Question

In a solid AB having the NaCl structure, A atom occupies the corners of the cubic unit cell. If all the face-centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is:

• A. $A{B}_2$
• B. $A_{2}B$
• C. $A_4{B}_3$
• D. $A_3{B}_4$

Answer 1

The correct option is D

$A_3{B}_4$

AB $\to$ NaCl type Structure (fcc system)
$\therefore$ No. of A atoms $=8\times \frac{1}{8}+6\times \frac{1}{2}=4$ (per unit cell)
No. of B atoms $=12\times \frac{1}{4}+1=4$ (per unit cell)
No. of A atoms removed $=2\left(faces\right)\times \frac{1}{2}=1$ (per unit cell)
No. of A atoms left = 4 - 1 =3
Formula $\Rightarrow A_3{B}_4$