Resonance Structures Practice: 14 Questions with Answers

Resonance questions reward discipline more than memory. Examiners set traps: two drawings that look like resonance forms but aren’t, “average” bond orders that need a moment’s arithmetic, and pairs of structures where you must say which one matters more. The fourteen questions below cover all three, beginning with small inorganic ions and moving on to the organic examples that appear in more advanced courses.

The ground rules

Keep these rules in front of you while you work:

1. Atoms never move. Resonance structures have identical atom positions and connectivity. Only electrons (lone pairs and π bonds) are redistributed.
2. The total number of electrons, and the overall charge, stay the same.
3. Every structure must be a valid Lewis structure. Second-period atoms (C, N, O) may not exceed an octet, and hydrogen never has more than two electrons.
4. The real species is a hybrid. It does not flip between the forms; it is a single structure whose electron distribution is a weighted blend of them.
5. Not all contributors are equal. A structure contributes more (it is a major contributor) when it has more atoms with full octets, fewer formal charges, no like charges on neighbouring atoms, and any negative charge on the more electronegative atom. Equivalent structures contribute equally.

Bond order in the hybrid = (total bonds between the two atoms, summed over the equivalent structures) ÷ (number of equivalent structures).

A curly arrow shows the movement of one pair of electrons, from a lone pair or bond to a new bond position or atom. For the theory behind all of this, read resonance structures explained.

Questions

Total: 40 marks.

1. For each pair, state whether the two structures are resonance forms of each other, and give a reason. (4 marks) (a) The two Kekulé structures of benzene. (b) Ethanol, CH₃CH₂OH, and methoxymethane, CH₃OCH₃. (c) Propanone, CH₃COCH₃, and its enol form, CH₂=C(OH)CH₃. (d) O=O–O and O–O=O for ozone, with the atoms in the same positions.
2. The nitrite ion, NO₂⁻, has nitrogen as the central atom with one lone pair. Describe its two resonance structures, give the formal charges in one of them, and state the N–O bond order in the hybrid. (3 marks)
3. Draw (or describe) the resonance structures of the carbonate ion, CO₃²⁻. State the C–O bond order and the average charge on each oxygen. (3 marks)
4. The O–O bond length is about 121 pm in O₂ (a double bond) and about 148 pm in hydrogen peroxide (a single bond). Both O–O bonds in ozone are about 128 pm. Explain these data using resonance. (3 marks)
5. In one nitrate resonance structure, the N=O double bond is on the top oxygen. Describe the two curly arrows needed to move the double bond to the bottom-left oxygen. (2 marks)
6. The C–C bonds in benzene are all about 139 pm, compared with about 154 pm for a C–C single bond and 134 pm for a C=C double bond. Explain this and state the C–C bond order. (3 marks)
7. The ethanoate ion, CH₃COO⁻, has two C–O bonds of equal length. (a) Explain why. (b) Use resonance to explain why ethanoic acid (pKa about 4.8) is a much stronger acid than ethanol (pKa about 16). (4 marks)
8. Draw the resonance structures of the allyl cation, CH₂=CH–CH₂⁺. On which carbons is the positive charge found? (2 marks)
9. The ion formed by removing a hydrogen from the CH₃ group of ethanal has the resonance forms ⁻CH₂–CH=O and CH₂=CH–O⁻. Which is the major contributor? Explain. (2 marks)
10. Ethanamide, CH₃CONH₂, has two resonance structures. (a) Describe the second structure. (b) Which is the major contributor? (c) Give two experimental consequences of the minor contributor. (4 marks)
11. For the cyanate ion, OCN⁻, three octet structures can be drawn: O=C=N⁻ (−1 on N), ⁻O–C≡N (−1 on O) and O≡C–N²⁻ (+1 on O, −2 on N). Rank them from major to minor contributor, with reasons. (3 marks)
12. A student proposes each of the following as resonance structures. Explain what is wrong with each. (3 marks) (a) For methanal, H₂C=O, a structure in which carbon forms a double bond to oxygen and also carries a lone pair. (b) For ethanoate, a structure in which the hydrogen of the CH₃ group has moved onto an oxygen. (c) For the nitrite ion, a structure with a double bond to each oxygen and a lone pair on nitrogen.
13. Explain, using resonance, why phenol (pKa about 10) is more acidic than ethanol, and state which ring positions carry negative charge in the phenoxide ion. (3 marks)
14. A textbook says “ozone rapidly switches between its two resonance structures”. Explain why this statement is wrong. (1 mark)

Answer key

1. (a) Yes: all atoms stay put; only the positions of the three π bonds change (1). (b) No: the atoms are connected differently, so these are structural isomers (1). (c) No: a hydrogen atom has moved from carbon to oxygen, so these are two different compounds in equilibrium (tautomers), not resonance forms (1). (d) Yes: the same three atoms in the same positions, with only a lone pair and a π bond swapped between ends (1).

2. Nitrogen: 5 electrons; each oxygen: 6; plus 1 for the charge = 18 valence electrons. Structure 1: O=N–O⁻, with a lone pair on N; structure 2 is its mirror, ⁻O–N=O (1). Formal charges in structure 1: N: 5 − 2 − 3 = 0; double-bonded O: 6 − 4 − 2 = 0; single-bonded O: 6 − 6 − 1 = −1. Total −1 ✓ (1). Bond order: (2 + 1) ÷ 2 = 1.5 for each N–O bond (1).

3. Three equivalent structures, each with one C=O and two C–O⁻; the double bond sits on a different oxygen in each (1). Bond order: each C–O bond is double in one structure and single in two, so (2 + 1 + 1) ÷ 3 = 4/3 ≈ 1.33 (1). The −2 charge is shared equally over three oxygens: −2/3 each (1). All three C–O bonds are the same length, and the ion is trigonal planar.

4. Ozone has two equivalent resonance structures, O=O–O and O–O=O (1). The real molecule is the hybrid, so each O–O bond has a bond order of (2 + 1) ÷ 2 = 1.5 (1). A bond order of 1.5 predicts a length between single (148 pm) and double (121 pm), and 128 pm fits, as does the fact that both bonds are identical (1). See the Lewis structure of ozone for the full drawing.

5. Arrow 1: from a lone pair on the bottom-left oxygen (which carries −1) into the N–O bond, forming the new N=O double bond (1). Arrow 2: from the old N=O π bond up onto the top oxygen, which becomes a lone pair and gives that oxygen the −1 charge (1). Two arrows are needed because nitrogen already has four bonds and cannot gain a fifth.

6. Benzene has two equivalent Kekulé structures, with alternating single and double bonds in opposite positions (1). Every C–C bond is single in one and double in the other, so all six bonds are identical in the hybrid, with a bond order of 1.5 (1). A length of 139 pm, between 154 and 134 pm, fits a bond order of 1.5. The six π electrons are delocalised around the whole ring (1). More detail: bonding in benzene.

7. (a) Two equivalent structures exist: CH₃C(=O)O⁻ with the double bond to either oxygen (1). Each C–O bond is therefore a hybrid of single and double (bond order 1.5), so both have the same length (1). (b) When ethanoic acid loses H⁺, the negative charge in the ethanoate ion is delocalised over two oxygens, which stabilises the ion (1). In the ethoxide ion, CH₃CH₂O⁻, there is no π system to delocalise into, so the charge stays on one oxygen. The more stable the anion, the more readily the acid ionises, so ethanoic acid is the stronger acid (1).

8. CH₂=CH–CH₂⁺ ↔ ⁺CH₂–CH=CH₂: the curly arrow moves the π bond from C1=C2 to C2=C3 (1). The positive charge is shared equally between C1 and C3, the two end carbons; the middle carbon never carries it (1). This delocalisation makes the allyl cation more stable than a simple propyl cation.

9. CH₂=CH–O⁻ is the major contributor (1). Both structures have full octets and one formal charge, but this one places the negative charge on oxygen (electronegativity 3.44) rather than carbon (2.55). The more electronegative atom stabilises negative charge better (1).

10. (a) The nitrogen lone pair moves into the C–N bond to form C=N, while the C=O π electrons move onto oxygen: CH₃C(–O⁻)=N⁺H₂ (1). (b) The neutral structure, CH₃C(=O)NH₂, is the major contributor, because it has no formal charges (1). (c) Any two of: the C–N bond is shorter than a normal C–N single bond; there is restricted rotation about the C–N bond; the nitrogen and its neighbours are planar (nitrogen behaves as sp², not sp³); amide nitrogen is only a very weak base, because its lone pair is tied up in delocalisation (2). The same effect keeps the peptide bonds in proteins flat.

11. Major: ⁻O–C≡N (only one formal charge, on the most electronegative atom, O = 3.44) (1). Next: O=C=N⁻ (only one formal charge, but on N = 3.04, less electronegative than O) (1). Minor: O≡C–N²⁻, with larger charges and a positive charge on oxygen (1). The formal charges are worked out in formal charge practice, question 13.

12. (a) Carbon would have 10 electrons around it (four bonds plus a lone pair), breaking the octet rule for a second-period atom, and the electron count no longer matches the molecule (1). (b) A hydrogen atom has moved; resonance moves only electrons, so this describes a different compound (1). (c) Two N=O bonds plus a lone pair would give nitrogen 10 electrons, which is not allowed for nitrogen, and the structure has the wrong electron count (it describes 20 electrons, not 18) (1).

13. In the phenoxide ion, one lone pair on the oxygen can be delocalised into the benzene ring (1). The resulting resonance structures place the negative charge on the oxygen and on the ring carbons at the two ortho positions and the para position (1). Spreading the charge stabilises phenoxide relative to ethoxide, whose charge is fixed on one oxygen, so phenol ionises more readily. It is still much weaker than ethanoic acid, because in phenoxide most of the delocalisation puts the charge on carbon, whereas in ethanoate it is shared between two oxygens (1).

14. Resonance structures are not real, separate species. Ozone exists all the time as a single hybrid structure with two identical bonds; the drawings are only a way of describing it within the limits of Lewis notation (1).

Exam tips

• Check the atoms first. If any atom (especially H) has moved, it is not resonance.
• Count electrons in every form. An extra or missing pair is the quickest way to lose the mark.
• Use double-headed arrows (↔) between resonance forms, never equilibrium arrows (⇌).
• Rank contributors with a checklist: full octets, then fewest formal charges, then negative charge on the more electronegative atom.
• Link resonance to data. Equal bond lengths, intermediate bond orders and extra acidity are the evidence examiners want you to mention.

Key takeaways

• Resonance forms differ only in the positions of electrons; atoms, electron count and charge stay the same.
• The real species is a hybrid, with bond orders and charges averaged over the contributing structures.
• Equivalent structures contribute equally (NO₂⁻, CO₃²⁻, O₃, benzene, ethanoate); non-equivalent ones are weighted by octets, formal charges and electronegativity.
• Delocalisation stabilises species, which explains the acidity of carboxylic acids and phenols and the flat, rigid amide bond.
• Practise the related skills in hybridisation practice problems.