Question

In a structure, $A$ atoms have FCC arrangement, $B$ atoms occupy all the tetrahedral sites and $C$ atoms occupy half the octahedral sites. The formula of the compound is:

• A. $A_2{B}_2{C}_2$
• B. $A_2{B}_{4}C$
• C. $A{B}_{2}C$
• D. $A_4{B}_{2}C$

Answer 1

Answer: (B)

$A_2{B}_{4}C$

$A$ has FCC arrangement. So, the number of atoms of $A$ is $4$.

$B$ atoms occupy all the tetrahedral sites. So, the number of atoms of $B$ is $8$.

$C$ atoms occupy half the octahedral sites. So, the number of atoms of $C$ is $2$.

Hence, the formula is $A_4{B}_8{C}_2$ or $A_2{B}_{4}C$.

Option B is correct.