Resonance

Question

The carbonate ion (CO₃²⁻) has three equivalent C–O bonds of length 129 pm, which is intermediate between a C–O single bond (143 pm) and a C=O double bond (121 pm). This observation is best explained by:

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1. **Observation**: All three C–O bonds in CO₃²⁻ are identical (129 pm), neither purely single (143 pm) nor purely double (121 pm).
2. **Draw Canonical Structures**: Three resonance structures can be drawn for CO₃²⁻, each with one C=O double bond and two C–O single bonds. The double bond rotates among the three oxygen atoms.
3. **Resonance Hybrid**: The actual structure is a weighted average (resonance hybrid) of these three structures. Each C–O bond has a bond order of $\frac{4}{3} \approx 1.33$, explaining the intermediate bond length.
4. **Why Not Others?**
   - sp² hybridisation explains the trigonal planar shape but not the bond length equivalence on its own.
   - Ionic character would predict different bond lengths for ionic vs covalent bonds.
   - Coordinate bonds would create non-equivalent bonds (donor–acceptor differs from shared pair).
5. **Verification**: Bond order = $\frac{	ext{total bonding electrons}}{	ext{number of bonds}} = \frac{4}{3}$, consistent with 129 pm being between single and double bond lengths.
6. **Conclusion**: Resonance delocalises the π electrons equally over all three C–O bonds, making them equivalent.

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