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Resonance Pre-foundation Career Care Programmes
(PCCP) Division
WORKSHOP TAPASYA
SHEET
CHEMISTRY
COURSE : IJSO (STAGE-) I
Subject : Chemistry
NTSE STAGE-I
S. No.
Topics
Page No.
1.
Structure of Atom
1 - 12
2.
Acids, Bases and Salts
13 - 23
3.
Periodic Table
24-38
4.
Mole Concept
39-57
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.13RPCCP
MOLE CONCEPT
Atomic
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
ATOMIC MASS UNIT
The atomic mass unit (amu) is equal to one-twelfth
(1/12) of the mass of an atom of carbon-12.The mass
of an atom of carbon-12 isotope was given the atomic
mass of 12 units, i.e. 12 amu or 12 u.
The atomic masses of all other elements are now
expressed in atomic mass units.
RELATIVE ATOMIC MASS
The atomic mass of an element is a relative quantity
and it is the mass of one atom of the element relative
to one -twelfth (1/12) of the mass of one carbon-12
atom. Thus, Relative atomic mass
=
Mass of one atom of the element
1
12
 mass of one C  12 atom
1 amu = 1.66 × 10

g = 1.66 × 10
Atomic
mass
1
4
7
9
11
12
14
16
19
20
23
24
27
28
31
32
35.5
40
39
40
RELATIVE MOLECULAR MASS
[1/12 the mass of one C-12 atom = 1 amu,
–24
Element
Symbol
Hydrogen
H
Helium
He
Lithium
Li
Beryllium
Be
Boron
B
Carbon
C
Nitrogen
N
Oxygen
O
Fluorine
F
Neon
Ne
Sodium
Na
Magnesium
Mg
Aluminium
Al
Silicon
Si
Phosphorus
P
Sulphur
S
Chlorine
Cl
Argon
Ar
Potassium
K
Calcium
Ca
–27
kg.]
Note :
The relative molecular mass of a substance is the
mass of a molecule of the substance as compared
to one-twelfth of the mass of one carbon -12 atom
i.e.,
Relative molecular mass
One amu is also called one dalton (Da).
GRAM-ATOMIC MASS
=
Mass of one molecule of the substance
12
The atomic mass of an element expressed in grams
is called the Gram Atomic Mass of the element.
The number of gram -atoms
=
Mass of the element in grams
Gram Atomic mass of the element
e.g.
Calculate the gram atoms present in (i) 16g of oxygen
GRAM MOLECULAR MASS
The molecular mass of a substance expressed in
grams is called the Gram Molecular Mass of the
substance . The number of gram molecules
Mass of the subs tance in grams
=
 Gram-Atomic Mass of oxygen (O) = 16 g.
16
No. of Gram-Atoms =
=1
16
 mass of one C  12 atom
The molecular mass of a molecule, thus, represents
the number of times it is heavier than 1/12 of the
mass of an atom of carbon-12 isotope.
and (ii) 64g of sulphur.
(i) The atomic mass of oxygen = 16.
1
Gram molecular mass of the subs t ance
e.g.
(i) Molecular mass of hydrogen (H2) = 2u.
 Gram Molecular Mass of hydrogen (H2) = 2 g .
(ii) The gram-atoms present in 64 grams of sulphur.
(ii) Molecular mass of methane (CH4) = 16u
 Gram Molecular Mass of methane (CH4) = 16 g.
64
64
= Gram Atomic Mass of sulphur =
=2
32
e.g. the number of gram molecules present in 64 g of
methane (CH4).
64
64
= Gram molecular mass of CH =
= 4.
4
16
PAGE # 1
(a) Calculation of Molecular Mass :
The molecular mass of a substance is the sum of
the atomic masses of its constituent atoms present
in a molecule.
Ex.1 Calculate the molecular mass of water.
(Atomic masses : H = 1u, O = 16u).
Sol. The molecular formula of water is H2O.
Molecular mass of water = ( 2 × atomic mass of
H) + (1 × atomic mass of O)
= 2 × 1 + 1 × 16 = 18
i.e., molecular mass of water = 18 amu.
Ex.2 Find out the molecular mass of sulphuric acid.
(Atomic mass : H = 1u, O = 16u, S = 32u).
(iv) Eq. mass of a salt
=
Mol. wt. of the salt
Number of metal atoms  valency of metal
(v) Eq. mass of an ion =
(vi) Eq. mass of an oxidizing/reducing agent
=
Mol wt. or At. wt
No. of electrons lost or gained by one
molecule of the substance
Equivalent weight of some compounds are given in
the table :
Sol. The molecular formula of sulphuric acid is H2SO4.
 Molecular mass of H2SO4
= (2 × atomic mass of H) + ( 1 × atomic mass of S)
+ ( 4 × atomic mass of O)
= (2 × 1) + (1× 32) + (4×16) = 2 + 32 + 64 = 98
i.e., Molecular mass of H2SO4= 98 amu.
S.No.
Compound
Equivalent
weight
1
HCl
36.5
2
H2SO4
49
3
HNO3
63
4
FORMULA MASS
45
COOH
The term ‘formula mass’ is used for ionic
compounds and others where discrete molecules
do not exist, e.g., sodium chloride, which is best
represented as (Na+Cl–)n, but for reasons of simplicity
as NaCl or Na+Cl–. Here, formula mass means the
sum of the masses of all the species in the formula.
Thus, the formula mass of sodium chloride = (atomic
mass of sodium) + (atomic mass of chlorine)
= 23 + 35.5
= 58.5 amu
EQUIVALENT MASS
5
COOH
.2H2O
63
6
NaOH
40
7
KOH
56
8
CaCO3
50
9
NaCl
58.5
10
Na2CO3
53
To find the relationship between molecular mass
and vapour density of a gas.
(a) Definition :
Equivalent mass of an element is the mass of the
element which combine with or displaces 1.008 parts
by mass of hydrogen or 8 parts by mass of oxygen or
35.5 parts by mass of chlorine.
(b) Formulae of Equivalent Masses of different
substances :
(i) Equivalent mass of an element =
Atomic wt. of the element
Valency of the element
Density of gas
Vapour density (V.D.) = Density of hydrogen
=
Mass of a certain volume of the gas
Mass of the same volume of hydrogen at
the same temp. and pressure
If n molecules are present in the given volume of a gas
and hydrogen under similar conditions of temperature
and pressure.
V.D. =
(ii) Eq. mass an acid =
Formula wt. of the ion
Charge on the ion
Mol. wt. of the acid
Basicity of the acid
Basicity is the number of replaceable H+ ions from
one molecule of the acid.
(iii) Eq. Mass of a base =
Mol. wt. of the base
Acidity of the base
Acidity is the number of replaceable OH– ions from
one molecule of the base
Mass of n molecules of the gas
Mass of n molecules of hydrogen
=
Mass of 1 molecule of the gas
Mass of 1 molecule of hydrogen
=
Molecular mass of the gas
Molecular mass of hydrogen
Molecular mass
2
(since molecular mass of hydrogen is 2)
Hence, Molecular mass = 2 × Vapour density
=
PAGE # 2
SOME IMPORTANT RELATIONS AND FORMULAE
In Latin, mole means heap or collection or pile. A
mole of atoms is a collection of atoms whose total
mass is the number of grams equal to the atomic
mass in magnitude. Since an equal number of moles
of different elements contain an equal number of
atoms, it becomes convenient to express the
amounts of the elements in terms of moles. A mole
represents a definite number of particles, viz, atoms,
molecules, ions or electrons. This definite number is
called the Avogadro Number (now called the Avogadro
constant) which is equal to 6.023 × 1023.
(i) 1 mole of atoms = Gram Atomic mass = mass of
6.023 × 1023 atoms
(ii) 1 mole of molecules = Gram Molecular Mass
= 6.023 x 1023 molecules
(iii) Number of moles of atoms
=
(iv) Number of moles of molecules
A mole is defined as the amount of a substance that
contains as many atoms, molecules, ions, electrons
or other elementary particles as there are atoms in
exactly 12 g of carbon -12 (12C).
=
(a) Moles of Atoms :
=
(i) 1 mole atoms of any element occupy a mass which
is equal to the Gram Atomic Mass of that element.
e.g. 1 Mole of oxygen atoms weigh equal to Gram
Atomic Mass of oxygen, i.e. 16 grams.
(ii) The symbol of an element represents 6.023 x 1023
atoms (1 mole of atoms) of that element.
e.g : Symbol N represents 1 mole of nitrogen atoms
and 2N represents 2 moles of nitrogen atoms.
(b) Moles of Molecules :
(i) 1 mole molecules of any substance occupy a mass
which is equal to the Gram Molecular Mass of that
substance.
Mass of element in grams
Gram Atomic Mass of element
Mass of substance in grams
Gram Molecular Mass of substance
(v) Number of moles of molecules
No. of molecules of element
N

Avogadro number
NA
Ex.3 To calculate the number of moles in 16 grams of
Sulphur (Atomic mass of Sulphur = 32 u).
Sol. 1 mole of atoms = Gram Atomic Mass.
So, 1 mole of Sulphur atoms = Gram Atomic Mass of
Sulphur = 32 grams.
Now, 32 grams of Sulphur = 1 mole of Sulphur
So, 16 grams of Sulphur
= (1/32) x 16 = 0.5 moles
Thus, 16 grams of Sulphur constitute 0.5 mole of
Sulphur.
22.4 litre
e.g. : 1 mole of water (H2O) molecules weigh equal to
Gram Molecular Mass of water (H2O), i.e. 18 grams.
In term of
volume
23
(ii) The symbol of a compound represents 6.023 x
10 23 molecules (1 mole of molecules) of that
compound.
6.023 × 10
(NA) Atoms
23
1 Mole
e.g. : Symbol H 2O represents 1 mole of water
molecules and 2 H2O represents 2 moles of water
molecules.

Note :
The symbol H2O does not represent 1 mole of H2
molecules and 1 mole of O atoms. Instead, it
represents 2 moles of hydrogen atoms and 1 mole
of oxygen atoms.

Note :
The SI unit of the amount of a substance is Mole.
(c) Mole in Terms of Volume :
Volume occupied by 1 Gram Molecular Mass or 1
mole of a gas under standard conditions of
temperature and pressure (273 K and 1atm.
pressure) is called Gram Molecular Volume. Its value
is 22.4 litres for each gas.
Volume of 1 mole
= 22.4 litre (at STP)

Note :
The term mole was introduced by Ostwald in 1896.
6.023 × 10
(NA) molecules
In terms of
particles
In terms of
mass
1 gram atom
of element
1 gram molecule
of substance
1 gram formula
mass of substance
PROBLEMS BASED ON THE MOLE CONCEPT
Ex.4 Calculate the number of moles in 5.75 g of
sodium. (Atomic mass of sodium = 23 u)
Sol. Number of moles
Mass of the element in grams
=
Gram Atomic Mass of element
=
5.75
23
= 0.25 mole
or,
1 mole of sodium atoms = Gram Atomic mass of
sodium = 23g.
23 g of sodium = 1 mole of sodium.
5.75 g of sodium =
5.75
23
mole of sodium = 0.25 mole
PAGE # 3
Ex.5 What is the mass in grams of a single atom of
chlorine ? (Atomic mass of chlorine = 35.5u)
Sol. Mass of 6.022 × 1023 atoms of Cl = Gram Atomic
Mass of Cl = 35.5 g.
 Mass of 1 atom of Cl =
35.5 g
6.022  10 23
= 5.9 × 10–23 g.
Ex.6 The density of mercury is 13.6 g cm–3. How many
moles of mercury are there in 1 litre of the metal ?
(Atomic mass of Hg = 200 u).
Sol. Mass of mercury (Hg) in grams = Density
(g cm–3)× Volume (cm3)
= 13.6 g cm–3 × 1000 cm3 = 13600 g.
 Number of moles of mercury
Mass of mercury in grams
Ex.12 Calculate the volume in litres of 20 g of hydrogen
gas at STP.
Sol. Number of moles of hydrogen
Mass of hydrogen in grams
13600
= Gram Atomic Mass of mercury =
200
= 68
Ex.7 The mass of a single atom of an element M is
3.15× 10–23 g . What is its atomic mass ? What
could the element be ?
Sol. Gram Atomic Mass = mass of 6.022 × 1023 atoms
= mass of 1 atom × 6.022 × 1023
= (3.15 × 10–23g) × 6.022 × 1023
= 3.15 × 6.022 g = 18.97 g.
 Atomic Mass of the element = 18.97u
Thus, the element is most likely to be fluorine.
Ex.8 An atom of neon has a mass of 3.35 × 10–23 g.
How many atoms of neon are there in 20 g of the
gas ?
Sol. Number of atoms
=
Ex.11 What mass in grams is represented by
(a) 0.40 mol of CO2,
(b) 3.00 mol of NH3,
(c) 5.14 mol of H5IO6
(Atomic masses : C=12 u, O=16 u, N=14 u,
H=1 u and I = 127 u)
Sol. Weight in grams = number of moles × molecular
mass.
Hence,
(a) mass of CO2 = 0.40 × 44 = 17.6 g
(b) mass of NH3 = 3.00 × 17 = 51.0 g
(c) mass of H5IO6 = 5.14 × 228 = 1171.92g
20
Total mass
=
= 5.97 × 1023
Mass of 1 atom
3.35  10 – 23
Ex.9 How many grams of sodium will have the same
number of atoms as atoms present in 6 g of
magnesium ?
(Atomic masses : Na = 23u ; Mg =24u)
Sol. Number of gram -atom of Mg
Mass of Mg in grams
1
6
= Gram Atomic Mass =
=
24
4
20
= 10
2
 Volume of hydrogen = number of moles × Gram
Molecular Volume.
= 10 ×22.4 = 224 litres.
=
=
Gram Molecular Mass of hydrogen
PERCENTAGE COMPOSITION
The percentage composition of elements in a
compound is calculated from the molecular formula
of the compound.
The molecular mass of the compound is calculated
from the atomic masses of the various elements
present in the compound. The percentage by mass
of each element is then computed with the help of the
following relations.
Percentage mass of the element in the compound
Total mass of the element
=
× 100
Molecular mass
Ex.13 W hat are the percentage compositions of
hydrogen and oxygen in water (H2O) ?
(Atomic masses : H = 1 u, O = 16 u)
Sol. Molecular mass of water, H2O = 2 + 16 = 18 amu.
H2O has two atoms of hydrogen.
So, total mass of hydrogen in H2O = 2 amu.
Percentage of H =
2  100
18
= 11.11 %
Similarly,
 Gram Atoms of sodium should be =
1
4
percentage of oxygen =
16  100
18
= 88.88 %
 1 Gram Atom of sodium = 23 g
1
4
gram atoms of sodium = 23 ×
1
= 5.75 g
4
Ex.10 How many moles of Cr are there in 85g of Cr2S3 ?
(Atomic masses : Cr = 52 u , S =32 u)
Sol. Molecular mass of Cr2S3 = 2 × 52 + 3 × 32 = 104
+ 96 = 200 u.
200g of Cr2S3 contains = 104 g of Cr.
 85 g of Cr2S3 contains =
104  85
200
Thus, number of moles of Cr =
g of Cr = 44.2g
44.2
52
= 0.85 .
The following steps are involved in determining the
empirical formula of a compound :
(i) The percentage composition of each element is
divided by its atomic mass. It gives atomic ratio of the
elements present in the compound.
(ii) The atomic ratio of each element is divided by the
minimum value of atomic ratio as to get the simplest
ratio of the atoms of elements present in the
compound.
(iii) If the simplest ratio is fractional, then values of
simplest ratio of each element is multiplied by
smallest integer to get the simplest whole number
for each of the element.
PAGE # 4
(iv) To get the empirical formula, symbols of various
elements present are written side by side with their
respective whole number ratio as a subscript to the
lower right hand corner of the symbol.
(v) The molecular formula of a substance may be
determined from the empirical formula if the molecular
mass of the substance is known. The molecular
formula is always a simple multiple of empirical
formula and the value of simple multiple (n) is
obtained by dividing molecular mass with empirical
formula mass.
Molecular Mass
n =
Empirical Formula Mass
Ex-14 A compound of carbon, hydrogen and nitrogen
contains these elements in the ratio of 9:1:3.5 respectively.
Calculate the empirical formula. If its molecular mass is
108, what is the molecular formula ?
Sol.
Element
Mass Atomic Relative Number
Ratio Mass
of Atoms
Carbon
9
12
Hydrogen
1
1
Nitrogen
3.5
14
9
 0.75
12
1
 1
1
3.5
 0.25
14
Simplest
Ratio
0.75×4 = 3
1×4=4
0.25 ×4 = 1
Whole of the hydrogen is present in the form of water
of crystallisation. Thus, 10 water molecules are
present in the molecule.
So, molecular formula = Na2SO4. 10H2O
CONCENTRATION OF SOLUTIONS
(a) Strength in g/L :
The strength of a solution is defined as the amount of
the solute in grams present in one litre (or dm3) of the
solution, and hence is expressed in g/litre or g/dm3.
Weight of solute in gram
Strength in g/L =
Volume of solution in litre
(b) Molarity :
Molarity of a solution is defined as the number of
moles of the solute dissolved per litre (or dm3) of
solution. It is denoted by ‘M’. Mathematically,
Number of moles of solute
M=
Volume of the solution in litre
Mass of solute in gram/Gram Molecular Mass of solute
Volume of solution in litre
Empirical ratio = C3H4N
Empirical formula mass = (3 × 12) + (4× 1) + 14 = 54
n=
Molecular Mass
108
2
=
Empirical Formula Mass
54
M=
Thus, molecular formula of the compound
= (Empirical formula)2
= (C3H4N)2
= C6H8N2
Ex.15 A compound on analysis, was found to have the
following composition :
(i) Sodium = 14.31%, (ii) Sulphur = 9.97%, (iii) Oxygen
= 69.50%, (iv) Hydrogen = 6.22%. Calculate the
molecular formula of the compound assuming that
whole hydrogen in the compound is present as water
of crystallisation. Molecular mass of the compound
is 322.
Sol.
Element Percentage
Sodium
14.31
Atomic
mass
Relative Number
of atoms
23
14 .31

23
9 .97

32
Sulphur
9.97
32
Hydrogen
6.22
1
Oxygen
69.50
16
6 . 22
1
69. 50

16

Simplest ratio
0.622
0.622
2
0.311
0.311
0.311
1
0.311
6.22
6.22
 20
0.311
4.34
4.34
 14
0.311
The empirical formula = Na2SH20O14
Empirical formula mass
= (2 × 23) + 32 + (20 × 1) + (14 × 16)
= 322
Molecular mass
= 322
Molecular formula = Na2SH20O14
M can be calculated from the strength as given below :
Strength in grams per litre
Molecular mass of solute
If ‘w’ gram of the solute is present in V cm3 of a given
solution , then
1000
w
M=
×
Molecular mass
V
e.g. a solution of sulphuric acid having 4.9 grams of it
dissolved in 500 cm3 of solution will have its molarity,
w
M=
M=
1000
Molecular mass
4.9
98
1000
×
500
×
V
= 0.1
(c) Formality :
In case of ionic compounds like NaCl, Na2CO3 etc.,
formality is used in place of molarity. The formality of
a solution is defined as the number of gram formula
masses of the solute dissolved per litre of the
solution. It is represented by the symbol ‘F’. The term
formula mass is used in place of molecular mass
because ionic compounds exist as ions and not as
molecules. Formula mass is the sum of the atomic
masses of the atoms in the formula of the compound.
Mass of solute in gram/Formula Mass of solute
Volume of solution in litre
PAGE # 5
(d) Normality :
Normality of a solution is defined as the number of
gram equivalents of the solute dissolved per litre (dm3)
of given solution. It is denoted by ‘N’.
Mathematically,
Number of gram equivalent s of solute
N =
N=
Volume of the solution in litre
Ex.17 100 ml of
=
S
N3 =

E
0.49
49
1000
×
250
Number of moles of the solute
× 1000 ‘m’ can
Weight of the solvent in gram
be calculated from the strength as given below :
1000
m=
V
= 0.04
m=
(Eq. mass of sulphuric acid = 49).
Solution
Semi
normal
Normality
Deci
normal
1
2
m=
1
100
(i) Milli equivalent of substance = N × V
where , N  normality of solution
V  Volume of solution in mL
m=

Where, W  Weight of substance in gram
E  Equivalent weight of substance
(iv) Calculation of normality of mixture :
N
HCl is mixed with 50 ml of
H SO .
10
5 2 4
Find out the normality of the mixture.
Sol. Milli equivalent of HCl + milli equivalent H 2SO 4
= milli equivalent of mixture
N1 V1 + N2 V2 = N3 V3
{ where, V3 =V1 + V2 )
 1
 1

 100     50   N3 × 150

 10
 5

N3 =
20
150
=
2
15
Note :
Ex.18 Calculate the molarity and normality of a solution
containing 0.4 g of NaOH dissolved in 500 cm3 of
solvent.
Sol. Weight of NaOH dissolved = 0.5 g
Volume of the solution = 500 cm3
(iii) S = N × E
S  Strength in g/L
N  Normality of solution
E  Equivalent weight
Ex.16 100 ml of
1.325 1000

= 0.05
106 250
Relationship Between Normality and Molarity of a
Solution :
Normality of an acid = Molarity × Basicity
Normality of base = Molarity × Acidity
w  1000
E
N
w
1000
×
Mol. mass of the solute
W
e.g. A solution of anhydrous sodium carbonate
(molecular mass = 106) having 1.325 grams of it,
dissolved in 250 gram of water will have its molality -
Some Important Formulae :
(ii) If weight of substance is given,
Strength per 1000 gram of solvent
Molecular mass of solute
If ‘w’ gram of the solute is dissolved in ‘W’ gram of the
solvent then
Centi
normal
1
10
milli equivalent (NV) =
Note :
Molality of a solution is defined as the number of
moles of the solute dissolved in 1000 grams of the
solvent. It is denoted by ‘m’.
Mathematically,
e.g. A solution of sulphuric acid having 0.49 gram of
it dissolved in 250 cm 3 of solution will have its
normality,
N=
1
25
(e) Molality :
1000
w
N = Equivalent mass of the solute ×
V
w
(i) Calculation of molarity :
Molecular weight of NaOH = 23 + 16 + 1 = 40
Weight of solute/ molecular weight of solute
Volume of solution in litre
0.5/40
Molarity =
=
500/1000
= 0.025
(ii) Calculation of normality :
Normality
Weight of solute/ equivalent weight of solute
Volume of solution in litre
0.5/40
=
= 0.025
500/1000
=
= 0.133
NaOH.
1 milli equivalent of an acid neutralizes 1 milli
equivalent of a base.
If ‘w’ gram of the solute is present in V cm3 of a given
solution.
N = Equivalent mass of the solute ×
5
 1
 1

 100  –   25  = N3 × 125

10
5

 

Volume of the solution in litre
Equivalent mass of solute
N
Sol. Milli equivalent of HCl – milli equivalent of NaOH
= milli equivalent of mixture
N1 V1 – N2 V2 = N3 V3
{ where, V3 =V1 + V2 )
Weight of solute in gram / equivalent weight of solute
Strength in grams per litre
HCl is mixed with 25 ml of
10
Find out the normality of the mixture.
N can be calculated from the strength as given below :
N=
N
PAGE # 6
Ex.19 Find the molarity and molality of a 15% solution
of H2SO4 (density of H2SO4 solution = 1.02 g/cm3)
(Atomic mass : H = 1u, O = 16u , S = 32 u)
Sol. 15% solution of H2SO4 means 15g of H2SO4 are
present in 100g of the solution i.e.
Wt. of H2SO4 dissolved = 15 g
Weight of the solution = 100 g
Density of the solution = 1.02 g/cm3 (Given)
Calculation of molality :
Weight of solution = 100 g
Weight of H2SO4 = 15 g
Wt. of water (solvent) = 100 – 15 = 85 g
Molecular weight of H2SO4 = 98
 15 g H2SO4 =
15
98
0.153
× 1000 = 1.8 mole
Hence ,the molality of H2SO4 solution = 1.8 m
15 g of H2SO4 = 0.153 moles
Wt. of solution
=
100
1.02
Density of solution
= 98.04 cm3
This 98.04 cm3 of solution contain H2SO4 = 0.153
moles
1000 cm3 of solution contain H2SO4
0.153
=
98.04
=
w/m  W/M
=
Moles of solution
10 / 60
10 / 60  90 / 18
= 0.032
Note :
Sum of mole fraction of solute and solvent is always
equal to one.
(a) Quantitative Relations in Chemical
Reactions :
Stoichiometry is the calculation of the quantities of
reactants and products involved in a chemical
reaction.
It is based on the chemical equation and on the
relationship between mass and moles.
N2(g) + 3H2(g)  2NH3(g)
Calculation of molarity :
Vol. of solution =
w/m
STOICHIOMETRY
Thus ,85 g of the solvent contain 0.153 moles .
85
Moles of solute
mole fraction of solute =

= 0.153 moles
1000 g of the solvent contain=
Ex.20 Find out the mole fraction of solute in 10% (by weight)
urea solution.
weight of solute (urea) = 10 g
weight of solution = 100 g
weight of solvent (water) = 100 – 10 = 90g
A chemical equation can be interpreted as follows 1 molecule N2 + 3 molecules H2  2 molecules
NH 3 (Molecular interpretation)
1 mol N2 + 3 mol H2  2 mol NH3
(Molar interpretation)
28 g N2 + 6 g H2  34 g NH3
(Mass interpretation)
1 volume N2 + 3 volume H2  2 volume NH3
(Volume interpretation)
Thus, calculations based on chemical equations are
divided into four types -
× 1000 = 1.56 moles
(i) Calculations based on mole-mole relationship.
Hence the molarity of H2SO4 solution = 1.56 M
(ii) Calculations based on mass-mass relationship.
(f) Mole Fraction :
(iii) Calculations based on mass-volume relationship.
The ratio between the moles of solute or solvent to
the total moles of solution is called mole fraction.
Moles of solute
mole fraction of solute =
Moles of solution

n
nN
(i) Calculations based on mole-mole relationship :
In such calculations, number of moles of reactants
are given and those of products are required.
Conversely, if number of moles of products are given,
then number of moles of reactants are required.

N
nN
Ex.21 Oxygen is prepared by catalytic decomposition
of potassium chlorate (KClO 3). Decomposition
of potassium chlorate gives potassium chloride
(KCl) and oxygen (O2). How many moles and how
many grams of KClO 3 are required to produce
2.4 mole O2.
Sol. Decomposition of KClO3 takes place as,
2KClO3(s)  2KCl(s) + 3O2(g)
2 mole KClO3  3 mole O2
 3 mole O2 formed by 2 mole KClO3
w/m
=
w/m  W/M
Moles of solvent
Mole fraction of solvent =
Moles of solution
W/M
=
w/m  W/M
where,
n  number of moles of solute
N  number of moles of solvent
m  molecular weight of solute
M  molecular weight of solvent
w  weight of solute
W  weight of solvent
(iv) Calculations based on volume -volume
relationship.
2

 2.4 mole O 2 will be formed by  3  2.4 


mole KClO3 = 1.6 mole KClO3
Mass of KClO3 = Number of moles × molar mass
= 1.6 × 122.5 = 196 g
PAGE # 7
(ii) Calculations based on mass-mass relationship:
In making necessary calculation, following steps are
followed (a) Write down the balanced chemical equation.
(b) Write down theoretical amount of reactants and
products involved in the reaction.
(c) The unknown amount of substance is calculated
using unitary method.
Ex.22 Calculate the mass of CaO that can be prepared
by heating 200 kg of limestone CaCO3 which is
95% pure.
Sol. Amount of pure CaCO 3 =
95
 200 = 190 kg
100
= 190000 g
CaCO3(s)  CaO(s) + CO2(g)
100 g CaCO3  56 g CaO
100 g CaCO3 give 56 g CaO
56
 190000 g CaCO3 will give=
× 190000 g CaO
100
4FeS2 + 11O2  2Fe2O3 + 8SO2
2SO2 + O2  2SO3
SO3 + H2O  H2SO4
Sol. Final balanced equation,
Ex.23 How many grams of oxygen are required to burn
completely 570 g of octane ?
Sol. Balanced equation
16CO2 + 18H2O
4 mole
4 × 120 g
8 mole
8 × 98 g
4 × 120 g of FeS2 yield H2SO4 = 8 × 98 g
1000 g of FeS2 will yield H2SO4 =
8  98
× 1000
4  120
= 1633.3 g
(iii) Calculations involving mass-volume relationship :
In such calculations masses of reactants are given
Mass of a gas can be related to volume according to
the following gas equation PV = nRT
w
RT
m
Ex-25. What volume of NH3 can be obtained from 26.75 g
of NH4Cl at 27ºC and 1 atmosphere pressure.
Sol. The balanced equation is -
NH4Cl(s)
25 mole
25 × 32
First method : For burning 2 × 114 g of the octane,
oxygen required = 25 × 32 g
For burning 1 g of octane, oxygen required =
2Fe2O3 + 8H2SO4
4FeS2 + 15O2 + 8H2O
PV =
= 106400 g = 106.4 kg
2 mole
2 × 114
obtained from 1 kg of iron pyrites (FeS2) according to
the following reactions ?
and volume of the product is required and vice-versa.
1 mole of a gas occupies 22.4 litre volume at STP.
1 mole CaCO3  1 mole CaO
2C8H18 + 25O2
Ex.24 How many kilograms of pure H 2SO 4 could be
25  32
g
2  114
Thus, for burning 570 g of octane, oxygen required
25  32
=
× 570 g = 2000 g
2  114
Mole Method : Number of moles of octane in 570
grams
NH3(g) + HCl(g)
1 mol
53.5 g
1 mol
 53.5 g NH4Cl give 1 mole NH3
1
 26.75 g NH4Cl will give
× 26.75 mole NH3
53.5
= 0.5 mole
PV = nRT
1 ×V = 0.5 × 0.0821 × 300
V = 12.315 litre
570
= 5.0
114
(iv) Calculations based on volume volume
relationship :
For burning 2.0 moles of octane, oxygen required
= 25 mol = 25 × 32 g
For burning 5 moles of octane, oxygen required
These calculations are based on two laws :
(i) Avogadro’s law
(ii) Gay-Lussac’s Law
=
25  32
× 5.0 g = 2000 g
2 .0
Proportion Method : Let x g of oxygen be required for
burning 570 g of octane. It is known that 2 × 114 g of
the octane requires 25 × 32 g of oxygen; then, the
proportion.
x
25  32 g oxygen
=
570
g
oc
tane
2  114 g oc tan e
25  32  570
x=
= 2000 g
2  114
e.g.
N 2(g)
+
1 mol
1 × 22.4 L
3H 2(g)
2NH 3(g) (Avogadro's law)
3 mol
3 × 22.4 L
2 mol
2 × 22.4 L
(under similar conditions of temperature and
pressure, equal moles of gases occupy equal
volumes)
N2(g)
1 vol
+
3H2(g)
3 vol
2NH3(g) (Gay- Lussac's Law)
2 vol
(under similar conditions of temperature and
pressure, ratio of coefficients by mole is equal to ratio
of coefficient by volume).
PAGE # 8
Ex-26 One litre mixture of CO and CO2 is taken. This is
passed through a tube containing red hot charcoal.
The volume now becomes 1.6 litre. The volume are
measured under the same conditions. Find the
composition of mixture by volume.
Sol. Let there be x mL CO in the mixture , hence, there will
be (1000 – x) mL CO2. The reaction of CO2 with red
hot charcoal may be given as -
CO2(g) + C(s)
2CO(g)
2 vol.
2(1000 – x)
1 vol.
(1000 –x)
Total volume of the gas becomes = x + 2(1000 – x)
x + 2000 – 2x = 1600
x = 400 mL
 volume of CO = 400 mL and volume of CO2 = 600 mL
Ex-27 What volume of air containing 21% oxygen by volume
is required to completely burn 1kg of carbon containing
100% combustible substance ?
Sol. Combustion of carbon may be given as,
C(s) + O2(g)
1 mol
12 g
CO2(g)
1 mol
32 g
 12 g carbon requires 1 mole O2 for complete
combustion
1
 1000 mole O2 for
 1000 g carbon will require
12
combustion, i.e. , 83.33 mole O2
Volume of O2 at STP = 83.33 × 22.4 litre
= 1866.66 litre
 21 litre O2 is present in 100 litre air
100
× 1866.66 litre air
21
3
= 8888.88 litre or 8.89 × 10 litre
1866.66 litre O2 will be present in
VOLUMETRIC CALCULATIONS
The quantitative analysis in chemistry is primarily
carried out by two methods, viz, volumetric analysis
and gravimetric analysis.In the first method the mass
of a chemical species is measured by measurement
of volume, whereas in the second method it is determined by taking the weight.
The strength of a solution in volumetric analysis is
generally expressed in terms of normality, i.e., number of equivalents per litre but since the volume in the
volumetric analysis is generally taken in millilitres
(mL), the normality is expressed by milliequivalents
per millilitre.
USEFUL FORMULAE FOR
VOLUMETRIC CALCULATIONS
(v) Strength in grams per litre = normality × equivalent
weight.
(vi) (a) Normality = molarity × factor relating mol. wt.
and eq. wt.
(b) No. of equivalents = no. of moles × factor relat
ing mol. wt. and eq. wt.
Ex.28 Calculate the number of milli equivalent of H2SO4
present in 10 mL of N/2 H2SO4 solution.
1
× 10 = 5.
2
Ex.29 Calculate the number of m.e. and equivalents of
NaOH present in 1 litre of N/10 NaOH solution.
Sol. Number of m.e. = normality × volume in mL =
Sol. Number of m.e. = normality × volume in mL
=
1
× 1000 = 100
10
Number of equivalents =
no. of m.e.
100
=
= 0.10
1000
1000
Ex.30 Calculate number of m.e. of the acids present in
(i) 100 mL of 0.5 M oxalic acid solution.
(ii) 50 mL of 0.1 M sulphuric acid solution.
Sol. Normality = molarity × basicity of acid
(i) Normality of oxalic acid = 0.5 × 2 = 1 N
m.e. of oxalic acid = normality × vol. in mL = 1 × 100
= 100.
(ii) Normality of sulphuric acid = 0.1 × 2 = 0.2 N
m.e. of sulphuric acid
= 0.2 × 50 = 10
Ex.31 What is strength in gram/litre of a solution of H2SO4,
12 cc of which neutralises 15 cc of
N
NaOH
10
solution ?
Sol. m.e. of NaOH solution =
1
× 15 = 1.5
10
m.e. of 12 cc of H2SO4 = 1.5
1.5
12
Strength in grams/litre = normality × eq. wt.
 normality of H2SO4 =
1.5
× 49 grams/litre
12
= 6.125 grams/litre.
=


molecular wt. 98
 eq. wt. of H2 SO 4 

 49 
basicity
2


(i) milliequivalents = normality × volume in millilitres.
Ex.32 What weight of KMnO4 will be required to prepare
(ii) At the end point of titration, the two titrants, say 1
and 2, have the same number of milliequivalents,
i.e., N1V1 = N2V2, volume being in mL.
N
solution if eq. wt. of KMnO4 is 31.6 ?
10
Sol. Equivalent weight of KMnO4 = 31.6
250 mL of its
(iii) No. of equivalents =
m.e.
.
1000
(iv) No. of equivalents for a gas =
Normality of solution (N) =
Volumeat STP
equivalent volume( vol. of 1eq. at STP)
W
1
10
Volume of solution (V) = 250 ml
NEV ; W = 1  31 .6  250 31 .6
 0.79 g
10
1000
1000
40
PAGE # 9
Ex.33 100 mL of 0.6 N H2SO4 and 200 mL of 0.3 N HCl
were mixed together. What will be the normality of the
SOLUBILITY
The solubility of a solute in a solution is always
expressed with respect to the saturated solution.
resulting solution ?
Sol. m.e. of H2SO4 solution = 0.6 × 100 = 60
m.e. of HCl solution = 0.3 × 200 = 60
(a) Definition :
The maximum amount of the solute which can be
dissolved in 100g (0.1kg) of the solvent to form a
saturated solution at a given temperature.
Suppose w gram of a solute is dissolved in W gram
of a solvent to make a saturated solution at a fixed
temperature and pressure. The solubility of the solute
will be given by -
 m.e. of 300 mL (100 + 200) of acidic mixture
= 60 + 60 = 120.
m.e.
total vol.
120
2
=
=
N.
300
5
Normality of the resulting solution =
Ex.34 A sample of Na2CO3. H2O weighing 0.62 g is added
w
to 100 mL of 0.1 N H2SO4. Will the resulting solution
W
0.62
= 0.01
62
124


 62 
 eq. wt. of Na 2 CO 3 .H2 O 
2


Sol. Equivalents of Na2CO3. H2O =
m.e. of H2SO4 = 0.1 × 100 = 10
The solubility of a substance in liquids generally
increases with rise in temperature but hardly changes
with the change in pressure. The effect of temperature
depends upon the heat energy changes which
accompany the process.
Since the m.e. of Na2CO3. H2O is equal to that of H2SO4,
the resulting solution will be neutral.
Indicators show change in colour or turbidity at the
stage of completion of titration.

(c) Concentraion representation of solution
(A) Strength of solution : Grams of solute dissolved
per litre of solution is called strength of solution'
dissolved per 10 6 grams of solvent is called
concentration of solution in the unit of Parts Per Million
(ppm). This unit is used to represent hardness of
water and concentration of very dilute solutions.
(C) Percentage by mass : Grams of solute dissolved
(ii) Effect of temperature on exothermic dissolution
process : Few salts like lithium carbonate, sodium
carbonate monohydrate, cerium sulphate etc.
dissolve in water with the evolution of heat. This
means that the process is of exothermic nature. In
these salts the solubility in water decreases with rise
in temperature.
per 100 grams of solution is called percentage by
mass.
(D) Percentage by volume : Millilitres of solute per
100 mL of solution is called percentage by volume.
For example, if 25 mL ethyl alcohol is diluted with
water to make 100 mL solution then the solution thus
mass by volume.
For example, let 25 g glucose is dissolved in water to
make 100 mL solution then the solution is 25% mass
by volume glucose.
Note :
If heat energy is needed or absorbed in the process,
it is of endothermic nature. If heat energy is evolved
or released in the process, it is of exothermic nature.
(i) Effect of temperature on endothermic dissolution
process : Most of the salts like sodium chloride,
potassium chloride, sodium nitrate, ammonium
chloride etc. dissolve in water with the absorption of
heat. In all these salts the solubility increases with
rise in temperature. This means that sodium chloride
becomes more soluble in water upon heating.
(B) Parts Per Million (ppm) : Grams of solute
present per 100 mL of solution is called percentage
× 100
(b) Effect of Temperature and Pressure on
Solubility of a Solids :
m.e. of Na2CO3. H2O = 0.01 × 1000 = 10
(E) Mass by volume percentage :Grams of solute
Mass of the solvent
For example, the solubility of potassium chloride in
water at 20ºC and 1 atm. is 34.7 g per 100g of water.
This means that under normal conditions 100 g of
water at 20ºC and 1 atm. cannot dissolve more than
34.7g of KCl.
be acidic, basic or neutral ?
obtained is 25% ethyl alcohol by volume.
Mass of the solute
× 100 =

Note :
1. While expressing the solubility, the solution must
be saturated but for expressing concentration (mass
percent or volume percent), the solution need not to
be saturated in nature.
2. While expressing solubility, mass of solvent is
considered but for expressing concentration the
mass or volume of the solution may be taken into
consideration.
PAGE # 10
(c) Ef fect of Temperature on the Solubility
Ex.36 4 g of a solute are dissolved in 40 g of water to
form a saturated solution at 25ºC. Calculate the
of a Gas
(i) The solubility of a gas in a liquid decreases with
solubility of the solute.
Mass of solute
the rise in temperature.
Sol. Solubility =
(ii) The solubility of gases in liquids increases on
increasing the pressure and decreases on decreas-
Mass of solvent
× 100
Mass of solute = 4 g
Mass of solvent = 40 g
ing the pressure.
Solubility =
SAMPLE PROBLEMS
4
× 100 = 10 g
40
Ex.35 12 grams of potassium sulphate dissolves in
Ex.37 (a) What mass of potassium chloride would be
75 grams of water at 60ºC. What is the solubility
needed to form a saturated solution in 50 g of
of potassium sulphate in water at that temperature ?
water at 298 K ? Given that solubility of the salt is
Sol. Solubility =
46g per 100g at this temperature.
mass of solute
 100
mass of solvent
=
(b) What will happen if this solution is cooled ?
12
75
Sol. (a) Mass of potassium chloride in 100 g of water
×100 = 16 g
in saturated solution = 46 g
Thus, the solubility of potassium sulphate in
Mass of potassium chloride in 50 g of water in
water is 16 g at 60ºC.
saturated solution.
=
46
× (50g) = 23 g
100
(b) When the solution is cooled, the solubility of
salt in water will decrease. This means, that upon
cooling, it will start separating from the solution
in crystalline form.

PAGE # 11
ACIDS AND BASES
(a) Classification of Acids :
ACIDS
Substances with sour taste are regarded as acids.
Lemon juice, vinegar, grape fruit juice and spoilt milk
etc. taste sour since they are acidic. Many substances
can be identified as acids based on their taste but
some of the acids like sulphuric acid have very strong
action on the skin which means that they are corrosive
in nature. In such cases it would be according to
modern definition An acid may be defined as a substance which
releases one or more H+ ions in aqueous solution.
Acids are mostly obtained from natural sources.
(i) Classification of acids on the basis of their
Source
On the basis of their source, acids can be classified
in two categories :
(A) Organic acids
(B) Inorganic acids
(A) Organic acids
The acids which are usually obtained from
organisms are known as organic acids. Oxalic acid
[(COOH)2], acetic acid (CH3COOH) etc. are very
common examples of organic acids. Some other
organic acids with their natural sources are given in
the following Table.
Some Organic Acids with Their Natural Sources
S.No. Organic acid
Natural sources
S.No. Organic acid Natural sources
1
Acetic acid
Vinegar
7
Oleic acid
Olive oil
2
Citric acid
Citrus fruits (like
orange and lemon)
8
Stearic acid
Fats
3
Butyric acid
Rancid butter
9
Amino acid
Proteins
4
Formic acid
Sting of bees and ants
10
Uric acid
Urine
5
Lactic acid
Sour milk
11
Tartaric acid
Tamarind
6
Malic acid
Apples
12
Oxalic acid
Tomatoes
It may be noted that all organic acids contain carbon
as one of their constituting elements. These are weak
acids and, therefore, do not ionise completely in their
aqueous solutions. Since these acids do not ionise
completely in their aqueous solutions, therefore, their
solutions contains both ions as well as
undissociated molecules. For example, formic acid’s
aqueous solution contains H3O+, HCOO– as well as
undissociated HCOOH molecules.
HCOOH + H2O
Formic acid
H3O+
+
HCOO–
Hydronium ion Formate ion
(B) Inorganic Acids.
The acids which are usually obtained from minerals
are known as inorganic acids. Since the acids are
obtained from minerals, therefore, these acids are
also called mineral acids. Some common examples
of inorganic acids are : Hydrochloric acid (HCl),
Sulphuric acid (H2SO4), Nitric acid (HNO3) etc.
It may be pointed out that except carbonic acid
(H2CO3), these acids do not contain carbon. Acids
like HCl, H2SO4 and HNO3 are strong acids which
ionise completely in their aqueous solutions and,
therefore, their aqueous solutions do not contain any
undissociated molecules.
(ii) Classification of acids on the basis of their
Basicity :
The basicity of an acid is defined as the number of
hydronium ions [H3O+ (aq.)] that can be produced by
the complete ionisation of one molecule of that acid
in aqueous solution.
For example, basicity of HCl, H2SO4, H3PO4 is 1, 2
and 3 respectively because one molecule of these
acids, on ionisation, produces 1, 2 and 3 hydronium
ions in aqueous solution respectively.
It may be pointed out here that the basicity of an acid
is determined by number of hydronium ions produced
per molecule of an acid on ionisation and not the
number of hydrogen atoms present in one molecule
of an acid. For example, basicity of acetic acid
(CH3COOH) is 1 because one molecule of acetic acid,
on ionisation in aqueous solution, produces one
hydronium ion although one molecule of acetic acid
contains four hydrogen atoms.
CH3COOH
Acetic acid
+ H2O
H3O+ + CH3COO–
Hydronium ion Acetate ion
On the basis of basicity, the acids can be classified
as under :
PAGE # 12
(C) Tribasic Acids :
When one molecule of an acid on complete ionisation
produces three hydronium ions (H3O+) in aqueous
solution, the acid is said to be a tribasic acid.
An example of tribasic acids is Phosphoric acid
(H3PO4).
(A) Monobasic Acids :
When one molecule of an acid on complete ionisation
produces one hydronium ion (H 3 O + ) in aqueous
solution, the acid is said to be a monobasic acid.
Examples of Monobasic Acids.
Some examples of monobasic acids are :
(i) Hydrochloric acid (HCl)
(ii) Hydrobromic acid (HBr)
(iii) Nitric acid (HNO3)
(iv) Acetic acid (CH3COOH)
(v) Formic acid (HCOOH)
(D) Tetrabasic Acids :
When one molecule of an acid on complete ionisation
produces four hydronium ions (H3 O+) in aqueous
solution, the acid is said to be a tetrabasic acid.
An example of tetrabasic acids is silicic acid (H4SiO4).
(iii) Classification of acids on the basis of their
strength :
Characteristics of Monobasic Acids.
Two important characteristics of monobasic acids
are :
We know that acids ionise in the aqueous solution to
produce hydronium ions. So, the strength of an acid
depends upon the degree of ionisation, usually
denoted by the letter alpha ().
Degree of ionisation of an acid ()
(i) A monobasic acid ionises in one step in aqueous
solution. For example,
HCl + H2O
H3O+ + Cl–
(Single step ionisation)
(ii) A monobasic acid forms only single salt or a normal
salt. For example,
HCl + NaOH  NaCl + H2O
Sodium chloride
(Normal salt)
(B) Dibasic Acids :
When one molecule of an acid on complete ionisation
produces two hydronium ions (H 3 O +) in aqueous
solution, the acid is said to be a dibasic acid.
Examples of Dibasic Acids :
Some examples of dibasic acids are :
(i) Sulphuric acid (H2SO4)
(ii) Sulphurous acid (H2SO3)
(iii) Carbonic acid (H2CO3)
(iv) Oxalic acid [(COOH)2]
(v) Hydrofluoric acid (HF)
Characteristics of Dibasic Acids :
Two important characteristics of dibasic acids are :
(i) A dibasic acid ionises in two steps in aqueous
solution. For example, sulphuric acid which is a dibasic
acid ionises to produce bisulphate ion
(HSO4–) in
the first step which further ionises to produce sulphate
ion (SO42–) in the second step.
H2SO4 + H2O
H3O+ + HSO4–
Sulphuric acid
–
HSO4 + H2O
Bisulphate ion
+
H3O + SO42–
Sulphate ion
\
(ii) Because of the presence of two replaceable
hydrogen ions, a dibasic acid forms two series of salts
i.e., an acid salt and a normal salt. For example, H2SO4
reacts with NaOH to form NaHSO4 (an acid salt) and
Na2SO4 (a normal salt)
NaOH + H2SO4 NaHSO4 + H2O
Sodium hydrogen
sulphate
(An acid salt)
2NaOH + H2SO4 Na2SO4 + 2H2O
Sodium sulphate
(Normal salt)
=
Number of molecules of the acid undergoing ionisation
 100
Total number of acid molecules
More the degree of ionisation () of an acid, more stronger
it will be. Generally, if the degree of ionisation () for an
acid is greater than 30%, it is considered to be a strong
acid. If it is less than 30%,it is considered to be a weak
acid.
On the basis of degree of ionisation, the acids can be
classified as under:
(A) Strong Acids :
The acids which undergo almost complete ionisation
in a dilute aqueous solution, thereby producing a high
concentration of hydronium ions (H3O+) are known as
strong acids.
Examples of strong acids :
Some examples of strong acids are :
(i) Hydrochloric acid (HCl)
(ii) Sulphuric acid (H2SO4)
(iii) Nitric acid (HNO3)
All these three mineral acids are considered to be
strong acids because they ionise almost completely
in their dilute aqueous solutions.
(B) Weak Acids :
The acids which undergo partial or incomplete
ionisation in a dilute aqueous solution, thereby
producing a low concentration of hydronium ions
(H3O+) are known as weak acids.
Examples of weak acids :
Some examples of weak acids are :
(i) Acetic acid (CH3COOH)
(ii) Formic acid (HCOOH)
(iii) Oxalic acid [(COOH)2]
(iv) Carbonic acid (H2CO3)
(v) Sulphurous acid (H2SO3)
(vi) Hydrogen sulphide (H2S)
(vii) Hydrocyanic acid (HCN)
The aqueous solution of weak acids contain both ions
as well as undissociated molecules.
HCl (aq) H+ (aq) + Cl– (aq)
CH3COOH (aq)
CH3COO–(aq) + H+(aq)
It may be noted that single arrow () is used to
represent complete ionization, while double half
headed arrow 
 is used to represent partial
ionization.
PAGE # 13
It must be mentioned here that concentration of an
acid simply tells the amount of water in the acid. It may
not be confused with strength of an acid, which is a
measure of concentration of hydronium ion it produces
in aqueous solution.
(iv) Classification on the basis of Concentration of
the Acid :
By the term concentration, we mean the amount of
water present in the given sample of acid solution in
water.
A concentrated acid may not necessarily be a strong
acid while a dilute acid may not necessarily be a weak
acid. A strong acid will remain strong even if it is dilute
because it produces a large concentration of hydronium
ions in aqueous solution. On the other hand, a weak
acid will remain weak even when concentrated
because it will produce lesser concentration of
hydronium ions in aqueous solution.
(A) Concentrated Acid :
The sample of an acid which contains very small or no
amount of water is called a concentrated acid.
(B) Dilute Acid :
The sample of an acid which contains far more amount
of water than its own mass is known as a dilute acid
CHEMICAL FORMULAE, TYPES AND USES OF SOME COMMON ACIDS
Name
Type
Chemical Formula
Where found or used
Carbonic acid
Mineral acid
H2CO3
In soft drinks and lends fizz.
Nitric acid
Mineral acid
HNO3
Used in the manufacture of explosives (TNT,
Nitroglycerine) and fertilizers (Ammonium nitrate,
Calcium nitrate, Purification of Au, Ag)
HCl
In purification of common salt, in textile industry
as bleaching agent, to make aqua regia, in
stomach as gastric juice, used in tanning industry
Commonly used in car batteries, in
the manufacture of fertilizers (Ammonium
sulphate, super phosphate) detergents etc, in
paints, plastics, drugs, in manufacture of artificial
silk, in petroleum refining.
Hydrochloric acid Mineral acid
Sulphuric acid
Mineral acid
H2SO4
Phosphoric acid
Mineral acid
H3PO4
Formic acid
Organic acid
HCOOH
Acetic acid
Organic acid
CH3COOH
Used in antirust paints and in fertilizers.
Found in the stings of ants and bees, used in
tanning leather, in medicines for treating gout.
Found in vinegar, used as solvent in the
manufacture of dyes and perfumes.
Responsible for souring of milk in curd.
Lactic acid
Organic acid
CH3CH(OH)COOH
Benzoic acid
Organic acid
C6H5COOH
Used as a food preservative.
Citric acid
Organic acid
C6H8O7
Present in lemons, oranges and citrus fruits.
Tartaric acid
Organic acid
C4H6O6
Present in tamarind.
(i) Action with metals :
Dilute acids like dilute HCl and dilute H2SO4 react with
certain active metals to evolve hydrogen gas.
(ii) Action with metal oxides :
Acids react with metal oxides to form salt and water.
These reactions are mostly carried out upon heating.
e.g.
ZnO(s) + 2HCl (aq)  ZnCl2(aq) + H2O()
2Na(s) + 2HCl (dilute)  2NaCl(aq) + H2(g)
MgO(s) + H2SO4(aq)  MgSO4(aq) + H2O()
(b) Chemical Properties of Acids :
Chemical Properties of Acids :
Mg(s) + H2SO4 (dilute)  MgSO4(aq) + H2(g)
Metals which can displace hydrogen from dilute acids
are known as active metals. e.g. Na, K, Zn, Fe, Ca, Mg
etc.
Zn(s) + H2SO4 (dilute)  ZnSO4(aq) + H2(g)
The active metals which lie above hydrogen in the
activity series are electropositive and more reactive in
nature. Their atoms lose electrons to form positive ions
and these electrons are accepted by H+ ions of the
acid. As a result, H2 is evolved.
e.g.
Zn(s)  Zn2+ (aq) + 2e–
+
2–
2–
2H (aq) + SO4 (aq) + 2e  H2(g) + SO4 (aq)
–
Zn(s) + 2H+(aq)  Zn++(aq) + H2(g)
CuO(s) + 2HCl(aq.)  CuCl2(aq) + H2O()
(Black)
(Bluish green)
(iii) Action with metal carbonates and metal
bicarbonates : Both metal carbonates and bicarbonates
react with acids to evolve CO2 gas and form salts.
e.g.
CaCO3(s)+ 2HCl(aq)  CaCl2(aq) + H2O() + CO2(g)
Calcium
Calcium
carbonate
chloride
2NaHCO3(s) + H2SO4(aq)  Na2SO4(aq) + 2H2O(aq) + 2CO2(g)
Sodium
bicarbonate
Sodium
sulphate
PAGE # 14
(iv) Action with bases :
Acids react with bases to give salt and water.
BASE
HCl (aq) + NaOH(aq)  NaCl + H2O
These reactions are mostly carried out upon heating.
3.
Action with metal carbonates and metal bicarbonates :
Both metal carbonates and bicarbonates react with
acids to evolve CO2 gas and form salts.
e.g.
CaCO3(s)+ 2HCl(aq)
 CaCl2(aq) + H2O()+ CO2(g)
Calcium
carbonate
 Na2SO4(aq) + 2H2O(aq) + 2CO2(g)
Sodium
bicarbonate
4.
(a) Alkalies :
Calcium
chloride
2NaHCO3(s) + H2SO4(aq)
Sodium
sulphate
Action with bases :
Acids react with bases to give salt and water.
Commercial Name
Chemical
Formula
NaOH
Uses
In manufacture of soap, paper, pulp, rayon,
refining of petroleum etc.
In alkaline storage batteries, manufacture of
soap, absorbing CO2 gas etc.
Sodium hydroxide
Caustic soda
Potassium hydroxide
Caustic potash
Calcium hydroxide
Slaked lime
Ca(OH)2
In manufacture of bleaching powder,
softening of hard water etc.
Magnesium hydroxide
Milk of magnesia
Mg(OH)2
As an antacid to remove acidity from stomach.
KOH
Aluminium hydroxide
–
Al(OH)3
As foaming agent in fire extinguishers.
Ammonium hydroxide
–
NH4OH
In removing grease stains from clothes and in
cleaning window panes.
(b) Chemical Properties :
1.
Some bases like sodium hydroxide and potassium
hydroxide are water soluble. These are known as
alkalies. Therefore water soluble bases are known as
alkalies eg. KOH, NaOH.
Bases like Cu(OH)2, Fe(OH)3 and Al(OH)3 these are
not alkalies.
A list of a few typical bases along with their chemical
formulae and uses is given below-
HCl + NaOH  NaCl + H2O
Name
Substances with bitter taste and soapy touch are
regarded as bases. Since many bases like sodium
hydroxide and potassium hydroxide have corrosive
action on the skin and can even harm the body, so
according to the modern definition A base may be defined as a substance capable of
releasing one or more OH¯ ions in aqueous solution.
(c) Strong and Weak Bases :
Action with metals :
Metals like zinc, tin and aluminium react with strong
(i) Strong base : A base contains one or more hydroxyl
alkalies like NaOH (caustic soda), KOH (caustic
potash) to evolve hydrogen gas.
upon ionisation. Bases which are almost completely
(OH–) groups which it releases in aqueous solution
ionised in water, are known as strong bases.
Zn(s) + 2NaOH(aq)  Na2ZnO2(aq) + H2(g)
Sodium zincate
e.g.
Sodium hydroxide (NaOH), potassium hydroxide
Sn(s) + 2NaOH(aq)  Na2SnO2(aq) + H2(g)
Sodium stannite
2Al(s)+ 2NaOH + 2H2O  2NaAlO2(aq) + 3H2(g)
Sodium meta
aluminate
2.
NaOH(s) + Water  Na+(aq) + OH–(aq)
KOH(s)
+ Water  K+(aq) + OH–(aq)
Both NaOH and KOH are deliquescent in nature which
means that they absorb moisture from air and get
Action with non-metallic oxides :
Acids react with metal oxides, but bases react with
oxides of non-metals to form salt and water.
e.g.
2NaOH(aq) + CO2(g)  Na2CO3(aq) + H2O()
Ca(OH)2(s) + SO2(g)  CaSO3(s) + H2O()
Ca(OH)2(s) + CO2(g)  CaCO3(s)
(KOH), barium hydroxide Ba(OH)2 are all strong bases.
liquefied.
(ii) Weak bases : Bases that are feebly ionised on
dissolving in water and produce less concentration of
hydroxyl ions are called weak bases.
e.g. Ca(OH)2, NH4OH
+ H2O()
PAGE # 15
(A) Limitations of Arrhenius theory :
Acids
Sour in taste.
Change Colours of indicators
e.g. litmus turns from blue to
red, phenolphthalein remains
colourless.
Show electrolytic conductivity
in aqueous solution.
Acidic properties disappear
when react with bases
(Neutralization)
Acids decompose carbonate
salts.
• It is applicable only to aqueous solutions. For the
acidic or basic properties, the presence of water is
absolutely necessary.
Bases
Bitter in taste.
Change colours of indicators e.g.
litmus turns from red to blue
phenolphthalein turns from
colourless to pink.
Show electrolytic conductivity in
aqueous solution
• The concept does not explain the acidic or basic
properties of acids or bases in non - aqueous solvents.
• It fails to explain the basic nature of compounds like
NH3, Na 2CO3 etc., which do not have OH– in their
molecules to furnish OH– ions.
Basic properties disappear when
react with acids (Neutralization)
• It fails to explain the acidic nature of non - protic
compounds like SO2, P2O5, CO2, NO2 etc., which do not
have hydrogen in their molecules to furnish H+ ions.
No decomposition of carbonate
salts by bases
• It fails to explain the acidic nature of certain salts like
AlCl3 etc., in aqueous solutions.
THEORIES OF ACIDS AND BASES
(b) Acid Base Concept of Bronsted and
Lowry :
(a) Arrhenius Theory :
This concept was given in 1884 .
This theory was given by Bronsted, a Danish chemist
and Lowry, an English chemist independently in 1923.
According to it, an acid is a substance, molecule or ion
which has a tendency to release the proton
(protogenic) and similarly a base has a tendency to
accept the proton (protophilic).
According to this theory all substances which give H+
ions when dissolved in water are called acids, while
those which ionise in water to give OH– ions are called
bases.
The main points of this theory are (i) An acid or base when dissolved in water, splits into
ions. This is known as ionisation.
e.g.
HCl + H2O
H3O+ + Cl–
In this reaction, HCl acts as an acid because it donates
a proton to the water molecule. Water, on the other
hand, behaves as a base by accepting a
proton.
(ii) Upon dilution, the ions get separated from each other.
This is known as dissociation of ions.
(iii) The fraction of the acid or base which dissociates
into ions is called its degree of dissociation and is
denoted by alpha  which can be calculated by the
following formula :
=

No. of molecules dissociate d at equilibriu m
total no. of molecules
(iv) The degree of dissociation depends upon the nature
of acid or base. Strong acids and bases are highly
dissociated, while weak acids and bases are
dissociated to lesser extent.
(v) The electric current is carried by the movement of
ions. Greater the ionic mobility more will be the
conductivity of the acid or base.
H3O+

HCl + H2O
H3O+ + Cl–
e.g.
HA + H2O
Acid
H3O+ + A¯
2H3O+ + SO4–2
H2SO4 + 2H2O
Acid
BOH
Base
Water
NaOH
Base
NH4OH
Base
(ii) NH4+ + H2O
H3O+ + NH3
(iii) NH3 + H2O
NH4+ + OH–
In the reactions (i) and (ii) water is acting as a base,
while in reaction (iii) it is acting as an acid.Thus water
can donate as well as accept H+ and hence can act as
both acid and base.
(vi) The H+ ions do not exist as such and exist in
combination with molecules of H 2O as H 3O + ions
(known as hydronium ion).
H+ + H2O
Note :
Bronsted and Lowry theory is also known as proton
donor and proton acceptor theory.
Other examples :
(i) CH3COOH + H2O
H3O+ + CH3COO–
Note :
The species like H2O, NH3, CH3COOH which can act
as both acid and base are called amphiprotic.
Moreover according to theory, an acid on losing a proton
becomes a base, called conjugate base, while the
base by accepting proton changes to acid called
conjugate acid.
+
accepts H
+
B + OH¯
–
Water
+
CH3COO + H3O
Base
Acid
CH3COOH + H2O
Base
Acid
Na+ + OH¯
+
loses H
Water
NH4+ + OH¯
Here CH3COO– ion is conjugate base of CH3COOH,
while H3O+ ion is conjugate acid of H2O.
PAGE # 16
(i) Merits :
(ii) Characteristics of species which can act as Lewis
bases :
(A) Besides water any other solvent, which has the
tendency to accept or lose a proton may decide the
acidic or basic behaviour of the dissolved substance.
(A) Neutral species having at least one lone pair of
electrons : For example, ammonia amines, alcohols
etc, act as Lewis bases as they contain a pair of
electrons.
(B) This theory states that the terms acid and base are
comparative. A substance may act as an acid in one
solvent, while as a base in another solvent.
e.g. Acetic acid acts as an acid in water while as a
base in HF.
(B) Negatively charged species or simple anions :
For example chloride (Cl–), cyanide (CN–), hydroxide
(OH–) ions etc. act as Lewis bases.
(ii) Demerits :
(A) Many acid - base reactions proceed without H+
transfer.
(C) Multiple bonded compounds : The compounds
such as CO, NO, ethylene, acetylene etc. can act as
Lewis bases.
SO2+ + SO42-
e.g. SO2 + SO3

(c) Lewis theory :
The theory was given by G.N. Lewis in 1938. According
to it, an acid is a species which can accept a pair of
electrons, while the base is one which can donate a
pair of electrons.

Note :
It is also known as electron pair donor and electron
pair acceptor theory.
e.g.
(i) FeCl3 and AlCl3 are Lewis acids, because the central
atoms have only six electrons after sharing and need
two more electrons.
(iv) Limitations of Lewis theory :
(A) Lewis theory fails to explain the relative strength of
acids and bases.
INDICATORS
,
(ii) NH3 is a Lewis base as it has a pair of electrons
which can be easily donated.
Lewis acids :- CH3+, H+, BF3, AlCl3, FeCl3 etc.
Lewis base :- NH3, H2O, R–O–R, R – OH, CN¯, OH¯
etc.
(A) Molecules in which the central atom has
incomplete octet : Lewis acids are electron deficient
molecules such as BF3, AlCl3, GaCl3 etc.
[H3N
An indicator indicates the nature of a particular solution
whether acidic, basic or neutral. Apart from this, indicator
also represents the change in nature of the solution
from acidic to basic and vice versa. Indicators are
basically coloured organic substances extracted from
different plants. A few common acid base indicators are(a) Litmus :
Litmus is a purple dye which is extracted from ‘lichen’
a plant belonging to variety Thallophyta. It can also be
applied on paper in the form of strips and is available
as blue and red strips. A blue litmus strip, when dipped
in an acid solution acquires red colour. Similarly a red
strip when dipped in a base solution becomes blue.
(i) Characteristics of species which can act as Lewis
acids :
H3N •• + AlCl3
Note :
It may be noted that all Bronsted bases are also Lewis
bases, but all Lewis acids are not Bronsted acids.
(b) Phenolphthalein :
It is also an organic dye and acidic in nature. In neutral
or acidic solution, it remains colourless while in the
basic solution, the colour of indicator changes to pink.
AlCl3]
(B) Molecules in which the central atom has empty
d-orbitals : The central atom of the halides such as
TiCl4, SnCl4, PCl3, PF5, SF4, TeCl4. etc., have vacant dorbitals. These can, therefore, accept an electron pair
and act as Lewis acids.
(c) Methyl Orange :
Methyl orange is an orange or yellow coloured dye and
basic in nature. In the acidic medium the colour of
indicator becomes red and in the basic or neutral
medium, its colour remains unchanged.
(d) Red Cabbage Juice :
It is purple in colour in neutral medium and turns red
or pink in the acidic medium. In the basic or alkaline
medium, its colour changes to green.
(C) Simple cations : All cations are expected to act as
Lewis acid, since they are electron deficient in nature.
(e) Turmeric Juice :
(D) Molecules having a multiple bond between
atoms of dissimilar electronegativity : Typical
examples of molecules belonging to this class of
Lewis acids are CO2, SO2 and SO3.
It is yellow in colour and remains as such in the neutral
and acidic medium. In the basic medium its colour
becomes reddish or deep brown.

Note :
Litmus is obtained from LICHEN plant.
PAGE # 17
concentration are inversely proportional to each other.
NEUTRALISATION
The relation between them can also be expressed as-
It may be defined as a reaction between acid and
base present in aqueous solution to form salt and
water.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O()
Basically neutralisation is the combination between
–
H+ ions of the acid with OH ions of the base to form H2O.
e.g.
–
 1 

 H 
pH = – log [H+] = log 
–
H+(aq) + Cl (aq) + Na+(aq) + OH (aq)  Na+(aq) + Cl–(aq) + H2O()
+
–
H (aq) + OH (aq)  H2O()
Neutralisation reaction involving an acid and base is
of exothermic nature. Heat is evolved in all
neutralisation reactions. If both acid and base are
strong, the value of heat energy evolved remains same
irrespective of their nature.

e.g .
HCl (aq) + NaOH (aq)
Strong
acid
 NaCl (aq) + H2O () + 57.1 KJ
Mathematically , pOH = – log [OH–] = log
Strong
base
HNO3 (aq) + KOH (aq)
Strong
Strong
acid
base
1
[OH – ]
Moreover, pH + pOH = 14.
Thus, if pH value of solution is known, its pOH value
can be calculated.
 KNO3 (aq) + H2O () + 57.1 KJ
Strong acids and strong bases are completely ionised
of their own in the solution. No energy is needed for
their ionisation. Since the cation of base and anion of
acid on both sides of the equation cancel out
completely, the heat evolved is given by the following
reaction H+ (aq) + OH– (aq)  H2O () + 57.1 KJ
So, negative logarithm of hydrogen ion concentration
is known as pH.
e.g.
Let the [H+] of an acid solution be 10–3 M. Its pH can be
calculated as pH = – log [H+]
= – log [10–3]
= (–) (–3) log 10
= 3 ( log 10 = 1)
Note :
Just as the [H+] of a solution can be expressed in terms
of pH value, the [OH–] can be expressed as pOH.

Note :
There are some solutions which have definite pH i.e.,
their pH do not change on dilution or on standing for
long. Such solutions are called buffer solutions.
APPLICATIONS OF NEUTRALISATION
(i) People particularly of old age suffer from acidity
problems in the stomach which is caused mainly due
to release of excessive gastric juices containing HCl.
The acidity is neutralised by antacid tablets which
contain sodium hydrogen carbonate (baking soda),
magnesium hydroxide etc.
(ii) The stings of bees and ants contain formic acid. Its
corrosive and poisonous effect can be neutralised by
rubbing soap which contains NaOH (an alkali).
(iii) The stings of wasps contain an alkali and its
poisonous effect can be neutralised by an acid like
acetic acid (present in vinegar).
(iv) Farmers generally neutralize the effect of acidity in
the soil caused by acid rain by adding slaked lime
(Calcium hydroxide) to the soil.
According to Arrhenius theory, an acid releases H+ ion
in aqueous solution. The concentration of these ions
is expressed by enclosing H+ in square bracket i.e. as
[H+]. Thus, greater the [H+] ions, stronger will be the
acid. However, according to pH scale, lesser the pH
value, stronger will be the acid. From the above
discussion, we can conclude that pH value and H+ ion
PAGE # 18
METALS AND NON-METALS
(iii)  Most reactive metals occur as salts as
carbonates, sulphates, halides etc. (e.g., Ca, Mg, K
etc.).
OCCURRENCE OF METALS
All metals are present in the earth’s crust either in the
free state or in the form of their compounds. Aluminium
is the most abundant metal in the earth’s crust. The
second most abundant metal is iron and third one is
calcium.
Some common ores are listed in the table
Nature of
ore
(a) Native and Combined States of Metals :
Metals occur in the crust of earth in the following two
states (i) Native state or free state : A metal is said to occur
in a free or a native state when it is found in the crust
of the earth in the elementary or uncombined form.
The metals which are very unreactive (lying at the
bottom of activity series) are found in the free state.
These have no tendency to react with oxygen and are
not attacked by moisture, carbon dioxide of air or other
non-metals. Silver, copper, gold and platinum are
some examples of such metals.

Metal
Name of the ore
Composition
Aluminium
Bauxite
Al2O3.2H2O
Copper
Oxide
ores
Iron
Haematite
Fe2O3
Copper
CuFeS2
Zinc
Lead
Mercury
Calcium
Zinc blende
Galena
Cinnabar
Limestone
Cu2S
ZnS
PbS
HgS
CaCO3
Zinc
Sodium
Magnesium
Calamine
Rock salt
Carnallite
ZnCO3
NaCl
KCl.MgCl2.6H2O
Calcium
Silver
Calcium
Fluorspar
Horn silver
Gypsum
CaF2
AgCl
CaSO4.2H2O
Magnesium
Epsom salt
MgSO4.7H2O
Barium
Barytes
BaSO4
Lead
Anglesite
PbSO4
Carbonate
ores
Halide
ores
Sulphate
ores
Cu2O
Fe3O4
Copper pyrites
Copper glance
Sulphide
ores
(ii) Combined state : A metal is said to occur in a
combined state if it is found in nature in the form of its
compounds. e.g. Sodium , magnesium etc.
Note :
Copper and silver are metals which occur in the free
state as well as in the combined state.
Cuprite
Magnetite
Important Minerals/Ores of Different elements
MINERALS AND ORES
The natural substances in which metals or their
compounds occur either in native state or combined
state are called minerals.
The minerals are not pure and contain different types
of other impurities. The impurities associated with
minerals are collectively known as gangue or matrix.
The mineral from which the metal can be conveniently
and profitably extracted, is called an ore.
For example, aluminium occurs in the earth’s crust
in the form of two minerals, bauxite (Al2O3.2H2O) and
clay (Al2O3.2SiO2.2H2O). Out of these two, aluminium
can be conveniently and profitably extracted from bauxite
. So, bauxite is an ore of aluminium.
Note : Oxygen is the most abundant element in earth’s
crust.
(a) Types of Ores :
The most common ores of metals are oxides,
sulphides, carbonates, sulphates, halides, etc. In
general, very unreactive metals (such as gold, silver,
platinum etc.) occur in elemental form or free state.
(i) Metals which are only slightly reactive occur as
sulphides (e.g., CuS, PbS etc.).
(ii)Reactive metals occur as oxides (e.g., MnO2, Al2O3
etc.).
Name
Formula
1.
Alumina
Al2O3
2.
Azurite
2CuCO3.Cu(OH)2
3.
Anhydrite
CaSO4
4.
Argentite
Ag2S
5.
Anglesite
PbSO4
6.
Bauxite
Al2O3 .2H2O
7.
Borax
Na2B4O7.10H2O
8.
Chile salt petre
NaNO3
9.
Cinnabar
HgS
10. Calcia
CaO
11. Carnallite
KCl.MgCl26H2O
12. Calamine
ZnCO3
13. Cassiterite
SnO2
14. Copper pyrites
CuFeS2
(Chalcopyrite)
15. Copper glance
Cu 2S
16. Cuprite
Cu2O
(Ruby copper)
17. Corundum
Al2O3
19
19
PAGE # 19
18. Cryolite
Na3AlF6
19. Zinc blende
ZnS
20. Chalk
CaCO3
(marble, Aragonite)
21. Siderite
Au, Pt, Ag
21. White gold :
Pt
22. Metalloids elements :
B, Si, Ge, As, Sb, Te
23. Poorest conductor of current :
Pb (metal), S (non-metal)
24. Hardest naturally occurring element: Diamond
FeCO3
22. Dolomite
CaCO3.MgCO3
23. Epsom salt
MgSO4.7H2O
(Epsomite)
24. Fluorspar
CaF2
25. Gypsum
CaSO4.2H2O
26. Galena
PbS
27. Horn silver
AgCl
(Chlorargyrite)
28. Haematite
20. Noble metals :
Fe2O3
(Red)
29. Iron pyrite
FeS2
30. Limonite
Fe2O3.3H2O
(Brown)
31. Magnesite
MgCO3
32. Malachite
CuCO3.Cu(OH)2
33. Magnetite
Fe3O4
34. Pyrolusite
MnO2
35. Salt petre (Indian)
KNO3
25. Most abundant gas :
N2
26. Amphoteric oxides :
ZnO, PbO, Al2O3, SnO,
BeO, As2O3, Sb2O3
27. Neutral oxides of non-metals :
CO, N2O, NO, H2O
28. Dry ice :
CO2 (solid)
Some important Alloys
Alloy
Compounds
1.
Brass :
Cu + Zn
2.
Bronze :
Cu + Sn
3.
Bell metal :
Cu + Sn
4.
Duralumin :
Al + Cu + Mg + Mn
5.
German silver :
Cu + Zn + Ni
6.
Gun metal :
Cu + Sn + Zn
7.
Magnalium :
Al + Mg
8.
Solder :
Pb + Sn
9.
Type metal :
Pb + Sb + Sn
1.
Aqua - regia :
Conc. HNO3 + Conc. HCl (1 : 3)
2.
Blue vitriol :
CuSO4. 5H2O
3.
Baking powder :
NaHCO3
4.
Bleaching powder :
CaOCl2
5.
Brine
NaCl solution
6.
Coinage metals :
Cu, Ag and Au
7.
Carborundum :
SiC
8.
Caustic soda :
NaOH
9.
Caustic potash :
KOH
Important facts to remember
1.
Lowest electronegativity:
Cs
2.
Highest electronegativity :
F
3.
Three most abundant elements :
O, Si, Al
4.
Lowest m.pt. (metal) :
Mercury (Hg)
:
(m.pt = –38.9ºC)
5.
Highest m.pt. (metal) :
Tungsten (W)
6.
Lowest m.pt. and b.pt. (non-metal):He (b.pt -268.9ºC)
7.
Smallest atomic size :
H
8.
Largest atomic size :
Cs
9.
Total number of gaseous elements:11 (H2, He, N2, F2, Ne,
(m.pt) = 3410ºC)
in periodic table
10. Total number of liquid elements :
O2Cl2, Ar, Kr, Xe, Rn)
4 (Ga, Br, Cs, Hg) (Fr
10. Green vitriol
:
FeSO4.7H2O
11. Glauber salt :
Na2SO4.10H2O
and Uub are also liquid)
11. Total number of solid elements :
89 in periodic table
12. Hydrolith :
CaH2
12. Element containing no neutron :
H1
1
13. Heavy water :
D2O
14. King of chemicals :
H2SO4
15. Lime (or quick :
CaO
13. Most abundant element on earth : Oxygen
14. Rarest element on earth :
Astatine (At)
15. Most abundant metal on earth :
Al
16. Element having maximum
tendency for catenation :
lime or burnt lime
Carbon
17. Non-metal having highest m.pt. b.pt. : Diamond
16. Lime water
:
18. Liquid silver or quick silver :
Hg
17. Laughing gas :
19. Liquid non-metal :
Br2
Ca(OH)2
N2O
20
20
PAGE # 20
18. Lunar caustic :
AgNO3
These steps are briefly discussed below -
19. Milk of lime :
Ca(OH)2 in water
20. Milk of magnesia :
Paste of Mg(OH)2 in water
(a) Crushing and Grinding of Ore :
Most of the ores occur as big rocks in nature. They are
broken into small pieces with the help of crushers.
These pieces are then reduced to fine powder with the
help of a ball mill or a stamp mill.
(Antacid)
21. Magnesia :
MgO
22. Marsh gas :
CH4
23. Oil of vitriol :
Conc. H2SO4
24. Oleum :
H2S2O7
25. Plaster of paris :
CaSO4.
26. Washing soda
Na2CO3.10H2O
:
1
HO
2 2
27. Phosgene :
COCl2
28. Phosphine :
PH3
29. Producer gas :
A mixture of CO + N2 + H2
30. Rust :
Fe2O3xH2O
31. Soda-lime :
NaOH + CaO
32. Soda ash :
Na2CO2
33. Slaked lime :
Ca(OH)2
34. Water gas :
CO + H2
35. White vitriol :
ZnSO4.7H2O
METALLURGY
The process of extracting metals from their ores and
then refining them for use is called metallurgy.
The ores generally contain unwanted impurities such
(a) Crushing in a hammer mill
(B) Pulverisation in a stamp mill
Crushing and pulverisation of an ore
(b) Concentration of Ore or Enrichment of
Ore :
The process of removal of unwanted impurities
(gangue) from the ore is called ore concentration or
ore enrichment.
(i) Hydraulic washing (washing with water) :
Principle : This method is based upon the difference
in the densities of the ore particles and the impurities
(gangue).
Ores of iron, tin and lead are very heavy and, therefore,
they are concentrated by this method.
as sand, stone, earthy particles, limestone, mica etc.,
these are called gangue or matrix.
The process of metallurgy depends upon the nature of
the ore, nature of the metal and the types of impurities
present. Therefore, there is not a single method for the
extraction of all metals. However, most of the metals
can be extracted by a general procedure which involves
the following steps.
Various steps involved in metallurgical processes are (a) Crushing and grinding of the ore.
A hydraulic classifier
(b) Concentration of the ore or enrichment of the ore.
(ii) Froth floatation process :
(c) Extraction of metal from the concentrated ore.
Principle : This method is based on the principle of
(d) Refining or purification of the impure metal.
difference in the wetting properties of the ore and
gangue particles with oil and water respectively.
This method is commonly used for sulphide ores.
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21
PAGE # 21
(iii) Magnetic separation :
Principle : This method depends upon the difference
in the magnetic properties of the ores and gangue.
For example :
Calcination
ZnCO3(s) 
+ CO2(g)

 ZnO(s)
Calamine
Zinc oxide Carbon dioxide
This method is used for the concentration of haematite,
an ore of iron.
Calcination
 FeO(s)
FeCO3(s)    
Siderite
Ore
+
CO2(g)
Iron (II)
Carbon
oxide
dioxide
(ii) Roasting: It is the process of heating the concentrated
ore strongly in the presence of excess air.
Electromagnet
Nonmagnetic
particles
This process is used for converting sulphide ores to
metal oxide. In this process, the following changes
take place :
Moving belt
• the sulphide ores undergo oxidation to their oxides.
• moisture is removed.
• volatile impurities are removed.
Magnetic
particles
Magnetic separation method

Note :
The froth floatation process is commonly used for the
sulphide ores of copper, zinc, lead etc.
For example :
Roasting
2ZnS(s) + 3O2 (g) 
 2ZnO(s) + 2SO2(g)
Zinc
Oxygen
Zinc oxide Sulphur
sulphide (from air)
dioxide
(Zinc blende)
EXTRACTION OF METAL FROM
THE CONCENTRATED ORE
Roasting
4FeS2(s) + 11O2(g) 
  2Fe2O3(s) + 8SO2(g)
Iron
Oxygen
Ferric
Sulphur
pyrites
oxide
dioxide
Metal is extracted from the concentrated ore by the
following steps :
(a) Conversion of the concentrated ore into its oxide :
The production of metal from the concentrated ore
mainly involves reduction process. This can be usually
done by two processes known as calcination and
roasting . The method depends upon the nature of the
ore.
(b) Conversion of oxide to metal by reduction process
(a) Conversion of Ore into Metal Oxide :
These are briefly discussed below :
(i) Calcination: It is the process of heating the concentrated
ore in the absence of air.
The calcination process is used for the following
changes :
• to convert carbonate ores into metal oxide.
• to remove water from the hydrated ores.
• to remove volatile impurities from the ore.

Note :
Calcination is used for hydrated and carbonate ores
and roasting is used for sulphide ores.
(b) Conversion of Metal Oxide into Metal :
The metal oxide formed after calcination or roasting is
converted into metal by reduction. The method used
for reduction of metal oxide depends upon the nature
and chemical reactivity of metal.
The metals can be grouped into the following three
categories on the basis of their reactivity :
• Metals of low reactivity.
• Metals of medium reactivity.
• Metals of high reactivity.
These different categories of metals are extracted by
different techniques. The different steps involved in
separation are as follows :
22
22
PAGE # 22
(i) Reduction by heating : Metals placed low in the
reactivity series are very less reactive. They can be
obtained from their oxides by simply heating in air.
opening at the bottom of the furnace. The molten iron
is allowed to solidify in casts or moulds. It is called Pig
Iron or Cast Iron.
Roasting
The function of adding limestone in the extraction of
iron is to remove earthy impurities like sand from the
blast furnace by forming fusible slag.
2HgS(s) + 3O2(g)      2HgO(s) + 2SO2(g)
Mercuric Oxygen
Mercuric Sulphur
sulphide
oxide
dioxide
Heat

 2Hg()
2HgO(s)
Mercuric
oxide
+
O2(g)
Mercury metal
Flux is the chemical substance used to remove impurity from an ore and slag is the chemical substance
formed by the combination of flux with impurity.
Oxygen
(ii) Chemical Reduction (For metals in the middle of
the reactivity series) :
(B) Reduction with carbon monoxide : Metals can be
obtained from oxides by reduction with carbon
monoxide in the furnace.
The metals in the middle of the reactivity series, such
as iron, zinc, lead, copper etc. are moderately reactive.
These are usually present as sulphides or carbonates.
Therefore, before reduction the metal sulphides and
carbonates must be converted to oxides. This is done
by roasting and calcination. The oxides of these metals
cannot be reduced by heating alone. Therefore, these
metal oxides are reduced to free metal by using
chemical agents like carbon, aluminium, sodium or
calcium.
Heat
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
Ferric
Carbon
Iron Carbon
oxide
monoxide
dioxide
(C) Reduction with aluminium : Certain metal oxides
are reduced by aluminium to metals.
Heat
3MnO2 (s) + 4Al(s)  3Mn()
Manganese
(A) Reduction with carbon : The oxides of moderately
Cr2O3 (s) +
series) like zinc, copper, nickel, tin, lead etc. can be
Chromium
oxide
reduced by using carbon as reducing agent.
Zinc
Heat

Carbon
Zn (s)
+
Zinc
2Fe(s)
Ferric
Iron Metal
+ 3CO(g)
oxide
PbO(s)
+ C(s)
Lead oxide
Carbon
Pb(s)
Fe2O3 (s) +
Iron
oxide
monoxide
Fe2O3(s) + 3C(s)
Carbon
CO(g)
Carbon
oxide (Reducing agent) metal
+
Lead metal
2Al2O3 (s)
Aluminium
oxide
2Al (s)
Heat

 2Cr()
Aluminium
2Al (s)
Aluminium
+
Chromium
Heat

 2Fe()
+
Iron
Al2O3 (s)
Aluminium
oxide
Al2O3 (s)
Aluminium
oxide
Carbon
Note :
Reduction of metal oxides with aluminium is known
monoxide
as aluminothermy or thermite process.

CO(g)
(iii) Reduction by electrolysis or electrolytic
reduction: The oxides of active metals (which are high
up in the activity series) are very stable and cannot be
reduced by carbon or aluminium. These metals are
commonly extracted by the electrolysis of their fused
salts using suitable electrodes. This is also called
electrolytic reduction i.e. reduction by electrolysis.
For example, aluminium oxide is very stable and
aluminium cannot be prepared by reduction with
carbon. It is prepared by the electrolysis of molten
alumina (Al2O3).
Carbon
monoxide
One disadvantage of using carbon as reducing agent
is that small traces of carbon are added to metal as
impurity. Therefore, it contaminates the metals.

+
Manganese
dioxide
reactive metals (occurring in the middle of reactivity
ZnO(s) + C(s)
Aluminium
Note :
(i) Coke is very commonly used as a reducing agent
because it is cheap.
(ii) Formation of slag :
Heat
 Al
Al 3+
+ 3e–
Aluminium ion Electrons
Aluminium
(From molten (From cathode)
(At cathode)
alumina)
It may be noted that during electrolytic reduction of
molten salts, the metals are always obtained at the
cathode (negative electrode).
Calcium Silicate
The molten slag flows down the furnace. It is lighter
than molten iron. It does not mix with iron. Therefore,
the slag floats on the surface of molten layer of iron in
the blast furnace. It thus protects the freshly prepared
iron from oxygen, which otherwise would be converted
into its oxide. It is also drawn off through a separate

Note :
The process of extraction of metals by electrolysis
process is called electrometallurgy.
23
23
PAGE # 23

Note :
In electrolytic refining impure metal is made anode
and pure metal is made cathode.

Note :
Zone refining and Van Arkel method are used for
obtaining metals (Si, Ge etc.) of very high purity for
certain specific applications.
PURIFICATION OR REFINING OF METALS
The metal obtained by any of the above methods is
usually impure and is known as crude metal. The
process of purifying the crude metal is called refining.
(a) Liquation :
This is based on the principle that the metal to be
refined is easily fusible (melt easily) but the impurities
do not fuse easily. The impure metal is placed on the
sloping hearth of a furnace and gently heated. The
metal melts and flows down leaving behind the
impurities on the hearth. This method is used for
refining the metals having low melting points, such as
tin, lead, bismuth etc.
(b) Distillation :
This method is used for the purification of volatile
metals (which form vapours readily). Impure metal is
heated and its vapours are separately condensed in a
receiver. The non-volatile impurities are left behind.
This is used for mercury, cadmium and zinc.
(c) Electrolytic Refining :
This is most general and widely used method for the
refining of impure metals. Many metals such as copper,
zinc, tin, nickel, silver, gold etc. are refined electrolytically.
It is based upon the phenomenon of electrolysis. In
this method, the crude metal is cast into thick rods and
are made as anodes, while the thin sheets of pure
metal are made as cathodes. An aqueous solution of
salt of the same metal is used as an electrolyte. On
passing current through the electrolyte, the pure metal
from the anode dissolves into the electrolyte. An
equivalent amount of pure metal from the electrolyte is
deposited on the cathode. The soluble impurities go
in the solution whereas the insoluble impurities settle
down at the bottom of the anode and are known as
anode mud. In this way, the pure metal from anode
goes into electrolyte and from electrolyte it goes to the
cathode.
At anode : Cu Oxidation

 Cu+2 + 2e–
Copper
Copper
(from impure anode)
ion
At cathode : Cu2+ + 2e– Reduction
   Cu
Copper
ion
Copper
(deposited at cathode)
CORROSION OF METALS
Surface of many metals is easily attacked when
exposed to atmosphere. They react with air or water
present in the environment and form undesirable
compounds on their surfaces. These undesirable
compounds are generally oxides.
Thus, corrosion is a process of deterioration of metal
as a result of its reaction with air or water (present in
environment) surrounding it.
(a) Corrosion of Iron :
Iron corrodes readily when exposed to moisture and
gets covered with a brown flaky substance called rust.
This is also called Rusting of Iron. Chemically, the rust
is hydrated iron (III) oxide, Fe2O3.xH2O. Rusting is an
oxidation process in which iron metal is slowly oxidized
by the action of air (in presence of water). Therefore,
rusting of iron takes place under the following
conditions:
• Presence of air (or oxygen)
• Presence of water (moisture)
• More the reactivity of the metal, the more will be the
possibility of the metal getting corroded.
(i) Experiment to show that rusting of iron requires
both air and water We take three test tubes and put one clean iron nail in
each of the three test tubes:
(A) In the first test tube containing iron nail, we put
some anhydrous calcium chloride to absorb water (or
moisture) from the damp air present in the test tube
and make it dry.
(B) In the second test tube containing iron nail, we put
boiled water because boiled water does not contain
any dissolved air or oxygen in it. A layer of oil is put over
boiled water in the test tube to prevent the outside air
from mixing with boiled water.
(C) In the third test tube containing an iron nail, we put
unboiled water so that about two-third of the nail is
immersed in water and the rest is above water exposed
to damp air.
After one week, we observe the iron nails kept in all the
three test tubes.
24
24
PAGE # 24
(ii) We will obtain the following observations from
the experiment :
(A) No rust is seen on the surface of iron nail kept in
dry air in the first test tube. This tells us that rusting of
iron does not takes place in air alone.
(B) No rust is seen on the surface of iron nail kept in air
free boiled water in the second test tube. This tells us
that rusting of iron does not take place in water alone.
(C) Red brown rust is seen on the surface of iron nail
kept in the presence of both air and water in the third
test tube. This tells us that rusting of iron takes place
in the presence of both air and water together.
(iii) Prevention of rusting :
(A) Corrosion of metals can be prevented by coating
the metal surface with a thin layer of paint, varnish or
grease.
(B) Iron is protected from rusting by coating it with a
thin layer of another metal which is more reactive than
iron. This prevents the loss of electrons from iron
because the active metal loses electrons in preference
to iron. Zinc is commonly used for covering iron surface
of iron. The process of covering iron with zinc is called
galvanization. Iron is also coated with other metals
such as tin known as tin coating.
(C) By alloying : Some metals when alloyed with other
metals become more resistant to corrosion. For
example, when iron is alloyed with chromium and
nickel, it forms stainless steel. This is resistant to
corrosion and does not rust at all.
(D) To decrease rusting of iron, certain antirust
solutions are used. For example, solutions of alkaline
phosphates are used as antirust solutions.
(b) Corrosion of Aluminium :
Due to the formation of a dull layer of aluminium oxide
when exposed to moist air, the aluminium metal loses
its shine very soon after use. This aluminium oxide
layer is very tough and prevents the metal underneath
from further corrosion (because moist air is not able to
pass through this aluminium oxide layer). This means
sometimes corrosion is useful.
(c) Corrosion of Copper :
When a copper object remains in damp air for a
considerable time, then copper reacts slowly with
carbon dioxide and water of air to form a green coating
of basic copper carbonate [CuCO3.Cu(OH)2] on the
surface of the object. Since copper metal is low in the
reactivity series, the corrosion of copper metal is very,
very slow.
(d) Corrosion of Silver :
Silver is a highly unreactive metal, so it does not reacts
with oxygen of air easily. But, air usually contains a little
of sulphur compounds such as hydrogen sulphide gas
(H2S), which reacts slowly with silver to form a black
coating of silver sulphide (Ag2S). Silver ornaments
gradually turn black due to the formation of a thin silver
sulphide layer on their surface and silver is said to be
tarnished.
ALLOY
An alloy is a homogenous mixture of two or more metals
or a metal and a non-metal.
For example, iron is the most widely used metal. But it
is never used in the pure form. This is because iron is
very soft and stretches easily when hot. But when it is
mixed with a small amount of carbon (about 0.5 to
1.5%), it becomes hard and strong. The new form of
iron is called steel.
(a) Objectives of Alloy Making :
Alloys are generally prepared to have certain specific
properties which are not possessed by the constituent
metals. The main objects of alloy-making are:
(i) To increase resistance to corrosion : For example,
stainless steel is prepared which has more resistance
to corr osion than iron.
(ii) To modify chemical reactivity : The chemical
reactivity of sodium is decreased by making an alloy
with mercury which is known as sodium amalgam.
(iii) To increase the hardness : Steel, an alloy of iron
and carbon is harder than iron.
(iv) To increase tensile strength : Magnalium is an
alloy of magnesium and aluminium. It has greater
tensile strength as compared to magnesium and
aluminium.
(v) To produce good casting : Type metal is an alloy of
lead, tin and mercury.
(vi) To lower the melting point : For example, solder is
an alloy of lead and tin (50% Pb and 50% Sn). It has a
low melting point and is used for welding electrical
wires together.
(b) Some Important Alloys :
The approximate composition and uses of some
important alloys are given below:
(i) Steel : Steel is an alloy of iron and carbon containing
0.5 to 1.5% carbon. Steel is very hard, tough and strong.
It is used for making rails, screws, girders, bridges
etc. Steel can also be used for the construction of
buildings, vehicles, ships, etc.
(ii) Alloy Steels : Steel obtained by the addition of some
other elements such as chromium, vanadium, titanium,
molybdenum, manganese, cobalt or nickel to carbon
steel are called Alloy Steels.
(iii) Alloys of Aluminium : The common alloys of
aluminium are:
(A) Duralumin : It is an alloy containing aluminium,
copper and traces of magnesium and manganese. Its
percentage composition is - Al=95%, Cu=4%,
Mg=0.5%, Mn=0.5% It is stronger than pure aluminium.
Since duralumin is light and yet strong, it is used for
making bodies of aircrafts, helicopters, jets and
kitchenwares like pressure cookers etc.
25
25
PAGE # 25
(B) Magnalium : It is an alloy of aluminium and
magnesium having the composition: Al=95%, Mg=5% It
is very light and hard. It is more hard than pure
aluminium. It is used for making light instruments,
balance beams, pressure cookers etc.
(C) Alnico : It is an alloy containing aluminium, iron,
nickel, and cobalt. It is highly magnetic in nature and
can be used for making powerful magnets.
(iv) Alloys of Copper : The important alloys of copper
are Brass and Bronze.
(A) Brass : It is an alloy of copper and zinc having the
composition - Cu=80%, Zn=20%. Brass is more
malleable and more strong than pure copper. It is used
for making cooking utensils, pipes, hardware, nuts,
bolts, screws, springs etc.
(B) Bronze : It is an alloy of copper and tin having the
composition: Cu=90%, Sn=10% Bronze is very tough
and highly resistant to corrosion. It is used for making
utensils, statues, coins, hardware etc.
(C) German Silver : It is an alloy of copper, zinc and nickel
having the composition: Cu=60%, Zn=20%, Ni=20%. It is
used for making silverware, utensils and for
electroplating.
(v) Alloying of Gold : Pure gold is very soft and cannot
be used as such for jewellery. Therefore, it is generally
alloyed with other metals commonly copper or silver to
make it harder and modify its colour. The purity of gold
is expressed as carats. Pure gold is of 24 carat. A 18
carat gold means that it contains 18 parts of gold in 24
parts by weight of alloy. Most of the jewellery is made of
22 carat gold.
AMALGAM
Amalgams are homogenous mixtures of a metal and
mercury. For example, sodium amalgam contains
sodium and mercury.
Different amalgams are prepared according to their
uses. For example,
(i) Sodium amalgam is produced to decrease the
chemical reactivity of sodium metal. It is also used as
a good reducing agent.
(ii) Tin amalgam is used for silvering cheap mirrors.
(iii) The process of amalgamation is used for the
extraction of metals like gold or silver from their native
ores.
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26
PAGE # 26
STRUCTURE OF ATOM
ELECTRON
Electrons are the fundamental particles of all
substances.
(a) Cathode Rays - Discovery of Electron :
The nature and existence of electron was established
by experiments on conduction of electricity through
gases.

(iii) They consist of negatively charged particles. When
the cathode rays pass through an electric field, they
bent towards the positive plate of the electric field.
This indicates that cathode rays are negatively
charged.
Note :
In 1859, Julius Plucker started the study of conduction
of electricity through gases at low pressure in a
discharge tube.
A number of interesting things happen when a high
voltage (say, 10,000 V) is applied across the
electrodes of the discharge tube, and the pressure of
the gas inside the tube is lowered.
Thus, some sort of invisible rays travel from the
negative electrode to the positive electrode. Since the
negative electrode is called cathode, these rays were
called cathode rays. The colour of glow depends upon
the nature of the glass used. For soda glass the
(iv) Cathode rays can affect the photographic plate.
(v) The nature of cathode rays is independent of the
nature of gas used in discharge tube or material of
cathode.
fluorescence is of yellowish green colour.
(b) Properties of cathode rays :
(i) Cathode rays travel in a straight line at a high
velocity and generate normally from the surface of
the cathode. If an opaque object is placed in the path
of cathode rays its shadow falls on opposite side of
the cathode. It shows that cathode rays travel in
straight lines.
To vacuum
pump
(vi) Cathode rays are deflected in the magnetic field
also.
S
+
–
N
Shadow
High
voltage
Deflection of
cathode rays
in magnetic field
Cathode
(vii) If cathode rays are focused on a thin metal foil,
the metal foil gets heated up to incandescence.
–
Object
High
voltage
+
(viii) When cathode rays fall on materials having high
atomic mass, new type of penetrating rays of very
small wavelength are emitted which are called X rays.
Anode
(ii) They are a beam of minute material particles having
definite mass and velocity. When a light paddle wheel
is placed in the path of the cathode rays, the blades
of the paddle wheel begin to rotate. This also proves
that cathode rays have mechanical energy.
Thus, investigations on cathode rays showed that
these consisted of negatively charged particles.

Note :
The negatively charged particles of cathode rays were
called ‘negatrons‘ by Thomson. The name negatron
was changed to ‘electron‘ by Stoney.
27
27
PAGE # 27
(c) Characteristics of electron :
(i) Electrons are sub - atomic particles which
constitute cathode rays.
(ii) In 1897, J.J.Thomson determined the charge to
mass (e/m) ratio of electron by studying the deflections
of cathode rays in electric and magnetic fields. The
value of e/m has been found to be 1.7588 × 108
coulombs/g. The e/m for electrons from different
gases was found to be the same. This indicates that
atoms of all kinds have the same kind of negatively
charged particles. Thus electrons are the common
constituents of all atoms.

Anode rays
Fluorescence
Anode
High voltage source

Note :
A cathode ray tube is used to measure the charge to
mass ratio of the electrons.
(iii) Charge on the electron :
The charge (e) on an electron was determined by
Robert Millikan in 1909. Millikan found the charge on
oil drops to be -1.6 × 10-19 C or its multiples. So, the
charge on an electron is to be -1.6 × 10-19 coulombs / unit.
(i) Anode rays travel in straight lines.
(ii) These rays rotate the light paddle wheel placed in
their path. This shows that anode rays are made up
of material particles.
By Thomson’s experiment e/m = 1.76 × 1011 C/kg
By Millikan’s experiment e = – 1.6 × 10-19 C
Mass of an electron in amu
(iii) Anode rays are deflected by magnetic or electric
field. In the electric field they get deflected towards
negatively charged plate. This indicates that these
rays are positively charged.
1.6  10 19
11
1.76  10
= 9.1096 × 10-31 kg
= 0.000549
(iv) The anode rays affect photographic plate.
(v) Mass of electron in comparison to that of
hydrogen :
Mass of hydrogen = 1.008 amu
= 1.008 × 1.66 × 10-24 g ( since 1 amu = 1.66 × 10-24 g )
= 1.673 × 10-24 g
(v) The nature of anode rays depend upon the type of
gas used.
(c) Discovery of Proton :
J.J.Thomson in 1906, found that particles obtained
in the discharge tube containing hydrogen have e/m
value as 9.579 × 10 4 coulomb/g. This was the
maximum value of e/m observed for any positive
particle. It was thus assumed that the positive
particles given by hydrogen represent fundamental
particle of positive charge. This particle was named
proton.
24
1.6 73  10
Mass of hydrogen atom
=
= 1837
9.1096  10 – 28
Mass of electron

Note :
Thus, the mass of an electron is
Note :
Anode rays are called canal rays because they pass
through the canals or holes of the cathode. These
rays are also called anode rays since they originate
from the anode side. Anode rays are produced from a
positively charged electrode, therefore these were
named positive rays by J.J.Thomson.
(b) Characteristics of Anode Rays :
(iv) Mass of an electron :
So mass of electron (m) =
Perforated cathode
1
times the
1837
mass of a hydrogen atom.
e
 H+ (Proton)
H 
PROTON

(a) Anode Rays (Canal rays) :
It has been established that electron is a negatively
charged particle and present in all the atoms. As an
atom is electrically neutral, there must be some
positively charged particles present in the atom to
neutralize the negative charges of the electrons. It
has been confirmed by experiments. Scientist
Goldstein in 1886 discovered the existence of a new
type of rays in the discharge tube. He carried out the
experiment in discharge tube containing perforated
cathode. It was observed that when high potential
difference was applied between the electrodes, not
only cathode rays were produced but also a new type
of rays were produced simultaneously from anode,
moving from anode towards cathode and passed
through the holes of cathode.
Note :
The name ‘proton’ was given by Rutherford in 1911.
(d) Characteristics of Proton :
(i) A proton is a sub - atomic particle which constitute
anode rays produced when hydrogen is taken in the
discharge tube.
(ii) Charge of a proton :
Proton is a positively charged particle. The charge on
a proton is equal but opposite to that on an electron.
Thus, the charge on a proton is +1.602 × 10 –19
coulombs/ unit.
(iii) Mass of a proton :
The mass of a proton is equal to the mass of a
hydrogen atom.
m p = 1.0073 amu
= 1.673 × 10-24 g
= 1.673 × 10-27 kg
28
28
PAGE # 28
(iv) Mass of proton relative to mass of electron :
(ii) About 99.0% of the -particles passed undeflected
through the gold foil and caused illumination of zinc
sulphide screen.
1.673 10 g
Mass of a proton
=
= 1837
Mass of an electron
9.110  28 g
24
(iii) Very few -particles underwent small and large
deflections after passing through the gold foil.
(iv) A very few (about 1 in 20,000) were deflected
backward on their path at an angle of 180º.
Thus, the mass of a proton is 1837 times larger than
the mass of an electron.
(v) Charge to mass ratio for a proton : The e/m of
particles constituting the anode rays is different for
different gases.
1.602  10 19
e
of proton =
= 9.579 × 104 C/g
1.673  10  24
m
THOMSON MODEL OF AN ATOM
J.J. Thomson (1898) tried to explain the structure of
atom. He proposed that an atom consists of a sphere
of positive electricity in which electrons are embedded
like plum in pudding or seeds evenly distributed in
red spongy mass in case of a watermelon. The radius
of the sphere is of the order 10–8 cm.
Rutherford was able to explain these observations
as follows:
(i) Since a large number of -particles pass through
the atom undeflected, hence, there must be large
empty space within the atom.
(ii) As some of the -particles got deflected, therefore,
there must be something massive and positively
charged structure present in the atom.
(iii) The number of -particles which get deflected is
very small, therefore, the whole positive charge in the
atom is concentrated in a very small space.
(a) Merits :
(i) Thomson’s model could explain the electrical
neutrality of an atom.
(iv) Some of the -particles retracted their path i.e.
came almost straight back towards the sources as a
result of their direct collisions with the heavy mass.
(ii) Thomson’s model could explain why only
negatively charged particles are emitted when a metal
is heated as he considered the positive charge to be
immovable by assuming it to be spread over the total
volume of the atom.
(iii) He could explain the formation of ions and ionic
compounds.
(b) Demerits :
This model could not satisfy the facts proposed by
Rutherford through his alpha particle scattering
experiment and hence was discarded.
RUTHERFORD MODEL OF AN ATOM
(a) Rutherford’s Alpha Particle Scattering
Experiment (1909) :
Ernest Rutherford and his coworkers performed
numerous experiments in which - particles emitted
from a radioactive element such as polonium were
allowed to strike thin sheets of metals such as gold or
platinum.
(i) A beam of -particles (He2+) was obtained by placing
polonium in a lead box and letting the alpha particles
come out of a pinhole in the lead box. This beam of rays was directed against a thin gold foil (0.0004 cm).
A circular screen coated with zinc sulphide was placed
on the other side of the foil.

Note :
 - particles are made up of two protons and two
neutrons and are Helium (He) nuclei.
(b) Rutherford Nuclear Model of Atom (1911) :
Rutherford proposed a new picture of the structure of
atom.
Main features of this model are as follows(i) The atom of an element consists of a small
positively charged “Nucleus” which is situated at the
centre of the atom and which carries almost the entire
mass of the atom.
(ii) The electrons are distributed in the empty space
of the atom around the nucleus in different concentric
circular paths (orbits).
29
29
PAGE # 29
(iii) The number of electrons in the orbits is equal to
the number of positive charges (protons) in the
nucleus.
(iv) Volume of nucleus is very small as compared to
the volume of atom.
(v) Most of the space in the atom is empty.

Note :
Rutherford’s model is also called “Planetary model’.
(v) Negatively charged electrons revolves around the
nucleus in circular path. The force of attraction
between the nucleus and the electron is equal to
centrifugal force of the moving electron.
Force of attraction towards nucleus = Centrifugal force
(vi) Out of infinite number of possible circular orbits
around the nucleus, the electron can revolve only in
those orbits whose angular momentum is an integral
h
h
, i.e. mvr = n
2
2
multiple of
(c) Defects in Rutherford’s Model :
(i) Rutherford did not specify the number of electrons
in each orbit.
(ii) According to electromagnetic theory, if a charged
particle (like electron) is accelerated around another
charged particle (like protons in nucleus) then there
would be continuous loss of energy due to continuous
emission of radiations. This loss of energy would
slow down the speed of electron and eventually the
electron would fall into the nucleus. But such a
collapse does not occur. Rutherford’s model could
not explain this theory.
(iii) If the electron loses energy continuously, the
observed spectrum should be continuous but the
actual observed spectrum consists of well defined
lines of definite frequencies. Hence the loss of energy
is not continuous in an atom.
where :
m = mass of the electron
v = velocity of electron
r = radius of the orbit, and
n =1,2,3 ---- number of the orbit.
The angular momentum can have values such as
h 2h 3h
,
,
, but it cannot have a fractional value.
2 2 2
Thus, the angular momentum is quantized. The
specified circular orbits (quantized) are called
stationary orbits.
RADII OF VARIOUS ORBITS
rn = 0.529 ×
BOHR MODEL OF AN ATOM (1913)
To overcome the objections to Rutherford’s model
and to explain the hydrogen spectrum, Bohr proposed
a quantum mechanical model of the atom.
The important postulates on which Bohr’s model is
based are the following (i) The atom has a nucleus where all the protons are
present. The size of the nucleus is very small. It is
present at the centre of the atom.
(ii) Each stationary orbit is associated with a definite
amount of energy. The greater is distance of the orbit
from the nucleus, more shall be the energy associated
with it. These orbits are also called energy levels and
are numbered as 1, 2, 3, 4 ------or K, L, M, N ---- from
nucleus to outwards.
(iii) By the time, the electron remains in any one of the
allowed stationary orbits, it does not lose energy.
Such a state is called ground or normal state.
(iv) The emission or absorption of energy in the form
of radiation can only occur when an electron jumps
from one stationary orbit to another.
E = Efinal - Einitial = h
Where h is Planck’s constant (h = 6.625 × 10–34 Js)
Energy is absorbed when the electron jumps from
lower to higher orbit and is emitted when it moves
from higher to lower orbit.
When the electron moves from inner to outer orbit by
absorbing definite amount of energy, the new state of
the electron is said to be excited state.
E=–
= –
=–
=–
n2
Å (for hydrogen like species).
Z
21.79  10 –19
n2
13.6
Z2 eV per atom (1 J = 6.2419 × 1018 eV)
n2
313.6
Z2 kcal/mol (1 eV = 23.06 kcal/mol)
n2
1312
n2
Z2 J per atom
Z2 kJ/mol
VELOCITY OF AN ELECTRON IN BOHR'S ORBIT
v=
Z
× 2.188 × 108 cm/sec
n
NEUTRONS
In 1932, James Chadwick bombarded the element
beryllium with  - particles. He observed the emission
of a radiation with the following properties (i) The radiation was highly penetrating.
(ii) The radiation remained unaffected in the electric
or magnetic field i.e. the radiation was neutral.
(iii) The particle constituting the radiation had the same
mass as that of the proton. These neutral particles
were called neutrons.
9
4 Be
+
4
2 He
(Beryllium) (á  particle)

12
6C
(Carbon)
+
1
0n
(Neutron)
30
30
PAGE # 30
COMPARATIVE STUDY OF ELECTRON,
PROTON AND NEUTRON
Property
Electron
Proton
Neutron
Symbol
e
p
n
Nature
Negatively charged Positively charged
Relative
charge
-1
+1
0
Absolute
charge
–1.602 × 10-19 C
+1.602 × 10 -19 C
0
Relative
mass
1
18 37
1
1
1.6725 × 10 -24 g
1.6748 × 10 -24 g
Absolute
mass
9.109 × 10 -28 g
(ii) Each energy level is further divided into subshells
designated as s,p,d,f .
1st shell (K) contains 1 subshell (s)
2nd shell (L) contains 2 subshells (s,p)
3rd shell (M) contains 3 subshells (s,p,d)
4th shell (N) contains 4 subshells (s,p,d,f).
ATOMIC STRUCTURE
An atom consists of two parts (a) Nucleus :
Nucleus is situated at the centre of an atom. All the
protons & neutrons are situated in the nucleus,
therefore, the entire mass of an atom is almost
concentrated in the nucleus. The overall charge of
nucleus is positive due to the presence of positively
charged protons (neutrons have no charge). The
protons & neutrons are collectively called nucleons.
(iii) Shells are divided into sub-shells, sub shells
further contain orbitals.
(A) An orbital may be defined as
“A region in the three - dimensional space around the
nucleus where the probability of finding the electron is
maximum.”
 Note :
The radius of the nucleus of an atom is of the order of
10–13 cm and its density is of the order of 1014 g/cm3.
(B)The maximum capacity of each orbital is that of two
electrons.
(b) Extra Nuclear Region :

In extra nuclear part or in the region outside the
nucleus, electrons are present which revolve around
the nucleus in orbits of fixed energies. These orbits
are called energy levels. These energy levels are
designated as K, L, M, N & so on.
(i) The maximum number of electrons that can be
accommodated in a shell is given by the formula
2n2.(n = number of shells i.e. 1,2,3 -------)
n
1
2
3
4
max. no.of electrons
2
8
18
32
u cl e
us
First energy level
Second energy level
Third energy level
Fourth energy level
2n2
2
2(1)
2
2(2)
2
2(3)
2
2(4)
N
Shell
K
L
M
N
+
Neutral
K L M N
Electron shells
2
Maximum number of
electrons which can be
accommodated in the
various shells
8 18 32
Note :
The maximum number of orbitals that can be present
in a shell is given by the formula n2.
(C) Types of orbitals :
(1) s-orbitals : The s-subshell contains just one orbital
which is non-directional & spherically symmetrical in
shape. The maximum number of electrons which can
be accommodated in s-orbital is 2.
Y
Z
X
s- orbital
(2) p - orbitals : The p-subshell contains three
orbitals which have dumb-bell shape and a directional
character. The three p-orbitals are designated as px,
py & pz which are oriented in the perpendicular axis
(x,y,z). The maximum number of electrons which can
be accommodated in the p subshell is 6 (2 electrons
in each of three orbitals).
31
31
PAGE # 31
z
z
x
y
px
z
x
x
y
y
z
z
x
y
pz
py
y
dx – y
2
(3) d - orbitals : The d-subshells contains 5 orbitals which
are double dumb-bell in shape. These orbitals are
designated as dxz, dxy, dyz, d x 2  y 2 , d 2 . The d-subshell
dz
2
(4) f-orbitals : The f-subshell contains 7 orbitals which
are complex in structure.The f-subshell can
z
can accommodate a maximum of 10 electrons.
z
z
z
y
x
2
accommodate a maximum of 14 electrons.

Note :
Letters s, p, d & f have originated from the words
sharp, principle, diffused & fundamental respectively.
y
x
y
x
x
dxz
dxy
y
dyz
(iv) Differences between orbit and orbital :
S.No.

Orbit
Orbital
It is a region in three dimensional space
around the nucleus where the probability
of finding electron is maximum.
s,p and d-orbitals are spherical, dumb-bell
and double dumb-bell in shape respectively.
It represents that an electron can move around
nucleus along three dimensional space (x,y and z
axis).
1
It is well defined circular path around the
nucleus in which the electron revolves.
2
It is circular in shape.
3
It represents that an electron moves
around the nucleus in one plane.
4
It represents that position as well as
momentum of an electron can be known
simultaneously with certainty. It is against
Heisenberg's uncertainty principle.
It represents that position as well as momentum
of an electron cannot be known simultaneously
with certainty. It is in accordance with Heisenberg's
uncertainty principle.
5
The maximum number of electrons in an
2
orbit is 2n where 'n' is the number of the
orbit.
The maximum number of electrons in an orbital is
two.
Note :
Heisenberg’s uncertainty principle - “It is impossible
to determine exactly both the position and momentum
(or velocity) of an electron or of any other moving
particle at the same time.”
ELECTRONIC CONFIGURATION OF AN ATOM

Note :
If the outermost shell has its full quota of 8 electrons
it is said to be an octet. If the first shell has its full
quota of 2 electrons, it is said to be duplet.
The pictorial representation of Bohr’s model of
hydrogen, helium, carbon, sodium and calcium atoms
having 1, 2, 6, 11 and 20 electrons respectively are
shown in the figure where the centre of the circle
represents the nucleus.
(i) The arrangement of the electrons in different shells
is known as the electronic configuration of the
element.
(ii) Each of the orbits can accommodate a fixed
number of electrons. Maximum number of electrons
in an orbit is equal to 2n2, where ‘n’ is the number of
the orbit.
(iii) Electrons are filled in the increasing order of
energy, i.e. K < L < M < N ......
(iv) In the outermost shell of any atom, the maximum
possible number of electrons is 8, except in the first
shell which can have at the most 2 electrons.
32
32
PAGE # 32
(b) Pauli’s Exclusion Principle :
According to Pauli’s exclusion principle “an orbital
cannot accommodate more than two electrons. If
there are two electrons in an orbital they must have
opposite spins.”
(a) Significance of Electronic Configuration :
The electronic configuration of an atom plays an
important role in determining the chemical behaviour
of an element.
(i) When the atoms of an element have completely
filled outermost shell, the element will be chemically
unreactive. For example the noble gases (He, Ne, Ar,
Kr, Xe and Rn) have completely filled outermost shell
i.e. contains 8 electrons (except helium which has
two valence electrons) in outermost shell.
(ii) When the atom of an element has less than 8
electrons in its outermost shell, the element will be
reactive.
ORDER OF FILLING OF ELECTRONS IN SUBSHELLS
There are different rules governing the filling of
subshells. They are described briefly as follows (a) Aufbau Principle :

The filling of subshells in atoms is based on their
energies. Electrons first occupy the subshell with
lowest energy and progressively fill the other
subshells in increasing order of energy.
Note :
(c) Hund’s Rule of Maximum Multiplicity :
According to this rule :
“no electron pairing takes place in the orbitals with
equivalent energy until each orbital in the given
subshell contains one electron & the spins of all
unpaired electrons are parallel i.e. in the same
direction”.
Electronic configuration of some elements Atom ic
Sym bol of
Nam e of
num ber the elem ent the elem ent
Electronic
configuration
1
H
Hydrogen
1s 1
2
He
Helium
1s
3
Li
Lithium
1s , 2s
4
Be
Beryllium
1s , 2s
2
2
1
2
2
2
2
1
5
B
Boron
1s , 2s , 2p
6
C
Carbon
1s 2, 2s 2 , 2p 2
7
N
Nitrogen
1s 2, 2s 2 , 2p 3
8
O
Oxygen
1s 2, 2s 2 , 2p 4
9
F
Fluorine
1s 2, 2s 2 , 2p 5
10
Ne
Neon
1s 2, 2s 2 , 2p 6
11
Na
Sodium
1s , 2s , 2p ,3s
12
Mg
Magnes ium
1s , 2s , 2p ,3s
13
Al
Alum inium
14
Si
15
P
Silicon
2
2
6
1
2
2
6
2
2
2
6
2
1
2
2
6
2
2
1s , 2s , 2p ,3s ,3p
1s , 2s , 2p ,3s ,3p
Phos phorus 1s 2, 2s 2 , 2p 6 ,3s 2 ,3p 3
The subshell with lowest energy is filled first.
16
S
Sulphur
1s 2, 2s 2 , 2p 6 ,3s 2 ,3p 4
The order of energy of different subshells of an atom is -
17
Cl
Chlorine
1s 2, 2s 2 , 2p 6 ,3s 2 ,3p 5
18
Ar
Argon
1s 2, 2s 2 , 2p 6 ,3s 2 ,3p 6
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p
< 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p.
19
K
Potas s ium
1s 2, 2s 2 , 2p 6 ,3s 2 ,3p 6, , 4s 1
20
Ca
Calcium
1s , 2s , 2p ,3s ,3p , 4s
21
Sc
Scandium
1s , 2s , 2p ,3s ,3p 3d , 4s
The number present before the subshells like 1,2,3
------ represents the number of the shell i.e. n.
The order of filling of different sub-shells is
represented diagrammatically as follows :
22
Ti
Titanium
1s , 2s , 2p ,3s ,3p 3d , 4s
23
V
Vanadium
1s , 2s , 2p ,3s ,3p 3d , 4s
24
Cr
C hrom ium
1s 2, 2s 2 , 2p 6 ,3s 2 ,3p 6, 3d 5, 4s 1
25
Mn
Manganes e 1s 2, 2s 2 , 2p 6 ,3s 2 ,3p 6, 3d 5, 4s 2
26
Fe
Iron
1s 2, 2s 2 , 2p 6 ,3s 2 ,3p 6, 3d 6, 4s 2
27
Co
Cobalt
1s 2, 2s 2 , 2p 6 ,3s 2 ,3p 6, 3d 7, 4s 2
28
Ni
Nickel
1s 2, 2s 2 , 2p 6 ,3s 2 ,3p 6, 3d 8, 4s 2
29
Cu
Copper
1s , 2s , 2p ,3s ,3p 3d , 4s
30
Zn
Zinc
1s , 2s , 2p ,3s ,3p 3d , 4s
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
6s
6p
6d
7s
7p
2
2
6
2
6,
2
2
6
2
6,
1
2
2
2
2
6
2
6,
2
2
2
2
6
2
6,
3
2
2
2
6
2
6,
10
1
2
2
6
2
6,
10
2
VALENCE SHELL AND VALENCE ELECTRONS
The outermost shell of an atom is known as the
valence shell. The electrons present in the valence
shell of an atom are known as valence electrons.
The remainder of the atom i.e. the nucleus and other
electrons is called the core of the atom. Electrons
present in the core of an atom are known as core
electrons.
e.g.
The electronic configuration of the sodium (Na) atom
is :Na (11)
K
L
M
2
8
1
Thus, valence electron in Na atom = 1 and core
electrons in Na atom = 2 + 8 = 10
33
33
PAGE # 33
(a) Significance of Valence Electrons :
(i) The valence electrons of an atom are responsible
for, and take part in, chemical changes.
(ii) The valence electrons in an atom determine the
mode of chemical combination.
(iii) The valence electrons determine the combining
capacity or the valency of the atom. The number of
electrons in an atom that actually take part in bond
formation is known as the valency of the element.
e.g. In the carbon atom, there are four valence
electrons.
C(6)
K L
2 4

The number of electrons gained, lost or contributed
for sharing by an atom of the element gives us directly
the combining capacity or valency of the element.
The carbon atom is capable of forming four bonds.
Hence, the valency of carbon is four.
Valency of an element is determined by the number
of valence electrons in an atom of the element.
The valency of an element = number of valence
electrons (if the number of valence electrons is1 to 4)
The valency of an element = 8– number of valence
electrons. (if the number of valence electrons is 4 to 8)
(iv) If the outermost shell of an atom is completely
filled, its valency is zero.
The outermost shells of helium, neon, argon, krypton
etc. are completely filled. Hence the valency of these
elements is zero.
(v) Elements having the same number of valence
electrons in their atoms possess similar chemical
properties.
e.g. All alkali metals have one valence electron in
their atoms. Hence, their chemical properties are
similar.
(vi) Elements having different number of valence
electrons in their atoms show different chemical
properties.
The isotopes of an element have the same atomic
number but different atomic masses.


The difference in their masses is due to the presence
of different number of neutrons.
e.g. Isotopes of hydrogen :
Hydrogen
isotopes
Note :
Whether the atom while combining with other atoms
can form ionic or covalent bonds is determined by
the number of valence electrons present in the atom.
Pr ot iu m
1H
1
De u t e r iu m
2
1H
T r itium
3
1H
1. Atomic number
1
1
1
2. No. of protons
1
1
1
3. No. of electrons
1
1
1
4. Mass number
1
2
3
5. No. of neutrons
0
1
2
Isotopes of oxygen :
Note :
Hydrogen and helium are exceptions to this rule.
Hydrogen and helium atoms have 1 and 2 valence
electrons respectively, but they are non-metals.
(ix) Elements with 4, 5, 6, 7 or 8 valence electrons in
their atoms are non metals.
e.g. carbon (C), nitrogen (N) and oxygen (O) are non
metals.
C = 2,4
6
N = 2,5
7
O = 2,6
8
Note :
The term isotope was given by Margaret Todd.
e.g. Let us consider the electronic configuration of
alkali metals and halogens. Alkali metal atoms have
single valence electron whereas halogen atoms have
seven valence electrons. It is observed that the
chemical properties of the alkali metals are entirely
different from those of halogens.
(vii) The number of the valence shell in the atom of an
element determines the period number of the
element in the periodic table.
e.g. Sodium (Na) :
Valence shell number = 3.
period number = 3
Calcium (Ca)
Valence shell number = 4
period number = 4
(viii) Elements with 1, 2 or 3 valence electrons in their
atoms are metals.

Valency of an element is the combining capacity of
the atoms of the element with atoms of the same or
different elements. The combining capacity of the
atoms was explained in terms of their tendency to
attain a fully filled outermost shell (stable octet or
duplet)
Note :
Oxygen isotopes
16
8O
17
8O
18
8O
1. Atomic number
8
8
8
2. No. of protons
8
8
8
3. No. of electrons
8
8
8
4. Mass number
16
17
18
5. No. of neutrons
8
9
10

Note :
All the isotopes of an element have identical chemical
properties.
34
34
PAGE # 34
(a) Characteristics of Isotopes :
(i) The physical properties of the isotopes of an
element are different. This is due to the fact that
isotopes have different numbers of neutrons in their
nuclei. Hence mass, density and other physical
properties of the isotopes of an element are different.
The atoms of different elements with different atomic
numbers, which have same mass number are called
isobars.
(ii) All the isotopes of an element contain the same
number of electrons. So, they have the same
electronic configuration with the same number of
valence electrons. Since the chemical properties of
an element are determined by the number of valence
electrons in its atom, all the isotopes of an element
have identical chemical properties.
e.g.
and
40
18 Ar
and
14
7N
are isobars.
40
20 Ca
are isobars.
40
18
Isobars
40
20
Ar
Ca
1. Atomic number
18
20
(b) Reason for the Fractional Atomic
2. Mass number
40
40
Masses of Elements :
3. No. of electrons
18
20
The atomic masses of many elements are in fraction
and not whole number. The fractional atomic
masses of elements are due to the existence of
their isotopes having different masses.
e.g.
The atomic mass of chlorine is 35.5 u. Chlorine has
4. No. of protons
18
20
5. No. of neutrons
22
20
2, 8, 8
2, 8, 8, 2
two isotopes
35
17 Cl
and
37
17 Cl
6. Electronic configuration

with abundance of 75%
and 25% respectively. Thus the average mass of a
chlorine atom will be 75% of 35 and 25% of 37,
which is 35.5 u.
So, Average atomic mass of chlorine
=
e.g.
2625
925
+
100
100
(ii) In industry : Isotopes are used for coating on the
arm of clock to see in dark. To identify the cracks in
metal casting.
(iii) In medicine :Thyroid, bone diseases, brain
tumors and cancer are controlled or destroyed with
24
32
the help of radioactive isotope 11
Na, 131
53 É, 15P etc.
(iv) Determination of the mechanism of chemical
reaction by replacing an atom or molecule by its
isotopes.
(v) In carbon dating : W ill and Libby (1960)
developed the technique of radiocarbon to
determine the age of plant, fossil and archeological
sample.
Note :
Isotopes (Like Uranium-238) are used in Nuclear
reactor to produce energy and power.
13
6C
and
30
31
14 Si , 15 P
= 26.25 + 9.25
= 35.5 u.
Thus, the average atomic mass of chlorine is 35.5
u.
(c) Applications of Radioactive Isotopes:
(i) In agriculture : Certain elements such as boron,
cobalt, copper, manganese, zinc and molybdenum
are necessary in very minute quantities for plant
nutrition. By radioactive isotopes we can identify the
presence and requirement of these element in the
nutrition of plants.
Note
Isobars contain different number of electrons,
protons and neutrons.
Isotones may be defined as the atoms of different
elements containing same number of neutrons.
75
25
= 35 ×
+ 37 ×
100
100

14
6C
14
7N
and
(Both contain 7 neutrons)
32
16 S
(All three contain 16 neutrons)
Ion or atom or molecule which have the same number
of electrons are called as isoelectronic species.
e.g.
17
No. of electrons
Cl
–
18
18 Ar
19 K
18
18

20 Ca
2
18
QUANTUM NUMBERS
To describe the position and energy of electron in an
atom, four numbers are required, which are known
as quantum numbers.
Four quantum numbers are :
(a) Principal quantum number
(b) Azimuthal quantum number
(c) Magnetic quantum number
(d) Spin quantum number
(a) Principal Quantum Number :
(i) It is denoted by ‘n’.
(ii) It represents the name, size and energy of the
orbit or shell to which the electron belongs.
(iii) Higher is the value of ‘n’ , greater is the distance
of the shell from the nucleus.
r1 < r2 < r3 < r4 < r5 < ----
35
35
PAGE # 35
(iv) Higher is the value of ‘n’, greater is the magnitude
of energy.
E1 < E2 < E3 < E4 < E5 ---(v) Maximum number
by 2n2.
Shell
First (n =1)
Second (n = 2)
Third ( n = 3)
Fourth ( n = 4)
of electrons in a shell is given
Max. number of electrons
2 × 12 = 2
2 × 22 = 8
2 × 32 = 18
2 × 42 = 32
(vi) Angular momentum can also be calculated using
principal quantum number.
mvr 
nh
2ð
(vii) Value of n is from 1 to 
(viii) Every shell is given a specific alphabetic name.
First shell (n = 1) is known as K shell.
Second shell (n = 2) is known as L shell.
Third shell (n = 3) is known as M shell and so on.

Note :
Principal quantum number was given by Bohr.
(b) Azimuthal Quantum Number :
(i) It is represented by ‘’.

Note :
Azimuthal quantum number is also called angular
quantum number, subsidiary quantum number or
secondary quantum number.
(ii) For a given value of n values of  is 0 to n – 1
Value of n
Values of 
1 (1st shell)
0
2 (2nd shell)
0,1
3 (3rd shell)
0,1,2
4 (4th shell)
0,1,2,3
(iii) It represents the sub-shell present in shell.
 = 0 represents s sub shell.
 = 1 represents p sub shell.
 = 2 represents d sub shell.
 = 3 represents f sub shell.

(viii) Maximum no. of electrons present in a subshell
= 2 (2 +1)
Subshell
Max. electrons
s ( = 0)
2 (2 × 0 +1) = 2
p ( = 1)
2 (2 × 1 +1) = 6
d ( = 2)
2 (2 × 2 +1) = 10
f ( = 3)
2 (2 × 3 +1) = 14
Note :
s,p,d and f signify sharp, principal, diffused and
fundamental respectively.
(iv) Number of sub-shell in a shell = Principal quantum
number of shell.
(v) Maximum value of  is always less than the value
of n. So 1p, 1f, 2d, 2f, 3f subshells are not possible.
s will start from 1s
p will start from 2p
d will start from 3d
f will start from 4f
(vi) Relative energy of various sub-shell in a shell are
as follows s<p<d<f
(vii) Subshells having equal values but with different n
values have similar shapes but their sizes increases
as the value of ‘n’ increases. 2s-subshell is greater in
size than 1s- subshell. Similarly 2p, 3p, 4p subshells
have similar shapes but their sizes increase in
order 2p < 3p < 4p.

Note :
Azimuthal quantum number was given by
Sommerfeld.
(c) Magnetic quantum number :
(i) It is denoted by ‘m’.
(ii) It represents the orbitals present in sub-shell. An
orbital can be defined as :
“Region in the three - dimensional space around the
nucleus where the probability of finding an electron is
maximum”.
(iii) For a given value of , values of m are from –
through 0 to +.

m
0
0
1
–1, 0, +1
2
–2, –1, 0, +1, +2
3
–3, –2, –1, 0, +1, +2, +3
(iv) Maximum number of orbitals in a sub-shell
= (2+1)
Sub shell
Orbitals
s ( = 0)
(2 × 0 +1) = 1
p ( = 1)
(2 × 1 +1) = 3
d ( = 2)
(2 × 2 +1) = 5
f ( = 3)
(2 × 3 +1) = 7
(v) Maximum number of orbitals in a shell = n2
Shell
Max. orbitals
First (n = 1)
12 = 1
Second (n = 2)
22 = 4
Third (n = 3)
32 = 9
Fourth (n = 4)
42 = 16
(vi) It represents the orientation of orbital in three
dimensional space.
When l = 0, m = 0, i.e. one value implies that ‘s’
subshell has only one space orientation and hence,
it can be arranged in space only in one way along x,y
or z axis. Thus, ‘s’ orbital has a symmetrical spherical
shape.
Y
Z
X
s- orbital
When  = 1,’m’ has three values –1, 0, +1 . It implies
that ‘p’ subshell of any energy shell has three space
orientations, i.e. three orbitals. Each p-orbital has
dumb-bell shape. Each one is disposed
symmetrically along one of the three axis. p orbitals
have directional character.
36
36
PAGE # 36
orbital
m
z
Pz
0
Px
±1
z
Py
±1
z
x
y
y
x
x
y
y
px
pz
py
When  = 2 ‘m’ has five values –2, –1, 0, +1, +2. It
implies that d-subshell of any energy shell has five
orientations, i.e. five orbitals. All the five orbitals are
not identical in shape. Four of the d-orbitals
dxy, dyz, dzx, d
x 2 – y2
contain four lobes while fifth orbital
dz2 consists of only two lobes.
z
z
y
x
z
y
x
y
x
dyz
dxz
dxy
z
z
x
y
2
dx – y
y
x
2
dz
2
There are seven f-orbitals designated as
f
, fx ( x2 – y2 ), fy ( x2 – y2 ), fz( x2 –y2 ), and fxyz .
Their shapes are complicated ones.
f
f
yz2 , xz2 , z3
(vii) Characteristics of orbitals :
(A) All orbitals of a subshell possess same energy
i.e., they are degenerate.
(B) All orbitals of the same shell differ in the direction
of their space orientation.
(C) Total number of orbitals in a shell is equal to n2.

Note :
Magnetic quantum number was given by Zeeman.
(d) Spin Quantum Number :
(i) It is denoted by ‘s’.
(ii) It represents the direction of spin of electron around
its own axis.
(iii) Clockwise spin is represented by +1/2 or  and
anticlockwise by –1/2 or .
(iv) Maximum two electrons with opposite spin can
be placed in an orbital.
(v) Electrons with same spin are called spin parallel
and those with opposite spin are called spin paired.

Note :
Spin quantum number was given by Gold Schmidt.
37
37
PAGE # 37
PERIODIC TABLE
(ii) Limitations of Dobereiner’s Classification :
DEFINITION
A periodic table may be defined as the table giving
the arrangement of all the known elements according
to their properties so that elements with similar
properties fall within the same vertical column and
elements with dissimilar properties are separated.
(A) Atomic mass of the three elements of some triads
are almost same.
e.g. Fe, Co, Ni
(B) It was restricted to few elements, therefore
discarded.
(c) Newlands’ Law of Octaves :
EARLY ATTEMPTS TO CLASSIFY ELEMENTS
(a) Metals and Non-Metals :
Among the earlier classifications, Lavoisier classified
the elements as metals and non-metals. However,
this classification proved to be inadequate. In 1803,
John Dalton published a table of relative atomic
weights (now called atomic masses). This formed
an important basis of classification of elements.
(b) Dobereiner’s Triads:
In 1866, an English chemist, John Newlands,
proposed a new system of grouping elements with
similar properties. He tried to correlate the properties
of elements with their atomic masses. He arranged
the then known elements in the order of increasing
atomic masses. He started with the element having
the lowest atomic mass (hydrogen) and ended at
thorium which was the 56th element. He observed
that every eighth element had properties similar to
that of the first.
(i) In 1817, J.W. Dobereiner a German Chemist gave
this arrangement of elements.
Newlands called this relation as a law of octaves
due to the similarity with the musical scale.
(A) He arranged elements with similar properties in
the groups of three called triads.
(i) Newlands’ arrangement of elements into ‘Octaves’:
(B) According to Dobereiner the atomic mass of the
central element was merely the arithmetic mean of
atomic masses of the other two elements.
e.g.
Elements of
the triad
Symbol Atomic mass
Lithium
Li
7
Sodium
Na
23
Potassium
K
39
Atomic mass of sodium =
Atomi c mass of lithium  Atomic mass of potassium
2
=
7  39
2
= 23
Some examples of triads are given in the table :
Notes of
Music
Elements
sa (do)
re (re)
ga (mi)
ma (fa)
H
Li
Be
B
pa (so) dha (la) ni (ti)
C
N
F
Na
Mg
Al
Si
P
S
Cl
K
Ca
Cr
Ti
Mn
Fe
Co and Ni
Cu
Zn
Y
In
As
Se
Br
Rb
Sr
Ce and La
Zr
–
–
O
(ii) Limitations of law of octaves : The law of octaves
has the following limitations :
(A) The law of octaves was found to be applicable
only upto calcium. It was not applicable to elements
of higher atomic masses.
(B) Position of hydrogen along with fluorine and
chlorine was not justified on the basis of chemical
properties.
(C) Newlands placed two elements in the same
slot to fit elements in the table. He also placed some
unlike elements under the same slot. For example,
cobalt and nickel are placed in the same slot and in
the column of fluorine, chlorine and bromine. But
cobalt and nickel have properties quite different from
fluorine, chlorine and bromine. Similarly, iron which
has resemblances with cobalt and nickel in its
properties has been placed far away from these
elements.
Thus, it was realized that Newlands’ law of octaves
worked well only with lighter elements. Therefore,
this classification was rejected.
38
38
PAGE # 38
(d) Lother Meyer’s Classification :
In 1869, Lother Meyer studied the physical properties
like volume, melting point, boiling point etc. of different
elements.
He plotted a graph between atomic masses against
their respective atomic volumes for a number of
elements. He found the following observations (i) Elements with similar properties occupied
similar positions on the curve.
(ii) Alkali metals (Li, Na, K, Rb, Cs etc.) having larger
atomic volumes occupied the crests .
(iii) Transition elements (V, Fe, Co, Cu etc.) occupied
the troughs.
(iv) The halogens (F, Cl, Br,  etc.) occupied the
ascending portions of the curve before the inert
gases.
(v) Alkaline earth metals (Mg, Ca, Sr, Ba etc.)
occupied positions at about the mid points of
descending portions of the curve.
Cs
3
Atomic Volume (cm per mole of atoms)
70
•
60
Rb
•
50
K
40
•
Li
30
20
10
•
• • Sr
•
• Ba
Na
I•
Br •
•
Ca
•
• Cl • •
•
•
Mg
• •Be •• • •• VFe Co ••••Zn •
•
••• •••• ••••Cu •• •• • •
•
•
••
••
•
••••
••
F
0
20
40
60 80
100 120
140
Atomic mass
Change of Atomic Volume with Atomic Mass.
Drawback of Lother Meyer’s classification : This
was a hypothetical classification and it was difficult to
remember the positions of different elements.
(e) Mendeleev’s Periodic Table :
The major credit for a systematic classification of
elements goes to Mendeleev. He tried to group the
elements on the basis of some fundamental property
of the atoms. When Mendeleev started his work, only
63 elements were known. He examined the
relationship between atomic masses of the elements
and their physical and chemical properties.
Among chemical properties, Mendeleev concentrated
mainly on the compound formed by elements with
oxygen and hydrogen. He selected these two
elements because these are very reactive and formed
compound with most of the elements known at that
time. The formulae of the compounds formed with
these elements (i.e. oxides and hydrides) were
regarded as one of the basic properties of an element
for its classification.
(i) Mendeleev’s periodic law : This law states that
the physical and chemical properties of the elements
are the periodic function of their atomic masses.
This means that when the elements are arranged in
the order of their increasing atomic masses, the
elements with similar properties recur at regular
intervals. Such orderly recurring properties in a cyclic
fashion are said to be occurring periodically. This is
responsible for the name periodic law or periodic
table.
(ii) Merits of Mendeleev’s periodic table : Mendeleev’s
periodic table was one of the greatest achievements in
the development of chemistry. Some of the important
contributions of his periodic table are given below :
(A) Systematic study of elements : He arranged
known elements in order of their increasing atomic
masses considering the fact that elements with
similar properties should fall in the same vertical
column.
(B) Correction of atomic masses : The Mendeleev’s
periodic table could predict errors in the atomic
masses of elements based on their positions in the
table. Therefore atomic masses of certain elements
were corrected. For example, atomic mass of
beryllium was corrected from 13.5 to 9. Similarly, with
the help of this table, atomic masses of indium, gold,
platinum etc. were corrected.
(C) Mendeleev predicted the properties of those
missing elements from the known properties of the
other elements in the same group. Eka-boron, eka aluminium and eka -silicon names were given for
scandium , gallium and germanium (not discovered
at the time of Mendeleev ).
(D) Position of noble gases : Noble gases like helium
(He), neon (Ne) and argon (Ar) were mentioned in
many studies. However, these gases were discovered
very late because they are very inert and are present
in extremely low concentrations. One of the
achievements of Mendeleev’s periodic table was that
when these gases were discovered, they could be
placed in a new group without disturbing the existing
order.
(iii) Limitations of Mendeleev’s periodic table :
Inspite of many advantages, the Mendeleev’s periodic
table has certain defects also. Some of these are
given below (A) Position of hydrogen : Position of hydrogen in
the periodic table is uncertain. It has been placed in
1A group with alkali metals, but certain properties of
hydrogen resemble those of halogens. So, it may be
placed in the group of halogens as well.
39
39
PAGE # 39
(B) Position of isotopes : Isotopes are the atoms of
the same element having different atomic masses.
Therefore, according to Mendeleev’s classification
these should be placed at different places depending
upon their atomic masses. For example, hydrogen
isotopes with atomic masses 1, 2 and 3 should be
placed at three places. However, isotopes have not
been given separate places in the periodic table
because of their similar properties.
• The positions of cobalt and nickel are not in proper
order. Cobalt (at. mass = 58.9) is placed before nickel
( at. mass = 58.7).
Tellurium (at. mass = 127.6) is placed before
iodine (at. mass = 126.9).
•
(D) Some similar elements are separated, in the
periodic table. For example copper (Cu) and mercury
(Hg). On the other hand, some dissimilar elements
have been placed together in the same group.
e.g. Copper (Cu), silver (Ag) and gold (Au) have been
placed in group 1 along with alkali metals. Similarly,
manganese (Mn) is placed in the group of halogens.
(C) Anomalous pairs of elements : In certain pairs
of elements, the increasing order of atomic masses
was not obeyed. In these, Mendeleev placed
elements according to similarities in their properties
and not in increasing order of their atomic masses.
 For example :
• The atomic mass of argon is 39.9 and that of
(E) Cause of periodicity : Mendeleev could not
explain the cause of periodicity among the elements.
Z r = 90
51
potassium 39.1. But argon is placed before
potassium in the periodic table.
40
40
PAGE # 40
(e) Long Form of Periodic Table :
MODERN PERIODIC TABLE
(i) The long form of periodic table is based upon
Modern periodic law. Long form of periodic table is
the contribution of Range, Werner, Bohr and Bury.
(a) Introduction :
In 1913, an English physicist, Henry Moseley showed
(ii) This table is also referred to as Bohr’s table since
it follows Bohr’s scheme of the arrangement of
elements into four types based on electronic
configuration of elements.
that the physical and chemical properties of the atoms
of the elements are determined by their atomic
number & not by their atomic masses. Consequently,
the periodic law was modified.
(b) Modern Periodic
Periodic Law) :
Law
(iii) Long form of periodic table consists of horizontal
rows (periods) and vertical columns (groups).
(Moseley’s
(f) Description of Periods :
(i) A horizontal row of periodic table is called a period.
“Physical and chemical properties of an element are
(ii) There are seven periods numbered as 1, 2 , 3 , 4,
5, 6 and 7.
the periodic function of its atomic number’’.The
atomic number gives us the number of protons in
(iii ) Each period consists of a series of elements
having the same outermost shell.
the nucleus of an atom and this number increases
by one in going from one element to the next.
(iv) Each period starts with an alkali metal having
outermost shell electronic configuration ns1.
Elements, when arranged in the order of increasing
atomic number Z, lead us to the classification known
(v) Each period ends with a noble gas with outermost
shell electronic configuration ns2 np6 except helium
having outermost electronic configuration 1s2.
as the Modern Periodic Table. Prediction of properties
of elements could be made with more precision when
elements were arranged on the basis of increasing
atomic number.
(vi) Each period starts with the filling of a new energy
level.
(c) Periodicity :
(A) 1st period : This period is called very short period
because this period contains only 2 elements H and
He.
The repetition of elements with similar properties after
certain regular intervals, when the elements are
arranged in order of increasing atomic number, is
(B) 2nd and 3rd periods : These periods are called
short periods because these periods contain 8
elements. 2nd period starts from 3Li to 10Ne and 3rd
period starts from 11Na to 18Ar.
called periodicity.
(d) Cause of Periodicity :
(C) 4th and 5th periods : These periods are called
long periods because these periods contain 18
elements. 4th period starts from 19K to 36Kr and 5th
period starts from 37Rb to 54Xe.
The periodic repetition of the properties of the
elements is due to the recurrence of similar valence
shell (outermost shell) electronic configuration after
certain regular intervals.
(D) 6th period : This period is called very long period.
This period contains 32 elements. Out of the 32
elements 14 elements belong to Lanthanoid series
(58Ce to 71Lu). 6th period starts from 55Cs to 86Rn.
e.g. Alkali metals have similar electronic configuration
(ns1) and therefore, have similar properties.
Alkali Metals
Atomic
number
Element
Symbol
Electronic
configuration
3
Lithium
Li
2,1
11
Sodium
Na
2,8,1
(E) 7th period : This period is called as incomplete
period. It contains 25 elements. Out of the 25
elements 14 elements belong to Actinoid series (90Th
to 103Lr). 7th period starts from 87Fr to 111Rg.

Note : Modern periodic table consists of seven periods
and eighteen groups.
Periods
19
Potassium
K
2,8,8,1
st
(1 ) n = 1
nd
37
Rubidium
Rb
2,8,18,8,1
55
Caesium
Cs
2,8,18,18,8,1
87
Francium
Fr
2,8,18,32,18,8,1
(2 ) n = 2
rd
(3 ) n = 3
(4 th ) n = 4
(5 th) n = 5
th
(6 ) n = 6
th
(7 ) n = 7
No. of
Elements
Called as
2
8
8
18
18
32
25
Very short period
Short period
Short period
Long period
Long period
Very long period
Incomplete period
41
41
PAGE # 41
(ii) Position of isotopes : Isotopes are atoms of the
same element having different atomic masses, but
same atomic number. All the isotopes of an element
will be given different positions, if atomic mass is
taken as a basis. This shall disturb the symmetry of
the table. In modern table, one position is fixed for
one atomic number and since all the isotopes of an
element have the same atomic number, these are
assigned only one position.
Different elements belonging to a particular period
have different electronic configurations and have
different number of valence electrons. That is why
elements belonging to a particular period have
different properties.
(g) Description of Groups :
(i) A vertical column of elements in the periodic table
is called a group.
(iii) Elements with similar properties were placed
together and elements with dissimilar properties were
separated in modern periodic table.
(ii) There are eighteen groups numbered as 1, 2 , 3,
4, 5, ------------ 13, 14, 15, 16,17,18.

(iii) A group consists of a series of elements having
similar valence shell electronic configuration and
hence exhibit similar properties.
(iv) Cause of periodicity : Modern periodic table
explains the cause of periodicity among the elements.
e.g. Li, Na, K belong to the same group and have 1
electron in their valence shell.
Following are the demerits of modern periodic table -
(iv) The group 18 is also known as zero group because
the valency of the elements of this group is zero.
uncertain in the periodic table.
Note : The elements of 18th or zero group are called
noble gases.
(ii) Position of lanthanides and actinides : The
positions of lanthanides and actinides were also
uncertain in the periodic table.
(v) The elements present in groups 1,2,13 to 17 are
called normal or representative elements.
(i) Demerits of Modern Periodic Table :
(i) Position of hydrogen : Position of hydrogen was
CLASSIFICATION OF THE ELEMENTS
(vi) Elements of group 1 and 2 are called alkali metals
and alkaline earth metals respectively.
It is based on the type of subshells which receives
the differentiating electron (i.e. last electron).
(vii) Elements present in group 17 are called
halogens.
(a) s- Block Elements :

Note : Elements present in a period have different
properties, while elements present in a group have
similar properties.

Note : Modern periodic table is based on atomic
W hen last electron enters the s- orbital of the
outermost (nth) shell, the elements of this class are
called s- block elements.
Characteristics :
(i) Group 1 & 2 elements constitute the s - block.
(ii) General electronic configuration is ns1–2 .
number, not on atomic mass.
(iii) s - block elements lie on the extreme left of the
periodic table.
(h) Merits of Modern Periodic Table :
(i) Anomalous pairs : The original periodic law based
on atomic masses is violated in case of four pairs of
elements in order to give them positions on the basis
of properties. The elements having higher atomic
masses have been assigned position before the
elements having lower atomic masses at four places
as shown below -
(iv) This block includes metals only, except H.

Note : The total number of elements in s-block is 13
(including hydrogen).
(b) p-Block Elements :
When differentiating electron enters the p - orbital of
the nth orbit, elements of this class are called p - block
elements.
(a)
Ar
(b)
K
Co Ni
(c)
Te
(d)
I

Characteristics :
(i) Elements of group 13 to 18 constitute the p - block.
Th Pa
(ii) General electronic configuration is ns2np1-6 .
At. Mass
40 39 60 59 128 127 232 231
At. No.
18 19 27 28 52 53 90 91
The discrepancy disappears, if the elements are
arranged in order of increasing atomic numbers.
(iii) p - block elements lie on the extreme right of the
periodic table.
(iv) This block includes some metals, all non-metals
and metalloids.

Note : The total number of elements in p-block is 31.
42
42
PAGE # 42
(c) d - Block Elements:

(i) All f - block elements belong to 3rd group.
When differentiating electron enters the (n–1)d orbital,
then elements of this class are called d - block
elements.

(ii) General electronic configuration is (n – 2)f 1 – 14
(n– 1)d0-1 ns2.
(iii) f-block elements are present in two separate
rows below the periodic table.
Characteristics :
(i) Elements of group 3 to 12 constitute the d - block.
(iv) All f-block elements are metals only.
(ii) General electronic configuration is (n – 1)d1–10 ns0- 2.
The elements of f- block have been classified into
two series :
(iii) d - block elements lie in the centre of the
periodictable.
(A) Lanthanides : 14 elements present after element
lanthanum (57La) are called lanthanides.1 st inner
transition series of metals or 4f - series, contains 14
elements i.e. 58Ce to 71Lu .
(iv) All the d - block elements are metals and most of
them form coloured complexes or ions.

Note : The total number of elements in d-block is 39.
(d) f - block elements :
When last electron enters into f - orbital of (n – 2)th
shell, elements of this class are called f - block
elements.
Characteristics :
(B) Actinides : 14 elements present after element
actinium (89Ac) are called actinides. 2nd inner transition
series of metals or 5f- series, also contains 14 elements
i.e. 90Th to 103Lr.

Note : The total number of elements in f-block is 28.

Note : Zn, Cd and Hg are d-block elements , but not
transition elements, because they do not contain
partially filled d-orbitals.
Division of the periodic table into various blocks :
Different Types of Elements :
(i) Noble gases : The elements belonging to group
18 are called noble gases or aerogenous. They are
also known as inert gases because their outermost
orbitals are completely filled. Helium (He) is an
exception. It has only two electrons which are present
in s-orbital.
(ii) Representative elements : Elements in which
atoms have all shells complete except outermost shell
(iv) Inner transition elements :- They contain three
incomplete outermost shell and were also referred
to as rare earth elements, since their oxides were
rare in earlier days.
(v) Diagonal relationship : Some elements of 2nd
and 3 rd period show diagonal relationship among
them. They represent the same properties of two
periods. This relation is known as diagonal relation.
which is incomplete. Except 18th group, all s - block
and p - block elements are collectively called normal
or representative elements.
(iii) Transition elements : Those elements which
have partially filled d - orbitals in neutral state or in
any stable oxidation state are called transition
elements.
(vi) Transuranium elements : The elements with
atomic number greater than 92 (Z > 92) are known as
transuranium elements.
43
43
PAGE # 43
PAGE # 44
44
44
Li
Hydrogen 2
6.941 4
K
4
7
6
Mg
87.62
Sr
85.468 38
Rb
Calcium
4
Ti
V
Cr
8
Mn
54.938 26
7 VIIB
51.996 25
6 VIB
50.942 24
5 VB
47.867 23
IVB
B
Boron
Fe
55.845 27
9
10
Co
58.933 28
VIIIB
Ni
58.963 29
11
12
Cu
63.546 30
IB
Zn
65.39
IIB
Ra
Radium
(226)
Barium
Fr
88
Ba
Francium
Caesium
87 (223)
Cs
N
O
F
18.998 10
17 VIIA
15.999 9
16 VIA
8
Ne
20.180
Helium
He
4.0026
18 VIIIA
2
Si
Ga
Ge
Aluminium Silicon
31 69.723 32 72.64
Al
33
As
S
Se
78.96
Sulphur
74.922 34
Phosphorus
P
Ar
Br
Kr
Argon
Chlorine
35 79.904 36 83.80
Cl
Oxygen
Fluorine
Boron
Nitrogen
Carbon
Neon
13 26.982 14 28.086 15 30.974 16 32.065 17 35.453 18 39.948
C
B
14.007
15 VA
12.011 7
14 IVA
10.811 6
13 IIIA
5
p–Block Elements
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Ac
Rf
Rutherfordium
ACTINOIDES
LANTHANOIDES
Actinium
104 (261)
Hafnium
(227)
Lanthanum
89
Hf
La
Sg
Rhenium
Bh
Nd
Protactinium
Hs
Ds
110 (281)
Platinum
Pt
Rg
Gold
111 (272)
Au
Meitnerium Darmstadtium Roentgenium
Mt
109 (268)
Iridium
Ir
Mercury
Hg
Thallium
Ti
Lead
Pb
83
(145)
Neptunium
Np
(237)
Promethium
Pm
(244)
Pu
Plutonium
94
(243)
Am
Americium
95
(247)
Cm
Curium
96
(247)
Bk
Berkelium
97
(251)
Cf
Californium
98
(252)
Es
Einsteinium
99
Er
Fermium
Fm
100 (257)
Te
Mendelevium
Md
101 (258)
Thulium
Tm
(210)
At
86
Nobelium
No
102 (259)
Ytterbium
Yb
Xe
Lu
174.97
Radon
Rn
(222)
Xenon
Lawrencium
Lr
103 (262)
Lutetium
173.04 71
Astatine
85
I
Iodine
168.93 70
Polonium
Po
(209)
Tellurium
197.26 69
Bismuth
Erbium
164.93 68
Ho
Holmium
162.50 67
Dy
Dysprosium
158.93 66
Tb
Terbium
157.25 65
Gd
Gadolinium
151.96 64
Eu
Europium
150.36 63
Sm
Samarium
62
208.98 84
Bi
Inner - Transition Metals (f-Block Elements)
Hassium
238.03 93
U
Uranium
231.04 92
Pa
Osmium
Os
108 (277)
144.24 61
Bohrium
Praseodymium Neodymium
Pr
Re
107 (264)
140.91 60
Seaborgium
232.04 91
Th
Thorium
90
Tungsten
W
106 (266)
140.12 59
Ce
Cerium
58
Dubnium
Db
105 (262)
Tantalum
Ta
Antimony
Sb
Iron
Selenium
Scandium
Titanium
Vanadium
Chromium Manganese
Copper
Zinc
Gallium Germanium
Arsenic
Bromine
Cobalt
Nickel
Krypton
39 88.906 40 91.224 41 92.906 42 95.94 43 (98)
44 101.07 45 102.91 46 106.42 47 107.87 48 112.41 49 114.82 50
118.71 51 121.76 52
127.60 53
126.90 54
131.29
Sc
44.956 22
IIIB
IIIA
10.811
Element name
5
13
Transition Metals (d –Block Elements)
Symbol
Atomic number
Group IUPAC
Zirconium
Niobium
Yttrium
Rhodium
Palladium
Cadmium
Strontium
Molybdenum Technetium Ruthenium
Silver
Indium
Tin
Rubidium
55 132.91 56 137.33 57 138.91 72 178.49 73 180.95 74 183.84 75 186.21 76 190.23 77 192.22 78 195.08 79 196.97 80 200.59 81 204.38 82 207.2
37
Potassium
Ca
Magnesium 3
Sodium
19 39.098 20 40.078 21
Na
5
Be
9.0122
IIA
Beryllium
Lithium
11 22.990 12 24.305
3
H
1.0079
1
3
2
Period 1
IA
Elements
1
Group
s–Block
Relative atomic mass
PERIODIC TABLE OF THE ELEMENTS
(iv) Variation of atomic radii in a period : As we move
PERIODICITY IN PROPERTIES
from left to right across a period, there is a regular
(a) Atomic Radius :
decrease in atomic radii of the representative
(i) Covalent radius : It may be defined as one - half of
the distance between the centres of the nuclei of two
similar atoms bonded by a single covalent bond.
X
elements. This is due to the fact that number of energy
shells remains the same in a period, but nuclear
charge increases gradually as the atomic number
X
increases. This increases the force of attraction
A
B
towards nucleus which brings contraction in size.
This can also be explained on the basis of effective
nuclear charge which increases gradually in a period
1
2 AB = rcovalent
(of element X)
i.e. electron cloud is attracted more strongly towards
nucleus as the effective nuclear charge becomes
e.g. The internuclear distance between two hydrogen
atoms in H 2 molecule is 74 pm. Therefore, the
covalent radius of hydrogen atom is 37 pm.

Note : Covalent radius is generally used for
non - metals.
(ii) Vander Waal’s radius : It may be defined as half
of the internuclear distance between two adjacent
atoms of the same element belonging to two nearest
neighbouring molecules of the same substance.
X
X
X
X
more and more as we move in a period. The increased
force of attraction brings contraction in size.
(v) Variation of atomic radii in a group : Atomic radii
in a group increase as the atomic number increases.
The increase in size is due to extra energy shells
which outweigh the effect of increased nuclear charge.
The following table illustrates the periodicity in atomic
radii (covalent radii) of representative elements. The
radii given in the table are in angstrom (Å).
Periodicity in atomic radii (covalent radii)

Characteristics :
1 EH = r
vander Waals
2
Period/
Group
1
1
H
0.32
2
3
4
5
6
7
2
Li
Be
B
C
N
O
F
1.23 0.89 0.80 0.77 0.75 0.73 0.72
(B) The formation of covalent bond involves
overlapping of atomic orbitals, as a result of this, the
internuclear distance between the covalently bonded
atoms is less than the internuclear distance between
the non bonded atoms.
3
Na Mg
Al
Si
P
S
Cl
1.54 1.36 1.20 1.17 1.10 1.04 0.99
4
K
Ca Ga Ge As Se
Br
2.03 1.74 1.26 1.22 1.20 1.16 1.14
e.g. Vander Waals radius of helium is 1.20 Å
5
(iii) Metallic radius (Crystal radius) : Metallic radius
may be defined as half of the internuclear distance
between two adjacent atoms in a metallic lattice.
X
X
I
Rb Sr
In
Sn
Sb Te
2.16 1.91 1.44 1.41 1.40 1.36 1.33
6
Cs Ba
Tl
Pb
Bi
Po
2.35 1.98 1.48 1.47 1.46 1.46
(A) Covalent radius of the elements is shorter than
its Vander Waal’s radius.
Increases
H
E
Decreases
C
D
1
CD = rcrystal
2
(of element X)
•
The metallic radius of an atom is always larger
Atoms of zero group elements do not form chemical
bonds among themselves. Hence for them Vander
Waals radii are considered.
Element
Vander Waals
radii (in Aº )
He
Ne
Ar
Kr
Xe
1.20 1.60 1.91 2.00 2.20
than its covalent radius.

Note : The order of different radius is - r Vander Waals > rMetallic
> rCovalent
The sudden increase in atomic radii in comparison
to the halogens (the elements of 7th group) in case
of inert gases, is due to the fact that, Vander Waals
radii are considered which always possess higher
values than covalent radii.
45
45
PAGE # 45
The decrease in the size of transition elements is
small since the differentiating electrons enter into
inner ‘d’ levels. The additional electrons into (n–1)d
levels effectively screen much of increased nuclear
charge on the outer ns electrons and therefore, size
remains almost constant.
Conclusions
However, in vertical columns of transition elements,
there is an increase in size from first member to
second member as expected, but from second
member to third member, there is very small change
in size and sometimes sizes are same. This is due
to Lanthanide contraction (in the lanthanide
elements differentiating electrons enter into 4f-levels).
Since these electrons do not effectively screen the
valence electrons from the increased nuclear charge,
the size gradually decreases. This decrease is termed
lanthanide contraction.
(iii) The ionic radii in a particular group increase in
moving from top to bottom.
Conclusions
(i) The alkali metals which are present at the extreme
left of the periodic table have the largest size in a
period.
(ii) The halogens which are present at the extreme
right of the periodic table have the smallest size.
(iii) The size of the atoms of inert gases are, however,
larger than those of preceding halogens because in
inert gases van der Waals' radii are taken into
consideration.
(iv) In a group of transition elements, there is an
increase in size from first member to second member
as expected but from second member to third
member, there is very small change in size and
sometimes sizes are same. This is due to Lanthanide
contraction.
•
Ionic radius : It is the distance between the nucleus
and outermost shell of an ion or it is the distance
between the nucleus and the point where the nucleus
exerts its influence on the electron cloud.
(a) The radius of the cation is always smaller than
the atomic radius of its parent atom. This is due to
the fact that nuclear charge in the case of a cation is
acting on a lesser number of electrons and pulls them
closer.
(b) The radius of the anion is always larger than the
atomic radius of its parent atom. In an anion as
electron or electrons are added to the neutral atom,
the nuclear charge acts on more electrons so that
each electron is held less tightly and thereby the
electron cloud expands.
Comparative sizes of atoms and their cations
Atom
Li
Na
K
Mg
Ba
Al
Pb
Atomic radii Corresponding
(crystal, Å)
cations
Ionic radii (Å)
+
Li
1.52
0.59
+
Na
1.86
0.99
+
K
2.31
1.33
2+
Mg
1.60
0.65
2+
Ba
2.22
1.35
3+
Al
1.43
0.50
2+
Pb
1.75
1.32
(i) The radius of cation (positive ion) is always smaller
than that of the parent atom.
(ii) The radius of anion (negative ion) is always larger
than that of the parent atom.
(iv) In a set of species having the same number of
electrons (isoelectronic), the size decreases as the
charge on the nucleus increases.
(v) The size of the cations of the same element
decreases with the increase of positive charge.
(b) Ionisation Energy (IE) :
Ionisation Energy (IE) of an element is defined as the
amount of energy required to remove an electron from
an isolated gaseous atom of that element resulting
in the formation of a positive ion.
Characteristics :
(A) The energy required to remove the outermost
electron from an atom is called first ionisation energy
(IE)1.
After removal of one electron, the atom changes into
monovalent positive ion.
M(g) + IE1  M+(g) + e–
(B) The minimum amount of energy required to
remove an electron from monovalent positive ion of
the element is known as second ionisation energy
(IE)2.
M+(g) + IE2  M2+(g) + e–
(C) The first, second etc. ionisation energies are
collectively known as successive ionisation energies.
M2+(g) + IE3  M3+(g) + e–
In general (IE)1 < (IE)2 < (IE)3 so on, because, as the
number of electrons decreases, the attraction
between the nucleus and the remaining electrons
increases considerably and hence subsequent
ionisation energies increase.
(D) Units : Ionisation energy is expressed either in
terms of electron volts per atom (eV/atom) or Kilojoules
per mole of atoms (KJ mol – 1) or K cal mol – 1.
1 eV/atom = 96.49 KJ/mol = 23.06 Kcal/mol = 1.602 × 10–19
J/atom
Factors influencing ionisation energy :
(A) Size of the atom : Ionisation energy decreases with
increase in atomic size. As the distance between the
outermost electrons and the nucleus increases, the force
of attraction between the valence shell electrons and the
nucleus decreases. As a result, outermost electrons are
held less firmly and lesser amount of energy is required
to knock them out.
For example, ionisation energy decreases in a group
from top to bottom with increase in atomic size.
46
46
PAGE # 46
(B) Nuclear charge : The ionisation energy
increases with increase in the nuclear charge. This
is due to the fact that with increase in the nuclear
charge, the electrons of the outermost shell are more
firmly held by the nucleus and thus greater amount of
energy is required to pull out an electron from the atom.
For example, ionisation energy increases as we move
from left to right along a period due to increase in nuclear
charge.
As noble gases have completely filled electronic
configurations, they have highest ionisation energies
in their respective periods.
(C) Shielding effect : The electrons in the inner shells
act as a screen or shield between the nucleus and
the electrons in the outermost shell. This is called
shielding effect or screening effect. Larger the number
of electrons in the inner shells, greater is the
screening effect and smaller the force of attraction
and thus ionisation energy decreases.
(A) nuclear charge increases regularly.
Variation of ionisation energy in a period :
In general, the value of ionisation energy increases
with increase in atomic number across a period. This
can be explained on the basis of the fact that on
moving across the period from left to right-
(B) addition of electrons occurs in the same shell.
(C) atomic size decreases.
Variation of ionisation energy in a group :
In general, the value of ionisation energy decreases
while moving from top to bottom in a group.This is
because (A) effective nuclear charge decreases regularly.
(B) addition of electrons occurs in a new shell.
These electrons
shield the outer
electrons from the
nucleus
This electron does
not feel the full inward pull
of the positive charge of
the nucleus
(D) Penetration effect of the electrons : The
ionisation energy increases as the penetration effect
of the electrons increases. It is a well known fact that
the electrons of the s-orbital have the maximum
probability of being found near the nucleus and this
probability goes on decreasing in case of p, d and f
orbitals of the same energy level.
Greater the penetration effect of electrons more firmly
the electrons will be held by the nucleus and thus
higher will be the ionisation energy of the atom.
For example, ionisation energy of aluminium is
comparatively less than magnesium as outermost
electron is to be removed from p-orbital (having less
penetration effect) in aluminium, whereas in
magnesium it will be removed from s-orbital (having
larger penetration effect) of the same energy level.

Note : With in the same energy level,the penetration
effect decreases in the order s > p > d > f
(E) Electronic Configuration : If an atom has exactly
half-filled or completely filled orbitals, then such an
arrangement has extra stability.The removal of an
electron from such an atom requires more energy
than expected. For example,
E1 of Be > E1 of B
(C) atomic size increases.
Conclusions :
(i) In each period, alkali metals show lowest first
ionisation enthalpy. Caesium has the minimum value.
(ii) In each period, noble gases show highest first
ionisation enthalpy. Helium has the maximum value
of first jonisation enthalpy.
(iii) The representative elements show a large range
of values of first ionisation enthalpies, metals having
low while non-metals have high values.
(iv) Generally. ionisation enthalpies of transition
metals increase slowly as we move from left to right
in a period. The f-block elements also show only a
small variation in the values of first ionisation
enthalpies.
(c) Electron Affinity (EA) :
Electron affinity is defined as the energy released in
the process of adding an electron to a neutral atom in
the gaseous state to form a negative ion.
X(g) + e–  X–(g) + Energy (E.A.)
Cl(g) + e–  Cl– (g) + 349 KJ/mol
The electron affinity of chlorine is 349 KJ/mol.
The addition of second electron to an anion is
opposed by electrostatic repulsion and hence the
energy has to be supplied for the addition of second
electron.
1s 2 , 2s 2



Be (Z = 4) Completely filled
orbital (more stable)
O(g) + e–  O– (g) + Energy (EA -)
1s 2 , 2s 2 , 2 p1


B (Z = 5)
Partially filled
orbital (less stable )
(EA -) is exothermic whereas, (EA-) is endothermic.
O–(g) + e–  O2– (g) – Energy (EA-)
(i) Units : Kilo joules per mole (KJ/mol) of atoms or
electron volts per atom (eV/atom).
47
47
PAGE # 47
(ii) Factors affecting electron affinity:
(A) Nuclear charge : Greater the magnitude of
nuclear charge greater will be the attraction for the
incoming electron and as a result, larger will be the
value of electron affinity.
Electron affinity  Nuclear charge.
(B) Atomic size : Larger the size of an atom is, more
will be the distance between the nucleus and the
incoming electron and smaller will be the value of
electron affinity.


1
 E.A. 

Atomic size 

(C) Electronic configuration : Stable the electronic
configuration of an atom lesser will be its tendency to
accept the electron and lower will be the value of its
electron affinity.
(iii) Variation of electron affinity in a period : On
moving across the period the atomic size decreases
and nuclear charge increases. Both these factors

Conclusion
(i) The electron gain enthalpies, in general, become
less negative in going down from top to bottom in a
group. This is due to increase in size on moving down
a group. This factor is predominant in comparison to
other factor, i.e., increase in nuclear charge.
Na
–53
Cl
–349
K
Rb
–48 –47
Br
I
–325 –295
Cs
–1
–46 KJ mol
At
– 270KJ mol–1
(ii) The electron gain enthalpies of oxygen and fluorine,
the members of the second period, have less
negative values than the elements sulphur and
chlorine of the third period. This is due to small size
of the atoms of oxygen and fluorine. As a result, there
is a strong inter-electronic repulsion when extra
electron is added to these atoms, i.e., electron density
is high and the addition of electron is not easy. Thus,
the electron gain enthalpies of third period elements,
sulphur and chlorine, have more negative values than
corresponding elements oxygen and fluorine.
–1
O –141 kJ mol
S –200 kJ mol–1
F
Cl
–328 kJ mol–1
–349 kJ mol–1
Similar trend is observed in nitrogen & phosphorous
N +31 kJ mol–1
P – 74 kJ mol–1
result into greater attraction for the incoming electron,
therefore electron affinity in general increases in a
period from left to right.
(iv) Variation of electron affinity in a group : On
moving down a group, the atomic size as well as
nuclear charge increase, but the effect of increase in
atomic size is much more pronounced than that of
nuclear charge and thus, the incoming electron feels
less attraction consequently, electron affinity
decreases on going down the group.
(v) Some irregularities observed in general trend:
(A) Halogens have the highest electron affinities in
their respective periods. This is due to the small size
and high effective nuclear charge of halogens.
Halogens have seven electrons in their valence shell.
By accepting one more electron they can attain stable
electronic configuration of the nearest noble gas. Thus
they have maximum tendency to accept an additional
electron.
(B) Due to stable electronic configuration of noble
gases electron affinities are zero.
(C) Be, Mg, N and P also have exceptionally low
values of electron affinities due to their stable
electronic configurations.
2
Be = 1s , 2s
2
2
2
3
N = 1s , 2s , 2p
Mg = 1s2, 2s2 , 2p6, 3s2 P = 1s2, 2s2, 2p6, 3s2, 3p3
(iii) In general, electron gain enthalpy becomes more
and more negative from left to right in a period. This
is due to decrease in size and increase in nuclear
charge as the atomic number increases in a period.
Both these factors favour the addition of an extra
electron due to higher force of attraction by the nucleus
for the incoming electron.
(iv) Electron gain enthalpies of some of the members
of alkaline earth metals, noble gases and nitrogen
are positive.
This is because they have stable configurations.
Alkaline earth metals have stable configurations due
to completely filled ns orbital while nitrogen has extra
stability due to half filled p-orbitals (1s2, 2s2, 2p3) i.e.,
symmetrical configuration. These atoms resist the
addition of extra electron as they do not want to disturb
their stability.
Noble gases have ns2np6 configuration, i.e., no place
for incoming electron. In case the extra electron is to
be accommodated, it will occupy its position on a
new principal shell, i.e., it will be weakly attracted
towards nucleus. Such anion will be extremely
unstable. Helium has also stable 1s2 configuration
and cannot accommodate the incoming electron.
(v) Halogens have highest negative electron gain
enthalpies.
Following two factors are responsible for this:
(a) Small atomic size and high nuclear charge of
halogens in a period.
(b) Halogens have the general electronic
configuration of ns2 np5, i.e., one electron less than
stable noble gas (ns2 np6) configuration.
48
48
PAGE # 48
Thus, halogens have very strong tendency to accept
an additional electron and their electron gain
enthalpies are, therefore, high.
•
(d) Electronegativity :
Electronegativity is a measure of the tendency of an
Importance of Electron Gain Enthalpy : Certain
properties of the elements can be predicted on the
basis of values of electron gain enthalpies.
(i) The elements having high negative values of
electron gain enthalpy are capable of accepting
electron easily. They form anions and thus form ionic
(electrovalent) compounds. These elements are
electronegative in nature.
element to attract electrons towards itself in a
(ii) The elements having high negative electron gain
enthalpies act as strong oxidising agents, for example,
F, CI, Br, O, S, etc.
On the basis of the general trend of ionisation
enthalpy and electron gain enthalpy, the following
properties can be predicted:
affinity. Higher ionisation potential & electron affinity
covalently bonded molecule .
(i) Factors influencing electronegativity :
(A) The magnitude of electronegativity of an element
depends upon its ionisation potential & electron
values indicate higher electronegativity value.
(B) W ith increase in atomic size the distance
between nucleus and valence shell electrons
increases, therefore the force of attraction between
(i) Metallic nature decreases in a period while nonmetallic nature increases. Metallic nature increases
in a group while non-metallic nature decreases. The
arrow () represents a group and () represents a
period.
the nucleus and the valence shell electrons
decreases and hence the electronegativity values
also decrease.
(C) In higher oxidation state, the element has higher
Metallic
magnitude of positive charge. Thus, due to more
Metallic
(Electro + ve)
positive charge on element, it has higher polarising
Decreases
power. Thus, with increase in the oxidation state of
Increases
element, its electronegativity also increases.
Non-metallic
Non-metallic (Electro-ve)
Increases
Decreases
(ii) Reducing nature decreases in a period while
oxidising nature increases. The reducing nature
increases in a group while oxidising nature
decreases.

Note :
Polarising power is the power of an ion (cation) to
distort the other ion.
(D) With increase in nuclear charge, force of attraction
between nucleus and the valence shell electrons
increases and, therefore electronegativity value
increases.
Reducing nature
(E) The electronegativity of the same element
Decreases
Reducing
nature
;
increases.
Increases
Oxidising nature
Increases
Oxidising
nature
increases as the s-character in the hybrid orbitals
Hybrid orbital
sp 3
sp 2
sp
s-character
25%
33%
50%
Electronegativity increases
Decreases
(iii) Stability of metal increases while activity of the
metal decreases in a period and in a group stability
decreases.
Stability of the metal
(ii) Variation of Electronegativity in a group : On
moving down the group atomic number increases,
so nuclear charge also increases. Number of shells
also increases, so atomic radius increases.
Stability of the metal
Increases
Decreases
Therefore electronegativity decreases on moving
down the group.
(iii) Variation of Electronegativity in a period : While
Activity of the metal
;
Activity of the metal
Decreases
moving across a period left to right atomic number,
nuclear charge increases & atomic radius decreases.
Therefore electronegativity increases along a period.
Increases
49
49
PAGE # 49
Difference between Electron gain enthalpy and Electronegativity
S.No.
Electron gain enthalpy
1
It is the tendency of an isolated atom to
attract electron.
2
It can be measured experimentally.
Its units are electron volts per atom
or kilo joules per mole or kilo calories per
mole.
It is a constant quantity for a
particular element.
3
4
It has no units but merely a number.
It has no units but merely a number.
Electronegativity of an element is not constant.
It depends on a number of factors such as
hybridised state. Oxidation state, etc. The
periodicity is regular in a period but not
so regular in groups.
Its periodicity is not regular in a
period or a group.
5
Electronegativity
It is the tendency of an atom in a combined
state, i.e., in a molecule to attract the shared
pair of electrons.
It is a relative number and cannot be
determined experimentally.
If a non-metal forms a number of oxy-acids, the
strength increases with the increase of percentage
of oxygen.
(e) Nature of Oxides :
In a period, the nature of the oxides varies from basic
to acidic.
Na2O MgO
AI2O3
SiO2 P2O5 Cl2O7
Strongly Basic Amphoteric
Weakly Strongly
basic
acidic
acidic
Sulphur forms two oxy-acids H2SO3 and H2SO4.H2SO4
is stronger acid than H2SO3.
Chlorine forms a number of oxy-acids:
P2 O3
P2 O4
P2 O5
Acidic nature
increases
N2 O3
NO2
N2 O5
Neutral
oxide
Acidic nature
increases
N2 O
NO
Acidic nature
increases
In a group, basic nature increases or acidic nature
decreases. Oxides of the metals are generally basic
and oxides of the nonnmetals are acidic. The oxides
of the metalloids are amphoteric. The oxides of AI,
Zn, Sn, As and Sb are amphoteric. We can summarise
that as the electronegativity of element increases,
acidic character of oxides increases. W hen an
element forms a number of oxides, the acidic nature
increases as the percentage of oxygen increases.
MnO
Mn2O3
MnO2
MnO3
Mn2O7
In a period, the strength of the oxy-acids formed by
non-metals increases from ieft to right.
III Period
strength increases
strength
increases
In a group, the strength of the oxy-acids of non-metals
decreases.
V group
VII group
( g ) Nature of Hydrides :
The nature of the hydrides changes from basic to
acidic in a period from left to right.
NH 3
H2O
HF
weak base netural weak acid
Basic
Basic
Neutral
Acidic
Acidic
CO, N2O, NO and H2O are neutral oxides. The oxides
CO2, N2O5, P2O3, P2O5, SO2, SO3, Cl2O7, etc., are called
acid anhydrides as these combine with water to form
oxy-acids.
CO2  H2CO3
P2O3  H3PO3
SO3  H2SO4
N2O5  HNO3
P2O5  H3PO4
Cl2O7  HClO4
N2O3  HNO2
SO2  H2SO3
(f) Nature of Oxy-Acids :
II Period
Greater is the oxidation state of central atom more
will be the acidic strength.
strength decreases
strength decreases
PH3
very weak
base
H2 S
HCl
weak acid strong acid
In a group, the acidic nature of the hydrides of nonmetals increases. The reducing nature also increases
but stability decreases from top to bottom
•
Anomalous behaviour of the elements of second
period :
It has been obesrved that in the case of representative
elements, the first element in each, i.e. lithium in the
first group, beryllium in the second group and boron
to fluorine in the group 13 to 17, differ in many respect
from the other member of their respective group. The
anomalous behaviour of the first member of each
group is attributed to following reasons :
(a) small atomic radius of the atom and ionic radius
of its ion.
(b) high electronegativity
(c) non-availability of d-orbitals in their valence shell.
(d) tendency to form multiple bonds by carbon,
nitrogen and oxygen
(e) high charge/radius ratio
50
50
PAGE # 50
NUCLEAR CHEMISTRY
(b) On the basis of stability :
(i) Stable nuclei : Those nuclei which are permanent
& their proton and neutron contents remain
unchanged forever (can be changed only under severe
conditions of bombardment by external radition) are
called stable nuclei.
INTRODUCTION
Atoms have three fundamental particles that are
electrons, protons and neutrons. Protons and
neutrons are present inside the nucleus and electrons
are present in the extranuclear region. Changes
occurring in the nucleus which are a source of
tremendous energy are called nuclear reactions. The
branch of science which deals with the study of atomic
nucleus and nuclear changes is called nuclear
chemistry.
Nucleus term was firstly introduced by Rutherford . It is
defined as the central part of an atom which contain
all the protons and neutrons.
rn  A1/ 3
(ii) Unstable nuclei : Those nuclei in which no. of
protons and neutrons change with time.
STABILITY OF NUCLEUS
Stability of a nucleus can be explained by following
two factors.
(a) On the basis of (n/p) ratio :
Neutrons help to hold protons together within the
nucleus. The number of neutrons necessary to create
a stable nucleus increases rapidly as the number of
protons increases. The number of neutron to proton
ratio (n/p) of stable nuclei increases with increasing
atomic number . The area of graph in which all stable
nuclei are found is known as the belt of stability.
Radioactive nuclei occur outside this belt.
 rn = RoA1/3
Ro = 1.2 × 10–15 m
where : A =mass number,
rn = radius of nucleus
R0 = Rutherford constant
Density of nucleus = 1017 kg/m3
(a) On the basis of their Z and n values :
(i) Isotopes : The isotopes of an element have the
same atomic number but different atomic masses
due to the presence of different number of neutrons.
e.g.
1
2
3
22
23
24
1H , 1 H , 1 H ; 11 Na , 11 Na , 11 Na
(ii) Isobars : The atoms of different elements with
different atomic numbers, but same mass number
are called isobars.
e.g.
14
14
6 C , 7N
;
40
20 Ca
and
40
18 Ar
are isobars.
(iii) Isotones : The isotones may be defined as the
atoms of different elements containing same
number of neutrons.
e.g.
13
6C
and
3
14
7N ; 1
(iv) Mirror nuclei :
H and 24 He
A1
X
z1
&
A2
Y
z2
Z1 = n2 and Z2 = n1
Z1 = A1 – n1 = n2 ..... (1)
Z2 = A2 – n2 = n1 ...... (2)
from equation(1) n1 = A1 – n2
by putting the value of n1 in equation (2)
A2 – n2 = A1 – n2
A1 = A2
So these are defined as those isobars in which the
Z and n values are interchanged
e.g.
3
1H
and
3
13
2 He , 6 C
and
13
7 N
The type of radioactive decay that a particular radio
isotope will undergo depends to a large extent on its
neutrons to protons ratio compared to those of nearby
nuclei that are within the belt of stability.
are mirror nuclei if
(i) A nucleus whose high n/p ratio places it above the
belt of stability emits a  -particle in order to lower n/p
ratio and move towards the belt of stability.
1
0n
11 p  –10 e  –
( – is Anti neutrino)

Note :
Antineutrino is the antiparticle of neutrino, which is
neutral particle produced in nuclear beta decay.
51
51
PAGE # 51
interaction is possible between two particles which
can exist in a state capable of sharing some
common property, thereby, the total energy level gets
lowered and the system becomes more stable. A
similar situation is proposed for the atomic nucleus
by assuming a ceaseless exchange of common
property (may be charge, spin or position) between
neighb ouring nucleons in the same state of motion.
Yukawa’s prediction (1935) of the existence of
mesons (particles with mass intermediate between
an electron and a nucleon) and very soon, the meson or pions (m= 237 me) were recognized as
exchange of common property.
(ii) A nucleus which has lower n/p ratio , is placed below
the belt of stability either emits positrons or undergoes
electron capture. Both modes of decay decrease the
number of protons and increase the number of neutrons
in the nucleus and thus, positron emission or electron
capture results in an increase in n/p ratio.
e.g.
1
1P
 10 n  01 e  
(Positron emission)
1
1p
1
 –10 e  0 n  X  ray
(Electron capture)

Note :
A positron has same mass as electron but carries
opposite charge. The positron has a very short life
because it is annihilated when it collides with
an electron, producing gamma rays. This phenomenon
is known as pair production.
(iii) The nuclei with atomic number > 83, outside the
belt of stability, undergo -emissions. Emission of an
-particle decreases both the number of protons and
neutrons and thereby increases n/p ratio.
Thus,
For lighter element if Z = n then nuclei will be stable
For nuclie Z > 20, stability condition is 1.6  n/p > 1
Unstable nuclei emit radiations to achieve stability, this
property is known as radioactivity.
(b) On the basis of even and odd nature of the
number of protons and neutrons :
(i) The number of stable nuclides is maximum when
both Z and n are even numbers. About 60% of stable
nuclides have both Z and n even.
(ii) The number of stable nuclides in which either the Z
or n is odd is about one third of those, where both are
even.
MAGIC NUMBERS
Just as certain numbers of electrons (2,8,18,36,54 and
86) correspond to stable closed shell electron
configuration, certain number of nucleons leads to
closed shell in nuclei. The protons and neutrons can
achieve closed shell. Nuclei with 2, 8, 20, 28, 50 or 82
protons or 2,8,20, 28, 50, 82, or 126 neutrons
correspond to closed nuclear shell. Closed shell nuclei
are more stable than those that do not have closed
shells. These numbers of nucleons that correspond to
closed nuclear shells are called magic numbers.
NUCLEAR EXCHANGE FORCE-MESON THEORY
If only the short range charge independent attractive
forces were to operate between the nucleons, the
attractive force would grow limitlessly till the nucleus
ultimately collapsed. There is a need, therefore, for
some repulsive forces to lead to the saturation of
attractive forces.
The nucleus stability has been explained in terms of
existence of nuclear exchange force. An exchange
n
p + – ;
p
n + +
p
p + 0 ;
n
n + 0.
where –, + 0 are pions.
The range of pions being of the same order as the
nuclear radius and thus, emission of pion by a
nucleon are reabsorption by another nucleon goes
on incessantly.
Ex-1
14
6C
nuclide undergoes  -decay. Which stable
nuclide is formed ? Give equation.
Sol.
14
6C

14
0
7 N + –1 e
RADIOACTIVITY
Radioactivity is a process in which nuclei of certain
elements undergo spontaneous disintegration
without excitation by any external means.
All heavy elements from bismuth (Bi) to uranium
and a few of lighter elements have naturally
occurring isotopes which possess the property of
radioactivity. All those substances which have the
tendency to emit these radiations are termed as
radioactive materials. Radioactivity is a nuclear
phenomenon i.e., the kind of intensity of the radiation
emitted by any radioactive substance is absolutely
the same whether the element is present as such
or in any one of its compounds.
e.g. Elements like uranium (U) , thorium (Th) ,
polonium (Po), radium (Ra) etc. are radioactive in
nature.
(a) History of the Discovery of Radioactivity :
In 1895, Henri Becquerel was studying the effect of
sunlight on various phosphorescent minerals, one
of the substance being studied was uranium ore.
He accidently left a crystal of uranium sample ;
Potassium uranyl sulphate [K2UO2 (SO4)2. 2H2O] in
a drawer along with some photographic plate
wrapped in black paper. Much to his surprise, he
discovered that the photographic plate had been
fogged by exposure to some invisible radiations
from uranium. He called this mysterious property of
the ore as ‘radioactivity’ (Radioactivity means rayemitting activity). A year later, in 1896, Marie Curie
found that besides uranium and its compounds,
thorium was another element which possessed the
property of radioactivity. 1898 Marie Curie and her
husband Pierrie Curie isolated two new radioactive
elements polonium and radium.
52
52
PAGE # 52
(b) Natural & Artificial Radioactivity :
(c) Analysis of Radioactive Radiations :
If a substance emits radiations by itself it possesses
In 1904, Rutherford and his co-workers observed that
when radioactive radiations were subjected to a
natural radioactivity but if a substance does not
possess radioactivity and starts emitting radiations
magnetic field or a strong electric field, these were
split into three types, as shown in the figure. The rays
on exposure to rays from a natural radioactive
which are attracted towards the negative plate, are
substance, it is called induced or artificial radioactivity.
positively charged , and called alpha () rays. The
e.g.
rays which are deflected towards the positive plate
When aluminium is bombarded with - particles , a
are negatively charged and are called beta ( ) rays.
radioactive isotope of phosphorus is formed which
disintegrate spontaneously with the emission of
The third type of rays which are not deflected on any
side but move straight are known as gamma () rays.
-rays
-rays
ys
ra
-
positrons (which are positively charged electron, +1e0).
ys
ra
-
+
+
+
+
+
+
+
+
+
-rays
-rays
Magnetic
field
Radioactive
substances

Note :
(A)
Natural radioactivity was discovered by Becquerel
while artificial radioactivity was discovered by Irene
Curie and Joliot.
(B)
Figure :
(A) Deflection of radioactive rays in electric field and,
(B) Emission of radioactive rays and their deflection
in a magnetic field. (The direction of magnetic field is
inward perpendicular to the page).
T The important properties of- rays ,  rays and -rays are as follows :
53
53
PAGE # 53
(ii) _ decay produces isodiaphers i.e.parent and the
daughter nuclides have same isotopic mass (which
is the difference between number of neutrons and
protons) .
( d ) Units of Radioactivity :
(i) SI unit is Becquerel (Bq) which is defined as
disintegration per sec (dps).
(ii) Earlier radioactivity was given in terms of Curie
(Ci).
1 Ci refers to the activity of Radium.
1 Ci = 3.7 × 1010 dps = 3.7 × 1010 Bq.
1 Milli Ci = 3.7 × 107 Bq.
1 Micro Ci = 3.7 × 104 Bq.
No. of neutrons:
No. of protons:
Difference
6
(1911_1913).The element emitting the  or  particle
is called parent element and the new element formed
is called daughter element.

e.g.
92 U
 90 Th
234
 2 He
4
(ii) When a  _ particle is emitted the new element
formed is displaced one position to the right in the
periodic table than that of parent element (because
the atomic number increases by 1).
e.g.
The emission of  _ particle by 6C 14 may be
represented as follows:
6
C14 
7
N14 +-1e0
(a) Explanation :
The results of the group displacement laws may be
explained as follows:
Since an _ particle is simply a helium nucleus
(containing two neutrons and two protons) therefore,
loss of _ particle means loss of two neutrons and
two protons . Thus, the new element formed has
atomic number less by 2 unit and mass number less
by 4 unit.
The  _ particle is simply an electron and there are no
electrons present in nucleus .However , the loss of  _
particle is also found to be a nuclear phenomenon
because the change in external conditions
(temperature etc.) has no effect on the rate of the
emission of  _ particle. It is therefore, believed that
for emission of  _ particle to occur, a neutron changes
to a proton and an electron i.e.
_
Neutron  Proton + Electron ( particle)
As a result ,the number of protons in the nucleus
increases by 1 and so does the atomic number.

Note :
(i) Increase or decrease in the number of protons in
the nucleus (due to loss of  particle or  _ particle)
is accompanied simultaneously by the loss or gain
of electrons in the extranuclear part (from the
surroundings) so that the electrical neutrality is
maintained in the new atom formed.
146
92
54
144
90
54
C14 
7
N14 +-1e0
(iv) Emission of 1 and 2 particles produces an
isotope of parent element.
e.g.



 92U234
 90Th234 
U238 
 91Pa234 
92
These laws were given by Soddy,Fajans and Russel
238
 90 Th 234  2 He 4
(iii)  _ decay produces isobars i.e.parent and the
daughter nuclides have different atomic numbers but
same mass number .
E.g.
(iii) Another unit is Rutherford (Rd).
1 Rd = 106 dps
(i) When an _ particle is emitted, the new element
formed is displaced two positions to the left in the
periodic table than that of parent element (because
the atomic number decreases by 2).
238
92 U
E.g.
Note :
The emission of  and  _ particles is also known as
_decay and  _ decay.
Ex-2 Calculate number of  and  - particles emitted when
238
92 U
changes into radioactive 206
82 Pb .
Sol. No. of -particles =
=
Change in mass number
4
238 – 206
32
=
=8
4
4
No. of  -particles = 2 × -particles - (ZA – ZB)
Here ZA and ZB are atomic no. of parent and daugther
nuclei respectively.
= 2 × 8 – (92 – 82)
16 – 10
=6
Thus, no. of -particles emitted out = 8
No. of  -particles emitted out = 6
Ex.3 90Th234 disintegrates to give 82Pb206 as final product .
How many alpha and beta particles are emitted during
this process ?
Sol. Suppose the no. of  particles emitted = x and no. of
 _ particles emitted = y.
Then
90
Th234 
82
Pb206 + x 2He4 + y-1e0
Equating the mass number on both sides ,we get
234 = 206 + 4x + 0y
or 4x = 28 or
x=7
Equating the atomic number on both sides ,we get
90=82+2x-y
y=6
Ans. 7 and 6 particles will be emitted.
Alternative Method:
No. of -particles =
Difference in atomic mass of reactants and products
4
=
234 – 206
4
=
28
=7
4
No. of -particle=( 2 × no. of -particles) – (ZA – ZB)
where , ZA = Atomic number of reactant
ZB = Atomic number of product
= (2 × 7) – (90 – 82) = 14 – 8 = 6
Ans.7 and 6 particles will be emitted.
54
54
PAGE # 54
(b) Significance of Half-Life Period :
RATE OF RADIOACTIVE DECAY
Radioactive disintegration is an example of first order
reaction, i.e., the rate of decay is directly proportional
to the no. of atoms (amount) of the element present
at the particular time.
A  Decay product
No. of atoms at t = 0  N0
No. of atoms left after t = t  N
Hence, rate  [N0 – N]/t because rate continuously
decreases with time. Let dN be the change in no. of
atoms in an infinitesimal small time dt, then rate of
decay can be written as -
–
dN
 [N]1 = N. The negative sign indicates the
dt
decreasing trend of N with increasing time.
where  is the proportionality constant.
Integration of this equation finally gives or  =

2.303
N0
log10
t
N
Note :  is also known as decay or disintegration or
(i) Stability of nuclei : The value of half-life period can
give an idea about relative stability of radio isotopes.
All isotopes with longer t1/2 are more stable.
(ii) The amount of substance left after ‘n’ number of
half lives can be given as :-
 1n

2
N = No 
Where ;
N = Amount of the substance left after ‘n’ half-lives.
No = Initial amount of the substance.
Ex.4 The half-life period of 53I125 is 60 days .What percent
of the original radioactivity would be present after 180
days ?
Sol. t½ = 60 days, t = 180 days
n=
180
Total time( t )
=
=3
60
Half  life peiod( t ½ )
Applying the formula N=
radioactive constant.
we get N =
CHARACTERISTICS OF RATE OF DISINTEGRATION
(i) Rate of disintegration continuously decreases with
time.
(ii) Rate of disintegration as well as  are independent
of P and T.
(iii) (a) Unit of rate of decay : disintegration per time
(b) Unit of decay constant : time–1
(iv) Time required to complete a definite fraction is
independent of initial no. of atoms (amount) of
radioactive species.
1
8
× 100 = 12.5%
 1n

2
N = No 
According to question , N =
1
1
or
1
N
8 0
 1n
1
N0 = No  
8
2
HALF-LIFE PERIOD
(2)
3
=
n
(2)
 n = 3
(a) Characteristics of Half-Life Period :
So, time taken by the sample to reduce to 1/8th of its
reactivity will be -
• It is denoted by t1/2 .
T= n×t½
• Each radioactive element has a characteristic halflife period .
T= 3×27.96
• Half-life period for an element is a constant.
• t1/2 =
0.693

Where  is a constant known as disintegration
constant or decay constant. It is the characteristic of
the nature of the radioactive element.

2
=
n
2n
Ex.5 The half -life period of a radioactive element is 27.96
days . Calculate the time taken by a given sample to
reduce to 1/8th of its activity.
Sol. The amount of substance left after ‘n’ number of half
lives can be given as :-
or
The time required for the decay of radioactive element
to half of the original amount is called half-life period.
N0
N0
Note :
Half-life period does not depend upon initial amount
= 83.88 days
Ex.6 Half- life period of a radioactive element is 100
seconds. Calculate the disintegration constant.
Sol. t1/2 = 100 Seconds,
0.693
 =
t
0.693
=
100
= 0.00693 s–1
1/2
= 6.93 × 10–3 sec–1
of element.
55
55
PAGE # 55
AVERAGE LIFE
NUCLEAR REACTIONS
Evidently , the whole of the radioactive element can
never disintegrate or in other words
, the time required for the disintegration of the whole of
a radioactive element will be infinity.
Thus, it is meaningless to talk of the total life of a
radioactive element . However, sometimes another
term is used ,called average life () which is the
reciprocal of the disintegration constant () i.e.
Average life ()=
t
1
½ = 1.44 t .
 0.693
1/2

The reactions in which nuclei of atoms interact with
other nuclei or elementary particles such as alpha
particle, proton, deutron, neutron etc. resulting in the
formation of a new nucleus and one or more
elementary particles are called nuclear reactions.
Nuclear reactions are expressed in the same fashion
as chemical reactions. In a nuclear reaction ,atomic
number and mass number are conserved.
e.g. the nuclear reaction :
14
4
 178O  11H
7 N  2He 
(a) Nuclear Fission :
ISODIAPHERS
(i) Atoms having the same difference of neutrons and
proton.
(ii) Nuclide and its decay product after -emission
are called isodiaphers.
m
(iii) e.g.,
Z
m–4
A
Z–2
(i)
238
92 U
is converted into
B
p=Z
n=m–Z
n – p = m – 2Z

The process of artificial transmutation in which heavy
nucleus is broken down into two lighter nuclei of nearly
comparable masses with release of large amount of
energy is termed as nuclear fission.
e.g.
p=Z–2
n=m–Z–2
n – p = m – 2Z
Note :
238
92 U
239
(i) Nuclides having identical atomic no. and mass
no. but differing in radioactive properties are known
as nuclear isomers.
(ii) Nuclear isomers differ in their energy state and
spins.
60m
CO,
69
Zn and
69m
Zn,
80
Br and
80m
Br
The symbol m with mass no. represents the
metastable state of parent element.
Isomeric
Transition
60
Co + -rays
(iii) Nuclear isomers, thus have different rate of decay,
decay constant, half life, average life and binding
energy.

144
90
1
 56 Ba 36 Kr 2 0 n +Energy
During fission, there is always loss of mass, known
as mass defect ,which is converted into energy
according to Einstein equation i.e.
E = mc2.
e.g.
235
92 U
NUCLEAR ISOMERS
Co
236
92 U

Unstable
(i) Molecules having same no. of atoms and same
no. of electrons are called isosters.
e.g., CO2 and N2O (There are three atoms and 22
electrons in both the molecules.)
60m
Plutonium
captures slow neutron and splits up into
ISOSTERS
CO and
239
Neptunium
235
92 U
235
1
92 U + 0 n
e.g.,
etc.
239
94 Pu
fragments.
Isotopic no. n – p = m – 2Z
60
and
–
–
10 n 
 93 Np 
 94 Pu
Uranium
(ii)
239
93 Np
Note :
In a metastable state, a system is in equilibrium (not
changing with time),but is susceptible to fall into lower
energy states with only slight interaction.
+
1
0n

235
118
.009
.
1


236.127amu
144
56 Ba
+
90
36 Kr
1
+ 20 n


143
.881
89
2.018


.947



235.846
 m = 236.127 – 235.846
= 0.281 amu
E(in MeV) = 0.281 × 931.5 = 261.75 MeV
Energy released in one fission is equal to 261.75
MeV.
(i) Chain reaction : Whatever are the primary products
of fission of uranium, it is certain that neutrons are
set free.If the conditions are so arranged that each of
these neutrons can, in turn, bring about the fission,
the number of neutrons will increase at a continuously
accelerating rate until whole of the material is
exhausted. Such type of reaction is called chain
reaction. It takes very small time and is uncontrolled.
It ends in terrible explosion due to release of
enormous amount of energy.
235
1
92 U + 0 n
92
1
  141
56 Ba  36 Kr  3 0 n + Energy
56
56
PAGE # 56
(B) Nuclear reactor :
The chain reaction is shown in the figure .
Ba
n
E
n
235
92U
Ba
n
n
E
U
235
U
92
Kr
n
n
235
Kr
Ba
n
Kr
n
235
U
235
92
U
Ba
n
n
n
• In a nuclear reactor, fission is based on the fact that
cadmium and boron can absorb neutrons thus
forming corresponding isotopes which are not
radioactive.
n
92
E
n
E
Kr
Kr
Ba
n
n
235
92U
n
E
113
1
 114
48 Cd  0n 
48 Cd 
10
1
 11
5 B  0n 
5B 
Kr
(ii) Critical mass : The minimum mass which the
fissionable material must have so that one of the
neutrons released in every fission hits another
nucleus and causes fission so that the chain reaction
continues at a constant rate is called critical mass .If
the mass is less than the critical mass , it is called
sub-critical mass . If the mass is more than critical
mass, it is called super-critical mass.
(iii) Applications of Nuclear Fission : Three practical
applications of nuclear fission are as follows (A) Atomic bomb
(B) Nuclear reactor
(C) Nuclear power plants
(A) Atomic Bomb :
• The basic principle of atomic bomb is uncontrolled
nuclear fission reaction (chain reaction).
• It requires several small samples of U-235 or Pu-239.
• An explosive like TNT (Trinitrotoluene) is placed
behind the samples which explodes to initiate the
reaction which causes the small samples to join and
form large mass.
• Neutron from Ra-Ba source (s) initiate the reaction
which starts the chain reaction finally leading to
explosion and release of large amount of energy.
• The rapid release of energy raises the temperature
enormously and generates a very high pressure front
in the atmosphere.
Atomic Bomb

• The energy thus liberated can be used for
constructive purposes like generation of steam to run
turbines and produce electricity.
• In a nuclear reactor, fission is controlled by controlling
the number of neutrons released.
Ba
n
n
n
E Kr
92
E
• An equipment in which nuclear chain reaction is
carried out in a controlled manner is called a nuclear
reactor.
Ba
n
n
235
92 U
n
Note :
The first atomic bomb dropped over Hiroshima city
during the second world war in 1945 utilized 235U and
the second atomic bomb dropped on Nagasaki
made use of 239Pu. India exploded her first atomic
bomb at Rajasthan in May 1974,and used 239Pu as
the fissionable material.

  rays
  rays
Note :
The first nuclear reactor was assembled by Fermi
and his coworkers at the University of Chicago in the
United states of America, in 1942. In India, the first
nuclear reactor was put into operation at Trombay
(Mumbai) in 1956.
(C ) Nuclear Power Plants :
When a nuclear reactor is used for the production of
electricity it is termed as a nuclear power plant.The
heat produced during a nuclear reaction is utilized in
generating steam which runs the steam turbines. The
electric generator is connected to the turbine. The
electric power is then obtained from the generator.
Thus, a nuclear power plant consists essentially of
the following four parts:
1. Reactor core
2. Heat exchanger
3. Steam turbine
4. Steam condensing system
Reactor core is the main part of nuclear reactor. It
consists of the following parts :
Fuel rod :
The fissionable material used in the reactor is called
fuel. The fuel used is enriched uranium -235 . This is
obtained from the naturally occurring U-235
(containing about 0.7% of U-235 ) by raising the
percentage of U-235 to about 2-3%.
• Control rods : Cadmium or boron rods are used to
raise or lower and control the fission process.
Because they can absorb neutrons.
• Moderator : The material used to slow down the
neutrons (without absorbing them so that they can be
easily captured by the fuel, is known as moderator.
Heavy water (D2O) or graphite is used as moderator
material in nuclear power plant.
57
57
PAGE # 57
• Coolant : To carry away the heat produced during
fission, a liquid is used. This liquid is known as coolant.
Usually heavy water is used as coolant so that it also
acts as a moderator.
• Shield : To prevent the losses of heat and to protect
the persons operating the reactor from the radiation
and heat, the entire reactor core is enclosed, in a
heavy steel or concrete dome, called the shield.
Steam
Electricity
Generator
Turbine
Reactor
D2O
Heat
exchanger
Primary
coolant
• The reactions occurring are :
Fission (in the centre)  heat +
1
0n
6
3 Li
+ 10 n 
3
1H
+ 24 He + 3.78 MeV
2
1H
+ 13 H 
4
2He
+ 10n + 17.6 MeV
2
1H
+ 12 H 
3
2 He
+ 10n + 3.2 MeV
3
1H
+ 13 H 
4
2 He
+ 2 10n + 13.14 MeV
(B) Fusion in sun : Among the celestial bodies in which
energy is produced, the sun is relatively cooler. There
are stars with temperature around 108 K inside. In
sun and other stars, where the temperature is less
than or around 107 K, fusion takes place dominantly
by proton-proton cycle as follows -
Condenser
D2O
Pump
Heavy water
Nuclear Reactor
note that the first two reactions should occur twice to
(b) Nuclear Fusion :
A nuclear reaction in which two lighter nuclei are fused
together to form a heavier nucleus is called nuclear
fusion. A fusion reaction is difficult to occur because
positively charged nuclei repel each other. At very high
temperature of the order of 106 to 107 K, the nuclei
may have sufficient energy to overcome the repulsive
forces and fuse.Therefore, fusion reactions are also
called thermonuclear reactions. Fusion reaction are
highly exothermic in nature because loss of mass
occurs when heavier nucleus is formed from the two
lighter nuclei.
4
2
2
1 H 1
H 2 He  23.85 MeV
3
3
1 H 1
H  2 He  210 n  11.3 MeV
1
3
1H 1
H  2 He  20.0 MeV
4
4
4
2
3
 2 He
1 H 1 H 
10 n  17 .6 MeV
Hydrogen bomb is based on fusion reaction. Energy
released is so enormous that it is about 1000 times
that of an atomic bomb.
It is believed that the high temperatures of stars
including the sun is due to fusion reactions.
(i) Applications of Nuclear Fusion :
produce two 32 He nuclei and initiate the third reaction.
As a result of this cycle, effectively, four hydrogen nuclei
combine to form a helium nucleus. About 26.7 MeV
energy is released in the cycle. Thus, hydrogen is the
fuel which ‘burns’ into helium to release energy. The
sun is estimated to have been radiating energy for
the last 3.5 × 109 years and will continue to do so till
all the hydrogen in it is used up. It is estimated that
the present store of hydrogen in the sun is sufficient
for the next 5 × 109 years.
In hotter stars where the temperature is  10 8K,
another cycle known as CNO (Carbon-nitrogenoxygen cycle) cycle takes places.
1
1H
13
7 N
the centre surrounded by a mixture of deuterium (12 H)
and lithium isotopes ( 36 Li) .
• The nuclear fission provides heat and neutrons.
• Neutrons convert
6
3 Li
3
1 H and the heat
between 12 H & 13 H .
to tritium
liberated is used for fusion

13
7
N   + 1.95 MeV
13 C  0e
1
6
  + 2.22 MeV
1
1H

13
6 C
1
1H
 158 O  
14
7 N
15
8 O
(A) Hydrogen bomb :
• Its principle is nuclear fusion.
• It consists of an arrangement of nuclear fission in
12

6 C 
1
1H
15
 7 N
14
7 N
 + 7.54 MeV
+ 7.35 MeV
0
+ 1 e ++ 2.75 MeV
12
4
15
 6 C  2 He + 4.96 MeV
7 N 
411H  24He  21e0   rays + 26.7 MeV
The end result of this cycle is again the fusion of four
hydrogen nuclei into a helium nucleus. Carbon
nucleus acts only as a catalyst.
58
58
PAGE # 58
DIFFERENCES BETWEEN NUCLEAR FISSION AND NUCLEAR FUSION
S.No.
Nuclear fission
Nuclear fusion
1
This process occurs in heavy nuclei.
This process occurs in lighter nuclei.
2
The heavy nucleus splits into lighter nuclei
of comparable masses.
The lighter nuclei fuse together to form a
heavy nucleus.
3
The reaction occurs at ordinary temperature. This occurs at very high temperature.
4
The energy liberated in one fission is about
200 MeV.
The energy liberated in one fusion is about
24 MeV.
5
This can be controlled.
This cannot be controlled.
6
Products of fission are usually unstable
and radioactive in nature.
Products of fusion are usually stable and
non-radioactive in nature.
7
The links of fission reactions are neutrons.
The links of fusion reactions are protons.
DIFFERENCES BETWEEN NUCLEAR REACTIONS AND CHEMICAL REACTIONS
Some of the characteristics that differentiate between nuclear reactions and ordinary chemical reactions are
summarized ahead :
Nuclear reactions
Chemical reactions
Involve conversion of one nuclide into Involve rearrangement of atoms and not
another.
change in the nucleus.
Particles within the nucleus are
involved .
Only outermost electrons participate.
Often accompanied by release
of tremendous amount of energy.
Accompanied by release or absorption of
relatively small amount of energy.
Rate of reaction is independent of
Rate of reaction is influenced by external
external factors such as temperature,
factors.
pressure and catalyst.
No breaking or making of bonds
involved.
Involves breaking or making of bonds.
Irreversible.
Can be reversible or irreversible.
(c) In Industry :
APPLICATIONS OF RADIOACTIVITY
AND RADIOISOTOPES
(a) In Medicine :
Radioisotopes are used to diagnose many diseases.
E.g. arsenic - 74 tracer is used to detect the presence
of tumour, sodium -24 tracer is used to detect the
presence of blood clots , iodine-131 tracer is used to
study the activity of the thyroid glands and cobalt-60
is used in the treatment of cancer . It should be noted
that the radioactive isotopes used in medicine have
very short half life periods.
(b) In Agriculture :
The use of radioactive phosphorus 32P in fertilizers has
revealed how phosphorus is absorbed by plants. This
study has led to an improvement in the preparation of
fertilizers. 14C is used to study the kinetics of
photosynthesis.
(i) The thickness of a material (e.g. cigarettes, metal
plates etc.) can be determined by placing a radioactive
source on one side of the material and a counting
device on the other. From the amount of radiation
reaching the counter, the thickness of the material
can be calculated.
(ii) When a single pipe line is used to transfer more
than one petroleum derivative, a small amount of
radioactive isotope is placed in last portion of one
substance to signal its end and the start of another.
(d) In Geological Dating :
The age of the earth and rocks can be predicted by
geological dating. Age of a rock sample can be
calculated by finding out the amounts of the parent
radioactive element and the isotope of lead (e.g.
92 U
238
and
82 Pb
206
) in rock sample.
(e) In Radio Carbon Dating :
The age of a fallen tree or dead animal can be
predicted by measuring the amount of C-14 in dead
plants or animals.
59
59
PAGE # 59
DATING
(i) The determination of age of minerals and rocks,
an important part of geological studies involves
determination of either a species formed during a
radioactive decay or the residual activity of an isotope
which is undergoing decay. For example
238
undergoes
92 U
a decay (t1/2 =4.5 × 109 years) series
206
82 Pb
forming a stable isotope
and He. Helium
obtained as a result of decay of 238
92 U has almost
certainly been formed from -particles. Thus, if
238 and
92 U
He contents are known in a rock we can
determine the age of rock sample (1g of 238
in
92 U
equilibrium with its decay products produces about
10–7 g He in a year). Also by assuming that initially
rock does not contain
206
82 Pb
and it is present in rocks
238 ,
92 U
due to decay of
we can calculate the age of
rocks and minerals by measuring the ratio of
238
and 206
82 Pb
92 U
.The amount of
to be obtained by decay of
238
92 U

Mole of
Mole of
206
82 Pb
+
238
8 24 He
206
82 Pb
U . Thus,
+ 6 –01e
238
U left = N at time t i.e., Nt
206
Pb formed = N’ at time t
 Initial mole of
238
is supposed
U = N + N’ (at time 0) i.e., (N0)
Thus, time t can be evaluated byN0
2.303
t=
log N

t
(ii) To determine the age of animals or objects of
vegetable origin such as wood, charcoal and textiles
by radio carbon dating technique.

Note :
Radio carbon dating technique was given by W.F.
Libby and was awarded Nobel Prize.
Carbon-14 has been used to determine the age of
organic material. The procedure is based on the
formation of 14C by neutron capture in the upper
atmosphere.
14
1
7 N + 0n

14
1
6 C + 1H
This reaction provides a small, but reasonably
constant source of 14 C . The 14 C isotope is
radioactive, undergoing  - decay with a half life of
5730 years.
14
6C

14
0
7 N + –1 e
In using radio carbon dating, we generally assume
that the ratio of 14C to 12C in the atmosphere has been
constant for at least 50,000 years. The 14 C is
incorporated into CO2, which is in turn incorporated
through photosynthesis, into more complex carbon
containing molecules within plants. When these
plants are eaten by animals, the 14C becomes
incorporated within them. Because a living plant or
animal has a constant intake of carbon compounds,
it also has to maintain 14C to 12C ratio that is identical
with that of atmosphere . However, once the organism
dies, it no longer ingests carbon compounds to
compensate 14C which is lost through radioactive
decay. The ratio of 14 C to 12 C therefore,
decreases.Thus, by knowing the equilibrium
concentration of 14C in a living matter as well as in a
dead piece of matter at a particular time, the age of
material can be determined.
HAZARDS OF RADIATIONS
(i) Radioactive radiations cause atmospheric
pollution.
(ii) When living organisms are exposed to radiations,
the complex organic molecules get ionized, break up
and disrupt the normal functioning of the organisms.
(iii) Effects of radiations :
(A) Pathological damage : i.e. permanent damage
to living body which causes death and development
of diseases e.g. cancers or leukemia etc.
(B) Genetic damage : i.e. effect on chromosomes
causing mutations.
RADIOACTIVE POLLUTION
Radioactive pollution is a special form of physical
pollution, relating to all systems air, water and soil.
This type of pollution is not only harmful for the present
generation but also for future generations. The
radioactive substances with long half-life are usually
the main sources of environmental concern.
Neutrons released during nuclear tests make other
materials radioactive in the surrounding. These
materials include 90Sr, 137Cr and 131. The radioactive
materials are converted into gases. These gases and
fine particles are thrown high up into the air and carried
away by wind to distant areas. They ultimately settle
down and cause pollution to water and soil. From
soil the radioactive substances enter in the food chain
and thus affect all forms of life including man. Cosmic
radiations and explosion of a hydrogen bomb produce
14
C in air.
Nuclear power plants and reprocessing plants
discharge 90Sr, 137Cs, 131, 140 Ba, 140La, 144Rh, etc. Coal
based thermal power stations release radioactive
gases such as 85Kr, 133Xe and particulates such as
137
, 60Co, 54Mn and 137 Cs through chimney.
Nuclear dumping within land or in ocean leads to
radiation pollution.
(a) Effects of Radioactive Pollution :
(i) Radiations induce mutations and breaks in
chromosomes particularly at the time of cell division.
(ii) Higher doses of radiations can cause cancer,
leukaemia, anaemia and sterility. Excessive use of
X-rays causes death of tissues.
(iii) Radiations induce mutations in plants also.
Morphological deformities occur.
(b) Control of Radiation Pollution :
(i) Manufacture and use of nuclear weapons should
be stopped.
(ii) Nuclear tests and further development should be
suspended.
(iii) Ocean dumping of nuclear wastes should be
suspended.
(iii) Proper handling of radio isotopes during their
use in various fields should be done.
60
60
PAGE # 60