In a solid 'AB' having structure, 'A' atoms occupy corners and face centres. If all face-centred ato — Solid State Chemistry Question

💡 Solution & Explanation

Initially, A atoms in FCC contribute $8\times\frac18+6\times\frac12=4$ atoms per unit cell. Removing the two face-centred A atoms along one principal axis removes $2\times\frac12=1$ A atom, so A becomes $3$. The B atoms in the $NaCl$ arrangement remain $4$. Hence the formula becomes $A_3B_4$.