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

Data:
Diffraction angle,
2θ = 14.8⁰ ∴ 8 = 7.4°
Interplanar distance, d = 0.400 nm
Order of reflection, n = 2
Formula:
nλ = 2dsinθ
∴ wavelength,
Solution:
\(\lambda =\frac { 2dsin\theta }{ 2 } \)
= dsinθ
= 0.400 sin 7.4^o
= 0.400 x 1.288 nm
= 0.0512 nm
λ = 0.0512 nm

Simple Cubic Lattice
100,101,111

\({ d }_{ hkl }=\frac { a }{ \sqrt { { h }^{ 2 }+{ k }^{ 2 }+{ l }^{ 2 } } } \)
\({ d }_{ 100 }=\frac { 1 }{ \sqrt { { 1 }^{ 2 }+0^{ 2 }+0^{ 2 } } } =1\)
\({ d }_{ 101 }=\frac { 1 }{ \sqrt { { 1 }^{ 2 }+0^{ 2 }+0^{ 2 } } } =\frac { 1 }{ \sqrt { 2 } } \)
\({ d }_{ 111 }=\frac { 1 }{ \sqrt { { 1 }^{ 2 }+1^{ 2 }+1^{ 2 } } } =\frac { 1 }{ \sqrt { 3 } } \)
\(\\ { d }_{ 100 }:{ d }_{ 101 }:{ d }_{ 111 }=1:\frac { 1 }{ \sqrt { 2 } } :\frac { 1 }{ \sqrt { 3 } } (or)\)
=1:0.707:0.577

(i) (a) When AgCl is doped with CdCI₂, two Ag⁺ ions will be replaced by one Cd²⁺ ion to maintain electrical neutrality.
(b) Thus, a hole is created at the lattice site for every Cd²⁺ ion introduced. Hence the crystal becomes conductor.
(ii) p-type semiconductor is obtained, when Si is doped with boron.

(i) Yellow colour of ZnO is due to metal excess defect. Zinc oxide which is white in colour on heating loses oxygen and turns yellow.
\(ZnO\overset { Heating }{ \longrightarrow } { Zn }^{ + }+\frac { 1 }{ 2 } { O }_{ 2 }+{ 2e }^{ -1 }\)
(ii) The excess Zn²⁺ ion thus, formed get trapped into vacant interstitial sites and the electrons in the neighbouring interstitial sites.

(i) The given defect is Stoichiometric since the number of cations and anions are missing from lattice sites unequal.
(ii) The defect is Schottky defect.
These defects are vacancy defects shown by ionic solids in which the anion and cation are nearly of the same size.
Eg: NaCI and KCI shows Schottky defects

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.

Given:
1 mole of AI = Avogadro's constant = 6.023 x 10²³
Mass of aluminium = 8.1 g
Solution:
1 mole of aluminium (AI) = 27 g
= 6.02 X 10²³ atoms
∴ No. of atoms present in 1g of AI = \(\frac{6.02\times 10^{23}}{27}\)
Face centered cubic unit cell contains 4 atoms.
∴ No. of unit cells \(=\frac { 6.02\times { 10 }^{ 23 } }{ 27\times 4 } \times 8.1=4.52\times { 10 }^{ 22 }\)

Number of tetrahedral voids formed = 2 x No. of atoms of element Y
No. of atoms of element Y in the ccp unit cell = 4
No. of tetrahedral voids by atoms of Y = 2 x 4 = 8
∴ No. of tetrahedral voids occupied by atoms of 
X = \(\frac{2}{3}\times 8=\frac{16}{3}\)
Ratio of the no. of atoms of X and Y is = \(\frac{16}{3}:4\)
= 16:12
= 4:3
Hence, formula would be X₄Y₃.

(i) Covalent Solids: Silicon carbide, graphite.
(ii) Molecular Solids: Urea, benzene, ammonia, water and argon.
(iii) Ionic Solids: Zinc sulphide, potassium sulphate.
(iv) Metallic solids: Rubidium and tin.

(i) Compressibility is the ability of the substance to change its shape by applying external pressure.
(ii) Solids form closed packed structure with negligible intermolecular space and possesses very strong intermolecular forces of attraction.
(iii) Thus, they do not change their shape in the presence of external pressure.

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
\(=\frac { { 10 }^{ -3 } }{ 100 } \times 6.02\times { 10 }^{ 23 }=6.02\times 10^{ 18 }\)

\(M=\frac { d\times { a }^{ 3 }\times NA }{ g } \)
d = Density of the material
a = Length of the edge of the cell.
N_A = Avogadro number
Z = No. of atoms
\(M=\frac { 10.5{ gcm }^{ -3 }{ (4.07\times { 10 }^{ -6 }cm) }^{ 3 }\times \left( 6.023\times { 10 }^{ 23 }{ mol }^{ -1 } \right) }{ 4 } \)
Atomic mass of silver M = 107.08 g mol^-1.

(i) The intermolecular forces of attraction that are present in solids are very strong.
(ii) The constituent particles of solids cannot: move from their positions. They have fixed positions.
(iii) However, they can oscillate about their mean positions.
(iv) This is the reason solids are rigid.

Formula is Ni_o.96 O_1.00
So the ration of Ni = O = 96.00
So if there are 100 atom of oxygen, as atoms of Ni
Let the number of atoms of Ni⁺² = x
The number of atoms of Ni⁺² = 96 - x
Charge on Ni = charge on O
So that oxygen has charge = 2
3 (96 - x) + 2x = 2(100)
288 - 3x + 2x = 200
-x = -88
x = 88
Percentage of Ni + 2 = (atom of Ni⁺² / total number of atoms of Ni 100.)
= 100.(94/98) x 100 = 96%
Percentage of Ni⁺³ = 100 - Ni⁺²
= 100 - 96 = 4%