IB Chemistry HLTopic 2 — Models of Bonding & StructurePaper 1 & 2Higher level~9 min read
Resonance Structures
Some ions refuse to be drawn. Try a Lewis formula for the nitrate ion and you will find three equally good answers, none of which matches what the ion actually looks like when you measure it. Resonance is chemistry admitting that a single dot-and-cross diagram is not always good enough.
📘 What you need to know
Resonance occurs when more than one valid Lewis formula can be drawn for the same species.
It happens because some electrons are delocalised — not confined to one bond or one atom.
The molecule does not flip between the structures. It is a single fixed thing called a resonance hybrid.
The hybrid is the average of the contributing structures, and it is more stable than any of them.
The strongest evidence is equal bond lengths, intermediate between single and double.
Resonance needs a π bond that can sit in more than one place, next to atoms of similar electronegativity.
Recognise it in: NO3−, CO32−, O3, RCOO− and benzene.
The problem with a single Lewis formula
Take the nitrate ion, NO3−. Count the valence electrons: 5 from nitrogen, 6 from each of three oxygens, plus one for the negative charge, giving 24 in total.
Draw it and you end up with one N=O double bond and two N–O single bonds. Fine — except there is no reason at all why the double bond should be on that oxygen. It could equally well be on either of the other two. Three drawings, all obeying the rules, all identical in energy.
Three equally valid drawings of the nitrate ion
The purple double headed arrow is reserved for resonance. Do not confuse it with the equilibrium arrows you meet in Reactivity — there is no interconversion happening here.
The most common misunderstanding in this whole topic: the ion is not rapidly switching between the three forms. It is one unchanging structure that none of the three drawings quite captures. Think of a mule — it is not flickering between horse and donkey.
What the ion is really like
The truth is that the extra pair of π electrons is not stuck on one bond at all. It is spread evenly over all three N–O positions. That spread-out arrangement is called delocalisation, and the single real structure is the resonance hybrid.
You can test this. If one bond really were double and two were single, they would have different lengths. Measure them and all three come out identical, at a value that sits neatly between a single and a double bond.
The hybrid, and the evidence for it
Equal bond lengths are the killer evidence. No single Lewis formula predicts them, but delocalisation explains them immediately.
🤔 Why does delocalisation make things more stable?
Electrons repel each other, so being crowded into one small region costs energy. Spreading the same electrons over a larger region lets them stay further apart, which lowers the energy. That is why the real hybrid is always more stable than any of the individual structures you can draw, and why delocalised species like the carboxylate ion are so unreactive.
Bond order
A neat way of describing a delocalised bond is its bond order: how many bonding pairs are shared per bond position on average.
Bond order
bond order = total number of bonds between the atomsnumber of bond positions
In the nitrate ion there are four bonds in total (one double plus two single) shared over three positions, giving a bond order of 4 ÷ 3 = 1.33. That fractional value is exactly why the measured length falls between a single and a double bond.
Where resonance turns up
Species
Structures
Delocalised over
Bond order
NO3−
3
three N–O bonds
1.33
CO32−
3
three C–O bonds
1.33
O3
2
two O–O bonds
1.5
RCOO−
2
two C–O bonds
1.5
C6H6
2
six C–C bonds in the ring
1.5
Why carboxylic acids are acidic. When RCOOH loses its H+, the negative charge left behind is spread evenly over both oxygens by resonance. A spread-out charge is a stable charge, so the ion is happy to exist — which is exactly why carboxylic acids give up their proton far more readily than alcohols do.
Worked examples
WE 1
Show that the nitrate ion has 24 valence electrons and explain why three resonance structures exist
Step 1: count, remembering the chargeN: 5, O: 3 × 6 = 18, charge: +15 + 18 + 1 = 24 electronsStep 2: build the structure
Three N–O bonds use 6. The remaining 18 cannot give every oxygen an octet and satisfy nitrogen, so one bond must become double.
Step 3: why three?
All three oxygens are equivalent, so the double bond has three equally valid homes.
24 electrons, three equivalent structuresthe +1 for the negative charge catches people out every time
WE 2
All three C–O bonds in CO32− are 129 pm. Explain, given C–O is 143 pm and C=O is 122 pm. [3]
Mark 1: what a single structure would predict
One double and two singles would give two different lengths, which is not what is observed.
Mark 2: name the explanation
The π electrons are delocalised over all three C–O positions.
Mark 3: link to the number
Each bond is identical and intermediate: 122 < 129 < 143, matching a bond order of 1.33.
Delocalisation gives three identical, intermediate bondsquote the numbers — “in between” without values rarely gets full marks
WE 3
Deduce the bond order in ozone, O3
Step 1: draw one resonance structure
Ozone is bent, with one O=O double and one O–O single bond.
Step 2: count bonds and positionsTotal bonds = 2 + 1 = 3Bond positions = 2Step 3: divide3 ÷ 2 = 1.5Bond order 1.5, so both bonds are equala bond order of exactly 1.5 also applies to benzene and the carboxylate ion
💡 Exam tips
Use the double headed arrow between resonance structures, and never the equilibrium arrow.
Show all lone pairs and the charge in brackets on every structure you draw, not just the first one.
Say “delocalised”. It is the word the mark scheme is looking for.
Quote equal bond lengths as your evidence, with numbers if you are given them.
Draw the hybrid with dashed lines for the partial bonds, and remember lone pairs are usually left off hybrids.
Explain increased stability by saying the electrons or charge are spread over a larger region.
⚠ Common mix-ups
Saying the molecule flips between structures. It does not. There is one real structure at all times.
Using equilibrium arrows. Resonance is not a reaction and nothing is interconverting.
Forgetting the charge in the electron count. Add one electron per negative charge before you start.
Thinking the hybrid is a 50:50 mixture of molecules. Every single ion in the sample is the hybrid.
Drawing resonance where there is none. You need a π bond that can genuinely move between equivalent positions.
Assuming the hybrid is less stable because it looks like a compromise. Delocalisation always lowers the energy.
Up next: Benzene — the most famous delocalised molecule of all, and the one where the experimental evidence for resonance is strongest.
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