JEE Chemistry 2024 Question with Solution

Given: We need the total number of electrons present in $\left(\pi^*\right)$ molecular orbitals of $\mathrm{O}_2$, $\mathrm{O}_2^+$, and $\mathrm{O}_2^-$.

Find: The sum of electrons in the antibonding $\pi^*$ molecular orbitals for all three species.

For $\mathrm{O}_2$, the molecular orbital configuration contains:

$$(\pi^*_{2p_x})^1(\pi^*_{2p_y})^1$$

So, total electrons in $\pi^*$ orbitals $= 2$.

For $\mathrm{O}_2^+$, one electron is removed from a $\pi^*$ orbital, so electrons in $\pi^*$ orbitals $= 1$.

For $\mathrm{O}_2^-$, one electron is added to a $\pi^*$ orbital, so electrons in $\pi^*$ orbitals $= 3$.

Adding these:

$$2 + 1 + 3 = 6$$

Therefore, the total number of electrons present in $\left(\pi^*\right)$ molecular orbitals is $6$.

Given: The species are $\mathrm{O}_2$, $\mathrm{O}_2^+$, and $\mathrm{O}_2^-$.

Find: Total electrons occupying $\pi^*$ molecular orbitals.

The relevant molecular orbital configuration for $\mathrm{O}_2$ is:

$$(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^*_{2p_x})^1(\pi^*_{2p_y})^1$$

Hence, $\mathrm{O}_2$ has $2$ electrons in $\pi^*$ orbitals.

For $\mathrm{O}_2^+$:

$$(\pi^*_{2p_x})^1(\pi^*_{2p_y})^0$$

So, it has $1$ electron in $\pi^*$ orbitals.

For $\mathrm{O}_2^-$:

$$(\pi^*_{2p_x})^2(\pi^*_{2p_y})^1$$

So, it has $3$ electrons in $\pi^*$ orbitals.

Now sum them:

$$2 + 1 + 3 = 6$$

Therefore, the required numerical value is $6$.