Question

In a cubic packed structure of mixed oxides, the lattice is made up of oxide ions, one-fifth of tetrahedral voids are occupied by X ion while half of the octahedral voids are occupied by Y ion. What will be the formula of oxide mixture?

• A. $X_5{Y}_4{O}_{10}$
• B. $X_4{Y}_5{O}_{10}$
• C. $X{Y}_2{O}_4$
• D. $X_{2}Y{O}_4$

Answer 1

The correct option is B $X_4{Y}_5{O}_{10}$

The given oxide mixture is in cubic closed packing, where the lattice is made of oxide ion.
$\therefore$
$\text{Number of}{O}^{2-}\text{ions in the unit cell}=n$
$\text{Number of tetrahedral voids}=2n$
$\text{Number of octahedral voids}=n$

Here,
X ion occupies one-fifth of tetrahedral voids. Thus,
$\text{Number of X ion}=\frac{2n}{5}$

Also, Y ion occupies half of the octahedral voids. Thus,
$\text{Number of Y ion}=\frac{n}{2}$

$\begin{array}{c}X:Y:O\\ \frac{2n}{5}:\frac{n}{2}:n\\ 4:5:10\end{array}\}\text{Simplest formula}=X_4{Y}_5{O}_{10}$