Question

The density of iridium is 22.4 g/cm³. The unit cell of iridium is fcc. Calculate the radius of iridium atom. Molar mass of iridium is 192.2 g/mol.

Answer 1

Given : Crystal structure of iridium = fcc

Molar mass of iridium = 192.2 gmol^-1

Density = ρ = 22.4 gcm^-3

Radius of iridium = ?

In fcc structure, there are 8 Ir atoms at 8 corners and 6 Ir atoms at 6 face centres.

Total number of Ir atoms = \(\frac{1}{8}×8 +\frac{1}{2}×6 \) = 1+3 = 4

Mass of Ir atom = = 31.92 x 10^-23 g

∴ Mass of 4 Ir atoms = 4 x 31.92 x 10^-23 g = 1.277 x 10^-21 g

Mass of unit cell = 1.277 x 10^-21 g

Density of unit cell = \(\frac{\text{mass of unit cell}}{\text{volume of unit cell}}\)

22.4 = \(\frac{1.277×10^{-21}}{a^3}\)

a³ = \(\frac{1.277×10^{-21}}{22.4}\) = 57 x 10^-24 cm³

∴ a = \((57×10^{-24})^{\frac{1}{3}}\) = 3.848 x 10^-8 cm

If r is the radius of iridium atom, then for fcc structure,

r = \(\frac{a}{2\sqrt{2}}=\frac{3.848×10^{-8}}{2\sqrt{2}}\) = 1.36 x 10^-8 cm = 136 pm

Ans: Radius of iridium atom = 136 pm.