation states (Sn

in +2 state is a reducing agent). Lead because of a protective oxide film formation.
re
compounds in +2 state are stable and in +4 (iii) Reactivity towards halogen
o R

state are strong oxidising agents. In tetravalent

These elements can form halides of formula
state the number of electrons around the
MX2 and MX4 (where X = F, Cl, Br, I). Except
central atom in a molecule (e.g., carbon in CCl4)
tt E

is eight. Being electron precise molecules, they carbon, all other members react directly with
are normally not expected to act as electron halogen under suitable condition to make
halides. Most of the MX4 are covalent in nature.
C

acceptor or electron donor species. Although

carbon cannot exceed its covalence more than The central metal atom in these halides
3
4, other elements of the group can do so. It is undergoes sp hybridisation and the molecule
no N

because of the presence of d orbital in them. is tetrahedral in shape. Exceptions are SnF4
Due to this, their halides undergo hydrolysis and PbF4, which are ionic in nature. PbI4 does
and have tendency to form complexes by not exist because Pb—I bond initially formed
accepting electron pairs from donor species. For during the reaction does not release enough
2– 2– 2
©

example, the species like, SiF6 , [GeCl6] , energy to unpair 6s electrons and excite one
2–
[Sn(OH)6] exist where the hybridisation of the of them to higher orbital to have four unpaired
3 2
central atom is sp d . electrons around lead atom. Heavier members
(i) Reactivity towards oxygen Ge to Pb are able to make halides of formula
MX2. Stability of dihalides increases down the
All members when heated in oxygen form
group. Considering the thermal and chemical
oxides. There are mainly two types of oxides,
i.e., monoxide and dioxide of formula MO and stability, GeX4 is more stable than GeX2,
MO2 respectively. SiO only exists at high whereas PbX2 is more than PbX4. Except CCl4,
temperature. Oxides in higher oxidation states other tetrachlorides are easily hydrolysed
of elements are generally more acidic than by water because the central atom can

316
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 THE p-BLOCK ELEMENTS 317

accommodate the lone pair of electrons from Carbon also has unique ability to form
oxygen atom of water molecule in d orbital. pπ– pπ multiple bonds with itself and with other
Hydrolysis can be understood by taking atoms of small size and high electronegativity.
the example of SiCl4. It undergoes hydrolysis Few examples of multiple bonding are: C=C,
by initially accepting lone pair of electrons C ≡ C, C = O, C = S, and C ≡ N. Heavier elements
from water molecule in d orbitals of Si, finally do not form pπ– pπ bonds because their atomic
leading to the formation of Si(OH)4 as shown orbitals are too large and diffuse to have
below : effective overlapping.

d
Carbon atoms have the tendency to link
with one another through covalent bonds to

he
form chains and rings. This property is called
catenation. This is because C—C bonds are
very strong. Down the group the size increases
and electronegativity decreases, and, thereby,
tendency to show catenation decreases. This

is
can be clearly seen from bond enthalpies
values. The order of catenation is C > > Si >

bl
Ge ≈ Sn. Lead does not show catenation.
–1
Bond Bond enthalpy / kJ mol
pu
Problem 11. 6
2– 2–
[SiF6] is known whereas [SiCl6] not.
C—C
Si —Si
348
297
Give possible reasons. Ge—Ge 260
be T

Solution Sn—Sn 240

re
The main reasons are :
o R

(i) six large chloride ions cannot be Due to property of catenation and pπ– pπ
accommodated around Si
4+
due to bond formation, carbon is able to show
limitation of its size. allotropic forms.
tt E

(ii) interaction between lone pair of 11.7 ALLOTROPES OF CARBON

4+
chloride ion and Si is not very strong.
Carbon exhibits many allotropic forms; both
C

crystalline as well as amorphous. Diamond

11.6 IMPORTANT TRENDS AND and graphite are two well-known crystalline
ANOMALOUS BEHAVIOUR OF
no N

forms of carbon. In 1985, third form of carbon

CARBON known as fullerenes was discovered by
Like first member of other groups, carbon H.W.Kroto, E.Smalley and R.F.Curl. For this
also differs from rest of the members of its discovery they were awarded the Nobel Prize
group. It is due to its smaller size, higher
©

in 1996.
electronegativity, higher ionisation enthalpy
and unavailability of d orbitals. 11.7.1 Diamond
In carbon, only s and p orbitals are It has a crystalline lattice. In diamond each
3
available for bonding and, therefore, it can carbon atom undergoes sp hybridisation and
accommodate only four pairs of electrons linked to four other carbon atoms by using
around it. This would limit the maximum hybridised orbitals in tetrahedral fashion. The
covalence to four whereas other members can C–C bond length is 154 pm. The structure
expand their covalence due to the presence of extends in space and produces a rigid three-
d orbitals. dimensional network of carbon atoms. In this

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 318 CHEMISTRY

d
he
is
Fig. 11.3 The structure of diamond Fig 11.4 The structure of graphite

bl
structure (Fig. 11.3) directional covalent bonds therefore, graphite conducts electricity along
are present throughout the lattice.
pu the sheet. Graphite cleaves easily between the
It is very difficult to break extended covalent layers and, therefore, it is very soft and slippery.
bonding and, therefore, diamond is a hardest For this reason graphite is used as a dry
substance on the earth. It is used as an lubricant in machines running at high
be T

abrasive for sharpening hard tools, in making temperature, where oil cannot be used as a
dyes and in the manufacture of tungsten lubricant.
re
filaments for electric light bulbs.
o R

11.7.3 Fullerenes
Problem 11.7 Fullerenes are made by the heating of graphite
Diamond is covalent, yet it has high in an electric arc in the presence of inert gases
tt E

melting point. Why ? such as helium or argon. The sooty material

n
formed by condensation of vapourised C small
Solution molecules consists of mainly C60 with smaller
C

Diamond has a three-dimensional quantity of C 70 and traces of fullerenes

network involving strong C—C bonds, consisting of even number of carbon atoms up
which are very difficult to break and, in to 350 or above. Fullerenes are the only pure
no N

turn has high melting point. form of carbon because they have smooth
structure without having ‘dangling’ bonds.
11.7.2 Graphite Fullerenes are cage like molecules. C 60
Graphite has layered structure (Fig.11.4). molecule has a shape like soccer ball and
©

Layers are held by van der Waals forces and called Buckminsterfullerene (Fig. 11.5).
distance between two layers is 340 pm. Each It contains twenty six- membered rings and
layer is composed of planar hexagonal rings twelve five membered rings. A six membered
of carbon atoms. C—C bond length within the ring is fused with six or five membered rings
layer is 141.5 pm. Each carbon atom in but a five membered ring can only fuse with
2
hexagonal ring undergoes sp hybridisation six membered rings. All the carbon atoms are
2
and makes three sigma bonds with three equal and they undergo sp hybridisation.
neighbouring carbon atoms. Fourth electron Each carbon atom forms three sigma bonds
forms a π bond. The electrons are delocalised with other three carbon atoms. The remaining
over the whole sheet. Electrons are mobile and, electron at each carbon is delocalised in

318
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 THE p-BLOCK ELEMENTS 319

molecular orbitals, which in turn give aromatic filters to remove organic contaminators and in
character to molecule. This ball shaped airconditioning system to control odour.
molecule has 60 vertices and each one is Carbon black is used as black pigment in
occupied by one carbon atom and it also black ink and as filler in automobile tyres. Coke
contains both single and double bonds with is used as a fuel and largely as a reducing
C–C distances of 143.5 pm and 138.3 pm agent in metallurgy. Diamond is a precious
respectively. Spherical fullerenes are also called stone and used in jewellery. It is measured in
bucky balls in short. carats (1 carat = 200 mg).

d
11.8 SOME IMPORTANT COMPOUNDS OF
CARBON AND SILICON

he
Oxides of Carbon
Two important oxides of carbon are carbon
monoxide, CO and carbon dioxide, CO2.
11.8.1 Carbon Monoxide

is
Direct oxidation of C in limited supply of
oxygen or air yields carbon monoxide.
Δ
2C(s) + O2 (g) ⎯⎯ ⎯
→ 2CO(g)

bl
On small scale pure CO is prepared by
dehydration of formic acid with concentrated
H2SO4 at 373 K
pu
Fig.11.5 The structure of C 60 , Buckminster-
373K
HCOOH ⎯⎯⎯⎯⎯
conc.H SO→ H2 O + CO
2 4
fullerene : Note that molecule has the
be T

shape of a soccer ball (football). On commercial scale it is prepared by the

passage of steam over hot coke. The mixture
re
It is very important to know that graphite of CO and H2 thus produced is known as water
o R

is thermodynamically most stable allotrope of gas or synthesis gas.

V
carbon and, therefore, Δf H of graphite is taken 473−1273K
V
as zero. Δf H values of diamond and fullerene, C ( s ) + H2 O ( g ) ⎯⎯⎯⎯⎯⎯⎯ → CO ( g ) + H2 ( g )
tt E

–1
C60 are 1.90 and 38.1 kJ mol , respectively. Water gas
Other forms of elemental carbon like carbon When air is used instead of steam, a mixture
C

black, coke, and charcoal are all impure forms of CO and N2 is produced, which is called
of graphite or fullerenes. Carbon black is producer gas.
obtained by burning hydrocarbons in a limited 1273K
2C(s) + O2 (g) + 4N 2 (g) ⎯⎯⎯⎯⎯ → 2CO(g)
no N

supply of air. Charcoal and coke are obtained

by heating wood or coal respectively at high + 4N 2 (g)
temperatures in the absence of air. Producer gas
11.7.4 Uses of Carbon Water gas and producer gas are very
©

Graphite fibres embedded in plastic material important industrial fuels. Carbon monoxide
form high strength, lightweight composites. in water gas or producer gas can undergo
The composites are used in products such as further combustion forming carbon dioxide
tennis rackets, fishing rods, aircrafts and with the liberation of heat.
canoes. Being good conductor, graphite is used Carbon monoxide is a colourless,
for electrodes in batteries and industrial odourless and almost water insoluble gas. It
electrolysis. Crucibles made from graphite are is a powerful reducing agent and reduces
inert to dilute acids and alkalies. Being highly almost all metal oxides other than those of the
porous, activated charcoal is used in alkali and alkaline earth metals, aluminium
adsorbing poisonous gases; also used in water and a few transition metals. This property of

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 320 CHEMISTRY

CO is used in the extraction of many metals atmosphere, is removed from it by the process
from their oxides ores. known as photosynthesis. It is the process
Δ by which green plants convert atmospheric
Fe2 O3 ( s ) + 3CO ( g ) ⎯⎯⎯→ 2Fe ( s ) + 3CO2 ( g )
CO2 into carbohydrates such as glucose. The
Δ
ZnO ( s ) + CO ( g ) ⎯⎯⎯→ Zn ( s ) + CO2 ( g ) overall chemical change can be expressed as:
In CO molecule, there are one sigma and hν
6CO2 +12H2 O ⎯⎯⎯⎯⎯⎯→ C6 H12 O6 + 6O2
two π bonds between carbon and oxygen, Chlorphyll
:C ≡ O: . Because of the presence of a lone pair + 6H2 O
on carbon, CO molecule acts as a donor and

d
By this process plants make food for
reacts with certain metals when heated to form themselves as well as for animals and human
metal carbonyls. The highly poisonous beings. Unlike CO, it is not poisonous. But the

he
nature of CO arises because of its ability to increase in combustion of fossil fuels and
form a complex with haemoglobin, which decomposition of limestone for cement
is about 300 times more stable than the manufacture in recent years seem to increase
oxygen-haemoglobin complex. This prevents the CO2 content of the atmosphere. This may

is
haemoglobin in the red blood corpuscles from lead to increase in green house effect and
carrying oxygen round the body and ultimately thus, raise the temperature of the atmosphere
resulting in death. which might have serious consequences.

bl
11.8.2 Carbon Dioxide Carbon dioxide can be obtained as a solid
in the form of dry ice by allowing the liquified
It is prepared by complete combustion of
CO2 to expand rapidly. Dry ice is used as a
carbon and carbon containing fuels in excess
pu
of air.
Δ
refrigerant for ice-cream and frozen food.
Gaseous CO2 is extensively used to carbonate
C(s) + O2 (g) ⎯⎯⎯→ CO2 (g) soft drinks. Being heavy and non-supporter
be T

of combustion it is used as fire extinguisher. A

Δ
CH4 (g) + 2O2 (g) ⎯⎯⎯→ CO2 (g) + 2H2 O(g) substantial amount of CO 2 is used to
re
manufacture urea.
o R

In the laboratory it is conveniently

prepared by the action of dilute HCl on calcium In CO2 molecule carbon atom undergoes
carbonate. sp hybridisation. Two sp hybridised orbitals
tt E

of carbon atom overlap with two p orbitals of

CaCO3(s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) +
oxygen atoms to make two sigma bonds while
H2O(l)
other two electrons of carbon atom are involved
C

On commercial scale it is obtained by in pπ– pπ bonding with oxygen atom. This

heating limestone. results in its linear shape [with both C–O bonds
It is a colourless and odourless gas. Its low of equal length (115 pm)] with no dipole
no N

solubility in water makes it of immense bio- moment. The resonance structures are shown
chemical and geo-chemical importance. With below:
water, it forms carbonic acid, H2CO3 which is
a weak dibasic acid and dissociates in two
©

steps: Resonance structures of carbon dioxide

– +
H2CO3(aq) + H2O(l) HCO3 (aq) + H3O (aq)
– 2– + 11.8.3 Silicon Dioxide, SiO2
HCO3 (aq) + H2O(l) CO3 (aq) + H3O (aq)
– 95% of the earth’s crust is made up of silica
H 2 CO 3 /HCO 3 buffer system helps to and silicates. Silicon dioxide, commonly known
maintain pH of blood between 7.26 to 7.42. as silica, occurs in several crystallographic
Being acidic in nature, it combines with alkalies forms. Quartz, cristobalite and tridymite are
to form metal carbonates. some of the crystalline forms of silica, and they
Carbon dioxide, which is normally present are interconvertable at suitable temperature.
to the extent of ~ 0.03 % by volume in the Silicon dioxide is a covalent, three-dimensional

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 THE p-BLOCK ELEMENTS 321

network solid in which each silicon atom is substituted chlorosilane of formula MeSiCl3,
covalently bonded in a tetrahedral manner to Me2SiCl2, Me3SiCl with small amount of Me4Si
four oxygen atoms. Each oxygen atom in turn are formed. Hydrolysis of dimethyl-
covalently bonded to another silicon atoms as dichlorosilane, (CH 3 ) 2 SiCl 2 followed by
shown in diagram (Fig 11.6 ). Each corner is condensation polymerisation yields straight
shared with another tetrahedron. The entire chain polymers.
crystal may be considered as giant molecule
in which eight membered rings are formed with
alternate silicon and oxygen atoms.

d
he
is
bl
pu
Fig. 11.6 Three dimensional structure of SiO2
Silica in its normal form is almost non-
reactive because of very high Si — O bond
be T

enthalpy. It resists the attack by halogens,

re
dihydrogen and most of the acids and metals
o R

even at elevated temperatures. However, it is

attacked by HF and NaOH.
The chain length of the polymer can be
tt E

SiO2 + 2NaOH → Na2SiO3 + H2O

controlled by adding (CH3)3SiCl which blocks
SiO2 + 4HF → SiF4 + 2H2O the ends as shown below :
Quartz is extensively used as a piezoelectric
C

material; it has made possible to develop extremely

accurate clocks, modern radio and television
broadcasting and mobile radio communications.
no N

Silica gel is used as a drying agent and as a support

for chromatographic materials and catalysts.
Kieselghur, an amorphous form of silica is used
in filtration plants.
©

11.8.4 Silicones
They are a group of organosilicon polymers,
which have (R2SiO) as a repeating unit. The
starting materials for the manufacture of
silicones are alkyl or aryl substituted silicon
chlorides, RnSiCl (4–n), where R is alkyl or aryl
group. When methyl chloride reacts with
silicon in the presence of copper as a catalyst
at a temperature 573K various types of methyl

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 322 CHEMISTRY

Silicones being surrounded by non-polar

alkyl groups are water repelling in nature.
They have in general high thermal stability,
high dielectric strength and resistance to
oxidation and chemicals. They have wide
applications. They are used as sealant, greases,
electrical insulators and for water proofing of
fabrics. Being biocompatible they are also used
in surgical and cosmetic plants.

d
Problem: 11.8

he
What are silicones ? (a) (b)
4–
Solution Fig. 11.7 (a) Tetrahedral structure of SiO 4
4–
anion; (b) Representation of SiO4 unit
Simple silicones consist of

is
neutralised by positively charged metal ions.
chains in which alkyl or phenyl groups If all the four corners are shared with other
occupy the remaining bonding positions tetrahedral units, three-dimensional network

bl
on each silicon. They are hydrophobic is formed.
(water repellant) in nature. Two important man-made silicates are
pu glass and cement.
11.8.5 Silicates
11.8.6 Zeolites
A large number of silicates minerals exist in
nature. Some of the examples are feldspar, If aluminium atoms replace few silicon atoms
be T

zeolites, mica and asbestos. The basic in three-dimensional network of silicon dioxide,
4–
structural unit of silicates is SiO4 (Fig.11.7) overall structure known as aluminosilicate,
re
o R

in which silicon atom is bonded to four acquires a negative charge. Cations such as
+ +
oxygen atoms in tetrahedron fashion. In Na , K or Ca2+ balance the negative charge.
silicates either the discrete unit is present or Examples are feldspar and zeolites. Zeolites are
tt E

a number of such units are joined together widely used as a catalyst in petrochemical
via corners by sharing 1,2,3 or 4 oxygen industries for cracking of hydrocarbons and
atoms per silicate units. When silicate units isomerisation, e.g., ZSM-5 (A type of zeolite)
C

are linked together, they form chain, ring, used to convert alcohols directly into gasoline.
sheet or three-dimensional structures. Hydrated zeolites are used as ion exchangers
Negative charge on silicate structure is in softening of “hard” water.
no N

SUMMARY
©

p-Block of the periodic table is unique in terms of having all types of elements – metals,
non-metals and metalloids. There are six groups of p-block elements in the periodic
2 1–6
table numbering from 13 to 18. Their valence shell electronic configuration is ns np
(except for He). Differences in the inner core of their electronic configuration greatly
influence their physical and chemical properties. As a consequence of this, a lot of
variation in properties among these elements is observed. In addition to the group oxidation
state, these elements show other oxidation states differing from the total number of valence
electrons by unit of two. While the group oxidation state is the most stable for the lighter
elements of the group, lower oxidation states become progressively more stable for the
heavier elements. The combined effect of size and availability of d orbitals considerably

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 THE p-BLOCK ELEMENTS 323

influences the ability of these elements to form π-bonds. While the lighter elements form
pπ–pπ bonds, the heavier ones form dπ–pπ or dπ–dπ bonds. Absence of d orbital in
second period elements limits their maximum covalence to 4 while heavier ones can
exceed this limit.
Boron is a typical non-metal and the other members are metals. The availability of 3
2 1
valence electrons (2s 2p ) for covalent bond formation using four orbitals (2s, 2px, 2py and
2pz) leads to the so called electron deficiency in boron compounds. This deficiency
makes them good electron acceptor and thus boron compounds behave as Lewis acids.
Boron forms covalent molecular compounds with dihydrogen as boranes, the simplest of

d
which is diborane, B2H6. Diborane contains two bridging hydrogen atoms between two
boron atoms; these bridge bonds are considered to be three-centre two-electron bonds.
The important compounds of boron with dioxygen are boric acid and borax. Boric acid,

he
B(OH)3 is a weak monobasic acid; it acts as a Lewis acid by accepting electrons from
hydroxyl ion. Borax is a white crystalline solid of formula Na2[B4O5(OH)4]·8H2O. The borax
bead test gives characteristic colours of transition metals.
Aluminium exhibits +3 oxidation state. With heavier elements +1 oxidation state gets

is
progressively stabilised on going down the group. This is a consequence of the so called
inert pair effect.
Carbon is a typical non-metal forming covalent bonds employing all its four valence
2 2
electrons (2s 2p ). It shows the property of catenation, the ability to form chains or

bl
rings, not only with C–C single bonds but also with multiple bonds (C=C or C≡C). The
tendency to catenation decreases as C>>Si>Ge ~ Sn > Pb. Carbon provides one of the
best examples of allotropy. Three important allotropes of carbon are diamond, graphite
pu
and fullerenes. The members of the carbon family mainly exhibit +4 and +2 oxidation
states; compouds in +4 oxidation states are generally covalent in nature. The tendency
to show +2 oxidation state increases among heavier elements. Lead in +2 state is stable
be T

whereas in +4 oxidation state it is a strong oxidising agent. Carbon also exhibits negative
oxidation states. It forms two important oxides: CO and CO2. Carbon monoxide is neutral
re
whereas CO2 is acidic in nature. Carbon monoxide having lone pair of electrons on C
o R

forms metal carbonyls. It is deadly poisonous due to higher stability of its haemoglobin
complex as compared to that of oxyhaemoglobin complex. Carbon dioxide as such is not
toxic. However, increased content of CO2 in atmosphere due to combustion of fossil fuels
tt E

and decomposition of limestone is feared to cause increase in ‘green house effect’. This,
in turn, raises the temperature of the atmosphere and causes serious complications.
Silica, silicates and silicones are important class of compounds and find applications
C

in industry and technology.

no N

EXERCISES

11.1 Discuss the pattern of variation in the oxidation states of

(i) B to Tl and (ii) C to Pb.
©

11.2 How can you explain higher stability of BCl3 as compared to TlCl3 ?
11.3 Why does boron triflouride behave as a Lewis acid ?
11.4 Consider the compounds, BCl 3 and CCl 4. How will they behave with
water ? Justify.
11.5 Is boric acid a protic acid ? Explain.
11.6 Explain what happens when boric acid is heated .
11.7 Describe the shapes of BF3 and BH4–. Assign the hybridisation of boron in
these species.
11.8 Write reactions to justify amphoteric nature of aluminium.

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 324 CHEMISTRY

11.9 What are electron deficient compounds ? Are BCl 3 and SiCl 4 electron
deficient species ? Explain.
2– –
11.10 Write the resonance structures of CO3 and HCO3 .
2–
11.11 What is the state of hybridisation of carbon in (a) CO 3 (b) diamond
(c) graphite?
11.12 Explain the difference in properties of diamond and graphite on the basis
of their structures.
11.13 Rationalise the given statements and give chemical reactions :

d
• Lead(II) chloride reacts with Cl2 to give PbCl4.
• Lead(IV) chloride is highly unstable towards heat.

he
• Lead is known not to form an iodide, PbI4.
–
11.14 Suggest reasons why the B–F bond lengths in BF 3 (130 pm) and BF 4
(143 pm) differ.
11.15 If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero
dipole moment.

is
11.16 Aluminium trifluoride is insoluble in anhydrous HF but dissolves on
addition of NaF. Aluminium trifluoride precipitates out of the resulting
solution when gaseous BF3 is bubbled through. Give reasons.

bl
11.17 Suggest a reason as to why CO is poisonous.
pu 11.18 How is excessive content of CO2 responsible for global warming ?
11.19 Explain structures of diborane and boric acid.
11.20 What happens when
(a) Borax is heated strongly,
be T

(b) Boric acid is added to water,

re
(c) Aluminium is treated with dilute NaOH,
o R

(d) BF3 is reacted with ammonia ?

11.21 Explain the following reactions
tt E

(a) Silicon is heated with methyl chloride at high temperature in the

presence of copper;
(b) Silicon dioxide is treated with hydrogen fluoride;
C

(c) CO is heated with ZnO;

(d) Hydrated alumina is treated with aqueous NaOH solution.
11.22 Give reasons :
no N

(i) Conc. HNO3 can be transported in aluminium container.

(ii) A mixture of dilute NaOH and aluminium pieces is used to open
drain.
(iii) Graphite is used as lubricant.
©

(iv) Diamond is used as an abrasive.

(v) Aluminium alloys are used to make aircraft body.
(vi) Aluminium utensils should not be kept in water overnight.
(vii) Aluminium wire is used to make transmission cables.
11.23 Explain why is there a phenomenal decrease in ionization enthalpy from
carbon to silicon ?
11.24 How would you explain the lower atomic radius of Ga as compared to Al ?
11.25 What are allotropes? Sketch the structure of two allotropes of carbon namely
diamond and graphite. What is the impact of structure on physical
properties of two allotropes?

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 THE p-BLOCK ELEMENTS 325

11.26 (a) Classify following oxides as neutral, acidic, basic or amphoteric:

CO, B2O3, SiO2, CO2, Al2O3, PbO2, Tl2O3
(b) Write suitable chemical equations to show their nature.
11.27 In some of the reactions thallium resembles aluminium, whereas in others
it resembles with group I metals. Support this statement by giving some
evidences.
11.28 When metal X is treated with sodium hydroxide, a white precipitate (A) is
obtained, which is soluble in excess of NaOH to give soluble complex (B).
Compound (A) is soluble in dilute HCl to form compound (C). The compound

d
(A) when heated strongly gives (D), which is used to extract metal. Identify
(X), (A), (B), (C) and (D). Write suitable equations to support their identities.

he
11.29 What do you understand by (a) inert pair effect (b) allotropy and
(c) catenation?
11.30 A certain salt X, gives the following results.
(i) Its aqueous solution is alkaline to litmus.

is
(ii) It swells up to a glassy material Y on strong heating.
(iii) When conc. H2SO4 is added to a hot solution of X,white crystal of an
acid Z separates out.

bl
Write equations for all the above reactions and identify X, Y and Z.
11.31 Write balanced equations for:
(i) BF3 + LiH →
pu (ii) B2H6 + H2O →
(iii) NaH + B2H6 →
be T

Δ
(iv) H3BO3 ⎯
→
(v) Al + NaOH →
re
o R

(vi) B2H6 + NH3 →

11.32. Give one method for industrial preparation and one for laboratory
preparation of CO and CO2 each.
tt E

11.33 An aqueous solution of borax is

(a) neutral (b) amphoteric
C

(c) basic (d) acidic

11.34 Boric acid is polymeric due to
(a) its acidic nature (b) the presence of hydrogen bonds
no N

(c) its monobasic nature (d) its geometry

11.35 The type of hybridisation of boron in diborane is
(a) sp (b) sp2 (c) sp3 (d) dsp2
11.36 Thermodynamically the most stable form of carbon is
©

(a) diamond (b) graphite

(c) fullerenes (d) coal
11.37 Elements of group 14
(a) exhibit oxidation state of +4 only
(b) exhibit oxidation state of +2 and +4
(c) form M2– and M4+ ions
(d) form M2+ and M4+ ions
11.38 If the starting material for the manufacture of silicones is RSiCl3, write the
structure of the product formed.

325
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27.7.6 (reprint)
326 CHEMISTRY

UNIT 12

ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES

AND TECHNIQUES

In the previous unit you have learnt that the element

carbon has the unique property called catenation due to
which it forms covalent bonds with other carbon atoms.
After studying this unit, you will be It also forms covalent bonds with atoms of other elements
able to like hydrogen, oxygen, nitrogen, sulphur, phosphorus and
• understand reasons for halogens. The resulting compounds are studied under a
tetravalence of carbon and separate branch of chemistry called organic chemistry.
shapes of organic molecules; This unit incorporates some basic principles and
• write structures of organic techniques of analysis required for understanding the
molecules in various ways; formation and properties of organic compounds.
• classify the organic compounds;
• name the compounds according 12.1 GENERAL INTRODUCTION
to IUPAC system of
nomenclature and also derive Organic compounds are vital for sustaining life on earth
their structures from the given and include complex molecules like genetic information
names; bearing deoxyribonucleic acid (DNA) and proteins that
• understand the concept of constitute essential compounds of our blood, muscles and
organic reaction mechanism; skin. Organic chemicals appear in materials like clothing,
• explain the influence of fuels, polymers, dyes and medicines. These are some of
electronic displacements on the important areas of application of these compounds.
structure and reactivity of Science of organic chemistry is about two hundred
organic compounds; years old. Around the year 1780, chemists began to
• recognise the types of organic distinguish between organic compounds obtained from
reactions; plants and animals and inorganic compounds prepared
• lear n the techniques of from mineral sources. Berzilius, a Swedish chemist
purification of organic proposed that a ‘vital force’ was responsible for the
compounds; formation of organic compounds. However, this notion
• write the chemical reactions was rejected in 1828 when F. Wohler synthesised an
involved in the qualitative
organic compound, urea from an inorganic compound,
analysis of organic compounds;
ammonium cyanate.
• understand the principles
involved in quantitative analysis NH4 CNO 
Heat
 NH2 CONH2
of organic compounds. Ammonium cyanate Urea
The pioneering synthesis of acetic acid by Kolbe (1845)
and that of methane by Berthelot (1856) showed
conclusively that organic compounds could be synthesised
from inorganic sources in a laboratory.
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 327

The development of electronic theory of necessary for a proper sideways overlap.

covalent bonding ushered organic chemistry Thus, in H2C=CH2 molecule all the atoms
into its modern shape. must be in the same plane. The p orbitals
are mutually parallel and both the p orbitals
12.2 TETRAVALENCE OF CARBON:
are perpendicular to the plane of the
SHAPES OF ORGANIC COMPOUNDS
molecule. Rotation of one CH2 fragment with
12.2.1 The Shapes of Carbon Compounds respect to other interferes with maximum
The knowledge of fundamental concepts of overlap of p orbitals and, therefore, such
molecular structure helps in understanding rotation about carbon-carbon double bond
and predicting the properties of organic (C=C) is restricted. The electron charge cloud
compounds. You have already learnt theories of the π bond is located above and below the
of valency and molecular structure in Unit 4. plane of bonding atoms. This results in the
Also, you already know that tetravalence of electrons being easily available to the
carbon and the formation of covalent bonds attacking reagents. In general, π bonds provide
by it are explained in terms of its electronic the most reactive centres in the molecules
configuration and the hybridisation of s and containing multiple bonds.
p orbitals. It may be recalled that formation
and the shapes of molecules like methane Problem 12.1
(CH 4 ), ethene (C 2 H 4 ), ethyne (C 2 H 2 ) are How many σ and π bonds are present in
explained in terms of the use of sp3, sp2 and each of the following molecules?
sp hybrid orbitals by carbon atoms in the
(a) HC≡CCH=CHCH3 (b) CH2=C=CHCH3
respective molecules.
Hybridisation influences the bond length Solution
and bond enthalpy (strength) in organic (a) σC – C: 4; σC–H : 6; πC=C :1; π C≡C:2
compounds. The sp hybrid orbital contains
(b) σC – C: 3; σC–H: 6; πC=C: 2.
more s character and hence it is closer to its
nucleus and forms shorter and stronger Problem 12.2
bonds than the sp3 hybrid orbital. The sp2 What is the type of hybridisation of each
hybrid orbital is intermediate in s character carbon in the following compounds?
between sp and sp3 and, hence, the length
and enthalpy of the bonds it forms, are also (a) CH3Cl, (b) (CH3)2CO, (c) CH3CN,
intermediate between them. The change in (d) HCONH2, (e) CH3CH=CHCN
hybridisation affects the electronegativity of
carbon. The greater the s character of the Solution
hybrid orbitals, the greater is the (a) sp3, (b) sp3, sp2, (c) sp3, sp, (d) sp2, (e)
electronegativity. Thus, a carbon atom having sp3, sp2, sp2, sp
an sp hybrid orbital with 50% s character is
Problem 12.3
more electronegative than that possessing sp2
or sp 3 hybridised orbitals. This relative Write the state of hybridisation of carbon
electronegativity is reflected in several in the following compounds and shapes
physical and chemical properties of the of each of the molecules.
molecules concerned, about which you will (a) H2C=O, (b) CH3F, (c) HC≡N.
learn in later units.
Solution
12.2.2 Some Characteristic Features of π
Bonds (a) sp2 hybridised carbon, trigonal planar;
(b) sp3 hybridised carbon, tetrahedral; (c)
In a π (pi) bond formation, parallel orientation sp hybridised carbon, linear.
of the two p orbitals on adjacent atoms is
328 CHEMISTRY

12.3 STRUCTURAL REPRESENTATIONS Similarly, CH3CH2CH2CH2CH2CH2CH2CH3

OF ORGANIC COMPOUNDS can be further condensed to CH3(CH2)6CH3.
12.3.1 Complete, Condensed and Bond-line For further simplification, organic chemists
Structural Formulas use another way of representing the
structures, in which only lines are used. In
Structures of organic compounds are
this bond-line structural representation of
represented in several ways. The Lewis
organic compounds, carbon and hydrogen
structure or dot structure, dash structure,
atoms are not shown and the lines
condensed structure and bond line structural
representing carbon-carbon bonds are drawn
formulas are some of the specific types. The
in a zig-zag fashion. The only atoms
Lewis structures, however, can be simplified
specifically written are oxygen, chlorine,
by representing the two-electron covalent
nitrogen etc. The terminals denote methyl
bond by a dash (–). Such a structural formula
(–CH3) groups (unless indicated otherwise by
focuses on the electrons involved in bond
a functional group), while the line junctions
formation. A single dash represents a single
denote carbon atoms bonded to appropriate
bond, double dash is used for double bond
number of hydrogens required to satisfy the
and a triple dash represents triple bond. Lone-
valency of the carbon atoms. Some of the
pairs of electrons on heteroatoms (e.g.,
examples are represented as follows:
oxygen, nitrogen, sulphur, halogens etc.) may
or may not be shown. Thus, ethane (C2H6), (i) 3-Methyloctane can be represented in
ethene (C2H4), ethyne (C2H2) and methanol various forms as:
(CH3OH) can be represented by the following (a) CH3CH2CHCH2CH2CH2CH2CH3
structural for mulas. Such structural |
representations are called complete structural CH3
formulas.

(b)

Ethane Ethene

(c)

Ethyne Methanol

These structural formulas can be further

abbreviated by omitting some or all of the
dashes representing covalent bonds and by (ii) Various ways of representing 2-bromo
indicating the number of identical groups butane are:
attached to an atom by a subscript. The
resulting expression of the compound is called
a condensed structural formula. Thus, ethane, (a) CH3CHBrCH2CH3 (b)
ethene, ethyne and methanol can be written
as:
CH3CH3 H2C=CH2 HC≡≡CH CH3OH
(c)
Ethane Ethene Ethyne Methanol
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 329

In cyclic compounds, the bond-line formulas

may be given as follows:
(b)

Solution
Condensed formula:
Cyclopropane (a) HO(CH2)3CH(CH3)CH(CH3)2
(b) HOCH(CN)2
Bond-line formula:
(a)

Cyclopentane

(b)

chlorocyclohexane
Problem 12.6
Problem 12.4
Expand each of the following bond-line
Expand each of the following condensed formulas to show all the atoms including
formulas into their complete structural carbon and hydrogen
formulas. (a)
(a) CH3CH2COCH2CH3
(b) CH3CH=CH(CH2)3CH3

Solution (b)
(a)
(c)

(b) (d)

Solution

Problem 12.5
For each of the following compounds,
write a condensed formula and also their
bond-line formula.

(a) HOCH2CH2CH2CH(CH3)CH(CH3)CH3
330 CHEMISTRY

Molecular Models
Molecular models are physical devices that
are used for a better visualisation and
perception of three-dimensional shapes of
organic molecules. These are made of wood,
plastic or metal and are commercially
available. Commonly three types of molecular
models are used: (1) Framework model, (2)
Ball-and-stick model, and (3) Space filling
model. In the framework model only the
bonds connecting the atoms of a molecule
and not the atoms themselves are shown.
This model emphasizes the pattern of bonds
of a molecule while ignoring the size of atoms.
In the ball-and-stick model, both the atoms
and the bonds are shown. Balls represent
atoms and the stick denotes a bond.
Compounds containing C=C (e.g., ethene) can
best be represented by using springs in place
12.3.2 Three-Dimensional
of sticks. These models are referred to as ball-
Representation of Organic and-spring model. The space-filling model
Molecules emphasises the relative size of each atom
The three-dimensional (3-D) structure of based on its van der Waals radius. Bonds
organic molecules can be represented on are not shown in this model. It conveys the
paper by using certain conventions. For volume occupied by each atom in the
example, by using solid ( ) and dashed molecule. In addition to these models,
computer graphics can also be used for
( ) wedge formula, the 3-D image of a
molecular modelling.
molecule from a two-dimensional picture
can be perceived. In these formulas the
solid-wedge is used to indicate a bond
projecting out of the plane of paper, towards
the observer. The dashed-wedge is used to
depict the bond projecting out of the plane of
the paper and away from the observer. Wedges
are shown in such a way that the broad end
of the wedge is towards the observer. The Ball and stick model
Framework model
bonds lying in plane of the paper are depicted
by using a normal line (—). 3-D representation
of methane molecule on paper has been
shown in Fig. 12.1.

Space filling model

Fig. 12.2
Fig. 12.1 Wedge-and-dash representation of CH4
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 331

12.4 CLASSIFICATION OF ORGANIC (homocyclic). Sometimes atoms other than

COMPOUNDS carbon are also present in the ring
(heterocylic). Some examples of this type of
The existing large number of organic
compounds are:
compounds and their ever -increasing
numbers has made it necessary to classify
them on the basis of their structures. Organic
compounds are broadly classified as follows:

Cyclopropane Cyclohexane

Cyclohexene Tetrahydrofuran
These exhibit some of the properties similar
to those of aliphatic compounds.
Aromatic compounds
Aromatic compounds are special types of
compounds. You will learn about these
compounds in detail in Unit 13. These include
benzene and other related ring compounds
(benzenoid). Like alicyclic compounds,
aromatic comounds may also have hetero
atom in the ring. Such compounds are called
I. Acyclic or open chain compounds hetrocyclic aromatic compounds. Some of the
examples of various types of aromatic
These compounds are also called as aliphatic
compounds are:
compounds and consist of straight or
branched chain compounds, for example: Benzenoid aromatic compounds

CH3CH3
Ethane

Isobutane
Benzene Aniline Naphthalene
Non-benzenoid compound

Acetaldehyde Acetic acid

II Alicyclic or closed chain or ring

compounds
Alicyclic (aliphatic cyclic) compounds contain
carbon atoms joined in the form of a ring Tropolone
332 CHEMISTRY

Heterocyclic aromatic compounds acid found in red ant is named formic acid
since the Latin word for ant is formica. These
names are traditional and are considered as
trivial or common names. Some common
names are followed even today. For example,
Furan Thiophene Pyridine Buckminsterfullerene is a common name
Organic compounds can also be classified given to the newly discovered C60 cluster
on the basis of functional groups, into families (a form of carbon) noting its structural
or homologous series. similarity to the geodesic domes popularised
by the famous architect R. Buckminster
Functional Group
Fuller. Common names are useful and in
The functional group may be defined as an many cases indispensable, particularly when
atom or group of atoms joined in a specific the alternative systematic names are lengthy
manner which is responsible for the and complicated. Common names of some
characteristic chemical properties of the organic compounds are given in Table 12.1.
organic compounds. The examples are
Table 12.1 Common or Trivial Names of Some
hydroxyl group (–OH), aldehyde group (–CHO) Organic Compounds
and carboxylic acid group (–COOH) etc.
Homologous Series
A group or a series of organic compounds each
containing a characteristic functional group
forms a homologous series and the members
of the series are called homologues. The
members of a homologous series can be
represented by general molecular formula and
the successive members differ from each other
in molecular formula by a –CH2 unit. There
are a number of homologous series of
organic compounds. Some of these are
alkanes, alkenes, alkynes, haloalkanes,
alkanols, alkanals, alkanones, alkanoic acids,
amines etc.
12.5 NOMENCLATURE OF ORGANIC
COMPOUNDS
Organic chemistry deals with millions of
compounds. In order to clearly identify them, a
systematic method of naming has been 12.5.1 The IUPAC System of Nomenclature
developed and is known as the IUPAC A systematic name of an organic compound
(International Union of Pure and Applied is generally derived by identifying the parent
Chemistry) system of nomenclature. In this hydrocarbon and the functional group(s)
systematic nomenclature, the names are attached to it. See the example given below.
correlated with the structure such that the
reader or listener can deduce the structure from
the name.
Before the IUPAC system of nomenclature,
however, organic compounds were assigned
names based on their origin or certain
properties. For instance, citric acid is named
so because it is found in citrus fruits and the
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 333

By further using prefixes and suffixes, the In order to name such compounds, the names
parent name can be modified to obtain the of alkyl groups are prefixed to the name of
actual name. Compounds containing carbon parent alkane. An alkyl group is derived from
and hydrogen only are called hydrocarbons. A a saturated hydrocarbon by removing a
hydrocarbon is termed saturated if it contains hydrogen atom from carbon. Thus, CH4
only carbon-carbon single bonds. The IUPAC becomes -CH3 and is called methyl group. An
name for a homologous series of such alkyl group is named by substituting ‘yl’ for
compounds is alkane. Paraffin (Latin: little ‘ane’ in the corresponding alkane. Some alkyl
affinity) was the earlier name given to these groups are listed in Table 12.3.
compounds. Unsaturated hydrocarbons are Table 12.3 Some Alkyl Groups
those, which contain at least one carbon-
carbon double or triple bond.
12.5.2 IUPAC Nomenclature of Alkanes
Straight chain hydrocarbons: The names
of such compounds are based on their chain
structure, and end with suffix ‘-ane’ and carry
a prefix indicating the number of carbon
atoms present in the chain (except from CH4
to C4H10, where the prefixes are derived from
trivial names). The IUPAC names of some
straight chain saturated hydrocarbons are
given in Table 12.2. The alkanes in Table 12.2 Abbreviations are used for some alkyl
differ from each other by merely the number groups. For example, methyl is abbreviated
of -CH 2 groups in the chain. They are as Me, ethyl as Et, propyl as Pr and butyl as
homologues of alkane series. Bu. The alkyl groups can be branched also.
Thus, propyl and butyl groups can have
Table 12.2 IUPAC Names of Some Unbranched branched structures as shown below.
Saturated Hydrocarbons
CH3-CH- CH3-CH2-CH- CH3-CH-CH2-
⏐ ⏐ ⏐
CH3 CH3 CH3
Isopropyl- sec-Butyl- Isobutyl-
CH3 CH3
⏐ ⏐
CH3-C- CH3-C-CH2-
⏐ ⏐
CH3 CH3
tert-Butyl- Neopentyl-
Common branched groups have specific
Branched chain hydrocarbons: In a
trivial names. For example, the propyl groups
branched chain compound small chains of
can either be n-propyl group or isopropyl
carbon atoms are attached at one or more
group. The branched butyl groups are called
carbon atoms of the parent chain. The small
sec-butyl, isobutyl and tert-butyl group. We
carbon chains (branches) are called alkyl
also encounter the structural unit,
groups. For example:
–CH2C(CH3)3, which is called neopentyl group.
CH3–CH–CH2–CH3 CH3–CH–CH2–CH–CH3
Nomenclature of branched chain alkanes:
⏐ ⏐ ⏐
We encounter a number of branched chain
CH3 CH2CH3 CH3
alkanes. The rules for naming them are given
(a) (b) below.
334 CHEMISTRY

1. First of all, the longest carbon chain in separated from the groups by hyphens and
the molecule is identified. In the example there is no break between methyl and
(I) given below, the longest chain has nine nonane.]
carbons and it is considered as the parent 4. If two or more identical substituent groups
or root chain. Selection of parent chain as are present then the numbers are
shown in (II) is not correct because it has separated by commas. The names of
only eight carbons. identical substituents are not repeated,
instead prefixes such as di (for 2), tri
(for 3), tetra (for 4), penta (for 5), hexa (for
6) etc. are used. While writing the name of
the substituents in alphabetical order,
these prefixes, however, are not considered.
Thus, the following compounds are
named as:
CH3 CH3 CH3 CH3
⏐ ⏐ ⏐ ⏐
CH3-CH-CH2-CH-CH3 CH3⎯C⎯CH2⎯CH⎯CH3
1 2 3 4 5 1 2⏐ 3 4 5
CH3
2. The carbon atoms of the parent chain are
numbered to identify the parent alkane and 2,4-Dimethylpentane 2,2,4-Trimethylpentane
to locate the positions of the carbon atoms H 3 C H2 C CH3
at which branching takes place due to the ⏐ ⏐
substitution of alkyl group in place of CH3⎯CH2⎯CH⎯C⎯CH2⎯CH2⎯CH3
hydrogen atoms. The numbering is done 1 2 3 ⏐4 5 6 7
in such a way that the branched carbon
CH3
atoms get the lowest possible numbers.
Thus, the numbering in the above example 3-Ethyl-4,4-dimethylheptane
should be from left to right (branching at
5. If the two substituents are found in
carbon atoms 2 and 6) and not from right
to left (giving numbers 4 and 8 to the equivalent positions, the lower number is
carbon atoms at which branches are given to the one coming first in the
attached). alphabetical listing. Thus, the following
compound is 3-ethyl-6-methyloctane and
1 2 3 4 5 6 7 8 9 not 6-ethyl-3-methyloctane.
C ⎯ C ⎯ C ⎯ C ⎯ C ⎯ C ⎯C ⎯ C ⎯ C
1 2 3 4 5 6 7 8
⏐ ⏐
CH3 — CH2—CH—CH2—CH2—CH—CH2 —CH3
C C⎯C
⏐ ⏐
9 8 7 6 5 4 3 2 1 CH2CH3 CH3
C⎯ C⎯C⎯C⎯C⎯C⎯C⎯C⎯C
⏐ ⏐ 6. The branched alkyl groups can be named
C C⎯C by following the above mentioned
3. The names of alkyl groups attached as a procedures. However, the carbon atom of
branch are then prefixed to the name of the branch that attaches to the root
the parent alkane and position of the alkane is numbered 1 as exemplified
substituents is indicated by the below.
appropriate numbers. If different alkyl 4 3 2 1
groups are present, they are listed in CH3–CH–CH2–CH–
alphabetical order. Thus, name for the ⏐ ⏐
compound shown above is: 6-ethyl-2- CH3 CH3
methylnonane. [Note: the numbers are 1,3-Dimethylbutyl-
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 335

The name of such branched chain alkyl group Cyclic Compounds: A saturated monocyclic
is placed in parenthesis while naming the compound is named by prefixing ‘cyclo’ to the
compound. While writing the trivial names of corresponding straight chain alkane. If side
substituents’ in alphabetical order, the chains are present, then the rules given above
prefixes iso- and neo- are considered to be are applied. Names of some cyclic compounds
the part of the fundamental name of alkyl are given below.
group. The prefixes sec- and tert- are not
considered to be the part of the fundamental
name. The use of iso and related common
prefixes for naming alkyl groups is also
allowed by the IUPAC nomenclature as long
as these are not further substituted. In multi-
substituted compounds, the following rules
may aso be remembered:
• If there happens to be two chains of equal
size, then that chain is to be selected
which contains more number of side 3-Ethyl-1,1-dimethylcyclohexane
chains. (not 1-ethyl-3,3-dimethylcyclohexane)
• After selection of the chain, numbering is
to be done from the end closer to the Problem 12.7
substituent. Structures and IUPAC names of some
hydrocarbons are given below. Explain
why the names given in the parentheses
are incorrect.

2,5,6- Trimethyloctane
[and not 3,4,7-Trimethyloctane]

5-(2-Ethylbutyl)-3,3-dimethyldecane
[and not 5-(2,2-Dimethylbutyl)-3-ethyldecane]

3-Ethyl-5-methylheptane
[and not 5-Ethyl-3-methylheptane]

Solution
(a) Lowest locant number, 2,5,6 is lower
than 3,5,7, (b) substituents are in
5-sec-Butyl-4-isopropyldecane equivalent position; lower number is
given to the one that comes first in the
name according to alphabetical order.

12.5.3 Nomenclature of Organic

Compounds having Functional
Group(s)
A functional group, as defined earlier, is an
atom or a group of atoms bonded together in a
5-(2,2-Dimethylpropyl)nonane unique manner which is usually the site of
336 CHEMISTRY

chemical reactivity in an organic molecule. suffix. In such cases the full name of the parent
Compounds having the same functional group alkane is written before the class suffix. For
undergo similar reactions. For example, example CH 2 (OH)CH 2 (OH) is named as
CH3OH, CH3CH2OH, and (CH3)2CHOH — all ethane–1,2–diol. However, the ending – ne of
having -OH functional group liberate hydrogen the parent alkane is dropped in the case of
on reaction with sodium metal. The presence compounds having more than one double or
of functional groups enables systematisation triple bond; for example, CH2=CH-CH=CH2 is
of organic compounds into different classes. named as buta–1,3–diene.
Examples of some functional groups with their
prefixes and suf fixes along with some Problem 12.8
examples of organic compounds possessing Write the IUPAC names of the compounds
these are given in Table 12.4. i-iv from their given structures.
First of all, the functional group present
in the molecule is identified which determines
the choice of appropriate suffix. The longest
chain of carbon atoms containing the
functional group is numbered in such a way
that the functional group is attached at the Solution
carbon atom possessing lowest possible • The functional group present is an
number in the chain. By using the suffix as alcohol (OH). Hence the suffix is ‘-ol’.
given in Table 12.4, the name of the compound • The longest chain containing -OH has
is arrived at. eight carbon atoms. Hence the
In the case of polyfunctional compounds, corresponding saturated hydrocarbon
one of the functional groups is chosen as the is octane.
principal functional group and the compound is • The -OH is on carbon atom 3. In
then named on that basis. The remaining addition, a methyl group is attached
functional groups, which are subordinate at 6th carbon.
functional groups, are named as substituents Hence, the systematic name of this
using the appropriate prefixes. The choice of compound is 6-Methyloctan-3-ol.
principal functional group is made on the basis
of order of preference. The order of decreasing
priority for some functional groups is:
-COOH, –SO3H, -COOR (R=alkyl group), COCl,
-CONH2, -CN,-HC=O, >C=O, -OH, -NH2, >C=C<,
-C≡≡C- . Solution
The –R, C6H5-, halogens (F, Cl, Br, I), –NO2, The functional group present is ketone
alkoxy (–OR) etc. are always prefix (>C=O), hence suffix ‘-one’. Presence of
substituents. Thus, a compound containing two keto groups is indicated by ‘di’,
both an alcohol and a keto group is named hence suffix becomes ‘dione’. The two
as hydroxyalkanone since the keto group is keto groups are at carbons 2 and 4. The
preferred to the hydroxyl group. longest chain contains 6 carbon atoms,
For example, HOCH2(CH2)3CH2COCH3 will be hence, parent hydrocarbon is hexane.
named as 7-hydroxyheptan-2-one and not as Thus, the systematic name is Hexane-
2-oxoheptan -7-ol. Similarly, BrCH2CH=CH2 2,4-dione.
is named as 3-bromoprop-1-ene and not 1-
bromoprop-2-ene.
If more than one functional group of the
same type are present, their number is
indicated by adding di, tri, etc. before the class
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 337

Table 12.4 Some Functional Groups and Classes of Organic Compounds

338 CHEMISTRY

Solution (iii) Six membered ring containing a

Here, two functional groups namely carbon-carbon double bond is implied by
ketone and carboxylic acid are present. cyclohexene, which is numbered as
The principal functional group is the shown in (I). The prefix 3-nitro means that
carboxylic acid group; hence the parent a nitro group is present on C-3. Thus,
chain will be suffixed with ‘oic’ acid. complete structural formula of the
Numbering of the chain starts from compound is (II). Double bond is suffixed
carbon of – COOH functional group. The functional group whereas NO2 is prefixed
keto group in the chain at carbon 5 is functional group therefore double bond
indicated by ‘oxo’. The longest chain gets preference over –NO2 group:
including the principal functional
group has 6 carbon atoms; hence the
parent hydrocarbon is hexane. The
compound is, therefore, named as
5-Oxohexanoic acid.

(iv) ‘1-ol’ means that a -OH group is

Solution present at C-1. OH is suffixed functional
The two C=C functional groups are group and gets preference over C=C
present at carbon atoms 1 and 3, while bond. Thus the structure is as shown
the C≡C functional group is present at in (II):
carbon 5. These groups are indicated by
suffixes ‘diene’ and ‘yne’ respectively. The
longest chain containing the functional
groups has 6 carbon atoms; hence the
parent hydrocarbon is hexane. The name
of compound, therefore, is Hexa-1,3-
dien-5-yne.
(v) ‘heptanal’ indicates the compound to
Problem 12.9
be an aldehyde containing 7 carbon
Derive the structure of (i) 2-Chlorohexane, atoms in the parent chain. The
(ii) Pent-4-en-2-ol, (iii) 3- Nitrocyclohexene, ‘6-hydroxy’ indicates that -OH group is
(iv) Cyclohex-2-en-1-ol, (v) 6-Hydroxy- present at carbon 6. Thus, the structural
heptanal. for mula of the compound is:
Solution CH3CH(OH)CH2CH2CH2CH2CHO. Carbon
atom of –CHO group is included while
(i) ‘hexane’ indicates the presence of
numbering the carbon chain.
6 carbon atoms in the chain. The
functional group chloro is present at
carbon 2. Hence, the structure of the 12.5.4 Nomenclature of Substituted
compound is CH3CH2CH2CH2CH(Cl)CH3. Benzene Compounds
(ii) ‘pent’ indicates that parent
For IUPAC nomenclature of substituted
hydrocarbon contains 5 carbon atoms in
benzene compounds, the substituent is
the chain. ‘en’ and ‘ol’ correspond to the
functional groups C=C and -OH at placed as prefix to the word benzene as
carbon atoms 4 and 2 respectively. Thus, shown in the following examples. However,
the structure is common names (written in bracket below)
of many substituted benzene compounds
CH2=CHCH2CH (OH)CH3. are also universally used.
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 339

Substituent of the base compound is

assigned number1 and then the direction of
numbering is chosen such that the next
substituent gets the lowest number. The
substituents appear in the name in
Methylbenzene Methoxybenzene Aminobenzene alphabetical order. Some examples are given
(Toluene) (Anisole) (Aniline)
below.

Nitrobenzene Bromobenzene 1-Chloro-2,4-dinitrobenzene

(not 4-chloro,1,3-dinitrobenzene)
If benzene ring is disubstituted, the
position of substituents is defined
b y n u m b e r i n g the carbon atoms of
the ring such that the substituents
are located at the lowest numbers
possible.For example, the compound(b) is
named as 1,3-dibromobenzene and not as
1,5-dibromobenzene.
2-Chloro-1-methyl-4-nitrobenzene
(not 4-methyl-5-chloro-nitrobenzene)

(a) (b) (c)

1,2-Dibromo- 1,3-Dibromo- 1,4-Dibromo-
benzene benzene benzene
In the trivial system of nomenclature the
terms ortho (o), meta (m) and para (p) are used 2-Chloro-4-methylanisole 4-Ethyl-2-methylaniline
as prefixes to indicate the relative positions
1,2- ;1,3- and 1,4- respectively. Thus,
1,3-dibromobenzene (b) is named as
m-dibromobenzene (meta is abbreviated as
m-) and the other isomers of dibromobenzene
1,2-(a) and 1,4-(c), are named as ortho (or just
o-) and para (or just p-)-dibromobenzene,
respectively.
For tri - or higher substituted benzene 3,4-Dimethylphenol
derivatives, these prefixes cannot be used and
the compounds are named by identifying When a benzene ring is attached to an
substituent positions on the ring by following alkane with a functional group, it is
the lowest locant rule. In some cases, common considered as substituent, instead of a
name of benzene derivatives is taken as the parent. The name for benzene as substituent
base compound. is phenyl (C6H5-, also abbreviated as Ph).
340 CHEMISTRY

different carbon skeletons, these are referred

Problem 12.10
to as chain isomers and the phenomenon is
Write the structural formula of: termed as chain isomerism. For example, C5H12
(a) o-Ethylanisole, (b) p-Nitroaniline, represents three compounds:
(c) 2,3 - Dibromo -1 - phenylpentane, CH3
(d) 4-Ethyl-1-fluoro-2-nitrobenzene. ⏐
Solution CH3CH2CH2CH2CH3 CH3−CHCH2CH3
Pentane Isopentane
(2-Methylbutane)

CH3
⏐
CH3⎯ C⎯ CH3
(a) (b) ⏐
CH3
Neopentane
(2,2-Dimethylpropane)

(ii) Position isomerism: When two or more

compounds dif fer in the position of
(c) (d) substituent atom or functional group on the
carbon skeleton, they are called position
12.6 ISOMERISM isomers and this phenomenon is termed as
position isomerism. For example, the
The phenomenon of existence of two or more molecular formula C 3H 8O represents two
compounds possessing the same molecular alcohols:
formula but different properties is known as
OH
isomerism. Such compounds are called as
⏐
isomers. The following flow chart shows
CH3CH2CH2OH CH3−CH-CH3
different types of isomerism.
Propan-1-ol Propan-2-ol
12.6.1 Structural Isomerism
Compounds having the same molecular (iii) Functional group isomerism: Two or
formula but different structures (manners in more compounds having the same molecular
which atoms are linked) are classified as formula but different functional groups are
structural isomers. Some typical examples of called functional isomers and this
different types of structural isomerism are given phenomenon is termed as functional group
below: isomerism. For example, the molecular
(i) Chain isomerism: When two or more formula C3H6O represents an aldehyde and a
compounds have similar molecular formula but ketone:
Isomerism

Structural isomerism Stereoisomerism

Chain Position Functional Metamerism Geometrical Optical

isomerism isomerism group isomerism isomerism
isomerism
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 341

O H understanding the reactivity of organic

  compounds and in planning strategy for their
CH3−C-CH3 CH3−CH2—C= O synthesis.
Propanone Propanal In the following sections, we shall learn
some of the principles that explain how these
(iv) Metamerism: It arises due to different alkyl reactions take place.
chains on either side of the functional group
12.7.1 Fission of a Covalent Bond
in the molecule. For example, C 4 H 10 O
represents methoxypropane (CH3OC3H7) and A covalent bond can get cleaved either by : (i)
ethoxyethane (C2H5OC2H5). heterolytic cleavage, or by (ii) homolytic
cleavage.
12.6.2 Stereoisomerism
In heterolytic cleavage, the bond breaks
The compounds that have the same in such a fashion that the shared pair of
constitution and sequence of covalent bonds electrons remains with one of the fragments.
but differ in relative positions of their atoms
After heterolysis, one atom has a sextet
or groups in space are called stereoisomers.
electronic structure and a positive charge and
This special type of isomerism is called as
the other, a valence octet with at least one
stereoisomerism and can be classified as
lone pair and a negative charge. Thus,
geometrical and optical isomerism.
heterolytic cleavage of bromomethane will give

12.7 FUNDAMENTAL CONCEPTS IN CH 3 and Br– as shown below.
ORGANIC REACTION MECHANISM
In an organic reaction, the organic molecule
(also referred as a substrate) reacts with an
appropriate attacking reagent and leads to the A species having a carbon atom possessing
formation of one or more intermediate(s) and sextext of electrons and a positive charge is
finally product(s) called a carbocation (earlier called carbonium

ion). The C H3 ion is known as a methyl cation
The general reaction is depicted as follows :
or methyl carbonium ion. Carbocations are
Attacking classified as primary, secondary or tertiary
Reagent [Intermediate] Product(s) depending on whether one, two or three
Organic
molecule carbons are directly attached to the positively
(Substrate) Byproducts charged carbon. Some+ other examples of
carbocations are: CH3C H2 (ethyl +
cation, a
Substrate is that reactant which supplies primary carbocation), (CH3)2C H (isopropyl+
carbon to the new bond and the other reactant cation, a secondary carbocation), and (CH3)3C
is called reagent. If both the reactants supply (tert-butyl cation, a tertiary carbocation).
carbon to the new bond then choice is Carbocations are highly unstable and reactive
arbitrary and in that case the molecule on species. Alkyl groups directly attached to the
which attention is focused is called substrate. positively charged carbon stabilise the
In such a reaction a covalent bond carbocations due to inductive and
between two carbon atoms or a carbon and hyperconjugation effects, which you will be
some other atom is broken and a new bond is studying in the sections 12.7.5 and 12.7.9.
formed. A sequential account of each step, The
+
observed+
order of carbocation
+
stability
+
is:
describing details of electron movement, C H3 < CH3CH2 < (CH3)2CH < (CH3)3C. These
energetics during bond cleavage and bond carbocations have trigonal planar shape with
formation, and the rates of transformation positively charged carbon +being sp 2
of reactants into products (kinetics) is hybridised. Thus, the shape of C H3 may be
referred to as reaction mechanism. The considered as being derived from the overlap
knowledge of reaction mechanism helps in of three equivalent C(sp2) hybridised orbitals
342 CHEMISTRY

with 1s orbital of each of the three hydrogen Alkyl radicals are classified as primary,
atoms. Each bond may be represented as secondary, or tertiary. Alkyl radical stability
C(sp 2)–H(1s) sigma bond. The remaining increases as we proceed from primary to
carbon orbital is perpendicular to the tertiary:
molecular plane and contains no electrons.
(Fig. 12.3). ,
Methyl Ethyl Isopropyl Tert-butyl
free free free free
radical radical radical radical
Organic reactions, which proceed by
homolytic fission are called free radical or
homopolar or nonpolar reactions.
12.7.2 Nucleophiles and Electrophiles
A reagent that brings an electron pair is called
Fig. 12.3 Shape of methyl cation a nucleophile (Nu:) i.e., nucleus seeking and
the reaction is then called nucleophilic. A
The heterolytic cleavage can also give a
reagent that takes away an electron pair is
species in which carbon gets the shared pair
called electrophile (E+) i.e., electron seeking
of electrons. For example, when group Z
and the reaction is called electrophilic.
attached to the carbon leaves without
During a polar organic reaction, a
nucleophile attacks an electrophilic centre of
the substrate which is that specific atom or
electron pair, the methyl anion is part of the electrophile that is electron
deficient. Similarly, the electrophiles attack at
formed. Such a carbon species carrying a nucleophilic centre, which is the electron
negative charge on carbon atom is called rich centre of the substrate. Thus, the
carbanion. Carbanions are also unstable and electrophiles receive electron pair from
reactive species. The organic reactions which nucleophile when the two undergo bonding
proceed through heterolytic bond cleavage are interaction. A curved-arrow notation is used
called ionic or heteropolar or just polar to show the movement of an electron pair from
reactions. the nucleophile to the electrophile. Some
In homolytic cleavage, one of the examples of nucleophiles are the negatively
electrons of the shared pair in a covalent bond charged ions with lone pair of electrons such
– –
goes with each of the bonded atoms. Thus, in as hydroxide (HO ), cyanide (NC ) ions and
–
homolytic cleavage, the movement of a single carbanions (R3C: ). Neutral molecules such
electron takes place instead of an electron
as etc., can also act as
pair. The single electron movement is shown
nucleophiles due to the presence of lone pair
by ‘half-headed’ (fish hook: ) curved arrow.
of electrons. Examples of electrophiles
Such cleavage results in the formation of 
neutral species (atom or group) which include carbocations ( C H 3 ) and neutral
contains an unpaired electron. These species molecules having functional groups like
are called free radicals. Like carbocations carbonyl group (>C=O) or alkyl halides
and carbanions, free radicals are also (R 3C-X, where X is a halogen atom). The
very reactive. A homolytic cleavage can be carbon atom in carbocations has sextet
shown as: configuration; hence, it is electron deficient
and can receive a pair of electrons from the
nucleophiles. In neutral molecules such as
Alkyl alkyl halides, due to the polarity of the C-X
free radical bond a partial positive charge is generated
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 343

on the carbon atom and hence the carbon 12.7.3 Electron Movement in Organic
atom becomes an electrophilic centre at Reactions
which a nucleophile can attack. The movement of electrons in organic
reactions can be shown by curved-arrow
Problem 12.11 notation. It shows how changes in bonding
Using curved-arrow notation, show the occur due to electronic redistribution during
formation of reactive intermediates when the reaction. To show the change in position
the following covalent bonds undergo of a pair of electrons, curved arrow starts from
heterolytic cleavage. the point from where an electron pair is shifted
(a) CH3–SCH3, (b) CH3–CN, (c) CH3–Cu and it ends at a location to which the pair of
electron may move.
Solution
Presentation of shifting of electron pair is
given below :

(i) from π bond to

adjacent bond position

(ii) from π bond to

adjacent atom
Problem 12.12
(iii) from atom to adjacent
Giving justification, categorise the
bond position
following molecules/ions as nucleophile
or electrophile: Movement of single electron is indicated
by a single barbed ‘fish hooks’ (i.e. half headed
curved arrow). For example, in transfer of
hydroxide ion giving ethanol and in the
dissociation of chloromethane, the movement
Solution of electron using curved arrows can be
depicted as follows:
Nucleophiles: HS ,C2H5O , CH3 3 N:,H2N:
  

These species have unshared pair of

electrons, which can be donated and
shared with an electrophile.
  
E l e c t r o p h i l e s : BF3,Cl,CH3 C  O,NO2 .
Reactive sites have only six valence 12.7.4 Electron Displacement Effects in
electrons; can accept electron pair from Covalent Bonds
a nucleophile. The electron displacement in an organic
molecule may take place either in the ground
Problem 12.13
state under the influence of an atom or a
Identify electrophilic centre in the
substituent group or in the presence of an
following: CH3CH=O, CH3CN, CH3I.
appropriate attacking reagent. The electron
Solution displacements due to the influence of
* * an atom or a substituent group present in
Among CH 3 HC =O, H 3 C C ≡N, and
* the molecule cause permanent polarlisation
H 3C –I, the starred carbon atoms are
electrophilic centers as they will have of the bond. Inductive ef fect and
resonance effects are examples of this type of
partial positive charge due to polarity of
electron displacements. Temporary electron
the bond.
displacement effects are seen in a molecule
344 CHEMISTRY

when a reagent approaches to attack it. This nitro (- NO2), cyano (- CN), carboxy (- COOH),
type of electron displacement is called ester (-COOR), aryloxy (-OAr, e.g. – OC6H5),
electromeric effect or polarisability effect. In etc. are electron-withdrawing groups. On the
the following sections we will learn about these other hand, the alkyl groups like methyl
types of electronic displacements. (–CH 3) and ethyl (–CH 2–CH 3) are usually
considered as electron donating groups.
12.7.5 Inductive Effect
When a covalent bond is formed between Problem 12.14
atoms of different electronegativity, the Which bond is more polar in the following
electron density is more towards the more pairs of molecules: (a) H3C-H, H3C-Br
electronegative atom of the bond. Such a shift (b) H 3 C-NH 2 , H 3 C-OH (c) H 3 C-OH,
of electron density results in a polar covalent H3C-SH
bond. Bond polarity leads to various electronic
effects in organic compounds. Solution
Let us consider cholorethane (CH3CH2Cl) (a) C–Br, since Br is more electronegative
in which the C–Cl bond is a polar covalent than H, (b) C–O, (c) C–O
bond. It is polarised in such a way that the Problem 12.15
+
carbon-1 gains some positive charge (δ ) and
– In which C–C bond of CH3CH2CH2Br, the
the chlorine some negative charge (δ ). The
fractional electronic charges on the two atoms inductive effect is expected to be the
in a polar covalent bond are denoted by least?
symbol δ (delta) and the shift of electron Solution
density is shown by an arrow that points from
+ – Magnitude of inductive effect diminishes
δ to δ end of the polar bond.
+ + − as the number of intervening bonds
δδ δ δ
increases. Hence, the effect is least in the
CH3 ⎯→⎯CH2⎯→⎯ ⎯→⎯Cl
bond between carbon-3 and hydrogen.
2 1
In turn carbon-1, which has developed 12.7.6 Resonance Structure
+
partial positive charge (δ ) draws some
There are many organic molecules whose
electron density towards it from the adjacent
behaviour cannot be explained by a single
C-C bond. Consequently, some positive charge
+ + Lewis structure. An example is that of
(δδ ) develops on carbon-2 also, where δδ
benzene. Its cyclic structure
symbolises relatively smaller positive charge
containing alternating C–C single
as compared to that on carbon – 1. In other
words, the polar C – Cl bond induces polarity and C=C double bonds shown is
in the adjacent bonds. Such polarisation of inadequate for explaining its Benzene
σ-bond caused by the polarisation of adjacent characteristic properties.
σ-bond is referred to as the inductive effect. As per the above representation, benzene
This effect is passed on to the subsequent should exhibit two different bond lengths, due
bonds also but the effect decreases rapidly to C–C single and C=C double bonds. However,
as the number of intervening bonds increases as determined experimentally benzene has a
and becomes vanishingly small after three uniform C–C bond distances of 139 pm, a
bonds. The inductive effect is related to the value inter mediate between the C–C
ability of substituent(s) to either withdraw or single(154 pm) and C=C double (134 pm)
donate electron density to the attached carbon bonds. Thus, the structure of benzene cannot
atom. Based on this ability, the substitutents be represented adequately by the above
can be classified as electron-withdrawing or structure. Further, benzene can be
electron donating groups relative to hydrogen. represented equally well by the energetically
Halogens and many other groups such as identical structures I and II.
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 345

unpaired electrons. Among the resonance

structures, the one which has more number
of covalent bonds, all the atoms with octet of
electrons (except hydrogen which has a
duplet), less separation of opposite charges,
(a negative charge if any on more
electronegative atom, a positive charge if any
Therefore, according to the resonance theory on more electropositive atom) and more
(Unit 4) the actual structure of benzene dispersal of charge, is more stable than others.
cannot be adequately represented by any of
these structures, rather it is a hybrid of the Problem 12.16
two structures (I and II) called resonance –
Write resonance structures of CH3COO
structures. The resonance structures and show the movement of electrons by
(canonical structures or contributing curved arrows.
structures) are hypothetical and
Solution
individually do not represent any real
molecule. They contribute to the actual First, write the structure and put
structure in proportion to their stability. unshared pairs of valence electrons on
Another example of resonance is provided appropriate atoms. Then draw the arrows
by nitromethane (CH 3NO 2) which can be one at a time moving the electrons to get
represented by two Lewis structures, (I and the other structures.
II). There are two types of N-O bonds in these
structures.

Problem 12.17
Write resonance structures of
CH2=CH–CHO. Indicate relative stability of
However, it is known that the two N–O the contributing structures.
bonds of nitromethane are of the same
length (intermediate between a N–O single Solution
bond and a N=O double bond). The actual
structure of nitromethane is therefore a
resonance hybrid of the two canonical
forms I and II.
The energy of actual structure of the
molecule (the resonance hybrid) is lower than
that of any of the canonical structures. The
difference in energy between the actual
structure and the lowest energy resonance
structure is called the resonance Stability: I > II > III
stabilisation energy or simply the [I: Most stable, more number of covalent
resonance energy. The more the number of bonds, each carbon and oxygen atom has
important contributing structures, the more an octet and no separation of opposite
is the resonance energy. Resonance is charge II: negative charge on more
particularly important when the contributing electronegative atom and positive charge
structures are equivalent in energy. on more electropositive atom; III: does
The following rules are applied while writing not contribute as oxygen has positive
resonance structures: charge and carbon has negative charge,
The resonance structures have (i) the same hence least stable].
positions of nuclei and (ii) the same number of
346 CHEMISTRY

The atoms or substituent groups, which

Problem 12.18
represent +R or –R electron displacement
Explain why the following two structures,
effects are as follows :
I and II cannot be the major contributors
to the real structure of CH3COOCH3. +R effect: – halogen, –OH, –OR, –OCOR, –NH2,
–NHR, –NR2, –NHCOR,
– R effect: – COOH, –CHO, >C=O, – CN, –NO2
The presence of alternate single and
double bonds in an open chain or cyclic
Solution system is termed as a conjugated system.
The two structures are less important These systems often show abnor mal
contributors as they involve charge behaviour. The examples are 1,3- butadiene,
separation. Additionally, structure I aniline and nitrobenzene etc. In such systems,
contains a carbon atom with an the π-electrons are delocalised and the system
develops polarity.
incomplete octet.
12.7.8 Electromeric Effect (E effect)
12.7.7 Resonance Effect It is a temporary ef fect. The organic
The resonance effect is defined as ‘the polarity compounds having a multiple bond (a double
produced in the molecule by the interaction or triple bond) show this effect in the presence
of two π-bonds or between a π-bond and lone of an attacking reagent only. It is defined as
pair of electrons present on an adjacent atom’. the complete transfer of a shared pair of
The effect is transmitted through the chain. π-electrons to one of the atoms joined by a
There are two types of resonance or multiple bond on the demand of an attacking
mesomeric effect designated as R or M effect. reagent. The effect is annulled as soon as the
(i) Positive Resonance Effect (+R effect) attacking reagent is removed from the domain
of the reaction. It is represented by E and the
In this effect, the transfer of electrons is away
shifting of the electrons is shown by a curved
from an atom or substituent group attached
to the conjugated system. This electron arrow ( ). There are two distinct types of
displacement makes certain positions in the electromeric effect.
molecule of high electron densities. This effect (i) Positive Eelctromeric Effect (+E effect) In
in aniline is shown as : this effect the π−electrons of the multiple bond
are transferred to that atom to which the
reagent gets attached. For example :

(ii) Negative Resonance Effect (- R effect)

This effect is observed when the transfer of
(ii) Negative Electromeric Effect (–E effect) In
electrons is towards the atom or substituent
this effect the π - electrons of the multiple
group attached to the conjugated system. For
bond are transferred to that atom to which
example in nitrobenzene this electron
the attacking reagent does not get attached.
displacement can be depicted as :
For example:
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 347

When inductive and electromeric effects In general, greater the number of alkyl
operate in opposite directions, the electomeric groups attached to a positively charged carbon
effect predominates. atom, the greater is the hyperconjugation
interaction and stabilisation of the cation.
12.7.9 Hyperconjugation
Thus, we have the following relative stability
Hyperconjugation is a general stabilising of carbocations :
interaction. It involves delocalisation of
σ electrons of C—H bond of an alkyl group
directly attached to an atom of unsaturated
system or to an atom with an unshared
p orbital. The σ electrons of C—H bond of the
alkyl group enter into partial conjugation with Hyperconjugation is also possible in
the attached unsaturated system or with the alkenes and alkylarenes.
unshared p orbital. Hyperconjugation is a Delocalisation of electrons by
permanent effect. hyperconjugation in the case of alkene can
To understand hyperconjugation effect, let be depicted as in Fig. 12.4(b).

us take an example of CH3 CH2 (ethyl cation)
in which the positively charged carbon atom
has an empty p orbital. One of the C-H bonds
of the methyl group can align in the plane of
this empty p orbital and the electrons
constituting the C-H bond in plane with this
p orbital can then be delocalised into the
empty p orbital as depicted in Fig. 12.4 (a).
Fig. 12.4(b) Orbital diagram showing
hyperconjugation in propene
There are various ways of looking at the
hyperconjugative effect. One of the way is to
regard C—H bond as possessing partial ionic
character due to resonance.

Fig. 12.4(a) Orbital diagram showing

hyperconjugation in ethyl cation
This type of overlap stabilises the
carbocation because electron density from the
adjacent σ bond helps in dispersing the
positive charge.
348 CHEMISTRY

The hyperconjugation may also be New methods of checking the purity of an

regarded as no bond resonance. organic compound are based on different
types of chromatographic and spectroscopic
Problem 12.19 + techniques.
Explain why (CH3)3C is more stable than
+ + 12.8.1 Sublimation
CH3CH2 and C H3 is the least stable
cation. You have learnt earlier that on heating, some
solid substances change from solid to vapour
Solution + state without passing through liquid state.
Hyperconjugation interaction in (CH3)3C The purification technique based on the above
+
is greater than in CH C H as the principle is known as sublimation and is used
+ 3 2 +
(CH3)3C has nine C-H bonds. In C H3 , to separate sublimable compounds from non-
sublimable impurities.
vacant p orbital is perpendicular to the
plane in which C-H bonds lie; hence 12.8.2 Crystallisation
+
cannot overlap with it. Thus, C H lacks This is one of the most commonly used
3
hyperconjugative stability. techniques for the purification of solid organic
compounds. It is based on the difference in
12.7.10 Types of Organic Reactions and the solubilities of the compound and the
Mechanisms impurities in a suitable solvent. The impure
Organic reactions can be classified into the compound is dissolved in a solvent in which
following categories: it is sparingly soluble at room temperature
(i) Substitution reactions but appreciably soluble at higher
temperature. The solution is concentrated to
(ii) Addition reactions
get a nearly saturated solution. On cooling
(iii) Elimination reactions
the solution, pure compound crystallises out
(iv) Rearrangement reactions
and is removed by filtration. The filtrate
You will be studying these reactions in (mother liquor) contains impurities and small
Unit 13 and later in class XII. quantity of the compound. If the compound
is highly soluble in one solvent and very little
12.8 METHODS OF PURIFICATION OF
soluble in another solvent, crystallisation can
ORGANIC COMPOUNDS
be satisfactorily carried out in a mixture of
Once an organic compound is extracted from these solvents. Impurities, which impart
a natural source or synthesised in the colour to the solution are removed by
laboratory, it is essential to purify it. Various adsorbing over activated charcoal. Repeated
methods used for the purification of organic crystallisation becomes necessary for the
compounds are based on the nature of the purification of compounds containing
compound and the impurity present in it. impurities of comparable solubilities.
The common techniques used for 12.8.3 Distillation
purification are as follows :
This important method is used to separate (i)
(i) Sublimation
volatile liquids from nonvolatile impurities and
(ii) Crystallisation
(ii) the liquids having sufficient difference in
(iii) Distillation their boiling points. Liquids having different
(iv) Differential extraction and boiling points vaporise at dif ferent
(v) Chromatography temperatures. The vapours are cooled and the
Finally, the purity of a compound is liquids so formed are collected separately.
ascertained by determining its melting or Chloroform (b.p 334 K) and aniline (b.p. 457
boiling point. Most of the pure compounds K) are easily separated by the technique of
have sharp melting points and boiling points. distillation (Fig 12.5). The liquid mixture is
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 349

taken in a round bottom flask and

heated carefully. On boiling, the
vapours of lower boiling component
are formed first. The vapours are
condensed by using a condenser and
the liquid is collected in a receiver. The
vapours of higher boiling component
for m later and the liquid can be
collected separately.
Fractional Distillation: If the
difference in boiling points of two
liquids is not much, simple distillation
cannot be used to separate them. The
vapours of such liquids are formed
within the same temperature range and
are condensed simultaneously. The
technique of fractional distillation is
used in such cases. In this technique,
vapours of a liquid mixture are passed
through a fractionating column before
Fig.12.5 Simple distillation. The vapours of a substance condensation. The fractionating
formed are condensed and the liquid is collected
column is fitted over the mouth of the
in conical flask.
round bottom flask (Fig.12.6).

Fig.12.6 Fractional distillation. The vapours of lower boiling fraction reach the
top of the column first followed by vapours of higher boiling fractions.
350 CHEMISTRY

Vapours of the liquid with higher boiling theoretical plate. Commercially, columns
point condense before the vapours of the with hundreds of plates are available.
liquid with lower boiling point. The vapours One of the technological applications of
rising up in the fractionating column become fractional distillation is to separate different
richer in more volatile component. By the time fractions of crude oil in petroleum industry.
the vapours reach to the top of the
Distillation under reduced pressure: This
fractionating column, these are rich in the
method is used to purify liquids having very
more volatile component. Fractionating
high boiling points and those, which
columns are available in various sizes and
decompose at or below their boiling points.
designs as shown in Fig.12.7. A fractionating
Such liquids are made to boil at a temperature
column provides many surfaces for heat
lower than their normal boiling points by
exchange between the ascending vapours
reducing the pressure on their surface. A
and the descending condensed liquid. Some
liquid boils at a temperature at which its
of the condensing liquid in the fractionating
vapour pressure is equal to the external
column obtains heat from the ascending
pressure. The pressure is reduced with the
vapours and revaporises. The vapours thus
help of a water pump or vacuum pump
become richer in low boiling component. The
(Fig.12.8). Glycerol can be separated from
vapours of low boiling component ascend to
spent-lye in soap industry by using this
the top of the column. On reaching the top,
technique.
the vapours become pure in low boiling
component and pass through the condenser Steam Distillation: This technique is
and the pure liquid is collected in a receiver. applied to separate substances which are
After a series of successive distillations, the steam volatile and are immiscible with
remaining liquid in the distillation flask gets water. In steam distillation, steam from a
enriched in high boiling component. Each steam generator is passed through a heated
successive condensation and vaporisation flask containing the liquid to be distilled.
unit in the fractionating column is called a The mixture of steam and the volatile
organic compound is condensed and
collected. The compound is later separated
from water using a separating funnel. In
steam distillation, the liquid boils when
the sum of vapour pressures due to the
organic liquid (p 1 ) and that due to water
(p 2 ) becomes equal to the atmospheric
pressure (p), i.e. p =p 1 + p 2 . Since p 1 is
lower than p, the organic liquid vaporises
at lower temperature than its boiling
point.
Thus, if one of the substances in the
mixture is water and the other, a water
insoluble substance, then the mixture will boil
close to but below, 373K. A mixture of water
and the substance is obtained which can be
separated by using a separating funnel.
Aniline is separated by this technique from
aniline – water mixture (Fig.12.9).
12.8.4 Differential Extraction
When an organic compound is present in an
Fig.12.7 Different types of fractionating columns. aqueous medium, it is separated by shaking
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 351

Fig.12.8 Distillation under reduced pressure. A liquid boils at a temperature below its
vapour pressure by reducing the pressure.

Fig.12.9 Steam distillation. Steam volatile component volatilizes, the vapours condense in
the condenser and the liquid collects in conical flask.
352 CHEMISTRY

it with an organic solvent in which it is more mixture get gradually separated from one
soluble than in water. The organic solvent and another. The moving phase is called the mobile
the aqueous solution should be immiscible phase.
with each other so that they form two distinct Based on the principle involved,
layers which can be separated by separatory chromatography is classified into different
funnel. The organic solvent is later removed categories. Two of these are:
by distillation or by evaporation to get back (a) Adsorption chromatography, and
the compound. Differential extraction is (b) Partition chromatography.
carried out in a separatory funnel as shown
in Fig. 12.10. If the organic compound is less a) Adsorption Chr omatography: Adsor-
ption chromatography is based on the fact
that different compounds are adsorbed on an
adsorbent to different degrees. Commonly
used adsorbents are silica gel and alumina.
When a mobile phase is allowed to move
over a stationary phase (adsorbent),
the components of the mixture move by
varying distances over the stationary
phase. Following are two main types of
chromatographic techniques based on the
principle of differential adsorption.
(a) Column chromatography, and
(b) Thin layer chromatography.
Column Chromatography: Column
chromatography involves separation of a
Fig.12.10 Differential extraction. Extraction of com- mixture over a column of adsorbent
pound takes place based on difference (stationary phase) packed in a glass tube. The
in solubility
column is fitted with a stopcock at its lower
soluble in the organic solvent, a very large end (Fig. 12.11). The mixture adsorbed on
quantity of solvent would be required to
extract even a very small quantity of the
compound. The technique of continuous
extraction is employed in such cases. In this
technique same solvent is repeatedly used for
extraction of the compound.
12.8.5 Chromatography
Chromatography is an important technique
extensively used to separate mixtures into
their components, purify compounds and also
to test the purity of compounds. The name
chromatography is based on the Greek word
chroma, for colour since the method was first
used for the separation of coloured
substances found in plants. In this technique,
the mixture of substances is applied onto a
stationary phase, which may be a solid or a
liquid. A pure solvent, a mixture of solvents, Fig.12.11 Column chromatography. Different
or a gas is allowed to move slowly over the stages of separation of components
stationary phase. The components of the of a mixture.
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 353

adsorbent is placed on the top of the adsorbent eluant rises up the plate, the components of
column packed in a glass tube. An appropriate the mixture move up along with the eluant to
eluant which is a liquid or a mixture of liquids different distances depending on their degree
is allowed to flow down the column slowly. of adsorption and separation takes place. The
Depending upon the degree to which the relative adsorption of each component of the
compounds are adsorbed, complete separation mixture is expressed in ter ms of its
takes place. The most readily adsorbed retardation factor i.e. Rf value (Fig.12.12 b).
substances are retained near the top and others Distance moved by the substance from base line (x)
come down to various distances in the column Rf =
Distance moved by the solvent from base line (y)
(Fig.12.11).
The spots of coloured compounds are visible
Thin Layer Chromatography: Thin layer
on TLC plate due to their original colour. The
chromatography (TLC) is another type of
spots of colourless compounds, which are
adsorption chromatography, which involves
invisible to the eye but fluoresce in ultraviolet
separation of substances of a mixture over a
light, can be detected by putting the plate under
thin layer of an adsorbent coated on glass
ultraviolet light. Another detection technique is
plate. A thin layer (about 0.2mm thick) of an
to place the plate in a covered jar containing a
adsorbent (silica gel or alumina) is spread over
few crystals of iodine. Spots of compounds,
a glass plate of suitable size. The plate is
which adsorb iodine, will show up as brown
known as thin layer chromatography plate or
spots. Sometimes an appropriate reagent may
chromaplate. The solution of the mixture to
also be sprayed on the plate. For example, amino
be separated is applied as a small spot about
acids may be detected by spraying the plate with
2 cm above one end of the TLC plate. The
ninhydrin solution (Fig.12.12b).
glass plate is then placed in a closed jar
containing the eluant (Fig. 12.12a). As the Partition Chromatography: Partition
chromatography is based on continuous
differential partitioning of components of a
mixture between stationary and mobile
phases. Paper chromatography is a type of
partition chromatography. In paper
chromatography, a special quality paper
known as chromatography paper is used.
Chromatography paper contains water
trapped in it, which acts as the stationary
phase.
A strip of chromatography paper spotted
Fig.12.12 (a) Thin layer chromatography. at the base with the solution of the mixture is
Chromatogram being developed. suspended in a suitable solvent or a mixture
of solvents (Fig. 12.13). This solvent acts as
the mobile phase. The solvent rises up the
paper by capillary action and flows over the
spot. The paper selectively retains different
components according to their differing
partition in the two phases. The paper strip
so developed is known as a chromatogram.
The spots of the separated coloured
compounds are visible at different heights
from the position of initial spot on the
chromatogram. The spots of the separated
Fig.12.12 (b) Developed chromatogram. colourless compounds may be observed either
354 CHEMISTRY

5H2O + CuSO4 ⎯→ CuSO4.5H2O

White Blue
12.9.2 Detection of Other Elements
Nitrogen, sulphur, halogens and phosphorus
present in an organic compound are detected
by “Lassaigne’s test”. The elements present
in the compound are converted from covalent
form into the ionic form by fusing the
compound with sodium metal. Following
reactions take place:
Na + C + N  NaCN

2Na + S  Na2S

Na + X  Na X
(X = Cl, Br or I)
C, N, S and X come from organic
compound.
Cyanide, sulphide and halide of sodium
so formed on sodium fusion are extracted from
the fused mass by boiling it with distilled
water. This extract is known as sodium fusion
extract.
(A) Test for Nitrogen
Fig.12.13 Paper chromatography.
The sodium fusion extract is boiled with
Chromatography paper in two different
shapes.
iron(II) sulphate and then acidified with
concentrated sulphuric acid. The formation
under ultraviolet light or by the use of an of Prussian blue colour confirms the presence
appropriate spray reagent as discussed under of nitrogen. Sodium cyanide first reacts
thin layer chromatography. with iron(II) sulphate and forms sodium
12.9 QUALITATIVE ANALYSIS OF hexacyanoferrate(II). On heating with
ORGANIC COMPOUNDS concentrated sulphuric acid some iron(II) ions
The elements present in organic compounds are oxidised to iron(III) ions which react with
are carbon and hydrogen. In addition to these, sodium hexacyanoferrate(II) to produce
they may also contain oxygen, nitrogen, iron(III) hexacyanoferrate(II) (ferriferrocyanide)
sulphur, halogens and phosphorus. which is Prussian blue in colour.
–
12.9.1 Detection of Carbon and Hydrogen 6CN + Fe2+ → [Fe(CN)6]4–
Carbon and hydrogen are detected by heating 3[Fe(CN)6]4– + 4Fe3+ Fe4[Fe(CN)6]3.xH2O
the compound with copper(II) oxide. Carbon Prussian blue
present in the compound is oxidised to carbon (B) Test for Sulphur
dioxide (tested with lime-water, which (a) The sodium fusion extract is acidified
develops turbidity) and hydrogen to water with acetic acid and lead acetate is added
(tested with anhydrous copper sulphate, to it. A black precipitate of lead sulphide
which turns blue). indicates the presence of sulphur.

C + 2CuO  2Cu + CO2 S2– + Pb2+⎯→ PbS

2H + CuO  Cu + H2O Black
(b) On treating sodium fusion extract with
CO2 + Ca(OH)2 ⎯→ CaCO3↓ + H2O sodium nitroprusside, appearance of a
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 355

violet colour further indicates the presence (D) Test for Phosphorus
of sulphur. The compound is heated with an oxidising
S2– + [Fe(CN)5NO]2– ⎯→ [Fe(CN)5NOS]4– agent (sodium peroxide). The phosphorus
Violet present in the compound is oxidised to
In case, nitrogen and sulphur both are phosphate. The solution is boiled with nitric
present in an organic compound, sodium acid and then treated with ammonium
thiocyanate is formed. It gives blood red molybdate. A yellow colouration or precipitate
colour and no Prussian blue since there are indicates the presence of phosphorus.
no free cyanide ions. Na3PO4 + 3HNO3 ⎯→ H3PO4+3NaNO3
Na + C + N + S ⎯→ NaSCN H3PO4 + 12(NH4)2MoO4 + 21HNO3 ⎯→
–
3+
Fe +SCN ⎯→ [Fe(SCN)]2+ Ammonium
Blood red molybdate
If sodium fusion is carried out with excess (NH4)3PO4.12MoO3 + 21NH4NO3 + 12H2O
of sodium, the thiocyanate decomposes to Ammonium
yield cyanide and sulphide. These ions give phosphomolybdate
their usual tests.
NaSCN + 2Na ⎯→ NaCN+Na2S 12.10 QUANTITATIVE ANALYSIS
(C) Test for Halogens The percentage composition of elements
present in an organic compound is
The sodium fusion extract is acidified with
determined by the methods based on the
nitric acid and then treated with silver nitrate.
following principles:
A white precipitate, soluble in ammonium
hydroxide shows the presence of chlorine, a 12.10.1 Carbon and Hydrogen
yellowish precipitate, sparingly soluble in Both carbon and hydrogen are estimated in
ammonium hydroxide shows the presence of one experiment. A known mass of an organic
bromine and a yellow precipitate, insoluble compound is burnt in the presence of excess
in ammonium hydroxide shows the presence of oxygen and copper(II) oxide. Carbon and
of iodine. hydrogen in the compound are oxidised to
X– + Ag+ ⎯→ AgX carbon dioxide and water respectively.
X represents a halogen – Cl, Br or I. CxHy + (x + y/4) O2 ⎯→ x CO2 + (y/2) H2O
If nitrogen or sulphur is also present in the The mass of water produced is determined
compound, the sodium fusion extract is first by passing the mixture through a weighed
boiled with concentrated nitric acid to U-tube containing anhydrous calcium
decompose cyanide or sulphide of sodium chloride. Carbon dioxide is absorbed in
formed during Lassaigne’s test. These ions another U-tube containing concentrated
would otherwise interfere with silver nitrate solution of potassium hydroxide. These tubes
test for halogens. are connected in series (Fig.12.14). The

Fig.12.14 Estimation of carbon and hydrogen. Water and carbon dioxide formed on oxidation of substance
are absorbed in anhydrous calcium chloride and potassium hydroxide solutions respectively
contained in U tubes.
356 CHEMISTRY

increase in masses of calcium chloride and

2  0.1014  100
potassium hydroxide gives the amounts of Percentage of hydrogen =
water and carbon dioxide from which the 18  0.246
percentages of carbon and hydrogen are = 4.58%
calculated.
Let the mass of organic compound be 12.10.2 Nitrogen
m g, mass of water and carbon dioxide
There are two methods for estimation of
produced be m1 and m2 g respectively;
nitrogen: (i) Dumas method and (ii) Kjeldahl’s
12  m 2  100 method.
Percentage of carbon= (i) Dumas method: The nitrogen containing
44  m
organic compound, when heated with copper
2  m1  100 oxide in an atmosphere of carbon dioxide,
Percentage of hydrogen =
18  m yields free nitrogen in addition to carbon
dioxide and water.
Problem 12.20
On complete combustion, 0.246 g of an CxHyNz + (2x + y/2) CuO ⎯→
organic compound gave 0.198g of carbon x CO2 + y/2 H2O + z/2 N2 + (2x + y/2) Cu
dioxide and 0.1014g of water. Determine Traces of nitrogen oxides formed, if any,
the percentage composition of carbon are reduced to nitrogen by passing the
and hydrogen in the compound. gaseous mixture over a heated copper gauze.
Solution The mixture of gases so produced is collected
over an aqueous solution of potassium
12  0.198  100 hydroxide which absorbs carbon dioxide.
Percentage of carbon =
44  0.246 Nitrogen is collected in the upper part of the
= 21.95% graduated tube (Fig.12.15).

Fig.12.15 Dumas method. The organic compound yields nitrogen gas on heating it with
copper(II) oxide in the presence of CO2 gas. The mixture of gases is collected
over potassium hydroxide solution in which CO2 is absorbed and volume of
nitrogen gas is determined.
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 357

Let the mass of organic compound = m g Calculate the percentage composition of

Volume of nitrogen collected = V1 mL nitrogen in the compound. (Aqueous
Room temperature = T1K tension at 300K=15 mm)
p1V1  273 Solution
Volume of nitrogen at STP=
760  T1 Volume of nitrogen collected at 300K and
(Let it be V mL) 715mm pressure is 50 mL
Where p1 and V1 are the pressure and volume Actual pressure = 715-15 =700 mm
of nitrogen, p 1 is dif ferent from the
273  700  50
atmospheric pressure at which nitrogen gas Volume of nitrogen at STP =
300  760
is collected. The value of p1 is obtained by
= 41.9 mL
the relation;
p1= Atmospheric pressure – Aqueous tension 22,400 mL of N2 at STP weighs = 28 g
22400 mL N2 at STP weighs 28 g. 28  41.9
41.9 mL of nitrogen weighs= g
22400
28  V
V mL N 2 at STP weighs = g 28  41.9  100
22400 Percentage of nitrogen =
22400  0.3
28  V  100 =17.46%
Percentage of nitrogen =
22400  m
(ii) Kjeldahl’s method: The compound
Problem 12.21 containing nitrogen is heated with
In Dumas’ method for estimation of concentrated sulphuric acid. Nitrogen in the
nitrogen, 0.3g of an organic compound compound gets converted to ammonium
gave 50mL of nitrogen collected at 300K sulphate (Fig. 12.16). The resulting acid
temperature and 715mm pressure. mixture is then heated with excess of sodium

Fig.12.16 Kjeldahl method. Nitrogen-containing compound is treated with concentrated H2SO4 to get
ammonium sulphate which liberates ammonia on treating with NaOH; ammonia is absorbed
in known volume of standard acid.
358 CHEMISTRY

hydroxide. The liberated ammonia gas is

Problem 12.22
absorbed in an excess of standard solution of
sulphuric acid. The amount of ammonia During estimation of nitrogen present in
produced is determined by estimating the an organic compound by Kjeldahl’s
amount of sulphuric acid consumed in the method, the ammonia evolved from 0.5
reaction. It is done by estimating unreacted g of the compound in Kjeldahl’s
sulphuric acid left after the absorption of estimation of nitrogen, neutralized 10 mL
ammonia by titrating it with standard alkali of 1 M H2SO4. Find out the percentage
solution. The difference between the initial of nitrogen in the compound.
amount of acid taken and that left after the Solution
reaction gives the amount of acid reacted with 1 M of 10 mL H2SO4=1M of 20 mL NH3
ammonia. 1000 mL of 1M ammonia contains 14 g
Organic compound + H2SO4 ⎯→ (NH4)2SO4 nitrogen
2 NaOH 20 mL of 1M ammonia contains

 Na 2 SO4  2NH3  2H2 O
14  20
g nitrogen
2NH3 + H2SO4 ⎯→ (NH4)2SO4 1000
Let the mass of organic compound taken = m g 14  20 100
Percentage of nitrogen =  56.0%
Volume of H2SO4 of molarity, M, 1000  05
.
taken = V mL
Volume of NaOH of molarity, M, used for 12.10.3 Halogens
titration of excess of H2SO4 = V1 mL Carius method: A known mass of an organic
V1mL of NaOH of molarity M compound is heated with fuming nitric acid
= V1 /2 mL of H2SO4 of molarity M in the presence of silver nitrate contained in
a hard glass tube known as Carius tube,
Volume of H 2SO 4 of molarity M unused
(Fig.12.17) in a furnace. Carbon and hydrogen
= (V - V1/2) mL
(V- V1/2) mL of H2SO4 of molarity M
= 2(V-V1/2) mL of NH3 solution of
molarity M.
1000 mL of 1 M NH3 solution contains 17g
NH3 or 14 g of N
2(V-V1/2) mL of NH3 solution of molarity M
contains:
14  M  2  V  V1 /2 
gN
1000
14  M  2  V-V1 /2  100
Percentage of N= 
1000 m
1.4  M  2 V -V1 /2 
=
m
Kjeldahl method is not applicable to
compounds containing nitrogen in nitro and
azo groups and nitrogen present in the ring Fig. 12.17 Carius method. Halogen containing
(e.g. pyridine) as nitrogen of these compounds organic compound is heated with fuming
does not change to ammonium sulphate nitric acid in the presence of silver
under these conditions. nitrate.
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 359

present in the compound are oxidised to 1 mol of BaSO4 = 233 g BaSO4 = 32 g sulphur
carbon dioxide and water. The halogen
32  m1
present forms the corresponding silver halide m1 g BaSO4 contains g sulphur
(AgX). It is filtered, washed, dried and weighed. 233
Let the mass of organic 32  m1  100
compound taken = m g Percentage of sulphur=
Mass of AgX formed = m1 g 233  m
1 mol of AgX contains 1 mol of X
Problem 12.24
Mass of halogen in m1g of AgX
In sulphur estimation, 0.157 g of an
atomic mass of X  m1g organic compound gave 0.4813 g of

molecular mass of AgX barium sulphate. What is the
Percentage of halogen percentage of sulphur in the compound?
Solution
atomic mass of X  m1  100
 Molecular mass of BaSO4 = 137+32+64
molecular mass of AgX  m = 233 g
Problem 12.23 233 g BaSO4 contains 32 g sulphur
In Carius method of estimation of 32  0.4813
halogen, 0.15 g of an organic compound 0.4813 g BaSO4 contains g
233
gave 0.12 g of AgBr. Find out the
sulphur
percentage of bromine in the compound.
Solution 32  0.4813  100
Percentage of sulphur=
Molar mass of AgBr = 108 + 80 233  0.157
= 188 g mol-1 = 42.10%
188 g AgBr contains 80 g bromine
12.10.5 Phosphorus
80  0.12
0.12 g AgBr contains g bromine A known mass of an organic compound is
188 heated with fuming nitric acid whereupon
phosphorus present in the compound is
80  0.12  100
Percentage of bromine= oxidised to phosphoric acid. It is precipitated
188  0.15 as ammonium phosphomolybdate, (NH4) 3
= 34.04% PO 4 .12MoO 3 , by adding ammonia and
ammonium molybdate. Alter natively,
12.10.4 Sulphur phosphoric acid may be precipitated as
A known mass of an organic compound is MgNH 4 PO 4 by adding magnesia mixture
heated in a Carius tube with sodium peroxide which on ignition yields Mg2P2O7.
or fuming nitric acid. Sulphur present in the Let the mass of organic compound taken
compound is oxidised to sulphuric acid. It is = m g and mass of ammonium phospho
precipitated as barium sulphate by adding molydate = m1g
excess of barium chloride solution in water.
Molar mass of (NH4)3PO4.12MoO3 = 1877 g
The precipitate is filtered, washed, dried and
weighed. The percentage of sulphur can be 31  m1  100
calculated from the mass of barium sulphate. Percentage of phosphorus = %
1877  m
Let the mass of organic
If phosphorus is estimated as Mg2P2O7,
compound taken = m g
and the mass of barium 62  m1  100
sulphate formed = m1g Percentage of phosphorus = %
222  m
 ∴

360 CHEMISTRY

where, 222 u is the molar mass of Mg2P2O7, equation (B) by multiplying the equations (A) and
m, the mass of organic compound taken, m1, (B) by 5 and 2 respectively; we find that each
the mass of Mg2P2O7 formed and 62, the mass mole of oxygen liberated from the compound will
of two phosphorus atoms present in the produce two moles of carbondioxide.
compound Mg2P2O7. Thus 88 g carbon dioxide is obtained if 32 g
12.10.6 Oxygen oxygen is liberated.
The percentage of oxygen in an organic compound Let the mass of organic compound taken be m g
is usually found by difference between the total Mass of carbon dioxide produced be m1 g
percentage composition (100) and the sum of the ∴ m1 g carbon dioxide is obtained from
percentages of all other elements. However, oxygen 32 m1
can also be estimated directly as follows: g O2
88
A definite mass of an organic compound is
decomposed by heating in a stream of nitrogen 32 m1 100
gas. The mixture of gaseous products containing ∴Percentage of oxygen = 88 m
%
oxygen is passed over red-hot coke when all the
The percentage of oxygen can be derived
oxygen is converted to carbon monoxide. This
from the amount of iodine produced also.
mixture is passed through warm iodine
Presently, the estimation of elements in an
pentoxide (I2O5) when carbon monoxide is
organic compound is carried out by using
oxidised to carbon dioxide producing iodine.
microquantities of substances and automatic
Compound ⎯⎯⎯→ O2 + other gaseous
heat
experimental techniques. The elements, carbon,
products hydrogen and nitrogen present in a compound
are determined by an apparatus known as CHN
2C + O2 ⎯⎯⎯⎯ → 2CO]× 5
1373 K
(A) elemental analyser. The analyser requires only
I2O5 + 5CO ⎯→ I2 + 5CO2]× 2 (B) a very small amount of the substance (1-3 mg)
and displays the values on a screen within a
On making the amount of CO produced in short time. A detailed discussion of such
equation (A) equal to the amount of CO used in methods is beyond the scope of this book.

SUMMARY

In this unit, we have learnt some basic concepts in structure and reactivity of organic
compounds, which are formed due to covalent bonding. The nature of the covalent bonding
in organic compounds can be described in terms of orbitals hybridisation concept, according
to which carbon can have sp3, sp2 and sp hybridised orbitals. The sp3, sp2 and sp hybridised
carbons are found in compounds like methane, ethene and ethyne respectively. The
tetrahedral shape of methane, planar shape of ethene and linear shape of ethyne can be
understood on the basis of this concept. A sp3 hybrid orbital can overlap with 1s orbital of
hydrogen to give a carbon - hydrogen (C–H) single bond (sigma, σ bond). Overlap of a sp2
orbital of one carbon with sp2 orbital of another results in the formation of a carbon–carbon
σ bond. The unhybridised p orbitals on two adjacent carbons can undergo lateral (side-by-
side) overlap to give a pi (π) bond. Organic compounds can be represented by various structural
formulas. The three dimensional representation of organic compounds on paper can be
drawn by wedge and dash formula.
Organic compounds can be classified on the basis of their structure or the functional
groups they contain. A functional group is an atom or group of atoms bonded together in a
unique fashion and which determines the physical and chemical properties of the compounds.
The naming of the organic compounds is carried out by following a set of rules laid down by
the International Union of Pure and Applied Chemistry (IUPAC). In IUPAC nomenclature,
the names are correlated with the structure in such a way that the reader can deduce the
structure from the name.
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 361

Organic reaction mechanism concepts are based on the structure of the substrate
molecule, fission of a covalent bond, the attacking reagents, the electron displacement effects
and the conditions of the reaction. These organic reactions involve breaking and making of
covalent bonds. A covalent bond may be cleaved in heterolytic or homolytic fashion. A
heterolytic cleavage yields carbocations or carbanions, while a homolytic cleavage gives
free radicals as reactive intermediate. Reactions proceeding through heterolytic cleavage
involve the complimentary pairs of reactive species. These are electron pair donor known as
nucleophile and an electron pair acceptor known as electrophile. The inductive, resonance,
electromeric and hyperconjugation effects may help in the polarisation of a bond making
certain carbon atom or other atom positions as places of low or high electron densities.
Organic reactions can be broadly classified into following types; substitution, addition,
elimination and rearrangement reactions.
Purification, qualitative and quantitative analysis of organic compounds are carried out
for determining their structures. The methods of purification namely : sublimation, distillation
and differential extraction are based on the difference in one or more physical properties.
Chromatography is a useful technique of separation, identification and purification of
compounds. It is classified into two categories : adsorption and partition chromatography.
Adsorption chromatography is based on differential adsorption of various components of a
mixture on an adsorbent. Partition chromatography involves continuous partitioning of the
components of a mixture between stationary and mobile phases. After getting the compound
in a pure form, its qualitative analysis is carried out for detection of elements present in it.
Nitrogen, sulphur, halogens and phosphorus are detected by Lassaigne’s test. Carbon and
hydrogen are estimated by determining the amounts of carbon dioxide and water produced.
Nitrogen is estimated by Dumas or Kjeldahl’s method and halogens by Carius method.
Sulphur and phosphorus are estimated by oxidising them to sulphuric and phosphoric
acids respectively. The percentage of oxygen is usually determined by difference between
the total percentage (100) and the sum of percentages of all other elements present.

EXERCISES

12.1 What are hybridisation states of each carbon atom in the following compounds ?
CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6
12.2 Indicate the σ and π bonds in the following molecules :
C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3
12.3 Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-
one.
12.4 Give the IUPAC names of the following compounds :

(a) (b) (c)

(d) (e) (f) Cl2CHCH2OH

12.5 Which of the following represents the correct IUPAC name for the compounds
concer ned ? (a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7-
Trimethyloctane or 2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or
4-Chloro-2-methylpentane (d) But-3-yn-1-ol or But-4-ol-1-yne.
362 CHEMISTRY

12.6 Draw formulas for the first five members of each homologous series beginning
with the following compounds. (a) H–COOH (b) CH3COCH3 (c) H–CH=CH2
12.7 Give condensed and bond line structural formulas and identify the functional
group(s) present, if any, for :
(a) 2,2,4-Trimethylpentane
(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid
(c) Hexanedial
12.8 Identify the functional groups in the following compounds

(a) (b) (c)

12.9 Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and
why ?
12.10 Explain why alkyl groups act as electron donors when attached to a π system.
12.11 Draw the resonance structures for the following compounds. Show the electron
shift using curved-arrow notation.

(a) C 6 H 5 OH (b) C 6 H 5NO 2 (c) CH 3 CH=CHCHO (d) C 6 H 5 –CHO (e) C6H5 CH2

(f) CH3CH  CHC H2

12.12 What are electrophiles and nucleophiles ? Explain with examples.

12.13 Identify the reagents shown in bold in the following equations as nucleophiles or
electrophiles:

(a) CH3COOH  HO  CH3COO  H2O
–

–
(b) CH3COCH3  C N   CH3 2 C  CN  OH 
+
(c) C6H6  CH3 C O  C6H5COCH3
12.14 Classify the following reactions in one of the reaction type studied in this unit.

(a) CH 3CH 2 Br  HS   CH 3CH 2 SH  Br 

(b)  CH3  C  CH2  HCl   CH3  ClC  CH3

2 2

(c) CH 3CH 2 Br  HO   CH 2  CH 2  H 2O  Br 

(d) CH3 C CH2OH HBr CH3 2 CBrCH2CH2CH3 H2O

3
12.15 What is the relationship between the members of following pairs of structures ?
Are they structural or geometrical isomers or resonance contributors ?

(a)
ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 363

(b)

(c)

12.16 For the following bond cleavages, use curved-arrows to show the electron flow
and classify each as homolysis or heterolysis. Identify reactive intermediate
produced as free radical, carbocation and carbanion.
(a)

(b)

(c)

(d)

12.17 Explain the terms Inductive and Electromeric effects. Which electron displacement
effect explains the following correct orders of acidity of the carboxylic acids?
(a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH
(b) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3C.COOH
12.18 Give a brief description of the principles of the following techniques taking an
example in each case.
(a) Crystallisation (b) Distillation (c) Chromatography
12.19 Describe the method, which can be used to separate two compounds with different
solubilities in a solvent S.
12.20 What is the difference between distillation, distillation under reduced pressure
and steam distillation ?
12.21 Discuss the chemistry of Lassaigne’s test.
12.22 Differentiate between the principle of estimation of nitrogen in an organic compound
by (i) Dumas method and (ii) Kjeldahl’s method.
12.23 Discuss the principle of estimation of halogens, sulphur and phosphorus present
in an organic compound.
12.24 Explain the principle of paper chromatography.
12.25 Why is nitric acid added to sodium extract before adding silver nitrate for testing
halogens?
12.26 Explain the reason for the fusion of an organic compound with metallic sodium
for testing nitrogen, sulphur and halogens.
12.27 Name a suitable technique of separation of the components from a mixture of
calcium sulphate and camphor.
12.28 Explain, why an organic liquid vaporises at a temperature below its boiling point
in its steam distillation ?
12.29 Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give
reason for your answer.
364 CHEMISTRY

12.30 Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved
during the estimation of carbon present in an organic compound?
12.31 Why is it necessary to use acetic acid and not sulphuric acid for acidification of
sodium extract for testing sulphur by lead acetate test?
12.32 An organic compound contains 69% carbon and 4.8% hydrogen, the remainder
being oxygen. Calculate the masses of carbon dioxide and water produced when
0.20 g of this substance is subjected to complete combustion.
12.33 A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s
method. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual
acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the
percentage composition of nitrogen in the compound.
12.34 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius
estimation. Calculate the percentage of chlorine present in the compound.
12.35 In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur
compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur
in the given compound.
12.36 In the organic compound CH2 = CH – CH2 – CH2 – C ≡ CH, the pair of hydridised
orbitals involved in the formation of: C2 – C3 bond is:
(a) sp – sp2 (b) sp – sp3 (c) sp2 – sp3 (d) sp3 – sp3
12.37 In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue
colour is obtained due to the formation of:
(a) Na4[Fe(CN)6] (b) Fe4[Fe(CN)6]3 (c) Fe2[Fe(CN)6] (d) Fe3[Fe(CN)6]4
12.38 Which of the following carbocation is most stable ?
+ + + +
(a) (CH3)3C. C H2 (b) (CH3)3 C (c) CH3CH2 C H2 (d) CH3 C H CH2CH3
12.39 The best and latest technique for isolation, purification and separation of organic
compounds is:
(a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography
12.40 The reaction:
CH3CH2I + KOH(aq) → CH3CH2OH + KI
is classified as :
(a) electrophilic substitution (b) nucleophilic substitution
(c) elimination (d) addition
 HYDROCARBONS 365

UNIT 13

HYDROCARBONS

d
he
Hydrocarbons are the important sources of energy.

is
After studying this unit, you will be
able to

bl
• name hydrocarbons according to The term ‘hydrocarbon’ is self-explanatory which means
IUPAC system of nomenclature; compounds of carbon and hydrogen only. Hydrocarbons
• recognise and write structures
pu play a key role in our daily life. You must be familiar with
of isomers of alkanes, alkenes, the terms ‘LPG’ and ‘CNG’ used as fuels. LPG is the
alkynes and aromatic
abbreviated form of liquified petroleum gas whereas CNG
hydrocarbons;
stands for compressed natural gas. Another term ‘LNG’
be T

• learn about various methods of

preparation of hydrocarbons; (liquified natural gas) is also in news these days. This is
also a fuel and is obtained by liquifaction of natural gas.
re
• distinguish between alkanes,
o R

alkenes, alkynes and aromatic Petrol, diesel and kerosene oil are obtained by the fractional
hydrocarbons on the basis of distillation of petroleum found under the earth’s crust.
physical and chemical properties; Coal gas is obtained by the destructive distillation of coal.
tt E

• draw and differentiate between Natural gas is found in upper strata during drilling of oil
various conformations of ethane; wells. The gas after compression is known as compressed
• appreciate the role of natural gas. LPG is used as a domestic fuel with the least
C

hydrocarbons as sources of pollution. Kerosene oil is also used as a domestic fuel but
energy and for other industrial
it causes some pollution. Automobiles need fuels like petrol,
applications;
diesel and CNG. Petrol and CNG operated automobiles
no N

• predict the formation of the

addition products of cause less pollution. All these fuels contain mixture of
unsymmetrical alkenes and hydrocarbons, which are sources of energy. Hydrocarbons
alkynes on the basis of electronic are also used for the manufacture of polymers like
mechanism; polythene, polypropene, polystyrene etc. Higher
©

• comprehend the structure of hydrocarbons are used as solvents for paints. They are also
benzene, explain aromaticity used as the starting materials for manufacture of many
and understand mechanism dyes and drugs. Thus, you can well understand the
of electrophilic substitution importance of hydrocarbons in your daily life. In this unit,
reactions of benzene;
you will learn more about hydrocarbons.
• predict the directive influence of
substituents in monosubstituted 13.1 CLASSIFICATION
benzene ring;
• learn about carcinogenicity and Hydrocarbons are of different types. Depending upon the
toxicity. types of carbon-carbon bonds present, they can be
classified into three main categories – (i) saturated
 366 CHEMISTRY

(ii) unsaturated and (iii) aromatic general formula for alkane family or
hydrocarbons. Saturated hydrocarbons homologous series? The general formula for
contain carbon-carbon and carbon-hydrogen alkanes is CnH2n+2, where n stands for number
single bonds. If different carbon atoms are of carbon atoms and 2n+2 for number of
joined together to form open chain of carbon hydrogen atoms in the molecule. Can you
atoms with single bonds, they are termed as recall the structure of methane? According to
alkanes as you have already studied in VSEPR theory (Unit 4), methane has a
Unit 12. On the other hand, if carbon atoms tetrahedral structure (Fig. 13.1) which is
form a closed chain or a ring, they are termed multiplanar, in which carbon atom lies at the

d
as cycloalkanes. Unsaturated hydrocarbons centre and the four hydrogen atoms lie at the
contain carbon-carbon multiple bonds – four corners of a regular tetrahedron. All

he
double bonds, triple bonds or both. Aromatic H-C-H bond angles are of 109.5°.
hydrocarbons are a special type of cyclic
compounds. You can construct a large number
of models of such molecules of both types
(open chain and close chain) keeping in mind

is
that carbon is tetravalent and hydrogen is
monovalent. For making models of alkanes,
you can use toothpicks for bonds and

bl
plasticine balls for atoms. For alkenes, alkynes
and aromatic hydrocarbons, spring models can
be constructed.
pu Fig. 13.1 Structure of methane
In alkanes, tetrahedra are joined together
13.2 ALKANES in which C-C and C-H bond lengths are
As already mentioned, alkanes are saturated 154 pm and 112 pm respectively (Unit 12). You
be T

open chain hydrocarbons containing have already read that C–C and C–H σ bonds
re
3
carbon - carbon single bonds. Methane (CH4) are formed by head-on overlapping of sp
o R

is the first member of this family. Methane is a hybrid orbitals of carbon and 1s orbitals of
gas found in coal mines and marshy places. If hydrogen atoms.
you replace one hydrogen atom of methane by
tt E

13.2.1 Nomenclature and Isomerism

carbon and join the required number of
hydrogens to satisfy the tetravalence of the You have already read about nomenclature
other carbon atom, what do you get? You get of different classes of organic compounds in
C

C 2 H 6 . This hydrocarbon with molecular Unit 12. Nomenclature and isomerism in

formula C2H6 is known as ethane. Thus you alkanes can further be understood with the
can consider C2H6 as derived from CH4 by help of a few more examples. Common names
no N

replacing one hydrogen atom by -CH3 group. are given in parenthesis. First three alkanes
Go on constructing alkanes by doing this – methane, ethane and propane have only
theoretical exercise i.e., replacing hydrogen one structure but higher alkanes can have
atom by –CH3 group. The next molecules will more than one structure. Let us write
©

be C3H8, C4H10 … structures for C4H10. Four carbon atoms of

C4H10 can be joined either in a continuous
chain or with a branched chain in the
following two ways :
I
These hydrocarbons are inert under
normal conditions as they do not react with
acids, bases and other reagents. Hence, they
were earlier known as paraffins (latin : parum,
little; affinis, affinity). Can you think of the Butane (n- butane), (b.p. 273 K)
 HYDROCARBONS 367

II structures, they are known as structural

isomers. It is also clear that structures I and
III have continuous chain of carbon atoms but
structures II, IV and V have a branched chain.
Such structural isomers which differ in chain
of carbon atoms are known as chain isomers.
2-Methylpropane (isobutane) Thus, you have seen that C4H10 and C5H12
(b.p.261 K) have two and three chain isomers respectively.

d
In how many ways, you can join five Problem 13.1
carbon atoms and twelve hydrogen atoms of Write structures of different chain isomers

he
C5H12? They can be arranged in three ways as of alkanes corresponding to the molecular
shown in structures III–V formula C6H14. Also write their IUPAC
III names.

Solution

is
(i) CH3 – CH2 – CH2 – CH2– CH2– CH3
n-Hexane
Pentane (n-pentane)

bl
(b.p. 309 K)

IV
pu 2-Methylpentane
be T

3-Methylpentane
re
o R

2-Methylbutane (isopentane)
(b.p. 301 K)
2,3-Dimethylbutane
tt E

V
C

2,2 - Dimethylbutane
no N

Based upon the number of carbon atoms

attached to a carbon atom, the carbon atom is
2,2-Dimethylpropane (neopentane)
termed as primary (1°), secondary (2°), tertiary
©

(b.p. 282.5 K)
(3°) or quaternary (4°). Carbon atom attached
Structures I and II possess same to no other carbon atom as in methane or to
molecular formula but differ in their boiling only one carbon atom as in ethane is called
points and other properties. Similarly primary carbon atom. Terminal carbon atoms
structures III, IV and V possess the same are always primary. Carbon atom attached to
molecular formula but have different two carbon atoms is known as secondary.
properties. Structures I and II are isomers of Tertiary carbon is attached to three carbon
butane, whereas structures III, IV and V are atoms and neo or quaternary carbon is
isomers of pentane. Since difference in attached to four carbon atoms. Can you identify
properties is due to difference in their 1°, 2°, 3° and 4° carbon atoms in structures I
 368 CHEMISTRY

to V ? If you go on constructing structures for compounds. These groups or substituents are

higher alkanes, you will be getting still larger known as alkyl groups as they are derived from
number of isomers. C6H14 has got five isomers alkanes by removal of one hydrogen atom.
and C7H16 has nine. As many as 75 isomers General formula for alkyl groups is CnH2n+1
are possible for C10H22. (Unit 12).
In structures II, IV and V, you observed Let us recall the general rules for
that –CH3 group is attached to carbon atom nomenclature already discussed in Unit 12.
numbered as 2. You will come across groups Nomenclature of substituted alkanes can
like –CH3, –C2H5, –C3H7 etc. attached to carbon further be understood by considering the

d
atoms in alkanes or other classes of following problem:

he
Problem 13.2
Write structures of different isomeric alkyl groups corresponding to the molecular formula
C5H11. Write IUPAC names of alcohols obtained by attachment of –OH groups at different
carbons of the chain.

is
Solution
Structures of – C5H11 group Corresponding alcohols Name of alcohol

bl
(i) CH3 – CH2 – CH2 – CH2– CH2 – CH3 – CH2 – CH2 – CH2– CH2 – OH Pentan-1-ol

(ii) CH3 – CH – CH2 – CH2 – CH3 CH3 – CH – CH2 – CH2– CH3 Pentan-2-ol
pu | |
OH
be T

(iii) CH3 – CH2 – CH – CH2 – CH3 CH3 – CH2 – CH – CH2– CH3 Pentan-3-ol
| |
re
OH
o R

CH3 CH3 3-Methyl-

| | butan-1-ol
tt E

(iv) CH3 – CH – CH2 – CH2 – CH3 – CH – CH2 – CH2– OH

CH3 CH3 2-Methyl-

C

| | butan-1-ol
(v) CH3 – CH2 – CH – CH2 – CH3 – CH2 – CH – CH2– OH
no N

CH3 CH3 2-Methyl-

| | butan-2-ol
(vi) CH3 – C – CH2 – CH3 CH3 – C – CH2 – CH3
| |
©

OH
CH3 CH3 2,2- Dimethyl-
| | propan-1-ol
(vii) CH3 – C – CH2 – CH3 – C – CH2OH
| |
CH3 CH3

CH3 CH3 OH 3-Methyl-

| | | | butan-2-ol
(viii) CH3 – CH – CH –CH3 CH3 – CH – CH –CH3
 HYDROCARBONS 369

Table 13.1 Nomenclature of a Few Organic Compounds

Structure and IUPAC Name Remarks

Lowest sum and

1 2 3 4 5 6
(a) CH3– CH – CH2 – CH – CH2 – CH3 alphabetical
(4 – Ethyl – 2 – methylhexane) arrangement

d
Lowest sum and
(b) CH3 – 7CH2 – 6CH2 – 5CH – 4CH –
8 3
C – 2CH2 – 1CH3 alphabetical

he
arrangement

(3,3-Diethyl-5-isopropyl-4-methyloctane)

is
sec is not considered
(c) CH3–2CH2–3CH2–4CH–5CH–6CH2–7CH2–8CH2–9CH2–10CH3
1
while arranging

bl
alphabetically;
isopropyl is taken
5-sec– Butyl-4-isopropyldecane
pu as one word
(d) CH3–2CH2–3CH2–4CH2–5CH–6CH2–7CH2–8CH2–9CH3
1
Further numbering
to the substituents
be T

of the side chain

re
o R

5-(2,2– Dimethylpropyl)nonane
(e) 1
CH3 – 2CH2 – 3CH – 4CH2 – 5CH – 6CH2 – 7CH3
Alphabetical
tt E

priority order
3–Ethyl–5–methylheptane
C

Problem 13.3 important to write the correct structure from

the given IUPAC name. To do this, first of all,
no N

Write IUPAC names of the following

the longest chain of carbon atoms
compounds :
corresponding to the parent alkane is written.
(i) (CH3)3 C CH2C(CH3)3 Then after numbering it, the substituents are
(ii) (CH3)2 C(C2H5)2 attached to the correct carbon atoms and finally
©

(iii) tetra – tert-butylmethane valence of each carbon atom is satisfied by

putting the correct number of hydrogen atoms.
Solution This can be clarified by writing the structure
(i) 2, 2, 4, 4-Tetramethylpentane of 3-ethyl-2, 2–dimethylpentane in the
(ii) 3, 3-Dimethylpentane following steps :
(iii) 3,3-Di-tert-butyl -2, 2, 4, 4 - i) Draw the chain of five carbon atoms:
tetramethylpentane C–C–C–C–C
If it is important to write the correct IUPAC ii) Give number to carbon atoms:
1 2 3 4 5
name for a given structure, it is equally C –C –C –C –C
 370 CHEMISTRY

iii) Attach ethyl group at carbon 3 and two Longest chain is of six carbon atoms and
methyl groups at carbon 2 not that of five. Hence, correct name is
CH3 3-Methylhexane.
| 7 6 5 4 3 2 1
1 2 3 4 5
C – C– C– C– C (ii) CH3 – CH2 – CH – CH2 – CH – CH2 – CH3
| |
CH3 C2 H5
iv) Satisfy the valence of each carbon atom by
putting requisite number of hydrogen Numbering is to be started from the end

d
atoms : which gives lower number to ethyl group.
Hence, correct name is 3-ethyl-5-
CH3

he
| methylheptane.
CH3 – C – CH – CH2 – CH3
| | 13.2.2 Preparation
CH3 C2H5
Petroleum and natural gas are the main
Thus we arrive at the correct structure. If

is
sources of alkanes. However, alkanes can be
you have understood writing of structure from prepared by following methods :
the given name, attempt the following
1. From unsaturated hydrocarbons

bl
problems.
Dihydrogen gas adds to alkenes and alkynes
Problem 13.4 in the presence of finely divided catalysts like
Write structural formulas of the following
pu platinum, palladium or nickel to form alkanes.
compounds : This process is called hydrogenation. These
metals adsorb dihydrogen gas on their surfaces
(i) 3, 4, 4, 5–Tetramethylheptane
and activate the hydrogen – hydrogen bond.
be T

(ii) 2,5-Dimethyhexane Platinum and palladium catalyse the reaction

re
at room temperature but relatively higher
Solution
o R

temperature and pressure are required with

nickel catalysts.
(i) CH3 – CH2 – CH – C – CH– CH – CH3 CH2 = CH2 + H2 ⎯⎯⎯⎯→
Pt/Pd/Ni
CH3 − CH3
tt E

(13.1)
Ethene Ethane
C

CH3 − CH = CH2 + H2 ⎯⎯⎯⎯ →CH3 − CH2 − CH3

Pt/Pd/Ni

Propene Propane
(ii) CH3 – CH – CH2 – CH2 – CH – CH3
no N

(13.2)
Problem 13.5
CH3 − C ≡ C − H + 2H2 ⎯⎯⎯⎯ → CH3 − CH2 − CH3
Pt/Pd/Ni
Write structures for each of the following
Propyne Propane
©

compounds. Why are the given names

incorrect? Write correct IUPAC (13.3)
names.
2. From alkyl halides
(i) 2-Ethylpentane
i) Alkyl halides (except fluorides) on
(ii) 5-Ethyl – 3-methylheptane reduction with zinc and dilute hydrochloric
Solution acid give alkanes.
(i) CH3 – CH – CH2– CH2 – CH3 +
CH 3 − Cl + H 2 ⎯⎯ ⎯
Zn, H
→ CH 4 + HCl (13.4)
Chloromethane Methane
 HYDROCARBONS 371

+
C 2 H 5 − Cl + H 2 ⎯⎯ ⎯
Zn, H
→ C 2 H 6 + HCl containing even number of carbon atoms
Chloroethane Ethane (13.5) at the anode.
Zn,H+
2CH3 COO− Na+ + 2H2 O
CH3 CH2 CH2 Cl + H2 ⎯⎯→ CH3 CH2 CH3 + HCl Sodium acetate
1-Chloropropane Propane
(13.6) ↓ Electrolysis
CH3 −CH3 + 2CO2 + H2 + 2NaOH (13.9)
ii) Alkyl halides on treatment with sodium
metal in dry ethereal (free from moisture) The reaction is supposed to follow the
following path :

d
solution give higher alkanes. This reaction
is known as Wurtz reaction and is used O

he
for the preparation of higher alkanes ||
−
containing even number of carbon i) 2CH3 COO Na+ 2CH3 −C − O− + 2Na+
atoms.
dry ether ii) At anode:
CH3 Br +2Na + BrCH3 ⎯⎯⎯⎯ →CH3 −CH3 +2NaBr
O O

is
Bromomethane Ethane || ||
• •
(13.7) –2e
2CH3 −C −O ⎯⎯
– −
→2CH3 −C −O: ⎯⎯
→2CH3 + 2CO2 ↑
••

bl
dry ether
C2 H5 Br + 2Na + BrC2 H5 ⎯⎯⎯⎯ →C2 H5 −C2 H5
Acetate ion Acetate Methyl free
Bromoethane n-Butane free radical radical
pu (13.8) • •
iii) H3 C + CH3 ⎯⎯
→ H3 C −CH3 ↑
What will happen if two different alkyl halides
are taken? iv) At cathode :
•
H2 O + e– → – OH + H
be T

3. From carboxylic acids

•
i) Sodium salts of carboxylic acids on heating 2H ⎯
→ H2 ↑
re
with soda lime (mixture of sodium
o R

Methane cannot be prepared by this

hydroxide and calcium oxide) give alkanes
method. Why?
containing one carbon atom less than the
carboxylic acid. This process of elimination 13.2.3 Properties
tt E

of carbon dioxide from a carboxylic acid is Physical properties

known as decarboxylation. Alkanes are almost non-polar molecules
C

because of the covalent nature of C-C and C-H

CH3 COO Na+ + NaOH ⎯
–
⎯→CH4 + Na2 CO3
CaO
Δ bonds and due to very little difference of
Sodium ethanoate electronegativity between carbon and
no N

hydrogen atoms. They possess weak van der

Problem 13.6 Waals forces. Due to the weak forces, the first
Sodium salt of which acid will be needed four members, C1 to C4 are gases, C5 to C17 are
for the preparation of propane ? Write liquids and those containing 18 carbon atoms
©

chemical equation for the reaction. or more are solids at 298 K. They are colourless
and odourless. What do you think about
Solution solubility of alkanes in water based upon non-
Butanoic acid, polar nature of alkanes? Petrol is a mixture of
−
CH3 CH2 CH2 COO Na + + NaOH ⎯⎯
CaO
→ hydrocarbons and is used as a fuel for
automobiles. Petrol and lower fractions of
CH3 CH2 CH3 + Na2 CO3
petroleum are also used for dry cleaning of
clothes to remove grease stains. On the basis
ii) Kolbe’s electrolytic method An aqueous of this observation, what do you think about
solution of sodium or potassium salt of a the nature of the greasy substance? You are
carboxylic acid on electrolysis gives alkane correct if you say that grease (mixture of higher
 372 CHEMISTRY

alkanes) is non-polar and, hence, hydrophobic reducing agents. However, they undergo the
in nature. It is generally observed that in following reactions under certain
relation to solubility of substances in solvents, conditions.
polar substances are soluble in polar solvents,
1. Substitution reactions
whereas the non-polar ones in non-polar
solvents i.e., like dissolves like. One or more hydrogen atoms of alkanes can
be replaced by halogens, nitro group and
Boiling point (b.p.) of different alkanes are
sulphonic acid group. Halogenation takes
given in Table 13.2 from which it is clear that
there is a steady increase in boiling point with place either at higher temperature

d
increase in molecular mass. This is due to the (573-773 K) or in the presence of diffused
fact that the intermolecular van der Waals sunlight or ultraviolet light. Lower alkanes do

he
forces increase with increase of the molecular not undergo nitration and sulphonation
size or the surface area of the molecule. reactions. These reactions in which hydrogen
atoms of alkanes are substituted are known
You can make an interesting observation
as substitution reactions. As an example,
by having a look on the boiling points of
chlorination of methane is given below:

is
three isomeric pentanes viz., (pentane,
2-methylbutane and 2,2-dimethylpropane). It Halogenation
is observed (Table 13.2) that pentane having a hν
CH4 + Cl2 ⎯⎯ → +

bl
CH3 Cl HCl
continuous chain of five carbon atoms has the
highest boiling point (309.1K) whereas Chloromethane (13.10)
2,2 – dimethylpropane boils at 282.5K. With
pu hν
increase in number of branched chains, the CH 3 Cl + Cl2 ⎯ ⎯⎯
→ CH 2 Cl2 + HCl
molecule attains the shape of a sphere. This Dichloromethane (13.11)
results in smaller area of contact and therefore
be T

weak intermolecular forces between spherical hν

CH2 Cl2 + Cl2 ⎯⎯⎯
→ CHCl3 + HCl
molecules, which are overcome at relatively
re
Trichloromethane (13.12)
o R

lower temperatures.
Chemical properties hν
CHCl3 + Cl2 ⎯⎯⎯
→ CCl4 + HCl
As already mentioned, alkanes are generally Tetrachloromethane (13.13)
tt E

inert towards acids, bases, oxidising and

Table 13.2 Variation of Melting Point and Boiling Point in Alkanes
C

Molecular Name Molecular b.p./(K) m.p./(K)

formula mass/u
no N

CH4 Methane 16 111.0 90.5

C2H6 Ethane 30 184.4 101.0
C3H8 Propane 44 230.9 85.3
C4H10 Butane 58 272.4 134.6
©

C4H10 2-Methylpropane 58 261.0 114.7

C5H12 Pentane 72 309.1 143.3
C5H12 2-Methylbutane 72 300.9 113.1
C5H12 2,2-Dimethylpropane 72 282.5 256.4
C6H14 Hexane 86 341.9 178.5
C7H16 Heptane 100 371.4 182.4
C8H18 Octane 114 398.7 216.2
C9H20 Nonane 128 423.8 222.0
C10H22 Decane 142 447.1 243.3
C20H42 Eicosane 282 615.0 236.2
 HYDROCARBONS 373

hν and may occur. Two such steps given below

CH3 -CH3 + Cl2 ⎯⎯⎯
→ CH3 − CH2 Cl + HCl
explain how more highly haloginated products
Chloroethane (13.14) are formed.
• •
It is found that the rate of reaction of alkanes CH3 Cl + Cl → CH2 Cl + HCl
with halogens is F2 > Cl2 > Br2 > I2. Rate of • •
replacement of hydrogens of alkanes is : CH2 Cl + Cl − Cl → CH2 Cl 2 + Cl
3° > 2° > 1°. Fluorination is too violent to be
controlled. Iodination is very slow and a (iii)Termination: The reaction stops after
some time due to consumption of reactants
reversible reaction. It can be carried out in the

d
and / or due to the following side reactions :
presence of oxidizing agents like HIO3 or HNO3.
The possible chain terminating steps are :

he
CH4 + I2 CH3 I + HI (13.15) • •
(a) Cl + Cl → Cl − Cl
HIO3 +5HI → 3I2 + 3H2 O (13.16) • •
(b) H3 C + CH3 → H3 C − CH3
Halogenation is supposed to proceed via
• •
free radical chain mechanism involving three (c) H3 C + Cl → H3 C − Cl

is
steps namely initiation, propagation and
termination as given below: Though in (c), CH3 – Cl, the one of the
products is formed but free radicals are

bl
Mechanism consumed and the chain is terminated. The
(i) Initiation : The reaction is initiated by above mechanism helps us to understand the
homolysis of chlorine molecule in the presence
pu reason for the formation of ethane as a
of light or heat. The Cl–Cl bond is weaker than byproduct during chlorination of methane.
the C–C and C–H bond and hence, is easiest to 2. Combustion
break.
be T

• •
Alkanes on heating in the presence of air or
hν dioxygen are completely oxidized to carbon
Cl −Cl ⎯⎯⎯⎯ → Cl + Cl
re
homolysis
dioxide and water with the evolution of large
o R

Chlorine free radicals

amount of heat.
(ii) Propagation : Chlorine free radical attacks
the methane molecule and takes the reaction CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(l);
tt E

in the forward direction by breaking the C-H Δc H V = − 890 kJ mol−1

bond to generate methyl free radical with the (13.17)
formation of H-Cl. C4 H10 (g)+13/2 O2 (g) → 4CO2 (g)+5H2O(l);
C

•
hν
•
Δc H V =−2875.84 kJ mol−1
( a) CH4 + Cl ⎯⎯⎯ →CH + H − Cl
3 (13.18)
no N

The methyl radical thus obtained attacks The general combustion equation for any
the second molecule of chlorine to form alkane is :
CH3 – Cl with the liberation of another chlorine
⎛ 3n +1⎞
free radical by homolysis of chlorine molecule. Cn H2n+2 + ⎜ ⎟O2 → nCO2 + (n +1) H2 O
©

⎝ 2 ⎠
• •
hν
(b) C H3 + Cl − Cl ⎯⎯⎯ → CH3 − Cl + C l (13.19)
Chlorine Due to the evolution of large amount of
free radical heat during combustion, alkanes are used
as fuels.
The chlorine and methyl free radicals
generated above repeat steps (a) and (b) During incomplete combustion of
respectively and thereby setup a chain of alkanes with insufficient amount of air or
reactions. The propagation steps (a) and (b) are dioxygen, carbon black is formed which is
those which directly give principal products, used in the manufacture of ink, printer ink,
but many other propagation steps are possible black pigments and as filters.
 374 CHEMISTRY

Incomplete
CH4 (g) + O2 (g) ⎯⎯⎯⎯⎯ pressure in the presence of oxides of
combustion→C(s) + 2H2 O(l)
vanadium, molybdenum or chromium
(13.20) supported over alumina get dehydrogenated
and cyclised to benzene and its homologues.
3. Controlled oxidation
This reaction is known as aromatization or
Alkanes on heating with a regulated supply of
reforming.
dioxygen or air at high pressure and in the
presence of suitable catalysts give a variety of
oxidation products.

d
Cu/523K/100atm
(i) 2CH4 + O2 ⎯⎯⎯⎯⎯⎯⎯→ 2CH3 OH
Methanol

he
(13.21)
Mo2O3
(ii) CH4 + O2 ⎯⎯⎯→ HCHO + H2 O (13.26)
Δ
Methanal Toluene (C7H8) is methyl derivative of

is
(13.22) benzene. Which alkane do you suggest for
(CH3COO)2 Mn
(iii)2CH3 CH3 + 3O2 ⎯⎯⎯⎯⎯⎯→ 2 CH3 COOH preparation of toluene ?
Δ
Ethanoic acid 6. Reaction with steam

bl
+ 2H2 O Methane reacts with steam at 1273 K in the
(13.23) presence of nickel catalyst to form carbon
pu monoxide and dihydrogen. This method is
(iv) Ordinarily alkanes resist oxidation but used for industrial preparation of dihydrogen
alkanes having tertiary H atom can be gas
oxidized to corresponding alcohols by
be T

Ni
potassium permanganate. CH4 + H2 O ⎯⎯ → CO + 3H2 (13.27)
Δ
re
KMnO
(CH3 )3 CH ⎯⎯⎯⎯
Oxidation→
4
(CH3 )3 COH 7. Pyrolysis
o R

2-Methylpropane 2-Methylpropan-2-ol Higher alkanes on heating to higher

(13.24) temperature decompose into lower alkanes,
alkenes etc. Such a decomposition reaction
tt E

4. Isomerisation into smaller fragments by the application of

n-Alkanes on heating in the presence of heat is called pyrolysis or cracking.
anhydrous aluminium chloride and hydrogen
C

chloride gas isomerise to branched chain

alkanes. Major products are given below. Some
minor products are also possible which you
no N

can think over. Minor products are generally

not reported in organic reactions. (13.28)
Anhy. AlCl3 /HCl Pyrolysis of alkanes is believed to be a
CH3 (CH2 )4 CH3 ⎯⎯⎯⎯⎯⎯→ free radical reaction. Preparation of oil gas or
©

n -Hexane petrol gas from kerosene oil or petrol involves

CH3 CH −(CH2 )2 −CH3 + CH3 CH2 −CH −CH2 −CH3 the principle of pyrolysis. For example,
| | dodecane, a constituent of kerosene oil on
CH3 CH3 heating to 973K in the presence of platinum,
2-Methylpentane 3-Methylpentane palladium or nickel gives a mixture of heptane
(13.25) and pentene.

5. Aromatization C12 H26 ⎯⎯⎯ → C7 H16 + C5 H10 + other

Pt/Pd/Ni
973K products
n-Alkanes having six or more carbon atoms Dodecane Heptane Pentene
on heating to 773K at 10-20 atmospheric (13.29)
 HYDROCARBONS 375

13.2.4 Conformations 1. Sawhorse projections

Alkanes contain carbon-carbon sigma (σ) In this projection, the molecule is viewed along
bonds. Electron distribution of the sigma the molecular axis. It is then projected on paper
molecular orbital is symmetrical around the by drawing the central C–C bond as a
internuclear axis of the C–C bond which is somewhat longer straight line. Upper end of
not disturbed due to rotation about its axis. the line is slightly tilted towards right or left
This permits free rotation about C–C single hand side. The front carbon is shown at the
bond. This rotation results into different lower end of the line, whereas the rear carbon
spatial arrangements of atoms in space which is shown at the upper end. Each carbon has

d
can change into one another. Such spatial three lines attached to it corresponding to three
arrangements of atoms which can be hydrogen atoms. The lines are inclined at an

he
converted into one another by rotation around angle of 120° to each other. Sawhorse projections
a C-C single bond are called conformations of eclipsed and staggered conformations of
or conformers or rotamers. Alkanes can thus ethane are depicted in Fig. 13.2.
have infinite number of conformations by
rotation around C-C single bonds. However,

is
it may be remembered that rotation around
a C-C single bond is not completely free. It is
hindered by a small energy barrier of

bl
–1
1-20 kJ mol due to weak repulsive
interaction between the adjacent bonds. Such
a type of repulsive interaction is called
pu
torsional strain.
Conformations of ethane : Ethane Fig. 13.2 Sawhorse projections of ethane
be T

molecule (C2H6) contains a carbon – carbon

single bond with each carbon atom attached 2. Newman projections
re
to three hydrogen atoms. Considering the In this projection, the molecule is viewed at the
o R

ball and stick model of ethane, keep one C–C bond head on. The carbon atom nearer to
carbon atom stationary and rotate the other the eye is represented by a point. Three
carbon atom around the C-C axis. This hydrogen atoms attached to the front carbon
tt E

rotation results into infinite number of spatial atom are shown by three lines drawn at an
arrangements of hydrogen atoms attached to angle of 120° to each other. The rear carbon
one carbon atom with respect to the hydrogen atom (the carbon atom away from the eye) is
C

atoms attached to the other carbon atom. represented by a circle and the three hydrogen
These are called conformational isomers atoms are shown attached to it by the shorter
(conformers). Thus there are infinite number lines drawn at an angle of 120° to each other.
no N

of conformations of ethane. However, there are The Newman’s projections are depicted in
two extreme cases. One such conformation in Fig. 13.3.
which hydrogen atoms attached to two
carbons are as closed together as possible is
©

called eclipsed conformation and the other

in which hydrogens are as far apart as
possible is known as the staggered
conformation. Any other intermediate
conformation is called a skew conformation.It
may be remembered that in all the
conformations, the bond angles and the bond
lengths remain the same. Eclipsed and the
staggered conformations can be represented
by Sawhorse and Newman projections. Fig. 13.3 Newman’s projections of ethane
 376 CHEMISTRY

Relative stability of conformations: As 13.3.1 Structure of Double Bond

mentioned earlier, in staggered form of ethane, Carbon-carbon double bond in alkenes
the electron clouds of carbon-hydrogen bonds consists of one strong sigma (σ) bond (bond
are as far apart as possible. Thus, there are –1
enthalpy about 397 kJ mol ) due to head-on
minimum repulsive forces, minimum energy 2
overlapping of sp hybridised orbitals and one
and maximum stability of the molecule. On the weak pi (π) bond (bond enthalpy about 284 kJ
other hand, when the staggered form changes –1
mol ) obtained by lateral or sideways
into the eclipsed form, the electron clouds of overlapping of the two 2p orbitals of the two
the carbon – hydrogen bonds come closer to

d
carbon atoms. The double bond is shorter in
each other resulting in increase in electron bond length (134 pm) than the C–C single bond
cloud repulsions. To check the increased (154 pm). You have already read that the pi (π)

he
repulsive forces, molecule will have to possess bond is a weaker bond due to poor sideways
more energy and thus has lesser stability. As overlapping between the two 2p orbitals. Thus,
already mentioned, the repulsive interaction the presence of the pi (π) bond makes alkenes
between the electron clouds, which affects behave as sources of loosely held mobile

is
stability of a conformation, is called torsional electrons. Therefore, alkenes are easily attacked
strain. Magnitude of torsional strain depends by reagents or compounds which are in search
upon the angle of rotation about C–C bond. of electrons. Such reagents are called

bl
This angle is also called dihedral angle or electrophilic reagents. The presence of
torsional angle. Of all the conformations of weaker π-bond makes alkenes unstable
ethane, the staggered form has the least
pu molecules in comparison to alkanes and thus,
torsional strain and the eclipsed form, the alkenes can be changed into single bond
maximum torsional strain. Thus it may be compounds by combining with the
inferred that rotation around C–C bond in electrophilic reagents. Strength of the double
be T

ethane is not completely free. The energy –1

bond (bond enthalpy, 681 kJ mol ) is greater
difference between the two extreme forms is of than that of a carbon-carbon single bond in
re
–1
the order of 12.5 kJ mol , which is very small.
o R

–1
ethane (bond enthalpy, 348 kJ mol ). Orbital
Even at ordinary temperatures, the ethane diagrams of ethene molecule are shown in
molecule gains thermal or kinetic energy Figs. 13.4 and 13.5.
tt E

sufficient enough to overcome this energy

–1
barrier of 12.5 kJ mol through intermolecular
collisions. Thus, it can be said that rotation
C

about carbon-carbon single bond in ethane is

almost free for all practical purposes. It has
not been possible to separate and isolate
no N

different conformational isomers of ethane.

13.3 ALKENES
Alkenes are unsaturated hydrocarbons
©

containing at least one double bond. What

should be the general formula of alkenes? If Fig. 13.4 Orbital picture of ethene depicting
there is one double bond between two carbon σ bonds only
atoms in alkenes, they must possess two
hydrogen atoms less than alkanes. Hence, 13.3.2 Nomenclature
general formula for alkenes is CnH2n. Alkenes For nomenclature of alkenes in IUPAC system,
are also known as olefins (oil forming) since the longest chain of carbon atoms containing
the first member, ethylene or ethene (C2H4) was the double bond is selected. Numbering of the
found to form an oily liquid on reaction with chain is done from the end which is nearer to
chlorine. the double bond. The suffix ‘ene’ replaces ‘ane’
 HYDROCARBONS 377

Fig. 13.5 Orbital picture of ethene showing formation of (a) π-bond, (b) π-cloud and (c) bond angles
and bond lengths

d
of alkanes. It may be remembered that first Solution

he
member of alkene series is: CH2 (replacing n
(i) 2,8-Dimethyl-3, 6-decadiene;
by 1 in CnH2n) known as methene but has a
very short life. As already mentioned, first (ii) 1,3,5,7 Octatetraene;
stable member of alkene series is C2H4 known (iii) 2-n-Propylpent-1-ene;
as ethylene (common) or ethene (IUPAC). (iv) 4-Ethyl-2,6-dimethyl-dec-4-ene;

is
IUPAC names of a few members of alkenes are
given below : Problem 13.8
Structure IUPAC name Calculate number of sigma (σ) and pi (π)

bl
CH3 – CH = CH2 Propene bonds in the above structures (i-iv).
CH3 – CH2 – CH = CH2 But – l - ene
pu Solution
CH3 – CH = CH–CH3 But-2-ene σ bonds : 33, π bonds : 2
CH2 = CH – CH = CH2 Buta – 1,3 - diene σ bonds : 17, π bonds : 4
be T

CH2 = C – CH3 2-Methylprop-1-ene σ bonds : 23, π bond : 1

|
σ bonds : 41, π bond : 1
re
CH3
o R

CH2 = CH – CH – CH3 3-Methylbut-1-ene

| 13.3.3 Isomerism
CH3 Alkenes show both structural isomerism and
tt E

geometrical isomerism.
Problem 13.7
Structural isomerism : As in alkanes, ethene
C

Write IUPAC names of the following

(C2H4) and propene (C3H6) can have only one
compounds:
structure but alkenes higher than propene
(i) (CH3)2CH – CH = CH – CH2 – CH
have different structures. Alkenes possessing
no N

y
C4H8 as molecular formula can be written in
CH3 – CH – CH
the following three ways:
|
C2H5 I. 1 2 3 4
©

(ii) CH2 = CH – CH2 – CH3

But-1-ene
(iii) CH2 = C (CH2CH2CH3)2 (C4H8)
(iv) CH3 CH2 CH2 CH2 CH2CH3
| | II. 1 2 3 4
CH3 – CHCH = C – CH2 – CHCH3 CH3 – CH = CH – CH3
|
CH3 But-2-ene
(C4H8)
 378 CHEMISTRY

III. 1 2 3 In (a), the two identical atoms i.e., both the

CH2 = C – CH3 X or both the Y lie on the same side of the
| double bond but in (b) the two X or two Y lie
CH3 across the double bond or on the opposite
2-Methyprop-1-ene sides of the double bond. This results in
different geometry of (a) and (b) i.e. disposition
(C4H8)
of atoms or groups in space in the two
Structures I and III, and II and III are the arrangements is different. Therefore, they are
examples of chain isomerism whereas stereoisomers. They would have the same

d
structures I and II are position isomers. geometry if atoms or groups around C=C bond
can be rotated but rotation around C=C bond
Problem 13.9

he
is not free. It is restricted. For understanding
Write structures and IUPAC names of this concept, take two pieces of strong
different structural isomers of alkenes cardboards and join them with the help of two
corresponding to C5H10. nails. Hold one cardboard in your one hand
and try to rotate the other. Can you really rotate

is
Solution
the other cardboard ? The answer is no. The
(a) CH2 = CH – CH2 – CH2 – CH3 rotation is restricted. This illustrates that the
Pent-1-ene restricted rotation of atoms or groups around

bl
(b) CH3 – CH=CH – CH2 – CH3 the doubly bonded carbon atoms gives rise to
Pent-2-ene different geometries of such compounds. The
stereoisomers of this type are called
pu
(c) CH3 – C = CH – CH3
|
geometrical isomers. The isomer of the type
(a), in which two identical atoms or groups lie
CH3
be T

on the same side of the double bond is called

2-Methylbut-2-ene cis isomer and the other isomer of the type
re
(d) CH3 – CH – CH = CH2 (b), in which identical atoms or groups lie on
o R

| the opposite sides of the double bond is called

CH3 trans isomer . Thus cis and trans isomers
3-Methylbut-1-ene have the same structure but have different
tt E

configuration (arrangement of atoms or groups

(e) CH2 = C – CH2 – CH3
in space). Due to different arrangement of
|
atoms or groups in space, these isomers differ
C

CH3
in their properties like melting point, boiling
2-Methylbut-1-ene point, dipole moment, solubility etc.
no N

Geometrical or cis-trans isomers of but-2-ene

Geometrical isomerism: Doubly bonded
are represented below :
carbon atoms have to satisfy the remaining two
valences by joining with two atoms or groups.
If the two atoms or groups attached to each
©

carbon atom are different, they can be

represented by YX C = C XY like structure.
YX C = C XY can be represented in space in the
following two ways :

Cis form of alkene is found to be more polar

than the trans form. For example, dipole
moment of cis-but-2-ene is 0.33 Debye,
whereas, dipole moment of the trans form
is almost zero or it can be said that
 HYDROCARBONS 379

trans-but-2-ene is non-polar. This can be (ii) CH2 = CBr2

understood by drawing geometries of the two
forms as given below from which it is clear that (iii) C6H5CH = CH – CH3
in the trans-but-2-ene, the two methyl groups (iv) CH3CH = CCl CH3
are in opposite directions, Threfore, dipole
moments of C-CH3 bonds cancel, thus making Solution
the trans form non-polar. (iii) and (iv). In structures (i) and (ii), two
identical groups are attached to one of the
doubly bonded carbon atom.

d
13.3.4 Preparation

he
1. From alkynes: Alkynes on partial
reduction with calculated amount of
cis-But-2-ene trans-But-2-ene dihydrogen in the presence of palladised
(μ = 0.33D) (μ = 0) charcoal partially deactivated with poisons

is
In the case of solids, it is observed that like sulphur compounds or quinoline give
the trans isomer has higher melting point alkenes. Partially deactivated palladised
charcoal is known as Lindlar’s catalyst.

bl
than the cis form.
Alkenes thus obtained are having cis
Geometrical or cis-trans isomerism
geometry. However, alkynes on reduction
is also shown by alkenes of the types
with sodium in liquid ammonia form trans
pu
XYC = CXZ and XYC = CZW
alkenes.
Problem 13.10
be T

Draw cis and trans isomers of the

re
following compounds. Also write their
o R

IUPAC names :
(i) CHCl = CHCl (13.30)
(ii) C2H5CCH3 = CCH3C2H5
tt E

Solution
C

(13.31)
no N

Pd/C
iii) CH ≡ CH + H2 ⎯⎯⎯ → CH2 = CH2 (13.32)
Ethyne Ethene
©

Pd/C
iv) CH3 − C ≡ CH + H2 ⎯⎯⎯ →CH3 − CH = CH2
Propyne Propene
(13.33)
Will propene thus obtained show
geometrical isomerism? Think for the
Problem 13.11 reason in support of your answer.
Which of the following compounds will
show cis-trans isomerism? 2. From alkyl halides: Alkyl halides (R-X)
on heating with alcoholic potash
(i) (CH3)2C = CH – C2H5
(potassium hydroxide dissolved in alcohol,
 380 CHEMISTRY

say, ethanol) eliminate one molecule of takes out one hydrogen atom from the
halogen acid to form alkenes. This reaction β-carbon atom.
is known as dehydrohalogenation i.e.,
removal of halogen acid. This is example of
β-elimination reaction, since hydrogen
atom is eliminated from the β carbon atom
(carbon atom next to the carbon to which
halogen is attached).

d
(13.37)
13.3.5 Properties

he
Physical properties
Alkenes as a class resemble alkanes in physical
properties, except in types of isomerism and
(13.34) difference in polar nature. The first three

is
members are gases, the next fourteen are
Nature of halogen atom and the alkyl
liquids and the higher ones are solids. Ethene
group determine rate of the reaction. It is
is a colourless gas with a faint sweet smell. All

bl
observed that for halogens, the rate is:
other alkenes are colourless and odourless,
iodine > bromine > chlorine, while for alkyl
insoluble in water but fairly soluble in non-
groups it is : tert > secondary > primary.
pu polar solvents like benzene, petroleum ether.
3. From vicinal dihalides: Dihalides in They show a regular increase in boiling point
which two halogen atoms are attached to with increase in size i.e., every – CH2 group
two adjacent carbon atoms are known as added increases boiling point by 20–30 K. Like
be T

vicinal dihalides. Vicinal dihalides on alkanes, straight chain alkenes have higher
treatment with zinc metal lose a molecule
re
boiling point than isomeric branched chain
o R

of ZnX2 to form an alkene. This reaction is compounds.

known as dehalogenation.
Chemical properties
CH2 Br − CH2 Br + Zn ⎯⎯
→CH2 = CH2 + ZnBr2 Alkenes are the rich source of loosely held
tt E

(13.35) pi (π) electrons, due to which they show

addition reactions in which the electrophiles
CH3 CHBr − CH2 Br + Zn ⎯⎯
→ CH3 CH = CH2 add on to the carbon-carbon double bond to
C

+ ZnBr2 form the addition products. Some reagents

also add by free radical mechanism. There are
(13.36)
no N

cases when under special conditions, alkenes

4. From alcohols by acidic dehydration: also undergo free radical substitution
You have read during nomenclature of reactions. Oxidation and ozonolysis reactions
different homologous series in Unit 12 that are also quite prominent in alkenes. A brief
alcohols are the hydroxy derivatives of description of different reactions of alkenes is
©

alkanes. They are represented by R–OH given below:

where, R is CnH2n+1. Alcohols on heating
1. Addition of dihydrogen: Alkenes add up
with concentrated sulphuric acid form
one molecule of dihydrogen gas in the
alkenes with the elimination of one water
presence of finely divided nickel, palladium
molecule. Since a water molecule is
or platinum to form alkanes (Section 13.2.2)
eliminated from the alcohol molecule in the
presence of an acid, this reaction is known 2. Addition of halogens : Halogens like
as acidic dehydration of alcohols. This bromine or chlorine add up to alkene to
reaction is also the example of form vicinal dihalides. However, iodine
β-elimination reaction since –OH group does not show addition reaction under
 HYDROCARBONS 381

normal conditions. The reddish orange

colour of bromine solution in carbon
tetrachloride is discharged when bromine
adds up to an unsaturation site. This
reaction is used as a test for unsaturation.
Addition of halogens to alkenes is an
example of electrophilic addition reaction
involving cyclic halonium ion formation (13.42)
which you will study in higher classes.

d
Markovnikov, a Russian chemist made a
generalisation in 1869 after studying such
reactions in detail. These generalisations led

he
Markovnikov to frame a rule called
Markovnikov rule. The rule states that
(13.38) negative part of the addendum (adding
molecule) gets attached to that carbon atom

is
(ii) CH3 − CH = CH2 + Cl − Cl ⎯⎯→CH3 − CH − CH2 which possesses lesser number of hydrogen
| | atoms. Thus according to this rule, product I
Cl Cl i.e., 2-bromopropane is expected. In actual

bl
Propene 1,2-Dichloropropane
practice, this is the principal product of the
(13.39) reaction. This generalisation of Markovnikov
3. Addition of hydrogen halides:
pu rule can be better understood in terms of
Hydrogen halides (HCl, HBr,HI) add up to mechanism of the reaction.
alkenes to form alkyl halides. The order of Mechanism
reactivity of the hydrogen halides is +
be T

Hydrogen bromide provides an electrophile, H ,

HI > HBr > HCl. Like addition of halogens which attacks the double bond to form
re
to alkenes, addition of hydrogen halides is carbocation as shown below :
o R

also an example of electrophilic addition

reaction. Let us illustrate this by taking
addition of HBr to symmetrical and
tt E

unsymmetrical alkenes
Addition reaction of HBr to symmetrical
C

alkenes
Addition reactions of HBr to symmetrical
(a) less stable (b) more stable
alkenes (similar groups attached to double
primary carbocation secondary carbocation
no N

bond) take place by electrophilic addition

mechanism. (i) The secondary carbocation (b) is more
stable than the primary carbocation (a),
CH2 = CH2 + H – Br ⎯⎯
→CH3 – CH2 – Br (13.40)
therefore, the former predominates because
©

CH3 – CH = CH – CH3 + HBr ⎯

→CH3 – CH2 – CHCH3 it is formed at a faster rate.
–
| (ii) The carbocation (b) is attacked by Br ion
Br to form the product as follows :
(13.41)
Addition reaction of HBr to
unsymmetrical alkenes (Markovnikov
Rule)
How will H – Br add to propene ? The two 2-Bromopropane
possible products are I and II. (major product)
 382 CHEMISTRY

Anti Markovnikov addition or peroxide

effect or Kharash effect
In the presence of peroxide, addition of HBr
to unsymmetrical alkenes like propene takes
place contrary to the Markovnikov rule. This
happens only with HBr but not with HCl and
The secondary free radical obtained in the
Hl. This addition reaction was observed
above mechanism (step iii) is more stable than
by M.S. Kharash and F.R. Mayo in 1933 at
the primary. This explains the formation of

d
the University of Chicago. This reaction 1-bromopropane as the major product. It may
is known as peroxide or Kharash effect be noted that the peroxide effect is not observed

he
or addition reaction anti to Markovnikov in addition of HCl and HI. This may be due
rule. to the fact that the H–Cl bond being
–1
(C H CO) O stronger (430.5 kJ mol ) than H–Br bond
CH3 –CH = CH2 + HBr ⎯⎯⎯⎯⎯⎯
6 5 2 2
→CH3 – CH2 –1
(363.7 kJ mol ), is not cleaved by the free
x radical, whereas the H–I bond is weaker

is
–1
CH2 Br (296.8 kJ mol ) and iodine free radicals
1-Bromopropane combine to form iodine molecules instead of

bl
(13.43) adding to the double bond.
Mechanism : Peroxide effect proceeds via free
radical chain mechanism as given below:
pu Problem 13.12
Write IUPAC names of the products
(i) obtained by addition reactions of HBr to
hex-1-ene
be T

(i) in the absence of peroxide and

re
(ii) in the presence of peroxide.
o R

Solution
tt E

• •
(ii) C6 H5 + H – Br ⎯⎯⎯⎯
Homolysis
⎯→C6 H6 + Br
C
no N
©

4. Addition of sulphuric acid : Cold

concentrated sulphuric acid adds to
alkenes in accordance with Markovnikov
rule to form alkyl hydrogen sulphate by
the electrophilic addition reaction.
 HYDROCARBONS 383

ketones and/or acids depending upon the

nature of the alkene and the experimental
conditions

(13.49)
(13.44) KMnO /H+
CH3 – CH = CH – CH3 ⎯⎯⎯⎯⎯
4
→2CH3 COOH

d
But -2-ene Ethanoic acid
(13.50)

he
7. Ozonolysis : Ozonolysis of alkenes involves
the addition of ozone molecule to alkene to
form ozonide, and then cleavage of the
ozonide by Zn-H2O to smaller molecules.

is
This reaction is highly useful in detecting
(13.45) the position of the double bond in alkenes
or other unsaturated compounds.

bl
5. Addition of water : In the presence of a
few drops of concentrated sulphuric acid
alkenes react with water to form alcohols,
pu
in accordance with the Markovnikov rule.
be T
re
o R

(13.51)
(13.46)
tt E

6. Oxidation: Alkenes on reaction with cold,

dilute, aqueous solution of potassium
permanganate (Baeyer’s reagent) produce
C

vicinal glycols. Decolorisation of KMnO4

solution is used as a test for unsaturation.
no N
©

(13.52)
(13.47) 8. Polymerisation: You are familiar with
polythene bags and polythene sheets.
Polythene is obtained by the combination
of large number of ethene molecules at high
temperature, high pressure and in the
presence of a catalyst. The large molecules
(13.48) thus obtained are called polymers. This
b) Acidic potassium permanganate or acidic reaction is known as polymerisation. The
potassium dichromate oxidises alkenes to simple compounds from which polymers
 384 CHEMISTRY

are made are called monomers. Other are named as derivatives of the corresponding
alkenes also undergo polymerisation. alkanes replacing ‘ane’ by the suffix ‘yne’. The
High temp./pressure
position of the triple bond is indicated by the
n(CH2 = CH2 ) ⎯⎯⎯⎯⎯⎯⎯⎯
Catalyst → —( CH2 –CH2 —
)n first triply bonded carbon. Common and
Polythene IUPAC names of a few members of alkyne series
(13.53) are given in Table 13.2.
High temp./pressure
You have already learnt that ethyne and
n(CH3 –CH = CH2 ) ⎯⎯⎯⎯⎯⎯⎯⎯
Catalyst → —
( CH–CH2 —)n propyne have got only one structure but there
|
are two possible structures for butyne –

d
CH3
Polypropene (i) but-1-yne and (ii) but-2-yne. Since these two
compounds differ in their structures due to the

he
(13.54) position of the triple bond, they are known as
Polymers are used for the manufacture of position isomers. In how many ways, you can
plastic bags, squeeze bottles, refrigerator dishes, construct the structure for the next homologue
toys, pipes, radio and T.V. cabinets etc. i.e., the next alkyne with molecular formula
Polypropene is used for the manufacture of milk

is
C5H8? Let us try to arrange five carbon atoms
crates, plastic buckets and other moulded with a continuous chain and with a side chain.
articles. Though these materials have now Following are the possible structures :
become common, excessive use of polythene

bl
Structure IUPAC name
and polypropylene is a matter of great concern
for all of us. 1 2 3 4
I. HC ≡ C– CH – CH – CH Pent–1-yne
5
pu 2 2 3

13.4 ALKYNES 1 2 3 4 5
II. H C– C ≡ C– CH – CH Pent–2-yne
Like alkenes, alkynes are also unsaturated 3 2 3

hydrocarbons. They contain at least one triple 4 3 2 1

be T

III. H C– CH – C ≡ CH 3-Methyl but–1-yne

bond between two carbon atoms. The number 3
|
re
of hydrogen atoms is still less in alkynes as
CH3
o R

compared to alkenes or alkanes. Their general

formula is CnH2n–2. Structures I and II are position isomers and
structures I and III or II and III are chain
The first stable member of alkyne series
tt E

isomers.
is ethyne which is popularly known as
acetylene. Acetylene is used for arc welding Problem 13.13
C

purposes in the form of oxyacetylene flame

Write structures of different isomers
obtained by mixing acetylene with oxygen gas. th
corresponding to the 5 member of
Alkynes are starting materials for a large
alkyne series. Also write IUPAC names of
no N

number of organic compounds. Hence, it is

all the isomers. What type of isomerism is
interesting to study this class of organic
exhibited by different pairs of isomers?
compounds.
13.4.1 Nomenclature and Isomerism Solution
©

th
In common system, alkynes are named as 5 member of alkyne has the molecular
derivatives of acetylene. In IUPAC system, they formula C6H10. The possible isomers are:
Table 13.2 Common and IUPAC Names of Alkynes (CnH2n–2)

Value of n Formula Structure Common name IUPAC name

2 C2H2 H-C≡CH Acetylene Ethyne
3 C3H4 CH3-C≡CH Methylacetylene Propyne
4 C4H6 CH3CH2-C≡CH Ethylacetylene But-1-yne
4 C4H6 CH3-C≡C-CH3 Dimethylacetylene But-2-yne
 HYDROCARBONS 385

(a) HC ≡ C – CH2 – CH2 – CH2 – CH3

Hex-1-yne
(b) CH3 – C ≡ C – CH2 – CH2 – CH3
Hex-2-yne
(c) CH3 – CH2 – C ≡ C – CH2– CH3
Hex-3-yne

d
3-Methylpent-1-yne

he
4-Methylpent-1-yne

is
bl
4-Methylpent-2-yne
pu Fig. 13.6 Orbital picture of ethyne showing
(a) sigma overlaps (b) pi overlaps.

orbitals of the other carbon atom, which

be T

undergo lateral or sideways overlapping to

re
3,3-Dimethylbut-1-yne form two pi (π) bonds between two carbon
o R

atoms. Thus ethyne molecule consists of one

Position and chain isomerism shown by C–C σ bond, two C–H σ bonds and two C–C π
different pairs. bonds. The strength of C≡C bond (bond
tt E

-1
enthalpy 823 kJ mol ) is more than those of
–1
13.4.2 Structure of Triple Bond C=C bond (bond enthalpy 681 kJ mol ) and
–1
C–C bond (bond enthalpy 348 kJ mol ). The
C

Ethyne is the simplest molecule of alkyne

series. Structure of ethyne is shown in C≡C bond length is shorter (120 pm) than those
Fig. 13.6. of C=C (133 pm) and C–C (154 pm). Electron
cloud between two carbon atoms is
no N

Each carbon atom of ethyne has two sp cylindrically symmetrical about the
hybridised orbitals. Carbon-carbon sigma (σ) internuclear axis. Thus, ethyne is a linear
bond is obtained by the head-on overlapping molecule.
of the two sp hybridised orbitals of the two
©

carbon atoms. The remaining sp hybridised 13.4.3 Preparation

orbital of each carbon atom undergoes 1. From calcium carbide: On industrial
overlapping along the internuclear axis with scale, ethyne is prepared by treating calcium
the 1s orbital of each of the two hydrogen atoms carbide with water. Calcium carbide is
forming two C-H sigma bonds. H-C-C bond prepared by heating quick lime with coke.
angle is of 180°. Each carbon has two Quick lime can be obtained by heating
unhybridised p orbitals which are limestone as shown in the following
perpendicular to each other as well as to the reactions:
plane of the C-C sigma bond. The 2p orbitals
of one carbon atom are parallel to the 2p CaCO3 ⎯Δ→ CaO + CO2 (13.55)
 386 CHEMISTRY

atoms in ethyne are attached to the sp

CaO + 3C ⎯→ CaC2 + CO (13.56)
hybridised carbon atoms whereas they are
Calcium 2
attached to sp hybridised carbon atoms in
carbide 3
ethene and sp hybridised carbons in ethane.
CaC2 2H2 O Ca(OH)2 C2 H2 (13.57) Due to the maximum percentage of s character
(50%), the sp hybridised orbitals of carbon
2. From vicinal dihalides : Vicinal atoms in ethyne molecules have highest
dihalides on treatment with alcoholic electronegativity; hence, these attract the
potassium hydroxide undergo shared electron pair of the C-H bond of ethyne

d
dehydrohalogenation. One molecule of to a greater extent than that of the sp
2

hydrogen halide is eliminated to form hybridised orbitals of carbon in ethene and the

he
alkenyl halide which on treatment with 3
sp hybridised orbital of carbon in ethane.
sodamide gives alkyne. Thus in ethyne, hydrogen atoms can be
liberated as protons more easily as compared
to ethene and ethane. Hence, hydrogen atoms
of ethyne attached to triply bonded carbon

is
atom are acidic in nature. You may note that
the hydrogen atoms attached to the triply
bonded carbons are acidic but not all the

bl
hydrogen atoms of alkynes.
–
pu HC ≡ CH + Na → HC ≡ C Na + + ½H2
13.4.4 Properties
Monosodium
Physical properties ethynide
Physical properties of alkynes follow the same
be T

trend of alkenes and alkanes. First three (13.59)

re
members are gases, the next eight are liquids
o R

– –
and the higher ones are solids. All alkynes are HC ≡ C – Na + + Na → Na + C ≡ C Na + + ½H2
colourless. Ethyene has characteristic odour. Disodium ethynide
Other members are odourless. Alkynes are
tt E

(13.60)
weakly polar in nature. They are lighter than
water and immiscible with water but soluble CH3 – C ≡ C − H + Na + NH2–
C

in organic solvents like ethers, carbon

↓
tetrachloride and benzene. Their melting point,
CH3 – C ≡ C – Na + + NH3 (13.61)
boiling point and density increase with
no N

increase in molar mass. Sodium propynide

Chemical properties These reactions are not shown by alkenes
Alkynes show acidic nature, addition reactions and alkanes, hence used for distinction
and polymerisation reactions as follows : between alkynes, alkenes and alkanes. What
©

A. Acidic character of alkyne: Sodium about the above reactions with but-1-yne and
metal and sodamide (NaNH2) are strong bases. but-2-yne ? Alkanes, alkenes and alkynes
They react with ethyne to form sodium follow the following trend in their acidic
acetylide with the liberation of dihydrogen gas. behaviour :
These reactions have not been observed in case i) HC ≡ CH > H2 C = CH2 > CH3 –CH3
of ethene and ethane thus indicating that
ethyne is acidic in nature in comparison to ii) HC ≡ CH > CH3 – C ≡ CH >> CH3 – C ≡ C – CH3
ethene and ethane. Why is it so ? Has it B. Addition reactions: Alkynes contain a
something to do with their structures and the triple bond, so they add up, two molecules of
hybridisation ? You have read that hydrogen dihydrogen, halogen, hydrogen halides etc.
 HYDROCARBONS 387

Formation of the addition product takes place

according to the following steps.

The addition product formed depends upon

stability of vinylic cation. Addition in

d
unsymmetrical alkynes takes place according
to Markovnikov rule. Majority of the reactions
of alkynes are the examples of electrophilic (13.66)

he
addition reactions. A few addition reactions are (iv) Addition of water
given below: Like alkanes and alkenes, alkynes are also
(i) Addition of dihydrogen immiscible and do not react with water.
Pt/Pd/Ni H2 However, one molecule of water adds to alkynes
HC ≡ CH+ H2 ⎯⎯⎯⎯ →[H2C = CH2 ] ⎯⎯→CH3 –CH3

is
on warming with mercuric sulphate and dilute
(13.62) sulphuric acid at 333 K to form carbonyl
Pt/Pd/Ni
CH3 – C ≡ CH + H2 ⎯⎯⎯⎯→[CH3 – CH = CH2 ] compounds.

bl
Propyne Propene
pu ↓ H2
CH3 – CH2 – CH3
Propane
(13.63)
be T

(ii) Addition of halogens

re
o R

(13.67)
tt E
C
no N

(13.64)
Reddish orange colour of the solution of
bromine in carbon tetrachloride is decolourised.
This is used as a test for unsaturation.
©

(13.68)
(iii) Addition of hydrogen halides
(v) Polymerisation
Two molecules of hydrogen halides (HCl, HBr,
HI) add to alkynes to form gem dihalides (in (a) Linear polymerisation: Under suitable
which two halogens are attached to the same conditions, linear polymerisation of ethyne
carbon atom) takes place to produce polyacetylene or
H – C ≡ C – H + H – Br ⎯→ [CH2 = CH – Br] ⎯→ CHBr2 polyethyne which is a high molecular weight
Bromoethene | polyene containing repeating units of
CH3 (CH = CH – CH = CH ) and can be represented
1,1-Dibromoethane as —( CH = CH – CH = CH )— n Under special
(13.65) conditions, this polymer conducts electricity.
 388 CHEMISTRY

Thin film of polyacetylene can be used as in a majority of reactions of aromatic

electrodes in batteries. These films are good compounds, the unsaturation of benzene ring
conductors, lighter and cheaper than the metal is retained. However, there are examples of
conductors. aromatic hydrocarbons which do not contain
(b) Cyclic polymerisation: Ethyne on passing a benzene ring but instead contain other highly
through red hot iron tube at 873K undergoes unsaturated ring. Aromatic compounds
cyclic polymerization. Three molecules containing benzene ring are known as
polymerise to form benzene, which is the benzenoids and those not containing a
starting molecule for the preparation of benzene ring are known as non-benzenoids.

d
derivatives of benzene, dyes, drugs and large Some examples of arenes are given
number of other organic compounds. This is below:

he
the best route for entering from aliphatic to
aromatic compounds as discussed below:

is
Benzene Toluene Naphthalene

bl
pu (13.69)

Problem 13.14
How will you convert ethanoic acid into Biphenyl
be T

benzene?
13.5.1 Nomenclature and Isomerism
re
Solution
The nomenclature and isomerism of aromatic
o R

hydrocarbons has already been discussed in

Unit 12. All six hydrogen atoms in benzene are
tt E

equivalent; so it forms one and only one type

of monosubstituted product. When two
hydrogen atoms in benzene are replaced by
C

two similar or different monovalent atoms or

groups, three different position isomers are
possible. The 1, 2 or 1, 6 is known as the ortho
no N

(o–), the 1, 3 or 1, 5 as meta (m–) and the 1, 4

as para (p–) disubstituted compounds. A few
examples of derivatives of benzene are given
below:
©

13.5 AROMATIC HYDROCARBON

These hydrocarbons are also known as
‘arenes’. Since most of them possess pleasant
odour (Greek; aroma meaning pleasant
smelling), the class of compounds was named
as ‘aromatic compounds’. Most of such
compounds were found to contain benzene Methylbenzene 1,2-Dimethylbenzene
ring. Benzene ring is highly unsaturated but (Toluene) (o-Xylene)
 HYDROCARBONS 389

Friedrich August Kekulé,a German chemist was born in 1829 at Darmsdt in

Germany. He became Professor in 1856 and Fellow of Royal Society in 1875. He
made major contribution to structural organic chemistry by proposing in 1858 that
carbon atoms can join to one another to form chains and later in 1865,he found an
answer to the challenging problem of benzene structure by suggesting that these
chains can close to form rings. He gave the dynamic structural formula to benzene
which forms the basis for its modern electronic structure. He described the discovery

d
of benzene structure later as:
FRIEDRICH
“I was sitting writing at my text book,but the work did not progress; my thoughts
AUGUST KEKULÉ

he
were elsewhere.I turned my chair to the fire, and dozed. Again the atoms were
(7th September
gambolling before my eyes. This time the smaller groups kept modestly in the
1829–13th July
background. My mental eye, rendered more acute by repeated visions of this kind, 1896)
could now distinguish larger structures of manifold conformations; long

is
rows,sometimes more closely fitted together; all twisting and turning in snake like motion. But look! What
was that? One of the snakes had seized hold of it’s own tail, and the form whirled mockingly before my eyes.
As if by a flash of lightning I woke;.... I spent the rest of the night working out the consequences of the

bl
hypothesis. Let us learn to dream, gentlemen, and then perhaps we shall learn the truth but let us beware of
making our dreams public before they have been approved by the waking mind.”( 1890).
One hundred years later, on the occasion of Kekulé’s centenary celebrations a group of compounds having
pu
polybenzenoid structures have been named as Kekulenes.
be T

found to produce one and only one

monosubstituted derivative which indicated
re
o R

that all the six carbon and six hydrogen atoms

of benzene are identical. On the basis of this
observation August Kekulé in 1865 proposed
tt E

the following structure for benzene having

cyclic arrangement of six carbon atoms with
alternate single and double bonds and one
C

1,3 Dimethylbenzene 1,4-Dimethylbenzene hydrogen atom attached to each carbon

(m-Xylene) ( p-Xylene) atom.
no N

13.5.2 Structure of Benzene

Benzene was isolated by Michael Faraday in
1825. The molecular formula of benzene,
C6H6, indicates a high degree of unsaturation.
©

This molecular formula did not account for its

relationship to corresponding alkanes, alkenes
and alkynes which you have studied in earlier
sections of this unit. What do you think about
its possible structure? Due to its unique The Kekulé structure indicates
properties and unusual stability, it took several the possibility of two isomeric
years to assign its structure. Benzene was 1, 2-dibromobenzenes. In one of the isomers,
found to be a stable molecule and found to the bromine atoms are attached to the doubly
form a triozonide which indicates the presence bonded carbon atoms whereas in the other,
of three double bonds. Benzene was further they are attached to the singly bonded carbons.
 390 CHEMISTRY

perpendicular to the plane of the ring as shown

below:

However, benzene was found to form only

one ortho disubstituted product. This problem
was overcome by Kekulé by suggesting the

d
concept of oscillating nature of double bonds
in benzene as given below.

he
The unhybridised p orbital of carbon atoms
are close enough to form a π bond by lateral
overlap. There are two equal possibilities of
forming three π bonds by overlap of p orbitals

is
of C1 –C2, C3 – C4, C5 – C6 or C2 – C3, C4 – C5,
Even with this modification, Kekulé
C6 – C1 respectively as shown in the following
structure of benzene fails to explain unusual

bl
figures.
stability and preference to substitution
reactions than addition reactions, which could
later on be explained by resonance.
pu
Resonance and stability of benzene
According to Valence Bond Theory, the concept
be T

of oscillating double bonds in benzene is now

explained by resonance. Benzene is a hybrid
re
of various resonating structures. The two
o R

structures, A and B given by Kekulé are the

main contributing structures. The hybrid
structure is represented by inserting a circle
tt E

or a dotted circle in the hexagon as shown in

(C). The circle represents the six electrons which
are delocalised between the six carbon atoms Fig. 13.7 (a)
C

of the benzene ring.

no N

(A) (B) (C)

The orbital overlapping gives us better
©

picture about the structure of benzene. All the

2
six carbon atoms in benzene are sp hybridized.
2
Two sp hybrid orbitals of each carbon atom
2
overlap with sp hybrid orbitals of adjacent
carbon atoms to form six C—C sigma bonds
which are in the hexagonal plane. The
2 Fig. 13.7 (b)
remaining sp hybrid orbital of each carbon
atom overlaps with s orbital of a hydrogen atom Structures shown in Fig. 13.7(a) and (b)
to form six C—H sigma bonds. Each carbon correspond to two Kekulé’s structure with
atom is now left with one unhybridised p orbital localised π bonds. The internuclear distance

390 C:\Chemistry XI\Unit-13\Unit-13(1)(reprint.pmd 27.7.6, 16.10.6 (reprint)

 HYDROCARBONS 391

between all the carbon atoms in the ring has (i) Planarity
been determined by the X-ray diffraction to be (ii) Complete delocalisation of the π electrons
the same; there is equal probability for the p in the ring
orbital of each carbon atom to overlap with the
(iii) Presence of (4n + 2) π electrons in the ring
p orbitals of adjacent carbon atoms [Fig. 13.7
where n is an integer (n = 0, 1, 2, . . .).
(c)]. This can be represented in the form of two
doughtnuts (rings) of electron clouds [Fig. 13.7 This is often referred to as Hückel Rule.
(d)], one above and one below the plane of the Some examples of aromatic compounds are
hexagonal ring as shown below: given below:

d
he
is
(electron cloud)

bl
Fig. 13.7 (c) Fig. 13.7 (d)
The six π electrons are thus delocalised and
pu
can move freely about the six carbon nuclei,
instead of any two as shown in Fig. 13.6 (a) or
(b). The delocalised π electron cloud is attracted
be T

more strongly by the nuclei of the carbon

re
atoms than the electron cloud localised
o R

between two carbon atoms. Therefore, presence

of delocalised π electrons in benzene makes
it more stable than the hypothetical
tt E

cyclohexatriene.
X-Ray diffraction data reveals that benzene
C

is a planar molecule. Had any one of the above

structures of benzene (A or B) been correct, two
13.5.4 Preparation of Benzene
types of C—C bond lengths were expected.
Benzene is commercially isolated from coal tar.
no N

However, X-ray data indicates that all the six

C—C bond lengths are of the same order However, it may be prepared in the laboratory
(139 pm) which is intermediate between by the following methods.
C— C single bond (154 pm) and C—C double (i) Cyclic polymerisation of ethyne:
©

bond (133 pm). Thus the absence of pure (Section 13.4.4)

double bond in benzene accounts for the (ii) Decarboxylation of aromatic acids:
reluctance of benzene to show addition Sodium salt of benzoic acid on heating with
reactions under normal conditions, thus sodalime gives benzene.
explaining the unusual behaviour of benzene.
13.5.3 Aromaticity
Benzene was considered as parent ‘aromatic’
compound. Now, the name is applied to all the
ring systems whether or not having benzene
ring, possessing following characteristics. (13.70)
 392 CHEMISTRY

(iii) Reduction of phenol: Phenol is reduced (ii) Halogenation: Arenes react with halogens
to benzene by passing its vapours over in the presence of a Lewis acid like anhydrous
heated zinc dust FeCl3, FeBr3 or AlCl3 to yield haloarenes.

(13.71) Chlorobenzene

d
13.5.5 Properties (13.73)
(iii) Sulphonation: The replacement of a

he
Physical properties
Aromatic hydrocarbons are non- polar hydrogen atom by a sulphonic acid group in
molecules and are usually colourless liquids a ring is called sulphonation. It is carried out
or solids with a characteristic aroma. You are by heating benzene with fuming sulphuric acid
also familiar with naphthalene balls which are (oleum).

is
used in toilets and for preservation of clothes
because of unique smell of the compound and
the moth repellent property. Aromatic

bl
hydrocarbons are immiscible with water but
are readily miscible with organic solvents. They
burn with sooty flame.
pu
Chemical properties (13.74)
Arenes are characterised by electrophilic
be T

substitution reactions. However, under special (iv) Friedel-Crafts alkylation reaction:

conditions they can also undergo addition and When benzene is treated with an alkyl halide
re
oxidation reactions. in the presence of anhydrous aluminium
o R

chloride, alkylbenene is formed.

Electrophilic substitution reactions
The common electrophilic substitution
tt E

reactions of arenes are nitration, halogenation,

sulphonation, Friedel Craft’s alkylation and
acylation reactions in which attacking reagent
C

+
is an electrophile (E )
(i) Nitration: A nitro group is introduced into (13.75)
no N

benzene ring when benzene is heated with a

mixture of concentrated nitric acid and
concentrated sulphuric acid (nitrating
mixture).
©

(13.76)

Why do we get isopropyl benzene on

treating benzene with 1-chloropropane instead
of n-propyl benzene?
(13.72)
(v) Friedel-Crafts acylation reaction: The
reaction of benzene with an acyl halide or acid
anhydride in the presence of Lewis acids (AlCl3)
Nitrobenzene yields acyl benzene.
 HYDROCARBONS 393

(13.77)

d
In the case of nitration, the electrophile,

he
+
nitronium ion, N O2 is produced by transfer
of a proton (from sulphuric acid) to nitric acid
in the following manner:
(13.78) Step I

is
If excess of electrophilic reagent is used,
further substitution reaction may take place

bl
in which other hydrogen atoms of benzene ring Step II
may also be successively replaced by the
electrophile. For example, benzene on
treatment with excess of chlorine in the
pu
presence of anhydrous AlCl 3 can be
chlorinated to hexachlorobenzene (C6Cl6)
Protonated Nitronium
nitric acid ion
be T

It is interesting to note that in the process

of generation of nitronium ion, sulphuric acid
re
o R

serves as an acid and nitric acid as a base.

Thus, it is a simple acid-base equilibrium.
(b) Formation of Carbocation
tt E

(arenium ion): Attack of electrophile

results in the formation of σ-complex or
(13.79) 3
arenium ion in which one of the carbon is sp
C

Mechanism of electrophilic substitution hybridised.

reactions:
According to experimental evidences, SE (S =
no N

substitution; E = electrophilic) reactions are

supposed to proceed via the following three
steps:
(a) Generation of the eletrophile sigma complex (arenium ion)
©

(b) Formation of carbocation intermediate The arenium ion gets stabilised by

(c) Removal of proton from the carbocation resonance:
intermediate
⊕
(a) Generation of electrophile E : During
chlorination, alkylation and acylation of
benzene, anhydrous AlCl3, being a Lewis acid
helps in generation of the elctrophile Cl⊕, R⊕,
RC⊕O (acylium ion) respectively by combining
with the attacking reagent.
 394 CHEMISTRY

Sigma complex or arenium ion loses its chemical equation:

aromatic character because delocalisation of
3
electrons stops at sp hybridised carbon. y y
CxHy + (x + ) O2 → x CO2 + HO
4 2 2
(c) Removal of proton: To restore the
aromatic character, σ -complex releases proton (13.83)
3
from sp hybridised carbon on attack by 13.5.6 Directive influence of a functional
–
[AlCl4] (in case of halogenation, alkylation and group in monosubstituted benzene
–
acylation) and [HSO4] (in case of nitration). When monosubstituted benzene is subjected

d
to further substitution, three possible
disubstituted products are not formed in equal
amounts. Two types of behaviour are observed.

he
Either ortho and para products or meta
product is predominantly formed. It has also
been observed that this behaviour depends on
the nature of the substituent already present

is
in the benzene ring and not on the nature of
the entering group. This is known as directive
Addition reactions influence of substituents. Reasons for ortho/

bl
Under vigorous conditions, i.e., at high para or meta directive nature of groups are
temperature and/ or pressure in the presence discussed below:
of nickel catalyst, hydrogenation of benzene
pu Ortho and para directing groups: The
gives cyclohexane. groups which direct the incoming group to
ortho and para positions are called ortho and
para directing groups. As an example, let us
be T

discuss the directive influence of phenolic

re
Cyclohexane (–OH) group. Phenol is resonance hybrid of
o R

following structures:
(13.80)
Under ultra-violet light, three chlorine
tt E

molecules add to benzene to produce benzene

hexachloride, C6H6Cl6 which is also called
gammaxane.
C
no N

Benzene hexachloride,
©

(BHC)
(13.81) It is clear from the above resonating structures
Combustion: When heated in air, benzene that the electron density is more on
burns with sooty flame producing CO2 and o – and p – positions. Hence, the substitution
H2O takes place mainly at these positions. However,
it may be noted that –I effect of – OH group
15 also operates due to which the electron density
C6 H 6 + O2 → 6CO2 + 3H2O (13.82)
2 on ortho and para positions of the benzene ring
General combustion reaction for any is slightly reduced. But the overall electron
hydrocarbon may be given by the following density increases at these positions of the ring
 HYDROCARBONS 395

due to resonance. Therefore, –OH group are also called ‘deactivating groups’. The
activates the benzene ring for the attack by electron density on o – and p – position is
an electrophile. Other examples of activating comparatively less than that at meta position.
groups are –NH2, –NHR, –NHCOCH3, –OCH3, Hence, the electrophile attacks on
–CH3, –C2H5, etc. comparatively electron rich meta position
In the case of aryl halides, halogens are resulting in meta substitution.
moderately deactivating. Because of their
13.6 CARCINOGENICITY AND TOXICITY
strong – I effect, overall electron density on
benzene ring decreases. It makes further Benzene and polynuclear hydrocarbons

d
substitution difficult. However, due to containing more than two benzene rings
resonance the electron density on o – and p – fused together are toxic and said to possess

he
positions is greater than that at the m-position. cancer producing (carcinogenic) property.
Hence, they are also o – and p – directing Such polynuclear hydrocarbons are formed
groups. on incomplete combustion of organic
materials like tobacco, coal and petroleum.
Meta directing group: The groups which
They enter into human body and undergo

is
direct the incoming group to meta position are
various biochemical reactions and finally
called meta directing groups. Some examples
damage DNA and cause cancer. Some of
of meta directing groups are –NO2, –CN, –CHO,
the carcinogenic hydrocarbons are given

bl
–COR, –COOH, –COOR, –SO3H, etc.
below (see box).
Let us take the example of nitro group. Nitro
group reduces the electron density in the
pu
benzene ring due to its strong– I effect.
Nitrobenzene is a resonance hybrid of the
following structures.
be T
re
o R
tt E
C
no N
©

In this case, the overall electron density on

benzene ring decreases making further
substitution difficult, therefore these groups

395 C:\Chemistry XI\Unit-13\Unit-13(1)(reprint.pmd 27.7.6, 16.10.6 (reprint)

 396 CHEMISTRY

SUMMARY

Hydrocarbons are the compounds of carbon and hydrogen only. Hydrocarbons are mainly
obtained from coal and petroleum, which are the major sources of energy.
Petrochemicals are the prominent starting materials used for the manufacture of a
large number of commercially important products. LPG (liquefied petroleum gas) and
CNG (compressed natural gas), the main sources of energy for domestic fuels and the
automobile industry, are obtained from petroleum. Hydrocarbons are classified as open
chain saturated (alkanes) and unsaturated (alkenes and alkynes), cyclic (alicyclic)

d
and aromatic, according to their structure.
The important reactions of alkanes are free radical substitution, combustion,

he
oxidation and aromatization. Alkenes and alkynes undergo addition reactions, which
are mainly electrophilic additions. Aromatic hydrocarbons, despite having unsaturation,
undergo mainly electrophilic substitution reactions. These undergo addition reactions
only under special conditions.

is
Alkanes show conformational isomerism due to free rotation along the C–C sigma
bonds. Out of staggered and the eclipsed conformations of ethane, staggered conformation
is more stable as hydrogen atoms are farthest apart. Alkenes exhibit geometrical
(cis-trans) isomerism due to restricted rotation around the carbon–carbon double bond.

bl
Benzene and benzenoid compounds show aromatic character. Aromaticity, the
property of being aromatic is possessed by compounds having specific electronic structure
characterised by Hückel (4n+2)π electron rule. The nature of groups or substituents
pu
attached to benzene ring is responsible for activation or deactivation of the benzene ring
towards further electrophilic substitution and also for orientation of the incoming group.
Some of the polynuclear hydrocarbons having fused benzene ring system have
be T

carcinogenic property.
re
o R

EXERCISES

13.1 How do you account for the formation of ethane during chlorination of methane ?
tt E

13.2 Write IUPAC names of the following compounds :

(a) CH3CH=C(CH3)2 (b) CH2=CH-C≡C-CH3
C

(c) (d) –CH2–CH2–CH=CH2

no N

(f ) CH3 (CH2 )4 CH(CH2 )3 CH3

(e) |
CH2 − CH (CH3 )2
©

(g) CH3 – CH = CH – CH2 – CH = CH – CH – CH2 – CH = CH2

|
C2H5
13.3 For the following compounds, write structural formulas and IUPAC names for all
possible isomers having the number of double or triple bond as indicated :
(a) C4H8 (one double bond) (b) C5H8 (one triple bond)
13.4 Write IUPAC names of the products obtained by the ozonolysis of the following
compounds :
(i) Pent-2-ene (ii) 3,4-Dimethylhept-3-ene
(iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene
 HYDROCARBONS 397

13.5 An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-

one. Write structure and IUPAC name of ‘A’.
13.6 An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C
π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass
44 u. Write IUPAC name of ‘A’.
13.7 Propanal and pentan-3-one are the ozonolysis products of an alkene?
What is the structural formula of the alkene?
13.8 Write chemical equations for combustion reaction of the following
hydrocarbons:

d
(i) Butane (ii) Pentene
(iii) Hexyne (iv) Toluene

he
13.9 Draw the cis and trans structures of hex-2-ene. Which isomer will have
higher b.p. and why?
13.10 Why is benzene extra ordinarily stable though it contains three double
bonds?
13.11 What are the necessary conditions for any system to be aromatic?

is
13.12 Explain why the following systems are not aromatic?

bl
(i) (ii) (iii)

13.13 How will you convert benzene into

pu (i) p-nitrobromobenzene
(iii) p - nitrotoluene
(ii) m- nitrochlorobenzene
(iv) acetophenone?
13.14 In the alkane H3C – CH2 – C(CH3)2 – CH2 – CH(CH3)2, identify 1°,2°,3° carbon
be T

atoms and give the number of H atoms bonded to each one of these.
13.15 What effect does branching of an alkane chain has on its boiling point?
re
13.16 Addition of HBr to propene yields 2-bromopropane, while in the presence
o R

of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain

and give mechanism.
13.17 Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene).
tt E

How does the result support Kekulé structure for benzene?

13.18 Arrange benzene, n-hexane and ethyne in decreasing order of acidic
behaviour. Also give reason for this behaviour.
C

13.19 Why does benzene undergo electrophilic substitution reactions easily

and nucleophilic substitutions with difficulty?
no N

13.20 How would you convert the following compounds into benzene?
(i) Ethyne (ii) Ethene (iii) Hexane
13.21 Write structures of all the alkenes which on hydrogenation give
2-methylbutane.
©

13.22 Arrange the following set of compounds in order of their decreasing

+
relative reactivity with an electrophile, E
(a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene
(b) Toluene, p-H3C – C6H4 – NO2, p-O2N – C6H4 – NO2.
13.23 Out of benzene, m–dinitrobenzene and toluene which will undergo
nitration most easily and why?
13.24 Suggest the name of a Lewis acid other than anhydrous aluminium
chloride which can be used during ethylation of benzene.
13.25 Why is Wurtz reaction not preferred for the preparation of alkanes
containing odd number of carbon atoms? Illustrate your answer by
taking one example.
 398 CHEMISTRY

UNIT 14

ENVIRONMENTAL CHEMISTRY

d
he
The world has achieved brilliance without wisdom, power
without conscience. Ours is a world of nuclear giants and

is
ethical infants.
After studying this unit, you will be
able to

bl
• understand the meaning of
environmental chemistry; You have already studied about environment in your earlier
• define atmospheric pollution, list
pu classes. Environmental studies deal with the sum of all
reasons for global warming. green social, economical, biological, physical and chemical
house effect and acid rain; interrelations with our surroundings. In this unit the focus
will be on environmental chemistry. Environmental
be T

• identify causes for ozone layer

depletion and its effects; chemistry deals with the study of the origin, transport,
reactions, effects and fates of chemical species in the
re
• give reasons for water pollution
o R

and know about international

environment. Let us discuss some important aspects of
standards for drinking water; environmental chemistry.
• describe causes of soil pollution; 14.1 ENVIRONMENTAL POLLUTION
tt E

• suggest and adopt strategies Environmental pollution is the effect of undesirable changes
for control of environmental in our surroundings that have harmful effects on plants,
C

pollution; animals and human beings. A substance, which causes

• appreciate the importance of green pollution, is known as pollutant. Pollutants can be solid,
chemistry in day to day life. liquid or gaseous substances present in greater
no N

concentration than in natural abundance and are

produced due to human activities or due to natural
happenings. Do you know, an average human being
requires nearly 12-15 times more air than the food. So,
©

even small amounts of pollutants in the air become

significant compared to similar levels present in the food.
Pollutants can be degradable, like discarded vegetables
which rapidly break down by natural processes. On the
other hand, pollutants which are slowly degradable, remain
in the environment in an unchanged form for many
decades. For example, substances such as dichlorodi-
phenyltrichloroethane (DDT), plastic materials, heavy
metals, many chemicals, nuclear wastes etc., once released
into the environment are difficult to remove. These
 ENVIRONMENTAL CHEMISTRY 399

pollutants cannot be degraded by natural sulphur dioxide, is a gas that is poisonous to

processes and are harmful to living organisms. both animals and plants. It has been reported
In the process of environmental pollution, that even a low concentration of sulphur
pollutants originate from a source and get dioxide causes respiratory diseases e.g.,
transported by air or water or are dumped into asthma, bronchitis, emphysema in human
the soil by human beings. beings. Sulphur dioxide causes irritation to
the eyes, resulting in tears and redness. High
14.2 ATMOSPHERIC POLLUTION concentration of SO2 leads to stiffness of flower
The atmosphere that surrounds the earth is buds which eventually fall off from plants.

d
not of the same thickness at all heights. There Uncatalysed oxidation of sulphur dioxide is
are concentric layers of air or regions and each slow. However, the presence of particulate

he
layer has different density. The lowest region matter in polluted air catalyses the oxidation
of atmosphere in which the human beings of sulphur dioxide to sulphur trioxide.
along with other organisms live is called
2SO2 (g) +O2 (g) → 2SO3(g)
troposphere. It extends up to the height of
~ 10 km from sea level. Above the troposphere, The reaction can also be promoted by

is
between 10 and 50 km above sea level lies ozone and hydrogen peroxide.
stratosphere. Troposphere is a turbulent,
SO2 (g) +O3 (g) → SO3(g) + O2 (g)
dusty zone containing air, much water vapour

bl
and clouds. This is the region of strong air SO2(g) + H2O2(l) → H2SO4(aq)
movement and cloud formation. The (b) Oxides of Nitrogen: Dinitrogen and
stratosphere, on the other hand, contains dioxygen are the main constituents of air.

vapour.
pu
dinitrogen, dioxygen, ozone and little water These gases do not react with each other at a
normal temperature. At high altitudes when
be T

Atmospheric pollution is generally studied lightning strikes, they combine to form oxides
as tropospheric and stratospheric pollution. of nitrogen. NO2 is oxidised to nitrate ion, NO3−
re
The presence of ozone in the stratosphere which is washed into soil, where it serves as a
o R

prevents about 99.5 per cent of the sun’s fertilizer. In an automobile engine, (at high
harmful ultraviolet (UV) radiations from temperature) when fossil fuel is burnt,
reaching the earth’s surface and thereby dinitrogen and dioxygen combine to yield
tt E

protecting humans and other animals from its significant quantities of nitric oxide (NO) and
effect. nitrogen dioxide ( NO2 ) as given below:
C

14.2.1 Tropospheric Pollution N2 (g) + O2 (g) ⎯⎯⎯⎯

1483K
→ 2NO(g)
Tropospheric pollution occurs due to the NO reacts instantly with oxygen to give NO2
presence of undesirable solid or gaseous
no N

2NO (g) + O2 (g) → 2NO2 (g)

particles in the air. The following are the major
gaseous and particulate pollutants present in Rate of production of NO2 is faster when
the troposphere: nitric oxide reacts with ozone in the
stratosphere.
©

1. Gaseous air pollutants: These are oxides

of sulphur, nitrogen and carbon, hydrogen NO (g) + O3 (g) → NO2 (g) + O2 (g)
sulphide, hydrocarbons, ozone and other The irritant red haze in the traffic and
oxidants. congested places is due to oxides of nitrogen.
Higher concentrations of NO2 damage the
2. Particulate pollutants: These are dust,
leaves of plants and retard the rate of
mist, fumes, smoke, smog etc.
photosynthesis. Nitrogen dioxide is a lung
1. Gaseous air pollutants irritant that can lead to an acute respiratory
(a) Oxides of Sulphur: Oxides of sulphur disease in children. It is toxic to living tissues
are produced when sulphur containing fossil also. Nitrogen dioxide is also harmful to
fuel is burnt. The most common species, various textile fibres and metals.
 400 CHEMISTRY

(c) Hydrocarbons: Hydrocarbons are atmosphere. With the increased use of fossil
composed of hydrogen and carbon only and fuels, a large amount of carbon dioxide gets
are formed by incomplete combustion of fuel released into the atmosphere. Excess of CO2
used in automobiles. Hydrocarbons are in the air is removed by green plants and this
carcinogenic, i.e., they cause cancer. They maintains an appropriate level of CO2 in the
harm plants by causing ageing, breakdown of atmosphere. Green plants require CO2 for
tissues and shedding of leaves, flowers and photosynthesis and they, in turn, emit oxygen,
twigs. thus maintaining the delicate balance. As you
(d) Oxides of Carbon know, deforestation and burning of fossil fuel

d
(i ) Carbon monoxide: Carbon monoxide (CO) increases the CO2 level and disturb the balance
is one of the most serious air pollutants. It is a in the atmosphere. The increased amount of

he
colourless and odourless gas, highly CO2 in the air is mainly responsible for global
poisonous to living beings because of its ability warming.
to block the delivery of oxygen to the organs Global Warming and Greenhouse Effect
and tissues. It is produced as a result of About 75 % of the solar energy reaching the

is
incomplete combustion of carbon. Carbon earth is absorbed by the earth’s surface, which
monoxide is mainly released into the air by increases its temperature. The rest of the heat
automobile exhaust. Other sources, which radiates back to the atmosphere. Some of the

bl
produce CO, involve incomplete combustion heat is trapped by gases such as carbon
of coal, firewood, petrol, etc. The number of dioxide, methane, ozone, chlorofluorocarbon
vehicles has been increasing over the years all
pu compounds (CFCs) and water vapour in the
over the world. Many vehicles are poorly atmosphere. Thus, they add to the heating of
maintained and several have inadequate the atmosphere. This causes global warming.
pollution control equipments resulting in the
We all know that in cold places flowers,
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release of greater amount of carbon monoxide

vegetables and fruits are grown in glass
and other polluting gases. Do you know why
re
covered areas called greenhouse. Do you
carbon monoxide is poisonous? It binds to
o R

know that we humans also live in a

haemoglobin to form carboxyhaemoglobin,
greenhouse? Of course, we are not surrounded
which is about 300 times more stable than the
by glass but a blanket of air called the
tt E

oxygen-haemoglobin complex. In blood, when

atmosphere, which has kept the temperature
the concentration of carboxyhaemoglobin
on earth constant for centuries. But it is now
reaches about 3–4 per cent, the oxygen
undergoing change, though slowly. Just as
C

carrying capacity of blood is greatly

the glass in a greenhouse holds the sun’s
reduced. This oxygen deficiency, results into
warmth inside, atmosphere traps the sun’s
headache, weak eyesight, nervousness and
heat near the earth’s surface and keeps it
no N

cardiovascular disorder. This is the reason why

warm. This is called natural greenhouse
people are advised not to smoke. In pregnant effect because it maintains the temperature
women who have the habit of smoking the and makes the earth perfect for life. In a
increased CO level in blood may induce greenhouse, solar radiations pass through
©

premature birth, spontaneous abortions and the transparent glass and heat up the soil
deformed babies. and the plants. The warm soil and plants emit
(ii) Carbon dioxide: Carbon dioxide (CO2) is infrared radiations. Since glass is opaque to
released into the atmosphere by respiration, infrared radiations (thermal region), it partly
burning of fossil fuels for energy, and by reflects and partly absorbs these radiations.
decomposition of limestone during the This mechanism keeps the energy of the
manufacture of cement. It is also emitted sun trapped in the greenhouse. Similarly,
during volcanic eruptions. Carbon dioxide gas carbon dioxide molecules also trap heat as
is confined to troposphere only. Normally it they are transparent to sunlight but not
forms about 0.03 per cent by volume of the to the heat radiation. If the amount of
 ENVIRONMENTAL CHEMISTRY 401

carbon dioxide crosses the delicate proportion

of 0.03 per cent, the natural greenhouse Think it Over
balance may get disturbed. Carbon dioxide is What can we do to reduce the rate of global
the major contributor to global warming. warming?
Besides carbon dioxide, other greenhouse If burning of fossil fuels, cutting down
forests and trees add to greenhouse gases
gases are methane, water vapour, nitrous
in the atmosphere, we must find ways to
oxide, CFCs and ozone. Methane is produced use these just efficiently and judiciously.
naturally when vegetation is burnt, digested One of the simple things which we can do
or rotted in the absence of oxygen. Large

d
to reduce global warming is to minimise the
amounts of methane are released in paddy use of automobiles. Depending upon the
fields, coal mines, from rotting garbage dumps situation, one can use bicycle, public

he
and by fossil fuels. Chlorofluorocarbons (CFCs) transport system, or go for carpool. We
are man-made industrial chemicals used in should plant more trees to increase the
air conditioning etc. CFCs are also damaging green cover. Avoid burning of dry leaves,
wood etc. It is illegal to smoke in public
the ozone layer (Section 14.2.2). Nitrous oxide
places and work places, because it is

is
occurs naturally in the environment. In recent
harmful not only for the one who is smoking
years, their quantities have increased but also for others, and therefore, we should
significantly due to the use of chemical avoid it. Many people do not understand

bl
fertilizers and the burning of fossil fuels. If the greenhouse effect and the global
these trends continue, the average global warming. We can help them by sharing the
temperature will increase to a level which may
pu information that we have.
lead to melting of polar ice caps and flooding
of low lying areas all over the earth. Increase Acid rain
in the global temperature increases the We are aware that normally rain water has a
be T

+
incidence of infectious diseases like dengue, pH of 5.6 due to the presence of H ions formed
malaria, yellow fever, sleeping sickness etc. by the reaction of rain water with carbon
re
o R
tt E
C
no N
©

Fig. 14.1 Acid deposition

 402 CHEMISTRY

dioxide present in the atmosphere.

Activity 1
H2O (l) + CO2 (g) H2CO3 (aq) You can collect samples of water from
+ – nearby places and record their pH values.
H2CO3 (aq) H (aq) + HCO (aq)
3
Discuss your results in the class. Let us
When the pH of the rain water drops below discuss how we can help to reduce the
5.6, it is called acid rain. formation of acid rain.
Acid rain refers to the ways in which acid This can be done by reducing the
from the atmosphere is deposited on the emission of sulphur dioxide and nitrogen
dioxide in the atmosphere. We should use

d
earth’s surface. Oxides of nitrogen and
sulphur which are acidic in nature can be less vehicles driven by fossil fuels; use less
blown by wind along with solid particles in the sulphur content fossil fuels for power

he
plants and industries. We should use
atmosphere and finally settle down either on
natural gas which is a better fuel than coal
the ground as dry deposition or in water, fog or use coal with less sulphur content.
and snow as wet deposition. (Fig. 14.1) Catalytic converters must be used in cars
Acid rain is a byproduct of a variety of to reduce the effect of exhaust fumes on

is
human activities that emit the oxides of the atmosphere. The main component of
sulphur and nitrogen in the atmosphere. As the converter is a ceramic honeycomb
mentioned earlier, burning of fossil fuels (which coated with precious metals — Pd, Pt and

bl
Rh. The exhaust gases containing unburnt
contain sulphur and nitrogenous matter) such
fuel, CO and NOx, when pass through the
as coal and oil in power stations and furnaces converter at 573 K, are converted into CO2
or petrol and diesel in motor engines produce
pu and N2. We can also reduce the acidity of
sulphur dioxide and nitrogen oxides. SO2 and the soil by adding powdered limestone to
NO2 after oxidation and reaction with water neutralise the acidity of the soil. Many
are major contributors to acid rain, because people do not know of acid rain and its
be T

polluted air usually contains particulate harmful effects. We can make them aware
matter that catalyse the oxidation. by passing on this information and save
re
the Nature.
o R

2SO2 (g) + O2 (g) + 2H2O (l) → 2H2SO4 (aq)

4NO2 (g) + O2 (g)+ 2H2O (l) → 4HNO3 (aq) Taj Mahal and Acid Rain
tt E

Ammonium salts are also formed and can The air around the city of Agra, where the
be seen as an atmospheric haze (aerosol of fine Taj Mahal is located, contains fairly high
particles). Aerosol particles of oxides or levels of sulphur and nitrogen oxides. It is
C

ammonium salts in rain drops result in wet- mainly due to a large number of industries
deposition. SO2 is also absorbed directly on and power plants around the area. Use of
both solid and liquid ground surfaces and is poor quality of coal, kerosene and firewood
no N

as fuel for domestic purposes add up to

thus deposited as dry-deposition.
this problem. The resulting acid rain
Acid rain is harmful for agriculture, trees reacts with marble, CaCO3 of Taj Mahal
and plants as it dissolves and washes away (CaCO 3 +H 2 SO 4 → CaSO 4 + H 2 O+ CO 2 )
nutrients needed for their growth. It causes causing damage to this wonderful
©

respiratory ailments in human beings and monument that has attracted people from
animals. When acid rain falls and flows as around the world. As a result, the
ground water to reach rivers, lakes etc. it affects monument is being slowly disfigured and
plants and animal life in aquatic ecosystem. It the marble is getting discoloured and
corrodes water pipes resulting in the leaching lustreless. The Government of India
announced an action plan in early 1995
of heavy metals such as iron, lead and copper
to prevent the disfiguring of this historical
into the drinking water. Acid rain damages
monument. Mathura refinery has already
buildings and other structures made of stone taken suitable measures to check the
or metal. The Taj Mahal in India has been emission of toxic gases.
affected by acid rain.
 ENVIRONMENTAL CHEMISTRY 403

This plan aims at clearing the air in herbicides and insecticides that miss their
the ‘Taj Trapezium’– an area that includes targets and travel through air and form
the towns of Agra, Firozabad, Mathura and mists.
Bharatpur. Under this plan more than
(d) Fumes are generally obtained by the
2000 polluting industries lying inside the
trapezium would switch over to the use of condensation of vapours during
natural gas or liquefied petroleum gas sublimation, distillation, boiling and
instead of coal or oil. A new natural gas several other chemical reactions. Generally,
pipeline would bring more than half a organic solvents, metals and metallic

d
million cubic metres of natural gas a day oxides form fume particles.
to this area. People living in the city will
also be encouraged to use liquefied
The effect of particulate pollutants are

he
petroleum gas in place of coal, kerosene or largely dependent on the particle size. Air-
firewood. Vehicles plying on highways in borne particles such as dust, fumes, mist etc.,
the vicinity of Taj would be encouraged to are dangerous for human health. Particulate
use low sulphur content diesel. pollutants bigger than 5 microns are likely to
lodge in the nasal passage, whereas particles

is
2. Particulate Pollutants of about 10 micron enter into lungs easily.
Particulates pollutants are the minute solid Lead used to be a major air pollutant

bl
particles or liquid droplets in air. These are emitted by vehicles. Leaded petrol used to be
present in vehicle emissions, smoke particles the primary source of air-borne lead emission
from fires, dust particles and ash from
pu in Indian cities. This problem has now been
industries. Particulates in the atmosphere overcome by using unleaded petrol in most of
may be viable or non-viable. The viable the cities in India. Lead interferes with the
particulates e.g., bacteria, fungi, moulds, development and maturation of red blood cells.
be T

algae etc., are minute living organisms that are Smog

re
dispersed in the atmosphere. Human beings The word smog is derived from smoke and fog.
o R

are allergic to some of the fungi found in air. This is the most common example of air
They can also cause plant diseases. pollution that occurs in many cities
Non-viable particulates may be classified throughout the world. There are two types of
tt E

according to their nature and size as follows: smog:

(a) Smoke particulates consist of solid or (a) Classical smog occurs in cool humid
C

mixture of solid and liquid particles formed climate. It is a mixture of smoke, fog and
during combustion of organic matter. sulphur dioxide. Chemically it is a
Examples are cigarette smoke, smoke from reducing mixture and so it is also called
no N

burning of fossil fuel, garbage and dry as reducing smog.

leaves, oil smoke etc. (b) Photochemical smog occurs in warm, dry
(b) Dust is composed of fine solid particles and sunny climate. The main components
(over 1μm in diameter), produced during
©

of the photochemical smog result from the

crushing, grinding and attribution of solid action of sunlight on unsaturated
materials. Sand from sand blasting, saw hydrocarbons and nitrogen oxides
dust from wood works, pulverized coal, produced by automobiles and factories.
cement and fly ash from factories, dust Photochemical smog has high
storms etc., are some typical examples of concentration of oxidising agents and is,
this type of particulate emission. therefore, called as oxidising smog.
(c) Mists are produced by particles of spray Formation of photochemical smog
liquids and by condensation of vapours in When fossil fuels are burnt, a variety of
air. Examples are sulphuric acid mist and pollutants are emitted into the earth’s
 404 CHEMISTRY

troposphere. Two of the pollutants that are to produce chemicals such as formaldehyde,
emitted are hydrocarbons (unburnt fuels) and acrolein and peroxyacetyl nitrate (PAN).
nitric oxide (NO). When these pollutants build 3CH4 + 2O3 → 3CH2 = O + 3H2O
up to sufficiently high levels, a chain reaction Formaldehyde
occurs from their interaction with sunlight in
which NO is converted into nitrogen dioxide CH2=CHCH=O CH3COONO2
(NO2). This NO2 in turn absorbs energy from Acrolein ⏐⏐
sunlight and breaks up into nitric oxide and O
free oxygen atom (Fig. 14.2). Peroxyacetyl nitrate (PAN)

d
NO2(g) NO(g) + O(g) (i) Effects of photochemical smog

he
Oxygen atoms are very reactive and The common components of photochemical
combine with the O2 in air to produce ozone. smog are ozone, nitric oxide, acrolein,
formaldehyde and peroxyacetyl nitrate (PAN).
O(g) + O2 (g) O3 (g) (ii) Photochemical smog causes serious health
The ozone formed in the above reaction (ii) problems. Both ozone and PAN act as powerful

is
reacts rapidly with the NO(g) formed in the eye irritants. Ozone and nitric oxide irritate the
reaction (i) to regenerate NO2. NO2 is a brown nose and throat and their high concentration
gas and at sufficiently high levels can causes headache, chest pain, dryness of the

bl
contribute to haze. throat, cough and difficulty in breathing.
Photochemical smog leads to cracking of
NO (g) + O3 (g) → NO2 (g) + O2 (g)
pu (iii) rubber and extensive damage to plant life. It
Ozone is a toxic gas and both NO2 and O3 also causes corrosion of metals, stones,
are strong oxidising agents and can react with building materials, rubber and painted
the unburnt hydrocarbons in the polluted air surfaces.
be T
re
o R
tt E
C
no N
©

Fig. 14.2 Photochemical smog occurs where sunlight acts on vehicle pollutants.
 ENVIRONMENTAL CHEMISTRY 405

How can photochemical smog be in the production of plastic foam and by the
controlled ? electronic industry for cleaning computer
Many techniques are used to control or reduce parts etc. Once CFCs are released in the
the formation of photochemical smog. If we atmosphere, they mix with the normal
control the primary precursors of atmospheric gases and eventually reach the
photochemical smog, such as NO 2 and stratosphere. In stratosphere, they get broken
hydrocarbons, the secondary precursors such down by powerful UV radiations, releasing
as ozone and PAN, the photochemical smog chlorine free radical.
will automatically be reduced. Usually catalytic •

d
CF2Cl2 (g) (g) + C F2Cl (g) (i)
converters are used in the automobiles, which
prevent the release of nitrogen oxide and The chlorine radical then react with

he
hydrocarbons to the atmosphere. Certain stratospheric ozone to form chlorine monoxide
plants e.g., Pinus, Juniparus, Quercus, Pyrus radicals and molecular oxygen.
• •
and Vitis can metabolise nitrogen oxide and C l (g) + O3 (g) → Cl O (g) + O2 (g) (ii)
therefore, their plantation could help in this
Reaction of chlorine monoxide radical with

is
matter.
atomic oxygen produces more chlorine
14.2.2 Stratospheric Pollution radicals.
• •
Formation and Breakdown of Ozone

bl
Cl O (g) + O (g) → C l (g) + O2 (g) (iii)
The upper stratosphere consists of The chlorine radicals are continuously
considerable amount of ozone (O3), which
pu regenerated and cause the breakdown of
protects us from the harmful ultraviolet (UV)
ozone. Thus, CFCs are transporting agents for
radiations (λ 255 nm) coming from the sun.
continuously generating chlorine radicals into
These radiations cause skin cancer
the stratosphere and damaging the ozone layer.
be T

(melanoma) in humans. Therefore, it is

important to maintain the ozone shield. The Ozone Hole
re
Ozone in the stratosphere is a product of In 1980s atmospheric scientists working in
o R

UV radiations acting on dioxygen (O 2 ) Antarctica reported about depletion of ozone

molecules. The UV radiations split apart layer commonly known as ozone hole over the
South Pole. It was found that a unique set of
tt E

molecular oxygen into free oxygen (O) atoms.

These oxygen atoms combine with the conditions was responsible for the ozone hole.
molecular oxygen to form ozone. In summer season, nitrogen dioxide and
C

methane react with chlorine monoxide

O2 (g) O(g) + O(g) (reaction iv) and chlorine atoms (reaction v)
O(g) + O2 (g) O3 (g) forming chlorine sinks, preventing much ozone
no N

depletion, whereas in winter, special type of

Ozone is thermodynamically unstable and
clouds called polar stratospheric clouds are
decomposes to molecular oxygen. Thus, a
formed over Antarctica. These polar
dynamic equilibrium exists between the
stratospheric clouds provide surface on which
production and decomposition of ozone
©

chlorine nitrate formed (reaction iv) gets

molecules. In recent years, there have been
hydrolysed to form hypochlorous acid
reports of the depletion of this protective ozone
(reaction (vi)). It also reacts with hydrogen
layer because of the presence of certain
chloride produced as per reaction (v) to give
chemicals in the stratosphere. The main
molecular chlorine.
reason of ozone layer depletion is believed to •
be the release of chlorofluorocarbon Cl O (g) + NO2 (g) → ClONO2(g) (iv)
• •
compounds (CFCs), also known as freons. C l (g) + CH4 (g) → C H3(g) + HCl(g) (v)
These compounds are nonreactive, non
ClONO2(g) + H2O (g) → HOCl (g) + HNO3 (g) (vi)
flammable, non toxic organic molecules and
therefore used in refrigerators, air conditioners, ClONO2(g) + HCl (g) → Cl2 (g) + HNO3 (g) (vii)
 406 CHEMISTRY

When sunlight returns to the Antarctica in where pollutants enter the water-source. Non
the spring, the sun’s warmth breaks up the point sources of pollution are those where a
clouds and HOCl and Cl2 are photolysed by source of pollution cannot be easily identified,
sunlight, as given in reactions (viii) and (ix). e.g., agricultural run off (from farm, animals
• • and crop-lands), acid rain, storm-water
HOCl (g) ⎯⎯⎯
hν
→ O H (g) + C l(g) (viii)
drainage (from streets, parking lots and lawns),
•
Cl2 (g) ⎯⎯⎯
hν
→ 2 C l (g) (ix) etc. Table 14.1 lists the major water pollutants
and their sources.
The chlorine radicals thus formed, initiate

d
the chain reaction for ozone depletion as 14.3.1 Causes of Water Pollution
described earlier. (i) Pathogens: The most serious water

he
Effects of Depletion of the Ozone Layer pollutants are the disease causing agents
called pathogens. Pathogens include bacteria
With the depletion of ozone layer, more UV
and other organisms that enter water from
radiation filters into troposphere. UV
domestic sewage and animal excreta. Human
radiations lead to ageing of skin, cataract,
excreta contain bacteria such as Escherichia

is
sunburn, skin cancer, killing of many
coli and Streptococcus faecalis which cause
phytoplanktons, damage to fish productivity
gastrointestinal diseases.
etc. It has also been reported that plant
(ii) Organic wastes: The other major water

bl
proteins get easily affected by UV radiations
which leads to the harmful mutation of cells. pollutant is organic matter such as
It also increases evaporation of surface water leaves, grass, trash etc. They pollute water as
through the stomata of the leaves and
pu a consequence of run off. Excessive
decreases the moisture content of the soil. phytoplankton growth within water is also a
Increase in UV radiations damage paints and cause of water pollution. These wastes are
biodegradable.
be T

fibres, causing them to fade faster.

The large population of bacteria
re
14.3 WATER POLLUTION decomposes organic matter present in water.
o R

Water is essential for life. Without water there They consume oxygen dissolved in water. The
would be no life. We usually take water as amount of oxygen that water can hold in the
granted for its purity, but we must ensure the solution is limited. In cold water, dissolved
tt E

quality of water. Pollution of water originates oxygen (DO) can reach a concentration up to
from human activities. Through different 10 ppm (parts per million), whereas oxygen in
paths, pollution reaches surface or ground air is about 200,000 ppm. That is why even a
C

water. Easily identified source or place of moderate amount of organic matter when
pollution is called as point source. e.g., decomposes in water can deplete the water of
no N

municipal and industrial discharge pipes its dissolved oxygen. The concentration of

Table 14.1 Major Water Pollutants

Pollutant Source
©

Micro-organisms Domestic sewage

Organic wastes Domestic sewage, animal excreta and waste, decaying animals
and plants, discharge from food processing factories.
Plant nutrients Chemcial fertilizers
Toxic heavy metals Industries and chemical factories
Sediments Erosion of soil by agriculture and strip mining
Pesticides Chemicals used for killing insects, fungi and weeds
Radioactive substances Mining of uranium containing minerals
Heat Water used for cooling in industries
 ENVIRONMENTAL CHEMISTRY 407

dissolved oxygen in water is very important The organic chemicals are another group
for aquatic life . If the concentration of dissolved of substances that are found in polluted water.
oxygen of water is below 6 ppm, the growth of Petroleum products pollute many sources of
fish gets inhibited. Oxygen reaches water water e.g., major oil spills in oceans. Other
either through atmosphere or from the process organic substances with serious impacts are
of photosynthesis carried out by many the pesticides that drift down from sprays or
aquatic green plants during day light. runoff from lands. Various industrial
However, during night, photosynthesis stops chemicals like polychlorinated biphenyls,
but the plants continue to respire, resulting (PCBs) which are used as cleansing solvent,

d
in reduction of dissolved oxygen. The detergents and fertilizers add to the list of
dissolved oxygen is also used by water pollutants. PCBs are suspected to be

he
microorganisms to oxidise organic matter. carcinogenic. Nowadays most of the detergents
available are biodegradable. However, their use
If too much of organic matter is added to
can create other problems. The bacteria
water, all the available oxygen is used up. This
responsible for degrading biodegradable
causes oxygen dependent aquatic life to die.
detergent feed on it and grow rapidly. While

is
Thus, anaerobic bacteria (which do not require
growing, they may use up all the oxygen
oxygen) begin to break down the organic waste
dissolved in water. The lack of oxygen kills all
and produce chemicals that have a foul smell

bl
other forms of aquatic life such as fish and
and are harmful to human health. Aerobic
plants. Fertilizers contain phosphates as
(oxygen requiring) bacteria degrade these
additives. The addition of phosphates in water
organic wastes and keep the water depleted
enhances algae growth. Such profuse growth
pu
in dissolved oxygen.
Thus, the amount of oxygen required by
of algae, covers the water surface and reduces
the oxygen concentration in water. This leads
be T

bacteria to break down the organic matter to anaerobic conditions, commonly with
present in a certain volume of a sample of accumulation of abnoxious decay and animal
re
water, is called Biochemical Oxygen Demand death. Thus, bloom-infested water inhibits the
o R

(BOD). The amount of BOD in the water is a growth of other living organisms in the
measure of the amount of organic material in water body. This process in which nutrient
the water, in terms of how much oxygen will enriched water bodies support a dense plant
tt E

be required to break it down biologically. Clean population, which kills animal life by depriving
water would have BOD value of less than it of oxygen and results in subsequent loss of
C

5 ppm whereas highly polluted water could biodiversity is known as Eutrophication.

have a BOD value of 17 ppm or more.
14.3.2 International Standards for
(iii) Chemical Pollutants: As we know that Drinking Water
no N

water is an excellent solvent, water soluble

The International Standards for drinking water
inorganic chemicals that include heavy metals
are given below and they must be followed.
such as cadmium, mercury, nickel etc
constitute an important class of pollutants. All Fluoride: For drinking purposes, water
©

these metals are dangerous to humans should be tested for fluoride ion concentration.
because our body cannot excrete them. Over Its deficiency in drinking water is harmful to
the time, it crosses the tolerance limit. These man and causes diseases such as tooth decay
metals then can damage kidneys, central etc. Soluble fluoride is often added to drinking
nervous system, liver etc. Acids (like sulphuric water to bring its concentration upto 1 ppm
–3 –
acid) from mine drainage and salts from many or 1 mg dm . The F ions make the enamel on
different sources including raw salt used to teeth much harder by converting
melt snow and ice in the colder climates hydroxyapatite, [3(Ca3(PO4)2.Ca(OH)2], the
(sodium and calcium chloride) are water enamel on the surface of the teeth, into much
soluble chemical pollutants. harder fluorapatite, [3(Ca 3 (PO 4 ) 2 .CaF 2 ].
 408 CHEMISTRY

–
However, F ion concentration above 2 ppm pollution levels. Ensure that appropriate
causes brown mottling of teeth. At the same action is taken. You can write to the press
time, excess fluoride (over 10 ppm) causes also. Do not dump waste into a
harmful effect to bones and teeth, as reported household or industrial drain which can
from some parts of Rajasthan. enter directly to any water body, such as,
Lead: Drinking water gets contaminated with river, pond, stream or lake. Use compost
lead when lead pipes are used for instead of chemical fertilizers in gardens.
transportation of water. The prescribed upper Avoid the use of pesticides like DDT,
malathion etc., at home and try to use

d
limit concentration of lead in drinking water
is about 50 ppb. Lead can damage kidney, dried neem leaves to help keep insects
liver, reproductive system etc. away. Add a few crystals of potassium

he
permanganate (KMnO 4) or bleaching
Sulphate: Excessive sulphate (>500 ppm) in powder to the water tank of your house.
drinking water causes laxative effect, otherwise
at moderate levels it is harmless.
14.4 SOIL POLLUTION
Nitrate: The maximum limit of nitrate in

is
India being an agriculture based economy
drinking water is 50 ppm. Excess nitrate in
gives high priority to agriculture, fisheries and
drinking water can cause disease such as
livestock development. The surplus

bl
methemoglobinemia (‘blue baby’ syndrome).
production is stored by governmental and
Other metals: The maximum concentration non-governmental organisations for the lean
of some common metals recommended in
pu season. The food loss during the storage also
drinking water are given in Table 14.2. needs special attention. Have you ever seen the
damages caused to the crops, food items by
insects, rodents, weeds and crop diseases etc?
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Table 14.2 Maximum Prescribed Concen- How can we protect them? You are acquainted
tration of Some Metals in
with some insecticides and pesticides for
re
Drinking Water.
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protection of our crops. However, these

Metal Maximum concentration insecticides, pesticides and herbicides cause
–3
(ppm or mg dm ) soil pollution. Hence, there is a need for their
tt E

Fe 0.2 judicious use.

Mn 0.05 14.4.1 Pesticides
C

Al 0.2 Prior to World War II, many naturally

Cu 3.0 occurring chemicals such as nicotine (by
Zn 5.0 planting tobacco plants in the crop field), were
no N

Cd 0.005 used as pest controlling substance for major

crops in agricultural practices.
Activity 2 During World War II, DDT was found to be
of great use in the control of malaria and other
©

You can visit local water sources and insect-borne diseases. Therefore, after the war,
observe if the river/lake/tank/pond are DDT was put to use in agriculture to control
unpolluted/slightly polluted/ moderately the damages caused by insects, rodents, weeds
polluted or severely polluted by looking and various crop diseases. However, due to
at water or by checking pH of water. adverse effects, its use has been banned in
Document the name of the river and the India.
nearby urban or industrial site from
where the pollution is generated. Inform Pesticides are basically synthetic toxic
about this to Pollution Control Board’s chemicals with ecological repercussions. The
office set up by Government to measure repeated use of the same or similar pesticides
give rise to pests that are resistant to that
 ENVIRONMENTAL CHEMISTRY 409

group of pesticides thus making the pesticides sodium chlorate (NaClO3), sodium arsinite
ineffective. Therefore, as insect resistance of (Na3AsO3) and many others. During the first
DDT increased, other organic toxins such as half of the last century, the shift from
Aldrin and Dieldrin were introduced in the mechanical to chemical weed control had
market by pesticide industry. Most of the provided the industry with flourishing
organic toxins are water insoluble and non- economic market. But one must remember that
biodegradable. These high persistent toxins these are also not environment friendly.
are, therefore, transferred from lower trophic Most herbicides are toxic to mammals but
level to higher trophic level through food chain are not as persistent as organo-chlorides.

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(Fig.14.3). Over the time, the concentration of These chemicals decompose in a few months.
toxins in higher animals reach a level which Like organo-chlorides, these too become

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causes serious metabolic and physiological concentrated in the food web. Some herbicides
disorders. cause birth defects. Studies show that corn-
fields sprayed with herbicides are more prone
to insect attack and plant disease than fields

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that are weeded manually.
Pesticides and herbicides represent only a
very small portion of widespread chemical

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pollution. A large number of other compounds
that are used regularly in chemical and
pu industrial processes for manufacturing
activities are finally released in the atmosphere
in one or other form.
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14.5 INDUSTRIAL WASTE

Industrial solid wastes are also sorted out as
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biodegradable and non-degradable wastes.
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Biodegradable wastes are generated by cotton

mills, food processing units, paper mills, and
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textile factories.
Non-biodegradable wastes are generated
by thermal power plants which produce fly
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Fig. 14.3 At each trophic level, the pollutant ash; integrated iron and steel plants which
gets 10 times concentrated. produce blast furnace slag and steel melting
In response to high persistence of slag. Industries manufacturing aluminium,
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chlorinated organic toxins, a new series of less zinc and copper produce mud and tailings.
persistent or more bio-degradable products Fertilizer industries produce gypsum.
called organo-phosphates and carbamates Hazardous wastes such as inflammables,
have been introduced in the market. But these composite explosives or highly reactive
©

chemicals are severe nerve toxins and hence substances are produced by industries
more harmful to humans. As a result, there dealing in metals, chemicals, drugs, pharma-
are reports of some pesticides related deaths ceuticals, dyes, pesticides, rubber goods etc.
of agricultural field workers. Insects have The disposal of non-degradable industrial
become resistant to these insecticides also. The solid wastes, if not done by a proper and
insecticide industry is engaged in developing suitable method, may cause serious threat to
new groups of insecticides. But one has to the environment. New innovations have led to
think, is this the only solution to pest menace? different uses of waste material. Nowadays,
These days, the pesticide industry has fly ash and slag from the steel industry are
shifted its attention to herbicides such as utilised by the cement industry. Large
 410 CHEMISTRY

quantities of toxic wastes are usually destroyed household discards, there are medical,
by controlled incineration, whereas small agricultural, industrial and mining wastes. The
quantities are burnt along with factory improper disposal of wastes is one of the major
garbage in open bins. Moreover, solid wastes causes of environmental degradation.
if not managed effectively, affect the Therefore, the management of wastes is of
components of the environment. utmost importance.
Collection and Disposal
Do you know about waste recycling?
Domestic wastes are collected in small bins,
•

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Fuel obtained from plastic waste has which are then transferred to community bins
high octane rating. It contains no lead by private or municipal workers. From these
and is known as “green fuel”. community bins, these are collected and

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• Due to recent developments made in carried to the disposable site. At the site,
chemical and textile industries, clothes garbage is sorted out and separated into
will be made from recycled plastic biodegradable and non-biodegradable
waste. These will be available soon in materials. Non-biodegradable materials such

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the global textile market. as plastic, glass, metal scraps etc. are sent for
• In India, our cities and towns face recycling. Biodegradable wastes are deposited
endless hours of power cut. We can also in land fills and are converted into compost.

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see piles of rotting garbage here and
The waste if not collected in garbage bins,
there. There is a good news that we can
finds its way into the sewers. Some of it is eaten
get rid from both these problems
pu by cattle. Non-biodegradable wastes like
simultaneously. Technology has now
been developed to produce electricity
polythene bag, metal scraps, etc. choke the
from the garbage. A pilot plant has been sewers and cause inconvenience. Polythene
bags, if swallowed by cattle can cost their lives
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set up, where after removing ferrous

metals, plastic, glass, paper etc. from also.
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garbage, it is mixed with water. It is then As a normal practice, therefore, all
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cultured with bacterial species for domestic wastes should be properly collected
producing methane, commonly known and disposed. The poor management causes
as biogas. The remaining product is health problems leading to epidemics due to
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used as manure and biogas is used to contamination of ground water. It is specially

produce electricity. hazardous for those who are in direct contact
with the waste such as rag pickers and workers
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14.6 STRATEGIES TO CONTROL involved in waste disposal, as they are the ones
ENVIRONMENTAL POLLUTION who handle waste materials mostly without
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protective device such as gloves or water proof

After studying air, water, soil and industrial boots and gas masks. What can you do for
waste pollution in this unit, by now you must them?
have started feeling the need of controlling
environmental pollution: How can you save 14.7 GREEN CHEMISTRY
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your immediate environment? Think of the 14.7.1 Introduction

steps/activities, which you would like to
It is well known fact that self-sufficiency in food
undertake for controlling air, water, soil and th
industrial waste pollution in your has been achieved in India since late 20
neighbourhood. Here, an idea about the century by using fertilizers and pesticides and
strategies for the management of waste is given. exploring improved methods of farming, good
quality seeds, irrigation etc. But over-
14.6.1 Waste Management exploitation of soil and excessive use of
Solid waste is not the only waste, which you fertilizers and pesticides have resulted in the
see in your household garbage box. Besides deterioration of soil, water and air.
 ENVIRONMENTAL CHEMISTRY 411

The solution of this problem does not lie in tetrachloride etc., are highly toxic. One should
stopping the process of development that has be careful while using them.
been set in; but to discover methods, which
As you know, a chemical reaction involves
would help in the reduction of deterioration of
reactants, attacking reagents and the medium
the environment. Green chemistry is a way of
in which the reaction takes place. Extent of any
thinking and is about utilising the existing
knowledge and principles of chemistry and reaction depends upon physical parameters
other sciences to reduce the adverse impact like temperature, pressure and use of catalyst.
on environment. Green chemistry is a In a chemical reaction, if reactants are fully

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production process that would bring about converted into useful environmental friendly
minimum pollution or deterioration to the products by using an environment friendly

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environment. The byproducts generated medium then there would be no chemical
during a process, if not used gainfully, add pollutants introduced in the environment.
to the environmental pollution. Such During a synthesis, care must be taken to
processes are not only environmental
choose starting materials that can be converted
unfriendly but also cost-ineffective. The

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into end products with yield approximately
waste generation and its disposal both are
economically unsound. Utilisation of existing upto 100 per cent. This can be achieved by
knowledge base for reducing the chemical arriving at optimum conditions of synthesis.

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hazards along with the developmental It may be worthwhile to carry out synthetic
activities is the foundation of green reactions in aqueous medium since water has
chemistry. Have you perceived the idea of green
pu high specific heat and low volatility. Water is
chemistry ? It is well known that organic cost effective, noninflammable and devoid of
solvents such as benzene, toluene, carbon any carcinogenic effects.
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Nobel goes to Green Chemists

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Yves Chauvin Robert H. Grubbs Richard R. Schrock

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Yves Chauvin, Institut Français du Pétrole, Rueil-Malmaison France, Robert H. Grubbs

California Institute of Technology (Caltech), Pasadena, CA, USA and Richard R. Schrock
Massachusetts Institute of Technology (MIT), Cambridge, MA, USA won the 2005 Nobel Prize
©

in chemistry for work that reduces hazardous waste in creating new chemicals. The trio won
the award for their development of the metathesis method in organic synthesis –a way to
rearrange groups of atoms within molecules that the Royal Swedish Academy of Sciences
likened to a dance in which couples change partners. The metathesis has tremendous
commercial potential in the pharmaceuticals, biotechnology and food stuffs production
industries. It is also used in the development of revolutionary environmentally-friendlier
polymers.
This represents a great step forward for ‘green chemistry’, reducing potentially hazardous
waste through smarter production. Metathesis is an example of how important application of
basic science is for the benefit of man, society and the environment.
 412 CHEMISTRY

14.7.2 Green Chemistry in day-to-day Life Green chemistry, in a nutshell, is a cost

(i) Dry Cleaning of Clothes effective approach which involves reduction in
material, energy consumption and waste
Tetra chlroroethene (Cl2C=CCl2) was earlier
generation.
used as solvent for dry cleaning. The
compound contaminates the ground water and
is also a suspected carcinogen. The process Think it Over
using this compound is now being replaced What is our responsibility as a human
by a process, where liquefied carbondioxide, being to protect our environment?

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with a suitable detergent is used. Replacement
of halogenated solvent by liquid CO2 will result Some concepts, if followed by an individual,
in less harm to ground water. contribute towards a better quality of our

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environment and human life. Always set up
These days hydrogen peroxide (H2O2) is
a compost tin in your garden or any other
used for the purpose of bleaching clothes in
place in your home and use it to produce
the process of laundary, which gives better
results and makes use of lesser amount of manure for your plants to reduce the use

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water. of fertilizers. Use a cloth bag and avoid
asking for plastic carry bags when you buy
(ii) Bleaching of Paper
groceries, vegetables or any other item. See

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Chlorine gas was used earlier for bleaching that all newspapers, glass, aluminum and
paper. These days, hydrogen peroxide (H2O2)
other items in your area are recycled. We
with suitable catalyst, which promotes the
pu might have to take little trouble to locate
bleaching action of hydrogen peroxide, is used.
such dealers. We must realize that we do
(iii) Synthesis of Chemicals not have solutions for every problem but
Ethanal (CH 3 CHO) is now commercially we can concentrate on issues, which we feel
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prepared by one step oxidation of ethene in strongly about and can do some thing
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the presence of ionic catalyst in aqueous about. We should take care to put into
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medium with a yield of 90%. practice whatever we preach. Always

CH2 = CH2 + O2 ⎯⎯⎯⎯⎯⎯⎯⎯ Catalyst
Pd(II)/ Cu ( II )( in water )
→ remember environment protection begins
with us.
CH3 CHO ( 90% )
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SUMMARY

Environmental chemistry plays a major role in environment. Chemical species present

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in the environment are either naturally occurring or generated by human activities.

Environmental pollution is the effect of undesirable changes in the surrounding that
have harmful effects on plants, animals and human beings. Pollutants exist in all the
three states of matter. We have discussed only those pollutants, which are due to human
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activities, and can be controlled. Atmospheric pollution is generally studied as

tropospheric and stratospheric pollution. Troposphere is the lowest region of the
atmosphere (~10 km) in which man along with other organisms including plants exist.
Whereas stratosphere extends above troposphere up to 50 km above sea level. Ozone
layer is one of the important constituents of stratosphere. Tropospheric pollution is
basically due to various oxides of sulphur, nitrogen, carbon, halogens and also due to
particulate pollutants. The gaseous pollutants come down to the earth in the form of
acid rain. 75% of the solar energy reaching earth is absorbed by the earth surface and
rest is radiated back to the atmosphere. These gases mentioned above trap the heat
which result into global warming. It is important to realise that these very gases are
also responsible for the life on the earth as they trap the requisite amount of solar
 ENVIRONMENTAL CHEMISTRY 413

energy for the sustainance of life. The increase in the greenhouse gases is raising the
temperature of the earth’s atmosphere which, if not checked, may eventually result in
melting of polar ice caps and consequently may submerge the costal land mass. Many
human activities are producing chemicals, which are responsible for the depletion of
ozone layer in the stratosphere, leading to the formation of ozone hole. Through the
ozone hole, ultraviolet radiations can penetrate into the earth’s atmosphere causing
mutation of genes. Water is the elixir of life but the same water, if polluted by pathogens,
organic wastes, toxic heavy metals, pesticides etc., will turn into poison. Therefore, one
should take care to follow international standards to maintain purity levels of drinking

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water. Industrial wastes and excessive use of pesticides, result into pollution of land
mass and water bodies. Judicious use of chemicals required for agricultural practices
can lead to sustainable development. Strategies for controlling environmental

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pollution can be: (i) waste management i.e., reduction of the waste and proper disposal,
also recycling of materials and energy, (ii) adopting methods in day-to-day life, which
results in the reduction of environmental pollution. The second method is a new branch
of chemistry, which is in its infancy known as green chemistry. It utilizes the existing
knowledge and practices so as to bring about reduction in the production of pollutants.

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EXERCISES

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14.1
pu Define environmental chemistry.
14.2 Explain tropospheric pollution in 100 words.
14.3 Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
14.4 List gases which are responsible for greenhouse effect.
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14.5 Statues and monuments in India are affected by acid rain. How?
14.6 What is smog? How is classical smog different from photochemical smogs?
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14.7 Write down the reactions involved during the formation of photochemical smog.
14.8 What are the harmful effects of photochemical smog and how can they be
controlled?
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14.9 What are the reactions involved for ozone layer depletion in the stratosphere?
14.10 What do you mean by ozone hole? What are its consequences?
14.11 What are the major causes of water pollution? Explain.
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14.12 Have you ever observed any water pollution in your area? What measures would
you suggest to control it?
14.13 What do you mean by Biochemical Oxygen Demand (BOD)?
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14.14 Do you observe any soil pollution in your neighbourhood? What efforts will you
make for controlling the soil pollution?
14.15 What are pesticides and herbicides? Explain giving examples.
14.16 What do you mean by green chemistry? How will it help decrease environmental
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pollution?
14.17 What would have happened if the greenhouse gases were totally missing in the
earth’s atmosphere? Discuss.
14.18 A large number of fish are suddenly found floating dead on a lake. There is no
evidence of toxic dumping but you find an abundance of phytoplankton. Suggest
a reason for the fish kill.
14.19 How can domestic waste be used as manure?
14.20 For your agricultural field or garden you have developed a compost producing
pit. Discuss the process in the light of bad odour, flies and recycling of wastes
for a good produce.