Resonance Structures

Some molecules cannot be drawn correctly with a single Lewis structure. Consider the carbonate ion, $\text{CO}_3^{2-}$ . Experiments show all three carbon–oxygen bonds are identical in length. Yet any single Covalent Bonding and Lewis Structures diagram forces one bond to be a double bond. The other two must then be single bonds. The fix is not to pick the "right" drawing. Instead, recognize that several equally valid drawings exist. These are called resonance structures. Together they describe one real, blended structure called a resonance hybrid. Resonance is bookkeeping for delocalized electrons. These are electrons not confined between two specific atoms. Instead, they are spread over three or more nuclei. Lewis structures can only show electrons as fixed dots or lines. So chemists draw multiple contributors connected by double-headed arrows ( $\leftrightarrow$ ). This signals that the true structure is an average. It is not a mixture or a rapid interconversion. Getting comfortable with resonance also sets up Formal Charge as a ranking tool. It also prepares the ground for VSEPR and Molecular Shapes. There, the hybrid — not any single contributor — determines geometry.

Key takeaways

Resonance structures are alternative Lewis diagrams for the same arrangement of atoms, differing only in electron placement.
The real molecule is a single resonance hybrid — a weighted blend, never an oscillating mixture of the contributors.
Only $\pi$ -electrons (and lone pairs adjacent to $\pi$ systems) move between resonance structures; atoms and $\sigma$ -bonds never move.
Formal charge and electronegativity rank contributors: lower formal charges, and negative charge on the more electronegative atom, mean a bigger contribution to the hybrid.
Delocalization lowers energy — the hybrid is always more stable than any single contributor would be if it existed alone.
Key idea: when one Lewis structure won't do, draw all the equivalent ways to place the double bonds and lone pairs, then blend them into one averaged, delocalized picture.

The core insight

A resonance hybrid is the true structure; individual resonance contributors are just incomplete snapshots drawn because a single Lewis diagram cannot show delocalized electrons.

The physical picture

Take the carbonate ion, $\text{CO}_3^{2-}$ . Carbon sits at the center bonded to three oxygens. Total valence electrons:

4 + 3(6) + 2 = 24 \text{ electrons}

A valid Lewis structure gives carbon one C=O double bond and two C–O single bonds, with the two singly-bonded oxygens each carrying three lone pairs and a $-1$ formal charge, the doubly-bonded oxygen carrying two lone pairs and formal charge $0$ , and carbon formal charge $0$ . But nothing distinguishes which oxygen gets the double bond — all three positions are chemically identical. That ambiguity is the signature of resonance: three structures, identical in energy and skeletal arrangement, differing only in which oxygen carries the double bond. Experimentally, all three C–O bond lengths in carbonate measure about $128\ \text{pm}$ — shorter than a typical C–O single bond ( $143\ \text{pm}$ ) but longer than a typical C=O double bond ( $120\ \text{pm}$ ). The bond order, defined for a resonance-averaged system as

\text{bond order} = \frac{\text{total bonds shared over all contributors}}{\text{number of equivalent contributors}}

comes out to $4/3 \approx 1.33$ for each C–O bond, exactly consistent with a length intermediate between single and double. No single contributor predicts this; the average of three equivalent contributors does.

Figure

The three equivalent resonance contributors of carbonate,

\text{CO}_3^{2-}

, and their hybrid

The pink oxygen carries the C=O double bond in each drawing; the blue oxygens carry the negative formal charge. Rotating which oxygen is pink generates all three contributors. None is the true structure — the true structure is their average, with each C–O bond one-third double-bond character and each oxygen carrying about $-\tfrac{2}{3}$ of the ion's total $-2$ charge. A second classic case is the nitrate ion, $\text{NO}_3^-$ , isoelectronic with carbonate in its $\pi$ system: three equivalent contributors, bond order $4/3$ per N–O bond, charge $-\tfrac{2}{3}$ per oxygen — the same arithmetic, different central atom.

The core relations

\text{Total valence electrons} = \sum_{\text{atoms}} (\text{group number electrons}) + \text{charge adjustment}

where: for anions add electrons equal to the magnitude of the negative charge; for cations subtract electrons equal to the magnitude of the positive charge. This total must be conserved and identical across every resonance contributor.

FC = V - N - \frac{B}{2}

where: $FC$ is formal charge on an atom, $V$ is the number of valence electrons the free atom has, $N$ is the number of non-bonding (lone-pair) electrons on that atom in the structure, and $B$ is the number of electrons in bonds around that atom (twice the number of bonds). This is exactly the Formal Charge formula, applied here to compare contributors rather than a single structure.

\text{Bond order (delocalized bond)} = \frac{\text{number of contributors sharing a bond at that position} \times 1 + \text{number sharing a double bond there} \times 2}{\text{total number of equivalent resonance contributors}}

where: this reduces, for a set of $n$ equivalent contributors in which every bond position is single in $(n-1)$ of them and double in exactly one, to bond order $= \dfrac{(n-1)(1) + 1(2)}{n} = \dfrac{n+1}{n}$ .

\Delta E_{\text{resonance}} = E_{\text{hybrid}} - E_{\text{best single contributor}}

where: $\Delta E_{\text{resonance}}$ (resonance stabilization energy) is always negative — the true molecule is lower in energy, hence more stable, than any individual contributor could be if it existed as an isolated, localized structure.

Derivation ABond order and average formal charge from equivalent resonance contributors

Goal. Show that for an ion such as $\text{CO}_3^{2-}$ or $\text{NO}_3^-$ , built from $n$ chemically equivalent resonance contributors each with exactly one double bond among $n$ equivalent positions, the average bond order at each position is $\dfrac{n+1}{n}$ and the average charge on each terminal atom is $\dfrac{q_{\text{total}}}{n}$ .

Step 1. Write the total number of bonding electron pairs contributed by the central atom to the $n$ terminal atoms in any one contributor. With $n$ equivalent positions and exactly one double bond (2 pairs) while the remaining $(n-1)$ positions are single bonds (1 pair each), the total bonding pairs in one contributor is

\text{pairs}_{\text{one contributor}} = 2 + (n-1)(1) = n+1

Step 2. Because all $n$ contributors are equivalent by symmetry (each terminal atom takes its turn as the double-bonded one), every position is double-bonded in exactly one contributor out of $n$ and single-bonded in the other $(n-1)$ contributors. Averaging over all $n$ contributors, the bond order at any one fixed position is

\text{bond order} = \frac{(n-1)(1) + (1)(2)}{n} = \frac{n+1}{n}

Step 3. Check total bonding pairs are conserved: multiplying the per-position bond order by the $n$ positions must recover the same total as Step 1.

n \times \frac{n+1}{n} = n+1

This matches Step 1 exactly, confirming the averaging is self-consistent.

Step 4. Now distribute the ion's total charge $q_{\text{total}}$ . In each contributor, the $(n-1)$ singly-bonded terminal atoms each carry formal charge $-1$ (three lone pairs, one bond) and the one doubly-bonded terminal atom carries formal charge $0$ (two lone pairs, one double bond), while the central atom carries formal charge $0$ . The sum of terminal formal charges in one contributor is

(n-1)(-1) + (1)(0) = -(n-1)

Step 5. For carbonate, $n=3$ and the total ionic charge is $q_{\text{total}} = -2$ ; check Step 4 gives $-(3-1) = -2$ , consistent. By the same equivalence argument as Step 2, averaging the charge on one fixed terminal atom over the $n$ contributors gives

\bar{q}_{\text{terminal}} = \frac{q_{\text{total}}}{n}

Step 5 (limiting case). As $n \to \infty$ (a hypothetical, highly delocalized system with many equivalent resonance positions), the bond order

\frac{n+1}{n} \to 1

meaning delocalization over very many equivalent sites washes each individual bond back down toward single-bond character, while the charge per site

\frac{q_{\text{total}}}{n} \to 0

spreads the total charge thin — both limits match the intuitive picture that delocalization dilutes any single bond or charge across more atoms.

Worked examples

Example 1 — Nitrate ion, $\text{NO}_3^-$ . Count valence electrons: N contributes 5, three O contribute $3(6)=18$ , plus 1 for the negative charge.

5 + 18 + 1 = 24 \text{ electrons}

Draw one contributor: nitrogen central, one N=O double bond, two N–O single bonds. Formal charges in this contributor:

FC_{\text{N}} = 5 - 0 - \frac{8}{2} = 5 - 0 - 4 = +1

FC_{\text{O(double)}} = 6 - 4 - \frac{4}{2} = 6 - 4 - 2 = 0

FC_{\text{O(single)}} = 6 - 6 - \frac{2}{2} = 6 - 6 - 1 = -1

Check the sum against the total ionic charge:

(+1) + (0) + (-1) + (-1) = -1

This matches the ion's charge of $-1$ , so the structure is valid. There are $n = 3$ equivalent contributors (double bond rotates among the three oxygens), giving bond order

\text{bond order} = \frac{n+1}{n} = \frac{3+1}{3} = \frac{4}{3} \approx 1.33

per N–O bond, and average terminal charge

\bar{q}_{\text{O}} = \frac{q_{\text{total}}}{n} = \frac{-1}{3} \approx -0.33

per oxygen — consistent with measured N–O bond lengths of about $124\ \text{pm}$ , intermediate between N–O single ( $\sim 136\ \text{pm}$ ) and N=O double ( $\sim 120\ \text{pm}$ ).

Example 2 — Benzene, $\text{C}_6\text{H}_6$ (an asymmetric numeric check on ring delocalization energy). Benzene has two equivalent Kekulé resonance contributors, alternating single and double bonds around the ring ( $n=2$ positions of alternation, though there are 6 bonds total, 3 double and 3 single in each contributor). By the bond-order formula applied per bond position, with each of the 6 bonds being double in exactly one of the 2 contributors and single in the other,

\text{bond order} = \frac{(2-1)(1) + (1)(2)}{2} = \frac{1+2}{2} = 1.5

matching the experimentally uniform benzene C–C bond length of $139\ \text{pm}$ , intermediate between a C–C single bond ( $154\ \text{pm}$ ) and a C=C double bond ( $134\ \text{pm}$ ). Now use real thermochemical numbers to quantify the resonance stabilization energy directly. The experimental enthalpy of hydrogenation of one isolated (hypothetical, localized) double bond in cyclohexene is

\Delta H_{\text{hydrog, one C=C}} = -120\ \text{kJ/mol}

If benzene behaved as a localized 1,3,5-cyclohexatriene with three independent, non-interacting double bonds, hydrogenating all three would be predicted to release

\Delta H_{\text{predicted}} = 3 \times (-120\ \text{kJ/mol}) = -360\ \text{kJ/mol}

The measured enthalpy of hydrogenating real benzene to cyclohexane is

\Delta H_{\text{measured}} = -208\ \text{kJ/mol}

The resonance (delocalization) stabilization energy is the difference:

\Delta E_{\text{resonance}} = \Delta H_{\text{measured}} - \Delta H_{\text{predicted}} = (-208) - (-360) = +152\ \text{kJ/mol}

The positive sign here means benzene releases $152\ \text{kJ/mol}$ less heat than the naive localized picture predicts — precisely because the real hybrid sits $152\ \text{kJ/mol}$ lower in energy (more stable) than any single Kekulé contributor would be on its own. As a limiting-case sanity check: if the ring had no delocalization at all, $\Delta H_{\text{measured}}$ would equal $\Delta H_{\text{predicted}}$ and $\Delta E_{\text{resonance}}$ would be exactly zero — the nonzero $152\ \text{kJ/mol}$ is the experimental signature that resonance is real, not just a drawing convention.

Limiting cases & common traps

Situation	What actually happens	Common mistake
Equivalent contributors (e.g. $\text{CO}_3^{2-}$ , $\text{NO}_3^-$ )	All contribute equally; hybrid bond orders and charges are simple averages	Treating one contributor as "the" structure and ignoring the others
Non-equivalent contributors (e.g. formamide, $\text{N,O}$ push-pull)	Lower-formal-charge, better-electronegativity-matched contributors dominate the hybrid unevenly	Assuming a 50/50 blend when contributors are not equivalent
Only one Lewis structure is possible	No resonance; bond lengths/orders are exactly as drawn	Inventing resonance forms that change atomic positions (not allowed — only electrons move)
$n \to$ large (long conjugated chains)	Bond order at any one site drifts toward 1; charge per site drifts toward 0 (see derivation)	Forgetting that delocalization dilutes, it does not eliminate, bond/charge character
Curved-arrow pushing lone pair onto an atom that would exceed an octet (for period-2 atoms)	Not a valid resonance structure	Drawing five bonds to carbon or nitrogen "because the arrow allows it"

The chemist behind it

Resonance theory grew out of quantum-mechanical valence-bond treatments of molecules in the 1920s–1930s, when it became clear that a single classical structure could not reproduce measured bond lengths and stabilities for species such as benzene. The theory was formalized and popularized chiefly through the work connecting valence-bond theory to observable thermochemistry — most visibly in the comparison between predicted and measured hydrogenation enthalpies used above for benzene, which became the standard classroom demonstration of "resonance energy" as a real, measurable quantity rather than an artifact of notation.

Origins · 1 minLinus Pauling and how this was discovered

Linus Pauling

1901–1994 · Developed valence-bond theory and popularized "resonance" as a rigorous quantum-mechanical concept, explaining benzene's extra stability and uniform bond lengths; awarded the Nobel Prize in Chemistry in 1954 for work on chemical bonding, including resonance and electronegativity

Ask the tutor: How do you decide, using formal charges alone, which of several non-equivalent resonance contributors of the thiocyanate ion ( $\text{SCN}^-$ ) contributes most to its hybrid?

Are resonance structures real, physically distinct molecules?

No. Only the hybrid is real. Resonance contributors are drawing conventions — snapshots that, individually, cannot be drawn to show delocalized electron density. No single contributor exists as an isolable species; the actual electron distribution is the weighted average of all valid contributors.

What is and is not allowed to change between resonance contributors?

Only the placement of $\pi$ -bonding electrons and lone pairs may change. Atomic positions, $\sigma$ -bond connectivity, and the total number of electrons must stay exactly fixed across every contributor. If atoms move, the two structures are isomers, not resonance forms.

How do I decide which resonance contributor matters most when they are not equivalent?

Use formal-charge rules: contributors with formal charges closest to zero, and with any necessary negative formal charge on the more electronegative atom, contribute most to the hybrid. Contributors requiring like charges on adjacent atoms, or negative charge on an electropositive atom, contribute little or nothing.

Does resonance apply outside ions like carbonate and nitrate?

Yes. Resonance applies whenever multiple valid Lewis structures share the same atomic skeleton — conjugated dienes, aromatic rings such as benzene, carboxylate anions, amide bonds, and many radicals and cations all show measurable delocalization effects captured only by drawing multiple contributors.

How does resonance connect to molecular shape?

Resonance does not change the shape prediction: VSEPR and Molecular Shapes treats a delocalized $\pi$ system as an unbroken set of bonding domains around the central atom, so carbonate is still trigonal planar and nitrate is still trigonal planar, regardless of which contributor is drawn.