In short, solid-state chemistry is crucial for advancing materials science, electronics, energy storage, catalysis, and nanotechnology. It underlies the development of new materials with specific properties, impacting fields ranging from electronic devices and renewable energy to environmental science and geology.

1. The Solid State
Lecturer: SMITA GUPTA (B.Sc., M.Sc., Ph.D.)
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2. Subtopics
• 1.1 General characteristics of solid state
• 1.2 Amorphous and Crystalline Solids
• 1.3 Classification of Crystalline solids
• 1.4 Crystal lattices and unit cells
• 1.5 Number of atoms in a unit cell
• 1.6 Close-packed structures
• 1.7 Packing efficiency
• 1.8 Calculations involving Unit cell Dimensions
• 1.9 Imperfections in solids
• 1.10 electrical properties
• 1.11 Magnetic properties
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3. States of matter
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4. Amorphous and Crystalline solids
Quartz (Silicon oxygen-SiO2) 75 percent silica, 10 percent lime, and 15 percent soda
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8. 1.8 Calculations involving unit cell dimensions
• Volume of a unit cell = 𝑎3
• Mass of the unit cell = (no. of atoms in unit cell)*(mass of each atom)
• = z * m
Mass of an atom (m) present in the unit cell =
Molar mass (M)
Avogadro constant ( NA)
• m=
M
NA
• For example;
• Atomic mass of sodium in unit cell =
22.989 g/𝑚𝑜𝑙
6.022 x 1023 atoms/mol
= 3.817 x 10-23 g
1 mole = 6.022 x 1023 atoms/mol
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9. Avogadro’s constant determination
• The charge on a mole of electrons had been known for some time and is the constant called the
Faraday.
• The best estimate of the value of a Faraday, according to the National Institute of Standards and
Technology (NIST), is 96,485.3383 coulombs per mole of electrons.
• The best estimate of the charge on an electron based on modern experiments is 1.60217653 x 10-
19 coulombs per electron.
• If you divide the charge on a mole of electrons by the charge on a single electron you obtain a value
of Avogadro’s number of 6.02214154 x 1023 particles per mole.
• Avogadro’s number =
charge on a mole of electrons
charge on a single electron
• Avogadros number =
96,485.3383 coulombs per mole of electrons
1.60217653 x 10−19 coulombs per electron
• NA = 6.022 x 1023 atoms/mol
• 1 mole = 6.022 x 1023 atoms
No. of moles = mass (g)/molar mass (g/mol)
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10. Packing efficiency
Type of
structure
No. of atoms a and r
relationship
Packing
efficiency
Void space
FCC 4 a=
4 𝑟
2
= 2 2 r 74% 26%
BCC 2 a =
4 𝑟
3
68.04% 31.96%
Simple cubic
cell
1 a = 2 r 52.4% 47.6%
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11. 6. Copper crystallizes in fcc with a unit cell of 361 pm. What is the radius of copper atom?
(a) 109 pm
(b) 127 pm
(c) 157 pm
(d) 181 pm
a=
4 𝑟
2
r=
2 a
4
=
2 (361 pm)
4
= (1.414*361)/4 =127 pm
6. Sodium metal crystallizes as a BCC lattice with the cell edge 4.29 Å. What is the radius of Sodium atom?
(a) 1.857 x 10-8 cm
(b) 2.371 x 10-7 cm
(c) 3.817 x 10-8 cm
(d) 9.312 x 10-7 cm
a=
4 𝑟
3
r=
3 a
4
=
3 (4.29 x 10−8 cm)
4
=
1.732 (4.29 x 10−8 cm)
4
= 1.857 x 10−8 cm
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12. • Density of unit cell =
mass of unit cell
volume of unit cell
=
z .m
a3 =
z. M
a3. NA
• a= edge length of unit cell of a cubic crystal (determined by x-ray
diffraction)
• d = density of the solid substance
• M = molar mass
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13. 1. Potassium has a bcc Structure with nearest neighbour distance 4.52 Å. Its atomic weight is 39. Its
density will be:
(a) 454 kg/m3 (b) 804 kg/m3 (c) 852 kg/m3 (d) 908 kg/m3
Bcc
Z= 2
M=39
D =?
Density of unit cell =
mass of unit cell
volume of unit cell
=
z .m
a3 =
z. M
a3. NA
Density of unit cell =
mass of unit cell
volume of unit cell
=
z .m
a3 =
z. M
a3. NA
=
(2)(39 g/mol)
(5.219 × 10−10 m)3 (6.022 x 1023 atoms/mol)
=
78
(1.4215 𝑋 10−28)((6.022 x 1023 atoms/mol)
=
78
8.560 𝑋 10−5
= 911214.9 g/m3 =911 kg/m3
In bcc structure, the nearest neighbor of body-centered atom is the atom present at
corner of the atom.
Given,
a 3
2
= (4.52 x 10-10 m)
a = (4.52 x 10-10 m)(2)/ 3 = 5.219 x 10-10 m
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14. 2. The number of atoms in 100 g of an FCC crystal with density d = 10 g/cm3 and cell edge as 200 pm is
equal to
(a) 3 x 1025 (b) 0.5 x 1025 (c) 1 x 1025 (d) 2 x 1025
Fcc; Z= 4
a = 200 pm =200x 10-10 cm
m= 100 g
d = 10 g/cm3
M =
d . a3. NA
z.
=
10 g/cm3) (200 X 10−10 cm 3
(6.022 x 1023 atoms/mol)
4
= 12 g
12 g contains------- 6.022 x 1023 atoms
100 g contains------ 5.01 x 1024 atoms (50.1 x 1023 atoms or 0.5 x 1025 atoms)
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15. 3. An element (atomic mass= 100 g/mol) having bcc structure has unit cell edge 400 pm. Then density of
the element is
(a) 10.376 g/cm3 (b) 5.188 g/cm3 (c) 7.289 g/cm3 (d) 2.144 g/cm3
BCC; Z= 2
a = 400 pm 4200 x 10-10 cm
M= 100 g/mol
d =
d =
z. M
a3. NA
=
(2)(100 g/mol)
(400 × 10−10 cm)3 (6.022 x 1023 atoms/mol)
= 5.189 g/cm3
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16. 4. The unit cell of aluminium is a cube with an edge length of 405 pm. The density of aluminium is 2.70 g/cm3. What is
the structure of unit cell of aluminium?
(a) Body-centred cubic cell (b) Face-centred cubic cell (c) end-centred cubic cell (d) simple cubic cell
Z= ?
a = 405 pm = 405 X 10-10 cm
d = 2.70 g/cm3
Molar mass of aluminium= 26.98 g/mol
Z =
d . a3. NA
M.
=
2.70 g/cm3) (405 X 10−10 cm 3
(6.022 x 1023 atoms/mol)
26.98 g/mol
= 4
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17. 5. How many lithium atoms are present in a unit cell with edge length 3.5 Å 𝐚𝐧𝐝 𝐝𝐞𝐧𝐬𝐢𝐭𝐲 𝟎. 𝟓𝟑𝐠/𝐜𝐦𝟑
? (𝐀𝐭𝐨𝐦𝐢𝐜 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐋𝐢 = 𝟔. 𝟗𝟒)
(a) 2
(b) 1
(c) 4
(d) 6
Z= ?
a = 3.5 Å = 3.5 X 10-8 cm
d = 0.53 g/cm3
Molar mass of Li = 6.94 g/mol
Z =
d . a3. NA
M.
=
0.53 g/cm3) (3.5 X 10−8 cm 3
(6.022 x 1023 atoms/mol)
6.94 g/mol
= 2
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18. Voids
• The empty spaces inside a sphere are called “Voids”. The size and
shape of voids depend upon the type of unit cell and packing.
• Radius ratio
• The size of void is expressed in terms of radius ratio of a sphere that
can exactly fit in the void to the radius of surrounding spheres. This is
expressed as
• Radius ratio =
𝑟
𝑅
•
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19. Types of Void
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20. 0.155 < 0.225 < 0.414 < 0.732
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22. Radius ratio =
𝑟 (𝑐𝑎𝑡𝑖𝑜𝑛)
𝑅 (𝐴𝑛𝑖𝑜𝑛)
= 180/187 = 0.962
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24. In which compound 4: 4 coordination is found
(a) ZnS (b) CuCl (c) AgI (d) All of these
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26. 1.9 Imperfections in solids
Stoichiometric
defects
• Vacancy defect
• Interstitial defect
• Frenkel Defect
• Schottky Defect
Impurity defects
Non-Stoichiometric
defects
• Metal excess
defect
• Metal deficiency
defect
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27. Stoichiometric defects
Decrease
in density
of a
substance
Increase in
density of
a
substance
Ionic solids
No change
in density
ZnS, AgCl,
AgBr, AgI
Ionic solids
Decrease in
density of a
substance
NaCl, KCl,
CsCl, AgBr
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28. Impurity Defects
• If molten NaCl containing a little
amount of SrCl2 is crystallized,
some of the sites of Na+ ions are
occupied by Sr2+ (Fig 1.27)
• Each Sr2+ replaces 2 Na+ ions
• Example : CdCl2 and AgCl
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29. Non-Stoichiometric defects
Metal Excess Defect
Due to anionic vacancies
Examples: Alkali halides NaCl and KCl
Due to cations at interstitial sites
Metal Deficiency
Defect
FeO
Mostly found-(Fe0.95O)
Range: Fe0.93O -Fe0.96O
Fe2+ is replaced by required no. of Fe3+
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Example: ZnO

30. Metal Excess Defect
Due to anionic vacancies
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F-center or Farbenzenter for colour center
LiCl ------Pink
KCl------Violet

31. Metal Excess Defect Due to cationic vacancies
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33. Electrical properties
Conductors Insulators Semiconductors
Solids with conductivities ranging
between 104 to 107 ohm-1 m-1
Solids with conductivities in the
intermediate range
Solids with very low
conductivities
Example: Metals are good
conductors
(conductivity: 107 ohm-1 m-1)
(conductivity: 10-6 to 104 ohm-1
m-1)
(conductivity: 10-20 to 10-10 ohm-1
m-1)
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