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JEE Main 2019 · Chemistry · Chemical Bonding and Molecular Structure · medium

The total number of lone pairs of electrons present on the central atoms in XeF2, XeO3 and XeOF4 taken together is

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The total number of lone pairs of electrons present on the central atoms in XeF2\mathrm{XeF_2}, XeO3\mathrm{XeO_3} and XeOF4\mathrm{XeOF_4} taken together is
  1. A. 4
  2. B. 6
  3. C. 7
  4. D. 5
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Answer: (D) 5

Xe has 8 valence electrons. XeF2\mathrm{XeF_2}: two bond pairs, so (8−2)/2=3(8-2)/2 = 3 lone pairs. XeO3\mathrm{XeO_3}: three Xe=O\mathrm{Xe{=}O} using six electrons, leaving 22 electrons = 1 lone pair. XeOF4\mathrm{XeOF_4}: four σ\sigma (Xe-F) and one Xe=O\mathrm{Xe{=}O} using 6 electrons, leaving 1 lone pair (square pyramidal). Total =3+1+1=5=3+1+1 = 5.
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