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2.2.12Additional lessons

Resonance Structures (HL) — IB Chemistry HL Revision Notes & Topical Questions

5 min read 6 parts Lesson 36 of 172

📚 IB Chemistry HL Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 6Why Resonance Is Needed

Resonance describes bonding that cannot be shown by a single Lewis formula.

Limitation
  • Limitation Some species have several valid Lewis structures that differ only in where the double bond is drawn.
Experimental facts
  • Experimental facts Measurements show equal bond lengths, so no single structure describes the real species.
Delocalisation
  • Delocalisation The electrons of the pi bond are spread over three or more atoms.
Resonance hybrid
  • Resonance hybrid The real structure is a combination of all the contributing structures, not a mixture that switches between them.

Part 2 of 6Examples

Examples
Ozone
  • Ozone O₃ has two equivalent structures, with the double bond on either oxygen–oxygen link, so both bonds are equal.
Nitrate ion
  • Nitrate ion NO₃⁻ has three structures with the double bond on a different oxygen each time.
Carbonate ion
  • Carbonate ion CO₃²⁻ has three equivalent structures, and all three C–O bonds are identical.
Ethanoate ion
  • Ethanoate ion The two C–O bonds in CH₃COO⁻ are the same length.

Part 3 of 6Properties of the Hybrid

1
Draw each contributing structure, joined by double-headed arrows to show they are resonance forms.
2
All the bonds involved have the same length, between a single and a double bond.
3
The bonds are also stronger than single bonds, so the ion is more stable than any single structure would suggest.
4
Delocalised electrons are drawn as dashed lines or partial bonds in the hybrid.
5
Bond order = total number of bonds ÷ number of bonding positions around the central atom.

Part 4 of 6Bond Order Values

Ozone
  • Ozone Three shared pairs spread over two bonds gives a bond order of 1.5.
Carbonate
  • Carbonate Four shared pairs spread over three bonds gives a bond order of 4/3, about 1.33.
Nitrate
  • Nitrate Four shared pairs over three N–O bonds gives 1.33, again between single and double.
SpeciesBonds in one structurePositionsBond order
O₃321.5
CO₃²⁻431.33
NO₃⁻431.33

Part 5 of 6Worked Example

Question Calculate the C–O bond order in the carbonate ion.

  1. One structure has two C–O single bonds and one C=O double bond.
  2. Total bonds = 1 + 1 + 2 = 4, shared over 3 C–O positions.
  3. Bond order = 4 ÷ 3.

Answer Bond order ≈ 1.33, so each C–O bond is between single and double in length.

Part 6 of 6Exam Checklist & Tips

2.2.12 – Key Points for the Exam

  • Resonance Description of bonding using several Lewis structures that differ only in electron positions.
  • Delocalisation Spreading of electrons over more than two atoms.
  • Resonance hybrid The real structure intermediate between the contributing structures.
  • Bond order The average number of bonding pairs between two atoms.

Common Mistakes

  • Thinking the molecule flips rapidly between the structures.
  • Believing the structures are real isomers; they are only drawing conventions.
  • Saying the bonds are a mix of long and short bonds; they are all equal.
  • Moving atoms rather than electrons when writing resonance forms.

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