2 SOLID STATE

Matter is anything that has mass and occupies space. There are three main phases
of matter
solids, liquids and gases. There is a fourth phase, plasma; this exists at very high temperature.
Characteristics of Solid State
(1) Solids are rigid and have definite shape.
(2) Solids maintain their shape independent of the size or shape of the container in which they
are placed.
(3) Solids are nearly incompressible.
(4) As compared to liquids or gases, solids diffuse very slowly.
(5) Solids occupy fixed positions and can oscillate about their mean positions.
Types of solids: There are two types of solids :
(1) Crystalline solid (2) Amorphous solid
Table 2.1: Comparison between Crystalline and Amorphous solids

Crystalline Solid Amorphous Solid

Crystalline solid is a substance whose constituent 1. Amorphous solid is a substance whose constituent
and orderly arrangement. particles do not possess a regular and orderly
particles have regular
arrangement.
2. It has no sharp melting point, they rather soften
It has sharp melting point.
gradually on heating.
They have definite geometry 3. They do not have definite geometry.
sharp tool, When an amorphous solid is cut with a sharp tool,
When a crystalline solid is cut with a

smaller crystals are obtained along particular an irregular fracture is obtained.

which are inclined to one
planes (cleavage planes)
another at a particular angle.
when a crystalline 5. They do not have crystal symmetry.
They have crystal symmetry i.e.,
solid is rotated about an axis its appearance e.g., Glass, Plastic, Rubber, etc.
remains the same.

e.g., NaCl, Diamond,

Metals, etc.

LATTICE
SPACE LATTICE OR CRYSTAL
to be generated from the regular repetition of some basic unit of
A crystal can be imagined
pattern such as atom, molecule,
ion or a group of ions
Space lattice
Space lattice may be defined
as a regular three-dimensional arrangement of identical points in
space.

27
 INTRODUCTION TO EIN

28 in Fig.
2.1.
OR P
shown
been
lattice has
An example ofspace SOL

of space lattice
Fig. 2.1. Representation

It must be noted here that:

as that of any other point in the latti
()each lattice point has the same environment Ca
by a lattice point, irresno
() the constituent particle has always to be represented espective ()
atom.
whether it contains single atom or more than one

UNIT CELL
Unit cell is the smallest, three-dimensional group of latice points that generates the wholelat:
by repeating its own dimensions in various directions.
A unit cell is characterized by lengths/edges of the unit cell (a, b and c) and by the ane
(a. B and y) as shown in Fig. 2.2.
g

ii

Fig. 2.2. Unit cell

For cubes,
a =
b =c and
It may be a= ß=y=90°
remembered that a is the
angle
Y-between 'a' and 'b'. between "b' and 'c, 'a and
B-between
Types of Unit Cell
There three types
are
of Unit Cells
(1) Simple cubic unit cell: A unit cell
simple, primitive or basic unit cell. having the lattice points only at the
the corners is calle
corners

simple crystal lattice. A crystal lattice having l

[Fig. 2.3(a)] primitive unit ce
(2)
Body-centred cubic unit cell: It is aunit cell
addition to
the
lattice points at the having lattice points
points at
at the
the centre
centre ofbody
corners. [Fig. 2.3(b)1
(3) Face-centred cubic unit cell:
It is a unit
face, in addition to the lattice points at cel
having lattice points
the corners. a
ints at the centre
o feact

[Fig. 2.3(c)]
 SOLID STATE 29

(a) b) (c)
Fig. 2.3. (a) Simple cubic unit cel (b) Body-centred cubic unit cel (c) Face-centred cubic unit cell

Calculation of numbers of atoms in unit cell

() Simple cubic unit cell
In a simple cubic unit cell there are eight atoms represented as spheres at the corners. Thus

1/ 8
atom
Fig. 2.4. Simple cubic unit cell
The contribution o f one sphere at the corner per unit cell = 1/8

C o n t r i b u t i o n o f 8 spheres a t the corners per unit cell = 1/8 x 8 =1

A simple cubic unit cell has one atom.

() Body-centred cubic unit cell

In such type of cubic unit cell, there are eight atoms at the corners and one at the body centre.
The contribution of one sphere at the corner per unit cell 1/8.
=

.. Contribution o f 8 spheres at the corners per unit cell = 1/8 x 8 = 1

of
Similarly, the contribution of one sphere at the body centre cube
A body-centred cubic unit cell has 1+1=2 atoms.

1/8
atom

I atom

Fig. 2.5. Body-centred cubic unit cell

(üi) Face-centred cubic unit cell
Ina face-centred cubic unit cell, there are cight atoms at the corners and one atom at the centre of
each face.
(There are eight corners of a cube, each corner has one sphere.
The contribution of 8 spheres at the corners per unit cell
8x atom
 MISTRY FOR
CHEA.
ENGINEERING

TO
INTRODUCTION

30
a cube
faces of 1/2
() There are six unil ccll
faces per
one sphere at the
Fach face has sphere unit cel
of one
faces per
at the
contribution
t the
spheres
he
contrihution ofsiN
ntoms
+34
has 1
unit cell
fence a face-centred
cubic

/8
lom

2 atom

Face-centred cubic unit cell

Fig. 2.6.

Caculation of Density of Unit Cell

The density of a solid is obtained by dividing mass of the atom with the volume of the cell

Mass of Unit Cell

Density of unit cell
Volume of Unit Cell1
Mass of unit cell = Number of atoms per unit cell x Mass of each atom

Atomic mass (M)

= Zx-
Avogadro's Number (N,)
If the edge length of a unit cell =a cm.
Volume ofunit cell (a)
ZM
Density of unit cell (p) = -
Ng .a
N, = 6.023 x 10 mol
where
For SC, Z = 1

For BCC, Z = 2
For FCC, Z =4
Unit of Density

ZM
P = 8 mol
Cm3
N, a mol X Cm
8

ATOMIC RADIUS OF A CUBIC LATTICE

Fdee length (a) is the distance between the
cube.
centres of two corner atoms or the
atoms of
Atomic radius (r) 1s hall the
distance
between two immediate neighbours
cell

n*
SOLID STATE
31
()For Simple Cubic Cell (SC): The atomic radius (r) and the length of a cube (a) are
related as
a 2 r or r= al2

1/8
atom
Fig. 2.7. Simple cubic cell
) For Body-Centred Cubic Cell (BCC):
(4r =(a +a)+a
(4r)3aF
3a
4

Fig. 2.8. Body-centred cubic cell

(ii) For Face-Centred Cubic Cell(FCC):

(4r = a +a
(4) 2a

Ar

Fig. 2.9. Face-centred cubic cell

 ENGINEERING CHEMISTRV
32 INTRODUCTION TO
radius of unit cell
FOR UPTU
and atomic
Table 2.1. Number of atoms
Number df atoms Atomic SOL
Typed Ca in Unit Cell (Z)
radius (r)
al2
SC

BCC 3al4
FCC 2a/4
RADIUS RATIO
Because of the difference in relative sizes and number of cations or anions ionic solido
different coordination number. have Inc
In ionic crystals large-sized anions have a definite close-packed structure and small-sized Catki
are accommodated in the voids between the om
large spheres
anions.
In an 1onic solid, the ratio of radius of cation to that of anion is called radius ratio. Di

Radius of cation (r+) rel

Radius ratio =
Radius of anion (r) Th
The effect of this ratio in
known determining the co-ordination number and shape of an ionic solid i
radius ratio effect.
as

Co-ordination number (C.N.) is the total number of immediate

ion. When we say C.N.
of
neighbours surrounding the centra
and each Cl is surrounded
NaCl is 6, it means in its structure, each Na is
surrounded by six CI ions D
by six Na ions.
Significance of radius ratio
Higher the radius ratio, larger is the size of cation and hence
(i)When the radius ratio is up to 0.155, a C.N. greater is its co-ordination numbe
(i) When the radius ratio is greater than 0.155 but
of two is usually observed.
observed. less than 0.225, a C.N. of three is usualy
(ii) When the radius ratio is
greater than 0.225 but less than
and C.N. of four is
observed. 0.414, a tetrahedral arrangemen
(iv) When the radius ratio is
greater than 0.414 but less than
(v)When the radius ratio is between 0.732-1, 0.732, a C.N. of six is observed.
observed. a BCC
arrangement with C.N. 8 is usuai)
Table 2.2. Ranges of Radius Ratio Values
and
Range of radius C.N. of the cation
Geometry of the Ionic Solid
ratio (rt/r) values
in the ionic Arrangement of
anions Examples of
solid around the cation ionic solid
0.0-0.155 2
(Geometry of ionic solid)
0.155 0.225 Linear
3
0.225 0.414 Planar Triangular BeF2
4 B,O. BN
0.414 0.732
6 Tetrahedral Zns, ZnO
0.732 1.0
8 Octahedral NaCl, AgCl
BCC
BRAGG'S LAW CsCl, CsBr
A crystal lattice is
regular. arrangement of
Laue (1912) s
suggested that X-rays can passlayers or planes of
come -100 pm (100
pm 100* 10 m equidistant atoms.
through crystals because
crystals (A) are comparable. 10m= 10*
cm =Å) and the wavelength once
the
interatomic al
SOLID STATE 33

Emergent ays in phase

Reinforcement
(Bright spot on
photographic plate)

Cancellation
(Dark area on
photographic plate)
Incdent X-rays Grating Emergent rays out of phase

Fig. 2.10

Diffraction patterns produced by erystals

In 1913. the duo of father and son. W.L. Bragg and W.H. Bragg, diseovered a mathematical
relation to determine interatomic distances from X-ray diffraction patterns, known as Bragg's equation.
They showed that
)the N-rays diffracted from atoms in crystal planes obey the laws of reflection.
() the two rays retlected by successive planes will be in phase ifthe extra distance travelled
by the second ray is an integral number of wavelengths.
Derivation of Bragg's equation
The horizontal lines in Fig. 2.11 represent parallel planes in the crystal structure separated by a
distance 'd". When the X-rays of wavelength A' strike the first plane at angle , some ofthe rays wil
be reflected at the same angle while some of the rays will penetrate and get reflected from the successive
(second) plane also at the same angle 0. These rays will reinforce those reflected from the first plane
if the extra distance travelled by them is equal to integral number 'n' of wavelength.
Extra distance = n.
then the waves will be
Drawing 0A and OB perpendiculars to the incident and reflected beams,
in phase provided the difference of path lengths ofwaves reflected by first two planes will be integral

multiple of wavelength.
Pathlength of MPM' > Pathlength of LOL' by AP +PB (Extra distance).

M M

d
AS B
P
d

Fig. 2.11
P a t h difference = AP + PB = n
...(i)
ZAOP= ZBOP =0
AP AP
In A AOP, sin 0=
OP d
AP d sin 0 ..(i)
 ENGINEERING CHEMISTRV
RY FOR UP
INTRODUCTION
TO
34
BP BP
Similarly in A BOP, sin 0 OP d
BP dsin 0
0
AP+ BP =d sin 0 + d sin 0 2d sin
na 2 dsin 0 ..

This equation (iv) is called Bragg's equation wnere

n order of reflection
n=1= First orderreflection, n =2 Second orderreflection)
(if
wavelength of X-rays
d=interplanar distance

angle
GRAPHITE : A TWO-DIMENSIONAL SOLID
Graphite is a second alotrope of carbon.
s 2s 2p
CGS configuration 1 1 1 1
CE.S. configuration
Sp
Hlybridization1 1|1
sp2
Thus, each carbon atom is sps hybridized with bond
of a large number of flat parallel
angle 120°. The structure of graphite cons
layers (or sheets) of carbon atoms. Each layer is made up of
hexagonal rings of C-atoms. In each layer, each C-atom is joined to
C-C covalent bonds. The C-C only three other C-atoms
covalent bond distance in each hexagonal ring is 1.42
Åwhic
intermediate between the C-C bond distance (1.54 and C
these threeC-C bonds has two-third
Ä) C bond distance (1.34 A).
=

Eacn
single bond character and one-third double bond charace
which is due to the resonance existing between
single and double bonds as shown below:
C
C
C C C
C
C-c C-c
The distance between the two
the two layers suggests that the
adjacent layers of C-atoms is 3.40 A. etwe
layers are joined together Such a large distance b
Properties and uses of Graphite by weak van der Waals
forces
() In each layer, each sp" hybridized carbon is
fourth valence electron on each carbon atom linked
is free toto carry
other heat
threeand electricity «andt
C-atoms ony
for making carbon electrodes or
(2 The graphite electrodes,
layers of carbon atoms are held
layers can easily slide over one together by weak van der Waals forces the
another
substance and it is used as lubricant for and hence graphite is soft,
softness and machines.
s flaky and
its abilit to
mark black on
pencils paper, graphite 1Si used for making lead"
 SOLID STATE 35

142A

40 A

Flat laves of Weak van der

carm atoms- Walls forces, holding
nade up ot the layers together
hexagonal nngs
3.40 A

Fig. 2.12. Structure of graphite crystal

FULLERENES
Kroto
Fullerenes is the third allotrope of carbon after diamond and graphite discovered by Curl,
whose
and Smalley in 198. They are a group of polymorphic form of pure carbon compounds
discrete molecules consist of a hollow. spherical cluster of a large even
number of carbon atoms.
known for the design of
Fullerenes were named after Richard Buckminster Fuller, an architect
fullerenes are often
geodesic domes which resemble spherical fullerenes in appearance. Spherical
called bucky balls and look like a soccer ball. The smallest fullerenes is Cwhile the others are C24
obtained is Cs0
C C C suchC40 Co C C ete. The largest of fullerenes studied fullerene is Cs C is the most
Despite large number of fullerenes the most widely
common and most stable.

Structure of C%o
60 vertices and 32 faces,
The structure of C is truncated icosahedron. It is polygon having
a

and 12-pentagonal rings. Each

Carbon is present at each vertex. C contains 20-Hexagonal rings
In the structure each pentagon is surrounded
carbon atom is bonded to three others and is sp* hybridized.
share a common side. The bond length in pentagon is 1.45 Å
by six hexagons and no two pentagons
while in hexagon it is 1.40 A.

Fig. 2.13. Ball and stick model of C60

 ENGINEERING CHEMIST.
36 INTRODUCTION TO RY
FOR UP
Preparation of fullerenes
vapourized at 0.05 0.1
od atmospheric pressure In inert atmosphoerethe
of arg
t e is
dioxide) and produc
uces
Ndaton of carbon to carbon monoxide
erenes
and carbon
On extraction of this black sooty dust with solvents or on sublimation
alumina-h
dust a
s.15fullet ,a mixture
using
fullerenesis obtained from which can be isolated chromatographically

Properties of fullerenes
i t i s mustard coloured powder. The colour changes from brown to black as the th
the thickne
the fibm increases.
2)ltis moderately soluble in organic solvents such as bcnzene, toulene and chlo

dissolved in toulene it forms magenta coloured solution.

chloroform. Whe
)ltis tough and
themallystable.
4) It is odourless.
(5) Pure C is insulator but on doping with alkali metals clectrical conductivityincrease
eases a
changes to conductor and superconductors.
(6) lt is compressible and volume can be reduced to 30%.

Applications of fullerenes
) I n the preparation of electronic and microelectronic devices.
2) In the
preparation of superconductors
which possess nearly no resistance to passage
electricity.
(3) In batteries as
charge carriers.
(4)In the preparation
of soft ferromagnets with zero remanence.
(5) Organofullerenes are used in the preparation of important polymers, nanomaterials
pharmaceuticals.
(6) Used as lubricant.
(7) In non-linear optical devices.

SOLVED NUMERICALS
Problem 1. X-ray analysis shows that the unit cell
length in NaCl is 5.628 A. Calculate
density.
(TU-20
Solution. ZM

N, .a
Z 4 (. NaCl is FCC Crystal)
M 58.5 (23 +35.5)
N 6.023x 10 g
mol (Avogadro's No.)
a 5.628 A =5.628 10 cm. x

Putting all these values in equation (i)

4x 58.5
6.023x10" x(5.628 x10*
P -2.179 g em*
Problem 2. Calculate the density of a BCC erystal if
side of cube is 4A and M- 60.
(UPTU-20
ZM
Solution.
N

.2(For BCC Crystal)

SOLID STATE 37
M 60
N,-6.023 10 g mol (Avogadro's No.)
a 4 A -4 x10 cm
Putting all these values in cquation (i)

2x6
P
6.023x10x(4x10*)
P 3.116g cm".
Problem 3. Molybdenum has body-centred cubic structure and a density of 10.2 gm/cc. Calculate
its atomic radius. The atomic weight of Mo is 95.94. (IU-2007)
ZM ...)
Solution. P
N.a
ZM ..ii)
a =

p..Na
Z2 (For BCC structure)
M 95.94

P 10.2 gm/cc
N = 6.023 x 10 g mo

Putting these values in equation (ii)

2x95.94
10.2x 6.023x103

a3.1232x1023

a = 3.1232 x 10-23
a = 3.1492 Å

For BCC lattice

3a
4
r=1.363 Å.
Problem 4. The density of NaCl is 2.163 g /c. Calculate the edge of its cubic cell, assuming that
four molecules of NaCl are associated per unit
cell. (UPTU-2004)
ZM
Solution. P =
...(i)
Ng.a
ZM
p N
Z4
M 58.5
N=6.023 x 10 g mol

p 2.163 g/cc
 INTRODUCTION TO ENGIINE

FOR
38
234
these values in
equation (1)
Putting 4x58.5 23x10 13.027x1023

=V179.6lx
x 1024
5.642 A th is 5.59 x 10
=

edgc-length
a
cm". Ifthe cm anda
3. 1876 g
Problem 5. A mctal has density
the lattice type.
mass S3.8, calculate 'Z' and explain
ZM
Solution. P 3
Ng.a
Z -P:N , a
M
p 3.1876 g cm
a =5.59 x 10 cm.

M 83.8
Putting these values in equation (ii).

Z
3.1876x6.023 x10x(5.59x10*)3
83.8
Z 4.001
Lattice is FCC (NaCl
type).
Problem 6. Gold has FCC structure with
unit length 4.07 Å and density of 19.3 g cm. Calci
Avogadro's number. Atomic weight of Au 197. =

Solution. ZM
P =

Ng.a
N =
ZM
p.03
Z=4 (For FCC Structure)
M= 197
P =19.3g em

Putting these values in

a =
4.07 Å 4.07 =

equation (ii).
x
10 em.

N, 4x197
19.3x(4.07 x10
N= 6.056 x 104°
Problems on
Bragg's law
Problem 1. X-rays of atoms/mol

Solution.
action/reflectwavelength
ion. Caleulate1.54the A are
falling at an
an
nA
separation betweenangle of 14°12' on
on a
a
Crystal to s

2d sin 6 two the ers

ofthe
parallel pla
planes

UPI
( U P T U - 0

2 sin 6
OLID STATE
39
n1(Firstorder reflection)
-1.54 A
14°12'
Putting these values in equation (ii)

IxI.54
d
2x sin 14°12'

d 3.14A
Problem 2. At what principle angle should first order diffraction occur, when copper radiations
f wavclength 1 54 pm interact with lattice planes that are 185 pm apart?
Solution. n 2d sin 0 .i)
n
sin ..(ii)
2d
n=1 (First order diffraction)
= 154 pm
d 185 pm
Putting these values in equation (ii),
1x154
sin = 0.4162
2x185
sin (0.4162)
24.590.
Problem 3. The distance between layers in a crystal is 282 pm. X-rays are diffracted from these
rays at an angle of 23.0°. If the diffraction is of first order, calculate the wavelength of X-rays in
Angstrom.
n = 2d sin 0
Solution.
= 2dsine
n
(ii)
n = 1 (First order diffraction)

d 282 pm
0 23.0
Putting these values in equation (ii),
2x 282xsin 23

220.34 pm
= 2.20 Å