Resonance
In covalent bonding, we often draw Lewis structures to show how electrons are shared between atoms. However, for many molecules and ions, a single Lewis structure fails to accurately represent reality. Resonance is the concept that describes molecules whose electron distribution cannot be captured by one structure alone. Instead, the true electronic structure is a quantum-mechanical blend — called a resonance hybrid — of two or more contributing structures known as resonance structures (or canonical forms). The classic example is ozone (O₃), where one Lewis structure would show one single bond and one double bond, implying two different bond lengths. In reality, both O–O bonds in ozone are identical in length (128 pm), a fact only explained by resonance. The key insight of resonance is electron delocalization: certain electrons, particularly π (pi) electrons, are not confined to a single bond between two atoms but are instead spread out over three or more atoms. This spreading out lowers the overall energy of the molecule, making it more stable than any single Lewis structure would predict. The difference in energy between the actual molecule and the most stable hypothetical resonance contributor is called the resonance energy (or delocalization energy). For benzene (C₆H₆), this stabilization is approximately 150 kJ/mol — a substantial effect that explains benzene's unusual resistance to addition reactions compared to simple alkenes. Resonance is not a process of the molecule 'switching' between structures — this is a common misconception. The molecule exists as a single, stable hybrid at all times. The resonance structures are simply the best representations we can draw using the limited tools of Lewis dot theory, each capturing part of the true picture. When drawing resonance structures, only electrons move (never atoms), and the positions of all atoms remain fixed. Curved arrows are used to show how electron pairs shift from one contributor to the next, and double-headed arrows (⟷) between structures signal that they are resonance contributors, not products of a reaction.
- Identify Candidates for Resonance
- Use Curved Arrows to Move Electrons
- Check Formal Charges
- Determine the Resonance Hybrid
- Resonance Energy and Stability
Identify Candidates for Resonance
A molecule or ion is likely to exhibit resonance when it has (a) a lone pair adjacent to a double bond, (b) a double bond adjacent to an empty p-orbital or a carbocation, or (c) a fully conjugated ring system. Examples: carbonate ion (CO₃²⁻), nitrate ion (NO₃⁻), benzene (C₆H₆), and ozone (O₃). Start by drawing a valid Lewis structure with correct formal charges.
Use Curved Arrows to Move Electrons
Draw curved arrows from a lone pair or a bonding pair toward an adjacent bond or atom. Each arrow represents the movement of two electrons. For ozone (O₃), in the first structure the central oxygen has a double bond to the right oxygen and a single bond to the left. A curved arrow from the right π bond moves to the right oxygen as a lone pair, and a curved arrow from a lone pair on the left oxygen moves to form a new π bond. This generates the second resonance structure, which is the mirror image of the first.
Check Formal Charges
Formal charge = (Valence electrons) − (Lone-pair electrons) − ½(Bonding electrons). For each resonance structure, calculate formal charges to verify they are valid. In NO₃⁻, the three resonance structures each show one N=O double bond and two N–O single bonds, with a formal charge of −1 on two oxygens and 0 on the third. The total formal charge must always match the overall charge of the species.
Determine the Resonance Hybrid
The true structure is the weighted average of all resonance contributors. In CO₃²⁻, all three C–O bonds are equivalent: each has a bond order of 4/3 ≈ 1.33, intermediate between a single bond (bond order 1) and a double bond (bond order 2). This is bond averaging. The bond length (129 pm) falls between a typical C–O single bond (143 pm) and C=O double bond (122 pm), consistent with the hybrid. Resonance contributors that are lower in energy (fewer formal charges, negative charges on electronegative atoms) contribute more to the hybrid.
Resonance Energy and Stability
The resonance energy is measured experimentally using heats of hydrogenation or combustion. For benzene, if it were a simple cyclohexatriene (3 isolated double bonds), the predicted heat of hydrogenation would be 3 × 120 kJ/mol = 360 kJ/mol. The measured value is only 208 kJ/mol, giving a resonance energy of 360 − 208 = 152 kJ/mol. This extra stability means benzene undergoes electrophilic substitution (preserving aromaticity) rather than addition reactions. Resonance stabilization also explains the acidity of carboxylic acids: the carboxylate anion (RCOO⁻) is stabilized by delocalization of the negative charge over both oxygens, making proton loss favorable.
The Key Equation
- \text{Bond Order} — Bond Order (dimensionless): A measure of the number of electrons shared between two atoms; correlates with bond length (higher order → shorter bond) and bond strength (higher order → stronger bond).
- \text{Number of bonds} — Cumulative bond count (dimensionless): Sum of individual bond orders for the bond of interest across all resonance contributors. For CO₃²⁻ with three contributors, the C–O bond appears as a double bond once and a single bond twice, giving a sum of 2 + 1 + 1 = 4.
- \text{Number of resonance structures} — Number of resonance contributors (dimensionless): The total number of valid Lewis structures contributing to the resonance hybrid. For CO₃²⁻ this is 3, giving bond order = 4/3 ≈ 1.33.
Worked Example
Problem. Draw all resonance structures for the nitrate ion (NO₃⁻) and calculate: (a) the formal charges in each structure, (b) the bond order of each N–O bond, and (c) the predicted N–O bond length given that a typical N–O single bond is 140 pm and a N=O double bond is 115 pm.
Answer. NO₃⁻ has three equivalent resonance structures, each with formal charges +1 on N, −1 on two single-bonded oxygens, and 0 on the double-bonded oxygen. Each N–O bond has a bond order of 4/3 ≈ 1.33, and the predicted bond length is approximately 132 pm (experimental: 124 pm).
Origins · 1 minLinus Pauling and how this was discovered
Linus Pauling developed the modern theory of resonance in the 1930s as part of his broader work on the nature of the chemical bond. Building on quantum mechanics, he introduced the concept of resonance to explain why real molecules are more stable and have more uniform bond lengths than any single Lewis structure suggests. His landmark textbook, 'The Nature of the Chemical Bond' (1939), formalized resonance theory alongside electronegativity scales and hybridization. Pauling was awarded the Nobel Prize in Chemistry in 1954 for this work. Pauling is one of only four individuals to have won two Nobel Prizes, and the only person to have won two unshared Nobel Prizes — the 1954 Nobel Prize in Chemistry and the 1962 Nobel Peace Prize for his anti-nuclear activism.
Related Concepts
- Lewis Structures — Resonance structures are drawn using Lewis dot notation; understanding how to draw valid Lewis structures with correct formal charges is a prerequisite for resonance.
- Bond Polarity and Electronegativity — In resonance hybrids, the distribution of electron density across bonds determines polarity; electronegativity differences drive how much each resonance contributor weighs toward the hybrid.
- Orbital Hybridization (sp, sp², sp³) — Delocalized π electrons in resonance systems arise from overlapping unhybridized p orbitals on sp² or sp hybridized atoms, making hybridization theory essential for understanding why resonance occurs.
- Aromaticity and Benzene — Benzene is the canonical example of resonance stabilization; aromaticity is defined by full delocalization of π electrons in a cyclic conjugated system, directly extending the concept of resonance.
- Bond Enthalpy — Resonance energy is quantified experimentally through differences in bond enthalpies and heats of combustion or hydrogenation, linking resonance theory to thermochemistry.