BO = (Nb - Na)/2 = 2. Why O₂ is paramagnetic.

Molecular Orbital Theory — Bond Order of O₂

Question

Using Molecular Orbital Theory, find the bond order of O₂ and explain why it is paramagnetic.

This one is a near-certain question in JEE Main and NEET. Bond order alone is a 1-mark MCQ; the paramagnetism explanation is what separates the 90-percentilers from the rest.

Solution — Step by Step

Oxygen has atomic number 8, so O₂ has 16 electrons total. We fill MOs in order of increasing energy:

\sigma_{1s}^2,\ \sigma_{1s}^{*2},\ \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}

The last two electrons go one each into the degenerate $\pi^*$ orbitals — Hund’s rule applies even in MO theory.

Count only the valence shell (from 2s onward) since the 1s core MOs cancel each other out:

Bonding electrons ( $N_b$ ): $\sigma_{2s}^2 + \sigma_{2p_z}^2 + \pi_{2p_x}^2 + \pi_{2p_y}^2 = 8$
Antibonding electrons ( $N_a$ ): $\sigma_{2s}^{*2} + \pi_{2p_x}^{*1} + \pi_{2p_y}^{*1} = 4$

\text{Bond Order} = \frac{N_b - N_a}{2}

\text{BO} = \frac{8 - 4}{2} = \frac{4}{2} = \mathbf{2}

O₂ has a double bond — consistent with what Lewis structure also predicts (though Lewis structure fails on the next step).

Look at those two $\pi^*$ antibonding electrons — they sit in separate, degenerate orbitals with parallel spins. Two unpaired electrons means O₂ is paramagnetic.

This is the big win of MO theory over Lewis dot structures. Lewis gives O=O with all electrons paired, predicting diamagnetism — which is experimentally wrong. Liquid O₂ is visibly attracted to a magnet.

Why This Works

The key insight is that when two atomic orbitals combine, they form one bonding MO (lower energy) and one antibonding MO (higher energy). Electrons in bonding MOs stabilise the molecule; electrons in antibonding MOs destabilise it. Bond order tells us the net bonding effect.

For O₂ specifically, the $\pi_{2p}$ bonding MOs fill completely, but the $\pi_{2p}^*$ antibonding MOs are half-filled. Each half-filled degenerate orbital gets one electron (Hund’s rule), and since these electrons have parallel spins, they cannot pair — paramagnetism is a direct consequence of the MO filling sequence.

The reason MO theory handles this correctly is that it treats electrons as delocalised across the entire molecule, not confined to a bond between two atoms. The antibonding $\pi^*$ orbitals simply have no Lewis structure equivalent.

Alternative Method — Quick MO Shortcut

For diatomic molecules of the second period, memorise this filling order once:

\sigma_{2s} < \sigma_{2s}^* < \sigma_{2p_z} < \pi_{2p} < \pi_{2p}^* < \sigma_{2p_z}^*

(Note: For N₂ and lighter, $\pi_{2p}$ fills before $\sigma_{2p_z}$ — this order flips at O₂.)

For O₂ specifically, just remember: 16 electrons, bond order = 2, two unpaired electrons. You can reconstruct the full configuration in the exam from these three facts alone.

Bond order 2 → double bond → bond length ≈ 121 pm, bond dissociation energy ≈ 498 kJ/mol. These numbers appear in thermodynamics and chemical bonding numericals. Having them ready saves time.

Common Mistake

Students often write the MO filling order for O₂ the same as N₂ — placing $\pi_{2p}$ orbitals above $\sigma_{2p_z}$ . This is correct for N₂ and lighter molecules, but wrong for O₂. For O₂, F₂, and Ne₂, the $\sigma_{2p_z}$ orbital drops below the $\pi_{2p}$ orbitals due to reduced s-p mixing. Using the wrong order changes which orbitals are filled last and can make you incorrectly predict O₂ as diamagnetic — a classic trap in JEE Main 2024 and NEET 2023 options.

The safest habit: whenever the question is about O₂ or anything to its right in period 2, mentally switch to the “post-nitrogen” filling order where $\sigma_{2p_z}$ fills first.

Question

Solution — Step by Step

Why This Works

Alternative Method — Quick MO Shortcut

Common Mistake

Try These Next