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Question #44187
The Question
According to the Coulson-Moffitt 'bent bond' model of cyclopropane () at , the bonds are formed by orbitals with high -character to minimize angle strain, forcing the bonding electron density to be highly localized outside the internuclear axis.
If this spatial confinement reduces the transverse positional uncertainty () of the bonding electrons by a factor of relative to an unstrained bond in ethane, by what factor does the minimum zero-point kinetic energy of these electrons increase (based on the Heisenberg Uncertainty Principle), and what is the idealized fractional -character of the bonding orbitals in this model?
Options
- A.
- B.
- C.
- D.
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Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Mechanism_Trap
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Concepts Behind This Question
Heisenberg Uncertainty Principle|Hybridization and Geometry|Alkane Conformation Energy Barriers|Cycloalkane Ring Strain
Why This Question Is Tricky
Common Mistakes
This question tests the 'electron density localization fallacy' by bridging Quantum Mechanics and Organic Chemistry.
Students often rigidly apply hybridization ( -character) to all saturated alkanes, ignoring that severe ring strain forces rehybridization (Bent/Banana bonds).
Furthermore, they rarely connect steric/angle strain to its quantum mechanical root: spatial confinement of electrons () leading to a quadratic spike in zero-point kinetic energy via the Heisenberg Uncertainty Principle.
Option B traps students who assume a linear relationship for kinetic energy and mistakenly assign the hybridization of the bonds to the bonds.
Option C traps those who correctly calculate the energy factor but default to standard hybridization.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Structure of Atom - Class 11 Part 1 |
| Concept | Heisenberg Uncertainty Principle |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |