Question

In a solid, oxide ions are arranged in ccp. One-sixth of tetrahedral voids are occupied by cation A while one-third of octahedral voids are occupied by cation B. What is the formula of compound?

• A. $A{B}_{2}O$
• B. $A_2{B}_{3}O$
• C. $AB{O}_3$
• D. $A_3{B}_{2}O$

Answer 1

Answer: (C)

The given solid has ccp arrangement.
Oxide ions are in ccp,
Thus,
$\text{Number of}{O}^{2-}$ $\text{ions}=n$,
$\therefore$
$\text{No. of octahedral voids}=n$
$\text{No. of tetrahedral voids}=2n$
Here,
$\frac{1}{6}$ of tetrahedral voids are occupied by cation A
$\text{No. of cation A}=\frac{1}{6}\times 2n=\frac{n}{3}$
$\frac{1}{3}$ of octahedral voids are occupied by cation B
$\text{No. of cation B}=\frac{1}{3}\times n=\frac{n}{3}$
Thus,
Ratio $\Rightarrow A:B:O^{2-}$
$=\frac{n}{3}:\frac{n}{3}:n$

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

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

$Theratioofcomponentsincrystalisasfollows$

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

Hence, the emperical formula is $AB{O}_3.$

$O_{3}becausethecomponentofOisintheratioof3$