Solid State

To determine the formula of the compound, we need to calculate the number of atoms of each type (A and B) in the unit cell of the crystal structure.

Firstly, let’s consider the A atoms which are located at the corners of the cube. In a cubic structure, each corner atom is shared among 8 adjacent unit cells. Thus, each corner contributes only $\frac{1}{8}$ of an atom to a single unit cell. Since there are 8 corners in a cube:

$\text{Number of A atoms} = 8 \times \frac{1}{8} = 1$

Next, consider the B atoms which are located at the face centers of the cube. Each face-centered atom is shared between 2 adjacent unit cells, therefore each face-centered atom contributes $\frac{1}{2}$ of an atom to a single unit cell. Since there are 6 faces in a cube:

$\text{Number of B atoms} = 6 \times \frac{1}{2} = 3$

Thus, in one unit cell, there is 1 A atom and 3 B atoms. Therefore, the formula of the compound is:

$\text{AB}_3$

So the correct answer is:

Option A: AB₃

Anti-fluorite structure refers to an anion array with tetrahedral cations and the compound having A₂B formula have anti-fluorite structure. So, Na₂O has an anti-fluorite structure in which Na⁺ ions occupy all the tetrahedral voids and O^2$-$ occupy half of the cubic holes.

Z_eff for fcc = 4

Z_eff for bcc = 2

Atomic volume of $\alpha$-form = ${{{{(3.68 \times {{10}^{ - 8}})}^3}} \over 4}$ $\times$ N_A

Atomic volume of $\beta$-form = ${{{{(2.92 \times {{10}^{ - 8}})}^3}} \over 2}$ $\times$ N_A

As atomic weight is same, element is same, so the density ratio is

${\rho _\alpha }:{\rho _\beta } = {V_\beta }:{V_\alpha } = {{{{(2.92)}^3}} \over 2}:{{{{(3.68)}^2}} \over 4}$

$ = 12.448:12.459 = 1:1$