Molecular Orbital Theory for Diatomics: Bond Order and Magnetism

Inorganic Chemistry · Bonding

Molecular Orbital Theory for Diatomics: Bond Order and Magnetism

MO diagrams answer three questions at once — is the molecule stable, how strong is the bond, and is it paramagnetic.

BSc & MSc · Inorganic Chemistry · Concept

The short answer: Atomic orbitals combine to give bonding and antibonding molecular orbitals. Fill them by the aufbau principle, then bond order is half the difference between bonding and antibonding electrons. Unpaired electrons mean paramagnetism — which is where MO theory succeeds and valence bond theory fails.

The core idea

When two atomic orbitals of comparable energy and correct symmetry overlap, they combine to give two molecular orbitals: one bonding, lower in energy than either parent, and one antibonding, higher. Electrons in bonding orbitals hold the molecule together; electrons in antibonding orbitals push it apart.

Bond order = ½ (bonding electrons − antibonding electrons)

A bond order of zero means the molecule does not exist as a stable species, which is the standard explanation for why the helium molecule is not observed.

Filling order, and the s–p mixing complication

For second-period diatomics the ordering depends on whether s–p mixing is significant:

ElementsOrderingReason
Li through Nπ(2p) below σ(2p)2s and 2p are close in energy, so mixing raises σ(2p)
O, F, Neσ(2p) below π(2p)The 2s–2p gap is large, so mixing is negligible
The crossover happens between nitrogen and oxygen, and it is examined constantly. Using the wrong ordering changes the predicted magnetism for several molecules, so establish which side of the crossover you are on before drawing anything.

The classic results

SpeciesValence electronsBond orderMagnetism
H221Diamagnetic
He240Does not exist
N2103Diamagnetic
O2122Paramagnetic — two unpaired electrons
F2141Diamagnetic

Oxygen is the reason MO theory is taught at all. Valence bond theory draws a double bond with all electrons paired and predicts diamagnetism. Oxygen is experimentally paramagnetic. MO theory places the last two electrons singly in two degenerate π* orbitals, predicting exactly two unpaired electrons. This single case is the standard argument for MO theory and appears in almost every syllabus.

Ions, and what removing an electron does

Adding or removing electrons changes bond order predictably. Removing a bonding electron lowers bond order and lengthens the bond; removing an antibonding electron raises bond order and shortens it.

So O2+ has a higher bond order than O2, because the electron removed came from an antibonding orbital, while O2 and O22− have progressively lower bond orders. Ordering a set of oxygen species by bond length is a routine question, and it is answered entirely by counting antibonding electrons.

Heteronuclear diatomics

When the two atoms differ, their atomic orbitals sit at different energies. The more electronegative atom's orbitals lie lower, so the bonding MO resembles them more closely and the antibonding MO resembles the less electronegative atom's. This unequal contribution is the MO description of bond polarity.

Carbon monoxide is the standard case. Its highest occupied orbital is concentrated on carbon, which is why CO binds to metals through carbon rather than oxygen — a fact usually memorised and rarely explained, and worth being able to justify.

Frequently asked questions

Why do only orbitals of similar energy combine?

Because the stabilisation from mixing falls as the energy gap widens. Orbitals far apart in energy interact negligibly, which is why core orbitals are ignored in these diagrams.

What decides whether overlap is allowed at all?

Symmetry. Orbitals must have matching symmetry with respect to the internuclear axis; if they do not, the positive and negative overlap regions cancel exactly and no bond results.

Does a higher bond order always mean a shorter bond?

Within a series of related species, yes. Comparing across different elements it does not hold reliably, since atomic size changes too.

How does s–p mixing actually change the ordering?

Mixing pushes the σ(2s) down and the σ(2p) up. Where mixing is strong, σ(2p) rises above the π(2p) pair, reversing the expected order.

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