Total number of electrons present in \(\left(\pi^*\right)\) molecular orbitals of \(\mathrm{O}_2,…
Total number of electrons present in \(\left(\pi^*\right)\) molecular orbitals of \(\mathrm{O}_2, \mathrm{O}_2{ }^{+}\) and \(\mathrm{O}_2^{-}\) is _______.
Solution
$\begin{aligned}
& \mathrm{O}_2(16 \mathrm{e}):\left(\sigma_{1 \mathrm{~s}}\right)^2\left(\sigma_{1 \mathrm{~s}}^*\right)^2\left(\sigma_{2 \mathrm{~s}}\right)^2\left(\sigma_{2 \mathrm{~s}}^*\right)^2 \\
& \left(\sigma_{2 \mathrm{p}}\right)^2\left[\left(\pi_{2 \mathrm{p}}\right)^2=\left(\pi_{2 \mathrm{p}}\right)^2\right],\left[\left(\pi_{2 \mathrm{p}}^*\right)^1=\left(\mathrm{m}_{2 \mathrm{p}}^*\right)^1\right]
\end{aligned}$ 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 $\left(\pi^*\right)=2+1+3=6$