Audited 31 Jul 2026·Last updated 28 Oct 2026·3 citations·Tier 3·0 uses

Bond Order Calculator

Compute molecular-orbital bond order from the number of electrons in bonding and antibonding orbitals, e.g. O₂ with 8 and 4 gives a double bond, order 2.

Bond Order Calculator

Molecular-orbital bond order
2
The molecular-orbital bond order, half the difference between bonding and antibonding electrons; a pure number where 1, 2 and 3 correspond to single, double and triple bonds.
Model scope
Simple molecular-orbital occupancy count; it does not construct an MO diagram, choose orbitals, handle fractional computational bond indices, resonance bond order, or establish stability and bond strength.

Background.

A Lewis structure counts bonds by drawing lines between atoms, one for each shared pair. Molecular orbital theory counts them differently. The electrons in a molecule occupy orbitals that belong to the whole molecule, and those orbitals come in two kinds: bonding orbitals, which pile electron density between the nuclei and hold them together, and antibonding orbitals, which have a node between the nuclei and work against the bond. Bond order is the balance between the two, half the difference between the electrons in each.

The count earns its place because it gets right some cases that line drawings struggle with. Oxygen comes out with a bond order of 2, a double bond as the Lewis structure says, but the same orbital filling puts two unpaired electrons in antibonding π* orbitals, which explains why liquid oxygen is drawn to a magnet. Helium has two bonding and two antibonding electrons, so He₂ has a bond order of zero and does not form a stable molecule. The hydrogen molecule ion H₂⁺, with a single electron, has a bond order of one half, something no line diagram can show.

Across a related series, a higher bond order goes with a shorter, stronger bond: removing an antibonding electron from O₂ to make O₂⁺ raises the order to 2.5 and shortens the bond. That is a trend, not a measurement. The page does the subtraction and halving; deciding which orbitals the electrons occupy, and reading the result for stability, is up to you.

What is bond order calculator?

Molecular-orbital bond order is half the difference between the number of electrons in bonding molecular orbitals and the number in antibonding molecular orbitals. A value of 1 corresponds to a single bond, 2 to a double and 3 to a triple; half-integer values arise when an odd number of electrons is involved, and zero means the bonding and antibonding effects cancel and no net bond is predicted. It is most often applied to diatomic molecules and ions.

How to use this calculator.

  1. Draw or look up the molecular orbital diagram for the molecule or ion and fill it with electrons, lowest energy first, adjusting the total for any charge.
  2. Count the electrons in bonding orbitals (σ and π without an asterisk) and enter that number as bonding electrons.
  3. Count the electrons in antibonding orbitals (σ* and π*) and enter it as antibonding electrons. Both must be whole numbers, and the antibonding count cannot exceed the bonding count here.
  4. You can count valence electrons only or include the core: filled inner shells add equally to both counts and leave the bond order unchanged.

The formula.

bond order = (N_bonding - N_antibonding) / 2

Bond order = (N_b − N_a) ÷ 2, where N_b is the number of electrons in bonding molecular orbitals and N_a the number in antibonding ones. The division by two converts electrons into pairs, so that two net bonding electrons make one bond, matching the shared-pair picture of a single bond. Each extra bonding electron raises the order by one half and each extra antibonding electron lowers it by the same amount, which is why adding or removing electrons from an ion shifts the order in steps of 0.5. The page accepts only whole-number counts and an antibonding count no larger than the bonding count, so the smallest result it gives is zero.

A worked example.

Example

Take the oxygen molecule, O₂, counting valence electrons only. Its twelve valence electrons fill σ2s², σ*2s², σ2p², π2p⁴ and π*2p². The bonding orbitals hold 2 + 2 + 4 = 8 electrons and the antibonding orbitals 2 + 2 = 4. The bond order is (8 − 4) ÷ 2 = 4 ÷ 2 = 2, a double bond, which is the result the page returns. Strip one electron from a π* orbital to make the O₂⁺ ion and the antibonding count drops to 3, giving (8 − 3) ÷ 2 = 2.5. Add one instead, for the superoxide ion O₂⁻, and the order falls to (8 − 5) ÷ 2 = 1.5. Nitrogen, N₂, with 8 bonding and 2 antibonding valence electrons, reaches (8 − 2) ÷ 2 = 3, its triple bond.

bonding Electrons8
antibonding Electrons4

Frequently asked questions.

Why is the bond order of He₂ zero, and what does that mean?
Each helium atom brings two electrons. Two go into the bonding σ1s orbital and two into the antibonding σ*1s orbital, so the count is (2 − 2) ÷ 2 = 0. The antibonding electrons cancel the benefit of the bonding ones, and molecular orbital theory predicts no stable He₂ molecule, which matches the fact that helium stays monatomic.
Do I count core electrons or just valence electrons?
Either works, as long as you are consistent. Filled inner shells contribute equally to bonding and antibonding orbitals. For O₂, adding the 1s electrons puts 2 more in σ1s and 2 more in σ*1s, turning 8 and 4 into 10 and 6, and (10 − 6) ÷ 2 is still 2.
Does the order of the σ2p and π2p orbitals change the bond order?
For the second-row diatomics, generally not. In B₂, C₂ and N₂ the π2p orbitals lie below σ2p, while in O₂ and F₂ the order reverses, but both σ2p and π2p are bonding, so the count of bonding electrons comes out the same. The ordering matters for which electrons are unpaired, which decides whether the molecule is magnetic.
What does a bond order of 1.5 or 2.5 mean?
It means an odd number of electrons sits in bonding or antibonding orbitals, giving a bond partway between single, double and triple. The superoxide ion O₂⁻ has order 1.5 and O₂⁺ has 2.5. Such bonds are real and measurable: their lengths fall between those of the neighbouring whole-number bonds.
Can I use this for benzene's 1.5 bonds?
Not directly. Benzene's carbon–carbon bond order of 1.5 is a resonance figure, the average of two structures that place the double bonds in alternate positions around the ring. It does not come from a bonding-minus-antibonding count for one pair of atoms, which is the simple count this page applies and which suits diatomic molecules and ions best.

How this page was produced

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Method
bond order = (N_bonding - N_antibonding) / 2
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