N_2 and O_2 are converted into monocations N_ | vedclass

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(D) The electronic configuration of $N_2$ 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 or

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$N_2^+$ becomes diamagnetic

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(D) The electronic configuration of $N_2$ 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 = $3$).When $N_2$ is converted to $N_2^+$,an electron is removed from the bonding molecular orbital $(\sigma 2p_z)$,resulting in a bond order of $2.5$. Thus,the $N-N$ bond weakens.$N_2^+$ has $13$ electrons,so its configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. Since it has one unpaired electron,it is paramagnetic.The electronic configuration of $O_2$ is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$ (Bond order = $2$).When $O_2$ is converted to $O_2^+$,an electron is removed from the antibonding molecular orbital $(\pi^*)$,increasing the bond order to $2.5$ and decreasing paramagnetism.Therefore,the statement that $N_2^+$ becomes diamagnetic is wrong.

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

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