JEE MAIN - Chemistry (2024 - 9th April Evening Shift - No. 24)
Total number of electrons present in $$\left(\pi^*\right)$$ molecular orbitals of $$\mathrm{O}_2, \mathrm{O}_2^{+}$$ and $$\mathrm{O}_2^{-}$$ is ________.
Answer
6
Explanation
$$\begin{aligned} & \mathrm{O}_2(16 \mathrm{e}):\left(\sigma_{1 \mathrm{~s}}\right)^2\left(\sigma_{1 \mathrm{~s}}^*\right)^2\left(\sigma_{2 \mathrm{~s}}\right)^2\left(\sigma_{2 \mathrm{~s}}^*\right)^2 \\ & \left(\sigma_{2 \mathrm{p}}\right)^2\left[\left(\pi_{2 \mathrm{p}}\right)^2=\left(\pi_{2 \mathrm{p}}\right)^2\right],\left[\left(\pi_{2 \mathrm{p}}^*\right)^1=\left(\pi_{2 \mathrm{p}}^*\right)^1\right] \end{aligned}$$
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 $$\left(\pi^*\right)=2+1+3=6$$
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