In the conversion of N_2 into N_2^+,the elect | vedclass

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(B) The electronic configuration of $N_2$ ($14$ electrons) is: $KK \sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^2 = \pi 2p_y^2 \sigma 2p_z^2$.When $N_2$ is conv

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$\sigma 2p_z$

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(B) The electronic configuration of $N_2$ ($14$ electrons) is: $KK \sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^2 = \pi 2p_y^2 \sigma 2p_z^2$.When $N_2$ is converted to $N_2^+$,one electron is removed from the highest occupied molecular orbital $(HOMO)$.In $N_2$,the $HOMO$ is the $\sigma 2p_z$ orbital.Therefore,the electron is lost from the $\sigma 2p_z$ molecular orbital.

In the conversion of $N_2$ into $N_2^+$,the electron will be lost from which of the following molecular orbitals?

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