Question

In a solid ‘AB’ having the $NaCl$ structure, ‘A’ atoms occupy the corners of the cubic unit cell. If all the face-centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is.

• A. $AB_{2}$
• B. $A_{2}B$
• C. $A_{4}B_{3}$
• D. $A_{3}B_{4}$

Answer 1

Answer: (D)

$A_{3}B_{4}$

Answer 2

There were 6 A atoms on the face-centres removing face-centred atoms along one of the axes means removal of 2 A atoms.

Now, number of A atoms per unit cell

$= \underset{(\text{corners})}{8 \times \frac{1}{8}} + \underset{(\text{face} - \text{centred})}{4 \times \frac{1}{2} = 3}$

Number of B atoms per unit cell

$= \underset{(\text{edgecentred})}{12 \times \frac{1}{4}}$+ $\underset{(\text{bodycentred})}{\underset{}{1 = 4}}$

Hence the resultant stoichiometry is $A_{3}B_{4}$