\(\operatorname{Density}(\rho)=\frac{\mathrm{nM}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}} \)
\(\rho=10 \mathrm{~g} \mathrm{~cm}^{-3} ; \mathrm{a}=100 \mathrm{pm}=1 \times 10^{-8} \mathrm{~cm} ; \mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23} ; \mathrm{n}=4 ; \mathrm{M}=? \)
\(M=\frac{\rho \mathrm{a}^{3} \mathrm{N_{A}}}{n} \)
\(=\frac{10 \times\left(1 \times 10^{-8}\right)^{3} \times 6.023 \times 10^{23}}{4}  \)
\(=\frac{6.023}{4} \)
= 1.505 g /mol
No. of moles \(=\frac{\text { Mass }}{\text { Molar mass }}=\frac{1}{1.505}\)
= 0.664 moles
Hence number of atoms = 0.664 x 6.023 x 10²³ = 3.99 x 10²³ atoms

The structural units of an ionic crystal are cations and anions. They are bound together by strong electrostatic attractive forces. To maximize the attractive force, cations are surrounded by as many anions as possible and vice versa. Hence they are hard and brittle.

Number of atoms in a fcc unit cell = \(\frac{N_{c}}{8}+\frac{N_{f}}{2}=\frac{8}{8}+\frac{6}{2}=1+3=4\)

The imperfection occurs due to missing atoms, displaced atoms or extra atoms, is named as a point defect. Such defects arise due to imperfect packing during the original crystallisation or they may arise from thermal vibrations of atoms at elevated temperatures.

a. P₄ -  Covalent solid
b. Brass - Metallic solid
c. Diamond - Covalent solid
d. NaCl - Ionic solid
e. Iodine - Covalent solid

(i) A basic repeating structural unit of a crystalline solid is called a unit cell.
(ii) A crystal is consisted of large number of unit cells.

density \(\rho=\frac{\mathrm{n} \mathrm{M}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}}\) = 4 (face centered cubic)
a = 352.4 pm = 3.524 x 10^-8 m
ρ = 8.9 gcm^-3
N^A = Avogadro number = 6.023 x 10^-23
\(M=\frac { r { a }^{ 3 } N_A }{ n } \)
\(M=\frac { 8.9 \times {( 3.524 \times 10^{-6}) }^{ 3 }\times 6.023 \times 10^{-23} }{ 4} \)
\(=\frac { 8.9 \times {( 3.524) }^{ 3 }\times 10^{-24} \times 6.023 \times 10^{-23} }{ 4} \)
= 586.47 x 10^-1 = 58.65 g
58.65 g of the element has 6.023 x 10^-23 atoms
100g of the element has \(=\frac { 6.023 \times 10^{-23} }{ 58.65} \times 100\)
= 0.1026 x 10²³ atoms
= 1.026 x 10²⁴ atoms

(i) Frenkel defect arises due to the dislocation of ions from its crystal lattice.
(ii) The ion which is missing from the lattice point occupies an interstitial position.
(iii) This defect is shown by ionic solids in which cation and anion differ in size.
(iv) Unlike Schottky defect, this defect does not affect the density of the crystal.
For example AgBr, in this case, small Ag⁺ ion leaves its normal site and occupies an interstitial position.

1. The number of nearest neighbours that surrounding a particle in a crystal is called the coordination number of that particle.
2. The coordination number of atoms in a bcc structure is '8'.

(i) Schottky defect arises due to the missing of equal number of cations and anions from the crystal lattice. This effect does not change the stoichiometry of the crystal.
(ii) Ionic solids in which the cation and anion are of almost of similar size show schottky defect.
Example: NaCl.
(iii) Presence of large number of schottky defects in a crystal, lowers its density.
(iv) Presence of Schottky defect in the crystal provides a simple way by which atoms or ions can move within the crystal lattice.

There are seven types of unit cell, Cubic, tetragonal, orthorhombic, hexagonal, monoclinic, triclinic and rhombohedral. They differ in the arrangement of their crystallographic axes and angles.
i) Cubic: a = b = c; α = β = ૪ = 90^o.

ii) Tetragonal: a = b ≠ c; α = β = ૪ = 90°.

iii) Orthorhombic:  a  ≠ b  ≠ c; α = β = ૪ = 90°.

iv) Hexagonal: a = b ≠ c; α = β = 90^o, ૪ = 120^o. 

v) Monoclinic: a ≠ b ≠ c; α = ૪ = 90^o, β ≠ 90^o, 

vi) Triclinic: a ≠ b ≠ c; α  ≠ β ≠ ૪ ≠ 90^o.

vii) Rhombohedral:  a = b = c; α = β = ૪ ≠ 90^o. 

(i) Ionic solids have high melting points.
(ii) These solids do not conduct electricity, because the ions are fixed in their lattice positions.
(iii) They are hard so strong external force can change the relative positions of ions.

In bcc unit cell, ΔABC
AC² = AB² + BC²
\(AC=\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \)
\(\\ AC=\sqrt { { a }^{ 2 }+{ a }^{ 2 } } =\sqrt { { 2a }^{ 2 } } =\sqrt { 2 } a\)

In ΔACG
AG² = AC² + CG²
\(AG=\sqrt { { AC }^{ 2 }+{ CG }^{ 2 } } \)
\(AG=\sqrt { { \left( \sqrt { 2a } \right) }^{ 2 }+{ a }^{ 2 } } \)
\(AG=\sqrt { { 2a }^{ 2 }+{ a }^{ 2 } } =\sqrt { { 3a }^{ 2 } } \)
\(AG=\sqrt { 3a } \)
\(\sqrt { 3 } a=4r\)
\(r=\frac { \sqrt { 3 } }{ 4 } a\)
∴ Volume of the sphere with radius 'r' \(=\frac { 4 }{ 3 } { \pi r }^{ 3 }\)
\(=\frac{4}{3}\pi { \left( \frac { \sqrt { 3 } }{ 4 } a \right) }^{ 3 }\)\(=\frac { \sqrt { 3 } }{ 16 } \pi { a }^{ 3 }\)
Number of spheres belong to a unit cell in BCC arrangement is equal to two and hence the total volume of all spheres.
(i) Packing fraction = \(=\frac{Total \quad volume \quad occupied \quad by \quad spheres \quad in \quad a \quad unit \quad cell}{volume \quad of \quad the \quad unit \quad cell}\times100\)
\(\therefore\)Volume of all spheres \(=2\times \left( \frac { \sqrt { 3 } \pi { a }^{ 3 } }{ 16 } \right) =\frac { \sqrt { 3 } \pi { a }^{ 3 } }{ 8 } \)
Packing fraction \(=\frac { \left( \frac { \sqrt { 3 } \pi { a }^{ 3 } }{ 8 } \right) }{ ({ a }^{ 3 }) } \times 100\)
\(=\frac { \sqrt { 3 } \pi }{ 8 } \times 100\)
\(\\ =\sqrt { 3 } \pi \times 12.5\)
= 1.732 x 3.14 x 12.5
= 68%

AAAA type of three dimensional packing:

1. This is simple cubic arrangement.
2. Three dimensional packing arrangement can be obtained by repeating the AAAA type two dimensional arrangements in three dimensions.
3. Spheres in one layer sitting directly on the top of in the previous layer so that all layers are identical.
4. All spheres of different layers of crystal are perfectly aligned horizontally and also vertically.
5. In simple cubic packing, each sphere is in contact with 6 neighbouring spheres
6. Four in its own layer, one above and one below and hence the coordination number of the sphere in simple cubic arrangement is 6.
ABABA type of three dimensional packing:

(i) This is body centered cubic arrangement.
(ii) The spheres in the first layer are slightly separated and the second layer is formed by arranging the spheres in the depressions between the spheres in layer A.
(iii) The third layer is a repeat of the first.
(iv) This pattern ABABAB is repeated throughout the crystal.
(v) Each sphere has a coordination number of 8, four neighbors in the layer above and four in the layer below.
ABCABC type of three dimensional packing:

(i) This is face centered cubic arrangement.
(ii) In this arrangement (FCC) second layer spheres are arranged at the dips of first layer. Third layer spheres are arranged in a manner such that it cover the octahedral void.
(iii) Then no longer third layer is similar to first or second layer.
(iv) Third layer gives different arrangement. Fourth layer spheres are similar to first layer.
(v) If the first, second and third layer are represented as A, B, C then this type of packing gives the arrangement of layers as ABCABC.. and the sequence is repeated.

Metal excess defect:
(i) It arises due to the presence of more number of metal ions as compared to anions.
(ii) Examples: NaCl, KCl
(iii) The electrical neutrality of the crystal can be maintained by the presence of anionic vacancies equal to the presence of extra cation.

(iii) For example, when NaCI crystals are heated in the presence of sodium vapour, Na⁺ ions are formed and are deposited on the surface of the crystal.
(iv) Chloride ions (Cl^-) diffuse to the surface from the lattice point and combines with Na⁺ ion.
(v) The electron lost by the sodium vapour diffuse into the vacancy created by the Cl^- ions.
(vi) Such anionic vacancies which are occupied by unpaired electrons are called F centers. Hence, the formula of NaCl can be written as Na_1+xCl.
Metal deficiency defect:

(i) Metal deficiency defect arises due to the presence of less number of cations than the anions. This defect is observed in a crystal in which, the cations have variable oxidation states.
(ii) For example, In FeO crystal, some of the Fe²⁺ ions are missing from the crystal lattice. To maintain the electrical neutrality, twice the number of other Fe²⁺ ions in the crystal is oxidized to Fe³⁺ ions. In such cases, overall number of Fe²⁺ and Fe³⁺ ions is less than the O^2- ions.

For a fcc structure = r_x+ / r_y- = 0.414
Given that r_x⁺ = 100 pm
r_y = 100pm/0.414

\(3 \sqrt{a} = r_{C_{s+}} + 2r_{C_{f-}} + r_{C_{s+}} \)
\(\left( \frac { \sqrt { 3 } }{ 2 } \right) a = ​​ r_{C_{s+}} + r_{C_{f-}} \)
\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \times 400\)= inter ionic distance

If the total number of M atoms is n, then the number of tetrahedral voids = 2n
Given that \(\left( \frac { 1 }{ 3 } \right) ^{rd}\) of tetrahedral voids are occupied.
i.,e \(\left( \frac { 1 }{ 3 } \right) \times 2n\) are occupied by N atoms
\(\therefore\)M : N = n : \(\left( \frac { 2 }{ 3 } \right) n\)
= 1 : \(\frac { 2 }{ 3 }\)
Hence M₃N₂ = 3 : 2

Lattice points are occupied by CO₂ molecules

Number of A ions = N_c/8 = 8/8 = 1
Number of B ions = N_f/2 = 6/2 = 3
Simplest formula = AB₃