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JEE Main 2016 · Chemistry · Solutions, Solid State and Surface Chemistry · medium

In a solid, atoms of element A form a cubic close-packed lattice. Atoms of element B occupy all the tetrahedral voids and exactly half of the…

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In a solid, atoms of element A form a cubic close-packed lattice. Atoms of element B occupy all the tetrahedral voids and exactly half of the octahedral voids. The formula of the compound is
  1. A. A5B2\mathrm{A_5B_2}
  2. B. A4B5\mathrm{A_4B_5}
  3. C. A2B5\mathrm{A_2B_5}
  4. D. AB2\mathrm{AB_2}
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Answer: (C) A2B5\mathrm{A_2B_5}

A ccp (fcc) unit cell contains 44 atoms of A, with 2×4=82\times4=8 tetrahedral voids and 44 octahedral voids. B occupies all 88 tetrahedral voids plus half of the 44 octahedral voids, i.e. 22, so B=8+2=10B=8+2=10. The ratio A:B=4:10=2:5\mathrm{A:B}=4:10=2:5, giving A2B5\mathrm{A_2B_5}. A5B2\mathrm{A_5B_2} results from inverting the ratio, and AB2\mathrm{AB_2} from filling only the tetrahedral voids.
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