Question

In a solid $AB$ having $NaCl$ structure atoms, occupy the corners of the unit cell. If all the face centred atoms along one of the axis 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 $A_3{B}_4$

A occupies $8$ corners which contributes $\frac{1}{8}\Rightarrow \frac{1}{8}\times 8=1$

and $6$ face centered contributes $\frac{1}{2}\Rightarrow \frac{1}{2}\times 6=3$

But face centered atom along any one axis is removed which will be $2$.

Now, face centre have $4\times \frac{1}{2}=2$

$Z_{effective}$ of $A$ after removing face centres at any axis $=4\times \frac{1}{2}+\frac{1}{8}\times 8=3$

$B$ occupies octahedral voids which contributes $\frac{1}{4}$

Therefore, $1+12\times \frac{1}{4}=4=Z_{effective}$ of $B$

Formula $=A_3{B}_4$