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JEE Main 2018 · Chemistry · Solutions, Solid State and Surface Chemistry · easy

In a crystal, anions X form a cubic close-packed (ccp) lattice. Cations Y occupy half of the tetrahedral voids and cations Z occupy half of the…

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In a crystal, anions X\mathrm{X} form a cubic close-packed (ccp) lattice. Cations Y\mathrm{Y} occupy half of the tetrahedral voids and cations Z\mathrm{Z} occupy half of the octahedral voids. The formula of the compound is
  1. A. X2Y2Z\mathrm{X_2Y_2Z}
  2. B. XY2Z\mathrm{XY_2Z}
  3. C. XYZ\mathrm{XYZ}
  4. D. X2YZ\mathrm{X_2YZ}
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Answer: (A) X2Y2Z\mathrm{X_2Y_2Z}

Take 44 anions per unit cell (ccp). There are 22 tetrahedral voids per anion, i.e. 88 per cell, and 11 octahedral void per anion, i.e. 44 per cell. Half of the tetrahedral voids give 44 Y\mathrm{Y}; half of the octahedral voids give 22 Z\mathrm{Z}. The ratio is X:Y:Z=4:4:2=2:2:1\mathrm{X:Y:Z}=4:4:2=2:2:1, so the formula is X2Y2Z\mathrm{X_2Y_2Z}.
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