This presentation is on Topic " Solid state Chemistry" which is based on the Chemistry Undergraduate syllabus of Mumbai University

1. Solid state Chemistry
Dr. P.U. Singare
Department of Chemistry,
N.M. Institute of Science, Bhavan’s College,
Andheri (West), Mumbai 400 058

2. Introduction
• Solid substances are broadly classified as amorphous and crystalline.
• A crystalline solids are having definite and regular geometrical arrangement of the
constituent atoms.
• This regular geometrical arrangement of the atoms in the crystal is responsible for its
fixed shape and definite properties of the crystals like refractive index, transparency,
melting point etc.
• The study of various physical properties, structure and geometry of a crystalline
substance is called crystallography.
• Every crystalline substance has certain elements of symmetry on the basis of which
they can be identified and differentiated.
• The different type of elements of symmetry in the cubic crystals are as follows:
1. Plane of symmetry
2. Axis of symmetry
3. Centre of symmetry

3. Plane of symmetry
• Plane of symmetry is an imaginary plane which divides the cubic crystal in to two equal parts such that one is
the mirror image of other.
• There are two type of planes of symmetry
Parallel plane of symmetry- in which the plane (dividing the crystal in to two equal parts) is parallel to the
opposite faces of the cubic crystal.
Diagonal plane of symmetry- in which the plane (dividing the crystal in to two equal parts) is passing
diagonally from one edge to the opposite edge of the cubic crystal.

4. Plane of symmetry
• https://www.youtube.com/watch?v=nmr46D5Cy9E

5. Plane & Axis of Symmetry
• https://www.youtube.com/watch?v=bvMBaXXErwo

6. Axis of symmetry
• Axis of symmetry is an imaginary line about which when a crystal is rotated by an angle of
360o will show same appearance more than once.
• If the crystal show same appearance after rotating about the axis through 180o, then the
axis is said to be of two-fold symmetry.
• If the crystal show same appearance after rotating about the axis through 120o, then the
axis is said to be of three-fold symmetry.
• If the crystal show same appearance after rotating about the axis through 90o, then the axis
is said to be of four-fold symmetry.

7. Axis of symmetry
• https://www.youtube.com/watch?v=Ch95sES5D9A

8. Centre of symmetry
• Centre of symmetry is an imaginary point within the cubic crystal such that every face has an
identical face at an equal distance but on the opposite side of this point.
• In other words any line drawn through this point will intersect two faces at an equal distance
but in opposite directions.
• The total number of plane, axis and centre of symmetries are called elements of
symmetries.
• In a cubic crystal there are 23 elements of symmetries of i.e. 9 plane of symmetries + 13 axis
of symmetries + 1 centre of symmetry.

9. Laws of Crystallography
• Crystallography is the study of various physical properties, structure and
geometry of a crystals and crystalline substances.
• There are 3 laws of Crystallography
1. Stenson’s Law of Constancy of interfacial angles
2. Law of crystal symmetry
3. Law of rational indices

10. Laws of Crystallography (Continued------)
1. Stenson’s Law of Constancy of interfacial angles
• According to this law “corresponding faces or planes which form the external surface
of
the crystal of a given substance always intersect at a definite angle and that this
angle is
constant irrespective of the way in which these faces develop”
• An interfacial angle is the angle between two corresponding faces of a crystal.
• Thus the size of a crystals of a given compound may vary depending on the conditions
under which crystallization take place, but the interfacial angle (θ) will always remain
same for any crystal of same substance.
• For example Quatrz crystal has hexagonal shape and the interfacial angle is always 120o.
120o

11. Laws of Crystallography (Continued------)
2. Law of Crystal Symmetry
• When a crystal is rotated about its axis, if it show same appearance
then the crystal is said to be symmetrical.
• Various type of symmetries in the crystal are called elements of
symmetry.
• Thus according to the law “all crystals of same substance possess
same elements of symmetry”
• Hence if two or more crystals have same elements of symmetry
they are the crystals of same substances.
• On the other hand if two or more crystals have different elements
of symmetry they are the crystals of different substances.

12. Laws of Crystallography (Continued------)
• 3. Law of Rational Indices • In the figure there are 3 axis X, Y & Z at a right angle
to each other and O is the origin.
• Consider a unit plane ABC having intercept a, b, c on
the 3 axis such that OA= a, OB= b,
OC =c .
• Suppose another plane DEF intercept the 3 axis such
that OD = 2a, OE = 2b and OF = 3c
• Then the ratio of intercept of the plane DEF along the
3 axis will be 2a:2b:3c.
• In general, the ratio of intercept of any plane along
the 3 axis can be given as ha:kb:lc
• Here h, k & l are the integral numbers also called
Weiss Coefficient of a plane.
• a, b, c are called Weiss indices of a plane.
• Thus according to Law of rational indices “ all planes
of the crystal will intercept the 3 axis at a distance
from the origin which bears a simple ratio to one
another”
X-axis
Y-axis
Z-axis
B
A
C
D
F
E
O

13. Miller’s Indices
• In the figure there are 3 axis X, Y & Z at a right angle to each
other and O is the origin.
• Consider a unit plane ABC having intercept a, b, c on the 3 axis
such that OA= a, OB= b,
OC =c .
• Suppose another plane DEF intercept the 3 axis such that OD =
2a, OE = 2b and OF = 3c
• Then the ratio of intercept of the plane DEF along the 3 axis will
be 2a:2b:3c.
• In general, the ratio of intercept of any plane along the 3 axis can
be given as ha:kb:lc
• Here h, k & l are the integral numbers also called Weiss
Coefficient of a plane. In case of plane DEF h =2, k=2, l=3
• a, b, c are called Weiss indices of a plane.
• The Miller indices of the plane DEF for which h =2, k=2, l=3 are
obtained by
 taking the reciprocal of Weiss coefficient as
1
ℎ
:
1
𝑘
:
1
𝑙
=
1
2
:
1
2
:
1
3
 multiplying by the least common multiple
6
2
:
6
2
:
6
3
= 2:2:3
Thus according to Miller indices, the DEF is (223) plane.
X-axis
Y-axis
Z-axis
B
A
C
D
F
E
O

14. Miller’s Indices (Continued----)
• Here plane ABCD is intersecting X-axis at
unit distance (h =1).
• The plane ABCD is parallel to Y & Z-axis.
• According to Miller indices, if the plane
is parallel to any axis, then the plane will
be intersecting that axis at infinity.
• Hence for plane ABCD which is parallel
to Y & Z-axis, k = ∞ & l = ∞.
• The Miller indices of the plane ABCD for
which h =1, k= ∞, l= ∞ are obtained by
taking the reciprocal of Weiss coefficient
as
1
ℎ
:
1
𝑘
:
1
𝑙
=
1
1
:
1
∞
:
1
∞
=1:0:0
• Thus according to Miller indices, the
ABCD is (100) plane.
D
A
B
C

15. Miller’s Indices (Continued----)
• Here plane ABCD is intersecting Y-axis at
unit distance (k =1).
• The plane ABCD is parallel to X & Z-axis.
• According to Miller indices, if the plane
is parallel to any axis, then the plane will
be intersecting that axis at infinity.
• Hence for plane ABCD which is parallel
to X & Z-axis, h = ∞ & l = ∞.
• The Miller indices of the plane ABCD for
which h = ∞, k= 1, l= ∞ are obtained by
taking the reciprocal of Weiss coefficient
as
1
ℎ
:
1
𝑘
:
1
𝑙
=
1
∞
:
1
1
:
1
∞
=0:1:0
• Thus according to Miller indices, the
ABCD is (010) plane.
D
A
B
C

16. Miller’s Indices (Continued----)
• Here plane ABC is intersecting X, Y &
Z-axis at unit distance.
• Hence for plane ABC, h = 1, k =1 & l =
1.
• The Miller indices of the plane ABC
for which h = 1, k= 1, l= 1 are
obtained by taking the reciprocal of
Weiss coefficient as
1
ℎ
:
1
𝑘
:
1
𝑙
=
1
1
:
1
1
:
1
1
=1:1:1
• Thus according to Miller indices, the
ABC is (111) plane.
A
B
C

17. Miller’s Indices (Continued----)
• Here plane ABCD is intersecting X & Y axis at
unit distance (i.e. h =1 & k =1) & the plane is
parallel to Z-axis.
• According to Miller indices, if the plane is
parallel to any axis, then the plane will be
intersecting that axis at infinity.
• Hence for plane ABCD which is parallel to Z-
axis l = ∞.
• Hence for plane ABCD, h = 1, k =1 & l = ∞ .
• The Miller indices of the plane ABCD for
which h = 1, k= 1, l= ∞ are obtained by
taking the reciprocal of Weiss coefficient as
1
ℎ
:
1
𝑘
:
1
𝑙
=
1
1
:
1
1
:
1
∞
=1:1:0
• Thus according to Miller indices, the ABCD is
(110) plane.
A
B
C
D

18. Unit Cell, Space Lattice & Lattice Planes
• Unit Cell:
 The crystal is built up of number of very small unit cells.
 In the unit cell, atoms are arranged in a definite
manner.
 The shape of the unit cell is related to the shape of the
crystal.
 The unit cells are repeated and extended in a three
dimensional manner to obtain macroscopic crystal. (Fig
1)
• Space lattice:
 In the unit cell, atoms are arranged in a definite
manner.
 Repetition of unit cells in the crystal results in regular
orderly arrangement of atoms.
 In 3 dimensional figure, the arrangement of atoms is
shown by the network of lines and atoms are supposed
to be present at every corner. (Fig. 2)
 Such network of lines used to represent the
arrangement of atoms in the crystal is called space
lattice.(Fig 2)
• Lattice planes:
 The atoms in the space lattice can form series of parallel
and equidistance planes.
 Such planes are called lattice planes. (Fig 2)
Fig.1
Fig.2

19. Cubic Lattice
• Cubic Lattice: is the arrangement of atoms in
the fundamental cubic unit cell.
• Depending on the arrangement of atoms in a
cube, there are 3 types of Cubic lattice.
1. Simple cubic lattice (Primitive Cubic lattice)
In this type of cubic lattice, there is 1 atom at
each corner of cubic unit cell.
Thus 8 corners is having 8 atoms.
Each corner atom is shared by 8 unit cell
including the original unit cell.
Hence 1/8 part of every corner atom belong to
the original unit cell.
Therefore 1/8 x 8 = 1 atom belong to the
original unit cell.
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8

20. Face centred Cubic Lattice
• In this type of cubic lattice, there are 6 atoms one ach at
the centre of each face.
• The atom at the centre of each face is shared by two cubic
unit cells including the original unit cell.
• Hence ½ part of each atom at the centre of the face belong
to the original unit cell.
• Thus ½ x 6 = 3 atoms at the centre of the face belongs to
the original unit cell.
• In addition to this there are 8 atoms one each at every
corner.
Each corner atom is shared by 8 unit cell including the
original unit cell.
Hence 1/8 part of every corner atom belong to the original
unit cell.
Therefore1/8 x 8 = 1 atom at the corner belong to the
original unit cell.
Thus in all 3 atoms at the centre of the face + 1 atom at the
corner = 4 atoms belong to the original unit cell.
1/8
1/8 1/8
1/8
1/8 1/8
1/8
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/2
1/8
1/8
1/8
1/8
1/8 1/8
1/8
1/8

21. Body centred Cubic Lattice
• In case of body centred cubic lattice, there are 8 atoms
one each at every corner.
Each corner atom is shared by 8 unit cell including the
original unit cell.
Hence 1/8 part of every corner atom belong to the
original unit cell.
Therefore 1/8 x 8 = 1 atom at the corner belong to the
original unit cell.
In addition to this, there is 1 atom at the centre of the
body which is not shared by any other unit cell.
This centre atom belongs exclusively to the original unit
cell.
Therefore total 1 atom at the corner + 1 atom at the
centre of the body = 2 atoms belong to the original unit
cell.
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1

22. Cubic lattice Video
• https://www.youtube.com/watch?v=y0c5WILbzQw

23. Cubic lattice Video
• https://www.youtube.com/watch?v=KNgRBqj9FS8

24. Bragg’s Equation
• Consider the crystalline planes P1, P2, P3 etc separated
by an interplanar distance ‘d’.
• A parallel beam of monochromatic x-rays ADI is
incident on the planes of the crystal.
• Ray A is scattered at point B along BC while ray D is
scattered at point F along FH.
• Path length travelled by ray A = AB + BC -------- (1)
• Path length travelled by ray D = DF + FH --------- (2)
• For a reflected beam to have maximum intensity, the
reflected beam should undergo constructive
interference i.e. they must reinforce with each other.
• For constructive interference to take place, the path
length of an interfering beam must differ by an integral
number of wavelength λ.
• i.e. Difference in path length = Path length travelled by
ray D - Path length travelled by ray A = nλ
• From eq. (1) & (2)
• (DF+FH) - (AB+BC) = nλ ---------- (3)
B
F
P1
P2
P3
C
H
L
θ d
D
I
A
E G
θ
θ

25. • Construction : Draw the two line BE and BG at a right angle to
the beam DFH such that
AB = DE & BC = GH --------- (4)
• Substitute the value of AB and BC from eq. (4) in eq. (3)
• (DF+FH) - (AB+BC) = nλ ---------- (3)
• (DF+FH) – (DE+ GH)= nλ
• DF-DE+FH-GH = nλ
• EF+FG = nλ ------------ (5)
• In ΔBEF, Sinθ =
𝐸𝐹
𝐵𝐹
EF = BF Sinθ -------- (6)
• Similarly in ΔBGF, Sinθ =
𝐹𝐺
𝐵𝐹
FG = BF Sinθ -------- (7)
• Substituting values of eq(6) & eq (7) in eq (5) we get
BF Sinθ + BF Sinθ = nλ ------ (8)
But BF = interplanar distance = d
Eq (8) will be
d Sinθ + d Sinθ = nλ
2d Sinθ = nλ --------- (9)
Eq (9) is called Bragg’s equation which gives relation between
the interplanar distance (d), wavelength of X-rays (λ) and angle
of maximum reflection also called glancing angle (θ)
B
F
P1
P2
P3
C
H
L
θ d
D
I
A
E G
θ
θ

26. • From the Bragg’s equation 2d Sinθ = nλ
Knowing the wavelength of x-rays (λ) and order of reflection
(n), glancing angle (θ) we can calculate interplanar distance
(d) in a crystal.
By using crystal of known substance (of which interplanar
distance ‘d’ is known to us) and by experimentally
determining the glancing angle (θ) and order of reflection
(n), we can calculate the Wavelength of X-rays.

27. Problem 1: The second order reflection with X-rays of wavelength 1.10Å occurs at an angle of 16o30’
from the 110 plane of a crystal. Calculate the interplanar distance.
Given: n =2 λ =1.10Å = 1.10x10-10 m θ = 16o30’
To Find : d
Formula: 2d Sinθ = nλ
Solution: 2d Sinθ = nλ
2d Sin (16o30’) = 2 (1.10x10-10 )
2d (0.2840) = 2 (1.10x10-10 )
d = 3.873x10-10 m
d = 3.873Å

28. Problem 2: The first order reflection from 100 plane of a crystal occurs at an angle of 9o30’.
Calculate the interplanar distance between 100 planes if the wavelength of X-rays used is 2 Å
Answer: d = 6.188x10-10 m

29. Problem 3: The first order reflection of a beam of X-ray from 100 plane of NaCl occurs at an
angle of 6.2o. Calculate the wavelength of X-rays used if d100 is 2.82x10-10 m
Given: n =1 d = 2.82x10-10 m θ = 6.2o
To find: λ
Formula: 2d Sinθ = nλ
Solution: 2d Sinθ = nλ
2(2.82x10-10 ) Sin (6.2o) = 1 x λ
λ = 5.64 x10-10 x 0.108
λ = 0.61 x10-10 m

30. Problem 4: Calculate the angle of reflection for first order reflection if the wavelength of X-rays used is
58x 10-14 m and d110 is 2.82x10-12 m.
Given: n =1 d = 2.82x10-12 m λ = 58x 10-14 m
To find: θ
Formula: 2d Sinθ = nλ
Solution: 2d Sinθ = nλ
2(2.82x10-12 ) Sinθ = 1 x 58x 10-14
5.64 x10-12 x Sinθ = 58x 10-14
Sinθ = 10.28 x10-2
θ = 5.9o

31. Problem 5: The first order reflection of a beam of X-ray from a crystal occurs at an angle of
5o15’. At what angle will be the third order reflection.
Given: θ1 = 5o15’ for n = 1 (first order reflection)
θ3 = ? for n = 3 (third order reflection)
To find: θ3
Formula: 2d Sinθ = nλ
Solution: Since X- rays used are same, their wavelength will remain same. Similarly since the
crystal used for study is same, the value of interplanar distance ‘d’ will also remain same.
2d Sin5o15’
2d Sinθ3
=
1
3
Sin(5o15’)
Sinθ3
=
1
3
3 Sin(5o15’) = Sin θ3
3 (0.0915) = Sin θ3
Sin θ3 = 0.2745
θ3 = 15o56’

32. Bragg’s X-ray spectrometer to determine the interplanar distance
• The x-rays from the x-ray tube are passed through the metal
screen (M), so that a homogeneous beam of x-rays having
definite wavelength are obtained.
• Slits S1 & S2 are used to get a narrow beam of x-rays.
• Ionization chamber (B) is attached to the turn table (T) having
circular scale.
• The ionization chamber is filled with the gases like methyl
iodide, ethyl iodide, or SO2 which strongly absorbs x-rays.
• When the x-rays enter the ionization chamber, the ionization of
gas take place causing ionic current.
• The intensity of ionic current thus produced is recorded on the
electrometer.
• The intensity of the current produced is proportional to the
ionization of gas which in turn is proportional to the intensity of
x-rays.
• Initially no crystal is placed on the turn table.
• As a result the narrow beam of x-rays pass in a straight line and
enters the ionization chamber.
• The turn table is adjusted in such a way that the electrometer
show maximum current intensity.
• The corresponding angle θ is recorded on the circular scale as
initial position.
X-ray tube
Ionization
Chamber
Turn table

33. Bragg’s X-ray spectrometer to determine the interplanar distance (Continued---)
• The crystal (D) is now mounted on the surface of the turn table.
• The x-ray beam are made to strike the (100) plane of a crystal and
x-rays get reflected.
• As a result, the intensity of current recorded at initial position will
now decrease.
• The turn table is now rotated so that the current intensity on the
electrometer will now increase and will become maximum.
• When the current intensity is maximum, the corresponding angle
(θ) at final position is recorded on the circular scale.
• The difference in the angle between the final and initial position is
θ1 which corresponds to first order reflection.
• The turn table is rotated in clockwise and anticlockwise direction to
get θ2 & θ3 which corresponds to second and third order reflection.
X-ray tube
Ionization
Chamber
Turn table

34. Bragg’s X-ray spectrometer to determine the interplanar distance (Continued---)
• Thus by using x-rays of known wavelength and by experimentally
measuring the glancing angle θ1, θ2 & θ3 values corresponding to
1st, 2nd & 3rd order reflection, the interplanar distance (d100) can be
calculated by using the Bragg’s equation
d100 =
nλ
2Sinθ
------------- (1)
• Similar experiment is repeated by making the x-rays to fall on (110)
(111) planes of a crystal.
• For every plane of a crystal, the glancing angle θ1, θ2 & θ3 values
corresponding to 1st, 2nd & 3rd order reflection are measured
experimentally.
• Measuring the values of glancing angle for (110) (111) planes of a
crystal, the interplanar distance d(110) and d(111) are calculated by
eq. (1).
X-ray tube
Ionization
Chamber
Turn table

35. Type of Cubic lattice Ratio of interplanar
distance
d100 : d110 : d111
Simple cubic lattice 1 : 0.707 : 0.577
Face center cubic lattice 1 : 0.707 : 1.154
Body center cubic lattice 1 : 1.414 : 0.577
• By calculating the values of d100, d110,
d111 we can calculate the ratio
d100 : d110 : d111
• From the calculated value of ratio of
d100 : d110 : d111, it is possible to identify
the type of cubic lattice present in the
crystal.

36. Problem 6: The angles of reflection for first order from (100), (110) &(111) planes of a cubic crystal are
13.4o, 9.533o and 23.833o. What type of cubic lattice it belongs to?
Given : n =1
θ1(100) = 13.4o θ1(110) = 9.533o θ1(111) = 23.833o
To find: type of cubic lattice
Formula: d =
nλ
2Sinθ
Solution: Since x-rays of uniform wavelength are used λ will be same (constant).
d100:d110:d111 =
1
2Sin(13.4o)
:
1
2Sin(9.533o)
:
1
2Sin(23.833o)
d100:d110:d111 =
1
2 (0.2317)
:
1
2(0.1656)
:
1
2(0.4041)
d100:d110:d111 =
1
0.4634
:
1
0.3312
:
1
0.8081
d100:d110:d111 = 2.158 : 3.019 : 1.2375
d100:d110:d111 = 1 : 1.40 : 0.573 (dividing by throughout by 2.158)
The above ratio match with the literature value Body centred cubic lattice ratio 1 : 1.414 : 0.577
Hence the crystal is having Body centred cubic lattice

37. Problem 7: The first order reflection from (100), (110) &(111) planes of a cubic crystal occurs at an
angle of 5.9o, 8.4o and 5.2o respectively. What type of cubic lattice it belongs to?
Given : n =1
θ1(100) = 5.9o θ1(110) = 8.4o θ1(111) = 5.2o
To find: type of cubic lattice
Formula: d =
nλ
2Sinθ
Solution: Since x-rays of uniform wavelength are used λ will be same (constant).
d100:d110:d111 =
1
2Sin(5.9o)
:
1
2Sin(8.4o)
:
1
2Sin(5.2o)
d100:d110:d111 =
1
2 (0.103)
:
1
2(0.146)
:
1
2(0.091)
d100:d110:d111 =
1
0.206
:
1
0.292
:
1
0.182
d100:d110:d111 = 4.85 : 3.425 : 5.49
d100:d110:d111 = 1 : 0.706 : 1.13 (dividing by throughout by 4.85)
The above ratio match with the literature value Face centred cubic lattice ratio 1 : 0.707 : 1.154
Hence the crystal is having Face centred cubic lattice