Topic 2.6: Resonance and Formal Charge Revision Notes

When One Lewis Diagram Is Not Enough

Some species cannot be represented honestly by a single Lewis diagram. Draw carbonate, CO₃²⁻, by the Topic 2.5 sequence and one C−O bond comes out double, the other two single, yet nothing picks out one oxygen for the double bond: any of the three could carry it. The three resulting diagrams are resonance structures, or contributors, of one and the same ion. Measured bond lengths prove it: all three C−O bonds are identical, each shorter than a single bond but longer than a double bond, which no single contributor matches.

What May and May Not Change Between Contributors

Resonance structures share everything except the placement of electrons: atoms, connectivity, total valence electrons and overall charge are all identical. Only multiple bonds and lone pairs trade places: a lone pair beside a multiple bond can join it, and a multiple bond can collapse into a lone pair on one of its atoms.

Contributors may be equivalent or nonequivalent. Carbonate’s three are equivalent: same arrangement, same charges, same energy, contributing equally. NCO⁻’s are nonequivalent, with different charge patterns, and contribute unequally; judging between them is the work of formal charge, below.

The Resonance Hybrid and Average Bond Order

The real species is a resonance hybrid: one static structure blending the contributors, its bonding spread over every position the contributors use. It does not switch or oscillate between them. The contributors are bookkeeping drawings; the hybrid is the molecule.

Because the bonding is shared, a bond that is double in one contributor and single in another has an intermediate, fractional bond order in the hybrid: total bond order across the equivalent positions, divided by the number of positions. Formate, HCOO⁻: (2 + 1) / 2 = 1.5 over its two C−O positions. Carbonate and nitrate: (2 + 1 + 1) / 3 = 4/3, about 1.33, over three positions. Benzene’s two contributors give every C−C bond an order of 1.5.

Bond order still predicts length and strength, as in Topic 2.2. Nitrate’s N−O bonds (order 1.33) are longer than nitrite’s (order 1.5, from (2 + 1) / 2), because the same bonding is spread over more positions. Delocalization also lowers the energy: the hybrid is more stable than any single contributor, its electrons spread over more atoms.

Formal Charge: The Bookkeeping Formula

Formal charge audits a finished diagram: the charge each atom would carry if every bonding pair were split exactly equally between the two atoms it joins:

Formal charge = valence electrons − (nonbonding electrons + ½ × bonding electrons)

The working shortcut counts lines instead of electrons: formal charge = valence electrons − dots − bonds, where dots are nonbonding electrons and bonds are the bond lines touching the atom. A double bond counts as two bonds, since the atom is assigned half of four bonding electrons.

Two checks make the arithmetic self-verifying. First, a diagram’s formal charges must sum to the species’ overall charge: zero for a neutral molecule, −1 for NO₃⁻. Any other sum exposes a counting error. Second, formal charge is a bookkeeping device, not a measured charge: it splits shared electrons equally, ignoring electronegativity, so never confuse it with oxidation state, which hands both bonding electrons to the more electronegative atom.

Worked Example: Nitrate, Atom by Atom

NO₃⁻ owns 5 + (3 × 6) + 1 = 24 valence electrons. In each contributor nitrogen is double bonded to one oxygen, single bonded to two, and carries no lone pair; the double bonded oxygen carries two lone pairs, each single bonded oxygen three.

Nitrogen: 5 − (0 + 4) = +1. Double bonded oxygen: 6 − (4 + 2) = 0. Each single bonded oxygen: 6 − (6 + 1) = −1. The sum, +1 + 0 − 1 − 1 = −1, matches the ion’s charge, and the placed electrons audit too: 8 bonding + 16 in lone pairs = 24. The same audit on ozone gives +1, 0 and −1 across its three oxygens, summing to zero as a neutral molecule must.

Judging Structures: The Stability Hierarchy

Formal charge exists to rank diagrams. Apply the tests in this order.

Complete the octets first. A contributor with complete octets, especially on period 2 atoms, beats one without, even at the price of formal charges. Carbon monoxide proves it: 10 electrons cannot give both atoms an octet through a double bond (that takes 12 placed electrons), so the accepted diagram is the triple bond, completing both octets at the price of −1 on carbon and +1 on oxygen.

Then minimize formal charges. Fewer and smaller formal charges mean a better contributor. Sulfate, SO₄²⁻, drawn with four single bonds puts +2 on sulfur and −1 on every oxygen. Converting two lone pairs into double bonds (legal for period 3 sulfur) puts 12 electrons around sulfur, with sulfur and the double bonded oxygens at 0 and only the single bonded oxygens at −1. Both diagrams place exactly 32 electrons; the second wins purely on smaller formal charges.

Then place unavoidable charges well. Negative formal charge belongs on the more electronegative atom, positive charge on the less electronegative one. In NCO⁻, the contributor [:N≡C−O]⁻ (charges 0, 0, −1, with the −1 on oxygen) dominates [N=C=O]⁻ (−1 on nitrogen), although both have complete octets and one charge each.

But formal charge never licenses a bad octet trade. BF₃ is the counterweight to sulfate. Its standard diagram leaves boron with 6 electrons and every formal charge zero. A B=F double bond completes boron’s octet only by putting +1 on fluorine, the most electronegative element there is, and −1 on boron. The incomplete octet with zero formal charges is the accepted structure. Octets first governs period 2 atoms and reasonable charges; it is not an instruction to complete an octet at any cost.

One limit case: NO, with 11 valence electrons, always leaves one electron unpaired, so no diagram completes every octet.

Traps That Cost Marks in Topic 2.6

Describing resonance as movement. The hybrid never flips between contributors. Saying the double bond “moves around” or the molecule “alternates” misdescribes a static blend; the credited language is delocalized electrons and equal bond lengths.

Offering a different skeleton as a resonance structure. Moved atoms, changed connectivity, or a broken and remade single bond all mean a different species. Resonance rearranges electrons only.

Formal charges that do not sum to the charge. The sum is a free error check, so run it before trusting any individual value: +1 with two −1s suits a −1 ion, not a neutral molecule.

Counting a double bond as one bond in the shortcut. Formal charge = valence − dots − bonds counts bond lines; a double bond is two lines, assigning the atom 2 of the 4 bonding electrons. Halving wrongly here is the commonest arithmetic slip in the topic.

Assuming the zero formal charge diagram always wins. CO keeps a triple bond with charges of −1 and +1 because the octet matters more; BF₃ keeps an incomplete octet because the alternative charges fluorine positively. Quote one rung as the whole rule and exactly these species catch you.

Key Relationships

Resonance (SAP-4.B): needed when one Lewis diagram cannot represent the species; contributors share connectivity, electron total and charge, differing only in where multiple bonds and lone pairs sit
Resonance hybrid: the single real structure, a static blend of the contributors; more stable than any contributor alone because its electrons are delocalized over more atoms
Average bond order: total bond order over the equivalent positions divided by the number of positions (HCOO⁻ 1.5, CO₃²⁻ and NO₃⁻ 4/3, PO₄³⁻ 5/4, benzene C−C 1.5); higher order means shorter, stronger bonds
Formal charge = valence electrons − (nonbonding electrons + ½ bonding electrons), or valence − dots − bonds; a diagram’s formal charges always sum to the species’ overall charge
Stability order: complete octets first (period 2 above all), then smallest formal charges, then least charge separation, then negative charge on the most electronegative atom (SO₄²⁻ two double bonds; NCO⁻ −1 on oxygen; CO and BF₃ mark the limits)