Question

In the following the correct bond order sequence is :

• A. \(\mathrm{O_2^{2-} \gt O_2^{+} \gt O_2^{-} \gt O_2}\)
• B. \(\mathrm{O_2^{+} \gt O_2^{-} \gt O_2^{2-} \gt O_2}\)
• C. \(\mathrm{O_2^{+} \gt O_2 \gt O_2^{-} \gt O_2^{2-}}\)
• D. \(\mathrm{O_2 \gt O_2^{-} \gt O_2^{2-} \gt O_2^{+}}\)

Hint:

Bond order = (bonding electrons - antibonding electrons)/2; adding electrons to the pi* orbitals lowers the bond order.

Answer 1

The Correct Option is C

Use MO theory (valence electrons): \(\mathrm{O_2}\) has 12 electrons, with bond order \(2\).
\(\mathrm{O_2^{+}}\): remove one \(\pi^*\) electron, B.O. \(= \dfrac{10-5}{2} = 2.5\).
\(\mathrm{O_2}\): B.O. \(= \dfrac{10-6}{2} = 2\).
\(\mathrm{O_2^{-}}\) (superoxide): B.O. \(= \dfrac{10-7}{2} = 1.5\).
\(\mathrm{O_2^{2-}}\) (peroxide): B.O. \(= \dfrac{10-8}{2} = 1\).
Order: \(\mathrm{O_2^{+} \gt O_2 \gt O_2^{-} \gt O_2^{2-}}\).
Answer: (3) \(\mathrm{O_2^{+} \gt O_2 \gt O_2^{-} \gt O_2^{2-}}\)