Atomic radius of a body-centred cubic metal

UKChO Round 2 · IChO · unreviewed

Density of a bcc metal known; find the cell edge, then the atomic radius.

Worked example

Given that iron is bcc (Z = 2) with ρ = 7.904 g cm⁻³ and M = 55.85 g mol⁻¹, find its metallic radius.

1. Density from a cubic unit cell: solve for Cell edge a
$a = \sqrt[3]{\dfrac{ZM}{\rho N_\mathrm{A}}}$
= 286.3 pm
2. Multiply by a coefficient: 0.4330127018922193 × Cell edge a → r = a√3 / 4 (atoms touch along the body diagonal)
= 124.0 pm

Answer: r(Fe) = 124 pm

Rearrange ρ = ZM/(N_A a³) for a, with Z = 2. In bcc the atoms touch along the body diagonal: 4r = a√3.

That uses the face diagonal (fcc). In body-centred cubic the atoms touch along the body diagonal: 4r = a√3.
A body-centred cubic cell contains two atoms: 8 corners × ⅛ + 1 in the centre.