Question

A crystallises in BCC arrangement and ions B occupy ${\frac{1}{3}}^{rd}$ of tetrahedral voids. What is the formula of the crystal?

• A. $A_3{B}_4$
• B. $A_3{B}_2$
• C. $A_5{B}_2$
• D. $A_3{B}_4$

Answer 1

The correct option is B

$A_3{B}_2$

In a BCC lattice, the total effective number of atoms = $8\times \frac{1}{8}+1=2$
So number of A = $2$
Again total number of tetrahedral voids = $4$
So effective number of B = $\frac{4}{3}$
So the formula of the crystal would be $A_2{B}_{\frac{4}{3}}=A_3{B}_2$