(b)\[C{{H}_{2}}=\overset{\Theta }{\mathop{C}}\,H\] is more basic than \[HC\equiv {{C}^{-}}\]. Explain why.
Answer:
(a)
(b)\[C{{H}_{2}}=\overset{s{{p}^{2}}}{\mathop{C{{H}^{-}}}}\,\] \[HC\equiv \overset{sp}{\mathop{{{C}^{-}}}}\,\]
Since, \[sp\]-carbon is more electronegative than \[s{{p}^{2}}\]-carbon, therefore, \[CH\equiv {{C}^{-}}\] is less willing to donate a pair of electrons than \[{{H}_{2}}C=C{{H}^{-}}\]. In other words, \[{{H}_{2}}C=C{{H}^{-}}\] is more basic than \[HC\equiv {{C}^{-}}\].
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