Question:

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

Updated On: Jun 23, 2023
  • $CH_{3}??H_{2}??overset{*}{C}H_{2} ??H_{3}$
  • $CH_{3} ?
  • $CH_{3} ??H_{2} ??\equiv\overset{*}{C}H$
  • $CH_{3} ??H_{2}??H = \overset{*}{C}H_{2}$
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The Correct Option is C

Solution and Explanation

The greater the $s$ character of the hybrid orbitals, the greater is the electronegativity. Thus, a carbon atom having an $sp$ hybrid orbital with $50\%$ $s$ character is more electronegative than those possessing $sp^{2}$ or $sp^{3}$ hybridised orbitals. The asterisk marked carbon in $CH_{3} ??H_{2} ??\equiv\overset{*}{C}H$ is $sp$ hybridised hence, it is most electronegative.
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