BThe formula of compound will be **AB**_{2}O_{4}

**Explanation:**

We all know that oxygen like bigger molecules occupy the lattice points. The effective number of atoms in a CCP unit cell is **8 x 1/8 + 6 x 1/2 = 4**

Also the total number of tetrahedral voids (X is presnt) in CCP is 8 so no. of divalent ions = 8/8 = 1

Number of Octahedral voids (Y is present) are 4 so 1/2 of no. of divalent ion will be 4/2 = 2

Number of oxide = present at corners + at face centre = 8 x 1/8 + 6 x 1/2 = 4

So formula of compound will be AB_{2}O_{4}