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Question #45106

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The Question

In highly strained small-ring carbocycles like cyclopropane, the hybridization of carbon atoms deviates significantly from the ideal $sp^3$ model, leading to the formation of 'bent bonds' (banana bonds). Assuming the H-C-H bond angle ( $\theta_1$ ) in cyclopropane is such that $\cos(\theta_1) = -0.4$ , determine the hybridization index $n$ of the carbon orbitals involved in the C-C bonds (represented as $sp^n$ ) and the cosine of the interorbital angle ( $\theta_2$ ) between the two C-C bonds. (Use Coulson's Theorem: $\cos(\theta) = -\frac{1}{i}$ for $sp^i$ hybridized orbitals).

Options

A.
$n = \frac{11}{3}$ , $\cos(\theta_2) = -\frac{3}{11}$
B.
$n = \frac{14}{3}$ , $\cos(\theta_2) = -\frac{3}{14}$
C.
$n = 3$ , $\cos(\theta_2) = -\frac{1}{3}$
D.
$n = 5$ , $\cos(\theta_2) = -\frac{1}{5}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Bond Angle and Strain|Valence Bond Theory Hybridization

Why This Question Is Tricky

Common Mistakes

Students often assume that all saturated carbons are strictly $sp^3$ hybridized with 109.5° bond angles.

This question tests the concept of 'hybridization as a continuum' in highly strained systems.

In cyclopropane, the severe angle strain forces the C-C bonds to adopt higher p-character (bent bonds) to minimize strain, which consequently forces the C-H bonds to adopt higher s-character.

Failing to apply Coulson's theorem and the conservation of s-character leads to the incorrect assumption of ideal $sp^3$ hybridization.

This framework accurately explains why cyclopropane exhibits alkene-like acidity and reactivity.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Organic Chemistry – Some Basic Principles and Techniques - Class 11 Part 2
Concept	Bond Angle and Strain
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free