The correct order of bond dissociation energy | vedclass

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Question

(C) The bond order is directly proportional to the bond dissociation energy. Calculating the bond order using Molecular Orbital Theory: For $N_2$ ($14

Vedclass · Accepted Answer

$N_2 > O_2 > O_2^-$

Vedclass · Answer

(C) The bond order is directly proportional to the bond dissociation energy. Calculating the bond order using Molecular Orbital Theory: For $N_2$ ($14$ electrons): Bond order = $(10-4)/2 = 3$. For $O_2$ ($16$ electrons): Bond order = $(10-6)/2 = 2$. For $O_2^-$ ($17$ electrons): Bond order = $(10-7)/2 = 1.5$. Since the bond order follows the order $N_2 (3) > O_2 (2) > O_2^- (1.5)$,the bond dissociation energy follows the same order: $N_2 > O_2 > O_2^-$.

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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