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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
book 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
Choice of the origin is arbitrary
Blue atom or orange atom or
even a space!
This cannot be a unit cell
Unit cells are not identical
This also cannot be a unit cell
Space between unit cells not allowed
Unit cells exist in only seven shapes
 Cubic
 Orthorhombic
 Rhombohedral
 Tetragonal
 Triclinic
 Hexagonal
 Monoclinic
Crystal intercepts & Angles
a
b
c



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
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
a = 2r
1
2
4
3
5
6
7
8
Unit Cell shape view
Unit Cell arrangement view
Layer arrangement view
Volume occupied by a sphere in the unit cell V
8
1

Total volume occupied by all the spheres in the unit
cell
V
V 1
8
8
1



Packing Fraction
Fraction of the Unit cell’s volume occupied by the
spheres
cell
unit
of
volume
cell
unit
the
inside
spheres
the
by
occupied
Volume

3
3
3
4
a
r


 3
3
2
3
4
r
r


52
.
0

Coordination number
6
Unit Cell shape view
Unit Cell arrangement view
Layer arrangement view
a
r 3
4 
2
3
2
a
r  a

a > 2r
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 V
1

Total Volume occupied by the spheres in the unit cell V
2

Packing Fraction
Coordination number
8
Unit Cell shape view
Unit Cell arrangement view
a
a
r 2
4 
2
2
a
r  a

r
2

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
Coordination number
x
y
z
y-z plane
x-z plane x-y plane
Coordination number
a/2
a/2
2
a
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
If we do the same with BCC & FCC we will get the
same result.
Lets try with End Centered
Like this 13 other arrangements in various shapes were
rejected.
We are left with only 14 unit cells
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
Layer A
Layer arrangement view
Layer B
Layer C
Layer A
Layer B
Layer C
Cubic Close Packing
(CCP)
Layer A
Layer arrangement view
Layer B
Layer A
Layer A
Layer B
Layer A
Hexagonal Close Packing
Unit Cell shape view
Unit Cell arrangement view
a
c
a = 2r
2r
2r
2r
r
O
A
B
3
0
o
OA = r
AOB = 30o
BD
r
OB 

3
2
O
B
c/2
D
D
E
2 r
2
3
2
2
3
2
)
2
(
2
2 c
r
r
r
EB 









3
2
4r
c 
Contribution of corner atom
6
1

Contribution of Face atom
2
1

Contribution of second layer
atoms
3

Total atoms per unit cell 6

Packing Fraction
 
3
2
4
2
4
3
6
3
4
6
2
3
r
r
r


 
 2
2
4
3
6 r

3
2
4r

74
.
0

Packing Fraction depends on:
1. Layout of each layer
2. Placement of one layer over the other
r
3
0
o
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
Two types of voids:
Octahedral
Tetrahedral
Found only in FCC & Hexagonal primitive unit cells
Octahedral void in FCC
Each octahedral void located at the edge center is shared by 4 unit
cells
Total contribution of edge centre voids = 3
12
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
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)
(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
















b
a
b
a




.
.
cos 1











16
3
16
cos 2
2
1
a
a









3
1
cos 1 o
268
.
109


With each corner as origin there are 8 tetrahedral voids in FCC unit
cell
 No. of tetrahedral voids = 2  no. of Octahedral voids
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
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 6
6
3
1
4 



In 3 layers 12
6
2 


Minimum rc/ra for various coordination numbers
2 r
B
O
A
3
2
30
sec 


 
a
a
c
r
r
r
AB
OA
155
.
0
1
3
2



a
c
r
r
3
0
o
Coordination number - 3
4
3
4
4
4
2
2
2
a
a
a
a
r
r
AB a
c 























a
r
a 4
2 
Coordination number - 4
z -
axis
A
B
(0,0,0)
(a/4, a/4,a/4)








2
4
4
3 a
a
c
r
r
r
a
a
c r
r
r
2
3


2
3
1 

a
c
r
r
225
.
0
1
2
3



a
c
r
r
Coordination number - 4 (square planar) or 6 (octahedron)
B
A 2
a
r
r
AB a
c 


a
r
a 2

a
a
a
c r
r
r
r 2
2
2




414
.
0
1
2 



a
c
r
r
Coordination number - 8 (cube)
a
c r
r
a


2
3
a
r
a 2

  a
c
a r
r
r 

 2
2
3
732
.
0
1
3 



a
c
r
r
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
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
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.
Consider the unit cell with Cl-
as FCC.

 
 Na
Cl
r
r
a 2
2

 Cl
r
a 4
2
Consider the unit cell with Na+
as FCC.

 

 Na
Cl
r
r
a 2
2


 Na
r
a 4
2
Similarly, rany alkali metal = rany halide

  Cl
Na
r
r
rany akaline earth metal = roxide
Comparing
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
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.
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
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
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
3
2
6
4
2
3
4
O
Al
O
Al
;
O
Al
is
formula

Coordination no. of Al3+ = 6
Coordination no. of O2- = 4
Other compounds which have this structure are: Fe2O3, Cr2O3, Mn2O3 etc.
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
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 CaTiO3
Coordination 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
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
(i) Lattice of atoms
(a) Vacancy  an atom is missing from its position
 density decreases
 percentage occupancy decreases
100
defect
without
present
atoms
of
no.
present
atoms
of
no.
occpancy
% 

100
density
l
theoretica
density
observed
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
(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
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)
2. In NaCl calculate:
The distance between the first 9 nearest neighbors
in a unit cell & their total number in all unit cells
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 a
2 12
8 a
2
3
24
9 a
3 8
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.
dis t ance bet ween t wo s uch
pla nes is a
distance bet ween t wo s uch
pla nes is a
distance between two such planes
is
A
B
C
A
3
3
3
distance
a
a


2
a
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
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
ABC
triangle
of
area 
6
1
3
ΔABC
in
marbles
of
no. 

 
0.0115
area
unit
per
marbles
of
no.
4
10
3
2
1
2 

18.4
1600
0.0115
figure
enclosed
this
in
marbles
total
the
this
on
based 


solid-state-1205589205798872-3.pdf
solid-state-1205589205798872-3.pdf
solid-state-1205589205798872-3.pdf

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solid-state-1205589205798872-3.pdf

  • 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 book 2
  • 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
  • 3.
  • 4. Choice of the origin is arbitrary
  • 5. Blue atom or orange atom or
  • 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
  • 15.
  • 16. Volume occupied by a sphere in the unit cell V 8 1  Total volume occupied by all the spheres in the unit cell V V 1 8 8 1   
  • 17. Packing Fraction Fraction of the Unit cell’s volume occupied by the spheres cell unit of volume cell unit the inside spheres the by occupied Volume  3 3 3 4 a r    3 3 2 3 4 r r   52 . 0 
  • 19. Unit Cell shape view Unit Cell arrangement view
  • 21. a r 3 4  2 3 2 a r  a  a > 2r
  • 22.
  • 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 V 1  Total Volume occupied by the spheres in the unit cell V 2  Packing Fraction
  • 25. Unit Cell shape view Unit Cell arrangement view
  • 26. a a r 2 4  2 2 a r  a  r 2 
  • 27.
  • 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
  • 31.
  • 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
  • 33.
  • 34.
  • 35.
  • 36.
  • 37.
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  • 58.
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  • 60.
  • 61.
  • 62.
  • 63.
  • 64. If we do the same with BCC & FCC we will get the same result. Lets try with End Centered
  • 65.
  • 66.
  • 67.
  • 68.
  • 69.
  • 70.
  • 71.
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  • 86.
  • 87.
  • 88.
  • 89.
  • 90.
  • 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
  • 95. Layer C Layer A Layer B Layer C Cubic Close Packing (CCP)
  • 96.
  • 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 2r 2r 2r r O A B 3 0 o OA = r AOB = 30o BD r OB   3 2 O B c/2 D D E 2 r 2 3 2 2 3 2 ) 2 ( 2 2 c r r r EB           3 2 4r c 
  • 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 4 2 4 3 6 3 4 6 2 3 r r 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 3 0 o 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
  • 106.
  • 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 = 3 12 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                 b a b a     . . cos 1            16 3 16 cos 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 6 6 3 1 4     In 3 layers 12 6 2   
  • 114. Minimum rc/ra for various coordination numbers 2 r B O A 3 2 30 sec      a a c r r r AB OA 155 . 0 1 3 2    a c r r 3 0 o Coordination number - 3
  • 115. 4 3 4 4 4 2 2 2 a a a a r r AB a c                         a r a 4 2  Coordination number - 4 z - axis A B (0,0,0) (a/4, a/4,a/4)         2 4 4 3 a a c r r r a a c r r r 2 3   2 3 1   a c r r 225 . 0 1 2 3    a c r r
  • 116. Coordination number - 4 (square planar) or 6 (octahedron) B A 2 a r r AB a c    a r a 2  a a a c r r r r 2 2 2     414 . 0 1 2     a c r r
  • 117. Coordination number - 8 (cube) a c r r a   2 3 a r a 2    a c a r r r    2 2 3 732 . 0 1 3     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.     Na Cl r r a 2 2   Cl r a 4 2 Consider the unit cell with Na+ as FCC.      Na Cl r r a 2 2    Na r a 4 2 Similarly, rany alkali metal = rany halide    Cl Na r r 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 3 2 6 4 2 3 4 O Al O Al ; O Al is formula  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 CaTiO3 Coordination 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 defect without present atoms of no. present atoms of no. occpancy %   100 density l theoretica density observed 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 a 2 12 8 a 2 3 24 9 a 3 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. dis t ance bet ween t wo s uch pla nes is a distance bet ween t wo s uch pla nes is a
  • 137. distance between two such planes is A B C A 3 3 3 distance a a   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 ABC triangle of area  6 1 3 ΔABC in marbles of no.     0.0115 area unit per marbles of no. 4 10 3 2 1 2   18.4 1600 0.0115 figure enclosed this in marbles total the this on based   