Rules for Drawing and Weighting Resonance Structures

Draw acetate ion with all the negative charge frozen on one oxygen. You have told a lie — a small, useful, but structurally misleading lie. The real acetate ion has two carbon–oxygen bonds of identical length (1.27 Å). This is exactly between a C–O single bond at 1.43 Å and a C=O double bond at 1.20 Å. The charge sits half on each oxygen. No single Lewis structure can show this. You need two, added together as a resonance hybrid. But the moment you allow more than one Lewis structure to represent a molecule, you need rules. Otherwise students (and, historically, chemists) draw structures that violate the octet rule. They invent atom movement that never happens. Or they treat every contributor as equally important when one dominates by orders of magnitude. This article is the rulebook. It is not new chemistry. It is bookkeeping discipline for the electron-pushing you already know from Resonance and Delocalization of Electrons and curved arrows. But the bookkeeping is what separates a correct resonance structure from nonsense. It also separates a correct ranking of structures from a guess. Get the rules wrong and you will predict that carbocations are stabilized by moving a proton. They are not — protons don't delocalize charge, electrons do. Or you will rank a structure with separated formal charges as dominant when it is actually a minor contributor. Such a structure is worth only a few percent of the true electron distribution. The big picture is this: a molecule has one real electron distribution. But that distribution is often too delocalized to draw with the localized-bond, localized-lone-pair language of a single Lewis structure. Resonance structures are alternative electron arrangements over a fixed nuclear skeleton. The true structure is the weighted superposition of all of them. This weighting is not equal — it depends on formal charge, electronegativity, bond count, and octet completeness. Learning to weight contributors correctly lets you predict which oxygen of a carboxylate is more nucleophilic (neither — they're equivalent). It also explains why an enolate reacts through carbon as often as oxygen. And it shows why the anilinium cation is destabilized relative to what naive charge-counting suggests. By the end of this article you will be able to generate the complete, legal set of resonance structures for any conjugated system. You will do this by applying five permissible curved-arrow moves. You will screen out illegal structures using the atom-movement and valence rules. You will rank the legal survivors using a four-criterion weighting hierarchy. These criteria are octet satisfaction, formal charge magnitude and placement, charge-electronegativity matching, and bond order/count. These same tools reappear, sharpened, in the Allyl System and in the analysis of Allylic Cations, Radicals and Anions. There, the ranking of contributors becomes the difference between correct and incorrect regiochemical predictions.

Key takeaways

Resonance structures differ only in electron placement; every nucleus stays in exactly the same position — no atom, bond angle, or connectivity ever moves.
Curved arrows always start at existing electron density (a lone pair or a π bond) and point to where those electrons end up; they never start on a positive charge or on a nucleus.
A structure is illegal if it exceeds an octet on a second-row atom, changes the total charge, or breaks/misplaces a σ bond.
Rank legal contributors by: (1) complete octets beat incomplete ones, (2) fewer and smaller formal charges beat more/larger ones, (3) negative charge on the more electronegative atom (and positive on the less electronegative atom) is better, (4) more bonds and less separation of charge is better.
Equivalent structures (identical connectivity, energy, and atom types related by symmetry) contribute equally — this is what makes acetate's two oxygens indistinguishable.
The true molecule is a single weighted hybrid, lower in energy than any single contributor — never an equilibrium or a mixture of separate species.
Key idea: resonance structures are alternative electron-counting schemes for one fixed nuclear framework, and only structures obeying the curved-arrow legality rules may be averaged, with weights set by octet completeness, formal-charge economy, and electronegativity matching.

The core insight

Resonance is not motion — it is bookkeeping. All resonance structures share identical atomic positions; only the assignment of electrons (bonding pairs, lone pairs) changes between them, and legal curved arrows are the only tool allowed to generate one structure from another.

Where this lives today

Resonance weighting is not a historical curiosity confined to textbook acetate — it is the everyday reasoning tool organic and medicinal chemists use to predict reactivity before running a single reaction. Drug designers use resonance-structure ranking to predict which nitrogen of an amidine or guanidine-containing drug (arginine side chains, many antihistamines, metformin) will be protonated and where the resulting positive charge truly sits, which governs salt-bridge formation with biological targets and membrane permeability. Enolate and enol chemistry, ranked exactly as in the acetone example above, underlies the regiochemistry of aldol reactions, Claisen condensations, and the design of soft versus hard electrophiles in asymmetric alkylation — industrially central to steroid, statin, and fragrance synthesis. Push-pull chromophores in dye and materials chemistry (merocyanine dyes, nonlinear-optical push-pull polyenes, and the color-tuning "cyanine limit" in photographic and OLED dyes) are explicitly designed using resonance-weighting logic: an electron-donor group and an electron-acceptor group at opposite ends of a conjugated chain create two resonance contributors of comparable energy (a neutral polyene form and a zwitterionic charge-separated form), and tuning substituents to bring these two contributors closer in energy is precisely how chemists engineer absorption wavelength and hyperpolarizability. The same weighting logic — Rank 1 through Rank 4 — explains why carboxylate, nitrate, and carbonate ions in biochemistry and geochemistry all show symmetric, delocalized bonding (identical bond lengths measured by X-ray crystallography), and it underlies computational chemistry's natural bond orbital (NBO) analysis, which numerically decomposes a molecule's true wavefunction into weighted Lewis-structure contributions, assigning each resonance structure an actual percentage weight rather than a qualitative "major/minor" label — the rigorous, computed descendant of the pencil-and-paper rules given here.

Ask the tutor: How do these same weighting rules predict which carbon of an unsymmetrical allylic cation an incoming nucleophile will attack preferentially?

The physical picture

Start from the orbital truth underneath the arrow-pushing formalism. A molecule like the carboxylate ion has a set of atomic p orbitals — one on each oxygen, one on the carbon — that overlap sideways to form a delocalized π system. Quantum mechanically, the true wavefunction of this π system is not well described by putting two electrons in a localized C=O π bond and leaving the other oxygen with a lone pair and a negative charge. The true molecular orbitals are combinations spread over all three atoms, and the electron density in the lowest occupied combination is symmetric across both oxygens. The Lewis-structure language, however, only knows how to draw localized two-center bonds and lone pairs. So chemists draw two Lewis structures, each localizing the π system differently, and declare that the true structure is their weighted average — a resonance hybrid. This is a genuine approximation to the delocalized molecular-orbital picture, and it works because a linear combination of two "bad" localized structures can reconstruct much of the delocalization energy that the true MO picture contains. The five legal curved-arrow moves you use to generate resonance structures are exactly the moves that keep this reconstruction valid: shifting a π bond, shifting a lone pair into a π system, shifting a lone pair to form a new π bond while pushing another pair off (as in enolate formation), and the radical single-barbed analogs for radical resonance. The weighting rules that follow — octet completeness, formal charge economy, electronegativity matching — are not arbitrary etiquette. They are quick, structure-based proxies for the actual energy of each contributor as a standalone (hypothetical) Lewis structure. A structure with an electronegative atom bearing extra electron density is lower in energy than one that dumps that density onto carbon, because oxygen's higher effective nuclear charge and lower-lying 2p orbital hold extra electrons more favorably. A structure with a broken octet at carbon is high in energy because it corresponds to an empty or half-filled orbital at an atom that "wants" eight electrons. The lower the intrinsic (hypothetical) energy of a contributor, the more heavily the true hybrid resembles it, and the greater its coefficient in the mixing.

Figure

Curved-arrow legality: acetate resonance, generated by one legal π-lone-pair shift

Figure

Legality test: two illegal "resonance" structures compared with the legal one

The core ideas

The complete legal move-set for generating one resonance structure from another consists of exactly these patterns, each conserving total electron count and total charge:

\text{(1) } \pi \to \pi: \quad \mathrm{C{=}C{-}C^{+}} \;\longleftrightarrow\; \mathrm{C^{+}{-}C{=}C}

\text{(2) lone pair} \to \pi: \quad \mathrm{:\!\ddot O{-}C{=}C} \;\longleftrightarrow\; \mathrm{O{=}C{-}\ddot C{:}^{-}}

\text{(3) lone pair} \to \pi \text{ with push-off}: \quad \mathrm{:\!\ddot N{-}C{=}O} \;\longleftrightarrow\; \mathrm{{}^{+}N{=}C{-}\ddot O{:}^{-}}

\text{(4) } \pi \to \text{lone pair (radical, single-barb)}: \quad \mathrm{C^{\bullet}{-}C{=}C} \;\longleftrightarrow\; \mathrm{C{=}C{-}C^{\bullet}}

where:

each arrow's tail sits on an existing bonding pair or lone pair (never on a bare positive charge or an atom with no electrons to give),
each arrow's head points to the new bond or new lone-pair location,
total number of electrons and total molecular charge are identical on both sides,
all nuclear positions, bond angles, and connectivities are identical on both sides — only π/lone-pair electron assignment differs.

The weighting of legal contributors is governed by a ranked checklist, applied in order until a decision is reached:

\text{Rank}_1: \; \text{complete octets (all non-H atoms)} \;>\; \text{one or more incomplete octets}

\text{Rank}_2: \; \text{fewer formal charges} \;>\; \text{more formal charges}, \quad \text{and smaller } |q| \;>\; \text{larger } |q|

\text{Rank}_3: \; \delta^{-} \text{ on more electronegative atom, } \delta^{+} \text{ on less electronegative atom} \;>\; \text{the reverse}

\text{Rank}_4: \; \text{more bonds (higher total bond order)} \;>\; \text{fewer bonds, for structures otherwise tied}

Reading these equations: Rank 1 is checked first and is nearly always decisive — a structure with a genuine octet violation (or, for carbon, more than four bonds) is not just "minor," it is usually outright illegal and should never even be drawn. Among legal structures, Rank 2 penalizes charge separation because separating charge costs the energy needed to pull an electron away from where it was electrostatically comfortable (Coulomb's law makes far-separated + and − charges lower in energy than close ones, but creating the separation from a neutral, all-bonds-satisfied structure always costs energy relative to no separation at all). Rank 3 is a direct electronegativity argument: an electron pair is more stable on an atom that holds electrons tightly. Rank 4 is a tie-breaker that rewards structures maximizing bonding (more bonds mean more shared, stabilizing electron pairs).

Derivation AMechanism 1 — Generating the complete resonance set for the acetate ion

Goal. Starting from one legal Lewis structure of acetate, CH₃CO₂⁻, generate every other legal resonance contributor using only permissible curved arrows, and show why no third distinct contributor exists.

Step 1 — Draw the starting Lewis structure. Place the carboxyl carbon with a double bond to one oxygen (call it O_a) and a single bond to the other (O_b), which carries three lone pairs and the formal negative charge. Verify: carbon has four bonds (to CH₃, O_a double, O_b single) — a complete octet, zero formal charge. O_a has two bonds (one σ, one π to C) plus two lone pairs — a complete octet, zero formal charge. O_b has one σ bond plus three lone pairs — a complete octet, formal charge $2-1-6=-1$ ... more precisely, formal charge $=$ (valence electrons of O, 6) $-$ (nonbonding electrons, 6) $-$ (bonding electrons/2, 1) $= 6-6-1=-1$ . This is legal: every atom's octet is satisfied, charges sum to $-1$ , matching the ion's overall charge.

Step 2 — Identify the source of a legal arrow. A lone pair on O_b is adjacent to the C=O_a π system (they share the carbon). This lone pair can, in principle, form a new π bond to carbon; but carbon already has four bonds (it would become five if the old C=O_a π bond simply stayed put). Therefore the arrow from O_b's lone pair must be coupled to a second arrow that relieves carbon's valence: push the C=O_a π electrons fully onto O_a as a new lone pair. This is the "lone pair → π with push-off" pattern.

Step 3 — Execute the two-arrow move. Arrow 1: lone pair on O_b → new π bond, O_b–C. Arrow 2: the existing π bond, C=O_a → lone pair on O_a. Track the bookkeeping: O_b loses one lone pair (from 3 to 2) and gains a π bond (from 1 σ bond to 1 σ + 1 π); O_a loses its π bond and gains a lone pair (from 2 to 3). Carbon's bond count is unchanged: it had one σ + one π to O_a and one σ to O_b (total 3 bonds, 4 bonding pairs counting the π as one bond... explicitly: σ(C–O_a), π(C–O_a), σ(C–O_b), σ(C–CH₃) = 4 bonds total) and after the move it has σ(C–O_a), σ(C–O_b), π(C–O_b), σ(C–CH₃) = still 4 bonds total. Carbon's octet is undisturbed throughout.

Step 4 — Verify the product structure is legal. New structure: C=O_b (π bond), C–O_a single bond, O_a bearing three lone pairs and formal charge $6-6-1=-1$ , O_b bearing two lone pairs and formal charge $6-4-2=0$ . Total charge still $-1$ . Every octet still complete. This structure is legal, and it is simply the mirror image (O_a and O_b swapped) of the structure in Step 1.

Step 5 — Check for a third, distinct contributor. Could a lone pair on O_a also push onto carbon while the O_b π bond pushes elsewhere? Any such arrow either regenerates one of the two structures already drawn or requires placing a second negative charge and a second π bond in a way that gives carbon five bonds (illegal) or breaks a C–O σ bond (illegal — σ bonds are never broken in resonance without breaking a rule, since resonance never changes connectivity). No third legal, distinct structure exists.

Step 6 — Assign weights. Structures 1 and 2 are legal, both have every atom's octet complete, both have exactly one formal charge (on the O bearing the "single-bond" role) of magnitude 1, and that charge sits on oxygen — the electronegative atom — in both cases. By symmetry (the two oxygens are related by the molecule's mirror plane and are chemically identical, both attached to the same CH₃CO group), the two structures are energetically identical and contribute with exactly equal weight, 50%/50%. This is why both C–O bonds in acetate measure the same 1.27 Å, intermediate between a single (1.43 Å) and double (1.20 Å) bond — consistent with a bond order of 1.5 on each oxygen, $\tfrac{1}{2}(1)+\tfrac{1}{2}(2)=1.5$ .

Limiting-case check. If instead of acetate you consider formate's asymmetric analog — an ester, CH₃CO₂CH₃, where one oxygen is tied up in a C–O–CH₃ single bond and cannot bear a negative charge without an additional positive charge appearing on that same oxygen (oxocarbenium-like) — the corresponding "second" resonance structure is now dramatically higher in energy (it places a positive charge on oxygen along with a broken octet consideration on the ether oxygen), so it contributes only a small amount. The 50/50 split is special to the symmetric acetate case; breaking that symmetry (as in an ester) breaks the equal weighting, exactly as the general weighting rules predict.

Go deeper · Derivation B · 3 minMechanism 2 — Ranking three legal contributors of the enolate of acetone

Derivation BMechanism 2 — Ranking three legal contributors of the enolate of acetone

Goal. Generate all legal resonance structures of the acetone enolate, $\mathrm{CH_3{-}C(O^-){=}CH_2} \leftrightarrow \mathrm{CH_3{-}C(=O){-}CH_2^-}$ , and rank them quantitatively using the four-criterion checklist, showing explicitly why one contributor dominates even though both are legal.

Step 1 — Draw structure A. Deprotonation of acetone at the α-carbon gives a carbanion adjacent to the carbonyl: $\mathrm{CH_3{-}C(=O){-}\overset{-}{C}H_2}$ . Here the central carbon has three σ bonds (to CH₃-bearing carbon, to the two H's are on the terminal carbon, not the central one — let me index clearly: label $C_1$ = carbonyl carbon, $C_2$ = the deprotonated methylene carbon). $C_1$ : σ to CH₃, σ to $C_2$ , σ+π to O — four bonds, octet complete, formal charge 0. O: σ+π to $C_1$ , two lone pairs — octet complete, formal charge 0. $C_2$ : σ to $C_1$ , two σ to H, one lone pair — three bonds plus a lone pair, octet complete ( $3\times2/2 \cdot$ ... explicitly formal charge $=4-2-3=-1$ ), formal charge $-1$ . This structure is legal: total charge $-1$ , matches the ion.

Step 2 — Identify the legal arrow. The lone pair on $C_2$ is adjacent (through the $C_1$ – $C_2$ σ bond) to the $C_1$ =O π system — a classic allylic-type arrangement (carbanion allylic to a π bond). The lone pair on $C_2$ can shift into the $C_1$ – $C_2$ bond to form a new π bond, provided the existing $C_1$ =O π bond is simultaneously pushed onto oxygen as a new lone pair (otherwise $C_1$ would gain a fifth bond).

Step 3 — Execute the arrows and produce structure B. Arrow 1: lone pair on $C_2$ → new π bond $C_1$ = $C_2$ . Arrow 2: π bond $C_1$ =O → lone pair on O. Resulting structure: $\mathrm{CH_3{-}C(O^-){=}CH_2}$ . Check bonds: $C_1$ now has σ(CH₃), σ(O), π( $C_2$ ), σ( $C_2$ ) — four bonds, octet complete, formal charge 0. O now has σ( $C_1$ ) and three lone pairs — formal charge $6-6-1=-1$ . $C_2$ now has σ( $C_1$ ), π( $C_1$ ), and two σ(H) — four bonds, octet complete, formal charge 0. Total charge still $-1$ . This structure — call it B — is legal.

Step 4 — Confirm no further legal structure exists. Could the negative charge now on oxygen in structure B migrate further, e.g., onto the CH₃ group? No: there is no π system or adjacent lone pair connecting the O⁻ lone pairs to the methyl C–H bonds; a σ(C–H) bond is not part of a conjugated π system here (no adjacent empty or π orbital to receive it, and pushing it would break a σ bond outright — illegal). The resonance set is exactly {A, B}.

Step 5 — Apply Rank 1 (octets). Both A and B have complete octets at every non-hydrogen atom. No decision yet; proceed to Rank 2.

Step 6 — Apply Rank 2 (formal charge count/magnitude). Structure A has one formal charge, $-1$ on carbon $C_2$ . Structure B has one formal charge, $-1$ on oxygen. Equal count, equal magnitude. No decision yet; proceed to Rank 3.

Step 7 — Apply Rank 3 (electronegativity matching). Structure A places the negative charge on carbon (Pauling electronegativity 2.55). Structure B places it on oxygen (Pauling electronegativity 3.44). Oxygen holds excess electron density far more comfortably — it is more electronegative, has a lower-energy 2p orbital, and its higher effective nuclear charge stabilizes the extra electron pair. Structure B is the major contributor.

Step 8 — Quantify the consequence. Because B dominates, the enolate's electron density (and therefore its HOMO, the orbital from which electrophiles are attacked) has substantially more amplitude at oxygen than the naive 50/50 guess would suggest — yet A is not negligible. Experimentally, enolates alkylate at both oxygen (O-alkylation, giving enol ethers) and carbon (C-alkylation, giving new C–C bonds), with the ratio depending on the electrophile's hardness and the counterion/solvent; the mere existence of competitive C-alkylation is direct evidence that structure A, while minor, is not zero. This is the resonance-weighting picture correctly predicting a mixture of reactivity, not an all-or-nothing outcome.

Limiting-case check. If we artificially replaced oxygen with carbon (a hypothetical symmetric bis-carbanion "enolate" of 1,3-butadiene's radical/anion analogs, i.e., the allyl anion itself, $\mathrm{CH_2{=}CH{-}\overset{-}{C}H_2} \leftrightarrow \mathrm{{}^{-}\!CH_2{-}CH{=}CH_2}$ ), both termini are carbon, electronegativity-matching gives no preference, and Rank 3 returns a tie — consistent with the allyl anion's two resonance structures being equal-weight, exactly the symmetric limiting case that validates the ranking rule (a rule that predicts a difference only when there is a genuine electronegativity or environment difference between the competing sites).

Worked examples

Example 1 (foundational) — Nitromethane, drawing the full legal set and assigning bond order. Nitromethane, $\mathrm{CH_3NO_2}$ , has a nitrogen bonded to two oxygens and a methyl group. Begin with a structure obeying the octet rule with formal charges: N is bonded to O_a by a double bond and to O_b by a single bond, with O_b carrying three lone pairs; nitrogen also bears a positive formal charge because it has four bonds (methyl, O_a double counts as one connection with a π, O_b single) — explicitly, N has bonds to: CH₃ (σ), O_a (σ+π), O_b (σ), totaling four bonds and zero lone pairs, formal charge $5-0-4=+1$ . O_a: σ+π to N, two lone pairs, formal charge $6-4-2=0$ . O_b: σ to N, three lone pairs, formal charge $6-6-1=-1$ . Total formal charge: $+1+0-1=0$ , matching the neutral molecule — legal. Apply the same lone-pair-push mechanism as in acetate: a lone pair on O_b shifts to form a new N=O_b π bond, while the N=O_a π bond simultaneously shifts to become a lone pair on O_a. Track nitrogen's bonds throughout: before, N has 4 bonds (σ to CH₃, σ+π to O_a, σ to O_b); after, N still has 4 bonds (σ to CH�3, σ to O_a, σ+π to O_b) — legal, nitrogen's octet (and its "four-bonds-plus-zero-lone-pairs" pattern, giving formal charge $+1$ throughout) is preserved. The product structure has O_a with formal charge $-1$ and O_b with formal charge $0$ — the mirror image of the start. By the symmetry of the two oxygens (both attached identically to the same N and both otherwise unsubstituted), these two structures are equivalent and weighted 50/50, exactly as in acetate. The predicted N–O bond order is

\text{bond order (each N–O)} = \frac{1}{2}(1) + \frac{1}{2}(2) = 1.5,

matching the experimental N–O bond length of 1.22 Å (compare N–O single bond ≈ 1.36 Å, N=O double bond ≈ 1.14 Å; 1.22 Å sits appropriately between these, close to the 1.5-bond-order prediction). Note also that nitrogen's formal $+1$ charge appears in every legal resonance structure — it never disappears — which correctly signals that nitro groups are strongly electron-withdrawing overall (both by induction, from electronegative oxygens, and by resonance, since nitrogen never becomes electron-rich in any legal contributor).

Example 2 (applied) — Why phenol's oxygen lone pair activates the ring toward electrophilic attack preferentially at ortho/*para, using resonance weighting to predict regiochemistry. Consider phenol, $\mathrm{C_6H_5OH}$ , attacked by an electrophile $\mathrm{E^+}$ at the para position (carbon 4) in electrophilic aromatic substitution. The Wheland intermediate (arenium ion) after E⁺ adds to C4 has the positive charge delocalized over C1, C3, and C5 by the ring's π system — three resonance structures from simple allylic-type shifting of the ring π bonds. But because the oxygen at C1 bears a lone pair conjugated directly into the ring, a fourth legal resonance structure exists: the lone pair on oxygen shifts to form a new π bond C1=C(ring), simultaneously pushing the ring's positive-charge-bearing empty p-orbital electron density (there is no lone pair to push in a carbocation lobe — instead, since C1 already carries a partial positive charge in one of the three ring contributors, the oxygen lone pair directly neutralizes it by forming O=C1, generating an oxocarbenium-stabilized structure with full octets everywhere and no ring-carbon left short). Apply the ranking rules: the three plain ring-only contributors each have a full ring octet everywhere except the carbocation carbon (only three bonds, empty p orbital — an octet-incomplete carbon; these are legal but not maximally stable). The fourth contributor — the oxocarbenium form — has every atom's octet complete (oxygen now has only one lone pair and a positive formal charge, ring carbon has four bonds). By Rank 1 (octet completeness), this oxygen-stabilized structure outranks the three plain ring structures decisively, even though it places a formal positive charge on the highly electronegative oxygen (which would normally be penalized by Rank 3). Octet completeness (Rank 1) is checked before electronegativity matching (Rank 3), so it wins the ranking outright: the resonance structure with the completed octet is the major contributor, despite the "wrong" placement of positive charge on oxygen — this is exactly why Rank 1 must be applied strictly before Rank 3, never averaged against it. This oxygen-stabilization pathway exists only when the electrophile attacks ortho or para to the OH group — attack at meta produces a Wheland intermediate where the positive charge, delocalized over three ring carbons, never lands on the carbon adjacent to oxygen, so the special oxocarbenium-stabilizing resonance structure is unavailable. This resonance-based accounting is the mechanistic reason phenol (and anilines, alkoxybenzenes, and other π-donor-substituted arenes) direct electrophiles ortho/para, not meta* — the intermediate's resonance hybrid is dramatically stabilized only for those two regiochemical outcomes.

Example 3 (exam/competition level) — Ranking five legal resonance structures of the diazonium-coupled azo dye precursor, an amidinium-type cation, $\mathrm{[H_2N{-}CH{=}NH_2]^+}$ (protonated formamidine), and predicting the C–N bond lengths. Protonated formamidine has a central carbon bonded to two nitrogens and one hydrogen, with overall charge $+1$ . Draw the reference structure: C=N_a (double bond) with N_a bearing two H's and no lone pair conjugated further (N_a: σ to two H's, σ+π to C — three bonds, one lone pair would be present on neutral =NH₂, but here we must track charge); C–N_b (single bond) with N_b bearing two H's and one lone pair, plus a formal positive charge if N_b has four bonds... Let's build this carefully. Central carbon $C$ bonded to: one H (σ), $N_a$ by a double bond (σ+π), $N_b$ by a single bond (σ). That's four bonds at carbon — octet complete, formal charge 0. $N_a$ : σ+π to C, two σ to H — that's three bonds and, for nitrogen to have a complete octet, one lone pair; formal charge $5-2-3=0$ . $N_b$ : σ to C, two σ to H, that's three bonds; for the overall ion to carry charge $+1$ somewhere, and since $N_a$ and $C$ are both neutral, $N_b$ must carry the formal $+1$ : with three bonds and only one lone pair the formal charge would be $5-2-3=0$ (neutral) — so instead $N_b$ must have three bonds and no lone pair to be electron-deficient enough for $+1$ : formal charge $5-0-3=+2$ , too high. The correct resolution is that $N_b$ , to be a legal trivalent-plus-lone-pair nitrogen cation, has three bonds and one lone pair removed relative to neutral amine nitrogen — i.e., $N_b$ has three σ bonds and zero lone pairs is wrong for octet completion (nitrogen would have only 6 electrons around it, violating the octet). The physically correct structure has $N_b$ with three σ bonds and one lone pair (octet complete, 8 electrons: 3 bonds ×2 + 1 lone pair ×2 = 8) giving formal charge $5-2-3=0$ — meaning the $+1$ charge of the ion must instead sit on $N_a$ , which if it has three bonds (σ+π to C, σ to one H — wait $N_a$ has two H's) needs re-examination: $N_a$ with σ+π(C) + 2σ(H) = 3 bonds, if it has one lone pair, formal charge 0; if it has zero lone pairs (all electrons in bonds, octet still complete via the double bond contributing... no, zero lone pairs plus 3 bonds gives only 6 electrons around N, an octet violation). The resolution: the positive charge must be placed by removing one lone pair's worth of ownership consistently with valence-electron bookkeeping applied to the whole cation, which is most cleanly done by recognizing protonated formamidine as isoelectronic with the guanidinium-type resonance system: the "extra" proton relative to neutral formamidine sits on one nitrogen, converting that nitrogen from an amine-type (three bonds, one lone pair, neutral) into an ammonium-type (four bonds, no lone pair, formal charge $+1$ ) only if it has four σ bonds, but here each N has only two H's plus one C — three bonds total, not four. Given three bonds and needing formal charge $+1$ : formal charge $=5-(\text{lone pair e}^-)-3$ ; setting this to $+1$ requires lone-pair electrons $=5-3-1=1$ , impossible (lone pairs come in pairs of 2, not 1). This shows the honest, structurally correct assignment is that the cationic charge is delocalized, and no single localized structure can put a clean integer +1 on one specific nitrogen while preserving both octets and integer bond counts simultaneously — which is exactly the pedagogical point: draw both legal resonance structures and let the hybrid carry the fractional charge. Structure I: $C{=}N_a H_2^+$ – $N_b H_2$ , i.e., the double bond and the formal $+1$ charge together on $N_a$ (three bonds to $N_a$ : σ+π to C and one σ to... recount with $N_a$ having two H's and the double bond to C: that is 2 (to H's) + 1 (σ to C) + 1 (π to C) = bonds "3" in the σ+π sense, 4 electron pairs total around N if you count the π — formal charge on $N_a$ here $=5-2(\text{one lone pair})-3(\sigma\text{-equivalent bonds, i.e., 3 groups})$ ... Using the standard, unambiguous formal-charge formula explicitly:

$q = Z_{\text{valence}} - n_{\text{lone-pair electrons}} - n_{\text{bonds}}$

For $N_a$ in Structure I (double bond to C, two single bonds to H, one lone pair): $n_{\text{bonds}}=3$ (double bond to C counts as one bond for this formula's bond-count term — the formula counts number of bonds, i.e., σ+π together counts as one "bond" to that neighbor), $n_{\text{lone-pair electrons}}=2$ : $q=5-2-3=0$ . For $N_b$ (single bond to C, two single bonds to H, two lone pairs): $n_{\text{bonds}}=3$ , $n_{\text{lone-pair electrons}}=4$ : $q=5-4-3=-2$ ?? That's wrong too — an amine nitrogen with three single bonds and one lone pair (not two) is neutral; giving it two lone pairs over-fills it (10 electrons, octet violated). So $N_b$ correctly has three single bonds and one lone pair, $q=5-2-3=0$ — both nitrogens neutral, and the cationic charge must sit on carbon: carbon here has 4 bonds (1 to H, 1σ+1π to $N_a$ , 1σ to $N_b$ ) $=4$ bond-count, 0 lone pairs, $q=4-0-4=0$ still neutral. The resolution is that this simplest "formamidine" skeleton as I've drawn it is neutral overall — to get the cation, one nitrogen must be protonated to bear an extra hydrogen (four σ bonds total: two H's plus C plus... no — an ammonium-like nitrogen needs four groups). The correctly drawn protonated formamidinium cation has, on one nitrogen, three hydrogens (not two): $N_a$ bonded to C by a double bond would then only allow one H (since N_a already uses 2 of its 3 valences on the C double bond formally counted as one connection plus one H = two connections, room for one more H only if it's a single-bonded amine-type N). Correctly, protonated formamidine is $\mathrm{[H_2N{-}CH{=}NH_2]^+}$ with $N_a$ = the doubly-bonded nitrogen bearing two H's (using its remaining single-bond valence twice: one to each of two H's is impossible for a doubly bonded N which has only one remaining valence slot for one substituent besides the double bond)...

Correcting explicitly and simply: the real, textbook-correct protonated formamidinium cation is

$\mathrm{H_2N^{+}=CH-NH_2 \;\longleftrightarrow\; H_2N-CH=NH_2^{+}}$

Here the doubly-bonded nitrogen has: the C=N double bond (one connection using 2 bonds) plus two N–H single bonds — total three σ-equivalent connections plus the π, i.e., four bonds total (σ to each of 2 H's, σ+π to C) and one lone pair: $q=5-2-4=-1$ ? That still isn't $+1$ . Try zero lone pairs on that nitrogen: $q=5-0-4=+1$ . Yes — with zero lone pairs, four bonds (two N–H, one σ and one π to C), nitrogen has $4\times2=8$ electrons around it purely from bonding, octet complete, and formal charge $5-0-4=+1$ . This is the correct assignment: the iminium-type nitrogen (C=N⁺H₂) has no lone pair (both its formal "lone pair slots" are used in bonds), exactly analogous to how an iminium carbon's nitrogen in any protonated imine bears the positive charge with zero lone pairs and four bonds. With this corrected structure I ( $\mathrm{H_2N{-}CH=\overset{+}{N}H_2}$ , positive charge and zero lone pair on the doubly bonded N, single-bonded $N$ neutral with one lone pair and three bonds), the legal curved arrow is: lone pair on the single-bonded (neutral) $NH_2$ nitrogen shifts to form a new π bond to carbon, while the existing C=N⁺ π bond shifts to become the (now regenerated) lone pair on that nitrogen — producing structure II, $\mathrm{H_2\overset{+}{N}=CH-NH_2}$ , the mirror image with the roles of the two nitrogens swapped.

Ranking: both structures are legal, both have complete octets everywhere, both have exactly one formal charge of magnitude 1 located on a nitrogen (electronegativity of nitrogen, 3.04, is the same atom type in both — this is the fully symmetric case, like acetate). By symmetry, structures I and II are equal-weight contributors (50/50), and the true C–N bond order in the hybrid is

$\text{bond order (each C–N)} = \tfrac12(1) + \tfrac12(2) = 1.5,$

predicting two equal, intermediate-length C–N bonds — confirmed experimentally in amidinium and guanidinium cations, where C–N bond lengths cluster around 1.32–1.33 Å (compare C–N single ≈ 1.47 Å, C=N double ≈ 1.28 Å), consistent with substantial but not full double-bond character on both nitrogens simultaneously. This is the same symmetric-hybrid outcome as acetate and nitromethane, and it is the electronic reason amidinium and guanidinium cations are exceptionally stable (and guanidine exceptionally basic) — the positive charge is shared equally across two (in guanidinium's case, three) nitrogens rather than localized on one, which is a direct, quantitative consequence of the equal-weighting rule for symmetric resonance contributors.

Limiting cases, scope & common traps

Situation	Rule outcome	Why
Two contributors related by a true symmetry element (mirror plane, rotation)	Equal weight (50/50, or 1/n for n-fold symmetry)	Rank 1–4 checklist returns an exact tie at every criterion
One contributor has an incomplete octet, the other complete	The complete-octet structure dominates overwhelmingly	Rank 1 is checked first and is nearly always decisive
Both contributors have one formal charge of equal magnitude, on different atoms	The structure with charge on the more electronegative atom dominates	Rank 3 (electronegativity matching)
A candidate structure requires moving a σ-bonded atom (e.g., an H)	Not resonance at all — it's a different molecule (tautomer)	Resonance structures share one fixed nuclear skeleton; moving atoms changes the molecule
A candidate arrow starts on a bare positive charge (no electrons there)	Illegal — arrows must start on real electron density	Curved arrows track real electron flow, not "attraction of positive charge"
A candidate structure gives a second-row atom (C, N, O) more than 4 bonding pairs	Illegal — octet cannot be exceeded for period-2 elements	No accessible d-orbitals at period 2 to house extra electron density
Charge separation created from a neutral, fully-bonded starting structure	The separated-charge structure is always a minor contributor, even if legal	Creating charge separation costs energy relative to no separation, per Rank 2

Common trapMisconception: "resonance structures are in equilibrium, interconverting over time"

This is false and is the single most common error. Acetate is not 50% of the time one Lewis structure and 50% of the time the other, flickering back and forth. There is exactly one real molecule with one real, static (time-independent, for a given geometry) electron distribution; the two Lewis structures are simultaneously present only as a mathematical device — a weighted sum used because our bonding language cannot draw the true delocalized distribution in one picture. No bond ever breaks and reforms during "resonance"; nothing happens on any timescale. Contrast this sharply with a genuine equilibrium (like keto–enol tautomerism), where atoms truly move, a real transition state truly exists, and the two forms truly are separate, interconverting molecules with separate physical existences.

The chemist behind it

Resonance theory's modern language — curved arrows, contributing structures, weighted hybrids — was built by Sir Robert Robinson and, in parallel and with more explicit quantum grounding, by Linus Pauling. Robinson, working in Britain in the 1920s–30s on natural-product synthesis (he determined the structure of morphine and synthesized tropinone in a legendary one-pot biomimetic synthesis), introduced the curved arrow as a bookkeeping device for electron movement in both mechanism and in delocalization, laying the graphical foundation later adopted universally. Pauling, working from the newly available quantum mechanics of Heitler and London's 1927 treatment of the hydrogen molecule, formalized resonance as a variational quantum-mechanical statement: the true wavefunction of a molecule like benzene is a linear combination of the wavefunctions corresponding to each Kekulé (and other) structure, and the true energy is always lower than that of any single contributing structure — the resonance (or delocalization) energy. Pauling's 1931–1933 papers "The Nature of the Chemical Bond" (later the title of his immensely influential 1939 book) gave the weighting rules a rigorous variational grounding: contributors that are lower in energy as isolated hypothetical structures mix in with larger coefficients, exactly the heuristic embodied in the octet/formal-charge/electronegativity ranking used by every organic chemist since.

Origins · 1 minLinus Pauling and how this was discovered

Linus Pauling

1901–1994 · Quantum theory of the chemical bond, resonance theory, electronegativity scale

Pauling trained as a chemical engineer at Oregon State before moving to Caltech, where he absorbed the new quantum mechanics directly from European sources during a 1926–27 fellowship in Munich, Zürich, and Copenhagen — sitting in on lectures by Sommerfeld, Schrödinger, and Bohr. Returning to Caltech, he synthesized quantum mechanics with classical structural chemistry to found the modern theory of the chemical bond, introducing hybridization, resonance, and the electronegativity scale that bears his name. He won the Nobel Prize in Chemistry in 1954 for this body of work, and the Nobel Peace Prize in 1962 for his campaign against atmospheric nuclear testing — the only person to win two unshared Nobel Prizes.

History

- 1916 — Gilbert N. Lewis introduces the shared electron-pair bond and the octet rule, the language resonance structures obey.

- 1926–1927 — Heitler and London's quantum treatment of H₂ shows a two-structure wavefunction lowers the calculated bond energy below either structure alone — the physical seed of resonance.

- 1928–1933 — Pauling formalizes resonance/valence-bond theory quantitatively; independently, Ingold and Robinson develop the curved-arrow electron-pushing formalism in Britain for mechanism and delocalization.

- 1933 — Pauling and Wheland apply resonance explicitly to benzene, quantifying the resonance (delocalization) stabilization energy.

- 1939 — Pauling's "The Nature of the Chemical Bond" cements resonance theory, formal-charge weighting, and electronegativity as standard tools of structural organic chemistry.

What exactly stays the same between two resonance structures, and what is allowed to change?

The positions of every nucleus, all σ bonds, all bond angles, and the overall molecular connectivity stay completely fixed. What is allowed to change is only the assignment of π-bonding electrons and lone-pair electrons — which atom "owns" a given pair, and whether a given pair is drawn as part of a π bond or as a nonbonding lone pair. Because only electron bookkeeping changes, resonance structures are not different molecules, not isomers, and not in equilibrium; they are alternative, individually incomplete descriptions of one single real electron distribution, and the real molecule is the weighted sum of all legal descriptions.

How do I know when to stop generating new resonance structures — how do I know I have found them all?

Keep applying the legal arrow patterns (π→π, lone pair→π, lone pair→π with push-off, and the radical single-barb analogs) to every new structure you generate, and stop when every possible legal arrow either regenerates a structure you already have or requires breaking a rule (exceeding an octet, moving a σ-bonded atom, or changing total charge). In practice, for typical conjugated systems (allylic cations/anions/radicals, enolates, carboxylates, aromatic and heteroaromatic rings, amidines/guanidines) the complete set is small — often two to five structures — because the conjugated π system is finite and every additional legal shift eventually cycles back to a structure already drawn.

Do all legal resonance structures contribute equally to the true hybrid?

No — only structures related by a genuine symmetry operation of the molecule (a mirror plane, rotation, or other symmetry element mapping one structure exactly onto the other, atom type for atom type) contribute equally. Whenever the local environment differs even slightly — different attached groups, different atom identity, different distance from an electronegative substituent — the four-criterion ranking (octet completeness, formal-charge count/magnitude, electronegativity matching, bond count) will, in general, break the tie and assign unequal weights, sometimes overwhelmingly so, as in the phenol arenium-ion example where one contributor dominates almost completely.

Why can't resonance structures ever have different total charge or different number of electrons from each other?

Resonance structures are alternative descriptions of the exact same molecule (or ion), and a molecule has one fixed total number of electrons and one fixed overall charge, set by its actual atomic composition. If a proposed "resonance structure" has a different total charge, it is describing a different chemical species altogether (for example, the molecule plus or minus an electron), which is not resonance — it may be a related redox species, but it cannot be averaged into the same hybrid. Legal curved arrows always conserve total electron count exactly, which is precisely why every legal resonance structure automatically preserves total charge.

Is it ever acceptable to draw a resonance structure with an expanded octet, like five bonds to phosphorus or sulfur?

For period-3-and-below elements (phosphorus, sulfur, silicon), the traditional teaching of readily accessible d-orbital expansion has been substantially revised: high-level calculations show true d-orbital participation is minor, and apparent "expanded octet" structures (like the classically drawn S with six bonds in sulfate) are better described using formal charges on a strict-octet structure or by ionic/ylidic bonding pictures instead. For period-2 elements (carbon, nitrogen, oxygen, fluorine), expanded octets are unambiguously illegal — there are no accessible low-energy d-orbitals at all — so any resonance structure putting five bonds on carbon or nitrogen is always wrong and should never be drawn, regardless of how attractive it might seem for lowering formal charge.

How does resonance weighting connect to what I'll see in the allyl cation, radical, and anion?

The allyl system is the cleanest possible test case for every rule given here: it has exactly two terminal carbons related by a genuine mirror symmetry, so its two resonance structures (for the cation, radical, or anion) are exactly equal-weight by the same symmetry argument used for acetate. Once a substituent breaks that symmetry — for instance, a methyl group on one terminus — the ranking rules (formal charge count is unchanged, but hyperconjugative/inductive stabilization differs) predict, correctly, that charge or spin density shifts toward the more substituted, more stabilized terminus, which is exactly the regiochemical logic used to explain Markovnikov-type selectivity in allylic cation chemistry.