Question

In a cubic packed structure of a compound, the lattice is made up of oxide ions, one-fifth of the tetrahedral voids are occupied by X ions while half of the octahedral voids are occupied by Y ions. Then the formula of the oxide will be

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

Answer 1

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

Number of $O^{2-}$ ions in the unit cell = 4
Number of $X$ ions in the unit cell = $\frac{1}{5}\times 8$ = 1.6
Number of $Y$ ions in the unit cell = $\frac{1}{2}\times 4$ = 2
We can condense the formula as $X_{1.6}{Y}_2{O}_4$
Upon stoichiometrically multiplying by 5/2, we get: $X_4{Y}_5{O}_{10}$