Test: JEE Previous Year Questions- Solid State - JEE MCQ

B.C.C - Points are at corners and one in the centre of the unit cell.
Number of atoms per unit cell = 8 x 18 + 1 = 2.

F.C.C - Points are at the corners and also centre of the six faces of each cell.
Number of atoms per unit cell = 8 x 18 + 6 x 12 = 4.

Mass (m) = density x volume = 1 g

Mol wt of NaCl = 58.5 g

Number of unit cell present in a cube shaped crystal of NaCl of mass 1 g = m x Na / (M x Z)

In NaCl each unit cell has 4 NaCl units.

Z = 4

Here the crystal defect is the paired vacancies caused by the ions missing from their lattice sites. 
 

Unit cell consists of A ions at the corners is 8. So we know that the 

contribution of corner is = 1/8th of corner = 1/8 x 8 = 1. 

Unit cell consists of B ions at the centre of the six faces 

Face = 1/2th of face 

= 1/2 x 6= 3 

So the empirical formula for this compound will be AB₃ 

Radius of the atom = a ÷ a√2

=361÷2√2

=127.56

No. of Y atoms per unit cell in ccp lattice = 4

 No. of tetrahedral voids = 2 × 4 = 8

 No. of tetrahedral voids occupied by X = 2/3 × 8 = 16/3

 Therefore, Formula of the compound = X_16/3Y₄

= X₁₆Y₁₂

= X₄ Y₃