In diamond, carbon atoms occupy FCC lattice points and alternate tetrahedral voids. If the edge leng — Solid State Chemistry Question

💡 Solution & Explanation

In diamond cubic structure, nearest neighbours lie along one fourth of the body diagonal, so the C-C distance is $\frac{\sqrt{3}a}{4}$. Since this equals $2r$, $r=\frac{\sqrt{3}a}{8}=\frac{1.732\times356}{8}\text{ pm}\approx77.1\text{ pm}$.