Using MO theory predict which of the following species has the shortest bond length?

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$O_{2}^{-}$

$O_{2}^{2 +}$

${O_{2}}^{2 -}$

answer is B.

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Detailed Solution

$Molecularorbitalconfigurationof O_{2}^{+} : σ_{1 s^{2}} > σ_{1 s^{2}}^{*} > σ_{2 s^{2}} > σ_{2 s^{2}}^{*} > σ_{2 p_{x}^{2}} > π_{2 p_{y}^{2}} = π_{2 {p_{z}}^{2}} > π_{2 {p_{y}}^{1}}^{*} = π_{2 {p_{z}}^{0}}^{*} Bondorderof O_{2}^{+} = \frac{N_{b} - N_{a}}{2} = \frac{10 - 5}{2} = 2 .5 Molecularorbitalconfigurationof O_{2}^{-} : σ_{1 s^{2}} > σ_{1 s^{2}}^{*} > σ_{2 s^{2}} > σ_{2 s^{2}}^{*} > σ_{2 {p_{x}}^{2}} > π_{2 {p_{y}}^{2}} = π_{2 {p_{z}}^{2}} > π_{2 {p_{y}}^{2}}^{*} = π_{2 p_{z} 1}^{*} Bondorderof O_{2}^{-} = \frac{10 - 7}{2} = 1 .5 Molecularorbitalconfigurationof O_{2}^{2 -} : σ_{1 s^{2}} > σ_{1 s^{2}}^{*} > σ_{2 s^{2}} > σ_{2 s^{2}}^{*} > σ_{2 {p_{x}}^{2}} > π_{2 p_{y} 2} = π_{2 {p_{z}}^{2}} > π_{2 {p_{y}}^{2}}^{*} = π_{2 p_{z} 2}^{*} Bondorderof O_{2}^{2 -} = \frac{10 - 8}{2} = 1 Molecularorbitalconfigurationof O_{2}^{2 +} : σ_{1 s^{2}} > σ_{1 s^{2}}^{*} > σ_{2 s^{2}} > σ_{2 s^{2}}^{*} > σ_{2 {p_{x}}^{2}} > π_{2 p_{y} 2} = π_{2 {p_{z}}^{2}} > π_{2 {p_{y}}^{0}}^{*} = π_{2 {p_{z}}^{0}}^{*} Bondorderof O_{2}^{2 +} = \frac{10 - 4}{2} = 3 Bondorder \propto \frac{1}{Bondlength} So, O_{2}^{2 +} hastheshortestbondlength .$