AIIMS 2012 Chemistry Hybridization and Stability of Molecules MCQ Question

The statement that H₂C=CH₂ (ethylene) is more stable than H₂C≡C–CH₃ (propyne) can be attributed to the hybridization and s-character of the carbon atoms involved in the bonding. In H₂C=CH₂, each carbon is sp² hybridized, which has 33% s-character. In contrast, in H₂C≡C–CH₃, the terminal carbon (the one bonded to CH₃) is sp hybridized, which has 50% s-character. Generally, greater s-character in hybrid orbitals leads to stronger bonds and thus greater stability. However, the overall stability of a molecule also depends on steric and electronic factors. The correct answer A states that H₂C=CH₂ has more s-character than H₂C≡C–CH₃, which is incorrect. Therefore, the stability comparison should be based on the overall energy and steric considerations, not just s-character. The correct answer is actually that H₂C≡C–CH₃ is less stable due to steric hindrance and the nature of the triple bond. Hence, the answer A is not correct in this context.