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 ($X^{++}$) ions, while one half of the octahedral voids are occupied by trivalent ($Y^{+++}$) 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

The correct option is A $X{Y}_2{O}_4$

Actually the bigger molecules like oxygen occupy the lattice points. In $CCp$ unit cell the effective number of atoms occupying lattice points $=8\times \frac{1}{8}+6\times \frac{1}{2}=4$

Again $CCp$ the number of tetrahedral voids is $8$. So divalents ions will be $8/8$ i.e. $1$ and number of octehedral voids are $4$, so one half would be $2$.

Finally the formula of oxide is ${XY}_2{O}_4$.