In a face centred cubic lattice, atoms of A form the corner points and atoms of B form the face centred points. If two atoms of A are missing from the corner points, the formula of the ionic compound is :

In a face centred cubic (FCC) lattice, atoms of A are located at the corner points, and atoms of B are located at the face centred points. Normally, an FCC unit cell has 8 corners and 6 faces. Each corner atom is shared by 8 adjacent unit cells, so the contribution of one corner atom to a unit cell is $$\frac{1}{8}$$. Similarly, each face centred atom is shared by 2 unit cells, so the contribution per face atom is $$\frac{1}{2}$$.

Without any missing atoms, the total contribution of A atoms (corners) would be $$\frac{1}{8} \times 8 = 1$$ atom per unit cell. However, the problem states that two atoms of A are missing from the corners. Therefore, only 6 corner atoms are present. The contribution of A atoms per unit cell is now $$\frac{1}{8} \times 6 = \frac{6}{8} = \frac{3}{4}$$.

For B atoms at the face centres, there are 6 faces, and no atoms are missing. So, the contribution of B atoms per unit cell is $$\frac{1}{2} \times 6 = 3$$.

Thus, in the unit cell, we have $$\frac{3}{4}$$ atom of A and 3 atoms of B. To find the simplest ratio of A to B, we write the ratio A : B as $$\frac{3}{4} : 3$$. Multiplying both sides by 4 to eliminate the fraction gives $$3 : 12$$. Dividing both sides by 3 simplifies the ratio to $$1 : 4$$. Therefore, the formula of the compound is AB₄.

Comparing with the options: A is AB₃, B is AB₄, C is A₂B₅, D is AB₂. Hence, the correct answer is Option B.