Molecular Orbital Diagrams Explained: Bond Order, Magnetism and the s–p Mixing Rule
Short answer: A molecular orbital (MO) diagram shows how atomic orbitals combine into bonding and antibonding orbitals. Fill the orbitals from the bottom with Hund's rule, then use bond order = (bonding − antibonding electrons) / 2. O₂ has bond order 2 and 2 unpaired electrons (paramagnetic). N₂ has bond order 3 and is diamagnetic. For 14 or fewer electrons, π2p lies below σ2p. The free MO diagram generator draws any diatomic from H₂ to Ne₂, plus CO, NO, CN⁻ and HF.
Key terms in one place
| Molecular orbital (MO) | An orbital that spreads over the whole molecule, formed by combining atomic orbitals. |
|---|---|
| Bonding MO | Lower in energy than the atomic orbitals. Electron density sits between the nuclei. Labels: σ, π. |
| Antibonding MO | Higher in energy, with a node between the nuclei. Labels: σ*, π*. |
| Bond order | (bonding electrons − antibonding electrons) / 2. Bond order 0 means no stable bond. |
| Paramagnetic | Has one or more unpaired electrons and is pulled into a magnetic field. O₂ is the classic case. |
| Diamagnetic | All electrons paired. Weakly pushed out of a magnetic field. N₂ and F₂ are examples. |
| HOMO and LUMO | Highest occupied and lowest unoccupied MO. A half-filled HOMO is a SOMO. |
The rules for drawing an MO diagram
- Count the electrons. Add the atomic numbers, subtract the charge for cations and add it for anions. O₂ has 16, O₂⁺ has 15, O₂²⁻ has 18.
- Choose the orbital order. With 14 or fewer electrons (Li₂ to N₂, CO, CN⁻, NO⁺) use σ2s, σ*2s, π2p, σ2p, π*2p, σ*2p. With more than 14 electrons (O₂, F₂, Ne₂, NO, O₂⁻) use σ2s, σ*2s, σ2p, π2p, π*2p, σ*2p.
- Fill from the bottom. Each orbital holds 2 electrons with opposite spins (Pauli principle).
- Apply Hund's rule. In the degenerate π or π* pair, place one electron in each orbital before pairing.
- Read the results. Bond order from the formula, magnetism from the unpaired electrons, HOMO and LUMO from the top of the filling.
Why the order switches after nitrogen
The switch is caused by s–p mixing. In B, C and N the 2s and 2p energies are close. The σ2s and σ2p orbitals have the same symmetry, so they interact and push σ2p up, above π2p. From oxygen onward the 2s–2p gap is larger (about 16 eV for O and 21 eV for F), the mixing is weak, and σ2p falls below π2p.
How to use the MO diagram generator step by step
- Open the molecular orbital diagram generator.
- Type a species in Molecule or ion, for example
O2,N2+,O2 2-,NO,COorCN-. Or pick Atom 1, Atom 2 and Charge. - Or press an example button such as B₂, C₂, O₂⁻ or HF.
- Keep Orbital order on Automatic (14-electron rule). To see why the rule matters, switch to No s–p mixing and watch B₂ become diamagnetic.
- Read the bond order, MO configuration, HOMO, LUMO and magnetism in the result table.
- Press Download SVG or Download PNG to save the diagram.
Worked examples
1. Oxygen, O₂
16 electrons: KK σ2s² σ*2s² σ2p² π2p⁴ π*2p². Bonding electrons = 10 (including σ1s²), antibonding = 6. Bond order = (10 − 6)/2 = 2. The two π* electrons sit in separate orbitals with parallel spins, so O₂ has 2 unpaired electrons and is paramagnetic. Liquid oxygen sticks between the poles of a magnet; liquid nitrogen does not.
2. Nitrogen, N₂
14 electrons: KK σ2s² σ*2s² π2p⁴ σ2p². Bond order = (10 − 4)/2 = 3. All electrons are paired, so N₂ is diamagnetic. The HOMO is σ2p and the LUMO is π*2p. The triple bond explains the high bond energy of 945 kJ/mol.
3. B₂ and C₂
B₂ has 10 electrons: KK σ2s² σ*2s² π2p¹ π2p¹. Bond order 1, 2 unpaired electrons, paramagnetic. C₂ has 12 electrons: KK σ2s² σ*2s² π2p⁴. Bond order 2, diamagnetic. The measured paramagnetism of B₂ is the experimental proof that π2p lies below σ2p for light elements.
4. The oxygen series
O₂⁺ (15 e) removes one π* electron: bond order 2.5, bond length 112 pm. O₂⁻, superoxide (17 e): bond order 1.5, 133 pm. O₂²⁻, peroxide (18 e): bond order 1, 149 pm, diamagnetic. More antibonding electrons give a longer, weaker bond.
5. NO and NO⁺
NO has 15 electrons with one electron in π*2p: bond order 2.5, paramagnetic. Removing that electron gives NO⁺ (14 e), isoelectronic with N₂: bond order 3, diamagnetic, and the bond shortens from 115 pm to 106 pm.
6. CO and HF (heteronuclear)
CO has 14 electrons like N₂: bond order 3, diamagnetic. Oxygen is more electronegative, so its atomic orbitals sit lower. The HOMO (σ2p, also called 3σ) is mostly a carbon lone pair, which is why CO binds metals through carbon. In HF the H 1s overlaps only with one F 2p orbital. The other F orbitals are nonbonding, so HF has bond order 1.
Summary table
| Species | Electrons | Valence configuration | Bond order | Magnetism |
|---|---|---|---|---|
| H₂ | 2 | σ1s² | 1 | diamagnetic |
| He₂⁺ | 3 | σ1s² σ*1s¹ | 0.5 | paramagnetic (1) |
| Li₂ | 6 | σ2s² | 1 | diamagnetic |
| Be₂ | 8 | σ2s² σ*2s² | 0 | not stable |
| B₂ | 10 | σ2s² σ*2s² π2p² | 1 | paramagnetic (2) |
| C₂ | 12 | σ2s² σ*2s² π2p⁴ | 2 | diamagnetic |
| N₂, CO, CN⁻, NO⁺ | 14 | σ2s² σ*2s² π2p⁴ σ2p² | 3 | diamagnetic |
| NO, O₂⁺ | 15 | … σ2p² π2p⁴ π*2p¹ | 2.5 | paramagnetic (1) |
| O₂ | 16 | σ2s² σ*2s² σ2p² π2p⁴ π*2p² | 2 | paramagnetic (2) |
| O₂⁻ | 17 | … π*2p³ | 1.5 | paramagnetic (1) |
| F₂, O₂²⁻ | 18 | … π*2p⁴ | 1 | diamagnetic |
| Ne₂ | 20 | … π*2p⁴ σ*2p² | 0 | not stable |
Common mistakes
- Using the O₂ order for B₂, C₂ or N₂. It gives the wrong magnetism for B₂ and C₂.
- Pairing electrons in π* too early. O₂ then looks diamagnetic, which is wrong.
- Forgetting to change the electron count for ions.
- Counting core electrons on one side of the formula but not the other. Count all electrons, or valence only, but do it the same way for bonding and antibonding.
- Thinking only odd-electron species are paramagnetic. O₂ (16) and B₂ (10) are even and still paramagnetic.
Exam tips
- JEE and NEET: learn the shortcut for second-row diatomics by total electrons: 10 gives 1, 12 gives 2, 14 gives 3, 15 gives 2.5, 16 gives 2, 17 gives 1.5, 18 gives 1 and 20 gives 0. Odd counts are always paramagnetic. Of the even counts, 10 (B₂) and 16 (O₂, NO⁻) are also paramagnetic.
- JEE and NEET: to rank bond lengths, rank bond orders in reverse: O₂²⁻ > O₂⁻ > O₂ > O₂⁺.
- AP Chemistry: MO theory explains what Lewis structures cannot: O₂ is paramagnetic even though its Lewis structure has all electrons paired.
- A-level and IB: be ready to define σ and π bonds by head-on and side-on overlap, and to explain why a π bond is weaker than a σ bond.
- University: use g and u labels for homonuclear diatomics (σg, σu*, πu, πg*), and link the MO diagram to photoelectron spectra and to CO back-bonding in metal carbonyls.
Check any answer in seconds with the MO diagram generator, then download the diagram for your notes.
Questions and answers
What is a molecular orbital diagram?
It is an energy diagram that shows the atomic orbitals of two atoms on the sides and the molecular orbitals they form in the middle. Electrons are drawn as arrows, and the diagram gives the bond order, magnetism, HOMO and LUMO.
How do you calculate bond order from an MO diagram?
Bond order = (electrons in bonding orbitals − electrons in antibonding orbitals) / 2. For O₂: (10 − 6)/2 = 2. For N₂: (10 − 4)/2 = 3.
Why is O₂ paramagnetic?
Its last two electrons occupy the two degenerate π*2p orbitals singly, with parallel spins (Hund's rule). Two unpaired electrons make O₂ paramagnetic.
When is σ2p below π2p?
For O₂, F₂, Ne₂ and other species with more than 14 electrons. For Li₂ to N₂ and species with 14 or fewer electrons, s–p mixing pushes σ2p above π2p.
Which diatomic molecules have bond order 3?
N₂, CO, CN⁻ and NO⁺. They all have 14 electrons and the configuration KK σ2s² σ*2s² π2p⁴ σ2p².
Why does He₂ not exist?
He₂ has 2 bonding and 2 antibonding electrons, so its bond order is 0. There is no net bond. He₂⁺ has bond order 0.5 and does exist.
Is B₂ paramagnetic or diamagnetic?
B₂ is paramagnetic with 2 unpaired electrons, one in each π2p orbital. This only works if π2p lies below σ2p, so B₂ is evidence for s–p mixing.