Density \(\rho=\frac{\mathrm{n} \mathrm{M}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}}\)
Here, n = 2 ; M = 137.3 gmol^-1 ;
a = 5.08pm = 5.08 x 10^-8cm
\(p=\frac{2\times 137.3}{(5.03\times10^{-8})(6.023\times 10^{23})}\)
\(=\frac{274.6}{(5.08)^{3}\times 10^{-24}\times 6.023\times 10^{23}}\)
\(=\frac{274.6}{(131.09)(6.023)}\times 10\)
\(=\frac{2746}{789.6}\)=3.5 g cm^-3

(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.

Given: NaCl is doped with 10^-2 mole % of SrCl₂
(i.e.) 100 moles of NaCl doped with 10^-2 moles of SrCl₂
\(\therefore \) 1 mole of NaCl is doped with 10^-4 moles of SrCl₂
1 Sr²⁺  ion creates 1 cation vacancy
The number of cation vacancies created by 10^-4 mole SrCl₂ = 10^-4 \(\times\) 6.023 \(\times\) 10²³
= 6.023 \(\times\) 10¹⁹ vacancies

For cubic closed packed structure
\(r =\frac{a \sqrt{2}}{4}  \)
\(\therefore a =\frac{4 r}{\sqrt{2}}  \)
\(=\frac{4 \times 1.25 \times 10^{-8}}{1.414}=3.53 \times 10^{-8} \mathrm{~cm} \)
= 353 pm

\(\operatorname{Density}(\rho)=\frac{\mathrm{nM}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}} \)
\(\mathrm{n}=2, \mathrm{~N}_{\mathrm{A}}=6.023 \times 10^{23} ; \mathrm{a}=288 \mathrm{pm}=2.88 \times 10^{-8} \mathrm{~cm}, \rho=7.2 \mathrm{~g} \mathrm{~cm}^{-3} \)
\(\therefore M=\frac{\rho \times a^{3} \times N_{A}}{n} \)
\(=\frac{7.2 \times\left(2.88 \times 10^{-8}\right)^{3} \times 6.023 \times 10^{23}}{2} \)
\(=517.95 \times 10^{-1} \)
\(=51.795 \mathrm{~g} \mathrm{~mol}^{-1} \)
Number of moles (n) \(=\frac{\text { Mass }}{\text { Molar mass }}=\frac{208}{51.795}\)=4.02 moles
No. of atoms = No. of moles \(\times\) Avogadro number
=n \(\times\) N_A
\(=4.01 \times 6.023 \times 10^{23} \)
\(=24.15 \times 10^{23} \text { atoms }\)

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.

  Tetrahedral voids Octahedral voids. 1. A void surrounded by 4 spheres occupying the corners of a tetrahedron. A void surrounded by 6 spheres along the corners of an Octahedron 2. It is formed when a sphere of second layer placed above the void of first layer. All the voids of first row which remain unoccupied from octahedral. 3. Co-ordination number is 4. Co-ordination number is 6. 4. The number of tetrahedral voids are equal to 2n. The number of Octahedral voids are equal to n.

	hcp structure	ccp structure
1.	This is 'aba' pattern of arrangement.	This is 'abc' pattern of arrangement.
2.	The spheres can be arranged so as to fit into the depression in such a way that the third layer is directly over a first layer.	The third layer may be placed over the second layer in such a way that all the spheres of the third layer fit in octahedral voids.
3.	The tetrahedral voids of the second layer are covered by the spheres of the third layer.	This arrangement of the third layer is different from other two layers and the stacking of layers continued.
4.	6 spheres are present	4 spheres are present

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.

For bcc structure \((r)=\frac{\sqrt{3}}{4} a\)
a = 4.3 \(\times\) 10^-8 cm, r = ?
\(=\frac{1.732 \times 4.3 \times 10^{-8}}{4}\)
\(r=1.86 \times 10^{-8} \mathrm{~cm}\)

Number of X type atoms in the unit cell \(=8 \times \frac{1}{8}=1\)
Number of Y type atoms in the unit cell \(=1 \times \frac{1}{1}=1\)
Hence the formula is XY (or) X₁Y₁