Exams › JEE Advanced › Chemistry
Correct answer: HC≡C(-)
A carbanion is stabilized when its lone pair occupies an orbital of higher s-character, because s-orbitals hold electrons closer to the nucleus (lower energy). The acetylide carbon HC≡C(-) is sp hybridized (50% s), the phenyl and vinyl-type carbons are sp2 (33% s), and the neopentyl carbanion carbon is sp3 (25% s). Highest s-character means the acetylide anion is the most stable.