N_2 and O_2 are converted into monoanions N_2 | vedclass

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(B) 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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In $O_2^-$,the $O-O$ bond order increases.

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(B) 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 = $(10-4)/2 = 3$. For $N_2^-$,the extra electron enters the $\pi^* 2p$ orbital. Bond order = $(10-5)/2 = 2.5$. Since bond order decreases,the $N-N$ bond weakens. $N_2^-$ has one unpaired electron,so 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 = $(10-6)/2 = 2$. For $O_2^-$,the extra electron enters the $\pi^* 2p$ orbital. Bond order = $(10-7)/2 = 1.5$. Since bond order decreases,the bond length increases. Therefore,the statement that the $O-O$ bond order increases is wrong.

$N_2$ and $O_2$ are converted into monoanions $N_2^-$ and $O_2^-$ respectively. Which of the following statements is wrong?

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