Question

In a face centred cubic lattice, atoms of $A$ form the comer points and atoms of $B$ form the face centred points. If two atoms of $A$ are missing from the comer points, the formula of the ionic compound is :

• A. $A{B}_3$
• B. $A{B}_4$
• C. $A_2{B}_5$
• D. $A{B}_2$

Answer 1

Answer: (B)

A form comer points and two atoms of $A$ are missing from comer
$\therefore$ Atoms at comer $\left(A\right)=6\times \frac{1}{8}=\frac{3}{4}$
Atoms at face centre $\left(B\right)=6\times \frac{1}{2}=3$
$\therefore A_{3/4}{B}_{3}i.e.,A{B}_4$