# AHL: Covalent bond order and electron delocalization

> IB Chemistry HL · IB Chemistry Higher Level
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u2-ahl-covalent-bond-order-and/

This subtopic explains how bond order describes the number of shared electron pairs between atoms, and how electron delocalization spreads π density across multiple bonds in resonance-stabilized molecules. You will connect bond order to measurable bond properties for exam questions.

**Prerequisites:** [Lewis structures and resonance](https://www.owlsprep.com/study/ib-chemistry-hl-u2-lewis-structures-resonance/); [Covalent bond properties (length, energy)](https://www.owlsprep.com/study/ib-chemistry-hl-u2-covalent-bond-properties/)

## Learning objectives

- Define covalent bond order and electron delocalization
- Calculate average bond order for resonance-stabilized molecules
- Explain how delocalization affects molecular stability and bond properties
- Relate bond order to bond length and bond energy

## 1. Bond Order: Definition and Calculation

**Covalent Bond Order** — The number of net bonding electron pairs shared between two atoms. For resonance-stabilized molecules, it is the average value across all equivalent bonding positions.

*Notation:* BO

*Example:* In diatomic O₂, bond order = 2

For molecules without resonance, bond order equals the number of bonding pairs between two atoms. For resonance hybrids, individual resonance forms do not represent the real structure, so we calculate the average bond order across all equivalent bonds.

**Worked example:** Calculate the average bond order of each oxygen-oxygen bond in ozone (O₃)

1. Draw the two valid resonance forms of ozone
2. Count the total number of O-O bonding pairs across all resonance positions:
3. $$1 \text{ single bond} + 1 \text{ double bond} = 3 \text{ total bonding pairs}$$
4. Count the number of equivalent O-O bonding domains: 2
5. Calculate average bond order:
6. $$BO = \frac{\text{Total bonding pairs}}{\text{Number of bonding domains}} = \frac{3}{2} = 1.5$$

> **Exam tip:** Never report bond order values from a single resonance form for delocalized molecules; always calculate the average for all equivalent bonds.

## 2. Electron Delocalization and Resonance Stabilization

**Electron Delocalization** — The spreading of π-bonding electrons across multiple adjacent atoms, enabled by overlapping p-orbitals, instead of being localized between two specific atoms.

*Example:* π electrons in benzene are delocalized across all 6 carbon atoms

Delocalization lowers the overall potential energy of the molecule, creating resonance stabilization. This makes the resonance hybrid more stable than any individual hypothetical resonance form would be. Only π electrons delocalize; σ bonding electrons remain localized between two atoms.

**Worked example:** Explain why all three carbon-oxygen bonds in the carbonate ion (CO₃²⁻) are identical

1. Carbonate has three equivalent resonance forms, each with one C=O double bond and two C-O single bonds
2. The π bonding electrons are not locked in one double bond; they delocalize evenly across all three C-O positions
3. Calculate average bond order:
4. $$BO = \frac{4 + 1}{3} = \frac{5}{3} 
  approx 1.67$$
5. Equal delocalization of electron density produces three identical bonds with the same length and energy

> **info**
>
> Resonance forms are not real structures that interconvert; the molecule only exists as the single delocalized resonance hybrid.

## 3. Bond Order and Bond Properties

Increased bond order means more shared electron density between two nuclei, leading to stronger electrostatic attraction. This creates two consistent, exam-tested relationships:

- Higher bond order → **shorter bond length**: stronger attraction pulls nuclei closer together
- Higher bond order → **higher bond energy**: more energy is required to break the stronger bond

**Worked example:** Predict which nitrogen-nitrogen bond is shorter and stronger: N₂ (BO = 3) or N₂H₄ (hydrazine, BO = 1)

1. Compare bond orders: N₂ has a much higher bond order than hydrazine
2. Apply the relationships: higher bond order = shorter bond, higher bond energy
3. Final result:
4. $$\text{Bond length: } N_2 < N_2H_4 \quad ; \quad \text{Bond energy: } N_2 > N_2H_4$$

**Check your understanding**

Test your understanding:

1. What is the relationship between C-C bonds in benzene (BO = 1.5) compared to single (BO=1) and double (BO=2) C-C bonds?

   - A. Benzene bonds are longer than C=C double bonds
   - B. Benzene bonds are shorter than C-C single bonds
   - C. Both A and B are correct
   - D. Neither A nor B are correct

   *Why:* Correct! 1.5 is between 1 and 2, so bond length is between single (longer, lower BO) and double (shorter, higher BO).

## 4. The Case of Benzene: Delocalization in Practice

Experimental evidence for delocalization in benzene comes from measured bond lengths: all six C-C bonds are identical at 139 pm, between a single C-C bond (154 pm) and a C=C double bond (134 pm). This matches the prediction of the delocalization model, contradicting the old Kekulé model of alternating single and double bonds.

**Worked example:** Calculate the average C-C bond order in benzene

1. Benzene has 6 C-C σ bonds and 3 delocalized π bonds, for a total of 9 bonding pairs across 6 C-C positions
2. Calculate average bond order:
3. $$BO = \frac{9}{6} = 1.5$$

> **tip**
>
> IB exams frequently ask you to explain how experimental bond length data for benzene supports the delocalization model over the Kekulé model.

## Common pitfalls

- **Wrong:** Claiming resonance forms are separate structures that rapidly interconvert in solution
  - Why it fails: Resonance forms are just hypothetical Lewis structure approximations; the molecule only exists as the single delocalized hybrid
  - Correct: Describe resonance forms as contributing models to the actual delocalized resonance hybrid structure
- **Wrong:** Reporting two different bond orders for the O-O bonds in ozone
  - Why it fails: Delocalization makes all O-O bonds identical; individual resonance forms are not real
  - Correct: Calculate the average bond order of 1.5 for all equivalent O-O bonds in ozone
- **Wrong:** Claiming σ electrons delocalize alongside π electrons
  - Why it fails: σ bonds form from end-on overlap of orbitals, so their electrons are locked between two nuclei
  - Correct: Attribute equal bond properties in delocalized molecules to delocalized π electron density only
- **Wrong:** Mixing up the relationship between bond order and bond length
  - Why it fails: Higher electron density increases attraction between nuclei, pulling them closer
  - Correct: Remember: higher bond order = shorter bond length = higher bond energy

## Cheatsheet

| Concept | Rule/Formula | Example |
| --- | --- | --- |
| Bond order (no resonance) | BO = number of bonding pairs between atoms | O₂: BO = 2 |
| Average BO (resonance) | BO = total bonding pairs ÷ number of equivalent bonds | Ozone: 3 ÷ 2 = 1.5 |
| Benzene C-C BO | BO = 1.5 | All 6 bonds identical |
| BO vs Bond Length | Higher BO → Shorter length | N₂ (BO=3): 110 pm; N₂H₄ (BO=1): 145 pm |
| BO vs Bond Energy | Higher BO → Higher energy | C=C (BO=2): 614 kJ/mol; C-C (BO=1): 346 kJ/mol |

## What's next

Understanding bond order and electron delocalization is foundational for studying the reactivity of aromatic compounds and conjugated systems, which you will explore in depth in organic chemistry topics. Delocalization also explains the stability of key species like carboxylate ions, which is central to understanding acid-base behavior of organic acids. These concepts also underpin advanced topics like hybridization and molecular orbital theory, which provides a quantum mechanical explanation for delocalization. Mastery of bond order calculations is required for many bonding and structure exam questions, which regularly appear in both Paper 1 and Paper 2 of IB Chemistry HL.

- [AHL: Advanced hybridization and delocalized pi systems](https://www.owlsprep.com/study/ib-chemistry-hl-u2-ahl-advanced-hybridization-and-delocalized/)
- [S3: Classification of matter](https://www.owlsprep.com/study/ib-chemistry-hl-u3-overview/)

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