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In a crystalline solid, having formula $\mathrm{XY}_2 \mathrm{O}_4$, oxide ions are arranged in cubic close packed lattice while, cations X are present is tetrahedral voids and cations Y are present in octahedral voids. The percentage of tetrahedral voids occupied by $X$ is
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$12.5 \%$
In a cubic close packed lattice of oxide ions, there would be two tetrahedral and one octahedral void per oxide ion.
Since, the formula shows the presence of 4 oxide ions, the number of tetrahedral voids is eight and that of octahedral voids is four. Out of the eight tetrahedral voids, one is occupied by X .
$\therefore$ Percentage of tetrahedral voids occupied
$=\frac{1}{8} \times 100=12.5 \%$
Since, the formula shows the presence of 4 oxide ions, the number of tetrahedral voids is eight and that of octahedral voids is four. Out of the eight tetrahedral voids, one is occupied by X .
$\therefore$ Percentage of tetrahedral voids occupied
$=\frac{1}{8} \times 100=12.5 \%$
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