Lewis structures: use formal charge and resonance together

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Use the electron budget to distinguish an invalid drawing, a minor contributor, and an equivalent representation of the same ion.

About 15 minutes, including the worked examples and checks.

Useful foundations

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Concept: keep an electron budget and an atom framework
Application: audit drawings of formate
Pattern: formal charge tests the allocation
Rule: contributors describe one delocalized structure
Changed application: track an electron pair into a new bond
Check your reasoning
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Concept: keep an electron budget and an atom framework

A Lewis structure makes a specific claim about where valence electrons are counted: in bonds or as nonbonding electrons. Before judging whether a drawing looks familiar, establish the molecular formula, total charge, and connectivity. An anion has more electrons than the corresponding neutral atoms; a cation has fewer. A line represents two shared electrons, whereas a dot represents one electron. The total must agree with the species being drawn.¹

For common closed-shell molecules built from second-row atoms, check an octet around carbon, nitrogen, oxygen, and fluorine, while hydrogen has two electrons. This is a useful constraint, not an exceptionless definition of a molecule. Electron-deficient boron compounds and odd-electron species exist. An expanded octet, however, is not a valid way to rescue a drawing with ten electrons around ordinary second-row carbon.

Application: audit drawings of formate

Consider HCOO⁻, with H attached to C and both O atoms attached to that same C. The following audit and drawings are original teaching examples. The electron budget is 1 + 4 + 2(6) + 1 = 18 valence electrons. Start with three single bonds, which use six electrons. Completing both oxygen octets uses the remaining twelve, but leaves carbon surrounded by only six.

Convert one oxygen lone pair into a second C–O bonding pair. This changes its bookkeeping location without adding electrons. Carbon now has four bonding pairs around it, while the double-bonded oxygen keeps two lone pairs and the single-bonded oxygen keeps three. Every atom meets its usual shell requirement, and the count remains eighteen.

The same atom framework appears twice. Dots are nonbonding electrons; each line is one shared pair. The arrow relates resonance contributors, not separately existing species.

Proposal	Total electrons	Carbon environment	Assessment
One C=O and one C–O	18	8 electrons	Complete octets
Two C=O bonds plus C–H	18	10 electrons	Reject: carbon exceeds its octet
Two C–O bonds plus C–H; no C lone pair	18	6 electrons	Octet-deficient, charge-separated contributor
Two C–O bonds plus C–H; add a C lone pair	20	8 electrons	Wrong electron budget for HCOO−

Audit of proposed formate drawings

The second proposal is a useful trap: its total electron count is correct, yet its local carbon environment is not. The third has the correct electron count and can be represented as a less important, electron-deficient contributor; it is not equivalent in importance to the complete-octet drawings. The fourth fixes carbon by inventing an extra pair, thereby changing the charge of the represented species.

Pattern: formal charge tests the allocation

For an atom, formal charge = neutral-atom valence electrons − nonbonding electrons − half the bonding electrons. Equivalently, subtract nonbonding electrons and the sum of its bond orders. Count a double bond as two in that final term. Sum all formal charges to recover the total charge of the species.²

In contributor A, the single-bonded O has formal charge 6 − 6 − 1 = −1. The double-bonded O has 6 − 4 − 2 = 0. Carbon has 4 − 0 − 4 = 0, and H has 1 − 0 − 1 = 0. The total is −1. In the all-single-bond drawing without a carbon lone pair, carbon instead has +1 and each O has −1: the total still agrees, but charge separation and an incomplete carbon octet make it less favorable.

Formal charge is an equal-sharing convention. It is not a measurement of the electron density on an atom and is not the same calculation as an oxidation state. A carbonyl oxygen can carry formal charge zero while having a partial negative charge because the actual bond is polarized.

Rule: contributors describe one delocalized structure

The two complete-octet formate drawings differ only in electron placement. They are resonance contributors to one electronic structure, not two molecules that alternate with a stopwatch. In an environment that treats the two O atoms equivalently, the actual C–O bonds are equivalent and intermediate between idealized single and double bonds. Describing an average bond order of 1.5 is a useful representation of these equivalent contributors, not evidence that a particular bond is single for half of each second.²³

Valid resonance comparisons preserve atom connectivity, electron count, and overall charge. Moving an H atom from one oxygen to another is a nuclear rearrangement, not merely drawing another resonance contributor. Likewise, rotating an entire drawing does not create an additional electronic arrangement.

When contributors are unequal, assess complete valence shells, unnecessary charge separation, and whether negative formal charge is reasonably placed on an electronegative atom. These are coordinated criteria. Do not select a drawing solely because its total charge is correct or because one charge sits on oxygen. A drawing must first respect the electron budget and allowable valence.

Changed application: track an electron pair into a new bond

A Lewis base donates an electron pair to a Lewis acid, which accepts it. For example, when NH₃ supplies its nitrogen lone pair to electron-deficient BH₃, that pair becomes the N–B bond. A curved electron-pair arrow begins at the nitrogen pair and ends at boron. The arrow follows electrons, not the direction in which an atom moves.⁴

After bond formation, nitrogen has four bonds and formal charge +1; boron has four bonds and formal charge −1. The adduct remains neutral. Those formal charges do not mean that the donor lost two electrons completely to the acceptor: the new pair is shared in a bond. Keeping the donor, acceptor, and formal-charge calculation separate prevents an acid–base step from being mislabeled as electron transfer.

This electron-accounting method connects the AAMC bonding and resonance scope to later mechanism questions.⁵ First count what exists, then test each atom, and finally ask what the representation predicts about equivalence or electron-pair availability. Geometry and reaction conditions provide additional information; neither can repair a missing electron budget.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Check 1

A proposed drawing of CO₃²⁻ has three C=O double bonds and two lone pairs on every O. It contains 24 valence electrons in total. Which criticism is decisive?

A. It has too few valence electrons for the specified ion.

Review this choice. Carbon contributes 4, the three oxygens 18, and the charge 2, for 24. The total count is correct.

B. It places twelve electrons around second-row carbon.

Correct. Three double bonds put six bonding pairs around C. Correct total counting does not permit a twelve-electron carbon valence shell.

C. It requires each oxygen to have an incomplete octet.

Review this choice. Each O has two lone pairs and one double bond, totaling eight electrons around that O.

D. It uses more atoms than the molecular formula allows.

Review this choice. There is one carbon and three oxygens. The defect is local electron occupancy, not atom count.

The principle to carry forward

Audit the whole-species budget and each atom separately.

Check 2

Two drawings of NO₂⁻ each use 18 electrons and the same O–N–O framework. Drawing I has one N=O, one N–O, and one lone pair on N. Drawing II has two N–O single bonds and one lone pair on N. Oxygen lone pairs complete each O octet. Which comparison is justified?

A. Drawing I is favored because it completes the N octet with less charge separation.

Correct. In I, N has eight surrounding electrons and formal charge 0; only the single-bonded O has −1. In II, N has six surrounding electrons and +1, with −1 on each O.

B. Drawing II is favored because single bonds always reduce formal charges.

Review this choice. Here the all-single-bond drawing creates a positive N and two negative O formal charges. Fewer drawn lines do not guarantee fewer formal charges.

C. The drawings are equally important because their electron totals match.

Review this choice. Equal totals are necessary for comparison but do not make incomplete and complete octets equally favorable.

D. Neither drawing is allowed because nitrogen cannot carry a lone pair.

Review this choice. Nitrogen can have a lone pair; the relevant checks are its total electron environment and formal charge.

The principle to carry forward

A valid budget does not imply equal contributor importance.

Check 3

Water binds to Zn²⁺ through oxygen in a Lewis acid–base step. Both O–H bonds remain intact, and zinc’s oxidation state remains +2. Which electron-pair movement correctly represents formation of the new O–Zn bond?

A. A bonding pair from an O–H bond moves toward zinc.

Review this choice. Moving an O–H bonding pair into a different bond would alter that O–H bond, contrary to the supplied intact-water coordination step.

B. A nonbonding pair from zinc moves toward oxygen.

Review this choice. In this Lewis acid–base interaction, Zn²⁺ accepts a ligand electron pair. Water oxygen supplies an available lone pair; zinc is not the pair donor.

C. An oxygen lone pair moves toward zinc.

Correct. The water oxygen donates an occupied nonbonding pair to Zn²⁺. The shared pair forms the O–Zn bond without a proton transfer or change in zinc oxidation state.

D. One electron from each participant forms the new pair.

Review this choice. A Lewis donor–acceptor bond is represented by movement of a pair from the donor. Combining one electron from each partner is not the supplied electron-pair-donation mechanism.

The principle to carry forward

Identify the source of the pair before assigning Lewis acid and base roles.