Question

In a compound $X{Y}_2{O}_4$ oxide ions are arranged in $CCP$ and cations $X$ are present in octahedral voids. Cations $Y$ are equally distributed between octahedral and tetrahedral voids. The fraction of the octahedral voids occupied is:

• A. $1/2$
• B. $1/4$
• C. $1/8$
• D. $1/6$

Answer 1

The correct option is A $1/2$

$\text{No of octahedral voids in a CCP unit cell}$

$=(\frac{1}{4}\times 12)+1=3+1=4$

Since, 12 Octahedral voids are present of cube shared by 4 more unit cells and

$\begin{array}{c}\text{1. No. of tetrahedral voids in}CCP=8\text{(2 tetrahedral}\\ \text{void at each body diagonal}\times 4\text{body diagonal)}\\ \text{Given:}-{XY}_2{O}_4\text{in CCP}\end{array}$

$\begin{array}{c}\text{X present in Octahedral void}\\ \text{and}Y\text{equally distributed in Octahedral and}\\ \text{tetrahedral void.}\end{array}$

$\begin{array}{c}\therefore \text{Octahedral void occupied}=2(\begin{array}{c}1\text{from}x\text{and}\text{from}y)\end{array}\end{array}$

$\begin{array}{c}\text{fraction of Octahedral void occupied}=\frac{2}{4}=\frac{1}{2}\\ \text{so correct answer is}\left(A\right)1/2\text{.}\end{array}$