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 divalent $\left(X^{+2}\right)$ ions, while one-half of the octahedral voids are occupied by trivalent ions $\left(Y^{+3}\right)$, then the formula of the oxide is:

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

Answer 1

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

In a cubic packed structure,

$O^{2-}$ present in ccp=4 $O^{2-}$ ions

Tetrahedral voids=8 and Octahedral voids=4

Thus, $X^{+2}$ ions=$\frac{8}{5}$

$Y^{+3}$ ions=$\frac{4}{2}=2$

So, the molecular formula is $X_{\frac{8}{5}}{Y}_2{O}_4$.

In simplest form, $X_4{Y}_5{O}_{10}$