In which of the following processes,the bond | vedclass

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(B) The electronic configuration of $NO$ ($15$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi

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$NO \rightarrow NO^{+}$

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(B) The electronic configuration of $NO$ ($15$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. It has one unpaired electron,so it is paramagnetic. Bond order $= (10-5)/2 = 2.5$. After losing one electron,$NO^{+}$ ($14$ electrons) has the configuration $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. It has no unpaired electrons,so it is diamagnetic. Bond order $= (10-4)/2 = 3$. Thus,in the process $NO \rightarrow NO^{+}$,the bond order increases from $2.5$ to $3$ and the character changes from paramagnetic to diamagnetic.

In which of the following processes,the bond order increases and paramagnetic character changes to diamagnetic one?

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