Question

In a solid AB having NaCl structure, A atoms forms cubic unit cell type structure. 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 $A_3{B}_4$

In NaCl structure:
$N{a}^{+}$ are present in body centre and edge centre.
$C{l}^{-}$ are present at corners and face centre
$\Rightarrow \left(C{l}^{-}\right)$ total 'A' removed $=\frac{1}{2}\times 2=1$
$\left(N{a}^{+}\right)$ total ‘B’ removed = 0
A remaining = 4 – 1 = 3
B remaining = 4
Formula $=A_3{B}_4$