• # question_answer 47)   Arrange the following carbanions in order of their decreasing stability : (A) ${{H}_{3}}C-C\equiv {{C}^{-}}$         (B) $H-C\equiv {{C}^{-}}$                           (C) ${{H}_{3}}C-CH_{2}^{-}$ (a) $A\text{ }>B>C~$                   (b)$B>A>C$                     (c) $C\text{ }>\text{ }B\text{ }>\text{ }A$          (d) $C>A>B$

(b) sp-Hybridized carbon is more electronegative than a $s{{p}^{3}}$-hybridized carbon, therefore (A) and (B) are more stable than (C) Since $C{{H}_{3}}$ group has $+I$ effect and hence it intensified the negative charge and so, destabilised (A) as compared to (B)