Molecular orbital theory

Oxygen liquefies pale blue and clings to a magnet, yet its Lewis structure $\text{O}=\text{O}$ draws every electron neatly paired — a molecule with no unpaired electrons should be pushed away from a magnet, not pulled in. Valence bond theory, which builds a bond from one localised electron pair shared between two atoms, has no way to explain this. Molecular orbital (MO) theory takes a different starting point: instead of electrons belonging to a bond between two specific atoms, it treats them as occupying orbitals that belong to the whole molecule, built by combining atomic orbitals mathematically. This article develops the LCAO (linear combination of atomic orbitals) method for period-2 homonuclear diatomics, builds the energy-level diagram and the bond-order formula, and uses them to derive — correctly, unlike valence bond theory — why oxygen is paramagnetic.

The core insight

Combining atomic orbitals constructively or destructively (LCAO) gives molecular orbitals shared by the whole molecule, and filling them by Aufbau, Pauli and Hund's rules yields a bond order and an unpaired-electron count that together predict bond strength and magnetism — succeeding for $\text{O}_2$ 's paramagnetism exactly where the simple Lewis picture fails.

The core idea

From atomic orbitals to molecular orbitals. When two atomic orbitals of comparable energy on neighbouring atoms overlap, they do not simply retain their identity — they combine mathematically (LCAO) into a matching number of new molecular orbitals delocalised over the whole molecule:

Bonding MO: the wavefunctions add constructively, raising electron density between the nuclei. It sits lower in energy than the parent atomic orbitals, and occupying it stabilises the molecule. Head-on ( $s$ or $p_z$ ) overlap gives a $\sigma$ bonding MO ( $\sigma_{1s}$ , $\sigma_{2s}$ , $\sigma_{2p_z}$ ); sideways $p$ overlap gives $\pi$ ( $\pi_{2p_x}$ , $\pi_{2p_y}$ ).
Antibonding MO: the wavefunctions subtract destructively, creating a node of zero electron density between the nuclei. It sits higher in energy than the parent orbitals and destabilises the molecule when occupied, marked with an asterisk ( $\sigma^*$ , $\pi^*$ ).

Filling rules are the same three principles used for atomic orbitals: Aufbau (fill lowest energy first), the Pauli exclusion principle (at most 2 electrons per MO, opposite spins), and Hund's rule (degenerate MOs — such as the two $\pi$ MOs — are singly occupied before any pairing).

Energy ordering flips across period 2. Because of $2s$ – $2p$ mixing, the relative position of $\sigma_{2p_z}$ and the $\pi_{2p}$ pair is not fixed:

\text{Li}_2\text{–}\text{N}_2:\quad \sigma_{2s} < \sigma^*_{2s} < (\pi_{2p_x}{=}\pi_{2p_y}) < \sigma_{2p_z} < (\pi^*_{2p_x}{=}\pi^*_{2p_y}) < \sigma^*_{2p_z}

\text{O}_2\text{–}\text{Ne}_2:\quad \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < (\pi_{2p_x}{=}\pi_{2p_y}) < (\pi^*_{2p_x}{=}\pi^*_{2p_y}) < \sigma^*_{2p_z}

where: each symbol names one molecular orbital, and " $<$ " means lower energy, filled first. The swap happens because $s$ – $p$ mixing (which pushes $\sigma_{2p_z}$ up) weakens as nuclear charge rises through the period, so $\sigma_{2p_z}$ drops below the $\pi_{2p}$ pair from oxygen onward. Getting this ordering wrong is the single most consequential mistake in the topic, because it changes which orbital receives the last electrons and therefore the predicted magnetism.

Figure

Two molecular orbital energy level diagrams side by side comparing the order before and after oxygen, showing atomic orbitals on both outer columns combining into bonding and antibonding molecular orbitals in the centre

Bond order and magnetism. Once an MO configuration is written down, two numbers follow directly. Bond order:

\text{Bond order} = \dfrac{N_b - N_a}{2}

where: $N_b$ is the number of electrons in bonding MOs and $N_a$ is the number of electrons in antibonding MOs. A bond order of zero (as for $\text{He}_2$ ) means the molecule does not exist; a higher bond order predicts a shorter, stronger bond. Separately, count unpaired electrons: if every MO is fully paired the species is diamagnetic (weakly repelled by a magnetic field); if one or more MOs are singly occupied it is paramagnetic (weakly attracted). The two numbers are independent facts read off the same filled diagram — bond order from the difference of bonding and antibonding electrons, magnetism from whether the last-filled MOs are singly or doubly occupied.

Derivation ABond order and magnetism of O2 from its molecular orbital configuration

The goal, spelled out in full because it is the flagship result of the whole topic, is to start from nothing but the atomic number of oxygen and the two filling rules above, build the complete MO configuration of $\text{O}_2$ , and show that the resulting bond order and unpaired-electron count correctly predict both its double-bond character and its well-documented paramagnetism — a result valence bond theory's single Lewis structure $\text{O}=\text{O}$ , with all electrons drawn paired, cannot produce.

Step 1. Count the electrons to place. Oxygen has atomic number 8, so $\text{O}_2$ has $8+8=16$ electrons total to distribute across the molecular orbitals, exactly as electrons fill atomic subshells in a single atom.

Step 2. Select the correct energy order. Oxygen lies in the $\text{O}_2$ – $\text{Ne}_2$ range, where reduced $2s$ – $2p$ mixing has already pushed $\sigma_{2p_z}$ below the $\pi_{2p}$ pair: $\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < (\pi_{2p_x}{=}\pi_{2p_y}) < (\pi^*_{2p_x}{=}\pi^*_{2p_y}) < \sigma^*_{2p_z}$ . Using the $\text{N}_2$ -side order here is the classic error and would place the last two electrons in the wrong orbital entirely.

Step 3. Fill by Aufbau and Pauli. Working up the list two electrons at a time: $\sigma_{1s}^2\ \sigma^{*2}_{1s}\ \sigma_{2s}^2\ \sigma^{*2}_{2s}\ \sigma_{2p_z}^2\ \pi_{2p_x}^2\ \pi_{2p_y}^2$ accounts for $2+2+2+2+2+2+2 = 14$ electrons, leaving 2 electrons still to place in the degenerate $\pi^*_{2p_x}, \pi^*_{2p_y}$ pair.

Step 4. Apply Hund's rule to the last two electrons. Because $\pi^*_{2p_x}$ and $\pi^*_{2p_y}$ are degenerate (equal energy), the two remaining electrons occupy them singly with parallel spins rather than pairing up in just one: $\pi^{*1}_{2p_x}\ \pi^{*1}_{2p_y}$ . This step, not any earlier one, is where the paramagnetism originates — pairing them by mistake is the most common error in the whole topic.

Step 5. Compute the bond order. Bonding electrons: $N_b = \sigma_{1s}(2)+\sigma_{2s}(2)+\sigma_{2p_z}(2)+\pi_{2p_x}(2)+\pi_{2p_y}(2) = 10$ . Antibonding electrons: $N_a = \sigma^*_{1s}(2)+\sigma^*_{2s}(2)+\pi^*_{2p_x}(1)+\pi^*_{2p_y}(1) = 6$ . Bond order $= \dfrac{10-6}{2} = 2$ .

Step 6. Read off the two predictions and check them. A bond order of 2 matches $\text{O}_2$ 's double-bond character (bond length about 121 pm, between typical single- and triple-bond lengths), while the two singly occupied $\pi^*$ orbitals from Step 4 give two unpaired electrons, correctly predicting paramagnetism — the result that makes $\text{O}_2$ the standard proof that MO theory captures physics the simple Lewis structure misses entirely. Check: $N_b + N_a = 10+6=16$ , matching the 16 electrons counted in Step 1; the configuration and arithmetic are self-consistent.

Worked example

Use MO theory to find the bond order of $\text{N}_2$ and predict its magnetic character, then compare it with $\text{O}_2$ .

Step 1. $\text{N}_2$ has $7+7=14$ electrons. Nitrogen lies in the $\text{Li}_2$ – $\text{N}_2$ range, so $\sigma_{2p_z}$ comes before the $\pi_{2p}$ pair: $\sigma_{1s}^2\ \sigma^{*2}_{1s}\ \sigma_{2s}^2\ \sigma^{*2}_{2s}\ (\pi_{2p_x}^2\ \pi_{2p_y}^2)\ \sigma_{2p_z}^2$ .

Step 2. Every orbital here is fully filled — 14 electrons exactly fill these seven MOs two at a time, with no leftover electrons to place singly, so there is no scope for Hund's rule to create unpaired electrons.

Step 3. Bonding electrons: $N_b = \sigma_{1s}(2)+\sigma_{2s}(2)+\pi_{2p_x}(2)+\pi_{2p_y}(2)+\sigma_{2p_z}(2) = 10$ . Antibonding electrons: $N_a = \sigma^*_{1s}(2)+\sigma^*_{2s}(2) = 4$ . Bond order $= \dfrac{10-4}{2} = 3$ .

Step 4. Since all electrons are paired, $\text{N}_2$ is predicted diamagnetic, matching its known triple bond ( $\text{N}{\equiv}\text{N}$ ), short bond length (110 pm) and high bond enthalpy ( $945\ \text{kJ mol}^{-1}$ ). Answer: bond order 3, diamagnetic — contrast this with $\text{O}_2$ 's bond order 2, paramagnetic, from the derivation above: the two molecules differ only in where the last electrons land (a fully paired $\sigma_{2p_z}$ pair for $\text{N}_2$ versus singly filled $\pi^*$ orbitals for $\text{O}_2$ ), yet that single difference in energy order changes both the bond order and the magnetism.

Common traps

Using the same MO energy order for every period-2 diatomic — $\sigma_{2p_z}$ and the $\pi_{2p}$ pair swap between $\text{N}_2$ and $\text{O}_2$ ; the wrong order for $\text{O}_2$ would place its last two electrons in a paired $\sigma_{2p_z}$ -type slot and wrongly predict it diamagnetic.
Forgetting Hund's rule when filling degenerate $\pi^*$ orbitals — pairing the last two electrons into one orbital instead of spreading them across both is the single most common error, and it is exactly the step that produces oxygen's paramagnetism.
Treating bonding and antibonding electrons as equally stabilising — bond order subtracts antibonding from bonding electrons, so equal numbers of each (as in $\text{He}_2$ , bond order 0) means the molecule is not stable at all.
Equating MO bond order with the Lewis-structure shared-pair count without checking — they usually agree for simple cases, but only the MO configuration reveals partially filled antibonding orbitals and the resulting magnetism.

Ask the tutor: Want to work through the MO configuration, bond order and magnetism of another period-2 diatomic or ion, like $\text{C}_2$ , $\text{F}_2$ , or $\text{O}_2^+$ ?

Why does the sigma 2pz and pi 2p order actually swap between N2 and O2?

It comes down to how much the $2s$ and $2p$ atomic orbitals mix before forming molecular orbitals. In lighter period-2 atoms ( $\text{Li}$ through $\text{N}$ ), the $2s$ and $2p$ orbitals are close enough in energy to mix significantly, and this mixing pushes $\sigma_{2p_z}$ up above the $\pi_{2p}$ pair. As nuclear charge increases across the period, the $2s$ – $2p$ energy gap widens and the mixing weakens, so from oxygen onward $\sigma_{2p_z}$ drops back below $\pi_{2p}$ , its "unmixed" expected position. The ISC syllabus asks you to know and apply both orders, not derive the mixing itself.

Does a higher bond order always mean a stronger, shorter bond?

Within the same pair of atoms, yes — bond order and bond strength (bond enthalpy) rise together while bond length falls, which is exactly the pattern MO theory reproduces for $\text{N}_2$ (order 3, short and strong) versus $\text{O}_2$ (order 2, longer and weaker) versus $\text{F}_2$ (order 1, longer still). Bond order zero, as calculated for $\text{He}_2$ , correctly predicts that no stable molecule forms at all. Comparing bond orders across different elements is less reliable, since bond strength also depends on atomic size and orbital overlap quality, but for a single family like the period-2 homonuclear diatomics the trend is dependable and is exactly what the syllabus expects you to state.

Is MO theory replacing valence bond theory, or do I need both?

You need both, for different jobs. Valence bond theory, with hybridisation and VSEPR, is far better at predicting molecular shape and is quicker for everyday bonding pictures. MO theory is better at predicting magnetic character and subtle bonding/antibonding effects that a localised electron-pair picture cannot see — oxygen's paramagnetism is the textbook case where valence bond theory gives the wrong answer and MO theory gives the right one. On the ISC syllabus, MO theory is examined specifically for homonuclear diatomics of period 1–2 elements, so keep the two toolkits separate rather than trying to merge them.

Molecular orbital theory

The core idea

Worked example

Common traps

Before this

Where this leads