N_2 and O_2 are converted into N_2^+ and O_2^ | vedclass

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(D) Bond order $= 1/2 	imes (N_B - N_{AB})$.For $N_2$ ($14$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_

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$N_2^+$ becomes diamagnetic

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(D) Bond order $= 1/2 	imes (N_B - N_{AB})$.For $N_2$ ($14$ 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 $= 1/2 	imes (10-4) = 3$.For $N_2^+$ ($13$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\pi 2p_x)^2(\pi 2p_y)^2(\sigma 2p_z)^1$. Bond order $= 1/2 	imes (9-4) = 2.5$. Since bond order decreases,the $N-N$ bond weakens.For $O_2$ ($16$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^1(\pi^* 2p_y)^1$. Bond order $= 1/2 	imes (10-6) = 2$.For $O_2^+$ ($15$ electrons): $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^1$. Bond order $= 1/2 	imes (10-5) = 2.5$. Since bond order increases,bond length decreases and paramagnetism decreases (from $2$ unpaired electrons to $1$).$N_2^+$ has $1$ unpaired electron,so it is paramagnetic,not diamagnetic. Thus,statement $D$ is incorrect.

$N_2$ and $O_2$ are converted into $N_2^+$ and $O_2^+$ respectively. Which of the following is not correct?

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