Answer:-

Given: Type of unit cell is fcc.

Density of iridium (ρ) = 22.4 g/cm³

Molar mass of iridium = 192.2 g/mol

To find: Radius of iridium atom (r)

Formula used:

1. Density (ρ) = M n / (a^3 N_A)

2. For fcc unit cell, r = 0.3535a

Calculation: For fcc unit cell, n = 4, using formula (i)

Density (ρ) = 22.4 g/cm³ = (192.2 g/mol × 4 atom) / (a^3 × 6.022 × 10²³ atom/mol)

a^3 = (192.2 × 4) / (22.4 × 6.022 × 10²³)

a = √³((192.2 × 4 × 10^-23) / (22.4 × 6.022))

a = √³((192.2 × 40 × 10^-24) / (22.4 × 6.022))

a = √³((192.2 × 40) / (22.4 × 6.022)) × 10^-8 cm

a = 3.849 × 10^-8 cm

a = 384.9 pm

Using formula (ii)

r = 0.3535a

r = 0.3535 × 384.9 pm

r = 135.7 pm (approximately 136 pm)

Radius of iridium atom (r) is 136 pm.