According to MO theory,which of the following | vedclass

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Question

(A) The molecular orbital configuration for $N_2$ ($14$ electrons) is: $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_y)

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

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(A) The molecular orbital configuration for $N_2$ ($14$ electrons) is: $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_y)^2(\sigma 2p_z)^2$. Bond order $(B.O.)$ $= \frac{10-4}{2} = 3$. For $N_2^-$ ($15$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_y)^2(\sigma 2p_z)^2(\pi^* 2p_x)^1$. $B.O. = \frac{10-5}{2} = 2.5$. For $N_2^{2-}$ ($16$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_y)^2(\sigma 2p_z)^2(\pi^* 2p_x)^1(\pi^* 2p_y)^1$. $B.O. = \frac{10-6}{2} = 2$. Comparing the bond orders: $2 Therefore,the increasing order is $N_2^{2-} < N_2^- < N_2$.

According to $MO$ theory,which of the following lists ranks the nitrogen species in terms of increasing bond order?

Similar Questions

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