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

The ratio of the bond order of O2^+ to that of O2^-, and a comparison of their unpaired electrons, is respectively

Single correct option

The ratio of the bond order of O2+\mathrm{O_2^+} to that of O2−\mathrm{O_2^-}, and a comparison of their unpaired electrons, is respectively
  1. A. 5/35/3; O2+\mathrm{O_2^+} has more unpaired electrons
  2. B. 3/53/5; O2−\mathrm{O_2^-} has more unpaired electrons
  3. C. 5/35/3; both have the same number of unpaired electrons
  4. D. 5/45/4; O2+\mathrm{O_2^+} has more unpaired electrons
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Answer: (C) 5/35/3; both have the same number of unpaired electrons

Valence MO electron counts: O2+\mathrm{O_2^+} has 11 (8 bonding, 3 antibonding): bond order =(8−3)/2=2.5=(8-3)/2 = 2.5. O2−\mathrm{O_2^-} has 13 (8 bonding, 5 antibonding): bond order =(8−5)/2=1.5=(8-5)/2 = 1.5. Ratio =2.5/1.5=5/3=2.5/1.5 = 5/3. Each ion has an odd electron count with exactly one unpaired electron in a π∗\pi^* orbital (O2+\mathrm{O_2^+}: π∗1\pi^{*1}; O2−\mathrm{O_2^-}: π∗3\pi^{*3}, one half-filled orbital). So both are paramagnetic with one unpaired electron each.
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