Question

In a solid AB that has a rock salt structure, B atoms occupy the corners of the cubic unit cell. If all the face centred atoms along one of the axis are removed, then the stoichiometry of the resultant structure will be:

• A. $A_2{B}_3$
• B. $A_{2}B$
• C. $A_4{B}_3$
• D. $A_3{B}_4$

Answer 1

The correct option is C $A_4{B}_3$

AB has a rock-salt structure i.e., structure like $N{a}^{+}C{l}^{-}$, in which four $N{a}^{+}$ and four $C{l}^{-}$ ions are present. So, in AB four $A^{+}$ and four $B^{-}$ ions will be present. In the unit cell of AB, when two face centred ions (i.e., $B^{-}$) are removed. The contribution of two face-centred $B^{-}$ ions: $=2\times \frac{1}{2}=1$
Thus, the no. of $B^{-}$ ions left $=4-1=3$
No. of $A^{+}$ ions (unchanged) $=4$
Thus, the resultant stoichiometry will be $A_4{B}_3$