theory of gases and solid state

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Chemistry (Theory of Gases and Solid State)
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1.THEORY OF GASES
Amongst the three common states of matter, the gaseous state is the simplest. And shows
greatest uniformity in behaviour.
Kinetic theory of gases
This theory was a generalization for about ideal gases. It was Presented by Bernoulli in 1738 and
developed in 1860 by Clausious, Maxwell, Kroning, and Boltzmann.
The kinetic theory is based on the following Postulates
1. A gas consists of a large number of minute particles, called atoms or molecules. Volume
of individual atom or molecule is considered negligible.
2. The molecules are in a state of constant rapid motion in all possible directions, colliding
with each other and with the wall of container.
3. Collision among gas molecules is perfectly elastic. i.e, there is no loss in kinetic energy
and moment during such collisions.
4. There are no attractive forces between molecules or between molecules and the walls of
the vessel in which the gas is contained. The molecules move completely independent of
one another.
5. Pressure excerted by gas is due to collisions of gas molecules with wall of the container.
6. Kinetic energy of gas molecule depends only on absolute temperature.
Kinetic energy  absolute temperature.
Derivation of kinetic gas equation :
From the basis of above postulate, it is possible to derive the mathematical expression, commonly
known as kinetic gas equation. i.e,
21
PV mnc
3

Where P = Pressure of the gas,
V = Volume of the gas,
m = mass of a molecule
n = number of molecules present in the given amount of gas
and c = root mean square speed.
Let us consider N molecules of gas, each having a mass m, enclosed in a cubical vessel of side
‘’ cm.
Consider one molecule of the gas having a velocity c. This velocity can be resolved into three
Components u, v, and w along three axis, x, y, and z.
then
c2
= u2
+ v2
+ w2
...(1)
Consider the movement of a molecule along the x - axis, in between opposite faces A and B,
striking the wall which is Perpendicular to its motion. Since the collision is elastic and the wall remains
stationary, on rebounding, only the sign of the velocity component changes.
Theory of Gases and Solid StateUNIT-3

62
The momentum of the molecule before collision = +mu
The momentum of the molecule after collision = -mu
The change in momentum in the x direction (Px) is
Px = mu – (–mu) = 2mu ...(2)
After Collision, the molecule takes time equal to l/u to collide with the opposite wall ( and time
equal to 2l/u to strike against the same wall again)
Hence
the frequency of collisions on the two opposite walls is given by u/l and the change in momentum
per unit time is given by -
Px/t = 2mu × u/ = 2mu2
/ ...(3)
The change of moment per second along x, y and z direction are 2mux2
/, 2muy2
/, and 2muz2
/
, respectively.
Hence,
total change of momentum per second on all the faces will be :
2 2 2
2mux 2muy 2muz
  
  
 2 2 22m
ux uy uz  

22m
c

...(4)
Change of momentum per second = force
Pressure =
Force
Area
Pressure created by one molecule
2
3
2m c
6


Pressure created by n molecules
2
3
2mnc
6


2
1 mnc
P
3 V
 ...(5)
or
21
PV mnc
3
 ...(6)
This equation is called kinetic gas equation.
Gas laws :
All the gas laws can be easily derived from the kinetic gas equation
1. Boyle’s Law
At constant temperature, the volume of gas is inversely proportional to the gas pressure.
1
V
P
 (T = Constant)
or
K
V
P

or PV = K
for two different gases :
P1
V1
= P2
V2
= K

63
Boyle’s law can be verified by any one of the following ways graphically.
P
V
(T Constant)
mass Constant
P
PV
Temperature 
Constant
mass Constant
V
Temperature 
Constant
mass Constant
Graph between P and V at constant temperature and constant mass is called isotherms.
According to Boyle’s law,
PV = Constant
So
log P + log V = Constant
log P = – log V + Constant
log P
log V
2. Charle’s Law
At constant pressure, the volume of a given amount of a gas is directly proportional to the
absolute temperature
V
V T (If P = Constant)
V = KT
or
V
K
T

or
1 2
1 2
V V
K
T T
  (for two or more gases)
3. Gay-Lussac’s Law
At constant volume the pressure of a given amount of a gas in directly proportional to the absolute
temperature
P T (If V = Constant)
or P = KT

64
or
P
K
T

or
1 2
1 2
P P
K
T T
4. Avogadro’s Law
According to this law “equal volumes of all gases under the same conditions of temperature and
Pressure Contain the same number of molecules”
thus V n (at Constant T and P)
or V = Kn
or
1V
K
n

or
1 2
1 2
V V
K
n n
for example :
2 H2
(g) + O2
(g)  2H2
O(g)
2 moles 1 moles 2 moles
2 litres 1 litre 2 litres
1 litre
1
litre
2
1 litre
1n litre
n
2
litre 1n litre
The volume of one mole of a gas is called molar volume (Vm
), which is 22.4 L mol-1
at S.T.P or
N.T.P
5. The Ideal Gas Equation
According to Boyle’s law
1
P
V

or
1
V
p
 ...(1)
According to Charle’s law
V  T ...(2)
On Combining T
V
P

or
PV
T
= Constant ...(3)
PV
R
T
 (R = ideal gas Constant)
PV = RT (for 1 mole gas)
PV nRT (for n mole gas) ...(4)

65
This equation is called ideal gas equation
where,
mass of the gas in gram
n
molarmass of the gas in gram

or n =
w
M
...(5)
hence, eq (4) becomes
PV =
w
M
RT ...(6)
or
w RT
P
M V
 
or p =
d
RT
M
w
d
v
 ...(7)
These equations (6) and (7) are modified forms of gas equation
The ideal gas Constant -
PV = RT
R =
P V
T

=
Force
volume
area
Temperature

=
 
 
3
2
Force
length
length
Temperature

=
Force length
Temperature

=
Work
Temperature
= Work per degree kelvin per mol
= Work mol–1
K–1
Numerical Values of R :
R = 0.0821 L atom mol–1
K–1
= 8.3143 Jmol–1
k–1
(SI Unit)
= 8.3143 Nm mol–1
k–1
= 8.3143 Kpa dm3
mol–1
k–1
= 8.3143 m pa cm3
mol–1
k–1
= 5.189 × 1019
ev mol–1
k–1
= 1.99 cal mol–1
k–1

66
6. Graham’s law of Diffusion
Ability of a gas to mix spontaneously and to form a homogenous mixture is called diffusion
According to this law at constant temperature and pressure, the rate of diffusion of a gas is inversely
proportional to the square root of the density of the gas.
rate of diffusion 
1
d
(at constant T and P)
1
r
d

or
1 2
2 1
r d
r d
 (for two gases)
or
1 2 2 2
2 1 1 1
r d 2 d M
r d 2 d M

  

M1
and M2
= molecular weight
Since rate of diffusion (r) =
volume of gas
Timetaken
then
1 1 1 2
2 2 2 1
r V / t d
r V / t d
 
1 1 2 2
2 2 1 1
V / t d M
V / t d M
 
1 2 2
1 2 1
V t M
t V M
 
for gases
1 1
2 2
V n
V n

1 2
2 1
n t
n t
 =
2
1
M
M
Case-I : If t1
= t2
and n =
w
M
1 2 2
1 2 1
w M M
M w M
 
or
2 1 2
1 2 1
M w M
M w M
 
or
1 2
2 1
w M
w M


67
Case-II : If V1
= V2
1 2 2
2 1 1
r t d
r t d
 
Case-III : When volume of the two gases diffused in the same time i.e. t1
= t2
1 1 2
2 2 1
r V d
r V d
 
Since r  P
1 1 2
2 2 1
r P M
r P M

P
r
M
 
 
 

7. Dalton’s law of Partial Pressure
According to this law when two or more gases, which do not react chemically are kept in a close
versel, the total pressure excerted by the mixture is equal to the sum of the partial pressure of individual
gases.
Thus,
PTotal
= P1
+ P2
+ P3
+ .............
This is dalton’s law of partial pressure,
This law is applicable only when the component gases in the mixture do not react with each other
for example
N2
and O2
, CO + CO2
, N2
+ Cl2
, CO + N2

68
2. EQUATION OF STATE OF IDEAL AND NON-IDEAL GASES
A number of equation of states have been suggested to describe the P-V-T. Relationship of real
gases. The earliest and the best known equation is that of Vander Walls.
The Vander Wall’s Equation of State
Gases which obey all the postulates of kinetic theory or which obey gas laws for the all values of
temperature and pressure or which follow the ideal gas equation (PV = nRT) is called ideal or perfect
gases.
Almost all gases deviate from the ideal behaviour i.e, no gas is perfect and Concept of ideal gas
is only theoretical.
Real gas obey this equation only approximately and that too under conditions of low pressure and
high temperature. The higher the pressure and the lower the temperature, the greater are the deviations
from the ideal behaviour.
The deviations from ideal behaviour are best represented in terms of the compressibility factor
(also called the Compression factor), Z, which is defined as
PV = ZnRT
or Z =
PV
nRT
or Z =
PVm
wRT
200 400 600 800 1000
CO
CH4
NH3
Ideal gas
NH3
CO CH4
H2
He
1.0
2.0
O
Z
P
It is necessary, therefore, to apply suitable corrections to the ideal gas equation so as to make
it applicable to real gases.
Where - Z = Compressibility factor
Greater is the departure of Z from unity, is the deviation from ideal behaviour, Thus when
(1) Z = 1 ; The gas is ideal at all temperatures and pressures.
(2) Z > 1 ; Shows positive deviation for example: O2
, N2
, CH4
, CO2
etc.
(3) Z < 1 ; Shows negative deviation for example : H2
, He
Some gases like CO2
show both negative and positive deviation

69
Vander Wall’s modified the ideal gas equation by introducing two corrections :
(i) Volume Correction
The ideal gas equation PV = nRT is derived on the assumption that the gas molecules are rigid
spherical particles which possess a definite volume ie, they do not have finite volume
Vander waal’s abandoned this assumption and suggested that a correction term nb should be
subtracted from the total volume V in order to get the ideal volume which is compressible.
Thus the compressible Volume n mole of gas would be (V – nb)
Corrected volume Vi
= V – nb ...(i)
(ii) Pressure Correction
In the derivation of the ideal gas equation it is assumed that there are no inter molecular forces
of attraction. Actually it is not so. It is, therefore, necessary to add a correction factor P, to the pressure
of the gas in order to get the ideal pressure.
Therefore,
The corrected Pressure ‘P’ideal
= Pobs
+ P ...(ii)
Where P is the Pressure Correction
Pressure correction depends upon two factors
(i) The force of attraction exerted on a single molecule which is about to strike the wall
evidently depends upon the number of molecules per unit volume in the bulk of the gas
(ii) The number of molecules striking the wall at any given instant also depends directly upon
the density of the gas.
Both these factors are proportional to the density of the gas,
Therefore, the attractive force is directly proportional to the square of the density of the gas
2
P d
P 2
1
V

or 2
a
P
V

where ‘a’ is a Constant depending upon the nature of the gas and V is Volume of 1 mole of the gas
Thus Corrected pressure,
Pideal
= Pobs
+ 2
a
V
...(iii)
for n mole of the gas Pideal
= Pobs
+
2
2
n a
V
...(iv)
making both the correction,
The ideal gas equation PV = nRT
may be written as -
 
2
2
n a
P V nb nRT
V
 
   
 
...(v)
depends where a and b are Vander Wall’s Constant, whose value depends on the nature of the
gas Normally for a >> b
Unit of a = Nm4
mol-1
Unit of b = m3
mol-1

70
Case - I : When pressure is low  when the pressure is not very high, the volume V - will be
sufficiently large and b may be ignored.
2
a
P V RT
V
 
  
 
or 2
aV
PV RT
V
 
  
 
or
a a
PV RT or PV RT
V V
    ...(vi)
Case - II : When Pressure is high :
When pressure is high than the volume (V) will be quite small. Now it may be not possible to
ignore b. But, as P is quite high, the Pressure correction factor a/V2
may become negligible in Comparison
to P
therefore -
PV = RT + Pb
or
PV Pb
1
RT RT
 
or
Pb
Z 1
RT
  ...(vii)
Case - III : When temperature is high :
At high temperature the volume will become sufficiently large to make the volume of a/V2
negligible
and b may also be negligible in comparision to V which is not sufficiently large.
therefore -
PV RT ...(viii)
Case - IV for hydrogen -
molecular mass of hydrogen is small hence value of ‘a’ will be small. Thus the terms
a
V
and 2
ab
V
may be ignored.
then the Vander Wall’s equation becomes.
PV = RT + Pb
or
PV Pb
1
RT RT
 
or
Pb
Z 1
RT
  ...(ix)
The boyle temperature :
The temperature at which a real gas obeys Boyle’s law is known as the boyle’s temperature (TB
).
It is given by the expression.
B
a
T
bR
 ...(x)

71
The Critical state
(1) Critical temperature (TC
) : The temperature above which the gas cannot be liquified
however, large pressure is applied. It is given by
C
8a
T
27Rb

(2) Critical pressure (PC
) : The minimum pressure which must be applied to a gas to liquefy
it at its critical temperature. It is given by
C 2
a
P
27b

(3) Critical volume (VC
) : The volume occupied by one mole of the substances at its critical
temperature and critical pressure It is given by
CV 3b
(4) Critical Compressibility factor (ZC
) :
C C
C
C
P V 3
Z 0.375
RT 8
  

72
3. MAXWELL - BOLTZMANN DISTRIBUTION LAW
As a result of random collisions of gaseous molecules velocity as well as direction of the movement
also changes. Hence, the molecules, move with different velocities and their kinetic energies are also
different.
In the 19th Century, Boltzmann had shown that the probability for a molecule to have an energy
E was proportional to e–E/kT
N  e–E
/kT
or
e
o
N
N = e–E/kT
...(1)
Hence,
No
is the total number of molecules and Ne
is the number of molecules having energy, more than
a reference point E,
k is Boltzmann’s Constant, a fundamental constant with the value k = 1.381 × 10-23
JK-1
The fraction of molecules that have speed in the range V to V + dv is proportional to the width
of the range, and is written f(v) dv, where f (v) is called the distribution of speed.
The precise form of f for molecules of a gas at a temperature T was derived by J.C. maxwell, and is
2
3 /2
2 mv /2RTm
f(v) 4 V e
2 RT
 
   
 
...(2)
This expression is called Maxwell distribution of speed.
The Maxwell’s distribution of molecular velocities is plotted as -
T1
T2
T3
Velocity
Most Probable Velocity
T > T > T3 2 1
FractionofMolecules
Fig. : The Maxwell’s Distribution of Molecular Velocities.
We see that the fraction of molecules having velocities greater than zero increases with an
increase in velocity, reaches a maximum and then falls off towards zero again at higher velocities.
The important features of the curves are :
(1) The fraction of molecules with too low or too high velocities is very small
(2) There are certain velocities for which the fraction of molecules is maximum. This is called
the most probable velocity.
Effect of temperature on distribution of molecular velocities :
The average speed, vmean
, of the molecules depends on the temperature and their molar mass.
as
vmean

1/ 2
T
M
 
 
 

73
That is, the average speed increases as the square root of the temperature and decrease as the
square root of the molar mass. Thus, the average speed is high for light molecules at high temperature
The distribution itself gives more information than the average value. For instance, the tail towards high
speed is longer at high temperature than at low, which indicates that at high temperatures more molecules
in a sample have speeds much higher than average
Relativenumberofmolecules
Low temperature
or
high molecular mass.
Intermediate temperature
or
molecular mass
High temperature
or
low molecular mass
Speed, V
Fig. : The distribution of molecular speeds with temperature and molar mass
Kinds of molecular speeds :
(i) The Rms speed
(ii) The average speed
(iii) The most probable speed
(i) The Root Mean Square Speed (Rms) : It is the square root of the mean of the squares
of the velocity of a large number of molecules of the same gas
vrms
=
22 2 2
1 2 3 nv v v ...v
n
  
vrms
=
3PV 3RT 3P
M M d
 
(ii) The Average Speed : It is the average of the various velocities possessed by the molecules
vav
=
1 2 3 nv v v .........v
n
  
vav
=
8RT 8KT
M M

 
It is related to Rms speed
Average speed = 0.9213 × Rms speed
Rms speed = 1.085 × average speed
(iii) Most Probable Speed : This is defined as the speed possessed by maximum number of
molecules of a gas at a given temperature
It is equal to =
2RT
M

74
Vmp
=
2RT 2PV 2P
M M d
 
Vmp
= 0.816 Rms
Vrms
= 1.224 Vmp
vmp
: vav
: vrms
2RT
M
:
8RT
M
:
3RT
M
2 :
8

: 3
1 : 1.128 : 1.224

75
4. EQUIPARTITION OF ENERGY
A molecule has a certain number of motional degree of freedom such as the ability to translate
(the motion of its centre of mass through space), rotate around its centre of mass, or vibrate ( as its
bond length and angles change, leaving its centre of mass unmoved) .
According to ‘equipartition theorem’, the average energy associated with each degree of freedom
of a molecule is 1/2 kT.
From the kinetic theory of gases we know that the average translation kinetic energy of a molecule
of an ideal gas is given by
E =
3
2
kT ...(i)
According to law of equipartition of energy ( or equipartition theorem), the total energy of a molecule is
divided equally amongst the various degree of freedom of the molecules
The distribution of kinetic energy along x, y and z direction is given by -
E = Ex
+ Ey
+ Ez
...(ii)
Since the motion of gas molecules is random and the motion along the three cartesian axis is equally
Probable, hence
Ex
= Ey
= Ez
=
1
3
rd of E =
1
2
kT ...(iii)
If a gaseous species has n1
translation degree of freedom, n2
rotation degree of freedom, and n3
vibration
degree of freedom, then the total energy of the species is given by
E = n1
(kT/2) + n2
(kT/2) + n3
(kT/2)...(iv)

76
5. SOLID STATE
In solid state molecules, atoms or ions are closely packed due to intramolecular attraction forces
and cannot move at random.
The characteristic properties of solid state
The following are the characteristic properties of the solid state
1. They have definite shape, size and volume.
2. They are rigid, incompressible and have high density.
3. Intermolecular forces are very strong.
4. Intermolecular distances are short.
5. Translation motion is absent.
6. Vibrational motion about their mean position is present.
Classification of Solids
Solids can be divided into two classes on the basis of nature of order present in the arrangement
of their constituent particles.
1. Crystalline solids
2. Amorphous solids
1. Crystalline Solids
In crystalline solids the arrangement of constituent particles is well ordered and regular. They
consist of a large number of small crystals.
Characteristics
1. They have definite geometry.
2. Definite heat of fusion.
3. Melt a sharp and characteristic temperature.
4. Crystalline solids are an isotropic. Some of their physical properties such as electrical
resistance, reflective index are different in different directions. Anisotropy in crystals is due
to different arrangement of particles along different direction.
AB
Fig : Anisotropic behaviour of crystals.
5. They are considered as “true solids”.
Examples - NaCl, quartz, diamond, sugar(sucrose) etc.
2. Amorphous Solids
The term ‘amorphous’ has been derived from a greek word ‘Omorphe’ meaning sharpless.
In amorphous solid, the arrangement of constituent particles is irregular and has only short range
order.

77
Characteristics
1. They does not have definite geometry due to irregular arrangement of particles.
2. Amorphous solids do not have sharp melting points and do not change abruptly into liquids.
3. Amorphous solids are isotropic in nature, their physical properties are same in all directions.
Isotropy in crystals is due to irregular arrangement along all directions.
4. These are considered as “pseudo solids” or “super cooled liquids” because they have a
tendency to flow, though very slowly.
Examples - Rubber, glass, plastic, starch, protein etc.
Classification of crystalline solids :
All crystalline solids are classified into following categories depending on the nature of intermolecular
forces.
TYPES OF SOLID FORCES PROPERTIES EXAMPLES
             
1. Molecular solids
(i) Non polar Dispersion or London Soft, very low M.P, Ar, CCl4
, H2
, I2
, CO2
forces insulator
(ii) Polar Dipole-dipole interaction Soft, low M.P, HCl, SO2
insulator
(iii) Hydrogen Hydrogen bonding Hard, low M.P H2
O (ice)
bonded insulator
2. Ionic solids Coulombic or Hard, High M.P NaCl, CsCl, Zns,
electrostatic Insulator CaF2
,
3. Metallic solids Metallic bonding Hard, fairly high Fe, Cu, Ag, Mg
M.P, conductors in
solid state as well
as in molten state
4. Covalent solids Covalent bonding Hard, very high M.P SiO2
, SiC,
Insulator but C(graphite)
diamond
conductor(exception)
Crystal Lattice and Unit Cell
A regular three dimensional arrangement of points in space is called a crystal lattice. It is also
called space lattice. It can also be defined as an array of identical points showing how molecules, atoms
or ions are arranged in different sites, in three dimensional space.
Unit Cells : A space lattice can be subdivided into a number of small cells known as unit cells.
It can be defined as the smallest portion of a crystal lattice which, when repeated in different directions,
generates the entire lattice.
Unit Cell
Fig : A portion of a Three Dimensional Cubic Lattice and Its Units Cell

78
Types of Unit Cells
Unit cells can be divided into four categories
1. Simple Cubic/Primitive Unit Cells
Lattice points are present only at each corner of the cube.
2. Face Centered Cubic Unit Cell (FCC)
Lattice points are present at the corners as well as at the centre of each face of the cube
3. Body Centered Cubic Cell (BCC) :
Lattice points present at the corners as well as at the centre of the cube.
4. End Centered Cubic Unit Cell (ECC)
Lattice points present at the corners as well as centre of alternate face.
Any Cube Contain
1. 8 corners –
1
8
per corner contribution
2. 6 faces –
1
2
per face contribution
3. 12 edges –
1
4
per edge contribution
4. 1 body centre – 1 per body centre contribution
5. 4 body diagonal – 1 per body diagonal contribution
6. Each face contain 2 face diagonal –
1
2
per face diagonal contribution

79
In a simple cubic unit cell, each corner
atom is shared between 8 unit cells.
An atom at face centre of unit cell
is shared between 2 unit cells.
Calculation of Number of Atoms Per Unit Cell
1. Primitive or simple cubic unit cell :
8 corners ×
1
8
per corner contribution = 1 atom/unit cell
2. Face centered cubic unit cell (FCC) :
(i) 8 corners ×
1
8
+
(ii) 6 faces ×
1
2
per face contribution = 3 atom/unit cell
Total number of atoms per unit cell = 4 atom / unit cell
3. Body centered cubic cell (BCC) :
(i) 8 corners ×
1
8
+
(ii) 1 body centre × 1 per body centre contribution = 1 atom/unit cell
Classification of Crystal System
Crystal system can be classified into seven categories depending upon edge length and axial
angles.
=90°
=90°
=90°
c
b
a
(Cubic)
a = b = c
= = = 90°  
(Tetragonal)
a = b c
= = = 90°  
=90°
=90°
=90°
c
b
a
(Orthorhombic)
a b = c
= = = 90°  
=90°
=90°
=90°
c
b
a

80
(Hexagonal)
a = b c
= = 90° 
= 120°
(Rhombohedral)
a = b = c
= = 90°  
(Monoclinic)
a b c 
= = 90° 
90°
=120°
=90 °=90°
a
b
c
90°

=90 °
=90°
c
b
a
(Triclinic)
a b c 
90°    
c
b
a



Seven primitive unit cells and their possible variations as centered unit cells has been listed in
following table.
Crystal System Edge length Axial Angles Unit cells Examples
Cubic a = b = c = = = 90° P, BCC, FCC NaCl, Zinc blende
Tetragonal a = b  c = = = 90° P, BCC TiO2
, CaSO4
, SnO2
white tin
Orthorhombic a  b  c = = = 90° P, BCC, Rhombic sulphur,
FCC, ECC KNO3
, BaSO4
Rhombohedral or a = b = c = =  90° P Calcite (CaCO3
),
Trigonal HgS (cinnabar)
Hexagonal a = b  c = = 90° P Graphite, ZnO, CdS
 120°
Monoclinic a  b  c  =  = 90° P, ECC Monoclinic sulphur,
 90° Na2
SO4
.10 H2
O
Triclinic a  b  c       90° P K2
Cr2
O7
,CuSO4
·5H2
O,
H3
BO3
All crystal systems do not have simple crystal lattice some are more complex. Bravais pointed
out that there can be 14-different ways in which similar points can be arranged in a three dimensional
space. Thus the total number of Bravais lattice are 14.

81
RELATIONSHIP BETWEEN EDGE LENGTH AND RADIUS
1. For a Primitive unit cell :
atoms at corners are in contact with adjacent atoms.
i.e.
a = 2r
2. For FCC unit cell :
Atoms on face diagonal are in contact
Length of face diagonal = r + 2r + r = 4r
In triangle ABC,
i.e. 4r = 2 a
(i) a = 4r / 2 a
(ii) d = 2r. (Here d = distance between two nearest atoms)
3. For BCC unit cell :
Atoms on body diagonal are in contact with each other
i.e.
Length of body diagonal = r + 2r + r = 4r
Also, body diagonal = 4r = 3 a
Illustration - 1 Al crystallizes in a CCP structure. Its metallic radius is 125 pm.
(i) What is the length of side of the unit cell?
(ii) How many unit cells are there in 1cm3
of Al?
Sol. (i) In CCP a = 2 2 r
= 2 × 4.14 × 125 pm = 354 pm
(ii) Volume of 1 unit cell = (3.54 × 10–10
)3
= 4.44 × 10–23
cm3
unit cell in 1cm3
= 23
1
4.44x10 = 2.25 × 1022
Illustration - 2 Lithium iodide crystal has a face-centred cubic unit cell. If the edge length of the unit cell
is 620 pm, determine the ionic radius of I-ion.
Sol. As LiI has fcc arrangement, I-ions will occupy the corners and face-centres. These ions will touch
each other along the face diagonal.

82
A
I
–
I
–
I
–
I
–
I
–
a
a
B
 Face diagonal AB = 4r1
= a2
+ a2
= 2a
 r1
= 2a / 4 = 1.414 × 620 pm / 4 = 219.17 pm
Illustration - 3 CsCl has bcc structure with at the centre and ion at each corner. If rCs+
= 1.69Å and
rCl–
= 1.81Å, what is the edge length "a" of the cube ?
Sol. Assuming the closest approach between and ions, the internuclear separation is one-half of the
cubic diagonal i.e.
1.69 + 1.81 = 3.50 = a3 / 2
a = 2 × 3.5 / 3 = 4.04Å
INTERSTITIAL VOIDS
These are of four types
(1) Trigonal void : This site is formed when three spheres lie at the vertices of an equilateral
triangle. Size of the trigonal site is given by the following relation,
r = 0.155 R
r = Radius of the spherical trigonal void
R = Radius of closely packed spheres
Trigonal
(2) Tetrahedral voids : In close packing arrangement, each sphere in the second layer rests on
the hollow (triangular void) in three touching spheres in the first layer. The centres of theses four spheres
are at the corners of a regular tetrahedral. The vacant space between these four touching spheres is
called tetrahedral void. In a close packing, the number of tetrahedral void is double the number of
spheres, so there are two tetrahedral voids for each sphere
Tetrahedral void
Radius of the tetrahedral void relative to the radius of the sphere is 0.225
i.e. rvoid
/ rsphere
= 0.225
In a multi layered close packed structure , there is a tetrahedral hole above and below each atom
he
(3) Octahedral Voids : The interstitial void formed by combination of two triangular voids of the
first and second layer is called octahedral void, because this is enclosed between six spheres, centres
of which occupy corners of a regular octahedron there is twice as many tetrahedral holes as there are
in close packed atoms

83
In close packing, the number of octahedral voids is equal to the number of spheres. Thus, there
is only one octahedral void associated with each sphere. Radius of the octahedral void in relation to the
radius of the sphere is 0.414 i.e.
(4) cubic void : The cubic void is formed when atoms /ions are placed i9n such a way they
forms a cube the space in the center of the cube is cubic void . Radius of the octahedral void in relation
to the radius of the sphere is 0.732
Cubic Void
The decreasing order of sizes of voids cubic void Octahedral Voids Tetrahedral voids Trigonal
void
CALCULATION INVOLVING UNIT CELL DIMENSIONS
From the unit cell dimensions, it is possible to calculate the volume of the unit cell. Knowing the
density of the metal. We can calculate the mass of the atoms in the unit cell. The determination of the
mass of a single atom gives an accurate determination of Avogadro constant.
Suppose edge of unit cell of a cubic crystal determined by X - Ray diffraction is a, d is density
of the solid substance and M is the molar mass, then in case of cubic crystal
Volume of a unit cell = a3
Mass of the unit cell = no. of atoms in the unit cell × mass of each atom = Z × m
Here Z = no. of atoms present in one unit cell
m = mass of a single atom
Mass of an atom present in the unit cell = m/NA
Density d = mass of unit cell / volume of unit cell = Z.M / a3
d = Z.M. / a3
× NA
Note :
Density of the unit cell is same as the density of the substance
Illustration 1 : An element having atomic mass 60 has face centred cubic unit cell. The edge length of
the unit cell is 400 pm. Find out the density of the element?
Sol. Unit cell edge length = 400 pm
= 400 × 10–10
cm
Volume of unit cell = (400 × 10–10
)3
= 64 × 10–24
cm3
Mass of the unit cell = No. of atoms in the unit cell × mass of each atom
No. of atoms in fcc unit cell = 8 × 1/8 + 6 × 1/2 = 4
 Mass of unit cell = 4 × 60 / 6.023 × 1023
Density of unit cell = mass of unit cell / volume of unit cell
= 4 × 60 / 6.023 × 1023
× 64 × 10–24
= 6.2 g/cm3

84
Illustration 2 : An element has a body centred cubic (bcc) structure with a cell edge of 288 pm. The
density of the element is 7.2 g/cm3
. How many atoms are present in 208 g of the element?
Sol. Volume of unit cell = (288×10–10
)3
cm3
= 2.39 × 10–23
cm3
Volume of 208 g of the element = mass / volume = 208 / 7.2 = 28.88cm3
No of unit cells in this volume = 28.88 / 2.39 × 10–23
= 12.08 × 1023
Since each bcc unit cell contains 2 atoms
no of atom in 208 g = 2 × 12.08 × 1023
= 24.16 × 1023
atom
Illustration 3 : A compound formed by elements X & Y, Crystallizes in the cubic structure, where X is
at the corners of the cube and Y is at the six face centers. What is the formula of the compound? If side
length is 5A°, estimate the density of the solid assuming atomic weight of X and Y as 60 and 90
respectively.
Sol. From eight corner atoms one atoms (X) contributes to one unit cell.
From six face centres, three atoms (Y) contributes to one unit cell.
So, the formula of the compound is XY3
.
As we know that,
p = n × Mm / NA × a3
= , here n = 1
Molar mass of XY3
Mm = 60 + 3 × 90 = 330 gm
p = 1 × 330 / 6.023 × 1023
× (5 × 10–8
)3
gm/cm3
a = 5Å = 5 × 10–8
cm
= 330 / 6.023 × 1023
× 125 × 10–24
gm/cm3
= 4.38 gm / cm3
Illustration 4 : Lithium borohydrides, LiBH4
, crystallizes in an orthorhombic system with 4 molecules per
unit cell. The unit cell dimensions are: a = 6.81 Å, b = 4.43 Å and c = 7.17 Å. Calculate the density of
the crystal.
Take atomic mass of Li = 7, B = 11 and H = 1 a.m.u.
Sol. Molar mass of LiBH4
= 7 + 11 + 4 = 22 g mol–1
Mass of the unit cell = 4 × 22 gmol–1
/ 6.02 × 1023
mol–1
= 14.62 × 10–23
g
Volume of the unit cell = a × b × c
= (6.81 × 10–8
cm) (4.43 × 10–8
cm) (7.17 × 10–8
cm)
= 21.63 × 10–23
cm3
Density of the unit cell = Mass / Volume
= 14.62 × 10–23
g / 21.63 × 10–23
cm3
= 0.676 gcm–3
Illustration 5 : An element crystallizes into a structure which may be described by a cubic type of unit
cell having one atom in each corner of the cube and two atoms on one of its face diagonals. If the volume
of this unit cell is 24 x 10–24
cm3
and density of the element is 7.20gm/cm3
, calculate no. of atoms present
in 200gm of the element.
Sol. Number of atoms contributed in one unit cell
= one atom from the eight corners + one atom from the two face diagonals
= 1 + 1 = 2 atoms

85
6. CLOSE PACKING IN CRYSTALS
In solids, the constituent particles are closely packed, leaving the minimum vacant space.
a. Close packing in one dimension
There is only one way of arranging spheres in a one dimensional close packed structure, that is
to arrange them in a row and touching each other.
Close packing of spheres in one dimension
Total number of atoms per unit cell = 2 atoms/unit cell
The number of nearest neighbour of a particle is called its coordination number.
Thus in one dimensional close packed arrangement the coordination number is 2.
b. Close Packing in Two Dimensions
There are two common ways in which sphere of equal size can be packed.
Arrangement -(i)
(Hexagonal close packing)
Arrangement -(ii)
(Square close packing)
Fig. : Two common ways of packing spheres of equal size.
In arrangement - (i), the spheres are packed in such a manner that the second row is placed
above the first one in a manner such that its spheres fit in the depression of the first row. In this type
of arrangement each sphere is surrounded by six other similar spheres and two dimensional coordination
number is 6.
The volume occupied by the spheres in arrangement-I is 52.4 %.
In this layer there are some empty space (voids). These are triangular in shape.
Triangular
void
In arrangement - (ii), the spheres are packed in such a way that the second row is placed in
contact with the first one such that the spheres of the second row are exactly above those of the first row.
The spheres of the two rows are aligned horizontally as well as vertically.
In this type of arrangement each sphere is surrounded by four other similar spheres and two
dimensional coordination number is 4. The volume occupied by the spheres in arrangement-(ii) is 60.4%.
c. Close packing in Three Dimensions
All the real structures are three dimensional structures. They can be obtained by stacking two
dimensional layers one above the other. Three dimensional close packing can be classified into two categories.
(A) Three Dimensional Close Packing from Two Dimensional Square Close Packing
Case - I :
 In this type of close packing the second layer is placed over the first layer such that the
spheres of the upper layer are exactly above those of the first layer.

86
 In this arrangement, spheres of both the layers are perfectly aligned horizontally as well
as vertically similarly, we may place more layers one above the other.
 If the arrangement of spheres in the first layer is called ‘A‘ type, all the layers have the
same arrangement is also called AAA...type pattern.
 The lattice thus generated is the simple cubic lattice and its unit cell is the primitive cubic
unit cell.
A
A
A
Simple Cubic lattice
or
‘ AAA....type pattern
Coordination number = 6
Packing fraction = 52.4
Case - II :
In this type of close packing the second layer is placed over the first layer such that the spheres
of second layer are placed in the depressions of the first layer and the third layer is placed over
the second layer such that the spheres of the third layer are exactly aligned with those of the first
layer, this pattern is also known as ABAB...type or body centered lattice.
B
A
Body centred cubic lattice
or
ABAB.... type pattern
Coordination number = 8
Packing fraction = 68%
(B) Three dimensional close packing from two dimensional hexagonal close packed layers
In this type of close packing, the second layer is placed over the first layer such that the spheres
fit in the depression of the first row. It can be observed that not all the triangular voids of the first layer are
covered by the spheres of the second layer. This gives the different arrangements wherever a sphere of
the second layer is above the void of the first layer or vice versa a tetrahedral is formed. These voids are
called tetrahedral voids because a tetrahedral is formed when the centres of these four spheres are joined.
Octahedral
void
Tetrahedral
void

87
Tetrahedral
void
Octahedral
void
At the other places, the triangular voids in the second layer are above the triangular voids in the
first layer, and the triangular shapes of these sphere do not overlap. Such voids are called octahedral
voids. These voids are surrounded by six spheres.
The number of these two types of voids depend upon the number of close packed spheres.
Let the number of close packed spheres be N, then :
The number of octahedral voids = N
The number of tetrahedral voids = 2N
To build up the third layer of spheres, there are two alternative ways
In the first way the spheres are of third layer are placed on the hollows of second layer. It is
observed that each sphere of third layer lies exactly above the spheres of first layer. When this is
continued indefinitely, the system obtained arrangement is found to possess hexagonal symmetry and
is called hexagonal close packing of spheres and is abbreviated as hcp-3D or ABAB.....
Alayer
B-layer
A-layer
A
B
A
In the second way, spheres are placed on the unoccupied hollows of the first layer. It is observed
spheres of the third layer do not come over those of first layer. This arrangement of close packing is
referred to as cubic close packing (CCP) or ABCABC.......
A
B
A
Alayer
B-layer
A-layer
C
 It is noted that in both the above systems hcp or ccp, each sphere is surrounded by
twelve others so the coordination number of hcp or ccp is 12 and packing fraction is 74%.
 Most of the metals belong to S-block and d-block elements possess any of the close
packing arrangements.
Structure Metals
hcp Mg, Zn, Mo, V, Cd
ccp or fcc Cu, Ag, Au, Ni, Pt
bcc Li, Na, K, Rb, Cs, Ba

88
7. X-RAYS
Atoms, molecules or ions are two small to be seen with naked eye. The arrangement of particles
in crystalline solid is determined indirectly by X-ray diffraction
The German physicist M.Von Laue (1879-1960), in 1913, suggested the possibility of diffraction
of X-ray by crystals
This observation has proved to be highly useful in determining structures and dimensions of
crystals as well as in the study of a number of properties of X-ray themselves.
The Bragg Equation
Bragg Pointed out that unlike reflection of ordinary light, the reflection of X-ray can take place only
at certain angles which are determined by the wave length of the X-rays and the distance between the
planes in the crystal. The fundamental equation which gives a simple relation between the wavelength
of the X-rays, the interplanar distance in the crystal and the angle of reflection, is known as the Bragg
equation.
Derivation of The Bragg Equation
Suppose a beam of X-rays falls on the crystal at glancing angle , as shown. Some of the rays
are reflected from the upper plane, some from the second and some from the lower planes. A strong
reflected beam will result only if all the reflected rays are in phase. The waves reflected by different
planes will be in phase with one another only if the difference in the path length of the waves reflected
from the success phase is equal to an integral number of wave length.
O
L N
M
D
E
F
A
B
C
(X-ray re ection from a Crystal)fl


OL and OM are the perpendicular drawn to the incident and reflected beam, it will be seen that
the difference in the path lengths (say, s) of the waves reflected from the first two planes is given by
S = LN + NM ...(1)
This should be equal to a whole number multiple of wave length , i.e.,
LN + NM = n ...(2)
Since the triangles OLN and OMN are Congruent,
hence LN = NM
2LN = n
or 2dsin n   ...(3)
This relation is called Bragg’s equation. Distance between two successive planes d can be
calculated from this equation

89
With X-rays of definite wavelength, reflections at various angles will be observed for a given set
of planes separated by a distance d. These reflections correspond to n = 1, 2, 3 and so on and are
spoken of as first order, second order, third order and so on. With each successive order, the angle 
increases and the intensity of the reflected beam weakens.
Atomic radii
It is defined as half the distance between nearest neighbouring atoms in a crystals. It is expressed
in terms of length of the edge (a) of the unit cell of the crystal.
S.No. Unit cell Atomic Radius
1. Simple cubic (SC) structure
a
r
2

2. Face centered (FCC) cubic structure
a
r
2 2

3. Body-centered cubic structure (BCC)
3a
r
4

Ionic Crystals
In ionic crystals, the units occupying lattice points are positive and negative ions. In sodium
chloride, for example, the units are Na+
ions and Cl–
ions. Each ion of a given sign in held by coulombic
force of attraction to all ions of opposite sign.
 The number of oppositely charged ions surrounding each ion is termed its coordination
number
 The Coordination number commonly encountered in ionic crystals are 8 (body-centred
cubic arrangement), 6 (octahedral arrangement) and 4 (tetrahedral arrangement)
In ionic Crystals, the Coordination numbers as well as the geometrical shapes of the crystals
depends mainly on the relative sizes of ions.
The ratio of the radii of the positive and negative ions is called radius ratio.
Radius of positive ion(cation)
Radius ratio
Radius of negative ion(anion)
 =
c
a
r
r
The following table shows the radius ratio values, permitted coordination numbers and the shapes
of ionic crystals :
Radius ratio Coordination Arrangement of anions Examples
number around the cations
0.155-0.225 3 Triangular 2 3B O
0.225-0.414 4 Tetrahedral ZnS
0.414-0.732 6 Octahedral NaCl
0.732-1.000 8 Cubic (body-centred) CsCl
STRUCTURE OF IONIC COMPOUNDS
Simple ionic compounds are of two types i.e. AB and AB2
type. From the knowledge of close
packed structures and the voids developed there in, we can have an idea about the structures of simple
ionic compounds.
Among the two ions, constituting the binary compounds, the anions usually constitute the space
lattice with hcp or ccp type of arrangements whereas the cations, occupy the interstitial voids

90
(a) If the anions (B–
) constitute the crystal lattice and all octahedral voids are occupied by
cations (A+
), then the formula of the ionic solid is AB.
(b) Similarly, if half of the tetrahedral voids are occupied by cations, then the formula of the
solid crystal becomes A+
B–
.
(c) When the anions (B–2
) are constituting space lattice and all the tetrahedral voids are
occupied by the cations (A+
), then the formula of the solid crystal will be A2
B.
Ionic compounds of the type AB
Ionic compounds of the type AB have three types of crystalline structures. (a) ZnS type (b) NaCl
types (c) CsCl types
1. Sodium chloride (Rock salt) type structure
The sodium chloride structure is composed of Na+
and Cl–
ions. The number of sodium ions is
equal to that of Cl–
ions. The radii of Na+
and Cl–
ions 95 pm and 181 pm giving the radius ratio of 0.524
rNA+
/ rCr
= 95 / 181 = 0.524
The radius ratio of 0.524 for NaCl suggest an octahedral void. Thus the salient features of this
structure are as follows:
(i) Chloride ions (In a typical unit cell) are arranged in cubic close packing (ccp). In this
arrangement, Cl–
ions are present at the corners and at the centre of each face of the
cube. This arrangement is also regarded as face centred cubic arrangement (fcc).
(ii) The sodium ions are present in all the octahedral holes.
(iii) Since, the number of octahedral holes in ccp structure is equal to the number of anions,
every octahedral hole is occupied by Na+
ions. So that the formula of sodium chloride is
NaCl i.e. stoichiometry of NaCl is 1:1.
Na ion are represented+
by hollow circles
Cl ion are represented
–
by lled circlesfi
Na ion Octahedrally
+
Surrounded by six Cl ions.
–
Cl ion Octahedrally
–
Surrounded by Six
Na ions+
(The rock salt structure)
Unit cell structure of NaCl
(iv) Since there are six octahedral holes around each chloride ions, each Cl–
ion is surrounded
by 6 Na+
ions. Similarly each Na+
ion is surrounded by 6 Cl–
ions. Therefore, the coordination
number of Cl- as well as of Na+ ions is six. This is called 6:6 coordination.
(v) It should be noted that Na+
ions to exactly fit the octahedral holes, the radius ratio rNA
+
/
rCr
should be equal to 0.414. However, the actual radius ratio (rNA
+
/ rCr
= 0.524) exceeds
this value. Therefore to accommodate large Na+
ions, the Cl–
ions move apart slightly i.e.
they do not touch each other and form an expanded face centred lattice.

91
(vi) The unit cell of sodium chloride has 4 sodium and 4 chloride ions as calculated below
No of sodium ions = 12 (at edge centres) × 1/4 + 1 (at body centre) × 1= 4
No of chloride ions = 8(at corner) × 1/8+6 (at face centres) × 1/2 = 4
Thus, the number of NaCl units per unit cell is 4.
(vii) The edge length of the unit cell of NaCl type of crystal is 2(r+R) (r = radii of Na+
ion)
(R = radii of Cl–
ion) a = 2 (rNA
+
+ rCr
)
Thus, the distance between Na+
and Cl–
ions = a/2
Most of the halides of alkali metals, oxides and sulphides of alkaline earth metals have this
type of structures. Some of the common examples are NaI, KCl, RbI, RbF, NH4Cl, NH4Br,
AgCl, AgBr and AgI.
Ferrous oxide also has sodium chloride, types structure in which O–2
ions are arranged
in ccp and Fe+2
ions occupy octahedral ions. However, this oxide is always non -
stoichiometric and has the composition. It can be explained on the assumption that some
of the Fe+2
ion are replaced by 2/3rd
as many Fe+3
ions in the octahedral voids.
2. Zinc blende (ZnS) type structures (sphelerite)
The zinc sulphide crystals are composed of equal number of Zn+2
and S2–
ions. The radii of the
two ions (Zn+2
= 74 pm and S–2
= 184 pm) led to the radius (r+
/ r–
) as 0.40 which suggests a tetrahedral
arrangement.
rZn
+2
/ rS
–2
= 0.40
The salient features of this structure are as follows
(i) The Zinc ions are arranged in ccp arrangement, i.e. sulphide ions are present at the
corners and the centres of each face of the cube
(ii) Zinc ions occupy tetrahedral hole. Only half of the tetrahedral holes are occupied by Zn+2
so that the formula of the zinc sulphide is ZnS i.e. the stoichiometry of the compound is
1:1 (Only alternate tetrahedral holes are occupied by Zn+2
)
Zinc blende structure zinc sulphide
S2–
Zn
+2
(iii) Since the void is tetrahedral, each zinc ion is surrounded by four sulphide ions and each
sulphide ion is surrounded tetrahedrally by four zinc ions. Thus zinc sulphide has 4:4 Co
- ordination.
(iv) For exact fitting of Zn+2
in the tetrahedral holes, formed by close packing of S-2 ions, the
ratio Zn+2
/S–2
should be 0.225. Actually this ratio is slightly large (0.40)
(v) There are four Zn+2
ions and four S–2
ions per unit cell as calculated below:
No. of S–2
ions = 8(at corners) × 1/8 + 6(at face centres) × 1/2 = 4
No. of Zn+2
ions = 4(within the body) × 1 = 4

92
Thus, the number of ZnS units per unit cell is equal to 4. Some more examples of ionic solids
having Zinc blende structures are CuC, CuBr, CuI, AgI, beryllium sulphide.
Illustration 6 : If silver iodide crystallizes in a zinc blende structure with I–
ions forming the lattice then
calculate fraction of the tetrahedral voids occupied by Ag+
ions.
Sol. In AgI, if there are nI–
ions, there will be nAg+
ions. As I–
ions form the lattice, number of tetrahedral
voids = 2n. As there are nAg+
ions to occupy these voids, therefore fraction of tetrahedral voids occupied
by Ag+
ions = n/2n = ½ = 50%.
3. The Wurtzite structures
It is an alternate form in which ZnS occurs in nature. The main features of this structure are
(i) Sulphide ions have HCP arrangement and zinc ions occupy tetrahedral voids.
(ii) Only half the alternate tetrahedral voids are occupied by Zn+2
ions.
(iii) Coordinate no. of Zn+2
ions as well as S–2
ions is 4. Thus, this structure has 4 : 4
coordination.
(iv) No. of Zn+2
ions per unit cell
= 4(within the unit cell) × 1 + 6(at edge centres) × 1/3 = 6
Zn+2
S–2
No. of S–2
ions per unit cell
= 12(at corners) × 1/6 + 2 (at face centres) × 1/2 + 3 (within the unit cell) × 1 = 6
Thus, there are 6 formula units per unit cell.
4. Caesium chloride (CsCl) structure
The caesium chloride crystal is composed of equal number of caesium (Cs+
) and Chloride Cl–
ions. The radii of two ions (Cs+
= 169 pm and Cl–
= 181 pm) led to radius ratio of rCS
+
to rCl
–
as 0.93
which suggest a body centred cubic structure having a cubic hole
rCS
+
to rCl
–
= 169 / 181 = 0.93
(i) The chloride ion form the simple cubic arrangement and the caesium ions occupy the
cubic interstitial holes. In other words Cl–
ions are at the corners of a cube whereas Cs+
ion is at the centre of the cube or vice versa
(ii) Each Cs+
ion is surrounded by 8 Cl–
ions and each Cl–
ion in surrounded by 8 Cs+
ions.
Thus the Co - ordination number of each ion is eight. The salient features of this structure
are as follows:

93
Caesium Chloride structure
(iii) For exact fitting of Cs+
ions in the cubic voids the ratio rCS
+
/ rCl
–
should be equal to 0.732.
However, actually the ratio is slightly larger (0.93). Therefore packing of Cl–
ions slightly
open up to accommodate Cs+
ions.
(iv) The unit cell of caesium chloride has one Cs+
ion and one Cl–
ion as calculated below
No. of Cl–
ion 8(at corners) × 1/8 = 1
No. of Cs+
ion = 1(at body centre) × 1 = 1
Thus, number of CsCl units per unit cell is 1
(v) Relation between radius of cation and anion and edge length of the cube, rCS
to rCl
= a3/2
Other common examples of this type of structure are CsBr, CsI, TlCl, TlBr
Higher coordination number in CsCl(8:8) suggest that the caesium chloride lattice is more stable
than the sodium chloride lattice in which Co - ordination number is 6:6. Actually the caesium chloride
lattice is found to be 1% more stable than the sodium chloride lattice. Then the question arises why NaCl
and other similar compounds do not have CsCl type
lattice - This is due to their smaller radius ratio. Any attempt to pack 8 anions around the relatively
small cation (Li+
, Na+
, K+
, Rb+
) will produce a state in which negative ions will touch each other, sooner
they approach a positive ion. This causes unstability to the lattice.
Effect of temperature on crystal structure
Increase of temperature decreases the coordination of number, e.g. upon heating to 760 K, the
CsCl type crystal structure having coordination 8:8 changed to NaCl type crystal structures having
coordination 6:6.
High temp
CsCl type crystal  NaCl type crystal
(8.8 coordination) (6.6 Co-ordination)
Effect of pressure on crystal structure
Increase of pressure increases the Co - ordination number during crystallization e.g. by applying
pressure, the NaCl type crystal structure having 6:6 coordination number changes to CsCl type crystal
having coordination number 8:8
highpressure
NaCl type crystal  CsCl type crystal
(8.8 coordination) (6.6 Co-ordination)
IONIC COMPOUND OF THE TYPE AB2
Calcium fluoride (Fluorite) structure s
The salient features of fluorite structure are
(i) The Ca+2
ions are arranged in ccp arrangement, i.e. these ions occupy all the corners and
the centres of each face of the cube
(ii) The F–
ions occupy all the tetrahedral holes.

94
(iii) Since there are two tetrahedral holes for each Ca+2
ion and F–
ions occupy all the tetrahedral
holes, there will be two F–
ions for each Ca+2
ions, thus the stoichiometry of the compound
is 1:2.
(iv) Each Ca+2
ion is surrounded by 8F–
ions and each F–
ions is surrounded by 4Ca+2
ions.
The Coordination number of Ca+2
ion is eight and that of F–
ion is four, this is called 8:4
Coordination.
Unit cell representation of Ca structure
+2
(v) Each unit cell has 4 calcium ions and 8 fluoride ions as explained below
No. of Ca+2
ions = 8(at corners) × 1/8 + 6 (at face centres)´1/2
No. of F ions = 8 (within the body) × 1 = 8
Thus the number of CaF2
units per unit cell is 4.
Other examples of structure are SrF2
, BaCl2
, BaF2
, PbF2
, CdF2
, HgF2
, CuF2
, SrCl2
, etc.
Ionic compound of A2
B type
The compound having A2
B formula are compounds having anti fluorite structure
Anti fluorite structure is having arrangement of cations and anions opposite to the fluorite structure
Li2
O has an anti fluorite structure.
(i) In the crystal structure of Li2
O, the O–2
ions constitute a cubic close packed lattice (fcc
structure) and the Li+
ions occupy all the tetrahedral voids
(ii) Each oxide ion, O–2
ion is in contact with 8 Li+
ions and each Li+
ions having contact with
4 oxide ion. Therefore, Li2
O has 4:8 coordination
Examples - Na2
O, K2
O, K2
S, Na2
S, Rb2
O, Rb2
S
Anti-1 fluorite structure
Normal spinel structure
Spinel is a mineral MgAl2
O4
. In it oxide ions are arranged in ccp with Mg+2
ions occupying
tetrahedral voids and Al+3
ions in a set of octahedral voids. Many ferrites (such as ZnFe2
O4
) also possess

95
spinel structure. These are very important magnetic materials and are used in telephone and memory
loops in computers.
Structure of Fe3
O4
(Magnetite)
In Fe3
O4
, Fe+2
and Fe+3
ions are present in the ratio 2:1. It may be considered as having composition
FeO.Fe2
O3
. In Fe3
O4
Oxide arranged in ccp. Fe+2
ions occupy octahedral voids while Fe+3
ions are
equally distributed between octahedral and tetrahedral voids
MgFe2
O4
also has structure similar to magnetite. In this Mg+2
ions are present in place of Fe+2
ion in Fe3
O4
. Magnetite has inverse spinet structure.
SUMMARY OF VARIOUS STRUCTURES OF IONIC SOLIDS
Crystal structure Brief description Coordination
number
No. of atoms
per unit cell
Examples
1. Rock salt
(NaCl – type)
Cl-
ions in ccp
Na+
ions occupy
all octahedral
voids
Na+
= 6
Cl-
= 6
4 Li, Na, KI, and Rb
halides
NH4Cl, NH4Br, NH4I,
AgF, AgCl, AgBr, MgO,
CaO, TiO, FeO, NiO
2. Zinc blende
(ZnS – types)
S-2
ions in ccp
Zn+2
ions occupy
alternate
tetrahedral voids
Zn+2
= 4
S-2
= 4
4 ZnS, BeS, CuCl, CuBr,
CuI, AgI, HgS
3. Wurtzite
(ZnS – type)
S-2
ions in hcp
Zn+2
ion occupy
alternate
tetrahedral voids
Zn+2
= 4
S-2
= 4
4 ZnS, ZnO, CdS, BeO
4. Caesium
chloride
(CsCl type)
Cl-
ions in bcc Cs+
ions in the body of
cube
Cs+2
= 8
Cl-
= 4
1 CsCl, CsBr, CsI,
CsCN,CaS
5. Fluorite
(CaF2 type)
Ca+2
ions in ccp,
F-
ions occupy all
tetrahedral voids
Ca+2
= 8
F-
= 4
4 CaF2, SrF2, BaF2, BaCl2,
SrCl2, CdF2, HgF2
6. Anti fluorite
(Li2O – type)
O-2
ions in ccp,
Li+
ions occupy
all tetrahedral
sites
Li+
= 4
O-2
= 8
4 K2O, Na2O, K2S, Na2S
Problem 19. Compute the percentage void space per unit volume of unit cell in zinc-fluoride structure.
Solution: Since anions occupy fcc positions and half of the tetrahedral holes are occupied by cations.
Since there are four anions and 8 tetrahedral holes per unit cell, the fraction of volume occupied
by spheres/unit volume of the unit cell is
= 4 × (4/3 ra
3
) + 1/2 × 8 × (4/3rc
3
) = 162ra
3
= /32 {1 + (rc
/ra
)3
}
for tetrahedral holes,
rc
/ ra
= 0.225 = /3 2 {1 + (0.225)3
} = 0.7496
Void volume = 1 – 0.7496 = 0.2504/unit volume of unit cell
% void space = 25.04%

96
IMPERFECTIONS IN SOLIDS: DEFECTS IN CRYSTALS
Atomic imperfections / point defects:
When deviations exist from the regular (or periodic) arrangements around an atom or a group of
atoms in a crystalline substance, the defects are called point defects.
Type of point defects - point defects in a crystal may be classified into three types
(a) Stoichiometric defects
(b) Non - stoichiometry defects
(c) Impurity defects
Stoichiometry defect
The compounds in which the number of cation and anions are exactly in the same ratio as
represented by their chemical formula are called stoichiometric compounds. The defects that do not
disturb the ratio of cations and anions are called stoichiometric defect.
These are of two types:
1. Schottky defect
If in an ionic crystal of the type A+
B–
, equal number of cations and anions are missing from their
lattice. It is called Schottky defect.
This type of defect is shown by highly ionic compounds which have
(i) High Co - ordination number and
(ii) Small difference in the sizes of cations and anions
A few examples of ionic compounds exhibiting Schottky defect are NaCl, KCl, KBr and CsCl.
Consequences of Frenkel defect
(a) As the number of ions decreases as a result of this defect, the mass decreases whereas
the volume remains the same. Hence density of the solid decreases
(b) The crystal begins to conduct electricity to a small extent by ionic mechanism
(c) The presence of too many voids lowers lattice energy and the stability of the crystal
2. Frenkel defect
If an ion is missing from its correct lattice sites (causing a vacancy or a hole) and occupies an
interstitial site, electrical neutrality as well as stoichiometry of the compounds are maintained.
This type of defect is called Frenkel defect. Since cations are usually smaller it is more common
to find the cations occupying interstitial sites.
This type of defect is present in ionic compounds which have
(i) Low co ordinations number
(ii) Larger difference in size of cation and anions
(iii) Compounds having highly polarising cation and easily polarisable anion. A few examples
of ionic compounds exhibiting this defect are AgCl, AgBr, AgI, ZnS etc.
Consequences of Frenkel defect
(a) As no ions are missing from the crystal lattice as a whole, therefore density of the solid
remains the same
(b) The closeness of like charges tends to increases the dielectric constant of the crystal
(c) The crystal conducts electricity to a small extent by ionic mechanism

97
A B A B
B B A
A B A B
B A A
Schottky defect
A B A B
B B A
A B A B
B A A
Frenkel defect
B
A
PROPERTIES OF SOLIDS
The three main properties of solids which depend upon their structure
(i) Electrical properties
(ii) Magnetic properties
(iii) Dielectric properties
ELECTRICAL PROPERTIES
Electrical conductivity of solids may arise through the motion of electrons and positive holes
(electronic conductivity) or through the motions of ions (ionic conductivity). The conduction through
electrons is called n-type conduction and through positive holes is called p - types conduction. Electrical
conductivity of metal is due to motion of electrons and it increases with the number of electrons available
to participate in the conduction process. Pure ionic solids where conduction can take place only through
motion of ions are insulators. However, the presence of defects in the crystal structure increases their
conductivity.
On the basis of electrical conductivity the solids can be classified into three types
(a) Metal (conductors): They allow the maximum portion of the applied electric field to flow
through them and have conductivities in order of 106
- 108
ohm–1
.
(b) Insulators: They have low conductivities i.e. they do not practically allow the electric circuit
to flow through them. The electrical conductivity is in order 10–10
–10–20
ohm–1
m–1
(c) Semi conductors: The solids with intermediate conductivities at the room temperature.
Semi conductors allow a portion of electric current to flow through them.
Actually semi conductors are those solids which are perfect insulators at absolute zero, but
conduct electric current at room temperature.
(1) Intrinsic semi conductors (semi-conductors due to thermal defects)
At zero Kelvin pure substance silicon and germanium act as insulators because electrons
fixed in covalent bonds are not available for conduction. However at higher temperature
some of the covalent bonds are broken and the electrons so released become free to
move in the crystal and thus conduct electric current. This type of conduction is known
as intrinsic conduction as it can be introduced in the crystal without adding an external
substance.
(2) Extrinsic semi conductors: (semi conductors due to impurity defects)
The conductivity of pure silicon and germanium is very low at room temperature. The
conductivity of silicon and germanium can be increased by doping with impurities producing
n-type semiconductors or p - type semi conductors.

98
MAGNETIC PROPERTIES
The magnetic properties of different materials are studies in terms of their magnetic moments
which arise due to the orbital motion and spinning motion of the electron. As electron is charged particle,
the circular motion of the electric charge causes the electron to act as a tiny electro magnet. The
magnetic moment of the magnetic field generated due to orbital motion of the electron is along the axis
of rotation. The electron also possesses magnetic moment due to the spin which is directed along the
spin axis.
Thus, magnetic moment of the electron is due to travelling in closed path (orbital motion) about
the nucleus and spinning on its axis. For each electron spin magnetic moment is ±B
Where B
, Bohr
Magneton is the fundamental unit of magnetic moment and is equal to 9.27 × 10–24
em2
. The magnetic
moment due to orbital motion is equal to Ml
B
where Ml
is the magnetic quantum number of the electron.
Electron
(a)
At. nucleus
Orbital magnetic moment
Electron
(b)
direction of spin
Spin magnetic moment
As magnetic moment is a vector quantity, the net magnetic moment of an electron may be
represented by an arrow. Thus a material may be considered to contain a number of magnetic dipoles
(similar to a bar magnet with north and south poles). Due to the magnetic moment of the electrons
different substances behave differently towards the external applied magnetic field. Based on the behaviour
in the external magnetic field, the substances are divided into different categories as explained below.
(i) Diamagnetic substance : Substances which are weakly repelled by the external magnetic
fields are called diamagnetic field e.g. TiO2
, NaCl, benzene etc. Diamagnetic substances have all their
electrons paired.
(ii) Paramagnetic substances : Substances which are weakly attracted by magnetic field are
called paramagnetic substances. These substance have permanent magnetic dipoles due to the presence
of some species (atoms, ions or molecules) with unpaired electron. The paramagnetic substances lose
their magnetism in the absence of magnetic field. For e.g. TiO, VO2
and CuO, O2
, Cu+2
, Fe+3
etc.
(iii) Ferromagnetic substances : Substances which show permanent magnetism even in the
absence of the magnetic field are called Ferromagnetic substances. e.g. Fe Ni. CO, CrO2
show
Ferromagnetism. Such substances remain permanently magnetised, once they have been magnetised.
This type of magnetism arises due to spontaneous alignment of magnetic moment due to unpaired
electrons in the same direction.
Ferromagnetism Antimagnetism Ferrimagnetism
(iv) Anti Ferromagnetic substance : Substances which are expected to possess paramagnetism
or Ferromagnetism on the basis of unpaired electron but actually they posses zero net magnetic moment
are called anti Ferromagnetic substances e.g. MnO, Mn2
O3
, MnO2
.

99
Anti Ferromagnetism is due to presence of equal number of magnetic moments in the opposite
direction.
(v) Ferrimagnetic substances : Substance which are expected to posses large magnetism on
the basis of the unpaired electrons but actually have small net magnetic moments are called Ferrimagnetic
substances e.g. Fe3
O3
(3) Dielectric properties :
A dielectric substance is, in which an electric field gives rise to no net flow of electric
charge. This is due to the reason that electrons in a dielectric substances are tightly held by individual
atoms. However when electric field is applied. Polarization takes place because nuclei are attracted to
one side and the electron cloud to the other side. In addition to these dipoles, there may also be
permanent dipoles in the crystal.
The alignment of these dipoles may be in compensatory way i.e. the net dipole moment is zero
or noncompensatory way i.e. has a net dipole moment. The net dipole moment leads to certain
characteristic properties to solids.
(a) Piezoelectricity (or pressure electricity) :
When mechanical stress is applied on crystals so as to deform them, electricity is produced due
to displacement of ions. The electricity thus produced is called piezoelectricity and the crystals are called
piezoelectric crystals. Conversely, if electric field is applied to such crystals, atomic displacement takes
place resulting into mechanical strain. This is sometimes called Inverse piezoelectric effect.
The crystals are used as pick - ups in record players where they produce electrical signals by
application of pressure. Examples of piezoelectric crystals include titanates of barium and lead, lead
zirconate (PbZrO3
), ammonium dihydrogen phosphate (NH4
H2
PO4
) and quartz. They are also used in
microphones, ultrasonic generators and sonar detectors.
(b) Pyroelectricity :
Some piezoelectric crystals when heated produce a small electric current. The electricity thus
produced is called pyroelectricity.
(c) Ferroelectricity :
In some of the piezoelectric crystals, the dipoles are permanently polarized even in the absence
of the electric field. However on applying electric field, the direction of polarization changes e.g. Barium
titanate (BaTiO3
) sodium potassium tartarate (Rochelle salt) and potassium dihydrogen phosphate (KH2
PO4
).
All ferroelectric solids are piezoelectric but the reverse is not true.
(d) Anti Ferroelectricity :
In some crystals, the dipoles align themselves in such a way, that alternately, they point up and
down so that the crystal does not posses any net dipole moment. Such crystal are said to be anti
Ferroelectric e.g. Lead zirconate (PbZrO3
)
Problem What happens when ferrimagnetic Fe3
O4
is heated to 850 K and why?
Ans : Fe3
O4
on heating to 850 K becomes paramagnetic. This is due to greater alignment of
domains in one direction on heating.
SUPER CONDUCTIVITY
A substance is said to be superconducting when it offers no resistance to the flow of electricity.
Electrical resistance decreases with decreases in temperature and becomes almost zero near the
absolute zero. The phenomenon was first discovered by Kammerlingh Onnesin 1913 when he found
that mercury becomes superconducting at 4 K. The temperature at which a substance starts behaving
as super conductor is called transition temperature. Most metals have transition temperatures between
2K -5K. Certain organic compounds also becomes superconducting below 5K. Such low temperature
can be attained only with liquid helium which is very expensive.

100
Certain alloys of niobium have been found to be superconducting at temperature as high as 23
K. Since 1987, many complex metal oxides have been found to possess super conductivity at some
what higher temperature e.g.
Super conductivity materials have great technical potentials. They can be used in electronics in
building magnets, in power transmission and levitation transportation (trains which move in air without
rails).
Problem : Out of SiO2
(s), Si(s), NaCl(s) and Br2
(l) which is the best electrical conductor?
Solution : Si(s) because only this is a semi-conductor, while others SiO2
(s), NaCl(s) and Br2
(l)
are insulators.
Ex Problem. plain:
(a) The basis of similarities and difference between metallic and ionic crystals.
(b) Unit cell is not simply a cube of 4Na+
ions and 4Cl–
ions.
(c) Can a cube consisting of Na+
and Cl–
ions at alternate corners serve as satisfactory unit
cell for the sodium chloride lattice?
(d) Ionic solids are hard and brittle.
Solution : (a) Similarities:
(i) Both involve electrostatic forces of attraction.
(ii) Both are non-directional.
Differences: Ionic bond is a strong bond due to electrostatic forces of attraction while metallic
bond may be weak or strong depending upon the kernels.
(b) Unit cell of NaCl has fcc arrangement of Cl–
ions and Na+
ions are present at the edge centres
and one at the body-cnetre.
Thus there are 14Cl–
ions and 13Na+
ions in the unit cell. However their net contribution towards
the unit cell is 4Na+
and 4Cl–
ions.
(c) Yes because its repetition in different directions produces the complete space lattice.
(d) Ionic solids are hard because there are strong electrostatic forces of attraction. However they
are brittle because the bond is non-directional.
Problem : Analysis shows that nickel oxide has the formula Ni0.98
O. What fractions of the nickel exist
as Ni2+
and Ni3+
ions?
Solution : 98 Ni atoms are associated with 100 O-atom. Out of 98 Ni atoms, suppose Ni present as
Ni2+
= x
Then Ni present as Ni3+
= 98 – x
Total charge on xNi2+
and (98 – x)Ni3+
should be equal to charge n 100O2–
. Hence
x × 2 + (98 – x) × 3 = 100 × 2
or 2x + 294 – 3x = 200
or x = 94
Fraction of Ni present as Ni2+
= 94 / 98 × 100 = 96%
Fraction of Ni present as Ni3+
= 4 / 98 × 100 = 4%

101
8. LATTICE ENERGY
It is defined as the amount of energy released when cations and anions in their gaseous state
are bought together from infinite separation to form a crystal.
U
M (g) X (g) MX(s)
 
; U = lattice Energy
The theoretical treatment of ionic lattice energy was given by M.Born and A. Lande. This treatment
has been discussed below.
2
A
0
o o
MN Z Z e 1
U 1
4 r n
   
    
 This is the Born-Lande equation for lattice energy of an ionic crystal.
here M = Modelung Constant
NA
= Avogadro’s Number
Z+ = Charge on Cation
Z– = Charge on Anion
o
= Dielectric Constant
ro
= Distance
n = Born exponent
 The Born exponent n depend upon the type of the ion involved. Large ions having relatively
higher densities have larger values of n.
 Experimental values of the lattice energy are obtained by using Born-Haber cycle,
A typical cycle, for potassium chloride, is shown in figure.
Consists of the following steps.
1. Sublimation of K(s) = H (kJ mol–1
) + 89 [dissociation enthalpy of K(s)]
2. Dissociation of
1
2
Cl2
(g) = +122 [
1
2
× dissociation enthalpy of Cl2
(g)]
3. Ionization of K(g) = +418 [ionization enthalpy of K(g)]
4. Electron attachment to Cl(g) = –349 [electron gain enthalpy Cl(g)
5. Formation of solid form gas = –HL
/(kJ mol–1
)
6. Decomposition of compound = +437 [negative of enthalpy of formation of KCl(s)]
Because the sum of these enthalpy changes is equal to zero, we can infer from
89 + 122 + 418 – 349 – H2
/(kJ mol–1
) + 437 = 0
that HL
= +717 kJ mol–1
Some lattice energies obtained in this way are listed in following table-
   HL
/(kJ/mol–1
)
NaF 787
NaBr 751
MgO 3850
MgS 3406

102
k (g) + e (g) + Cl(g)
+ –
– 349
k (g)+Cl (g)+ –
+418
k(g)+Cl(g)
+122
k(g)+½ Cl (g)2
+89
+437
kCl(s)
– H L
The Born-Haber cycle for KCl at 298 k. Enthalpy changes are in kilo joules per mole.
Isomorphism
Ionic crystals possess same crystalline structure i.e. show isomorphism if the ions (cations and
anions) have same electronic configuration.
For eg :
NaF and MgO are isomorphous because these have same electronic configuration
NaF  Na+
+ F–
(2,8) (2,8)
Heat Capacity of Solids :
Heat capacity of a system is defined as the quantity of heat required to raise the temperature of
the system by one degree.
Let a very small quantity of heat dq be given to a system and the temperature of the system rise
by dT
Thus, Heat Capacity =
dq
dT
The heat capacity of a system at Constant volume given by
V
V
E
C
T
 
   
In 1819, Dulong and Petit found that, at constant pressure, the molar heat capacity at constant
volume of most of the solid at room temperature is given by-
CV
 6 Cal k–1
mol–1

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