The document discusses the properties and arrangements of solid-state materials, particularly focusing on unit cells, their shapes, and arrangements in crystal structures. It details different crystal systems, packing fractions, and coordination numbers, including specific examples of various ionic compounds and their crystal structures. Key characteristics of unit cells, lattice parameters, and void arrangements in both face-centered and body-centered cubic systems are also highlighted.

1. Solid State
Gases: no definite shape and volume
Solids: definite shape, volume and order.
Order: definite pattern of arrangement of atoms or molecules or
ions.
Liquids: no definite shape but definite volume
Solids: definite shape and volume

2. Intensive properties: do not depend on the amount.
Unit Cells
• Smallest Repeating Unit
• Unit Cells must link-up  cannot have gaps between them
• All unit cells must be identical

4. Choice of the origin is arbitrary

5. Blue atom or orange atom or

6. even a space!

7. This cannot be a unit cell
Unit cells are not identical

8. This also cannot be a unit cell
Space between unit cells not allowed

9. Unit cells exist in only seven shapes
 Cubic
 Orthorhombic
 Rhombohedral
 Tetragonal
 Triclinic
 Hexagonal
 Monoclinic

10. Crystal intercepts & Angles
a
b
c




11. Crystal Systems Lattice Parameters
Crystal Intercepts Crystal Angles
Cubic a = b = c  =  =  = 90o
Orthorhombic a  b  c  =  =  = 90o
Rhombohedral a = b = c  =  =   90o
Tetragonal a = b  c  =  =  = 90o
Triclinic a  b  c       90o
Hexagonal a = b  c  =  = 90o,
 = 120o
Monoclinic a  b  c  =  = 90o,
  90o

12. There are not more than 4 ways of arranging
spheres in any shape of unit cell
These are Primitive, Body Centered, Face
Centered & End Centered

13. a = 2r
1
2
4
3
5
6
7
8
Unit Cell shape view
Unit Cell arrangement view

14. Layer arrangement view

16. Volume occupied by a sphere in the unit cell V
8
1

Total volume occupied by all the spheres in the unit
cell
VV 18
8
1


17. Packing Fraction
Fraction of the Unit cell’s volume occupied by the
spheres
cellunitofvolume
cellunittheinsidespheresthebyoccupiedVolume

3
3
3
4
a
r

 3
3
2
3
4
r
r

52.0

18. Coordination number
6

19. Unit Cell shape view
Unit Cell arrangement view

20. Layer arrangement view

21. ar 34 
2
3
2
a
r  a
a > 2r

23. Packing Fraction
3
3
3
4
3
4
2








r
r
68.0
Volume occupied by a corner sphere in the unit cell V
8
1

Volume occupied by the central sphere in the unit cell V1
Total Volume occupied by the spheres in the unit cell V2
Packing Fraction

24. Coordination number
8

25. Unit Cell shape view
Unit Cell arrangement view

26. a
ar 24 
2
2
a
r  a
r2

28. Packing Fraction
3
3
2
4
3
4
4








r
r
74.0
Volume occupied by a corner sphere in the unit cell V
8
1

Volume occupied by a face centered sphere in the unit
cell
V
2
1

Total Volume occupied by the spheres in the unit cell
4
2
1
6
8
1
8 
Packing Fraction
Highest Packing Fraction of all shapes and of
all arrangements

29. Coordination number
x
y
z
y-z plane
x-z plane x-y plane

30. Coordination number
a/2
a/2
2
a

32. Out of all the twenty eight possible unit cells only
14 exist !
Those arrangements in a given shape that violate even
one symmetry element of that shape do not exist in
that shape
90o axis of symmetry

64. If we do the same with BCC & FCC we will get the
same result.
Lets try with End Centered

91. Like this 13 other arrangements in various shapes
were rejected.
We are left with only 14 unit cells

92. Crystal Systems
Cubic
Orthorhombic
Rhombohedral
Tetragonal
Triclinic
Hexagonal
Monoclinic
Bravais Lattices
Primitive, FCC, BCC
Primitive, FC, BC, EC
Primitive
Primitive, BC
Primitive
Primitive
Primitive, EC

93. Layer A
Layer arrangement view

94. Layer B

95. Layer C
Layer A
Layer B
Layer C
Cubic Close Packing
(CCP)

97. Layer A
Layer arrangement view

98. Layer B

99. Layer A
Layer A
Layer B
Layer A
Hexagonal Close
Packing

100. Unit Cell shape view
Unit Cell arrangement view

101. a
c
a = 2r
2r2r
2r
r
O
OA = r
AOB = 30o
BD
r
OB 
3
2
O
c/2
DD
E
2r
23
2
2
3
2
)2(
2
2 c
r
r
rEB 






3
2
4rc 

102. Contribution of corner atom
6
1

Contribution of Face atom
2
1

Contribution of second layer
atoms
3
Total atoms per unit cell 6

103. Packing Fraction
 
3
2
42
4
3
6
3
4
6
2
3
rr
r

 
 2
2
4
3
6 r
3
2
4r
74.0

104. Packing Fraction depends on:
1. Layout of each layer
2. Placement of one layer over the other

105. r
O
A
B
AB = r
OA = r tan30o
3
r
 < r
Packing Fraction  same
Rank of unit cell  2
Volume of unit cell  1/3 of previous
mass of unit cell  1/3 of previous
density  same

107. Two types of voids:
Octahedral
Tetrahedral
Found only in FCC & Hexagonal primitive unit cells
Octahedral void in FCC

108. Each octahedral void located at the edge center is shared by 4 unit
cells
Total contribution of edge centre voids = 312
4
1

Contribution of central void 1
Total contribution of all octahedral voids per unit cell of FCC = 4
No. of Octahedral voids per unit cell =
Rank of unit cell

109. Tetrahedral void in FCC
(0,0,0)
x-axis
y-axis
z-axis
(a/2, a/2,0)
(a/2, 0,a/2)
(0, a/2,a/2)
(a/4, a/4,a/4)

110. (0,0,0)
(a/2, a/2,0)
(a/4, a/4,a/4)
a

b

k
a
j
a
i
a
a ˆ
4
ˆ
4
ˆ
4


k
a
j
a
i
a
b ˆ
4
ˆ
4
ˆ
4












ba
ba


.
.
cos 1









16
3
16cos 2
2
1
a
a




3
1
cos 1 o
268.109

111. With each corner as origin there are 8 tetrahedral voids in FCC unit
cell
 No. of tetrahedral voids = 2  no. of Octahedral voids

112. Voids in Hexagonal Primitive
Let us assume that this is the unit cell
then according to what we have done in FCC no. of
Octahedral voids = 6 & no. of tetrahedral voids = 12
Octahedral voids
Octahedral
void

113. Voids in Hexagonal Primitive
Let us assume that this is the unit cell
then according to what we have done in FCC no. of
Octahedral voids = 6 & no. of tetrahedral voids = 12
Tetrahedral voids
Contribution of tetrahedral voids formed inside the unit
cell is 1 each. The ones formed on the corners of the
hexagon have a contribution of 1/3.
Total contribution 66
3
1
4 
In 3 layers 1262 

114. Minimum rc/ra for various coordination numbers
2r
aB
O
A
3
2
30sec 

 
a
ac
r
rr
AB
OA
155.01
3
2

a
c
r
r
Coordination number - 3

115. 4
3
444
222
aaaa
rrAB ac 















ara 42 
Coordination number - 4
z-
axis
A
B
(0,0,0)
(a/4, a/4,a/4)







2
4
4
3 a
ac
r
rr
aac rrr
2
3

2
3
1 
a
c
r
r
225.01
2
3

a
c
r
r

116. Coordination number - 4 (square planar) or 6 (octahedron)
B
A 2
a
rrAB ac 
ara 2
a
a
ac r
r
rr 2
2
2

414.012 
a
c
r
r

117. Coordination number - 8 (cube)
ac rr
a

2
3
ara 2
  aca rrr  2
2
3
732.013 
a
c
r
r

118. Final Radius Ratios
Radius Ratio, rc/ra
Co-ordination No.
<0.155 2
[0.155, 0.225) 2 or 3
[0.225, 0.414) 2 or 3 or 4 Td
[0.414, 0.732) 2 or 3 or 4 Td, 4 sq. pl or 6 Oh
[0.732, 0.99) 2 or 3 or 4 Td, 4 sq. pl or 6 Oh
or 8

119. For ionic compounds of the general formula AxBy the ratio of the
coordination number of A to that of B will be the ratio of y:x.
1. Rock Salt Structure (NaCl)  Cl-
 Na+
Cl- is FCC
Na+ occupies Octahedral voids
No. of Cl- per unit cell = 4
No. of Na+ per unit cell = 4
 formula is NaCl
Coordination no. of Na+ = 6
Coordination no. of Cl- = 6

120. Other compounds which have this structure are: all halides of alkali metals
except cesium halide, all oxides of alkaline earth metals except beryllium
oxide, AgCl, AgBr & AgI.

121. Consider the unit cell with Cl- as FCC.
  NaCl
rra 22
 Cl
ra 42
Consider the unit cell with Na+ as FCC.
  NaCl
rra 22
 Na
ra 42
Similarly, rany alkali metal = rany halide
  ClNa
rr
rany akaline earth metal = roxide
Comparing

122. 2. Zinc Blende (ZnS)
 S2-
 Zn2+
S2- is FCC
Zn2+ occupies alternate tetrahedral
voids
No. of S2- per unit cell = 4
No. of Zn2+ per unit cell = 4
 formula is ZnS
Coordination no. of Zn2+ = 4
Coordination no. of S2- = 4
Other compound which have this structure is: BeO

123. 3. Fluorite (CaF2)
 F-
 Ca2+
Ca2+ is FCC
F- occupies all tetrahedral voids
No. of Ca2+ per unit cell = 4
No. of F- per unit cell = 8
 formula is CaF2
Coordination no. of F- = 4
Coordination no. of Ca2+ = 8
Other compounds which have this structure are: UO2,
ThO2, PbO2, HgF2 etc.

124. 4. Anti-Fluorite (Li2O)
 O2-
 Li+
O2- is FCC
Li+ occupies all tetrahedral voids
No. of O2- per unit cell = 4
No. of Li+ per unit cell = 8
 formula is Li2O
Coordination no. of Li+ = 4
Coordination no. of O2- = 8
Other compounds which have this structure are: Na2O,
K2O, Rb2O

125. 5. Cesium Halide
 Cl-
 Cs+
Cl- is Primitive cubic
Cs+ occupies the centre of the unit cell
No. of Cl- per unit cell = 1
No. of Cs+ per unit cell = 1
 formula is CsCl
Coordination no. of Cs+ = 8
Coordination no. of Cl- = 8
Other compounds which have this structure are: all halides
of Cesium and ammonium

126. 6. Corundum (Al2O3)
Oxide ions form hexagonal primitive unit cell and trivalent ions (Al3+) are
present in 2/3 of octahedral voids.
No. of O2- per unit cell = 2
No. of Al3+ per unit cell = 4/3
32
642
3
4
OAl
OAl;OAlisformula
Coordination no. of Al3+ = 6
Coordination no. of O2- = 4
Other compounds which have this structure are: Fe2O3, Cr2O3, Mn2O3 etc.

127. 7. Rutile (TiO2)
Oxide ions form hexagonal primitive unit cell and tetravalent ions (Ti4+)
are present in 1/2 of octahedral voids.
No. of O2- per unit cell = 2
No. of Ti4+ per unit cell = 1
Coordination no. of Ti4+ = 6
Coordination no. of O2- = 3
Other compounds which have this structure are: MnO2, SnO2, MgF2, NiF2
 formula is TiO2

128. 8. Pervoskite (CaTiO3)
 O2-
 Ca2+ (divalent ion)
Ca2+ is Primitive cubic
Ti4+ occupies the centre of the unit cell
No. of O2- per unit cell = 3
No. of Ca2+ per unit cell = 1
 formula is CaTiO3Coordination no. of O2- = 6
Coordination no. of Ti4+
Other compounds which have this structure are: BaTiO3,
SrTiO3
 Ti4+ (tetravalent ion)
O2- occupies face centres
No. of Ti4+ per unit cell = 1
= 6
Coordination no. of Ca2+ = 12

129. 9. Spinel & Inverse Spinel (MgAl2O4)
O2- ion is FCC
Mg2+(divalent ion) 1/8th of tetrahedral voids
Al3+ (trivalent ion) 1/2 of octahedral voids
O2- per unit cell = 4
Mg2+ per unit cell = 1
Al3+ per unit cell = 1
 formula is MgAl2O4
Spinel Inverse Spinel
O2- ion is FCC
divalent ion 1/8th of tetrahedral voids
trivalent ion 1/4th of octahedral voids & 1/8th of
tetrahedral voids
O2- per unit cell = 4
Divalent per unit cell = 1
Trivalent per unit cell = 1

130. (i) Lattice of atoms
(a) Vacancy  an atom is missing from its position
 density decreases
 percentage occupancy decreases
100
defectwithoutpresentatomsofno.
presentatomsofno.
occpancy% 
100
densityltheoretica
densityobserved
occpancy% 
(b) Self interstitial  an atom leaves its lattice site & occupies interstitial space
 density & percentage occupancy remains same
(c) Substitutional impurity  foreign atom substitutes a host atom & occupies its lattice
 density & percentage occupancy may change
(c) Interstitial impurity  foreign atom occupies occupies the interstitial space
 density & percentage occupancy increases

131. (i) Ionic structures
(a) Schottky Defect  Cation – anion pair are missing
 electro neutrality is maintained
 density decreases
(b) Frenkel Defect  ion leaves lattice position & occupies interstitial space
 electro neutrality is maintained
 density maintained
(c) Substitutional Impurity Defect  Ba2+ is replaced by Sr2+
 electro neutrality is maintained
 density changes
(d) Interstitial Impurity Defect  H2 is trapped in TiC
 electro neutrality is maintained
 density increases
(a) F-Centre
 electron replaces anion
 electro neutrality is maintained
 density decreases
 colour is imparted

132. 1. Assuming diamond to be FCC of carbon atoms
and that each carbon atom is sp3 hybridized
then which of the following statements is
correct.
(a) all voids are empty
(b) 100% octahedral voids are filled
(c) 50% octahedral voids are filled
(d) 100% tetrahedral voids are filled
(e) 50% tetrahedral voids are filled
Sol: If no void is filled then each carbon would be in contact with 12 carbon atoms. This is
not possible as each carbon is sp3 hybridized.
If octahedral voids are filled then those carbons in the voids would be in contact with
6 carbon atoms. This also is not possible.
If 100% tetrahedral voids are filled then the FCC carbons would be in contact with 8
carbon atoms as they are shared in 8 unit cells and would be in contact with 8
tetrahedral voids. Not possible.  (e)

133. 2. In NaCl calculate:
The distance between the first 9 nearest neighbors
in a unit cell & their total number in all unit cells

134. neighbor no. distance no. of neighbors
1
2
a
6
2
2
a
12
3
2
3a 8
4 a 6
5
2
5a 24
6 a
2
3 24
7 a2 12
8 a
2
3 24
9 a3 8

135. 3. Iron crystallizes in FCC lattice. The figures given below shows
the iron atoms in four crystallographic planes.
Draw the unit cell for the corresponding structure and identify
these planes in the diagram. Also report the distance between
two such crystallographic planes in each terms of the edge length
‘a’ of the unit cell.

136. distance between two such
planes is a
distance between two such
planes is a

137. distance between two such planes
is
A
B
C
A
33
3
distance
aa

2
a

138. 3. Marbles of diameter 10 mm are to be placed on a flat surface
bounded by lines of length 40 mm such that each marble has its
centre within the bound surface. Find the maximum number of
marbles in the bound surface and sketch the diagram. Derive an
expression for the number of marbles per unit area.
Interpretation:
1. count marbles as 1 each even if some portion
goes outside the bound surface
2. count marbles based on the portion that is
inside the bound surface
25

139. To calculate no. of marbles per unit area we
need to select the smallest repeating unit.
18
d
A B
C
 2
d
4
3
ABCtriangleofarea 
6
1
3ΔABCinmarblesofno. 
 
0.0115areaunitpermarblesofno.
4
103
2
1
2 
18.416000.0115figureenclosedthisinmarblestotalthethisonbased 