An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of the faces of the cube. The empirical formula for this compound would be

The empirical formula of an ionic solid is obtained by counting the effective number of each kind of ion present in one unit cell and then writing their simplest whole-number ratio.

Step 1: Contribution of A ions (at the eight corners)
• Each cubic unit cell has $$8$$ corners.
• An ion located at a corner is shared by $$8$$ neighbouring unit cells, so its contribution to one cell is $$\frac{1}{8}$$.
Effective number of A ions
$$N_A = 8 \times \frac{1}{8} = 1$$

Step 2: Contribution of B ions (at the six face centres)
• A cube has $$6$$ faces.
• An ion at the centre of a face is shared by $$2$$ adjacent unit cells, so its contribution to one cell is $$\frac{1}{2}$$.
Effective number of B ions
$$N_B = 6 \times \frac{1}{2} = 3$$

Step 3: Empirical formula
The ratio of A : B ions inside one unit cell is $$1:3$$, giving the empirical formula $$AB_3$$.

Option C which is: AB$$_3$$