See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: In organic molecules, the hybridization of carbon atoms determines the length of single bonds (C–C) between them. sp-hybridized carbons form shorter single bonds than sp3-hybridized carbons because sp orbitals have more s-character (50%) compared to sp3 orbitals (25%), making them closer to the nucleus and better at overlapping. The molecule is H–C≡C–C≡C–CH3 (penta-1,3-diyne). Bond a is the single bond between the two sp-hybridized carbons (C2 and C3, both part of triple bonds). This is an sp–sp single bond. Bond b is the single bond between an sp-hybridized carbon (C4, part of the terminal triple bond C≡C) and an sp3-hybridized carbon (CH3). This is an sp–sp3 single bond. Reasoning: Greater s-character in the hybrid orbitals leads to shorter, stronger bonds. An sp–sp single bond (bond a) involves two carbons each with 50% s-character, while an sp–sp3 single bond (bond b) involves one carbon with 50% s-character and one with 25% s-character. The sp3 carbon in bond b contributes a larger, more diffuse orbital, making bond b longer than bond a. Wait — re-examining: bond a (sp–sp) should be shorter than bond b (sp–sp3). So b > a means bond b is longer than bond a. That contradicts. Let me reconsider the direction: b > a means the length of b is greater than the length of a, i.e., bond b is longer. Since sp–sp3 (bond b) is longer than sp–sp (bond a), we have b > a. Why other options fail: - (a) a = b: incorrect, hybridization differences cause unequal lengths. - (b) a > b: incorrect, sp–sp bonds are shorter than sp–sp3 bonds, so a < b. - (d) a >>> b: incorrect direction and magnitude. Therefore, the correct answer is C.