Molecular Orbital Theory: Bond Order and Magnetism Explained for NEET and JEE

Lewis structures say oxygen, O₂, has all its electrons paired in a neat double bond. Yet liquid oxygen sticks to a strong magnet, which only unpaired electrons can explain. Molecular orbital (MO) theory is the model that gets this right, and it matters for both NEET and JEE, where bond order and magnetic behaviour are asked almost every year.

Quick Summary

• When atoms bond, their atomic orbitals combine to form molecular orbitals that belong to the whole molecule.

• Bonding orbitals hold electron density between the nuclei and lower the energy. Antibonding orbitals (marked with *) raise it.

• Bond order = (Nb − Na) ÷ 2, where Nb is the number of electrons in bonding orbitals and Na the number in antibonding orbitals.

• Higher bond order means a shorter, stronger bond. A bond order of zero means the molecule does not exist.

• Any unpaired electron in the MO diagram makes the species paramagnetic. All electrons paired means diamagnetic.

From waves to orbitals

An electron in an atom is described by a wave, the atomic orbital. When two atoms approach, their waves overlap, and waves can add or subtract. If they add (in phase), the amplitude builds up in the region between the nuclei. This gives a bonding molecular orbital, lower in energy than the original atomic orbitals, because electrons there are attracted by both nuclei at once.

If the waves subtract (out of phase), a node forms between the nuclei where the electron density drops to zero. This gives an antibonding molecular orbital, higher in energy, and an electron in it actually pushes the nuclei apart. The rule to remember is that n atomic orbitals always give n molecular orbitals: two combine into one bonding and one antibonding orbital, with no loss or gain.

Filling the diagram

Electrons enter molecular orbitals by the same three rules as atoms: lowest energy first (Aufbau), at most two electrons per orbital with opposite spins (Pauli), and singly occupy degenerate orbitals before pairing (Hund). For second-period homonuclear diatomics we use the valence electrons, and the order depends on the molecule.

• Li₂ to N₂ (up to 14 total electrons): σ2s < σ*2s < π2p(x) = π2p(y) < σ2p(z) < π*2p(x) = π*2p(y) < σ*2p(z)

• O₂ and F₂: σ2s < σ*2s < σ2p(z) < π2p(x) = π2p(y) < π*2p(x) = π*2p(y) < σ*2p(z)

The swap happens because in the lighter atoms the 2s and 2p orbitals are close in energy and mix, pushing σ2p above the π orbitals. In O₂ and F₂ the gap is larger, so no such mixing occurs.

Worked example: O₂ and its ions

Find the bond order and magnetic nature of O₂, O₂⁺ and O₂⁻ (superoxide), using valence electrons only.

1. O₂: each oxygen contributes 6 valence electrons, so 12 in total. Fill in order: σ2s² σ*2s² σ2p² π2p⁴ π*2p². The last two electrons go into the two degenerate π* orbitals one each, by Hund's rule. Nb = 2 + 2 + 4 = 8 and Na = 2 + 2 = 4. Bond order = (8 − 4) ÷ 2 = 2. Two unpaired electrons, so O₂ is paramagnetic. This explains the magnet.

2. O₂⁺: remove one electron, taken from a π* orbital. Now Nb = 8 and Na = 3. Bond order = (8 − 3) ÷ 2 = 2.5. One unpaired electron, so paramagnetic.

3. O₂⁻: add one electron, which pairs up in a π* orbital. Now Nb = 8 and Na = 5. Bond order = (8 − 5) ÷ 2 = 1.5. One unpaired electron, so paramagnetic.

Ranking bond order: O₂⁺ (2.5) > O₂ (2) > O₂⁻ (1.5) > O₂²⁻ (1). Bond length runs in exactly the opposite direction, so O₂⁺ has the shortest bond and O₂²⁻ the longest. For the peroxide ion, O₂²⁻, the 14 electrons fill both π* orbitals completely, bond order is (8 − 6) ÷ 2 = 1, and it is diamagnetic.

Mistakes students actually make

• Using the wrong energy order. Applying the O₂ order to N₂ or C₂ changes which orbital is filled and gives a wrong magnetic answer. Check the total number of electrons first: 14 or fewer uses the first order (so B₂, C₂, N₂), and more uses the second.

• Removing an electron from the wrong place. A positive ion loses the electron from the highest-energy occupied orbital, which for O₂ is π*, not a bonding orbital. That is why O₂⁺ is stronger than O₂, while N₂⁺ is weaker than N₂.

• Forgetting Hund's rule for the last two electrons of O₂ and pairing them in one π* orbital. That wrongly predicts diamagnetism.

• Counting core electrons or total electrons in the bond order formula when the diagram only includes valence electrons.

• Assuming bond order must be a whole number. Half-integer values like 1.5 and 2.5 are common and fully valid.

How it is asked in NEET and JEE

• Arrange a set of species (such as O₂, O₂⁺, O₂⁻, O₂²⁻) in order of bond length, bond energy or bond order.

• Identify which species in a list are paramagnetic, or count unpaired electrons.

• Decide whether a species such as He₂ or Be₂ can exist (bond order zero means no).

• Compare an ion with its parent, for example "N₂⁺ versus N₂, which has the longer bond?" (N₂⁺, because it has lost a bonding electron).

Pro Tip: Treat bond order as a sum with a sign. Adding an electron to a bonding orbital raises it by 0.5, and adding to an antibonding orbital lowers it by 0.5. So for any ion, start from the neutral molecule and ask only one question: "Did the electron move in or out of a bonding orbital or an antibonding one?" For N₂ the extra electron of N₂⁻ goes into π*, lowering bond order from 3 to 2.5, while for N₂⁺ the lost electron comes from σ2p, a bonding orbital, which also lowers it to 2.5. The same number arises for different reasons, and recognising that is what the toughest questions test.

Frequently Asked Questions

Why is O₂ paramagnetic when its Lewis structure shows no unpaired electrons?

Lewis structures treat electrons as localised pairs and cannot show degenerate orbitals. In the MO diagram the last two electrons occupy the two π* orbitals separately with parallel spins, as Hund's rule requires. Those two unpaired electrons make O₂ paramagnetic.

What does a bond order of zero mean?

It means the number of bonding and antibonding electrons is equal, so there is no net stabilisation. He₂ has 2 bonding and 2 antibonding electrons, giving bond order zero, and the molecule does not form under ordinary conditions.

Is a higher bond order always a stronger bond?

For comparing similar species, yes. Higher bond order generally means higher bond dissociation energy and shorter bond length. Be careful when comparing very different molecules, though, because atom size and other factors also play a part.

Why does the energy order of σ2p and π2p change?

In lighter atoms such as B, C and N the 2s and 2p orbitals have similar energies, so s-p mixing pushes the σ2p orbital above π2p. In O and F, the 2s–2p gap is bigger, mixing is negligible, and σ2p stays below π2p.

Which exam is this topic relevant for?

Chemical bonding, including molecular orbital theory, is part of Class 11 chemistry and is relevant to both NEET and JEE. Questions are usually short numerical or ordering problems.

If any step of the MO diagram still feels shaky, drop your doubt in the comments and I will walk through it. Share this with a fellow aspirant who keeps losing marks on bond order questions.