QJEEAdv02Ch5682: JEE Advanced 2002 Chemistry - Class 11 JEEAdv02Ch5682 Solution: Which of the following molecular species has unpaired…

To determine which molecular species have unpaired electrons, let's examine the electron configurations of each species using molecular orbital (MO) theory. The presence of unpaired electrons can be deduced from the electron filling in the molecular orbitals.

Option A: N₂

For nitrogen diatomic (N₂), each nitrogen atom has 7 electrons, so the total number of electrons is 14. According to the molecular orbital theory, the electrons fill the molecular orbitals in the following order: $\sigma_{1s}$ , $\sigma_{1s}^*$ , $\sigma_{2s}$ , $\sigma_{2s}^*$ , $\pi_{2p_x}$ = $\pi_{2p_y}$ , $\sigma_{2p_z}$ .

Filling these orbitals results in no unpaired electrons:

$\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2$

So, N₂ has no unpaired electrons.

Option B: F₂

For fluorine diatomic (F₂), each fluorine atom has 9 electrons, resulting in a total of 18 electrons. They fill the orbitals in the following order: $\sigma_{1s}$ , $\sigma_{1s}^*$ , $\sigma_{2s}$ , $\sigma_{2s}^*$ , $\pi_{2p_x}$ = $\pi_{2p_y}$ , $\sigma_{2p_z}$ , $\pi_{2p_x}^{*}$ = $\pi_{2p_y}^{*}$ , $\sigma_{2p_z}^*$ .

Filling these orbitals also results in no unpaired electrons:

$\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^{*2} = \pi_{2p_y}^{*2}$

So, F₂ has no unpaired electrons.

Option C: $O_2^-$

The dioxygen anion ( $O_2^-$ ) has one extra electron compared to neutral O₂, which normally has 16 electrons. Hence, $O_2^-$ has 17 electrons. These electrons fill the orbitals in the following order: $\sigma_{1s}$ , $\sigma_{1s}^*$ , $\sigma_{2s}$ , $\sigma_{2s}^*$ , $\pi_{2p_x}$ = $\pi_{2p_y}$ , $\sigma_{2p_z}$ , $\pi_{2p_x}^{*}$ = $\pi_{2p_y}^{*}$ , $\sigma_{2p_z}^*$ .

Filling these orbitals, the last electron will be placed in one of the degenerate $\pi_{2p_x}^{*}$ or $\pi_{2p_y}^{*}$ orbitals, resulting in one unpaired electron:

$\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^{*2} = \pi_{2p_y}^{*1}$

So, $O_2^-$ has one unpaired electron.

Option D: $O_2^{2-}$

The peroxide ion ( $O_2^{2-}$ ) has two extra electrons compared to neutral O₂, giving it a total of 18 electrons. These electrons fill the orbitals as follows: $\sigma_{1s}$ , $\sigma_{1s}^*$ , $\sigma_{2s}$ , $\sigma_{2s}^*$ , $\pi_{2p_x}$ = $\pi_{2p_y}$ , $\sigma_{2p_z}$ , $\pi_{2p_x}^{*}$ = $\pi_{2p_y}^{*}$ , $\sigma_{2p_z}^*$ .

Filling these orbitals results in no unpaired electrons:

$\sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \pi_{2p_x}^2 = \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^{*2} = \pi_{2p_y}^{*2}$

So, $O_2^{2-}$ has no unpaired electrons.

Based on the molecular orbital theory, the molecular species that has unpaired electron(s) is Option C: $O_2^-$ .