Concept:The length of a C-H bond decreases as the s-character of the carbon's hybrid orbital increases, because s-orbitals are spherical and held closer to the nucleus.
Formula:$$ \text{Bond Length} \propto \frac{1}{\text{\% s-character}} $$
Solution:- In ethane and methane (\( \text{sp}^3 \)), the s-character is 25%.
- In ethylene (\( \text{sp}^2 \)), the s-character is 33.3%.
- In acetylene (\( \text{HC}\equiv\text{CH} \), which is \( \text{sp} \) hybridized), the s-character is 50%.
- Because an \( \text{sp} \) orbital holds electrons closest to the nucleus, the resulting C-H bond in acetylene is the shortest and tightest among the options.
Why other options are incorrect:- Option A, Option C, Option D: These have lower s-character (\( \text{sp}^2 \) and \( \text{sp}^3 \)), resulting in more elongated, p-character dominant hybrid orbitals, leading to longer C-H bonds.
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