AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/2/2-6)
Unit 2 · Topic 2.6
2.6 Resonance and Formal Charge
Sometimes more than one Lewis diagram fits a molecule. If the diagrams are equivalent, as in ozone or nitrate, the real particle is a resonance hybrid, a blend of them. If they're different, formal charge helps you pick the best one.
Key terms
- resonance
- resonance hybrid
- formal charge
- bond order
- octet rule
Resonance
Draw a Lewis diagram for ozone, O₃, and you'll find you can put the double bond on either side. Neither picture alone is right. Experiments show both O–O bonds in ozone are the same length, partway between a single and a double bond.
When two or more equivalent Lewis diagrams can be drawn, the real structure is a resonance hybrid: an average of them. You show this by drawing every structure with double-headed arrows (↔) between them. The electrons in the hybrid are delocalized, spread over more than two atoms.
Resonance structures differ only in where electrons are placed. The atoms stay in the same positions.
Bond order in a resonance hybrid
Average bond order = total number of shared electron pairs in the bonds being compared ÷ number of those bonds.
| Species | Resonance structures | Average bond order |
|---|---|---|
| O₃ | 2 | 3 pairs over 2 O–O bonds = 1.5 |
| NO₃⁻ | 3 | 4 pairs over 3 N–O bonds = 1.33 |
| CO₃²⁻ | 3 | 4 pairs over 3 C–O bonds = 1.33 |
Formal charge
Formal charge is a bookkeeping tool. It pretends every bond is shared perfectly equally and asks whether an atom has more or fewer electrons than it brought.
Formal charge = (valence electrons of the free atom) − (lone-pair electrons) − ½(bonding electrons)
A quick version: valence electrons − dots − lines. The formal charges in a structure always add up to the overall charge of the particle.
Choosing the best structure
When several valid diagrams have different formal charges, use these guidelines to pick the best one.
- Prefer structures where atoms follow the octet rule.
- Prefer formal charges as close to zero as possible.
- If a negative formal charge is needed, it should be on the more electronegative atom.
- Avoid structures with like charges on neighboring atoms.
Formal charge in practice
Formal charge isn't the real charge on an atom, which depends on electronegativity. It's a tool for comparing diagrams.
Example: for CO₂, the diagram O=C=O gives every atom a formal charge of 0. The alternative O≡C–O puts +1 on the triple-bonded O and −1 on the single-bonded O. Both have octets, but O=C=O has formal charges closer to zero, so it's the better model, and it correctly predicts two identical C=O bonds.
Limits of the Lewis model
Lewis diagrams are a model, and models have limits. Molecules with an odd number of electrons, like NO₂ (17 valence electrons), can't give every atom an octet. A single Lewis diagram can't show delocalized electrons, which is why you need resonance. Experiments, such as measuring bond lengths, are the final test of any structure.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Resonance and bond length
Draw the resonance structures for the nitrate ion, NO₃⁻, and predict how its N–O bond lengths compare with a typical N–O single bond and N=O double bond.
Show the solutionHide the solution
- Step 1: Count: N = 5, three O = 18, plus 1 for the charge = 24 electrons.
- Step 2: One valid structure has N with one double bond and two single bonds to O. The double bond can be placed on any of the three O atoms, giving three equivalent structures, joined by ↔ arrows.
- Step 3: In the hybrid, there are 4 bonding pairs over 3 N–O bonds, so each bond has order 4/3 ≈ 1.33.
- Step 4: All three bonds are identical, and with a bond order between 1 and 2, they're shorter than an N–O single bond but longer than an N=O double bond.
Answer: Three equivalent resonance structures; all three N–O bonds are the same, with bond order 1.33, intermediate in length between single and double N–O bonds.
- Example 2
Using formal charge to choose
Three Lewis diagrams can be drawn for the cyanate ion, OCN⁻ (carbon in the center): [O=C=N]⁻, [O≡C–N]⁻ and [O–C≡N]⁻, each with octets on all atoms. Use formal charges to choose the best one.
Show the solutionHide the solution
- Step 1: 16 valence electrons in each. Compute formal charge = valence − dots − lines.
- Step 2: [O=C=N]⁻: O has 2 lone pairs and 2 lines: 6 − 4 − 2 = 0. C: 4 − 0 − 4 = 0. N: 5 − 4 − 2 = −1.
- Step 3: [O≡C–N]⁻: O: 6 − 2 − 3 = +1. C: 0. N: 5 − 6 − 1 = −2. This one has large formal charges, so reject it.
- Step 4: [O–C≡N]⁻: O: 6 − 6 − 1 = −1. C: 0. N: 5 − 2 − 3 = 0.
- Step 5: The first and third both have formal charges of 0, 0 and −1. Break the tie with electronegativity: O is more electronegative than N, so the −1 belongs on O.
Answer: [O–C≡N]⁻, with the −1 formal charge on oxygen
- Example 3
Resonance is not flipping (classic trap)
A student says ozone switches back and forth rapidly between its two resonance structures. Explain what's wrong.
Show the solutionHide the solution
- Step 1: Resonance structures are separate drawings of one real structure; the molecule doesn't flip between them.
- Step 2: The real ozone molecule always has two identical O–O bonds with bond order 1.5.
- Step 3: Evidence: experiments find one O–O bond length, not two.
Answer: Ozone is a single hybrid structure all the time, with both bonds identical (bond order 1.5); resonance structures are drawings, not states the molecule switches between.
Common mistakes
- Moving atoms between resonance structures. Only electrons move.
- Treating formal charge as the real charge on an atom.
- Placing a negative formal charge on the less electronegative atom when a better choice exists.
- Saying a resonance hybrid has some single and some double bonds. All the equivalent bonds are identical.
On the exam
- Expect questions comparing bond lengths or bond energies in species with resonance, such as which has longer C–O bonds, CO₂ or CO₃²⁻. Calculate the average bond order for each.
- When justifying a choice of Lewis diagram, show the formal charges on each atom and state the rule you used.
Connected topics
Videos
Check yourself
4 questions on 2.6 Resonance and Formal Charge. Pick an answer to see if you got it, and why.
The cyanate ion, OCN⁻, has carbon in the middle and 16 valence electrons. Three Lewis diagrams that give every atom an octet are described below.
Diagram I: O=C=N, with two lone pairs on O and two lone pairs on N.
Diagram II: O≡C–N, with one lone pair on O and three lone pairs on N.
Diagram III: O–C≡N, with three lone pairs on O and one lone pair on N.
Described Lewis diagrams
What is the formal charge on the nitrogen atom in Diagram II?
Based on formal charges, which diagram best represents the cyanate ion, and why?
The nitrate ion, NO₃⁻, can be drawn with three equivalent resonance diagrams, each with one N=O double bond and two N–O single bonds. What is the bond order of each N–O bond in the actual ion?
A typical O–O single bond is about 148 pm long and an O=O double bond is about 121 pm long. Both oxygen–oxygen bonds in ozone, O₃, measure about 128 pm. Which of the following best explains this?
0 of 4 answered