Theory — Molecular Orbital Theory
1. From atomic orbitals to molecular orbitals (LCAO)
When two atoms approach, their atomic orbitals (AOs) overlap and combine. The linear combination of atomic orbitals (LCAO) model says that N atomic orbitals combine to give N molecular orbitals. Two 1s orbitals combine two ways:
- In phase (constructive): electron density builds between the nuclei → a lower-energy bonding orbital (σ).
- Out of phase (destructive): a node appears between the nuclei → a higher-energy antibonding orbital (σ*).
Electrons in bonding orbitals hold the molecule together; electrons in antibonding orbitals push it apart. This single idea — count bonding versus antibonding electrons — is the engine of the whole subject.
2. Bond order, magnetism, and frontier orbitals
Once the molecular orbitals are filled (lowest energy first, two electrons per orbital, one electron in each degenerate orbital before pairing — the aufbau, Pauli, and Hund rules), three properties follow directly:
- Magnetism: if any orbital holds a single unpaired electron, the molecule is paramagnetic (attracted to a magnet). If all electrons are paired, it is diamagnetic.
- HOMO = highest occupied molecular orbital; LUMO = lowest unoccupied molecular orbital. These two — the frontier orbitals — control most of a molecule's chemistry: the HOMO donates electrons (nucleophile / base / reductant), the LUMO accepts them (electrophile / acid / oxidant).
3. Second-row diatomics and s–p mixing
For homonuclear diatomics of the second period, the valence 2s and 2p orbitals build eight molecular orbitals. There are two possible orderings, and which one applies depends on the nuclear charge:
| Molecules | Ordering (low → high energy) | Why |
|---|---|---|
| B₂, C₂, N₂ (and lighter) | σ2s < σ*2s < π2p < σ2p < π*2p < σ*2p | Strong s–p mixing pushes σ2p above the π2p pair. |
| O₂, F₂, Ne₂ | σ2s < σ*2s < σ2p < π2p < π*2p < σ*2p | The 2s–2p gap is large, mixing is weak, so σ2p drops below π2p. |
The two orderings differ only in the relative position of σ2p and π2p, but that difference is real and measurable: B₂ is paramagnetic precisely because the π2p orbitals lie below σ2p and fill first, one electron in each.
4. Heteronuclear diatomics: CO and NO
When the two atoms differ, the more electronegative atom's orbitals sit lower in energy, so the molecular orbitals are lopsided — bonding orbitals lean toward the electronegative atom, antibonding orbitals toward the other. In CO (isoelectronic with N₂, bond order 3) the HOMO is a σ orbital whose largest lobe sits on carbon; that carbon-based lone pair is what donates to metal centres. NO has one extra electron (eleven valence electrons), which must go into a π* orbital, giving a bond order of 2.5 and one unpaired electron — an odd-electron radical.
5. Cyclic π systems, the Frost circle, and aromaticity
For a planar, fully conjugated ring, the π molecular orbital energies can be read straight off a simple mnemonic. Inscribe the regular polygon inside a circle with one vertex pointing down; each point where the polygon touches the circle marks the energy of one π MO. Orbitals below the circle's mid-line are bonding, on the line are non-bonding, above it are antibonding.
Fill the π electrons into that pattern. A ring is:
- Aromatic if it is cyclic, planar, fully conjugated, and holds 4n+2 π electrons (2, 6, 10…) — every bonding MO is filled and the shell is closed. Exceptionally stable (benzene, 6 π e⁻).
- Antiaromatic if it meets the same geometric conditions but holds 4n π electrons (4, 8…) — electrons are forced, unpaired, into non-bonding orbitals. Strongly destabilised (cyclobutadiene, 4 π e⁻).
- Non-aromatic if it is not planar or not fully conjugated, so the counting rule does not apply.
6. Frontier orbitals, reactivity, acidity, and basicity
Frontier-orbital reasoning turns MO diagrams into predictions. A high-energy HOMO makes a molecule a good electron donor (base, nucleophile); a low-energy LUMO makes it a good electron acceptor (acid, electrophile). Aromatic stabilisation of a product ion is a powerful driving force:
- Cyclopentadiene is unusually acidic for a hydrocarbon because losing a proton gives the aromatic cyclopentadienyl anion (6 π e⁻).
- Cycloheptatriene readily loses hydride to give the aromatic tropylium cation (6 π e⁻).
- Pyrrole uses its nitrogen lone pair inside the aromatic π sextet, so that lone pair is not freely available — pyrrole is a very weak base, and its N–H is weakly acidic.
- Pyridine keeps its nitrogen lone pair in an in-plane sp² orbital, outside the π system, so it is fully available — pyridine is a genuine base.
Everything in this lab reduces to two moves: build the orbital picture, then read bond order, magnetism, and the frontier orbitals off it.
Apparatus
Molecular orbital theory is a pencil-and-paper (and computer) analysis rather than a bench experiment. These are the conceptual and computational tools you use to build and read an MO picture; in the simulation they are provided for you, but each corresponds to a real tool a chemist uses.
Instructions
This is the only activity this week, so treat it as a full working session. Two examples are worked out completely below; study them, then reproduce the same steps for your own assigned molecule and ring system in the Simulation.
Worked example A — He₂ (bond order zero)
Step 2 — fill the MOs (only 1s available) σ1s² σ*1s²
Step 3 — bond order BO = ½(2 bonding − 2 antibonding) = ½(2 − 2) = 0
Step 4 — magnetism & frontier orbitals All electrons paired → diamagnetic. HOMO = σ*1s, LUMO = none (no bonding).
Step 5 — connect to reality
Worked example B — O₂ (paramagnetism)
Step 2 — choose the ordering & fill O is in the heavy group (O, F, Ne), so σ2p sits below π2p (no s–p mixing):
σ2s² σ*2s² σ2p² π2p⁴ π*2p²
The last 2 electrons go one each into the two degenerate π*2p orbitals (Hund's rule).
Step 3 — bond order Bonding = 2 + 2 + 4 = 8; antibonding = 2 + 2 = 4. BO = ½(8 − 4) = 2
Step 4 — magnetism & frontier orbitals Two unpaired electrons in π*2p → paramagnetic. HOMO = π*2p, LUMO = π*2p (same, partly filled).
Step 5 — connect to reality
Now do it yourself
Learning outcomes. By the end you should be able to (1) construct MO diagrams for simple diatomics and for benzene, predict bond order and magnetic properties, and explain aromaticity, antiaromaticity, and non-aromaticity in MO terms; and (2) use HOMO–LUMO (frontier-orbital) reasoning to predict reactivity and classify the acidity and basicity of cyclic compounds.
Simulation
Four interactive parts. Predict first, then press Check to compare against the model. Use ↺ Reset Simulation to clear all answers.
Pick your assigned molecule, predict its four properties, and press Check. He₂ and O₂ (dashed) are the worked examples — open them to compare against Instructions.
Pick a ring, predict its π-system, and press Check. The Frost-circle diagram is revealed after you answer.
Six frontier-orbital problems: reactivity, acidity, and basicity of cyclic compounds. Pick an answer to reveal the explanation.
Five connections between an MO result and a real observation. Pick an answer to reveal the explanation.
Team Questions
Work these with your team. Type an answer and press Check for instant feedback, then use the peer-response prompts to verify each other's arithmetic.
Peer response — check each other's arithmetic
- Swap notebooks with a teammate who analysed a different molecule. Re-count their valence electrons from scratch: does the total match the group number sum (and ±1 for a charged species)?
- Recompute their bond order independently: BO = ½(bonding − antibonding). Do you get the same number? If not, find whether the disagreement is in the ordering (s–p mixing) or in the electron count.
- Check their magnetism call: count unpaired electrons in the highest partly-filled level. Paramagnetic needs at least one unpaired electron — confirm their claim against their own diagram.
- Verify their HOMO and LUMO are the highest filled and lowest empty levels in their ordering, not yours — B₂–N₂ use the mixed ordering, O₂–Ne₂ do not.
- For the ring systems, re-inscribe the Frost circle and re-count π electrons; confirm the 4n+2 / 4n verdict and whether the shell is open or closed.
Example Lab Notebook Entry
Use the format below as a template for your own molecule.
Molecular Orbital Theory — Lab Notebook Entry
Submitted by: [Student Name]
Course: Organic Chemistry · Section: 201-A · Date: May 10, 2026
Objective
To construct the molecular orbital diagram of my assigned molecule (worked here for N₂), determine its bond order, magnetism, and frontier orbitals, and connect the result to a real observation; and to extend MO reasoning to benzene, aromaticity, and the acidity/basicity of cyclic compounds.
Worked molecule — N₂
| Step | Result |
|---|---|
| Valence electrons | 5 + 5 = 10 |
| Ordering | N is light (Z = 7), so s–p mixing applies: π2p below σ2p |
| Configuration | σ2s² σ*2s² π2p⁴ σ2p² |
| Bond order | ½(8 bonding − 2 antibonding) = 3 |
| Magnetism | all paired → diamagnetic |
| HOMO / LUMO | σ2p / π*2p |
| Real observation | bond order 3 + large HOMO–LUMO gap → N₂ is very inert (the unreactive bulk of air) |
Discussion
Molecular orbital theory builds bonding from the whole-molecule orbitals formed by combining atomic orbitals. Counting bonding versus antibonding electrons gives the bond order directly, and the presence or absence of unpaired electrons gives the magnetism. For the second-row diatomics, the key subtlety is s–p mixing: for B₂ through N₂ the π2p orbitals lie below σ2p, whereas for O₂ through Ne₂ the order reverses. This is not a bookkeeping detail — it is why B₂ is observed to be paramagnetic and why the O₂ and N₂ diagrams look different.
The two worked examples frame the method. He₂ has four electrons filling σ1s and σ*1s equally: bond order 0, no molecule — helium stays monatomic. O₂ has twelve valence electrons ending in two singly-occupied π*2p orbitals: bond order 2 and, crucially, paramagnetic, which a Lewis O=O structure cannot explain. My molecule, N₂, sits between them with bond order 3 and a diamagnetic, closed-shell configuration; its inertness follows from the strong triple bond and the large gap between the σ2p HOMO and the π*2p LUMO.
Extending to cyclic π systems, the Frost circle gives the π-MO pattern of a conjugated ring. Benzene's six π electrons fill all three bonding orbitals for a closed aromatic shell (4n+2, n = 1), the origin of its exceptional stability and its preference for substitution over addition. Cyclobutadiene, with four π electrons, is forced to place two electrons unpaired in non-bonding orbitals — antiaromatic and highly unstable. The same 4n+2 rule explains why cyclopentadiene is unusually acidic (its anion is the aromatic six-π-electron cyclopentadienyl) and why cycloheptatriene gives the aromatic tropylium cation on loss of hydride.
Frontier-orbital reasoning ties the ideas together. A high HOMO makes a molecule a good donor (base/nucleophile); a low LUMO makes it a good acceptor (acid/electrophile). Pyridine's nitrogen lone pair sits in an in-plane sp² orbital outside the aromatic π system, so it is available for protonation — pyridine is a real base. Pyrrole's nitrogen lone pair is part of the aromatic sextet, so donating it would destroy aromaticity; pyrrole is therefore a very weak base and instead a weak N–H acid.
Conclusion
MO theory predicts bond order, magnetism, and frontier-orbital behaviour that Lewis structures miss, and the same framework scales from diatomics to aromatic rings. The method is always the same: build the orbital picture, count bonding versus antibonding electrons, then read bond order, magnetism, and the HOMO/LUMO off the diagram.
References
1. Miessler, G. L.; Fischer, P. J.; Tarr, D. A. Inorganic Chemistry, 5th ed., Pearson, 2014, Ch 5.
2. Albright, T. A.; Burdett, J. K.; Whangbo, M.-H. Orbital Interactions in Chemistry, 2nd ed., Wiley, 2013.
3. Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed., Oxford, 2012, Ch 4 & 7.
4. Fleming, I. Molecular Orbitals and Organic Chemical Reactions, Wiley, 2010.
Practice Questions
Work each out before opening the hint.