Molecular Orbital Theory

Draw a Lewis structure for O2\text{O}_2 and you get every electron neatly paired — yet pour liquid oxygen between the poles of a magnet and it clings there, visibly attracted. Valence bond theory has no explanation. Molecular orbital theory (MOT) does: instead of localised bonds, it combines atomic orbitals into new orbitals belonging to the whole molecule, some lower in energy (bonding) and some higher (antibonding), and fills them the same way electrons fill atomic shells. Do that correctly for O2\text{O}_2 and two electrons land unpaired, in exactly the orbitals that explain the magnet. This article builds MOT from the LCAO idea up: how bonding and antibonding orbitals form, the two energy orders you must choose between for period-2 diatomics, how bond order is read off an electron configuration, and how that configuration predicts whether a molecule is paramagnetic or diamagnetic — with N2\text{N}_2 and O2\text{O}_2 worked in full.

The core insight

Atomic orbitals of similar energy combine (LCAO) into bonding and antibonding molecular orbitals belonging to the whole molecule; filling these by Aufbau, Pauli and Hund's rules gives a bond order =(Nb−Na)/2= (N_b-N_a)/2 and correctly predicts magnetic behaviour, most famously O2\text{O}_2's paramagnetism.

The core idea

Linear combination of atomic orbitals (LCAO). When two atomic orbitals of comparable energy overlap, they do not simply stay as they were — they combine into two new molecular orbitals (MOs), one from constructive (in-phase) overlap and one from destructive (out-of-phase) overlap. Two atomic orbitals in, two molecular orbitals out, always.

Bonding versus antibonding. The constructive combination gives a bonding MO: electron density piles up between the two nuclei, energy is lower than either parent atomic orbital, and putting electrons here stabilises the molecule. The destructive combination gives an antibonding MO (written with an asterisk, σ∗\sigma^* or π∗\pi^*): there is a node — a plane of zero electron density — between the nuclei, energy is higher than the parent orbitals, and electrons here destabilise the molecule.

Filling rules. MOs fill exactly the way atomic orbitals do: the Aufbau principle (lowest energy first), the Pauli exclusion principle (at most 2 electrons per MO, with opposite spins), and Hund's rule (electrons occupy a set of degenerate MOs singly, with parallel spins, before any pairing occurs).

Two energy orders. For period-2 homonuclear diatomics, 2s–2p orbital mixing shifts where the σ2pz\sigma 2p_z orbital sits relative to the degenerate π2px,π2py\pi 2p_x, \pi 2p_y pair. Which order applies depends on the molecule:

Li2 to N2:σ1s<σ∗1s<σ2s<σ∗2s<(π2px=π2py)<σ2pz<(π∗2px=π∗2py)<σ∗2pz\text{Li}_2 \text{ to } \text{N}_2:\quad \sigma1s < \sigma^*1s < \sigma2s < \sigma^*2s < (\pi2p_x=\pi2p_y) < \sigma2p_z < (\pi^*2p_x=\pi^*2p_y) < \sigma^*2p_z
O2,F2,Ne2:σ1s<σ∗1s<σ2s<σ∗2s<σ2pz<(π2px=π2py)<(π∗2px=π∗2py)<σ∗2pz\text{O}_2, \text{F}_2, \text{Ne}_2:\quad \sigma1s < \sigma^*1s < \sigma2s < \sigma^*2s < \sigma2p_z < (\pi2p_x=\pi2p_y) < (\pi^*2p_x=\pi^*2p_y) < \sigma^*2p_z

where: each symbol names one MO; the subscript zz marks the orbital along the internuclear axis, and x,yx,y mark the two mutually perpendicular π\pi orbitals, which stay degenerate (equal energy) in both orders. The only thing that swaps is whether σ2pz\sigma2p_z sits above or below the π2p\pi2p pair — mixing up which order belongs to which molecule is the single most common MOT error.

Bond order and magnetism. Once the electrons are placed, bond order tells you how strong the bond is and magnetism follows from whether every MO is fully paired.

Figure
Bonding and antibonding sigma molecular orbitals formed from two 1s atomic orbitals showing constructive overlap building density between the nuclei and destructive overlap leaving a node between them
LCAO: two 1s orbitals combine into sigma bonding and sigma star antibonding MOs sigma 1s — bonding (in phase) density builds up between nuclei lower energy, stabilising sigma star 1s — antibonding (out of phase) node between nuclei higher energy, destabilising Two atomic orbitals always combine into two molecular orbitals one bonding, lower in energy, and one antibonding, marked with an asterisk
Derivation ABond order from a molecular orbital configuration

The goal is to show, using nothing beyond counting electrons in an MO configuration you already know how to write, that a single formula — bond order equals half the difference between bonding and antibonding electrons — reproduces the familiar single, double and triple bonds of Lewis structures, so that MOT and VBT agree wherever VBT already works while MOT additionally gets magnetism right.

Step 1. Start from the definition. Every electron placed in a bonding MO stabilises the molecule by one "unit" of bonding character; every electron placed in an antibonding MO cancels one unit of that stabilisation, since an antibonding electron pulls density away from between the nuclei rather than into it. The net stabilisation, per bonding pair, is therefore proportional to Nb−NaN_b - N_a, the bonding electrons minus the antibonding electrons.

Step 2. Convert that net count into a bond count. A single covalent bond, in Lewis terms, is one shared electron pair, i.e. 2 net bonding electrons. So dividing the net by 2 converts an electron count into a number of effective bonds:

Bond order=Nb−Na2\text{Bond order} = \frac{N_b - N_a}{2}

where: NbN_b is the total number of electrons occupying bonding MOs (σ\sigma, π\pi, without asterisks) and NaN_a is the total occupying antibonding MOs (marked with asterisks).

Step 3. Check it against a known case. For H2\text{H}_2 (2 electrons, configuration σ1s2\sigma1s^2), Nb=2N_b=2, Na=0N_a=0, giving bond order 11 — a single bond, matching H−H\text{H}-\text{H}.

Step 4. Check the null case. For the hypothetical He2\text{He}_2 (4 electrons, configuration σ1s2σ∗1s2\sigma1s^2\sigma^*1s^2), Nb=2N_b=2, Na=2N_a=2, giving bond order 2−22=0\dfrac{2-2}{2}=0. A bond order of zero means no net stabilisation — MOT correctly predicts that He2\text{He}_2 does not exist as a stable molecule, a result Lewis structures cannot produce directly.

Step 5. Read the general trend. Higher bond order means more net bonding electron density between the nuclei, so it correlates with a shorter, stronger bond — exactly the same trend as bond order defined from Lewis structures, but now derivable from an orbital picture that also predicts magnetism, which is what the worked example below uses for O2\text{O}_2.

Worked example

Construct the MO electron configuration of O2\text{O}_2 (16 electrons), find its bond order, and explain why liquid oxygen is attracted into a magnetic field.

Step 1. Each oxygen atom contributes 8 electrons, so the molecule has 8×2=168\times2=16 electrons to place in MOs.

Step 2. O2\text{O}_2 uses the σ2pz\sigma2p_z-below-π2p\pi2p order. Filling from the bottom: σ1s2 σ∗1s2 σ2s2 σ∗2s2 σ2pz2 π2px2 π2py2\sigma1s^2\,\sigma^*1s^2\,\sigma2s^2\,\sigma^*2s^2\,\sigma2p_z^2\,\pi2p_x^2\,\pi2p_y^2 places 14 electrons, leaving 2 electrons for the degenerate π∗2px,π∗2py\pi^*2p_x,\pi^*2p_y pair.

Step 3. By Hund's rule, the last 2 electrons occupy π∗2px\pi^*2p_x and π∗2py\pi^*2p_y singly, with parallel spins, rather than pairing up in just one of the two orbitals. Full configuration: σ1s2 σ∗1s2 σ2s2 σ∗2s2 σ2pz2 π2px2 π2py2 π∗2px1 π∗2py1\sigma1s^2\,\sigma^*1s^2\,\sigma2s^2\,\sigma^*2s^2\,\sigma2p_z^2\,\pi2p_x^2\,\pi2p_y^2\,\pi^*2p_x^1\,\pi^*2p_y^1.

Step 4. Count bonding and antibonding electrons. Bonding: σ1s(2)+σ2s(2)+σ2pz(2)+π2px(2)+π2py(2)=Nb=10\sigma1s(2)+\sigma2s(2)+\sigma2p_z(2)+\pi2p_x(2)+\pi2p_y(2) = N_b = 10. Antibonding: σ∗1s(2)+σ∗2s(2)+π∗2px(1)+π∗2py(1)=Na=6\sigma^*1s(2)+\sigma^*2s(2)+\pi^*2p_x(1)+\pi^*2p_y(1) = N_a = 6.

Step 5. Apply the bond-order formula from the derivation above: Bond order=10−62=2\text{Bond order} = \dfrac{10-6}{2} = 2, matching the O=O\text{O}=\text{O} double bond of the Lewis structure.

Step 6. Read off the magnetism. Two electrons, in π∗2px\pi^*2p_x and π∗2py\pi^*2p_y, are unpaired. A species with any unpaired electrons is paramagnetic — weakly drawn into a magnetic field. Answer: O2\text{O}_2 has bond order 22 and is paramagnetic because of its two unpaired π∗\pi^* electrons — exactly why liquid oxygen clings between the poles of a magnet, a fact plain Lewis structures cannot explain.

Competitive stretch

Repeat the same steps for N2\text{N}_2 (14 electrons, π2p\pi2p-below-σ2pz\sigma2p_z order) and you get σ1s2 σ∗1s2 σ2s2 σ∗2s2 π2px2 π2py2 σ2pz2\sigma1s^2\,\sigma^*1s^2\,\sigma2s^2\,\sigma^*2s^2\,\pi2p_x^2\,\pi2p_y^2\,\sigma2p_z^2, with Nb=10N_b=10, Na=4N_a=4, bond order 10−42=3\dfrac{10-4}{2}=3 — the N≡N\text{N}\equiv\text{N} triple bond, fully paired and hence diamagnetic, consistent with N2\text{N}_2's very high bond enthalpy and short bond length.

Common traps

  • Using the wrong energy order — swapping the "σ2pz\sigma2p_z below π2p\pi2p" order (for O2\text{O}_2, F2\text{F}_2, Ne2\text{Ne}_2) with the "π2p\pi2p below σ2pz\sigma2p_z" order (for Li2\text{Li}_2 through N2\text{N}_2) gives a wrong configuration and a wrong magnetic prediction.
  • Forgetting Hund's rule on the degenerate pair — pairing electrons prematurely instead of placing them singly first is the most common reason students miss O2\text{O}_2's paramagnetism.
  • Treating bond order as always a whole number — ions like O2+\text{O}_2^+ or O2−\text{O}_2^- have fractional or altered bond orders that are still physically meaningful.
  • Assuming any pair of atoms bonds — a calculated bond order of 00 (as for He2\text{He}_2) correctly predicts that no stable molecule forms.

Ask the tutor: Want to build the MO configuration and bond order for a specific diatomic molecule or ion step by step? Ask the AI tutor to work through it with you.

Why does N2 use a different MO energy order than O2?

Because of 2s–2p mixing. In lighter period-2 atoms (up to nitrogen), the 2s and 2p atomic orbitals are close enough in energy to mix, which pushes the σ2pz\sigma2p_z molecular orbital above the π2p\pi2p pair. From oxygen onward the 2s–2p gap is large enough that mixing is negligible, so σ2pz\sigma2p_z drops back below π2p\pi2p, its "expected" position. NCERT expects you to know which order goes with which molecules, not to derive the mixing itself.

If MOT and VBT both give O2 a double bond, why do textbooks call MOT better?

Bond order alone is not where MOT wins — VBT's Lewis structure also gives O2\text{O}_2 a double bond with all electrons paired. The difference is magnetism: that fully-paired Lewis structure predicts O2\text{O}_2 should be diamagnetic, but experimentally liquid oxygen is paramagnetic, clearly attracted into a magnetic field. MOT's electron configuration naturally places two electrons unpaired in the degenerate π∗2p\pi^*2p orbitals, correctly predicting the paramagnetism that VBT cannot explain.

Do I need to know MO diagrams for CO or NO for CBSE?

No. The NCERT Class 11 syllabus restricts MO diagrams to homonuclear diatomics of period 1 and period 2 — H2\text{H}_2, He2\text{He}_2, and Li2\text{Li}_2 through Ne2\text{Ne}_2. Heteronuclear diatomics like CO or NO, where the two atomic orbitals contribute unequally to each MO, are conceptually similar but are out of scope; at most expect a passing mention, not a full worked configuration.

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