Draw MOT diagram for O₂ molecule — explain why O₂ is paramagnetic

Question

Draw the Molecular Orbital Theory (MOT) diagram for the O₂ molecule and explain why O₂ is paramagnetic despite having an even number of electrons.

Solution — Step by Step

O₂ has 2 oxygen atoms, each with 8 electrons — so we have 16 electrons total to fill into molecular orbitals.

We fill these in increasing order of energy, following the Aufbau principle for MOs.

For O₂ (and molecules after it in the second period), the correct energy ordering is:

\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z}

Note: for O₂, N₂, and beyond, $\sigma_{2p_z}$ comes before the $\pi_{2p}$ orbitals — this is different from B₂, C₂, N₂ where $\pi$ comes first. This detail costs marks in JEE if you get it wrong.

\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 first 14 electrons fill everything up to the bonding $\pi_{2p}$ orbitals. The last 2 electrons go into the degenerate $\pi^*_{2p_x}$ and $\pi^*_{2p_y}$ orbitals — one each, by Hund’s rule.

\text{Bond Order} = \frac{N_b - N_a}{2} = \frac{10 - 6}{2} = \mathbf{2}

$N_b$ = 10 (electrons in bonding MOs), $N_a$ = 6 (electrons in antibonding MOs). Bond order 2 confirms the O=O double bond — consistent with Lewis structure.

We have 2 unpaired electrons — one in each $\pi^*_{2p}$ orbital. Any molecule with one or more unpaired electrons is paramagnetic (attracted to a magnetic field).

Since O₂ has 2 unpaired electrons, it is paramagnetic. This was experimentally confirmed when liquid O₂ is poured between poles of a magnet — it sticks.

Why This Works

The Lewis structure of O₂ shows a double bond with all electrons paired, which wrongly predicts diamagnetic behaviour. MOT fixes this because it considers all molecular orbitals simultaneously, including degenerate antibonding ones.

The two $\pi^*_{2p}$ orbitals are equal in energy (degenerate). When 2 electrons must be distributed across 2 equal-energy orbitals, Hund’s rule says they occupy separate orbitals with parallel spins — exactly like how p-orbitals fill in an atom. This produces 2 unpaired spins, giving paramagnetism.

This is one of the greatest triumphs of MOT: it correctly predicts the magnetic nature of O₂ where Lewis theory completely fails. NCERT Class 11 highlights this as a key argument for MOT over the valence bond approach.

Alternative Method — Using Bond Order to Verify

Once we have the electronic configuration, we can also verify stability:

Bond order = 2 → O₂ is stable (exists as a molecule) ✓
Bond length ≈ 121 pm (double bond, shorter than O–O single bond at 148 pm) ✓
Bond dissociation energy ≈ 498 kJ/mol ✓

All these are consistent with a double bond. If bond order had come out as zero or negative, the molecule wouldn’t exist — useful self-check in exam conditions.

Bond order shortcut for homonuclear diatomics: for O₂ specifically, just remember “10 bonding, 6 antibonding” — it’s faster than counting from scratch. Bond order = (10−6)/2 = 2. Works every time.

Common Mistake

Students often use the wrong MO energy order for O₂ — placing $\pi_{2p}$ before $\sigma_{2p_z}$ (which is correct only for B₂, C₂, N₂). For O₂, F₂, and Ne₂, the order flips: $\sigma_{2p_z}$ is lower in energy than $\pi_{2p}$ .

Using the wrong order doesn’t change the final answer for O₂ in this case (the number of unpaired electrons stays the same), but it will give wrong bond orders for other molecules like NO or CO. In JEE Main 2023 Shift 2, a question on bond order of NO specifically tested whether students knew this distinction.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Bond Order to Verify

Common Mistake

Try These Next