10.5 Bond Properties & Resonance

Key Takeaways

  • Bond order (single=1, double=2, triple=3, fractional for resonance) inversely relates to bond length and directly to bond energy: higher order means shorter, stronger bonds
  • Sigma (σ) bonds form from head-on orbital overlap (single bonds); pi (π) bonds form from side-on p-orbital overlap (the second and third components of double and triple bonds)
  • Resonance structures are alternative Lewis structures that delocalize electrons; the real hybrid averages bond orders and is more stable than any single contributor
  • Aromaticity requires a planar, cyclic, fully conjugated system with 4n+2 π electrons (Hückel rule); benzene (n=1, 6 π e⁻) is the archetypal aromatic ring
  • Formal charge = valence electrons − (nonbonding electrons + ½ bonding electrons); minimizing formal charge guides the best resonance contributor
Last updated: August 2026

Bond Properties & Resonance

Quick Answer: A covalent bond's length, energy, and order are tightly linked, and many real molecules cannot be drawn with a single Lewis structure — resonance delocalizes electrons and stabilizes the molecule. The PA-CAT Bulletin of Information (rev. 20240815) places bond properties and resonance squarely in the first organic-chemistry cluster (Table 5). Aromaticity, introduced here, is resonance taken to its cyclic limit.

Bond Order, Length, and Energy

Bond order is the number of shared electron pairs between two atoms: single = 1, double = 2, triple = 3. Across a row, as bond order rises, bond length decreases and bond energy increases.

BondOrderLength (pm)Energy (kJ/mol)
C–C1154347
C=C2134614
C≡C3120839
C–O1143358
C=O2123745
N–N1145163
N≡N3110945

A C=C is shorter and stronger than C–C but not twice as strong — the second bond (a π bond) is weaker than the first (σ) bond, so π bonds are more reactive.

Sigma (σ) and Pi (π) Bonds

A sigma (σ) bond arises from head-on (axial) overlap of orbitals (s-s, s-p, or hybrid orbitals). Every single bond is one σ. A double bond is one σ + one π; a triple bond is one σ + two π. A pi (π) bond arises from side-on overlap of unhybridized p orbitals, with electron density above and below the bond axis. π bonds prevent free rotation about the bond, which is why cis/trans (E/Z) isomers exist around double bonds (see 10.6).

Resonance Structures

A single Lewis structure cannot represent molecules where electrons are delocalized. Resonance structures are alternative valid Lewis structures; the real molecule is a resonance hybrid (weighted average), not an oscillator. Rules:

  • Only electrons move (especially π and lone pairs), never atoms.
  • Each resonance structure must obey the octet rule (for period-2 elements) and keep the same net charge.
  • The best contributor has minimal formal charges, negative charges on more electronegative atoms, and full octets.

Formal charge = valence e⁻ − (nonbonding e⁻ + ½ bonding e⁻).

Example — the nitrate ion NO₃⁻: three equivalent resonance structures, each with one N=O and two N–O⁻. The hybrid has each N–O bond at order 1⅓ and the charge spread equally over the three oxygens. This delocalization explains why nitrate is far more stable than a hypothetical localized structure.

Resonance Energy (Stabilization)

The hybrid is more stable than any single contributor by an amount called resonance energy. Benzene's resonance energy is about 150 kJ/mol — its actual heat of hydrogenation is less exothermic than three times cyclohexene's because the aromatic system is stabilized.

flowchart LR
    R1[Resonance structure A] --> H[Real hybrid]
    R2[Resonance structure B] --> H
    R3[Resonance structure C] --> H
    H --> S[Lower energy, delocalized bonds]

Aromaticity and the Hückel Rule

A molecule is aromatic if it meets all four criteria:

  1. Cyclic — ring closed.
  2. Planar — p orbitals can overlap continuously.
  3. Fully conjugated — every ring atom has an unhybridized p orbital.
  4. 4n + 2 π electrons (Hückel rule), n = 0, 1, 2, …

Benzene (C₆H₆): cyclic, planar, 6 sp² carbons fully conjugated, 6 π electrons → n = 1 → aromatic, unusually stable, non-reactive in addition reactions that destroy aromaticity (it undergoes electrophilic substitution instead).

Systemπ electronsnAromatic?
Benzene61Yes
Cyclobutadiene4No (antiaromatic, 4n)
Cyclopentadienyl anion61Yes
Cyclooctatetraene8No (adopts nonplanar tub)
Tropylium cation (C₇H₇⁺)61Yes

A 4n π-electron planar, cyclic, conjugated system is antiaromatic (destabilized) — cyclobutadiene is the classic example. Nonplanar or nonconjugated systems are simply nonaromatic.

Bond Polarity and Electronegativity

Bond polarity follows the electronegativity difference ΔEN between the bonded atoms. C–H (ΔEN 0.4) is essentially nonpolar; C–O (ΔEN 1.0) is polar covalent; Na–Cl (ΔEN 2.1) is ionic. Polar bonds create dipole moments; molecular polarity depends on geometry (vector sum of bond dipoles). CO₂ has polar C=O bonds but is nonpolar overall because the dipoles cancel linearly.

PA-CAT Tips

  • Compare bond lengths with bond order first, then atomic radius (C–C < Si–Si).
  • Count π electrons for Hückel by counting p orbitals contributing one electron each, plus lone pairs in the ring (cyclopentadienyl anion uses a lone pair).
  • Resonance does not mean rapid switching; the hybrid is the real structure.

Bond Energetics, Polarity Vectors, and Resonance vs Tautomerization

Average bond energies let you estimate a reaction's enthalpy change from a Lewis skeleton alone. The bookkeeping rule is ΔH° ≈ Σ(bonds broken) − Σ(bonds formed): sum positive bond-energy values for each bond the reactants lose, and subtract the energy released as each new bond forms. For the gas-phase reaction H₂ + Cl₂ → 2 HCl: break H–H (436 kJ/mol) and Cl–Cl (243), form two H–Cl (431 each) → ΔH° ≈ (436 + 243) − 2(431) = −183 kJ/mol, exothermic as observed. The estimate is rough because tabulated values are averages, but the sign is reliable for PA-CAT predictions. Remember bond energies are gas-phase, so add vaporization enthalpies for any liquid reactant or product.

Bond polarity is set by the electronegativity difference (ΔEN) along each bond, but molecular polarity is a vector sum of those bond dipoles, so geometry decides whether polar bonds yield a polar molecule. In linear CO₂ each C=O dipole points from C⁺ to O⁻, but the two equal dipoles point in opposite directions and cancel → net dipole zero, nonpolar. In bent H₂O the two O–H dipoles (104.5° apart) do not cancel → net dipole ≈ 1.85 D, polar. Symmetry elements (a center of inversion or perpendicular C₂ axes) generally force cancellation; their absence lets bond dipoles sum to a molecular dipole. For PA-CAT, when asked whether a molecule is polar, first assign VSEPR shape, then ask whether the vector sum of bond dipoles (and lone-pair contributions) is non-zero.

Resonance equalizes bond order across equivalent positions, which is why nitrate's three N–O bonds are the same length (order 1⅓) and benzene's six C–C bonds are identical (order 1½, intermediate between 154 and 134 pm). The best Lewis contributor is chosen by formal-charge minimization: prefer structures with zero formal charges on all atoms; if charges are unavoidable, place negative charge on the more electronegative atom and positive on the less electronegative one. Compare two NO₃⁻ drawings — one with N=O and two N–O⁻ (charges on oxygen, the electronegative atom) beats one placing positive charge on oxygen.

A classic PA-CAT trap confuses resonance with tautomerization. Resonance structures share the same atom connectivity — only π electrons and lone pairs move; the hybrid is one substance. Tautomers (e.g., keto–enol, imine–enamine) have different connectivity: a proton physically shifts from one atom to another, and the two forms are distinct, isolable compounds in equilibrium. If the skeleton changes, it is not resonance.

Test Your Knowledge

Which statement correctly compares a C–C single bond and a C=C double bond?

A
B
C
D
Test Your Knowledge

Which of the following meets all four Hückel criteria for aromaticity?

A
B
C
D
Carbon–Carbon Bond Energy Increases with Bond Order (kJ/mol)