Crystal defects are classified as follows
(i) Point defects
(ii) Line defects
(iii) Interstitial defects
(iv) Volume defects
Point defects are further classified as follows

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%

Volume of unit cell = (2.88\(\mathring { A } \))³ = 23.9 x 10^-24cm³
Volume of 100 g of the metal \(=\frac { m }{ A } =\frac { 100g }{ 7.2{ gcm }^{ -3 } } \) = 13.9 cm³
Number of unit cells in this volume \(=\frac { 13.9{ cm }^{ 3 } }{ 23.9\times { 10 }^{ -24 }{ cm }^{ 3 } } \) = 5.82 x 10²³

i) Crystals have long range repeated pattern of arrangement of constituent particles but in the process of crystallisation.
ii) Some deviations from the ideal arrangement may be introduced because the constituent particles may not get sufficient time to arrange themselves in a perfect order. Therefore, crystals are usually not perfect.

Doping of NaCI with 10^-3 mol% SrCl₂ means that 100 moles of NaCl are doped with 10^-3 mol SrCl₂.
∴ 1mole of NaCl is doped with SrCl₂
\(=\frac { { 10 }^{ -3 } }{ 100 } \times 6.02\times { 10 }^{ 23 }=6.02\times 10^{ 18 }\)

Gallium.

\(\operatorname{Density}(\rho)=\frac{\mathrm{nM}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}} \)
\(\rho=10 \mathrm{~g} \mathrm{~cm}^{-3} ; \mathrm{a}=100 \mathrm{pm}=1 \times 10^{-8} \mathrm{~cm} ; \mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23} ; \mathrm{n}=4 ; \mathrm{M}=? \)
\(M=\frac{\rho \mathrm{a}^{3} \mathrm{N_{A}}}{n} \)
\(=\frac{10 \times\left(1 \times 10^{-8}\right)^{3} \times 6.023 \times 10^{23}}{4}  \)
\(=\frac{6.023}{4} \)
= 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.

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.

sc ⇒ 2r = a
⇒ r = a/2
bcc ⇒ 4r = \( \sqrt{3a} \) ⇒ r = \(\frac { \sqrt{3a}} {4}\)
fcc ⇒ 4r ⇒  \( \sqrt{2a} \) ⇒ r = \(\frac { \sqrt{2a}} {4} = \frac { a} {2\sqrt{2a}}\)
\(\left( \frac { a }{ 2 } :\frac { \sqrt { 3 } }{ 4 } a:\frac { a }{ 2\sqrt { 2 } } \right) \)

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