The correct order of bond dissociation energy among $N_{2}, O_{2}, O_{2}^{-}$ is shown in which of the following arrangements?

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$N_{2} > O_{2}^{-} > N_{2}$

$O_{2}^{-} > O_{2} > N_{2}$

$N_{2} > O_{2} > O_{2}^{-}$

$O_{2} > O_{2}^{-} > N_{2}$

answer is C.

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Detailed Solution

The molecular electroinc configuration along with their bond orders (BO) are as follows:

$\begin{array}{lll} N_{2} KK (σ 2 s)^{2} (σ * 2 s)^{2} {(π 2 p_{x})}^{2} {(π 2 p_{y})}^{2} {(σ 2 p_{z})}^{2}; BO = 3 \\ O_{2} KK (σ 2 s)^{2} (σ * 2 s)^{2} {(σ 2 p_{z})}^{2} {(π 2 p_{x})}^{2} {(π 2 p_{y})}^{2} {(π^{*} 2 p_{x})}^{1} {(π^{*} 2 p_{y})}^{1}; BO = 2 \\ O_{2}^{-} KK (σ 2 s)^{2} {(σ^{*} 2 s)}^{2} {(σ 2 p_{z})}^{2} {(π 2 p_{x})}^{2} {(π 2 p_{y})}^{2} {(π^{*} 2 p_{x})}^{2} {(π^{*} 2 p_{y})}^{1}; BO = 1 \frac{1}{2} \end{array}$ The larger the bond order, larger the dissociation energy.