What is the size of the largest atom that could be fit into octahedral holes of the lattice without — Solid State Chemistry Question

💡 Solution & Explanation

For FCC, $4R = \sqrt{2}a$, so $R = \frac{a}{2\sqrt{2}} = \frac{0.5/\sqrt{2}}{2\sqrt{2}} = 0.125$ nm. Radius of octahedral void $r = 0.414 \times R = 0.414 \times 0.125 = 0.05175 \approx 0.052$ nm. Therefore, correct answer is D.