The bond stability of N_2,N_2^{-},and N_2^{2- | vedclass

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Question

(C) The stability of a molecule is directly proportional to its bond order: Stability $\propto$ Bond order. Using Molecular Orbital Theory $(MOT)$,the

Vedclass · Accepted Answer

$N_2^{2-} < N_2^{-} < N_2$

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(C) The stability of a molecule is directly proportional to its bond order: Stability $\propto$ Bond order. Using Molecular Orbital Theory $(MOT)$,the electronic configurations are: $N_2$ ($14$ electrons): $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$. Bond order = $(10-4)/2 = 3$. $N_2^{-}$ ($15$ electrons): $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2 (\pi^* 2p_x)^1$. Bond order = $(10-5)/2 = 2.5$. $N_2^{2-}$ ($16$ electrons): $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. Bond order = $(10-6)/2 = 2$. Since the bond order follows the order $N_2 > N_2^{-} > N_2^{2-}$,the stability follows the same order: $N_2^{2-} Thus,the correct option is $(C)$.

The bond stability of $N_2$,$N_2^{-}$,and $N_2^{2-}$ varies as:

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