(i) By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be: determined.
(ii) Let 'a' be the edge length of a unit cell of a crystal, 'd' be the density of the metal, 'm' be the atomic mass of the metal and 'z' be the number of atoms in the unit cell.
(iii) Now,
Density of the unit cell
\(=\frac{Mass\ of\ the\ unit\ cell}{Volume\ of\ the\ unit\ cell}\)
\(d=\frac{Z\times m}{a^3}\) ...(1)
[Since, mass of the unit cell = Number of atoms in the unit cell x Atomic mass]
[Volume of the unit cell = (edge length of the cubic unit cell)³]
(iv) From equation (1), We have
\(m=\frac{d\times a^3}{Z}\) ....(2)
(v) Now, 
Mass of the metal (M) \(=\frac{Atomic\ mass(M)}{Avogadro's\ number(N_A)}\)
M=\(\frac{d\times a^3 \times N_A}{Z}\)
(vi) From equation (3), we can determine the atomic mass of the unknown metal.

Molecular solids: 
In molecular solids, the constituents are neutral molecules. They are held together by weak Vander Waals forces. Generally molecular solids are soft and they do not conduct electricity. These molecular solids are further classified into three types.
(i) Non-polar molecular solids:
(a) In non-polar molecular solids constituent molecules are held together by weak dispersion forces or London forces.
(b) They have low melting points and are usually in liquids or gaseous state at room temperature.
Ex: Naphthalene, anthracene etc.,
(ii) Polar molecular solids:
(a) The constituents are molecules formed by polar covalent bonds. 
(b) They are held together by relatively strong dipole-dipole interactions.
(c) They have higher melting points than the nonpolar molecular solids. 
Ex: Solid CO₂, solid NH₃ etc.

(iii) Hydrogen bonded molecular solids:
(a) The constituents are held together by hydrogen bonds.
(b) They are generally soft solids under room temperature.
(c) Examples: solid ice (H₂O), glucose, urea etc.

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.

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.

\(M=\frac { d\times { a }^{ 3 }\times NA }{ g } \)
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) 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.

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%

  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.

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. 

Coordination number is defined as the number of nearest neighbours that an atom has in a unit cell.
Coordination no. of = 6
Coordination no. of fcc = 12
Coordination no. of bcc = 8

The empty spaces present between the metal atom or the ions when they are packed within the crystal are called voids.

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

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.

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