Explain Molecular Orbital Theory (MOT) and use MOT to explain the magnetic nature of the O₂ molecule

What MOT is trying to explain

Lewis structures describe bonding using localized pairs, but they cannot correctly predict some properties, like the observed paramagnetism of $O_2$. Molecular Orbital Theory fixes this by treating bonding electrons as occupying orbitals spread over both atoms.

Core ideas of Molecular Orbital Theory

When two atomic orbitals (AOs) of similar energy and proper symmetry overlap, they combine to form two molecular orbitals (MOs):

• A bonding MO (lower energy), with electron density between the nuclei
• An antibonding MO (higher energy), written with an asterisk $^*$, with a node between the nuclei

Electrons fill MOs using:

• Aufbau principle: fill lower-energy MOs first
• Pauli exclusion: max 2 electrons per MO with opposite spins
• Hund’s rule: in degenerate MOs (same energy), electrons stay unpaired as long as possible

MO energy order relevant for $O_2$

For second-row diatomic molecules from oxygen onward ($O_2, F_2, Ne_2$), the usual valence MO order is: $$\sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < \pi_{2p_x}=\pi_{2p_y} < \pi^*_{2p_x}=\pi^*_{2p_y} < \sigma^*_{2p_z}$$

Filling the MOs for $O_2$

Each oxygen has 6 valence electrons, so $O_2$ has 12 valence electrons to place in these MOs.

Filling in order gives: $$ (\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^*_{2p_x})^1(\pi^*_{2p_y})^1 $$

Notice the key result: the two $\pi^*$ orbitals are degenerate, so by Hund’s rule they each get one electron before pairing.

Why $O_2$ is paramagnetic

Magnetism depends on whether there are unpaired electrons:

• Paramagnetic: one or more unpaired electrons
• Diamagnetic: all electrons paired

In $O_2$, the last two electrons occupy separate $\pi^*_{2p}$ antibonding orbitals, so there are two unpaired electrons. That is why $O_2$ is paramagnetic and is attracted into a magnetic field.

Extra result MOT gives: bond order of $O_2$

Bond order in MOT is: $$\text{Bond order} = \frac{N_b - N_a}{2}$$ where $N_b$ is bonding electrons and $N_a$ is antibonding electrons.

For $O_2$ (valence MOs):

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

So, $$\text{Bond order} = \frac{8-4}{2} = 2$$ This matches the idea of a double bond, while also correctly predicting paramagnetism (which Lewis structures miss).