Resonance — delocalization, resonance energy (benzene, ozone, carbonate): Worked examples

Before anything, the two quantities we lean on constantly:

The scenario matrix

Every resonance problem you can be handed falls into one of these cells. The examples below are tagged with the cell they cover, and together they hit all of them.

Cell	Case class	What makes it tricky	Example
A	Symmetric, 2 equal forms	split one π over 2 links	E1 (ozone)
B	Symmetric, 3 equal forms + net charge	split charge too	E2 (carbonate)
C	Ring delocalization (aromatic)	count π over a loop	E3 (benzene)
D	Asymmetric — forms not equal	must weight them	E4 (OCN⁻)
E	Degenerate / zero — only one valid form	resonance energy $= 0$	E5 (CO₂ vs "fake")
F	Charged, non-equivalent oxygens	mix of B.O. on one molecule	E6 (nitrate vs nitrite)
G	Word / real-world — heats of hydrogenation	build RE from data	E7 (benzene RE)
H	Exam twist — limiting/edge reasoning	naive averaging fails	E8 (allyl vs butadiene ends)

E1 — Cell A: Ozone, the symmetric pair

Worked example Find the O–O bond order and charge on each O in

$O_{3}$ . Forecast: guess the bond order now — is it $1$ , $1.5$ , or $2$ ? And where does the negative charge sit?

1. Draw both structures. $O = O + - O - \leftrightarrow O - - O + = O$ Why this step? The extra π-bond can sit on the left or the right O. Both drawings obey the octet and have the same charges, so they are equally valid — a symmetric pair (Cell A).

2. Count bond-pairs on one O–O link across all forms. Left link: double (2) in form 1, single (1) in form 2. That's $1 + 2 = 3$ pairs over $2$ forms. Why this step? Bond order averages the pairs that occupy that link as we sweep through the forms.

3. Divide. $B.O. = \frac{1 + 2}{2} = 1.5$ Why this step? Delocalization means the real bond is literally the average of the two drawings.

4. Spread the charge. The two terminal O's are equivalent, each carries $- 1$ in exactly one of the two forms, so each holds $\frac{- 1}{2} = - \frac{1}{2} on average.$ The central O keeps $+ 1$ throughout. (See Formal Charge for how the $+ / -$ were assigned.)

Verify: predicted bond length is between single ( $148$ pm) and double ( $121$ pm) — real O₃ is $128$ pm ✔. Total charge $= (+ 1) + 2 (- \frac{1}{2}) = 0$ ✔ (ozone is neutral).

E2 — Cell B: Carbonate, three equal forms with charge

Worked example Bond order and per-oxygen charge in

$CO_{3}^{2 -}$ . Forecast: three oxygens, one double bond. Will B.O. be more or less than ozone's $1.5$ ?

1. Draw all three structures. The single C=O rotates among the three oxygens — three symmetric forms (Cell B, with a real $2 -$ charge to also spread). Why this step? Each drawing is one snapshot; because all three are equally valid, the averaging formula treats every C–O link on equal footing — we need every form on the page before we can average across them.

2. Count bond-pairs on one C–O link. Across the three forms a given C–O is: double (2) once, single (1) twice ⇒ $2 + 1 + 1 = 4$ pairs. Why this step? Same averaging idea, now over three equivalent links.

3. Divide. $B.O. = \frac{2 + 1 + 1}{3} = \frac{4}{3} \approx 1.33$ Why this step? One π-bond shared over three links gives a smaller extra share than one π over two — so it's less than ozone's 1.5, matching the forecast intuition.

4. Spread the charge. The $2 -$ sits equally on three oxygens: $\frac{- 2}{3} \approx - 0.67 per O.$

Verify: all three C–O equal, length between single ( $143$ ) and double ( $122$ ) — real is $\sim 129$ pm ✔. Charge check: $3 \times (- \frac{2}{3}) = - 2$ ✔.

E3 — Cell C: Benzene ring bond order

E4 — Cell D: The asymmetric case (cyanate, OCN⁻)

This is the cell students lose marks on: the forms are not equal, so you cannot just average blindly — you must weight them, and then average with those weights.

The figure below lays the three forms side by side, colour-coded by how much each contributes — read it top (best) to bottom (worst), following the up-arrow of increasing stability.

Figure — Resonance — delocalization, resonance energy (benzene, ozone, carbonate)

Worked example Rank the cyanate

$OCN^{-}$ forms, then estimate the weighted bond orders and charge spread. Forecast: three drawings differ in where the double bonds and the $- 1$ sit. Which is best — and will the C–O or the C–N bond come out stronger?

1. Draw the three forms (moving only π/lone pairs, nuclei fixed):

Form I: $O - - C \equiv N$ — negative on oxygen; C–O single, C–N triple.
Form II: $O = C = N -$ — negative on nitrogen; C–O double, C–N double.
Form III: $O + \equiv C - N 2 -$ — charges large and separated; C–O triple, C–N single.

Why this step? Unlike the symmetric cells, here the forms are not interchangeable, so we cannot just average equally — we must lay all candidates out first, then judge each one's stability to decide how much it contributes to the hybrid.

2. Apply the weighting rules (from the parent note): least charge, more bonds, negative on the more electronegative atom. Why this step? Unequal forms contribute unequally; the most stable drawing describes the real molecule best.

3. Score and assign approximate weights (these numbers are also written on figure s01, so you never need the colours):

Form I: $- 1$ on O (electronegativity $3.44$ ) — very good ✔. Weight $\approx 0.60$ (dominant).
Form II: $- 1$ on N ( $3.04$ ) — decent, less good. Weight $\approx 0.35$ .
Form III: a $+ 1$ and a $2 -$ , charges separated — worst. Weight $\approx 0.05$ (negligible).

Why this step? Oxygen is more electronegative than nitrogen, so it "wants" the negative charge more; the split-charge form barely counts.

4. Weighted bond orders. Multiply each form's bond order by its weight and add. With C–O bond orders $(1, 2, 3)$ and C–N bond orders $(3, 2, 1)$ for forms $(I, II, III)$ : $B.O._{C-O} = 0.60 (1) + 0.35 (2) + 0.05 (3) = 1.45$ $B.O._{C-N} = 0.60 (3) + 0.35 (2) + 0.05 (1) = 2.55$ Why this step? The hybrid is the weighted blend, not the equal average — this is exactly what makes Cell D different from Cells A–C.

5. Weighted charge spread. The $- 1$ sits on O in Form I, on N in Form II, and (as part of a $- 2/ + 1$ ) mostly on N in Form III: $q_{O} \approx - 0.60, q_{N} \approx - 0.40.$ Why this step? The dominant Form I puts more of the negative charge on oxygen — consistent with the ranking.

Verify: the C–N bond order ( $2.55$ ) exceeds the C–O bond order ( $1.45$ ), so the C–N bond is shorter/stronger — real OCN⁻ has C–N $\approx 117$ pm (near triple) vs C–O $\approx 126$ pm (near single–double) ✔. Charge check: $q_{O} + q_{N} = - 0.60 + (- 0.40) = - 1$ ✔ (matches the anion charge). Weights sum: $0.60 + 0.35 + 0.05 = 1$ ✔.

E5 — Cell E: The degenerate/zero case (no real resonance)

Worked example Does

$CO_{2}$ gain resonance energy? Compare its "single structure" to reality. Forecast: CO₂ is $O = C = O$ . Is there delocalization energy like benzene?

1. Draw the main structure: $O = C = O$ , symmetric, no formal charge, full octets. Why this step? RE is measured relative to the best single structure (see the definition up top). So we must first pin down that dominant drawing — it is the $E_{single structure}$ baseline against which any delocalization stabilization would be measured.

2. Are there other equal forms? You can draw $O - - C \equiv O +$ , but it has separated charges — a worse, minor form. There is no set of equivalent, equally-good structures like ozone's. Why this step? Resonance energy needs several forms of comparable stability to blend. When one dominant form vastly outweighs the rest, delocalization is minimal.

3. Conclusion. The best single structure already describes CO₂ almost perfectly, so $RE \approx 0.$ Why this step? $RE = E_{single} - E_{hybrid}$ ; if the hybrid ≈ the single structure, the difference vanishes — the zero / degenerate cell.

Verify: both C=O bonds are $116$ pm — very close to a pure C=O double ( $122$ pm elsewhere), not stretched toward single. No bond-length "smearing" ⇒ negligible resonance ✔. Contrast: ozone's $128$ pm sits between single and double, betraying real delocalization.

E6 — Cell F: Nitrate vs nitrite (different link counts)

E7 — Cell G: Word problem, resonance energy from data

The bar chart below turns the three data points into energies: the tall pink bar is the predicted release for three isolated double bonds, the yellow bar is what benzene actually releases, and the pink double-arrow between them is the resonance energy gap.

Worked example Given: hydrogenating

cyclohexene releases $120$ kJ/mol; hydrogenating benzene releases $208$ kJ/mol. Find benzene's resonance energy. Forecast: three double bonds — will "fake benzene" release about $360$ ? Will real benzene release more or less?

1. Cost of one localized C=C. Cyclohexene has exactly one double bond: $Δ H_{one} = - 120 kJ/mol, ∣Δ H_{one} ∣ = 120 kJ/mol .$ Why this step? It calibrates the energy released per isolated double bond.

2. Predict "cyclohexatriene" (fake benzene). If benzene were three separate C=C: $Δ H_{predicted} = 3 \times (- 120) = - 360 kJ/mol, ∣Δ H_{predicted} ∣ = 360.$ Why this step? Builds the $E_{single structure}$ baseline, assuming bonds don't interact.

3. Real benzene releases less. $Δ H_{actual} = - 208 kJ/mol, ∣Δ H_{actual} ∣ = 208.$ Why this step? Less energy released means benzene started lower (more stable) — see figure s02, the actual (yellow) bar is shorter than the predicted (pink) one.

4. Take the difference of magnitudes (keeps RE positive). $RE = ∣Δ H_{predicted} ∣ - ∣Δ H_{actual} ∣ = 360 - 208 = + 152 kJ/mol .$ Why this step? RE = "how much less heat the real molecule gave off" = the stability it already had from delocalization. This equals the boxed $E_{single} - E_{hybrid}$ because the more stable (lower-energy) hybrid releases less on hydrogenation. RE is a positive stabilization.

Verify: magnitude $152$ kJ/mol $\approx 150$ (the parent's recall value) ✔. Sign: benzene is more stable, and $RE = + 152 > 0$ as the definition demands ✔. Units: kJ/mol throughout ✔.

E8 — Cell H: Exam twist — the allyl anion's ends vs middle

The trap: "resonance means all atoms share equally." False. Some positions carry charge; others carry none. This edge reasoning separates top scorers.

Worked example In the allyl anion

$[CH_{2} -CH-CH_{2}]^{-}$ , on which carbons does the $- 1$ charge live? Forecast: all three carbons, or only some?

1. Draw the two forms.

Form 1: $C - H_{2} - CH = CH_{2}$ — negative on C1.
Form 2: $CH_{2} = CH - C - H_{2}$ — negative on C3. Why this step? Only the π-bond and the lone pair shift; the middle carbon is double-bonded in both forms.

2. Where does the charge appear? C1 in form 1, C3 in form 2 — the middle carbon C2 is never charged in either form. Why this step? Averaging: C2's charge $= \frac{0 + 0}{2} = 0$ . The ends: each is $- 1$ in one of two forms $\Rightarrow - \frac{1}{2}$ each.

3. Bond order of each C–C link. Each C–C is single in one form, double in the other: $B.O. = \frac{1 + 2}{2} = 1.5.$ Why this step? The bonds delocalize even though the charge does not sit on the centre.

4. The twist stated plainly. Delocalization spreads the charge to the terminal carbons only, while the bonding is equal across both links.

Verify: charge check $2 \times (- \frac{1}{2}) + 0 = - 1$ ✔ (matches the anion charge). Symmetry: swapping C1↔C3 maps form 1↔form 2, confirming ends are equivalent and the middle is special ✔. Molecular Orbital Theory confirms it: the non-bonding MO has a node exactly on C2, so C2 holds no extra electron density.

Recall grid

Recall Cover the answers first

O₃ bond order ::: 1.5 CO₃²⁻ bond order ::: 4/3 ≈ 1.33 NO₂⁻ vs NO₃⁻ — which shorter N–O? ::: nitrite (B.O. 1.5 > 1.33) Benzene resonance energy ::: +152 kJ/mol (positive stabilization) CO₂ resonance energy ::: ≈ 0 (one dominant structure) Allyl anion — charge on middle carbon? ::: zero; only the two ends carry −½ Which OCN⁻ form dominates, O⁻ or N⁻? ::: the O⁻ form (oxygen more electronegative)

Connections

Yeh Hinglish mein →
Formal Charge — the ranking tool for asymmetric cases (E4).
Bond Order & Bond Length — turns every B.O. above into a predicted length.
Molecular Orbital Theory — explains the allyl node in E8.
Aromaticity & Hückel's Rule — why benzene's RE (E7) is exceptionally large.