Question

A solid ${}^{\prime }A{B}^{\prime }$ has a $NaCl$ structure where $B^{-}$ ions occupy the corners of the unit cell. If all the face centred atoms along one axis are removed then the resultant stoichiometry of the solid is:

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

Answer 1

The correct option is C $A_3{B}_3$

$B^{-}$ ions are present at corners and face centres, and $A^{+}$ ions are in the octahedral voids, i.e., at the edge centres and the body centre. A unit cell will contain $4A^{+}$ and $4B^{-}$ ions. In this, one axis removed, which means that two face centred ions and the body centred ion is removed. $B^{-}$ lies on the face centres and $A^{+}$ is at the body centre.
Contribution of two face centred $B^{-}$ ions $=2\times \frac{1}{2}=1$
In the resultant unit cell;
$A^{+}$ ions present per unit cell = 4 - 1 = 3
$B^{-}$ ions present per unit cell = 4 - 1 = 3
$\therefore$ Stoichiometry = $A_3{B}_3$