The correct order to 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 $N_{2} > O_{2} > O_{2}^{-}$
The correct option is $C$ .
The bond dissociation energy of $N_{2}$ $= 946$
The bond dissociation energy of $O_{2}$ $= 493.6$
The bond dissociation energy of $O_{2}^{-}$ $= 395$
Hence according to dissociation energy the order is $N_{2} > O_{2} > O_{2}^{-}$ .

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Similar questions

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

N_{2}

N_{2}^{+}

O_{2}

O_{2}^{+}

N_{2}

{N_{2}}^{+}

O_{2}

{O_{2}}^{+}

N_{2}

O_{2}

O_{2}^{+}

{O_{2}}^{-}

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

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