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 $\left(X^{++}\right)$ ions, while one half of the octahedral voids occupied triavlent ions $\left(Y^{+++}\right),$ then the formula of the oxides.

• A. $X_2{Y}_2{O}_4$
• 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 $X_4{Y}_5{O}_{10}$

Number of Tetrahedral voids $=8$
Number of Octahedral voids $4$
$Z_{X^{2+}}=\frac{1}{5}\times 8$
$Z_{Y^{3+}}=\frac{1}{2}\times 4=2$
$O^{2-}$ forms FCC lattice.
$Z_{O^{2-}}=4$
$X_{8/5}{Y}_2{O}_4=X_4{Y}_5{O}_{10}$