Molecular orbital theory

A single Lewis dot structure cannot tell you why liquid oxygen sticks to a magnet. Molecular orbital theory (MOT) can, because it stops treating a bond as two atoms sharing a fixed pair of electrons and instead spreads every valence electron over orbitals that belong to the whole molecule. Two atomic orbitals of similar energy and matching symmetry combine to give one lower-energy bonding orbital and one higher-energy antibonding orbital; electrons then fill this new ladder exactly as they fill atomic orbitals, obeying the Aufbau order, the Pauli principle and Hund's rule. Counting how many electrons land in bonding versus antibonding orbitals gives the bond order directly, and — because MOT keeps track of unpaired electrons in degenerate orbitals — it correctly predicts magnetism, something valence-bond theory simply cannot do. This article builds the qualitative homonuclear and heteronuclear MO pictures you need for CY11001 and walks through the two flagship examples, N2\text{N}_2 and O2\text{O}_2.

The core insight

Combining nn atomic orbitals gives nn molecular orbitals split into lower-energy bonding and higher-energy antibonding sets, and the number of electrons in each set fixes both the bond order, BO=12(nb−na)\text{BO} = \tfrac12(n_b - n_a), and the molecule's magnetism.

The core idea

Orbitals combine, they don't just overlap. When two atomic orbitals ϕA\phi_A and ϕB\phi_B of comparable energy and correct symmetry combine (LCAO, linear combination of atomic orbitals), they produce exactly two molecular orbitals:

ψbonding=c1ϕA+c2ϕB,ψantibonding=c1ϕA−c2ϕB\psi_{\text{bonding}} = c_1\phi_A + c_2\phi_B, \qquad \psi_{\text{antibonding}} = c_1\phi_A - c_2\phi_B

where: ψbonding\psi_{\text{bonding}}, ψantibonding\psi_{\text{antibonding}} are the resulting molecular orbitals, and c1,c2c_1, c_2 are mixing coefficients (equal in size for a homonuclear diatomic, unequal for a heteronuclear one). The bonding combination reinforces electron density between the nuclei and sits lower in energy than either parent AO; the antibonding combination (σ∗\sigma^* or π∗\pi^*) has a node between the nuclei and sits higher in energy. Electrons fill this ladder from the bottom up, obeying the Pauli exclusion principle and Hund's rule — nothing new is invented here, it is the same filling logic you already use for atoms.

Bond order counts the tug-of-war. Once the configuration is written, bond order follows from simple bookkeeping:

Bond order=12(nb−na)\text{Bond order} = \tfrac{1}{2}(n_b - n_a)

where: nbn_b is the number of electrons in bonding MOs and nan_a is the number in antibonding MOs. A higher bond order means a shorter, stronger bond; a bond order of zero (as for He2\text{He}_2) means the molecule simply does not form, because bonding and antibonding effects cancel exactly.

The filling order flips partway across period 2. For Li2\text{Li}_2 through N2\text{N}_2, ss–pp mixing pushes σ2pz\sigma_{2p_z} above the π2p\pi_{2p} pair:

σ2s<σ2s∗<π2px=π2py<σ2pz<π2px∗=π2py∗<σ2pz∗\sigma_{2s} < \sigma_{2s}^{*} < \pi_{2p_x}=\pi_{2p_y} < \sigma_{2p_z} < \pi^{*}_{2p_x}=\pi^{*}_{2p_y} < \sigma^{*}_{2p_z}

For O2\text{O}_2, F2\text{F}_2, Ne2\text{Ne}_2, the wider 2s2s–2p2p gap suppresses that mixing and σ2pz\sigma_{2p_z} drops back below π2p\pi_{2p}:

σ2s<σ2s∗<σ2pz<π2px=π2py<π2px∗=π2py∗<σ2pz∗\sigma_{2s} < \sigma_{2s}^{*} < \sigma_{2p_z} < \pi_{2p_x}=\pi_{2p_y} < \pi^{*}_{2p_x}=\pi^{*}_{2p_y} < \sigma^{*}_{2p_z}

Heteronuclear diatomics keep the O2_2-type pattern but lose the symmetry. For CO and NO the two atoms have different AO energies, so c1≠c2c_1 \ne c_2: bonding MOs lean toward the more electronegative atom, antibonding MOs lean toward the less electronegative one.

Figure
Bonding and antibonding molecular orbital energy levels formed from two atomic orbitals of equal energy
Two atomic orbitals combine into one bonding and one antibonding MO AO phi A AO phi B antibonding sigma star (higher energy) bonding sigma (lower energy) combine
Derivation Abond order of N2 from its MO configuration

Predict the bond order and magnetism of the nitrogen molecule N2\text{N}_2 purely from its valence electron count and the MO filling order that applies before the 2s2s-2p2p crossover, then check the result against the familiar triple-bond, diamagnetic picture that Lewis and valence-bond theory already give you, so you can see exactly where MOT reproduces old results and where it adds new information.

Step 1. Count valence electrons. Each nitrogen atom contributes 5 valence electrons (2s22p32s^2 2p^3), giving 10 valence electrons total for N2\text{N}_2. The 1s core electrons form σ1s\sigma_{1s} and σ1s∗\sigma_{1s}^{*}, which cancel exactly in bond order and are dropped from the count.

Step 2. Fill the "before N2\text{N}_2" order, σ2s<σ2s∗<π2px=π2py<σ2pz\sigma_{2s} < \sigma_{2s}^{*} < \pi_{2p_x}=\pi_{2p_y} < \sigma_{2p_z}, with 10 electrons, two per orbital, lowest energy first: σ2s2 σ2s∗2 π2px2π2py2 σ2pz2\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\pi_{2p_x}^2\pi_{2p_y}^2\,\sigma_{2p_z}^2. Every orbital up to σ2pz\sigma_{2p_z} is now full, and no electrons remain for the antibonding π∗\pi^{*} or σ2pz∗\sigma^{*}_{2p_z} levels.

Step 3. Separate bonding from antibonding electrons. Bonding electrons come from σ2s\sigma_{2s}, both π2p\pi_{2p} orbitals and σ2pz\sigma_{2p_z}: nb=2+2+2+2=8n_b = 2 + 2 + 2 + 2 = 8. Antibonding electrons come only from σ2s∗\sigma_{2s}^{*}: na=2n_a = 2.

Step 4. Apply the bond order formula. Bond order=12(nb−na)=12(8−2)=3\text{Bond order} = \tfrac12(n_b - n_a) = \tfrac12(8-2) = 3. This matches the classical triple bond :N≡N::\text{N}\equiv\text{N}:, and it explains why N2\text{N}_2 is short (≈109.8 pm\approx 109.8\ \text{pm}), extremely strong and chemically inert. Because every filled orbital, including the degenerate π2p\pi_{2p} pair, holds two paired electrons, N2\text{N}_2 is diamagnetic — MOT reproduces the Lewis-structure conclusion exactly in this case.

Worked example

Problem: Use MOT to find the bond order of O2\text{O}_2 and explain why it is paramagnetic, a result plain Lewis structures and valence-bond theory get wrong.

Step 1. Count valence electrons: each oxygen atom contributes 6 (2s22p42s^2 2p^4), giving 12 valence electrons for O2\text{O}_2.

Step 2. Since oxygen lies after the 2s2s-2p2p crossover, use the "after N2\text{N}_2" order, σ2s<σ2s∗<σ2pz<π2px=π2py<π2px∗=π2py∗\sigma_{2s} < \sigma_{2s}^{*} < \sigma_{2p_z} < \pi_{2p_x}=\pi_{2p_y} < \pi^{*}_{2p_x}=\pi^{*}_{2p_y}. Filling 12 electrons gives σ2s2 σ2s∗2 σ2pz2 π2px2π2py2 π2px∗1π2py∗1\sigma_{2s}^2\,\sigma_{2s}^{*2}\,\sigma_{2p_z}^2\,\pi_{2p_x}^2\pi_{2p_y}^2\,\pi^{*1}_{2p_x}\pi^{*1}_{2p_y} — the last two electrons enter the degenerate π∗\pi^{*} pair singly, with parallel spins, by Hund's rule, rather than pairing up in one orbital.

Step 3. Split bonding and antibonding electrons: nb=2(σ2s)+2(σ2pz)+2+2(π2p)=8n_b = 2(\sigma_{2s}) + 2(\sigma_{2p_z}) + 2+2(\pi_{2p}) = 8; na=2(σ2s∗)+1+1(π∗)=4n_a = 2(\sigma_{2s}^{*}) + 1+1(\pi^{*}) = 4. Bond order =12(8−4)=2= \tfrac12(8-4) = 2, matching the classical double bond O=O\text{O}=\text{O}.

Step 4. Read off the magnetism directly from the configuration: two unpaired electrons sit in the degenerate π2px∗,π2py∗\pi^{*}_{2p_x}, \pi^{*}_{2p_y} orbitals, so O2\text{O}_2 is paramagnetic — it is drawn into a magnetic field, exactly as liquid oxygen is observed to cling between the poles of a magnet. Answer: bond order 2, with two unpaired electrons giving paramagnetism, a fact no simple electron-pair Lewis structure can predict.

Competitive stretch

Remove one electron to form O2+\text{O}_2^{+} (dioxygenyl cation): 11 valence electrons empty one electron from a π∗\pi^{*} orbital, giving nb=8n_b = 8, na=3n_a = 3, so bond order =12(8−3)=2.5= \tfrac12(8-3) = 2.5 — stronger and shorter than neutral O2\text{O}_2. Add an electron instead to form superoxide O2−\text{O}_2^{-}: 13 valence electrons fill both π∗\pi^{*} orbitals partly further, giving na=5n_a = 5 and bond order =12(8−5)=1.5= \tfrac12(8-5) = 1.5 — weaker than O2\text{O}_2. The trend bond order O2+(2.5)>O2(2)>O2−(1.5)\text{O}_2^{+}(2.5) > \text{O}_2(2) > \text{O}_2^{-}(1.5) tracks measured bond lengths exactly, since every added electron goes into an antibonding orbital.

Common traps

  • Using one filling order for every second-row diatomic. The σ2pz\sigma_{2p_z}/π2p\pi_{2p} ordering swaps between N2\text{N}_2 and O2\text{O}_2 because the 2s2s-2p2p energy gap widens across the period — check which side of the crossover your molecule sits on.
  • Ignoring degeneracy when predicting magnetism. Magnetism depends entirely on whether the degenerate π/π∗\pi/\pi^{*} pair holds its electrons singly or doubly; always apply Hund's rule explicitly to that pair before concluding paramagnetic or diamagnetic.
  • Treating heteronuclear MOs as 50:50 mixtures. In CO or NO the coefficients c1≠c2c_1 \ne c_2; bonding MOs skew toward the more electronegative atom and antibonding MOs skew toward the less electronegative one, which is why CO binds metals through carbon's slightly antibonding-character HOMO.
  • Confusing bond order with a Lewis dot count. Bond order comes purely from nbn_b and nan_a and can be a half-integer once ions like O2+\text{O}_2^{+} or O2−\text{O}_2^{-} are involved — it need not match a simple formal bond count.

Ask the tutor: if you want a step-by-step MO diagram for NO or another odd-electron heteronuclear molecule, ask the AI tutor to build it with you.

Why does the sigma 2p and pi 2p ordering swap between N2 and O2?

Across period 2, the energy gap between the 2s2s and 2p2p atomic orbitals widens as nuclear charge increases. For Li2\text{Li}_2 through N2\text{N}_2 that gap is small enough that the σ2s\sigma_{2s} and σ2pz\sigma_{2p_z} orbitals mix (s-p mixing), which pushes σ2pz\sigma_{2p_z} above the π2p\pi_{2p} pair in energy. For O2\text{O}_2, F2\text{F}_2 and Ne2\text{Ne}_2 the 2s2s-2p2p gap is large enough that this mixing becomes negligible, so σ2pz\sigma_{2p_z} falls back to its "expected" position below π2p\pi_{2p}. Always check which side of N2\text{N}_2 your molecule is on before filling electrons.

How do I predict whether a diatomic species is paramagnetic without drawing the full diagram?

Write the valence electron count, fill the correct order (checking the N2/O2 crossover), and look only at whether the last filled degenerate set — usually π2p\pi_{2p} or π2p∗\pi^{*}_{2p} — holds an odd number of electrons across its two orbitals. If any degenerate pair has unpaired electrons by Hund's rule, the species is paramagnetic; if every orbital is fully paired, it is diamagnetic. You rarely need to compute the full bond order just to answer a magnetism question.

Why can't valence-bond theory explain O2's paramagnetism?

Valence-bond theory (and the Lewis structure it is based on) builds a bond by pairing up all available electrons into two-electron bonds and lone pairs, so by construction it leaves no unpaired electrons for O2\text{O}_2's double bond. Molecular orbital theory instead fills a shared ladder of molecular orbitals following the Aufbau principle and Hund's rule, and for O2\text{O}_2 that ladder happens to place the last two electrons in a degenerate π∗\pi^{*} pair, which Hund's rule keeps unpaired. This is the single clearest case where MOT gives a physically correct prediction that the electron-pairing picture cannot.

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