Q. In a solid $‘AB’$ having the $NaCl$ structure, $‘A$’ atoms occupy 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) $AB_{2}$
B) $A_{2}B$
C) $A_{3}B_{4}$
D) $A_{4}B_{3}$
Explanation
(c)
There are 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
$= 8\ \times \frac{1}{8} + 4\ \times \ \frac{1}{2} = 3$
(corners) (face- centred)
Number of $B$-atoms per unit cell
$= 12 \times \frac{1}{4} + \ \ \ 1 = \ 4$
(edge centred) (body
Centred)
Hence, the resultant stoichiometry is $A_{3}B_{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.