Ncert Exercises & Solutions

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

According to molecular orbital theory electronic configurations of

O_{2}^{+}

and

O_{2}^{-}

species are as follows

O_{2}^{+} : {(σ 1 s)}^{2} {(σ * 1 s)}^{2} {(σ 2 s)}^{2} {(σ * 2 s)}^{2} {(σ 2 p_{z})}^{2} (π 2 {p_{x}}^{2}, π 2 {p_{y}}^{2}) (π * 2 {p_{x}}^{1})

Bond order of

O_{2}^{+} = \frac{10 - 5}{2} = \frac{5}{2} = 2.5

O_{2}^{-} : {(σ 1 s)}^{2} {(σ * 1 s)}^{2} {(σ 2 s)}^{2} {(σ * 2 s)}^{2} {(σ 2 p_{z})}^{2} (π 2 {p_{x}}^{2}, π 2 {p_{y}}^{2}) (π * 2 {p_{x}}^{2}, π * 2 {p_{y}}^{1})

Bond order of

O_{2}^{-} = \frac{10 - 7}{2} = \frac{3}{2} = 1.5

Higher bond order of

O_{2}^{+}

shows that it is more stable than

O_{2}^{-}

. Both the species have unpaired electrons. So, both are paramagnetic in nature.