Element B forms ccp structure and A occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is:

We have a crystal in which the atoms of element B are arranged in a cubic close-packed (ccp or fcc) lattice. It is well known that a ccp unit cell contains $$4$$ atoms of the parent element. Before proceeding, let us recall the standard relationships between the atoms present in a close-packed array and the number of interstitial (void) sites produced by that array.

In a ccp (fcc) lattice:

• Number of octahedral voids $$=$$ number of close-packed atoms present.

• Number of tetrahedral voids $$=$$ twice the number of close-packed atoms present.

Stating these facts algebraically for a single unit cell:

$$\text{Number of B atoms} = 4$$

$$\text{Number of octahedral voids} = 4$$

$$\text{Number of tetrahedral voids} = 2 \times 4 = 8$$

According to the question, element A occupies exactly one-half of all the octahedral voids. Therefore

$$\text{Number of A atoms} = \dfrac{1}{2}\times(\text{number of octahedral voids}) = \dfrac{1}{2}\times4 = 2$$

Next, every tetrahedral void is filled by an oxygen atom. So

$$\text{Number of O atoms} = \text{number of tetrahedral voids} = 8$$

Collecting the numbers obtained per unit cell, we have the ratios

$$A : B : O = 2 : 4 : 8$$

To express the composition as the simplest integral formula, divide each subscript by the highest common factor, here $$2$$:

$$A : B : O = \dfrac{2}{2} : \dfrac{4}{2} : \dfrac{8}{2} = 1 : 2 : 4$$

So the empirical formula of the bimetallic oxide becomes

$$AB_2O_4$$

Examining the options given, this matches Option B.

Hence, the correct answer is Option B.