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(C) Stability of a molecule is directly proportional to its Bond Order $(B.O.)$.For $N_2$ ($14$ electrons): $B.O. = 3$. For $N_2^+$ ($13$ electrons):

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$N_2^+ > N_2 > N_2^-$

Vedclass · Answer

(C) Stability of a molecule is directly proportional to its Bond Order $(B.O.)$.For $N_2$ ($14$ electrons): $B.O. = 3$. For $N_2^+$ ($13$ electrons): $B.O. = 2.5$. For $N_2^-$ ($15$ electrons): $B.O. = 2.5$. Since $N_2^+$ and $N_2^-$ have the same $B.O.$,stability is determined by the number of antibonding electrons. $N_2^+$ is more stable than $N_2^-$. Thus,$N_2 > N_2^+ > N_2^-$ is correct.For $O_2$ ($16$ electrons): $B.O. = 2$. For $O_2^+$ ($15$ electrons): $B.O. = 2.5$. For $O_2^{+2}$ ($14$ electrons): $B.O. = 3$. Thus,$O_2^{+2} > O_2^+ > O_2$ is correct.For $O_2^{-2}$ ($18$ electrons): $B.O. = 1$. Thus,$O_2^{+2} > O_2 > O_2^{-2}$ is correct.Option $C$ states $N_2^+ > N_2 > N_2^-$,which is incorrect because $N_2$ has the highest $B.O.$ $(3)$ and is the most stable.

Which of the following options is incorrect regarding the stability of diatomic molecules?

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