Which of the following has unpaired electron(s)?

To determine which species has unpaired electrons, we use Molecular Orbital Theory (MOT). This theory helps us understand the electron configuration in molecules and identify unpaired electrons by filling molecular orbitals according to their energy levels.

We analyze each option step by step, considering the total number of electrons and the molecular orbital energy order. For oxygen-based molecules (like O₂ and its ions), the energy order of molecular orbitals is: $$\sigma 2s < \sigma^* 2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z$$. For nitrogen-based molecules (like N₂ and its ions), the energy order is: $$\sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z$$. We focus on valence electrons, ignoring core electrons (1s orbitals), as they are filled and do not affect unpaired electrons.

Option A: O₂⁻

O₂ has 16 electrons (8 from each oxygen atom). O₂⁻ gains one extra electron, so total electrons = 17. Valence electrons: Each oxygen atom has 6 valence electrons (2s²2p⁴), so O₂ has 12 valence electrons, and O₂⁻ has 13 valence electrons. Using the energy order for oxygen-based molecules:

Fill the orbitals:

• $$\sigma 2s^2$$ (2 electrons)
• $$\sigma^* 2s^2$$ (2 electrons, total 4)
• $$\sigma 2p_z^2$$ (2 electrons, total 6)
• $$\pi 2p_x^2$$ and $$\pi 2p_y^2$$ (4 electrons, total 10)
• Remaining electrons: 13 - 10 = 3 electrons go to the $$\pi^*$$ orbitals. The $$\pi^*$$ orbitals are degenerate ($$\pi^* 2p_x$$ and $$\pi^* 2p_y$$). First, place one electron in each $$\pi^*$$ orbital: $$\pi^* 2p_x^1$$ and $$\pi^* 2p_y^1$$ (2 electrons, total 12). The last electron goes into one of them, say $$\pi^* 2p_x^2$$ (total 13). So the configuration is: $$\sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2 \pi^* 2p_x^2 \pi^* 2p_y^1$$.

Here, $$\pi^* 2p_y^1$$ has one unpaired electron. Thus, O₂⁻ has an unpaired electron.

Option B: N₂²⁺

N₂ has 14 electrons (7 from each nitrogen atom). N₂²⁺ loses two electrons, so total electrons = 12. Valence electrons: Each nitrogen atom has 5 valence electrons (2s²2p³), so N₂ has 10 valence electrons, and N₂²⁺ has 8 valence electrons. Using the energy order for nitrogen-based molecules:

Fill the orbitals:

• $$\sigma 2s^2$$ (2 electrons)
• $$\sigma^* 2s^2$$ (2 electrons, total 4)
• Remaining electrons: 8 - 4 = 4 electrons go to the $$\pi 2p$$ orbitals. The $$\pi 2p$$ orbitals are degenerate ($$\pi 2p_x$$ and $$\pi 2p_y$$). Place two electrons in each: $$\pi 2p_x^2$$ and $$\pi 2p_y^2$$ (4 electrons, total 8). So the configuration is: $$\sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^2 \pi 2p_y^2$$.

All orbitals are fully filled with paired electrons. Thus, N₂²⁺ has no unpaired electrons.

Option C: O₂²⁻

O₂ has 16 electrons. O₂²⁻ gains two extra electrons, so total electrons = 18. Valence electrons: O₂ has 12 valence electrons, so O₂²⁻ has 14 valence electrons. Using the energy order for oxygen-based molecules:

Fill the orbitals:

• $$\sigma 2s^2$$ (2 electrons)
• $$\sigma^* 2s^2$$ (2 electrons, total 4)
• $$\sigma 2p_z^2$$ (2 electrons, total 6)
• $$\pi 2p_x^2$$ and $$\pi 2p_y^2$$ (4 electrons, total 10)
• Remaining electrons: 14 - 10 = 4 electrons go to the $$\pi^*$$ orbitals. Place two electrons in each $$\pi^*$$ orbital: $$\pi^* 2p_x^2$$ and $$\pi^* 2p_y^2$$ (4 electrons, total 14). So the configuration is: $$\sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2 \pi^* 2p_x^2 \pi^* 2p_y^2$$.

All orbitals are fully filled with paired electrons. Thus, O₂²⁻ has no unpaired electrons.

Option D: N₂

N₂ has 14 electrons. Valence electrons: 10. Using the energy order for nitrogen-based molecules:

Fill the orbitals:

• $$\sigma 2s^2$$ (2 electrons)
• $$\sigma^* 2s^2$$ (2 electrons, total 4)
• $$\pi 2p_x^2$$ and $$\pi 2p_y^2$$ (4 electrons, total 8)
• $$\sigma 2p_z^2$$ (2 electrons, total 10). So the configuration is: $$\sigma 2s^2 \sigma^* 2s^2 \pi 2p_x^2 \pi 2p_y^2 \sigma 2p_z^2$$.

All orbitals are fully filled with paired electrons. Thus, N₂ has no unpaired electrons.

Only O₂⁻ (Option A) has an unpaired electron. Hence, the correct answer is Option A.