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

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

(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%

(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.

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. 

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.

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