All calculatorsBond Order Calculator
Compute bond order from either MO theory (diatomic) using \( \text{BO}=\frac{N_b - N_a}{2} \) or from Lewis/resonance structures as an average across equivalent bonds. Quick picks and step-by-step derivations included.
Background
Bond order estimates the strength/number of bonds between two atoms. In MO theory for diatomics, it’s half the difference between electrons in bonding and antibonding molecular orbitals. In resonance systems (e.g., carbonate, benzene), it’s the average number of bond pairs per equivalent bond (total bond-order units divided by the number of equivalent bonds).
How to use this calculator
MO (Diatomic)
- Enter bonding electrons Nb and antibonding electrons Na.
- We compute .
Lewis/Resonance (Average bond order)
- Enter the number of equivalent bonds k (e.g., 3 C–O in CO₃²⁻).
- Either (a) provide counts of singles/doubles/triples across resonance, or (b) enter the total bond-order units directly.
- We compute BO per bond = .
This tool gives idealized textbook values; real molecules can deviate due to delocalization.
Formula & Equation Used
MO (Diatomic):
BO = ( Nb − Na ) / 2
Average over resonance (equivalent bonds):
BO per bond = ( total bond-order units ) / k
Bond-order units from counts:
total bond-order units = (1 × singles) + (2 × doubles) + (3 × triples)
Example Problems & Step-by-Step Solutions
Example 1 — MO (Diatomic O₂)
Given MO electron counts: Nb = 10, Na = 6.
BO = (Nb − Na)/2 = (10 − 6)/2 = 2.0
Example 2 — MO (O₂⁺ vs O₂)
O₂ has BO = 2.0. Removing one electron from an antibonding orbital (O₂⁺) decreases Na by 1:
BO = (10 − 5)/2 = 2.5 (half-unit increases/decreases are common with ionization or electron addition).
Example 3 — Resonance Average (CO₃²⁻)
Total bond-order units across the three equivalent C–O bonds = 4
(conceptually “one double + two single” spread over the resonance forms).
BO per C–O = total units / number of equivalent bonds = 4 / 3 = 1.33
Example 4 — Aromatic Ring (Benzene, C₆H₆)
Across the six C–C bonds: 3 “double” (2 each) + 3 “single” (1 each) = 3×2 + 3×1 = 9 bond-order units.
BO per C–C = 9 / 6 = 1.5
Example 5 — No Bond Case (He₂)
MO counts (filled σ1s and σ1s*): Nb = 2, Na = 2 ⇒ BO = (2 − 2)/2 = 0 (no stable bond).
Quick Recipe (counting approach)
Total bond-order units = (1 × #single) + (2 × #double) + (3 × #triple).
If bonds are equivalent (resonance), BO per bond = (total units) / (number of equivalent bonds).
Common Pitfalls
- Fractional BO is normal for resonance-delocalized bonds (e.g., 1.33, 1.5).
- For MO problems, be sure which orbital (bonding vs antibonding) gains/loses electrons.
- Only average over equivalent bonds. Don’t average localized, non-equivalent bonds.
Frequently Asked Questions
Q: Why do some diatomics have fractional bond orders (e.g., O₂⁺)?
Adding or removing electrons changes the balance of bonding vs. antibonding occupancy, producing half-integer BO values.
Q: In resonance, which bonds are “equivalent”?
Bonds that are symmetry-equivalent (same environment in the resonance hybrid)—e.g., the three C–O bonds in CO₃²⁻ or the six C–C bonds in benzene.
Q: Does bond order equal “number of bonds” exactly?
It’s a useful index (correlates with bond strength and length) but not always an exact integer in delocalized systems.
