Question

In a solid, oxide ions $\left(O^{2-}\right)$ are arranged in ccp, cation $\left(A^{3+}\right)$ occupy one-sixth of tetrahedral void and cations $\left(B^{3+}\right)$ occupy one-third of the octahedral voids. What is the formula of the compound?

• A. $AB{O}_3$
• B. $A{B}_2{O}_3$
• C. $A_{2}B{O}_3$
• D. $AB{O}_2$

Answer 1

Answer: (A)

$AB{O}_3$

Number of $O^{2-}$ ions in packing $=4$

Number of octahedral voids $=4$

Number of tetrahedral voids $=8$

Number of $A^{3+}=\frac{1}{6}\times 8=\frac{4}{3}$

Number of $B^{3+}=\frac{1}{3}\times 4=\frac{4}{3}$

$A^{3+}:B^{3+}:O^{2-}=\frac{4}{3}:\frac{4}{3}:4=1:1:3$

Formula of the compound $=AB{O}_3$