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

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

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

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

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

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Solution

The correct option is C $N_{2} > O_{2} > O_{2}^{-}$
$N_{2}$ has bond order = 3
$O_{2}$ has bond order=2
$O_{2}^{-}$ has bond order = 1.5
As bond dissociation energy $\propto$ bond order
Thus order of bond dissociation energy is: $N_{2} > O_{2} > O_{2}^{-}$

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N_{2}^{+}

O_{2}

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{N_{2}}^{+}

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What is the effect of the following processes on the bond order in $N_{2}$ and $O_{2}$ ?

(i). $N_{2} \to N_{2}^{+} + e^{-}$

(ii) $O_{2} \to O_{2}^{+} + e^{-}$

$N_{2}$ and $O_{2}$ are converted into monocations, $N_{2}^{+}$ and $O_{2}^{+}$ respectively. Which of the following is wrong ?

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