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Correct trend in the bond order is

Question and options

Correct trend in the bond order is
  1. O2+>O22−>O2\mathrm{O_2^{+}} > \mathrm{O_2^{2-}} > \mathrm{O_2}
  2. O22−>O2−>O2+\mathrm{O_2^{2-}} > \mathrm{O_2^{-}} > \mathrm{O_2^{+}}
  3. O22−>O2+>O2−\mathrm{O_2^{2-}} > \mathrm{O_2^{+}} > \mathrm{O_2^{-}}
  4. O2+>O2−>O22−\mathrm{O_2^{+}} > \mathrm{O_2^{-}} > \mathrm{O_2^{2-}}

Answer and explanation

Correct answer: D - \(\mathrm{O_2^{+}} > \mathrm{O_2^{-}} > \mathrm{O_2^{2-}}\)
MO bond orders: O2+=2.5\mathrm{O_2^{+}} = 2.5, O2=2.0\mathrm{O_2} = 2.0, O2−=1.5\mathrm{O_2^{-}} = 1.5, O22−=1.0\mathrm{O_2^{2-}} = 1.0. So O2+>O2−>O22−\mathrm{O_2^{+}} > \mathrm{O_2^{-}} > \mathrm{O_2^{2-}} is the correct trend. The other orders either rank peroxide higher than O2\mathrm{O_2}, reverse the trend, or put peroxide highest.

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