Question

In a face centered cubic lattice, atom $\left(A\right)$ occupies the corner positions and atom $\left(B\right)$ occupies the face centre positions. If one atom of $\left(B\right)$ is missing from one of the face centered points, the formula of the compound is:

• A. $A_2{B}_5$
• B. $A_2{B}_3$
• C. $A{B}_2$
• D. $A_{2}B$

Answer 1

The correct option is A $A_2{B}_5$

In a face centered cubic lattice, atom $\left(A\right)$ occupies the corner positions. There are $8$ corner positions and each position contributes one eighth to the unit cell. Hence, total number of $\left(A\right)$ atoms per unit cell.
$=\frac{1}{8}\times 8=1$
Atom $\left(B\right)$ occupies face centre positions.
There are six face centre positions in one unit cell. One atom of $\left(B\right)$ is missing from one of the face centered points. Thus, there are $5$ face centre positions that are occupied with $\left(B\right)$. Each such positions contributes one half to the unit cell. Hence, total number of $\left(B\right)$ atoms per unit cell
$=\frac{1}{2}\times 5=2.5$
The formula of the compound is $A{B}_{2.5}\text{or}{A}_2{B}_5$