Resonance
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:
Select the correct option:
Solution
Resonance among three equivalent canonical structures
- Observation: All three C–O bonds in CO₃²⁻ are identical (129 pm), neither purely single (143 pm) nor purely double (121 pm).
- 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.
- Resonance Hybrid: The actual structure is a weighted average (resonance hybrid) of these three structures. Each C–O bond has a bond order of 34≈1.33, explaining the intermediate bond length.
- 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).
- Verification: Bond order = number of bondstotal bonding electrons=34, consistent with 129 pm being between single and double bond lengths.
- Conclusion: Resonance delocalises the π electrons equally over all three C–O bonds, making them equivalent.
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About This Question
- Subject
- chemistry
- Chapter
- chemical bonding and molecular structure
- Topic
- resonance
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
Resonance among three equivalent canonical structures
- Observation: All three C–O bonds in CO₃²⁻ are identical (129 pm), neither purely single (143 pm) nor purely double (121 pm).
- 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.
- Resonance Hybrid: The actual structure is a weighted average (resonance hybrid) of these three structures. Each C–O bond has a bond order of 34≈1.33, explaining the intermediate bond length.
- 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).
- Verification: Bond order = number of bondstotal bonding electrons=34, consistent with 129 pm being between single and double bond lengths.
- Conclusion: Resonance delocalises the π electrons equally over all three C–O bonds, making them equivalent.
This easy difficulty chemistry question is from the chapter chemical bonding and molecular structure, covering the topic of resonance. It appeared in the 2025 exam.
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