Bond order is the single number that tells you how strong a bond is, how short it is, and — for diatomics — whether the molecule is paramagnetic. A bond order of 3 in N₂ explains why nitrogen gas is so inert that we breathe 78% of it without consequence; a bond order of 2 in O₂ predicts a weaker bond, but the unpaired π* electrons hidden inside that “2” predict the paramagnetism that Lewis structures completely miss. The arithmetic is one subtraction and one division. The skill is knowing which model to use — molecular orbital theory for diatomics and ions, resonance averaging for delocalized systems — and getting the MO ordering right when you do.

What bond order tells you

bond order = (bonding e⁻ − antibonding e⁻) / 2

Higher bond order → shorter, stronger bond. N₂ (BO 3): 110 pm, 945 kJ/mol. O₂ (BO 2): 121 pm, 498 kJ/mol. F₂ (BO 1): 142 pm, 158 kJ/mol. The trend is monotone.
Bond order of 0 means as many antibonding as bonding electrons. The molecule won’t form (or will dissociate immediately). He₂ is the canonical example.
Fractional bond orders appear in resonance-stabilized systems (benzene 1.5, carbonate 1.33, nitrate 1.33) and in odd-electron diatomic ions like O₂⁻ (1.5).
Unpaired electrons in MO diagrams predict magnetism. O₂’s two unpaired π* electrons make it paramagnetic — a famous early win for MO theory over Lewis dot structures.

Method 1: MO diagram for second-row diatomics

Step 1: count valence electrons

Add the valence electrons of both atoms. Adjust for charge: +1 per negative charge, −1 per positive charge.

Step 2: choose the right MO ordering

This is the step everyone gets wrong. The 2p sigma and pi orbitals swap order between early and late second-row diatomics, because of s-p mixing.

Z ≤ 7 (Li₂, Be₂, B₂, C₂, N₂): π_2p sits below σ_2p. Order: σ_2s, σ_2s, π_2p (×2), σ_2p, π_2p (×2), σ*_2p
Z > 7 (O₂, F₂, Ne₂): s-p mixing weakens, σ_2p drops below π_2p. Order: σ_2s, σ_2s, σ_2p, π_2p (×2), π_2p (×2), σ*_2p

Step 3: fill, count, divide

Apply Aufbau, Pauli, Hund. Tally bonding (σ_2s, σ_2p, π_2p) and antibonding (σ*, π*). Subtract, divide by 2.

Worked examples

N₂

10 valence electrons. Use the π-before-σ ordering (Z = 7).

σ_2s: 2 (bonding)
σ*_2s: 2 (antibonding)
π_2p: 4 (bonding)
σ_2p: 2 (bonding)

Bonding = 8, antibonding = 2. BO = (8 − 2)/2 = 3. Triple bond, every electron paired → diamagnetic. Matches the N≡N Lewis structure and the famously inert behavior of N₂.

O₂

12 valence electrons. Use the σ-before-π ordering (Z = 8).

σ_2s: 2 (bonding)
σ*_2s: 2 (antibonding)
σ_2p: 2 (bonding)
π_2p: 4 (bonding)
π*_2p: 2 (antibonding, one in each degenerate orbital by Hund)

Bonding = 8, antibonding = 4. BO = 2. Double bond — but those two unpaired π* electrons are why liquid oxygen sticks to a magnet pole. Lewis structures cannot reproduce this; MO theory does it for free.

O₂⁻ (superoxide)

13 valence electrons. The 13th joins π*_2p.

Bonding = 8, antibonding = 5. BO = 1.5.

The bond is weaker and longer than O₂’s (134 pm vs 121 pm) — exactly what BO 1.5 predicts.

He₂

4 electrons. σ_1s: 2 (bonding), σ*_1s: 2 (antibonding). BO = 0. Doesn’t exist as a stable molecule. (The dimer formed by van der Waals forces at < 1 K is a different beast — not a covalent bond.)

Method 2: averaging over resonance structures

For polyatomic species with delocalized bonding, draw all equivalent resonance structures and average the bond orders for the bond of interest.

Nitrate (NO₃⁻)

Three resonance structures, each with one N=O and two N–O. Each individual N–O bond is double in one structure, single in the other two:

BO = (2 + 1 + 1) / 3 = 1.33

All three N–O bonds are equivalent in the real molecule (124 pm — between single 136 pm and double 122 pm).

Benzene

Each C–C bond is single in one Kekulé structure, double in the other:

BO = (1 + 2) / 2 = 1.5

Predicts a bond length (139 pm) between single C–C (154 pm) and double C=C (134 pm). Matches experiment.

Traps people fall into

Using the wrong MO ordering. For B₂, C₂, N₂ the π orbitals are below σ_2p. For O₂, F₂ they’re above. Use the wrong order and you’ll predict B₂ diamagnetic when it’s actually paramagnetic — a giveaway that you’ve muddled it.
Forgetting to adjust for charge. O₂⁺ and O₂⁻ differ from O₂ by one electron each; that one electron changes the bond order by 0.5.
Counting Lewis-structure resonance only once. All equivalent resonance structures contribute. For a CO₃²⁻ ion you average over three, not two.
Mixing models. Don’t compute a “bond order” for benzene from an MO diagram and then average it with a resonance answer. Pick a method and stick with it.

Practice

Try these against the Electron Configuration Calculator:

Bond order of F₂ from MO theory.
Bond order of NO⁺. Higher or lower than NO?
Bond order of B₂. Paramagnetic or diamagnetic?
Average bond order in CO₃²⁻ from resonance.
Bond order of He₂. Does the molecule exist?

How to Determine Bond Order