Using molecular orbital theory, compare the bond energy and magnetic character of O{_{2}}^{+} and O{_{2}}^{-} species.

Answers (1)

The molecular orbital configuration of O{_{2}}^{+} and  O{_{2}}^{-} has been specified as follows: -

O^{2+}(15):\sigma (1s)^{2}\; \sigma ^{*}(1s)^{2}\; \sigma (2s)^{2}\sigma ^{*}(2s)^{2}\; \sigma (2p_{z})^{2}\pi ( 2p_{x}^{2}=\pi\ 2p_{y}^{2})\pi ^{*}2p_{x}^{1}

O^{2+}(15):\sigma (1s)^{2}\; \sigma ^{*}(1s)^{2}\; \sigma (2s)^{2}\sigma ^{*}(2s)^{2}\; \sigma (2p_{z})^{2}\pi ( 2p_{x}^{2}=\pi\ 2p_{y}^{2})\(pi ^{*}2p_{x}^{2}\pi ^{*}2p_{y}^{1})

Therefore, the bond order for O{_{2}}^{+}= 10-5/2 = 2.5

And the bond order for O{_{2}}^{-}= 10-7/2 = 1.5

We know that as per the molecular orbital theory, greater is the bond order, greater is the bond energy. Therefore, O{_{2}}^{+} is more stable than O{_{2}}^{-} .

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