In a crystalline solid, anion A is arranged in FCC type unit cell arrangement. Cation B is equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied, what is the chemical formula of the solid? 

A₂B

The correct answer is option 2. A₂B.

Anion Arrangement: A occupies all FCC lattice points. In an FCC unit cell, there are 4 anions (one at each corner and one in the body center).

Cation Distribution: B is equally distributed between octahedral and tetrahedral voids. However, since we only care about the formula when all octahedral voids are occupied, the number of cations in tetrahedral voids is irrelevant for determining the formula.

Octahedral Voids: Given that all octahedral voids are occupied by cations B, the number of B cations filling these voids is x (where x represents the total number of cations).

Therefore, the ratio of cations (A) to anions (B) becomes:

Cations: x (from octahedral voids)

Anions: 4 (from FCC lattice points)

This translates to a 2:1 ratio of A:B, resulting in the chemical formula A₂B.

From the choices provided, only A₂B reflects this 2:1 ratio.