See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The strength of a C–H bond depends on the percent s-character of the hybrid orbital on carbon used to form the bond. A higher s-character means the electrons are held closer to the nucleus (s orbitals are closer to the nucleus than p orbitals), resulting in a shorter and stronger bond. Step 1: Determine the s-character of each hybridization. - sp hybridized carbon: 50% s-character (1 s + 1 p, so 1/2 = 50%) - sp^2 hybridized carbon: 33.3% s-character (1 s + 2 p, so 1/3 ≈ 33%) - sp^3 hybridized carbon: 25% s-character (1 s + 3 p, so 1/4 = 25%) Step 2: Relate s-character to bond strength. Higher s-character → electrons held closer to nucleus → shorter bond length → stronger C–H bond. Therefore: sp (50%) > sp^2 (33%) > sp^3 (25%) Step 3: Confirm with known bond energies. - C(sp)–H bond (e.g., in acetylene): ~≈558 kJ/mol - C(sp^2)–H bond (e.g., in ethylene): ~≈444 kJ/mol - C(sp^3)–H bond (e.g., in ethane): ~≈410 kJ/mol This confirms the order sp > sp^2 > sp^3. Step 4: Eliminate wrong options. - (b) sp^3 > sp^2 > sp: Incorrect, this is the reverse of the correct order. - (c) sp^2 > sp^3 > sp: Incorrect, sp is strongest, not weakest. - (d) sp^2 > sp > sp^3: Incorrect, sp should be the strongest, not second. Only option (a) sp > sp^2 > sp^3 correctly reflects increasing s-character leading to increasing bond strength. Therefore, the correct answer is A.