The unit cell of a binary alloy composed of A and B metals has a ccp structure with A atoms occupying the corners and B atoms occupying centres of each face of the cube. If during the crystallisation of this alloy in the unit cell, two A atoms are missed, the overall composition per unit cell is

Since the unit cell of the compound has CCP structure in which A atoms are present at corners and B are present at face centers. Also, 2 of the atoms of A are missing.

So, number of atoms of A present at the corner =8−2=6

Where 8 is the total corners of cube and 2 is the missing.

Number of atoms of B present at face centers =6

Now, contribution of each present atom at the corner = $\frac{1}{2}$

Contribution of each atom present at face center = $\frac{1}{2}$

So, number of atoms of A present in unit cell = $\frac{6}{8}=\frac{3}{4}$

Number of atoms of B present in unit cell = $\frac{6}{2}=3$

Now, $\mathrm{A}:\mathrm{B}=\frac{3}{4}:3$ 

So, the formula of the given compound is ${\mathrm{AB}}_4$