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

The orbital in which last electron is present is called the highest occupied molecular orbital(HOMO).

${N}_{2}$N2​ : $\sigma 1{s}^{2}{\sigma }^{*}1{s}^{2}\sigma 2{s}^{2}{\sigma }^{*}2{s}^{2}\pi 2{p}_{x}^{2}=\pi 2{p}_{y}^{2}\frac{\sigma 2{p}_{z}^{2}}{\mathrm{HOMO}}$σ1s2σ∗1s2σ2s2σ∗2s2π2px2​=π2py2​HOMOσ2pz2​​

$${N}_{2}^{+}:\sigma 1{s}^{2}{\sigma }^{*}1{s}^{2}{\sigma }^{*}2{s}^{2}{\sigma }^{*}2{s}^{2}\pi 2{p}_{x}^{2}=\pi 2{p}_{y}^{2}\frac{\sigma 2{p}_{z}^{1}}{\mathrm{HOMO}}$$N2+​:σ1s2σ∗1s2σ∗2s2σ∗2s2π2px2​=π2py2​HOMOσ2pz1​​

$${O}_{2}:\sigma 1{s}^{2}{\sigma }^{*}1{s}^{2}{\sigma }^{*}2{s}^{2}\sigma 2{p}_{z}^{2}\pi 2{p}_{x}^{1}={\pi }^{*}2{p}_{y}^{1}(\mathrm{HOMO})$$O2​:σ1s2σ∗1s2σ∗2s2σ2pz2​π2px1​=π∗2py1​(HOMO)

$${O}_{2}^{+}:\sigma 1{s}^{2}{\sigma }^{*}1{s}^{2}\sigma 2{s}^{2}{\sigma }^{*}2{s}^{2}\sigma 2{p}_{z}^{2}\pi 2{p}_{x}^{2}=\pi 2{p}_{y}^{2}{\pi }^{*}2{p}_{x}^{1}={\pi }^{*}2{p}_{y}^{0}(\mathrm{HOMO})$$O2+​:σ1s2σ∗1s2σ2s2σ∗2s2σ2pz2​π2px2​=π2py2​π∗2px1​=π∗2py0​(HOMO)

${N}_{2}\Rightarrow 0$N2​⇒0 unpaired ${e}^{–}$e– in $\mathrm{HOMO}$HOMO

${N}_{2}^{+}\Rightarrow 1$N2+​⇒1 unpaired ${e}^{-}$e− in $\mathrm{HOMO}$HOMO

${O}_{2}\Rightarrow 2$O2​⇒2 unpaired ${e}^{-}$e− in $\mathrm{HOMO}$HOMO

${O}_{2}^{+}\Rightarrow 1$O2+​⇒1 unpaired ${e}^{-}$e− in $\mathrm{HOMO}$HOMO