Answer: Radius of iridium atom \(r \approx 136\ \text{pm}\) (edge length \(a \approx 385\ \text{pm}\))
Given: \(\rho = 22.4\ \text{g cm}^{-3}\), \(M = 192.2\ \text{g mol}^{-1}\), fcc unit cell so \(n = 4\) atoms per unit cell, \(N_A = 6.022\times10^{23}\ \text{mol}^{-1}\).
Step 1: Find the edge length \(a\)
Density of a unit cell: \(\rho = \dfrac{nM}{a^3 N_A}\), so \(a^3 = \dfrac{nM}{\rho N_A}\).
\(a^3 = \dfrac{4\times192.2}{22.4\times6.022\times10^{23}} = \dfrac{768.8}{1.349\times10^{25}} = 5.70\times10^{-23}\ \text{cm}^3 = 57.0\times10^{-24}\ \text{cm}^3\)
\(a = \sqrt[3]{57.0}\times10^{-8}\ \text{cm} = 3.848\times10^{-8}\ \text{cm} = 384.8\ \text{pm}\)
Step 2: Relate radius to edge length
In an fcc cell the atoms touch each other along the face diagonal. The face diagonal \(\sqrt{2}\,a\) contains four radii, so \(4r = \sqrt{2}\,a\).
\(r = \dfrac{\sqrt{2}\,a}{4} = \dfrac{a}{2\sqrt{2}} = \dfrac{384.8\ \text{pm}}{2.828} \approx 136\ \text{pm}\)
Hence the radius of an iridium atom is about \(136\ \text{pm}\) (\(1.36\times10^{-8}\ \text{cm}\)).
