Worked example 1
Draw the resonance structures of ozone O₃ and state the O-O bond order.
Try it first: Draw one valid structure, then ask what else is equally valid.
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Molecular StructureLewis structuresContent level: Core 20 min
Resonance Structures
What you'll be able to do: Draw a full set of resonance structures and describe the real molecule as a hybrid with delocalised electrons.
Best after: Multiple Bonds and Polyatomic Ions
Introduction
Ozone has two equal bonds, yet no single Lewis structure can show that. Resonance is the fix.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Recognise when resonance structures are required
- Draw a complete set of equivalent contributors
- Describe the hybrid using average bond order
- Explain why the molecule does not flip between structures
Lesson
Why one structure is not enough
In ozone O₃ you can place the double bond on the left oxygen or the right one. Both drawings obey every rule, so neither can be preferred. Experiment shows the two O-O bonds are identical in length, intermediate between a single and a double bond, so neither drawing alone is the truth.
Resonance structures differ only in electron placement. If atoms move, you have drawn a different molecule, not a resonance form.
The hybrid
The real species is a single unchanging structure called the resonance hybrid, best pictured as a blend of the contributors. The double headed arrow between structures means blend, not equilibrium. Ozone does not oscillate; every molecule is the hybrid at all times.
Average bond order and delocalisation
Average bond order equals the total number of bonds between the two atoms across all contributors divided by the number of contributors. In nitrate, one N=O and two N-O give an average of 4 bonds over 3 positions, so each bond has order 1.33. Spreading electrons over more atoms lowers the energy, which is why delocalised species such as carbonate, nitrate and benzene are unusually stable.
Unequal contributors
When the contributors are not equivalent, the ones with lower formal charges and with negative formal charge on the more electronegative atom contribute more to the hybrid. That idea is developed further in the formal charge lessons.
Test whether two drawings are resonance contributors
Keep the nuclei and their connectivity fixed while moving only lone pairs and π electrons. Each contributor must use the same total number of electrons and retain the same overall charge. Equivalent contributors have equal weight; when two equivalent drawings alternate a single and double bond at one position, that bond has an average order of 1.5 in the resonance hybrid.
If an atom changes neighbors, the drawings are different structures, not resonance forms.
Key ideas
Definition
Resonance structures
Two or more valid Lewis structures for the same arrangement of atoms, differing only in electron placement.
Definition
Resonance hybrid
The single real structure, an average of all contributors.
Rule
Atoms stay put
Only electrons move between resonance forms; moving atoms creates an isomer instead.
Key concept
Carbonate
Three equivalent contributors give each C-O bond an average order of 1.33.
Equation
Average bond order
average bond order = total bonds between the pair across contributors / number of contributors
Worked examples
Worked example 2
How many equivalent resonance structures does the carbonate ion CO₃ 2- have, and what is each C-O bond order?
Try it first: Count electrons, then see how many positions the double bond can occupy.
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Worked example 3
Draw the two equivalent contributors for nitrite, NO₂⁻, and predict its N-O bond orders.
Try it first: Count 5 + 2(6) + 1 = 18 electrons and place N in the center.
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Common mistakes
Saying the molecule flips rapidly between resonance structures.
Why it's wrong: There is only one real structure, the hybrid, and it never changes.
Check instead: Describe the electrons as delocalised over the whole set of atoms.
Moving an atom to create a new resonance structure.
Why it's wrong: Different atom positions mean a different compound.
Check instead: Check that every atom sits in the same place in each drawing.
Expecting the bonds to have integer bond orders.
Why it's wrong: Delocalisation gives fractional averages such as 1.33 or 1.5.
Check instead:
After the lesson · Question bank
Independent practice
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What you should now know
When more than one valid Lewis structure differs only in where the multiple bonds and lone pairs sit, the real species is a resonance hybrid, an average of the contributors. Delocalisation makes all equivalent bonds identical in length and gives fractional bond orders and fractional average formal charges.
- Resonance is needed when several equally valid structures exist
- Only electrons move between contributors, never atoms
- The real species is one hybrid with delocalised electrons
- Average bond order is often fractional
- Delocalisation stabilises the species
Sources and further reading
- Chemistry 2e, Section 7.4: Formal Charges and ResonanceOpenStax · Rice University · 7.4View source
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