In a solid 'AB' having the NaCl structure, 'A' atoms occupy the corners and face centres of the cubi — Solid State Chemistry Question

💡 Solution & Explanation

Initially A is at corners ($8 \times \frac{1}{8} = 1$) and faces ($6 \times \frac{1}{2} = 3$). Removing face-centered atoms along one axis removes exactly 2 face atoms, leaving 4. Remaining A at faces = $4 \times \frac{1}{2} = 2$. Total A = $1 + 2 = 3$. B is in all octahedral voids (edges + body center), so total B = $12 \times \frac{1}{4} + 1 = 4$. The formula is $A_3B_4$.