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 = $(10-4)/2 = 3$.In $N_2^+$,one electron is removed from the bonding molecular orbital $(\sigma 2p_z)$,so the configuration becomes $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. Bond order = $(9-4)/2 = 2.5$. Since the bond order decreases,the $N-N$ bond weakens and $N_2^+$ is paramagnetic.For $O_2$,the configuration 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 = $(10-6)/2 = 2$.In $O_2^+$,one electron is removed from the antibonding molecular orbital $(\pi^*)$,so the configuration becomes $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order = $(10-5)/2 = 2.5$. The bond order increases,and paramagnetism decreases as the number of unpaired electrons reduces from $2$ to $1$.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 of the following statements is wrong?

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