In which of the following ionization processe | vedclass

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(C) The bond order $(BO)$ is calculated as $\frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibond

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$NO \rightarrow NO^+$

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(C) The bond order $(BO)$ is calculated as $\frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.For $NO$ ($15$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. $BO = \frac{1}{2}(10 - 5) = 2.5$. It is paramagnetic.For $NO^+$ ($14$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. $BO = \frac{1}{2}(10 - 4) = 3.0$. It is diamagnetic.In the process $NO \rightarrow NO^+$,the electron is removed from the antibonding $\pi^*$ orbital,which increases the bond order from $2.5$ to $3.0$ and changes the magnetic behavior from paramagnetic to diamagnetic.

In which of the following ionization processes,the bond order has increased and the magnetic behaviour has changed?

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