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Total number of electron present in $\left(\pi^*\right)$ molecular orbitals of $\mathrm{O}_2, \mathrm{O}_2^{+}$and $\mathrm{O}_2^{-}$is ............
$6$
$7$
$9$
$10$
Solution
$ \mathrm{O}_2(16 \mathrm{e}):\left(\sigma_{1 \mathrm{~s}}\right)^2\left(\sigma_{1 \mathrm{~s}}^*\right)^2\left(\sigma_{2 \mathrm{~s}}\right)^2\left(\sigma_{2 \mathrm{~s}}^*\right)^2 $
$ \left(\sigma_{2 \mathrm{p}}\right)^2\left[\left(\pi_{2 \mathrm{p}}\right)^2=\left(\pi_{2 \mathrm{p}}\right)^2\right],\left[\left(\pi_{2 \mathrm{p}}^*\right)^1=\left(\pi_{2 \mathrm{p}}^*\right)^1\right]$
Number of $\mathrm{e}^{-}$present in $\left(\pi^*\right)$ of $\mathrm{O}_2=2$
Number of $\mathrm{e}^{-}$present in $\left(\pi^*\right)$ of $\mathrm{O}_2^{+}=1$
Number of $\mathrm{e}^{-}$present in $\left(\pi^*\right)$ of $\mathrm{O}_2^{-}=3$
So total $\mathrm{e}^{-}$in $\left(\pi^*\right)=2+1+3=6$
