JEE Advanced 2020 · Paper 1, Chemistry question 12

Contents of the cell.

• M at the centres of 2 faces: $2\times\frac12=1$.
• X at the midpoints of 4 edges: $4\times\frac14=1$.

So the formula is MX, and (A) is true.

Coordination. Place M at $(0,\frac12,\frac12)$. Its nearest X are the edge midpoints $(\pm\frac12,\,0\text{ or }1,\,0\text{ or }1)$: eight of them at the corners of a cube. By symmetry each X also has eight M at the corners of a cube (a CsCl-type arrangement). The two geometries are the same, so (B) is false.

Bond length. From $(0,\frac12,\frac12)$ to $(\frac12,0,0)$:
$$\sqrt{\tfrac14+\tfrac14+\tfrac14}\,a=0.866\,a.$$
(C) is true.

Radius ratio. In eightfold (cubic) coordination the limiting ratio is 0.732, not 0.414 (which belongs to octahedral holes). (D) is false.

(A), (C)