What is the number of unpaired electron(s) in | vedclass

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(A) $N_2$ $(14 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $HOMO$ is $\sigma 2p_z$,whic

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$0, 1, 2, 1$

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(A) $N_2$ $(14 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. $HOMO$ is $\sigma 2p_z$,which has $0$ unpaired electrons.$N_2^{+}$ $(13 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^1$. $HOMO$ is $\sigma 2p_z$,which has $1$ unpaired electron.$O_2$ $(16 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. $HOMO$ is $\pi^* 2p$,which has $2$ unpaired electrons.$O_2^{+}$ $(15 \ e^-)$: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. $HOMO$ is $\pi^* 2p$,which has $1$ unpaired electron.Thus,the number of unpaired electrons is $0, 1, 2, 1$ respectively.

What is the number of unpaired electron$(s)$ in the highest occupied molecular orbital $(HOMO)$ of the following species: $N_2$,$N_2^{+}$,$O_2$,$O_2^{+}$?

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