Vsepr & Hybridisation
Approximate bond dissociation enthalpy of an average C–C single bond is closest to:
Select the correct option:
Solution
350 kJ mol⁻¹
- Bond Strength: Bond enthalpy is the energy required to break one mole of a specific bond in the gaseous state.
- Standard Values for Carbon:
- C–C (Single): ≈348 kJ/mol.
- C=C (Double): ≈612 kJ/mol.
- C≡C (Triple): ≈837 kJ/mol.
- Selection: Among the given options, 350 kJ/mol is the standard value for a saturated hydrocarbon C−C bond (e.g., in Ethane).
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About This Question
- Subject
- chemistry
- Chapter
- chemical bonding and molecular structure
- Topic
- vsepr & hybridisation
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
350 kJ mol⁻¹
- Bond Strength: Bond enthalpy is the energy required to break one mole of a specific bond in the gaseous state.
- Standard Values for Carbon:
- C–C (Single): ≈348 kJ/mol.
- C=C (Double): ≈612 kJ/mol.
- C≡C (Triple): ≈837 kJ/mol.
- Selection: Among the given options, 350 kJ/mol is the standard value for a saturated hydrocarbon C−C bond (e.g., in Ethane).
This medium difficulty chemistry question is from the chapter chemical bonding and molecular structure, covering the topic of vsepr & hybridisation. It appeared in the 2025 exam.
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