it is all about the solid state main part that is frequently asked in various examinations. it is must for the basics of solid state.

1. JFC Solid State Chemistry By Rawat Sir [M.Sc. Chemistry, 3 times NET (JRF), GATE ]
rwtdgreat@gmail.com
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S.No. Crystalline Solids Amorphous solids
1 Regular arrangement of particles
Long Range order
irregular arrangement of particles
Short Range order
2 Sharp melting point Melt over a rage of temperature
3 Regarded as true solids Regarded as super cooled liquids or pseudo solids
4 Undergo regular cut Undergo irregular cut
5 Anisotropic in nature Isotropic in nature
Based on binding forces:
Type Unit Particles Binding Forces Properties Examples
Atomic and
Molecular
Solids
Atoms, Polar or
non – polar
molecules
Vander Waal’s
forces (London
forces, dipole –
dipole interactions,
hydrogen bonds)
 Fairly soft,
 Low to moderately high melting pts,
 Poor thermal and electrical
conductors
Noble gases,
Dry ice (solid CO2),
CH4, CCl4, H2, I2, SO2,
HCl, H2O
Ionic Solids Negative and Positive
ions
Ionic bonds(i.e.
electrostatic force)
 Hard and brittle,
 High melting points,
 Poor thermal and electrical
conductors,
 Insulator in solid form but conductor
in molten and aqueous form
NaCl, NaF, CaO, MgO,
CaF2,ZnS
Covalent or
Network Solid
Atoms that are
connected in covalent
bond network
Covalent bonds  Very hard,
 Very high melting points,
 Poor thermal and electrical
conductors
 But Graphite is conductor
Diamond(C), Si,
Quartz (SiO2),
Silicon Carbide
(carborundum) SiC,
graphile (C), AlN
Metallic Solids Cations in electron
sea
Metallic bonds  Soft to very hard,
 Low to very high melting points,
 Good thermal and electrical
conductors,
 Malleable and ductile
All metallic elements,
for example Fe, Cu, Zn,
Mg, Ag etc
Crystal Systems:
Total number of crystal systems: 7 Total number of Bravais Lattices: 14
Crystal Systems Bravais Lattices Side length Interfacial angle Example
Cubic Primitive,
Face Centered,
Body Centered
a = b = c α = β = = 90o
Pb, Hg, Ag, Au, Cu, Diamond,
NaCl,ZnS (Zinc Blende)
Tetragonal Primitive,
Body Centered
a = b ≠ c α = β = = 90o
Rutile-TiO2, CaSO4, SnO2
Orthorhombic Primitive,
Face Centered,
Body Centered,
End Centered
a ≠ b ≠ c α = β = = 90o
KNO3, K2SO4,
Rhombic Sulphur (α- Sulphur)
Monoclinic Primitive,
End Centered
a ≠ b ≠ c α = = 90o
,
β ≠ 90o
β-Sulphur, CaSO4.2H2O,
Washing soda Na2SO4.10H2O
Triclinic Primitive a ≠ b ≠ c α ≠ β ≠ ≠900
K2Cr2O7, CuSO4.5H2O
Hexagonal Primitive a = b ≠ c α = β = 900
,
= 120o
Graphite, Mg, SiO2, ZnO, CdS
Rhombohedral Primitive a = b = c α = β = ≠ 90o
As, Sb, Bi, quartz, Calcite-CaCO3,
Cinnabar-HgS

2. JFC Solid State Chemistry By Rawat Sir [M.Sc. Chemistry, 3 times NET (JRF), GATE ]
rwtdgreat@gmail.com
www.slideshare.net/RawatDAgreatt
Google+/blogger/Facebook/Twitter-@RawatDAgreatt +919808050301, +917017378183
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Position of lattice point Contribution to one cubic unit cell
Corner
Edge
Face centre
Body centre/ tetrahedral void/ octahedral voids
1/8
1/4
1/2
1
Z= actual no. of particles in one unit cell or effective number of atoms
Z= No. of particles present at some position x contribution from that position + same + ……
Unit cell C.N. d a r r Z
SCC 6 r 1
BCC 8
x
r 2
DCC 4
x
r 4
ECC 6-corner and 2-edge
centre
x r 4
FCC(end
)
FCC,[hcp]-12
x r FCC-4, HCP-6,End-2
Body diagonal = a√3 face diagonal =a√2
Packing Efficiency
Packing Efficiency = (Volume occupied by all the atoms present in unit cell / Total volume of unit cell) × 100
Calculate % Packing fraction by using ( ) = ( )
©Rawat’s JFC (Just For Chemistry)
©Rawat’s JFC (Just For Chemistry)

3. JFC Solid State Chemistry By Rawat Sir [M.Sc. Chemistry, 3 times NET (JRF), GATE ]
rwtdgreat@gmail.com
www.slideshare.net/RawatDAgreatt
Google+/blogger/Facebook/Twitter-@RawatDAgreatt +919808050301, +917017378183
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 For FCC unit cell effective number of atoms (Z) =
8 corner atoms x (1/8) (each atom is shared by 8 unit cells) +
6 face centered atoms x1/2 (each shared by two unit cells) = 4
 Considering radius of atoms= R, The relation between R and the
FCC cell side ‘a’ as shown in the figure is 2 a  4 R (face diagonal)
 % Packing fraction by using ( ) = = 74%
 For BCC structure, effective number of atoms per unit cell (Z) is 8 x 1/8 + 1 = 2
 The relation between R and a is a  4 R (body diagonal)
 % Packing fraction by using ( ) = = 68 %
 In the Hexagonal unit cell, number of atoms (Z) =
12 corner atoms x 1/6 (shared by six unit cells) +
Two face centre atoms x 1/2 + 3 within the body = 6
Close structure Number of atoms
per unit cell ‘z’
Relation between edge length
‘a’ and radius of atom ‘r’
Packing Fraction
HCP and CCP or FCC 4 for CCP and FCC
6 for HCP
r = a/(2√2) 74%
BCC 2 r = (√3/4)a 68%
Simple cubic (SCC) 1 r = a/2 52.4%
Octahedral and Tetrahedral Voids:
Tetrahedral void (hole) – void surrounded by 4 spheres
Octahedral void (hole) – void surrounded by 6 spheres
Number of octahedral voids = Number of effective atoms present in unit cell
Number of tetrahedral voids = 2 × Number of effective atoms present in unit cell
So, Number of tetrahedral voids = 2 × Number of octahedral voids.
T = 2O
Lattice point (24) = Oh void (24) = Td void (48)
Radius of Td void = 0225 x radius of sphere
Radius of Oh void = 0.414 x radius of sphere

4. JFC Solid State Chemistry By Rawat Sir [M.Sc. Chemistry, 3 times NET (JRF), GATE ]
rwtdgreat@gmail.com
www.slideshare.net/RawatDAgreatt
Google+/blogger/Facebook/Twitter-@RawatDAgreatt +919808050301, +917017378183
4
Coordination numbers and radius ratio:
 For the stability of an ionic compound cation must be surrounded by maximum no. of anions and vice versa, for
greater electrostatic force of attraction.
 The no. of nearest anion (or cation) surrounding a particular cation (or anion) is called its coordination no. (C.N.)
 Radius ratio may be defined as the ration of radius of cation to the radius of anion. = radius ratio
 According to radius ratio rule, the coordination no. increases as the radius ratio increases and vice versa.
Coordination
numbers
Geometry Radius ratio (R.R.) Example
2 Linear R.R. < 0.155 BeF2
3 Planar Trigonal 0.155 ≤ R.R. < 0.225 AlCl3
4 Tetrahedron 0.225 ≤ R.R. < 0.414 ZnS
4 Square planar 0.414 ≤ R.R. < 0.732 PtCl4
2-
6 Octahedron 0.414 ≤ R.R. < 0.732 NaCl
8 Body centered cubic (BCC) 0.732 ≤ R.R. < 0.999 CsCl
To calculate the radius of Oh void consider this figure, for the sake of simplicity the above and below spheres have been
removed.
In triangle ACB, applying Pythagoras theorem-
(AB)2
= (AC)2
+(BC)2
(2R+2r)2
=(2R)2
+(2R)2
(2R+2r)2
= 8R2
2R+2r= 8𝑅 =2 2 R
2r= 2 2 R – 2R
2r = 2R ( 2 -1)
r = R(1.414-1)= 0.414R
r = 0.414R
r= radius of Oh void and R= radius of sphere
All spheres are touching one another,
Spheres are present at all eight corners,
Let the radius of spheres be R and that of tetrahedral void be r,
AB is face diagonal so, AB= 2 𝑎
Again as A sphere is touching B sphere, so AB = R+R = 2R
AC is body diagonal so, AC = 𝑎
Again as sphere A is touching sphere C,
so AC = R+r+r+R = 2R+2r=2(R+r)
𝐴𝐷
𝐴𝐶
[𝑣𝑎𝑙𝑢𝑒 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑠𝑖𝑑𝑒 𝑙𝑒𝑛𝑔𝑡ℎ] =
𝐴𝐷
𝐴𝐶
[𝑣𝑎𝑙𝑢𝑒 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑟𝑎𝑑𝑖𝑢𝑠]
𝑎
𝑎
=
(𝑅+𝑟)
𝑅
=1+
𝑟
𝑅
or − =
𝑟
𝑅
or 0.225 =
𝑟
𝑅

5. JFC Solid State Chemistry By Rawat Sir [M.Sc. Chemistry, 3 times NET (JRF), GATE ]
rwtdgreat@gmail.com
www.slideshare.net/RawatDAgreatt
Google+/blogger/Facebook/Twitter-@RawatDAgreatt +919808050301, +917017378183
5
Density formula-
d = Z is No. of particles in unit cell M is atomic mass or molecular mass in u or g/mol
a is cell parameter or edge length, d is density , NA is Avogadro’s no. 6.023x1023
d = M stands for mass and N stands for No. of Atoms
Packing in solids
 1D C.N. =2
 2D packing
 Square close AAAA… C.N. =4
 Hexagonal close ABABAB… C.N. =6
 3D close packing
 By using 2D square close packed structure,AAAAA…
SCC C.N. = 6
 By using 2D hexagonal close packed structure…..
HCP ABABAB… C.N. =12
CCP or FCC ABCABC… C.N. =12
Understanding HCP and FCC (CCP)-
 FCC and hexagonal crystal structures are most highly packed with packing
efficiency of 74%.
 There are two types of voids between the atoms – vertex up (b), and
vertex down (c). The atoms in the next layer sit on the b sites
 In FCC, atoms in the third layer sit over the c sites and this repeats giving
rise to ABC ABC ABC type of stacking.
 In HCP system,
centers of atoms
of the third layer
lie directly over
centers of atoms
of the first layer
(a positions)
giving rise to AB
AB AB type of
stacking.