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Electronegativity of Carbon atoms depends upon their state of hybridization. In which of the following compounds the carbon marked with asterisk is most electronegative?

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${CH}_{3} - {CH}_{2} - C \equiv \overset{*}{C} H$

${CH}_{3} - {CH}_{2} - CH = \overset{*}{C} H_{2}$

${CH}_{3} - {CH}_{2} - \overset{*}{C} H_{2} {-CH}_{3}$

${CH}_{3} - \overset{*}{C} {H=CH-CH}_{3}$

answer is C.

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Detailed Solution

Greater the s character of the hybrid orbitals, greater is the electronegativity. Thus, a carbon atom having sp hybrid orbital with 50% s character is more electronegative than those possessing sp² or sp³ hybridised orbitals.

The asterisk marked carbon in CH₃−CH₂−C≡CH is sp hybridized, hence, it is most electronegative.

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