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

The double-headed single arrow $\leftrightarrow$ means resonance: the two drawings are imperfect sketches of one single blended molecule (the hybrid). Nothing is moving between them.

The two half-arrows $\rightleftharpoons$ mean chemical equilibrium: two genuinely different molecules that really interconvert back and forth.

So: $\leftrightarrow$ = resonance (one real thing). $\rightleftharpoons$ = equilibrium (two real things flipping).

No. The first rule of resonance is that ==nuclei must stay put; only $\pi$ electrons and lone pairs may move==. In the second drawing an atom has physically moved (an O–H bond broke and the H's rebonded). That is a different molecule, not a resonance form. Resonance never breaks or makes bonds to different atoms — it only redistributes electrons over the same skeleton.

Bond order counts how many bonding electron-pairs are shared, on average, between two atoms. See the picture: the double bond sits on the left in one structure and on the right in the other.

Average the two bond types over the 2 equivalent O–O positions: $\text{B.O.}=\frac{1+2}{2}=1.5$ Both bonds are identical, halfway between single and double — matching the measured 128 pm (single 148, double 121). ✔

There is one double bond that "rotates" among the three C–O positions (see figure), plus two single bonds at any instant. Total bond-pairs shared over the three positions: $\text{B.O.}=\frac{1+1+2}{3}=\frac{4}{3}\approx 1.33$

The overall $2-$ charge is spread equally over the three oxygens: $\text{charge per O}=\frac{-2}{3}\approx -0.67$ All three C–O bonds are equal (~129 pm). ✔

Identical logic to carbonate — three equivalent positions sharing one double bond: $\text{B.O.}=\frac{1+1+2}{3}=\frac{4}{3}\approx 1.33$ The extra insight: isoelectronic species with the same skeleton give the same bond order. Nitrate and carbonate are geometric twins.

The weighting rule: the structure with negative charge on the more electronegative atom contributes more (and, secondarily, more bonds / smaller charges are better).

Oxygen is more electronegative than nitrogen. In structure (B) the $-1$ sits on oxygen; in (A) it sits on nitrogen.

Therefore (B) — negative charge on oxygen — is the major contributor. Both are legitimate and both appear in the hybrid, but the real molecule looks more like (B).

Apply the weighting hierarchy in order:

1. No formal charge is best ⇒ I is the strongest contributor.
2. When charges are unavoidable, opposite charges close together is better (they partly cancel electrostatically). So II (adjacent) beats III (far apart).

Ranking: $\text{I}>\text{II}>\text{III}$.

Step 1 — build the hypothetical "cyclohexatriene." If benzene were three independent localized C=C bonds, hydrogenation would release three times one double bond: $\Delta H_{\text{predicted}}=3\times (-120)=-360\text{kJ/mol}$

Step 2 — compare to reality. Real benzene releases only $-208\text{kJ/mol}$, i.e. less energy. Releasing less means benzene started lower (more stable) than the fake.

Step 3 — take the difference: $\text{RE}=\Delta H_{\text{predicted}}-\Delta H_{\text{actual}}=-360-(-208)=-152\text{kJ/mol}$

Meaning: benzene sits ~152 kJ/mol lower than the un-delocalized cyclohexatriene would. That gap is a static comparison (nothing oscillates) — it is the extra stability delocalization buys. See Aromaticity & Hückel's Rule for why 6 $\pi$-electrons make this so large.

The rule from Bond Order & Bond Length: higher bond order ⇒ shorter, stronger bond. Order the bond orders, then reverse for length: $\text{B.O.: }2>\frac{4}{3}>1$ So shortest → longest: $\text{C=O (double)}<\text{carbonate C–O}<\text{C–O (single)}$ The carbonate bond (~129 pm) lands between double (~122 pm) and single (~143 pm), exactly as its $4/3$ bond order predicts. ✔

Invalid. Oxygen is a period-2 element; its valence shell has only 2s and 2p orbitals — capacity 8 electrons. It has no accessible d-orbitals to expand into. A resonance structure must obey normal valence capacity (rule 3).

So even though it looks like it delocalizes more, this drawing violates the octet limit for a second-row atom and cannot contribute. The genuine forms keep every O at $\le 8$ electrons.

(a) Because the three structures are equivalent (the double bond rotates symmetrically among all three oxygens), the hybrid must treat all three positions identically ⇒ all three S–O bonds equal.

(b) Same averaging as carbonate/nitrate: $\text{B.O.}=\frac{1+1+2}{3}=\frac{4}{3}\approx 1.33$

(c) The central S has 3 bonding regions and no lone pair ⇒ trigonal planar, 120° angles, which also keeps all three p-orbital overlaps equivalent (see Hybridization ($sp^2$)). Delocalization requires this planar, symmetric geometry.

$\frac{152\text{kJ/mol}}{4.184\text{kJ/kcal}}=36.3\text{kcal/mol}$ Consistent with the ~36 kcal/mol figure. ✔ The two numbers are the same physical stabilization in different units — always check units before claiming two sources disagree.

• Ozone O–O bond order ::: 1.5
• Carbonate / nitrate / SO₃ X–O bond order ::: 4/3 ≈ 1.33
• Benzene C–C bond order ::: 1.5
• Benzene resonance energy ::: ~152 kJ/mol (≈ 36 kcal/mol)
• Charge per O in carbonate ::: −2/3 ≈ −0.67
• Meaning of $\leftrightarrow$ ::: one blended hybrid, not two flipping species