Q50 / 180 · JIPMER 2018 · Chemistry

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

  • Correct answer
    A
    $12.5 \%$
  • B
    $25 \%$
  • C
    $50 \%$
  • D
    $75 \%$

Solution

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 $=\frac{1}{8} \times 100=12.5 \%$

Hence, the correct answer is option: 'A'.

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