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 onehalf of the octahedral voids are occupied by trivalent ions $(Y^{3+})$, 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

Answer: (C)

In ccp, anions occupy primitives of the cube while cations occupy voids. In ccp, there are two tetrahedral voids and one octahedral hole per anion.
For one oxygen atom there are two tetrahedral holes and one octahedral hole.
Since one-fifth of the tetrahedral voids are occupied by divalent cations $(X^{2+})$.
$\therefore$ Number of divalent cations in tetrahedral voids $=2\times \frac{1}{5}$
Since one-half of the octahedral voids are occupied by trivalent cations $(Y^{3+})$
$\therefore$ Number of trivalent cations $=1\times \frac{1}{2}$
So, the formula of the compound is
${\left(X\right)}_{2\times \frac{1}{5}}{\left(Y\right)}_{\frac{1}{2}}{\left(O\right)}_1$ or $X_{\frac{2}{5}}{Y}_{\frac{1}{2}}{O}_1$ or $X_4{Y}_5{O}_{10}$