Explain why the carbon–oxygen bond length in the carbonate ion (CO₃²⁻) is intermediate between a typical C–O single bond and a typical C=O double bond.

IB DP Chemistry Higher Level (2023 syllabus) — S2.2 The covalent model · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (4 marks)

All three C–O bonds in CO₃²⁻ are equivalent because the π electrons are delocalised over the whole ion. The delocalisation gives each C–O bond a bond order of 4/3 (between 1 and 2). A bond order greater than 1 means more electron density between the nuclei, so the bonds are shorter and stronger than a C–O single bond but longer and weaker than a C=O double bond, giving an intermediate bond length.

Examiner tips

  • State bond equivalence first, then delocalisation, then bond order, finally link bond order to length

Common mistakes

  • Saying the bonds are all single or all double, ignoring delocalisation
  • Failing to mention the 4/3 bond order or how it relates to bond length

Mark scheme (4 marks)

  1. All three C–O bonds in CO₃²⁻ are equivalent / the electron density is spread equally across all three bonds
  2. The bonding is explained by delocalisation of electrons / the pi electrons are delocalised over all three C–O bonds
  3. Each C–O bond has a bond order between 1 and 2 (approximately 1.33 / 4/3)
  4. Greater bond order than a single bond means greater electron density between nuclei / shorter and stronger than C–O single bond, but less than a full C=O double bond, giving an intermediate bond length

Key terms in this question

bond length · carbonate ion

Related

More The covalent model questions

▶ Try answering this question with AI marking (free) →