In a solid AB, having the NaCl structure, A atoms occupy the corners of the cubic unit cell. If all — Solid State Chemistry Question

💡 Solution & Explanation

A atoms form the FCC lattice: $8 \times (1/8)$ from corners + $6 \times (1/2)$ from faces = 4. B atoms occupy octahedral voids = 4. Removing face-centred atoms along one axis removes 2 face-centred A atoms. Remaining A atoms = $1 (\text{corners}) + 4 \times (1/2) (\text{remaining faces}) = 3$. B atoms remain 4. The formula is $A_3B_4$. Therefore, correct answer is D.