Super course in chemistry
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Super Course in Chemistry
PHYSICAL
CHEMISTRY
for IIT-JEE
Volume 1
Trishna Knowledge Systems
A division of
Triumphant Institute of Management Education Pvt. Ltd
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Contents
Preface
Chapter 1
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Basic Concepts of Chemistry
1.1—1.63
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Chapter 2
States of Matter
2.1—2.67
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Chapter 3
Atomic Structure
3.1—3.80
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Chemical Bonding
4.1—4.80
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Chemical Energetics and Thermodynamics
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iv Contents
Chapter 6
Chemical and Ionic Equilibria
6.1—6.90
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Chapter 7
Dilute Solutions and Electrochemistry
7.1—3.77
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Chapter 8
Chemical Kinetics and Solid State
8.1—8.79
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Chapter 9
Surface Chemistry and Nuclear Chemistry
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Preface
The IIT-JEE, the most challenging amongst national level engineering entrance examinations, remains on the top of the
priority list of several lakhs of students every year. The brand value of the IITs attracts more and more students every year,
but the challenge posed by the IIT-JEE ensures that only the best of the aspirants get into the IITs. Students require thorough
understanding of the fundamental concepts, reasoning skills, ability to comprehend the presented situation and exceptional
problem-solving skills to come on top in this highly demanding entrance examination.
The pattern of the IIT-JEE has been changing over the years. Hence an aspiring student requires a step-by-step study
plan to master the fundamentals and to get adequate practice in the various types of questions that have appeared in the
IIT-JEE over the last several years. Irrespective of the branch of engineering study the student chooses later, it is important
to have a sound conceptual grounding in Mathematics, Physics and Chemistry. A lack of proper understanding of these
subjects limits the capacity of students to solve complex problems thereby lessening his / her chances of making it to the
top-notch institutes which provide quality training.
This series of books serves as a source of learning that goes beyond the school curriculum of Class XI and Class XII
and is intended to form the backbone of the preparation of an aspiring student. These books have been designed with the
objective of guiding an aspirant to his / her goal in a clearly defined step-by-step approach.
x Master the Concepts and Concept Strands!
This series covers all the concepts in the latest IIT-JEE syllabus by segregating them into appropriate units. The theories
are explained in detail and are illustrated using solved examples detailing the different applications of the concepts.
x Let us First Solve the Examples – Concept Connectors!
At the end of the theory content in each unit, a good number of “Solved Examples” are provided and they are designed
to give the aspirant a comprehensive exposure to the application of the concepts at the problem-solving level.
x Do Your Exercise – Daily!
Over 200 unsolved problems are presented for practice at the end of every chapter. Hints and solutions for the same are
also provided. These problems are designed to sharpen the aspirant’s problem-solving skills in a step-by-step manner.
x Remember, Practice Makes You Perfect!
We recommend you work out ALL the problems on your own – both solved and unsolved – to enhance the effectiveness of your preparation.
A distinct feature of this series is that unlike most other reference books in the market, this is not authored by an individual. It is put together by a team of highly qualified faculty members that includes IITians, PhDs etc from some of the
best institutes in India and abroad. This team of academic experts has vast experience in teaching the fundamentals and
their application and in developing high quality study material for IIT-JEE at T.I.M.E. (Triumphant Institute of Management Education Pvt. Ltd), the number 1 coaching institute in India. The essence of the combined knowledge of such an
experienced team is what is presented in this self-preparatory series. While the contents of these books have been organized
keeping in mind the specific requirements of IIT-JEE, we are sure that you will find these useful in your preparation for
various other engineering entrance exams also.
We wish you the very best!
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CHAPTER
1
BASIC
CONCEPTS OF
CHEMISTRY
Q Q Q C H A PT E R OU TLIN E
Preview
STUDY MATERIAL
Introduction
Definition and Description of Matter
Dalton’s atomic theory and model of the atom
Mass of an Atom
Valency of elements
Valency and Formula of Radicals
Formulae of Compounds
Mole Concept
Molar Mass
s Concept Strands (1-2)
Chemical Formulae
s Concept Strands (3-6)
Chemical Equation
s Concept Strand (7)
Excess and Limiting Reactants
s Concept Strand (8)
Oxidation Number or Oxidation State
s Concept Strands (9-10)
Balancing Redox Reactions
s Concept Strands (11-13)
Disproportionation Reactions
s Concept Strands (14-17)
Equivalent Weight
Strength of Solutions
s Concept Strands (18-32)
Volumetric Analysis
s Concept Strands (33-37)
Neutralization Reactions
s Concept Strands (38-50)
TOPIC GRIP
s
s
s
s
s
s
Subjective Questions (10)
Straight Objective Type Questions (5)
Assertion–Reason Type Questions (5)
Linked Comprehension Type Questions (6)
Multiple Correct Objective Type Questions (3)
Matrix-Match Type Question (1)
IIT ASSIGNMENT EXERCISE
s
s
s
s
s
Straight Objective Type Questions (80)
Assertion–Reason Type Questions (3)
Linked Comprehension Type Questions (3)
Multiple Correct Objective Type Questions (3)
Matrix-Match Type Question (1)
ADDITIONAL PRACTICE EXERCISE
s
s
s
s
s
s
Subjective Questions (10)
Straight Objective Type Questions (40)
Assertion–Reason Type Questions (10)
Linked Comprehension Type Questions (9)
Multiple Correct Objective Type Questions (8)
Matrix-Match Type Questions (3)
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1.2 Basic Concepts of Chemistry
INTRODUCTION
Chemistry is basically an experimental science and chemists are engaged in studying a variety of reactions in the
laboratory and many reactions find wide applications in
industry. Many industrial processes such as extraction of
metals from their ores, refining of petroleum, manufacture of drugs, pharmaceuticals and high polymers are of
great practical and commercial value. The success of all
such efforts is the result of the strong basic foundation laid
by early generations of chemists. This chapter deals with
some of the fundamental ideas out of which chemical
theory has evolved viz., the atomic theory of matter, mole
concept etc.
DEFINITION AND DESCRIPTION OF MATTER
Matter may be defined as anything which has mass and occupies space.
Some common examples of matter (or synonymously
also called materials) are salt, sand, water, oxygen etc. Matter may also be classified into the three states of aggregation, namely solid, liquid and gas.
The Alchemists discovered many new substances, developed new types of apparatus and perfected some chemical procedures like crystallization from solution, distillation
and separation of mixture etc. Thus, over a period of time,
it became evident that matter could be physically separated
into different pure substances each having a unique set of
properties (crystalline form, colour etc.).
For example, sea water can be separated by distillation into pure water and a residue consisting of a mixture
of several salts. This mixture can be further separated into
different components, each component having different
properties from the others.
The mixtures, consisting of more than one substance,
may again be divided into two categories
(i) Homogeneous mixtures
(ii) Heterogeneous mixtures
Heterogeneous mixtures
These are mixtures whose composition is not uniform
throughout and in which the individual components can be
distinguished from one another.
Some examples are,
(i) an insoluble solid like BaSO4 or AgCl placed in a
beaker containing water.
(ii) a gas in contact with a solid in an enclosed vessel, like
CO2 in contact with solid CaCO3.
(iii) A mixture of sand and salt (or Iron powder and Sulphur).
Elements
Elements are defined as fundamentally inseparable elemental
materials and consist of only one kind of atom.
Solid substances like carbon, liquids like mercury and
gases like helium are elements.
Compounds
Homogeneous mixtures
These are mixtures having uniform composition throughout
and in which the individual components cannot be physically
distinguished from one another.
Air is an example of a homogeneous mixture of several
gases (Nitrogen, Oxygen, Carbon dioxide, water vapour
etc.). So also are solutions, say, glucose in water.
The pure substances that are formed by the combination of
two or more elements (for example, residue from sea water)
are referred to as compounds.
Compounds contain two or more different types of atoms in them, as for example, in compounds like H2O, NH3,
CO2 etc. Further, the elements in a compound do not have
their individual characteristics.
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Basic Concepts of Chemistry
1.3
DALTON’S ATOMIC THEORY AND MODEL OF THE ATOM
Based on the above ideas of matter, elements and compounds, John Dalton put forward his atomic theory in
1803. The three laws of chemical combination, which were
already proposed prior to this period, further led Dalton in
1804 to formulate the existence of atoms on the basis of the
following postulates.
(i) Matter is discrete (i.e., discontinuous) and is made up of
atoms. An atom is the smallest (chemically) particle of
an element, which can take part in a chemical change.
(ii) Atoms of a given element are all similar in all respects,
size, shape, mass etc., but differ from atoms of other
elements.
(iii) An atom of an element has a definite mass.
(iv) Atoms are indestructible i.e., they can neither be
created nor destroyed and chemical reactions involve
only rearrangement of atoms.
A compound like SO2 is suggested to be composed of
a group of atoms. The group of atoms, which is capable of
independent existence, is known as molecule. Thus matter
was structured in terms of atoms and molecules. Some postulates of Dalton’s theory need to be modified in the light of
subsequent developments in this area.
(i) Atom is divisible as known from studies of nuclear
fission.
(ii) Atoms of an element may have different masses due to
existence of isotopes.
MASS OF AN ATOM
Atom cannot be seen or isolated by simple physical/chemical methods. Therefore, the mass of an atom as suggested
by Dalton is based on relative terms viz., the average mass
of one atom relative to the average mass of another. The
method that helped in the development of atomic mass
(relative to that of another) is based on Avogadro’s hypothesis proposed by Avogadro in 1811.
Avogadro’s hypothesis states that equal volumes of all
gases at the same temperature and pressure contain equal
number of particles.
Considering the formation of water from hydrogen
and oxygen, one can write
1 volume of oxygen + 2 volumes of hydrogen =
2 volumes of water vapour
— (1)
Assuming that a unit volume of gas contains x particles, equation (1) may be written as
x particles of oxygen + 2x particles of hydrogen = 2x
particles of water vapour
— (2)
Simplifying equation (2) further, we can write
1 particle of oxygen + 2 particles of hydrogen =
2 particles of water vapour
— (3)
Assuming that each particle of oxygen is a diatomic
oxygen molecule and each particle of hydrogen is a diatomic hydrogen molecule, equation (3) may be written as
1 molecule of oxygen + 2 molecules of hydrogen =
2 molecules of water vapour
or, O2 (g) + 2H2 (g) = 2H2O (g) (the suffix (g) stands for the
designation “gas”)
Considering that the mass ratio of oxygen to hydrogen
is 8:1 (in water) and that in water there are two atoms of
hydrogen and one atom of oxygen, it follows that an atom
of oxygen is 16 times heavier than a hydrogen atom.
The atomic mass scale is currently based on carbon
12-isotope standard since 1961. It assigns a mass of 12 unified mass (u) to the mass of one atom of 6 C12 isotope.
Atomic mass of an element
Average mass of 1 atom of the element
=
1
u mass of 1 atom of C 12
12
— (4)
Thus atomic mass is a relative mass.
For example, the atomic mass of carbon atom is
12.011 u. This is because the element carbon consists of
two isotopes, (i) 98.89 percent of C12 isotope whose mass is
12 u and (ii) 1.11 percent of C13 isotope whose mass is
13.00335 u.
The atomic mass of an element thus depends on the
fractional abundance of the isotopes of that element. The
fractional dependence, in turn, represents the fraction of
the total number of atoms of the isotope concerned.
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1.4 Basic Concepts of Chemistry
1. Relative atomic mass is a number.
2. Gram atomic mass (or) gram-atom is the atomic mass
expressed in grams.
Molecular mass
Molecular mass =
Mass number and atomic mass
Mass number refers to the total number of nucleons (protons
and neutrons) in an atom. It will be the same as atomic mass
if the element does not have an isotope.
19
F does not have isotopes. Thus its mass number is
equal to its atomic mass.
Mass of 1 molecule
1
u mass of 1 C 12 atom
12
(i) Molecular mass is not absolute but relative and is also
known as relative molecular mass.
(ii) Relative molecular mass is a number.
(iii) Molecular mass expressed in grams is the grammolecular mass (or) gram-mole.
VALENCY OF ELEMENTS
Table 1.1
Table 1.2 Elements with variable valency
Valency
Element
H, Li, Na, K, Rb, Cs, F,
Cl, Br, I, Ag
1
Fe
2 (Ferrous)
3 (Ferric)
Cu
1 (Cuprous)
2 (Cupric)
Be, Mg, Ca, Sr, Ba, O,
S, Ni, Zn, Cd
2
Au
1 (Aurous)
B, Al, N, As, Sb, Bi,
3
Sn
2 (Stannous)
4 (Stannic)
C, Si
4
Pb
2 (Plumbous)
4 (Plumbic)
Elements
Hg
Valencies
3 (Auric)
2+
2
1(Mercurous Hg ) 2 (Mercuric)
VALENCY AND FORMULA OF RADICALS
Table 1.3
Valency 2
Valency 1
Radical
Formula
Radical
Hypochlorite
Ammonium
NH4+
Hydroxide
OH
Iodate
Nitrate
NO3
Periodate
Nitrite
NO2
Meta aluminate
Permanganate
MnO4 Meta borate
Formula
ClO
IO3
IO4
AlO2
BO2
Bisulphate
HSO4
Cyanide
CN
Bisulphite
HSO3
Isocyanide
NC
Bicarbonate
HCO3
Cyanate
CNO
Dihydrogen
phosphate
Chlorate
Perchlorate
H2PO4 Isocyanate
ClO3
ClO4
Thiocyanate
Isothiocyanate
NCO
CNS
NCS
2
Carbonate
CO3
Dichromate
Cr2O72
Sulphate
SO42
Zincate
ZnO22
Sulphite
SO32
Tetrathionate
S4O62
Thiosulphate
S2O32
Oxalate
C2O42
Hydrogen
phosphate
HPO42
Silicate
SiO32
Manganate
MnO42
Tetraborate
B4O72
Chromate
CrO42
Stannate
SnO32
Valency 3
Phosphate
PO43
Arsenate
AsO43
Borate (Ortho
borate)
BO33
Arsenite
AsO33
Aluminate
AlO33
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Basic Concepts of Chemistry
1.5
FORMULAE OF COMPOUNDS
Table 1.4
2
1
1.
Calcium
chloride
Ca Cl
CaCl2
6.
Potassium
perchlorate
K1ClO41
KClO4
2.
Magnesium
sulphate
Mg2SO42
MgSO4
(simple ratio)
7.
Sodium meta
aluminate
Na1AlO21
NaAlO2
3.
Stannic sulphide
Sn4S2
SnS2
8.
Sodium zincate
Na1ZnO22
Na2ZnO2
4.
Mercurous
chloride
Hg222Cl1
Hg2Cl2 (dimeric)
9.
Magnesium
bicarbonate
Mg2HCO31
Mg(HCO3)2
5.
Sodium thiosulphate
Na1S2O32
Na2S2O3
10.
Ammonium
oxalate
NH41C2O42
(NH4)2C2O4
MOLE CONCEPT
A mole is the amount of any substance containing 6022 u 1023
particles.
A rigorous definition of mole in S, terminology is given, as ‘the mole is the amount of any substance that contains
as many elementary entities as there are atoms in 12 grams
of carbon-12 isotope.
Mole is the S, unit of amount of substance
The elementary entities may be atoms, molecules, ions,
electrons etc. A mole of any substance always contains the
same number of entities irrespective of what the substance
is. For example, a mole of nitrogen or a mole of a metal like
Li, contains 6022 u 1023 molecules or atoms respectively.
This number is also known as Avogadro’s constant and so
named in honour of the Italian scientist, A. Avogadro.
MOLAR MASS
The term ‘molar mass’ refers to the mass of one mole of any
substance, be it an element or a compound.
For example, the molar mass of NH3 represents the
mass of 6022 u 1023 molecules of NH3 and is equal to
17.064 g mol-1.
It may be noted from equation (4) that the molar mass
of any atom (or molecule) is numerically equal to the atomic mass (or molecular mass) in amu.
For example, from equation (4), for carbon-12
12.000
12amu =
× 6.022 × 1023 g = molar mass in
6.022 × 1023
grams of C-12
The number of atoms of an element in a molecule is
called its atomicity.
From Avogadro’s hypothesis mentioned earlier, it follows
that one mole of any gas at STP (0qC and 1atm) occupies
22.4litres of volume, which is known as the molar volume of
the gas at STP.
No. of moles atoms =
Calculation of Number of Moles
No. of moles of atoms =
weight (gram)
atomic mass
No. of moles molecules (or) No. of moles =
No. of moles =
weight (gram)
molecular mass
No. of molecules
6.022 u 1023
No. of atoms
No. of moles of gases =
6.022 × 1023
Volume at STP(L)
22.4
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1.6 Basic Concepts of Chemistry
C ON CE P T ST R A N D S
Concept Strand 1
Calculate the number of atoms in
(i) half a mole of nitrogen gas.
(ii) 5.6 litres of helium gas at S.T.P.
Solution
Since ‘He’ is monoatomic, number of atoms in 0.25
mole = 1.51 u 1023
Concept Strand 2
A molecule of a diatomic compound weighs 4.98 u 1023 g.
What is its molecular weight? How many molecules and
atoms are present in 30 kg of this compound?
(i) One mole of N2 = 6.022u1023 molecules
6.022 u 1023
= 3.015 u 1023 molecules.
2
Since N2 is diatomic, the number of atoms of N2 =
3.015 u 10 23 u2 = 6.022 u 1023 atoms
(ii) From Avogadro’s hypothesis, 22.4 litres of gas at S.T.P
are occupied by one mole.
5.6
= 0.25
5.6 litres of gas at S.T.P are occupied by
22.4
moles
No. of atoms in 0.25moles = 6.022 u 1023 u 0.25 = 1.51
u 1023 atoms.
0.5 moles of N2 =
Solution
Molecular weight of this compound = 4.98 u 1023 u 6.022
u 1023 = 30.
Number of moles in 30 kg of this compound
30 u 103
1000
=
30
No. of molecules in 30 kg = 1000 u 6.022 u 1023
= 6.022 u 1026
No. of atoms in 30 kg = 1000 u 6.022 u 1023 u 2
= 1.2046 u 1027
CHEMICAL FORMULAE
When dealing with chemical formulae, two types of formulae must be taken into account.
(i) empirical formulae, which give the relative number of
different types of atoms present in a molecule of the
compound.
(ii) molecular formulae, which give the exact number of
atoms in a molecule of the compound.
The empirical formulae of a compound can be obtained by using chemical analytical data on the compound
and the atomic masses of various elements present in the
compound. The given mass per cent composition of the
compound is converted into mole per cent composition
by dividing the composition of each element by its atomic
mass. The mole per cent composition of each element is
next divided by the least value among the composition obtained to get the relative number of atoms of different elements in the compound in the simplest ratio.
The molecular formula is an integral multiple of the empirical formula.
C ON CE P T ST R A ND S
Concept Strand 3
An ore of chromium contains 24.95 weight % of
iron, 46.46 weight % chromium and the rest is
made up of oxygen. Find the empirical formula of
the ore.
(Given: At mass: Fe = 55.85, Cr = 52; O = 16)
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Basic Concepts of Chemistry
Solution
In 100 grams of the ore
Element
Weight Atomic ratio Smallest Ratio
24.95
0.45
1
24.95
55.85
Iron
1.7
Empirical formula: MgSO11H14.
Since all H atoms are part of water, the above formula
can be revised as MgSO4.7H2O or (MgSO4.7H2O)n = 246.5.
? n = 1
Molecular formula of the compound = MgSO4. 7H2O
Concept Strand 5
Chromium
46.46
46.46
52.00
Oxygen
28.59
28.59
1.79
16.00
0.89
2
4
Ratio of Fe, Cr, O = 1: 2 : 4 Empirical formula = FeCr2O4
A compound contains carbon, hydrogen and nitrogen in
the weight ratio 10 : 1 : 2. If the molecular weight of the
compound is 93 g mol1, find its molecular formula.
Solution
10
71.5
= 6 g atom.
u 93 = 71.5 g;
13
12
1
7.15
Weight of hydrogen =
= 7 g atom.
u 93 = 7.15 g;
13
1
2
14.3
Weight of nitrogen =
= 1 g atom.
u 93 = 14.3 g;
13
14
? Molecular formula is C6H7N (C6H5NH2)
Weight of carbon =
Concept Strand 4
An inorganic compound, on analysis, gave the following weight percent composition. Mg = 9.86; S = 13.00;
O = 71.41, the rest being hydrogen. Assuming that all the
hydrogen is in the form of water, calculate its empirical
formula. If its molecular weight is 246.5, what is the molecular formula of the compound?
(Atomic mass of Mg = 24.3; S = 32.06; O = 16.00; H = 1)
Solution
Element
Weight%
Atomic ratio
9.86
0.4
24.3
Mg
9.86
S
13.00
13.00
32.06
0.4
O
71.41
71.41
16.00
4.4
H
5.73
5.73
1
5.7
Smallest ratio
Concept Strand 6
The ratio of weights of mercurous chloride to mercury is
2.354. Find the atomic mass of mercury. (This example illustrates the importance of the use of correct formula for
a compound.)
1
Solution
1
11
14
Mercurous chloride is Hg2Cl2. If x is the atomic mass of
2x (2 u 35.5)
2.354
Hg, then
x
Solving for x, we get atomic mass of Hg as 200.6
If we use the formula HgCl, the answer becomes
different.
CHEMICAL EQUATION
Chemical equations are a manifestation of chemical reactions both in a qualitative and quantitative manner. The
quantitative aspects of mass and volume relations between
reactants and products are known as “stoichiometry”. The
following points must be borne in mind when writing
chemical equation.
(i) The equation must be properly balanced as to
maintain conservation of mass of all the species
participating in the reaction. For example, the
reaction
CH4O + O2 o CO2 + H2O
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1.8 Basic Concepts of Chemistry
gives the products correctly but does not show the
mass conservation in hydrogen and oxygen. The
balanced equation is
3
CH4O + O2 o CO2 + 2H2O
2
multiplying throughout by 2, we get the correct
balanced equation
2CH4O + 3O2 o 2CO2 + 4H2O
— (5)
(ii) The physical state of all the reactants and products
i.e., solid (s), liquid (l) or gas (g) must be indicated
along the side of the constituents. Equation (5) is more
appropriately written as
2CH4O( ) + 3 O2 (g) o 2 CO2 (g) + 4 H2O ( )
(iii) In case of thermo chemical equation, it is important to
incorporate the 'H (enthalpy change) of the reaction
with the sign. A balanced chemical equation enables
calculation of mass of reactants consumed and products formed from the given data.
C ON CE P T ST R A N D
Concept Strand 7
Calculate the weight of hydrochloric acid required to react
with 5 g of calcium carbonate completely. How much of
CO2 is formed under the given conditions?
1mole of CaCO3 react with 2 moles of HCl for complete reaction
i.e., 100 g of CaCO3 react with 2 u 36.5 = 73.0 g of HCl
5 u 73.0
3.65g of HCl
5 g of CaCO3 react with
100
Mass of CO2 formed:
100 g of CaCO3 on reaction with HCl gives 44 g of CO2.
5 g of CaCO3 on complete reaction with HCl gives
Solution
Balanced equation:
5 u 44
100
CaCO3 + 2HCl o CaCl2 + H2O + CO2
2.20 g of CO2.
EXCESS AND LIMITING REACTANTS
In the reaction between lead nitrate and potassium iodide
to give lead iodide, lead nitrate and potassium iodide in the
mole ratio of 1:2 react to give 1 mole lead iodide as per the
stoichiometric equation.
Pb(NO3)2 + 2K, o Pb,2 + 2KNO3
— (6)
Sometimes, a reactant taken in excess amount, than
what is required by stoichiometry, will be left behind after the
reaction and it is known as the excess reactant.
In some cases, the amount of reactant taken, limits the
amount of the product formed because it is completely consumed when the reaction is complete. As this reactant limits
the amount of product formed, it is called a limiting reactant.
If for example, in the above reaction, 2 moles of lead
nitrate is allowed to react with 6 moles of potassium iodide,
2 moles of lead iodide are formed and 2 moles of potassium iodide are left behind under these conditions, lead nitrate is the
limiting reactant and potassium iodide is the excess reactant.
C ON CEP T ST R A N D
Concept Strand 8
20.0 g of silver nitrate in aqueous solution react with 6.0 g
sodium chloride to form silver chloride precipitate. Calculate the amount of silver chloride formed and specify
which among the reactants, is an (Atomic mass of Ag =
108, Cl = 35.5, Na = 23).
(i) excess reactant
(ii) limiting reactant.
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Basic Concepts of Chemistry
Solution
The reaction is AgNO3 + NaCl o AgCl p + NaNO3
1 mole of AgNO3 reacts with one mole of NaCl to give
one mole of AgCl.
i.e., 170 g of AgNO3 reacts with 58.5 g of NaCl to give
143.5 g of AgCl.
6 u 170
= 17.44
6.00 g of NaCl react completely with
58.5
g of AgNO3.
1.9
Since 20.00 17.44 = 2.56 g of AgNO3 remains
in solution, the excess reactant is AgNO3 and as NaCl
limits the amount of AgCl formed, it is the limiting
reactant.
6 u 143.5
58.5
= 14.72 g
Amount of AgCl formed =
OXIDATION NUMBER OR OXIDATION STATE
Oxidation number of an atom in a compound is the
number of electrons shifting towards or away from that
atom. Loss of electron is indicated by a positive sign and
gain of electron by a negative sign. In the case of sharing, the more electronegative atom involved in the sharing is given a negative sign and the less electronegative a
positive sign.
For example in the ionic compound, CaCl2 the oxidation number of Ca is +2 and that of Cl is 1 and in MgO
the oxidation number of Mg is +2 and that of O is 2 and
in the covalent compound HCl, the oxidation number
of H is +1 and that of Cl is 1.
Rules for assigning oxidation number
(i) The oxidation number of a free element (regardless
of whether it exists in mono atomic or poly atomic
form. example: Cu, Cl2, P4, S8) is zero.
(ii) Sum of oxidation numbers of all the atoms of a
charged species (e.g., SO42-, NH4+ etc) is equal to the
charge on the ion.
(iii) The oxidation number of F in all its compounds
is 1, for all the other halogens, oxidation number = 1
except in those with oxygen (e.g., ClO3) and with
other halogens (e.g., BrCl2–).
In the former case it is determined from oxygen, i.e.,
Oxidation number of Cl = +5 and in the latter case it
is determined from the more electronegative halogen
i.e., oxidation number of Br is +1.
(iv) Oxidation number of alkali metals = +1 and alkaline
earth metals = +2.
(v) Oxidation number of H = +1 except in metal hydrides
like CaH2, LiAlH4 etc., where it is 1.
(vi) Oxidation number of O = 2 except in peroxide and
OF2. In peroxides it is 1 and in OF2, it is +2.
(vii) Algebraic sum of the oxidation numbers of all the
atoms in a neutral molecule is equal to zero and for a
radical it is equal to the charge.
C ON CE P T ST R A N D
Concept Strand 9
Calculate the oxidation number of
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Mn in K2MnO4
Cr in K2Cr2O7
S in Na2S2O3
C in CH3COOH
N in HN3
Cr in CrO5
Solution
(i) Oxidation number of Mn in K2MnO4 is + 2 + x 8 =
0, x = +6
(ii) Oxidation number of Cr
S
in K2Cr2O7 is + 2 + 2x
14 = 0, x = +6
+
+
(iii) Oxidation number of S Na O S O Na
in Na2S2O3 is + 2 + 2x
O
6 = 0, x = +2
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1.10 Basic Concepts of Chemistry
In fact, the two sulphur atoms are in +6 and 2
oxidation states and the average works out to be +2.
(iv) Oxidation number of C in CH3COOH (C2H4O2) is
2x + 4 4 = 0, x = 0
1
(v). Oxidation number of N in HN3 is .
3
(vi) Oxidation number of Cr in CrO5 is +6. (since, 4 u 1 +
x + 1 u 2 = 0 x = +6)
Oxidation and Reduction in terms of
oxidation number
The species for which there is an increase in oxidation
number is said to be oxidized. It is or known as reducing
agent (reductant) and the species, which undergoes a decrease
in oxidation number, is reduced, and it is or known as oxidizing agent or oxidant.
An increase in oxidation number indicates oxidation and a
decrease in oxidation number indicates reduction.
O
O
Cr
O
O
O
Oxidants
Table 1.5
Oxidant
Radical or ele- O.N of effec- Reduction
ment involved tive element Product
New
O.N
Decrease
In O.N
Gain in
electrons
KMnO4
(acid medium)
MnO4
Mn = +7
Mn2+
+2
5
5
KMnO4
(alkaline medium)
MnO4
Mn = +7
MnO42
+6
1
1
KMnO4
(neutral medium)
MnO4
Mn = +7
MnO2
+4
3
3
K2Cr2O7
(acid medium)
Cr2O72
Cr = +6
Cr3+
+3
6
(for 2 Cr)
6
Cl2
Cl
Cl2 = 0
Cl
Fe
FeCl3
3+
Fe = +3
NaOCl
OCl
1
1
1
2+
+2
1
1
Fe
Cl = +1
1
2
2
Cl
KIO3
IO
I = +5
I
1
6
6
H2O2
O
O = 1
O2
2
2
2
O.N of the effective
element
Oxidation
product
New
O.N
Increase in O.N
Loss in
electrons
Sn = +2
Sn 4+
3
2
2
Reductants
Table 1.6
Reductant
SnCl2
FeSO4
H2C2O4
H2SO3
Radical or element involved
Sn2+
2+
Fe
2
C2O4
SO3
2
+4
2
2
Fe = +2
Fe
3+
+3
1
1
C = +3
CO2
+4
2 (for 2 carbons)
2
+6
2
2
S = +4
SO4
2
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Basic Concepts of Chemistry
1.11
C ON CE P T ST R A N D
Concept Strand 10
In the following reactions, identify the species oxidized,
reduced, oxidizing agent and reducing agent.
(i) H2SO3 + 2H2S o 3S + 3 H2O
(ii) H2O2 + H2O2 o 2H2O + O2
(ii) H2O2 + H2O2
+1 –1
Solution
(i) H2 SO3
+1 +4 –2
+ 2H2 S
3S + 3H2O
+1 –2
+1 –2
0
Oxidation states of oxygen and hydrogen remain the
same. Sulphur in H2SO3 is reduced to S (+4 o 0), while
sulphur in H2S is oxidized to S (2 o0). Hence H2SO3
is the oxidizing agent and H2S is the reducing agent.
+1 –1
2H2O + O2
+1 –2
0
Oxidation state of hydrogen remains the same.
Oxygen in one molecule of H2O2 is oxidized to O2
(1o0) where as in another molecule it is reduced to
H2O (1o 2). Thus H2O2 acts both as an oxidizing
and reducing agent.
BALANCING REDOX REACTIONS
Balancing redox equations is based on the fact that the
number of electrons gained during reduction must be equal
to the number of electrons lost during oxidation. There are
two methods for balancing the redox reactions.
I. Ion-electron method
Various steps involved in this method are:
(i) Split up the complete reaction into two halves—
oxidation half (change undergone by the reducing
agent) and reduction half (change undergone by the
oxidizing agent).
(ii) Balance the number of atoms of each element other
than oxygen and hydrogen in each half of the reaction.
(iii) For balancing oxygen, H2O may have to be added,
while for balancing hydrogen, H+ are added. But in
alkaline medium, H+ must be converted to H2O by
adding the same number of OH. While adding OH,
it must be added on both sides of the equation.
(iv) Equalize the charge on both sides by adding
electrons to the side where positive charge has to be
reduced.
(v) Next is to add the two half reactions, for which the
number of electrons on both the reactions must be the
same, so that the final equation does not contain any
electron. This is achieved by multiplying with suitable
numbers.
(vi) Similar terms on either side of the equation can be
cancelled.
C ON CE P T ST R A ND
Concept Strand 11
(i)
(ii)
(iii)
(iv)
Solution
Balance Fe + MnO4 o Fe + Mn in acid medium.
Balance Cr2O72 + Fe2+ o Cr3+ + Fe3+ in acid medium.
Balance the equation: MnO4 + C2O42 o CO2 + Mn2+.
Balance the equation: CrO42 + SO3 o CrO2 + SO42
in alkaline medium.
2+
3+
2+
(i) Fe2+ + MnO4o Fe3+ + Mn2+ in acid medium
Oxidation half
Fe2+ o Fe3+
Balancing charge Fe2+ o Fe3+ + 1e
— (7)
