Question

In a cubic close-packed structure of mixed oxides, the lattice is made up of oxide ions, one-tenth of the tetrahedral voids are occupied by divalent ($X^{2+}$) ions, while one-half of the octahedral voids are occupied by trivalent ions ($Y^{3+}$), then formula of the oxide is:

• A. $X{Y}_4{O}_4$
• B. $X_{2}Y{O}_4$
• C. $X_2{Y}_5{O}_{10}$
• D. $X_5{Y}_4{O}_{10}$

Answer 1

Answer: (C)

$X_2{Y}_5{O}_{10}$

Oxygen forms the CCP (fcc) lattice.

$Z_{O^{2-}}=4$

Effective number of $X^{2+}=\frac{1}{10}\times$ Tetrahedral voids $=\frac{1}{10}\times 8=\frac{4}{5}$

Effective number of $Y^{3+}=\frac{1}{2}\times$ Octahedral voids $=\frac{1}{2}\times 4=2$

Formula $\Rightarrow X_{\frac{4}{5}}{Y}_2{O}_4=X_{\frac{2}{5}}Y{O}_2=X_2{Y}_5{O}_{10}$