Resonance

Draw the carbonate ion, $\text{CO}_3^{2-}$ , on paper and you must pick one oxygen to carry the double bond — yet every real C–O bond in the ion measures exactly the same, about $128\ \text{pm}$ , halfway between a single and a double bond. No single Lewis structure can show that. Chemists resolve the mismatch not by inventing a new structure but by admitting that several valid structures, differing only in electron placement, must be blended together: the true species is a single, static resonance hybrid, a weighted average of every contributing canonical form. This idea — resonance — turns up wherever a Lewis structure forces an arbitrary choice among equally good alternatives, most visibly in $\text{CO}_3^{2-}$ and ozone. It explains equalised bond lengths, extra stability called resonance energy, and why some formal charges look "spread out" rather than sitting on one atom. This article builds the rules for valid resonance structures, ranks them by stability, and shows how to turn a set of structures into a single number: the average bond order.

The core insight

A molecule or ion whose true structure cannot be captured by one Lewis diagram is represented as a resonance hybrid — a weighted blend of several valid canonical forms that differ only in electron arrangement — and this delocalisation equalises bond lengths and lowers the energy below any single contributor.

The core idea

Why one Lewis structure sometimes fails. A single Lewis diagram fixes each bond as either single or double, but some real species show bond lengths and bond strengths that are identical across positions where a lone diagram would predict different bond orders. Resonance is the bookkeeping fix: write down every electronically distinct, equally valid diagram (moving only electrons, never nuclei) and treat the actual molecule as their blend.

Rules a set of canonical forms must obey. All contributing structures must keep the same nuclear positions — atoms never move between them, only bonding and lone-pair electrons do. They must also represent the same net charge and the same number of unpaired electrons; a structure with a different total electron count or a different charge is describing a different species, not an alternative resonance form.

Ranking contributors by stability. Not every valid structure contributes equally to the hybrid. A structure is a stronger (major) contributor when it has more covalent bonds and fewer formal charges, when any charges present are as close to zero and as little separated as possible, when a negative charge sits on the more electronegative atom, and when no two adjacent atoms carry like charges. Structures that are related by symmetry — as in $\text{CO}_3^{2-}$ or ozone — have identical energy and so contribute exactly equally; asymmetric or high-charge-separation structures contribute little or can be ignored.

What delocalisation buys you. Averaging several structures with the bond in different places spreads the pi-electron density over more than two atoms. Two measurable consequences follow directly: bond lengths that would otherwise differ (single vs. double) equalise to an intermediate value, and the hybrid sits lower in energy than any single contributing structure — the gap is the resonance energy, the extra stability from delocalisation.

Figure

Three resonance structures of the carbonate ion blending into one hybrid with three equal partial double bonds

Derivation AAverage bond order from a set of resonance structures

The goal is to turn a set of canonical forms, each a mix of single and double bonds at a given position, into one number — the average bond order at that position — that predicts the intermediate, equalised bond length actually measured, and to show why that number lands strictly between the single-bond and double-bond values rather than at either extreme.

Step 1. Identify every significant (equal-energy, symmetry-related) contributing structure and note the bond order contributed at the position of interest in each one. For a given C–O bond in $\text{CO}_3^{2-}$ , exactly one of the three canonical forms shows it as a double bond (bond order $2$ ) and the other two show it as a single bond (bond order $1$ each), because the double bond rotates among the three equivalent oxygens.

Step 2. Average the bond order across the equally weighted contributing structures, since each form contributes an equal share to the hybrid: $\text{bond order} = \dfrac{1+1+2}{3} = \dfrac{4}{3} \approx 1.33$ .

Step 3. Interpret the result. A bond order of $1.33$ lies strictly between a pure single bond (order $1$ , longer) and a pure double bond (order $2$ , shorter), so the predicted bond length must also lie strictly between the single- and double-bond lengths — matching the experimentally observed, uniform $128\ \text{pm}$ C–O bond length in carbonate, which is intermediate between the C–O single-bond length ( $\sim 143\ \text{pm}$ ) and the C=O double-bond length ( $\sim 120\ \text{pm}$ ).

Step 4. Generalise the formula: for $n$ equally weighted structures contributing a total of $S$ bond-order units to a given position, average bond order $= S/n$ . This single ratio is the quantitative signature of resonance and is the number examiners expect you to compute for $\text{CO}_3^{2-}$ and $\text{O}_3$ .

Worked example

Bond order and bond length prediction for ozone, $\text{O}_3$ , extending the Pass-1 result.

Step 1. Ozone has $18$ valence electrons ( $3 \times 6$ ) and a bent shape (VSEPR, one lone pair on the central O). Two equally valid, symmetry-related Lewis structures exist: one with the double bond to the left terminal oxygen and a single bond to the right, the other with the assignment swapped. Both have identical energy, so both contribute equally to the hybrid.

Step 2. Apply the bond-order averaging rule from the derivation above, for either O–O bond position: across the two structures, that bond is a double bond (order $2$ ) in one and a single bond (order $1$ ) in the other, so average bond order $= \dfrac{1+2}{2} = 1.5$ .

Step 3. Compare with experiment. Both O–O bonds in ozone are measured as equal, about $128\ \text{pm}$ , which sits between the O–O single-bond length ( $\sim 148\ \text{pm}$ ) and the O=O double-bond length ( $\sim 121\ \text{pm}$ ) — exactly consistent with a bond order of $1.5$ , roughly midway between $1$ and $2$ . The predicted bond order of $1.5$ and the observed equal, intermediate bond length of $128\ \text{pm}$ confirm that ozone is correctly described only by the resonance hybrid, not by either single canonical form.

Competitive stretch

If a species had three equal contributing structures with bond orders $1, 1, 2$ at one position and a fourth minor contributor (higher charge separation, contributing negligibly) showed bond order $3$ there, you would still average only over the significant contributors — minor, high-energy structures are excluded from the bond-order average, not blindly averaged in with equal weight.

Common traps

Averaging bond orders over ALL drawable structures, including minor ones. Only structures of comparable, near-equal energy get equal weight; a structure with large charge separation contributes far less and should not be averaged in as if equal.
Forgetting resonance structures must share identical nuclear positions. If atoms move, you have drawn an isomer, not a resonance form, and the bond-order averaging trick does not apply.
Reading the double-headed arrow ( $\leftrightarrow$ ) as an equilibrium. The molecule does not oscillate between structures; it has one static hybrid structure at all times.
Skipping the stability check before averaging. Always rank contributors by covalent bond count, formal charge, and charge placement first — only then decide which structures are "significant" enough to include in the average.

Ask the tutor: Want to practice ranking resonance structures by stability or computing bond order for a species like the nitrate ion? Ask the AI tutor to walk through it with you.

If the hybrid is a fixed structure, why do we bother drawing several Lewis structures at all?

Because a single Lewis structure, built from whole single and double bonds, simply cannot represent a bond order like $1.33$ or $1.5$ . Drawing the full set of valid canonical forms and averaging is a bookkeeping technique that lets you predict the real, intermediate bond order and bond length using only the simple single/double-bond rules you already know, without needing molecular-orbital theory.

How do I decide how many resonance structures to draw for a new ion I have not seen before?

Start from one valid Lewis structure that satisfies the octet and formal-charge rules. Then look for any electron pair or double bond that could equally well sit on a chemically equivalent neighbouring atom without moving any nucleus. Each such shift, if it gives a structure of comparable energy, is a new contributing structure; keep going until no further equal-energy shift is possible, then apply the stability criteria to weight them.

Does resonance energy mean the molecule stores extra energy that can be released?

No — resonance energy is stabilisation, not stored energy waiting to be released. It is the amount by which the true hybrid sits lower in energy than the most stable single contributing structure would predict, which is why species showing resonance are generally less reactive and harder to break apart than a non-delocalised structure of the same formula would be.

Resonance

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads