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Resonance, Formal Charge & Bond Order

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Resonance is delocalization, not flipping

When one Lewis structure cannot capture the bonding, we draw several resonance structures connected by a double-headed arrow (↔). The real molecule is not rapidly switching between them — it is a single, fixed resonance hybrid, the weighted average of all valid forms. The electrons in the multiple bonds are delocalized, spread smoothly over several atoms rather than pinned between two. Delocalization lowers energy, so a molecule with resonance is more stable than any one drawing suggests.

Formal charge
formal charge = valence electrons − nonbonding electrons − ½(bonding electrons)
Count the atom's group valence, subtract all of its lone-pair electrons, then subtract half of the electrons it shares in bonds. The formal charges of a species must sum to its overall charge.

Formal charge ranks the structures

When several structures are valid, the best (major) contributor obeys two rules: (1) formal charges are as close to zero as possible, and (2) any negative formal charge sits on the most electronegative atom. A structure that puts a large positive charge on an electronegative atom, or piles up charges, contributes little. Formal charge is bookkeeping — it is not a real charge — but it is the exam's tool for deciding which drawing dominates.

Average bond order (resonance hybrid)
bond order = (total bonds shared among the equivalent positions) ÷ (number of equivalent positions)
Equivalently, average the bond (single = 1, double = 2, triple = 3) for one linkage across all resonance structures. For NO₃⁻: 4 bonds over 3 equal N–O positions → 4/3 ≈ 1.33.

Bond order controls length and strength

A fractional bond order is the fingerprint of delocalization. As bond order rises, more electron density is packed between the nuclei, so the bond gets shorter and stronger (higher bond energy). This is why every N–O bond in NO₃⁻ is measured to be identical — about 124 pm, between a pure N–O single bond (≈136 pm) and an N=O double bond (≈115 pm) — exactly what an average bond order of 1.33 predicts. Equal bond lengths are the observable proof that resonance is real.

Worked example

Nitrous oxide, N₂O, is linear (N–N–O) with 16 valence electrons. Compare the two resonance structures [N≡N–O] and [N=N=O] using formal charge, and decide which is the major contributor.

  1. 1.Structure A, N≡N–O: terminal N has 1 lone pair and a triple bond → 5 − 2 − ½(6) = 0; central N has no lone pair, a triple and a single bond → 5 − 0 − ½(8) = +1; O has 3 lone pairs and a single bond → 6 − 6 − ½(2) = −1.
  2. 2.Check A: formal charges sum to 0 + (+1) + (−1) = 0, matching the neutral molecule. ✓
  3. 3.Structure B, N=N=O: terminal N has 2 lone pairs and a double bond → 5 − 4 − ½(4) = −1; central N has no lone pair and two double bonds → 5 − 0 − ½(8) = +1; O has 2 lone pairs and a double bond → 6 − 4 − ½(4) = 0.
  4. 4.Check B: formal charges sum to (−1) + (+1) + 0 = 0. ✓ Both structures are valid.
  5. 5.Apply the tiebreaker: the negative formal charge should rest on the most electronegative atom. In A the −1 sits on O (most electronegative); in B it sits on terminal N. Structure A is therefore the major contributor.
Answer: Both are legitimate resonance forms, but [N≡N–O] dominates because its −1 formal charge lies on the more electronegative oxygen. The true molecule is a hybrid weighted toward A.
Worked example

Calculate the average carbon–oxygen bond order in the carbonate ion, CO₃²⁻.

  1. 1.Draw the structure: C is central, bonded to three equivalent O atoms. Total valence electrons = 4 + 3(6) + 2 = 24.
  2. 2.One C=O double bond and two C–O single bonds satisfy every octet; formal charges are 0 on C, 0 on the double-bonded O, and −1 on each single-bonded O (sum = −2 ✓).
  3. 3.The double bond is not stuck on one oxygen — three equivalent resonance structures place it on each O in turn, so it is delocalized over all three positions.
  4. 4.Count the bonds shared among the three equal C–O linkages: two single bonds + one double bond = 4 bonds total, spread over 3 positions.
  5. 5.Average bond order = 4 ÷ 3 ≈ 1.33.
Answer: Each C–O bond has an average bond order of about 1.33, so all three are identical in length — intermediate between a single and a double bond.
Tip

Fast bond-order shortcut for a symmetric oxoanion: bond order = (number of bonding regions around the center) ÷ (number of terminal atoms). For CO₃²⁻ and NO₃⁻ that is 4/3; for O₃ it is 3/2 = 1.5. Fractional answers are the tell-tale sign that delocalization is present.

Checkpoint

In the tetrafluoroborate ion BF₄⁻, boron forms four B–F single bonds and has no lone pairs. What is the formal charge on boron?

Checkpoint

Ozone, O₃, is described by two equivalent resonance structures, each with one O=O double bond and one O–O single bond. What is the average O–O bond order?

Checkpoint

Experiments show all three N–O bonds in the nitrate ion, NO₃⁻, have exactly the same length, about 124 pm — shorter than a single N–O bond but longer than a double N=O bond. What is the best explanation?

On the exam

On free response, never justify equal bond lengths by saying the molecule "switches" between structures — that earns no credit. State that the bonding is delocalized into one resonance hybrid with a single fractional bond order. And always confirm your chosen Lewis structure by showing formal charges that sum to the species' charge.

Answer the 3 checkpoints as you read.

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