Bond order is one number that predicts bond length, bond strength, and whether a molecule can exist at all. There are two ways to get it depending on what you're given, and they answer slightly different questions.
Method 1: from an MO diagram
Fill the molecular orbitals with your electrons lowest-first, count what landed in bonding orbitals versus antibonding ones, subtract, halve.
| Species | Bonding | Antibonding | Bond order |
|---|---|---|---|
| H₂ | 2 | 0 | 1 |
| He₂ | 2 | 2 | 0 — doesn't exist |
| N₂ | 10 | 4 | 3 |
| O₂ | 10 | 6 | 2 |
| F₂ | 10 | 8 | 1 |
| O₂⁻ | 10 | 7 | 1.5 |
Our bond order calculator does both methods and has quick picks for common molecules.
Method 2: from resonance structures
For molecules where you've drawn Lewis structures instead:
Bond order = total bonds between the two atoms across all resonance forms ÷ number of resonance forms
Take the carbonate ion, CO₃²⁻. Three resonance structures; in each, one C–O is double and two are single. Any given C–O bond is double once and single twice: (2 + 1 + 1) ÷ 3 = 1.33.
That fractional answer is the point. It's why all three C–O bonds in carbonate are identical and measure between a single and a double bond — no individual Lewis structure is the real molecule, and the fraction is closer to the truth than any of them.
What bond order predicts
| Higher bond order means |
|---|
| Shorter bond |
| Stronger bond (higher dissociation energy) |
| Higher vibrational frequency (IR stretch) |
| More stable molecule |
N₂ is the demonstration. Bond order 3, dissociation energy around 945 kJ/mol, and a bond length of just 110 pm. That triple bond is why nitrogen is so inert that fixing it industrially takes high temperature, high pressure and a catalyst — and why the same reaction sustains a large fraction of the world's food supply.
MO theory predicts this exactly. Filling O₂'s orbitals puts two electrons into separate degenerate π* antibonding orbitals with parallel spins, per Hund's rule. Bond order still comes out at 2, matching the double bond — but now the two unpaired electrons are right there in the diagram.
Lewis theory simply cannot produce this result. It's the cleanest example of why MO theory earns its complexity.
Fractional bond orders are normal
Add an electron to O₂ and it goes into an antibonding orbital, dropping bond order to 1.5 — a longer, weaker bond. Remove one from an antibonding orbital and bond order rises to 2.5, giving a shorter, stronger bond.
That's a genuinely useful check on your understanding: whether ionisation strengthens or weakens a bond depends entirely on whether the electron came from a bonding or antibonding orbital. Nothing about "removing an electron" alone tells you.
Exam mistakes worth avoiding
Forgetting to halve. The most common arithmetic slip in the topic.
Miscounting valence electrons for ions. Anions gain, cations lose — get this wrong and everything downstream is wrong.
Using the wrong orbital ordering. For B₂, C₂ and N₂, the σ2p sits above the π2p due to s–p mixing. For O₂, F₂ and Ne₂ the normal order applies. This doesn't change bond order — but it absolutely changes your magnetic prediction, which is usually the other half of the question.
Assuming bond order must be a whole number. It frequently isn't, and the fractional values are often the interesting ones.
Working on a related calculation? Our ligation calculator covers molar ratios in molecular biology.
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Frequently Asked Questions
How do you calculate bond order?
Bond order = (bonding electrons − antibonding electrons) ÷ 2 from a molecular orbital diagram. From resonance structures, divide the total bonds between two atoms by the number of resonance forms.
What does a bond order of 0 mean?
The molecule doesn't exist. He₂ has two bonding and two antibonding electrons, giving a bond order of zero, which is why helium is monatomic.
Why is O₂ paramagnetic?
Its two highest electrons occupy separate degenerate π* antibonding orbitals with parallel spins, leaving them unpaired. MO theory predicts this; Lewis structures cannot.
Can bond order be a fraction?
Yes, and often it is. The carbonate ion has a C–O bond order of 1.33 across its resonance forms, and O₂⁻ has a bond order of 1.5.