Study Guide

AHL: Covalent bond order and electron delocalization

IB Chemistry HL· S2: Models of bonding and structure (AHL)· 25 min read

1. 1. Bond Order: Definition and Calculation★★☆☆☆HL only⏱ 8 min

📘 Definition

Covalent Bond Order

BOBO

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.

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. 1

    Draw the two valid resonance forms of ozone

  2. 2

    Count the total number of O-O bonding pairs across all resonance positions:

  3. 3
    1extsinglebond+1extdoublebond=3exttotalbondingpairs1 ext{ single bond} + 1 ext{ double bond} = 3 ext{ total bonding pairs}
  4. 4

    Count the number of equivalent O-O bonding domains: 2

  5. 5

    Calculate average bond order:

  6. 6
    BO = rac{\text{Total bonding pairs}}{\text{Number of bonding domains}} = rac{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. 2. Electron Delocalization and Resonance Stabilization★★★☆☆HL only⏱ 7 min

📘 Definition

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. 1

    Carbonate has three equivalent resonance forms, each with one C=O double bond and two C-O single bonds

  2. 2

    The π bonding electrons are not locked in one double bond; they delocalize evenly across all three C-O positions

  3. 3

    Calculate average bond order:

  4. 4
    BO = rac{4 + 1}{3} = rac{5}{3} approx 1.67
  5. 5

    Equal delocalization of electron density produces three identical bonds with the same length and energy

3. 3. Bond Order and Bond Properties★★☆☆☆HL only⏱ 6 min

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. 1

    Compare bond orders: N₂ has a much higher bond order than hydrazine

  2. 2

    Apply the relationships: higher bond order = shorter bond, higher bond energy

  3. 3

    Final result:

  4. 4
    Bond length: N2<N2H4;Bond energy: N2>N2H4\text{Bond length: } N_2 < N_2H_4 \quad ; \quad \text{Bond energy: } N_2 > N_2H_4
✓ Quick check

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

    Reveal answer
    C —

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

4. 4. The Case of Benzene: Delocalization in Practice★★★☆☆HL only⏱ 4 min

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. 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. 2

    Calculate average bond order:

  3. 3
    BO = rac{9}{6} = 1.5

5. Common Pitfalls

Wrong move:

Claiming resonance forms are separate structures that rapidly interconvert in solution

Why:

Resonance forms are just hypothetical Lewis structure approximations; the molecule only exists as the single delocalized hybrid

Correct move:

Describe resonance forms as contributing models to the actual delocalized resonance hybrid structure

Wrong move:

Reporting two different bond orders for the O-O bonds in ozone

Why:

Delocalization makes all O-O bonds identical; individual resonance forms are not real

Correct move:

Calculate the average bond order of 1.5 for all equivalent O-O bonds in ozone

Wrong move:

Claiming σ electrons delocalize alongside π electrons

Why:

σ bonds form from end-on overlap of orbitals, so their electrons are locked between two nuclei

Correct move:

Attribute equal bond properties in delocalized molecules to delocalized π electron density only

Wrong move:

Mixing up the relationship between bond order and bond length

Why:

Higher electron density increases attraction between nuclei, pulling them closer

Correct move:

Remember: higher bond order = shorter bond length = higher bond energy

6. Quick Reference 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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

  • 2022 · 1

    Calculate bond order in ozone

  • 2023 · 2

    Compare bond lengths in benzene

Going deeper

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.