What is the number of unpaired electrons in the highest occupied molecular orbital of the following species: N2, N2+,…

🧠 Identify the HOMO (highest occupied MO) for each species and count unpaired electrons in it.

$\ce{N2}$ (14e): MO config ends in ...\sigma_{2s}^2\sigma^*_{2s}^2\pi_{2p}^4\sigma_{2p}^2 (σ₂p is HOMO for N₂). The HOMO $\sigma_{2p}$ is fully occupied (2e, paired) → 0 unpaired.

$\ce{N2+}$ (13e): Remove 1e from the HOMO ($\sigma_{2p}$) → $\sigma_{2p}^1$ → 1 unpaired.

$\ce{O2}$ (16e): MO config ends in ...\pi_{2p}^4\pi^*_{2p}^2. HOMO = $\pi^*_{2p}$ (degenerate, 2 orbitals each with 1e by Hund's rule) → 2 unpaired.

$\ce{O2+}$ (15e): Remove 1e from $\pi^*_{2p}$ → \pi^*_{2p}^1 (only 1 orbital occupied) → 1 unpaired.

$\boxed{\text{Answer: (a) 0, 1, 2, 1}}$