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

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

Answer 1

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

A cubic packed structure has a FCC lattice.
Oxide ions:
$\text{Total number of oxide ions}=Z_{eff}\text{for FCC}=4$
There are 8 tetrahedral voids and 4 octahedral voids in a FCC lattice .
For $X^{2+}$:
$\text{Number of ions}=\frac{1}{5}\times 8=\frac{8}{5}$
For $Y^{3+}$:
$\text{Number of ions}=\frac{1}{2}\times 4=2$
Molecular formula is :
$X_{\frac{8}{5}}{Y}_2{O}_4$
Simplest formula:
$X_4{Y}_5{O}_{10}$