N_2 and O_2 are converted to monocations N_2^ | vedclass

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(D) Bond order $= 1/2 	imes (N_B - N_{AB})$.For $N_2$ ($10$ valence electrons): $(\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z

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$N_2^+$ becomes diamagnetic.

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(D) Bond order $= 1/2 	imes (N_B - N_{AB})$.For $N_2$ ($10$ valence electrons): $(\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$. Bond order $= 1/2 	imes (8 - 2) = 3$.For $N_2^+$ ($9$ valence electrons): $(\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^1$. Bond order $= 1/2 	imes (7 - 2) = 2.5$. Since bond order decreases,the $N-N$ bond weakens. $N_2^+$ has one unpaired electron,so it is paramagnetic.For $O_2$ ($12$ valence electrons): $(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. Bond order $= 1/2 	imes (8 - 4) = 2$.For $O_2^+$ ($11$ valence electrons): $(\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$. Bond order $= 1/2 	imes (8 - 3) = 2.5$. Since bond order increases,the $O-O$ bond strengthens. The number of unpaired electrons decreases from $2$ to $1$,so paramagnetism decreases.Statement $D$ is wrong because $N_2^+$ is paramagnetic,not diamagnetic.

$N_2$ and $O_2$ are converted to monocations $N_2^+$ and $O_2^+$ respectively. Which statement is wrong?

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