See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: sp-hybridization occurs when a carbon (or other atom) forms two sigma bonds and has two pi bonds, resulting in a linear geometry (bond angle 180°). This occurs in atoms involved in triple bonds or in cumulated double bonds (allenes, ketenes), or in carbocations with only two groups. Step-by-step analysis of each underlined atom: (u) CH2=CH-CH3: The underlined C is the first carbon (CH2=). This carbon is part of a C=C double bond, making it sp2 hybridized, not sp. (v) CH2=C=CHCl: The underlined C is the central carbon in an allene-type (cumulated diene) structure. A central carbon in a cumulated diene has two double bonds (C=C=C), meaning it forms two sigma bonds and participates in two pi bonds → sp hybridization. YES, this is sp. (w) CH3-CH2+: The underlined C is a carbocation (CH2+). A carbocation with three substituents (or two and an empty p orbital contributing to sp2) is sp2 hybridized, not sp. (x) H-C≡C-H (acetylene): Both carbons in a triple bond are sp hybridized. YES, these are sp. (y) CH3-CN: The underlined C is the carbon of the nitrile group (C≡N). Carbon in a triple bond (C≡N) is sp hybridized. YES, this is sp. (z) (CH3)2C=N-NH2: The underlined N is the nitrogen directly bonded to the carbon via a double bond (imine nitrogen, C=N). This nitrogen has a double bond to C and a single bond to NH2 plus a lone pair → sp2 hybridized, not sp. Summary: - v: sp ✓ - x: sp ✓ - y: sp ✓ - u: sp2 ✗ - w: sp2 ✗ - z: sp2 ✗ The underlined atoms with sp-hybridization are v, x, and y. Why other options fail: (a) x and z: z (N in C=N) is sp2, not sp. Incorrect. (b) x, y, and z: z is sp2. Incorrect. (c) u, w and x: u (sp2) and w (sp2) are not sp. Incorrect. (d) v, x and y: All three are sp hybridized. Correct. Therefore, the correct answer is D.