Chemistry - Solid State Question with Solution | TestHub

In $\mathrm{CCP}$, anions occupy the face-centered and the corner positions of the cube.

Based on the position and number of voids in $\mathrm{CCP}$, for one oxide ion there are two tetrahedral voids and one octahedral void.

Since one-fifth of the tetrahedral voids are occupied by divalent cations ${\mathrm{X}}^{2+}$-

Number of divalent cations for each oxide ion$=2\times \frac{1}{5}$

Also, since half of the octahedral voids are occupied by trivalent cations ${\mathrm{Y}}^{3+}$-

Number of trivalent cations for each oxide ion$=1\times \frac{1}{2}$

The empirical formula gives the simplest ratio of all the atoms present in a molecule. So, the formula of the compound is ${\mathrm{X}}_{\frac{2}{5}}{\mathrm{Y}}_{\frac{1}{2}}\mathrm{O}$ or ${\mathrm{X}}_4{\mathrm{Y}}_5{\mathrm{O}}_{10}$