Question
$\text{CH}_2=\stackrel{- \ \ \ }{\text{CH}}$ is more basic than $\text{HC}\equiv\text{C}^-.$ Explain why?

Answer

$\text{CH}_2=\stackrel{\text{sp}^2}{\text{CH}^-}$ $\text{HC}\equiv\stackrel{\text{sp}}{\text{C}^-}$
Since, sp-carbon is more electronegative than $\text{sp}^2$-carbon, therefore, $\text{CH}\equiv\text{C}^-$ is less willing to donate a pair of electrons than $\text{H}_2\text{C}=\text{CH}^-.$ In other words, $\text{H}_2\text{C}=\text{CH}^-$ is more basic than $\text{HC}\equiv\text{C}^-.$

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