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Question: Electronegativiety of carbon atoms depends upon their state of hybridisatio. In which of the followi...

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

A

CH3−CH2−C∗H2−CH3CH_{3} - CH_{2} - \overset{*}{C}H_{2} - CH_{3}

B

CH3−C∗H=CH−CH3CH_{3} - \overset{*}{C}H = CH - CH_{3}

C

CH3−CH2−C≡C∗HCH_{3} - CH_{2} - C \equiv \overset{*}{C}H

D

CH3−CH2−CH=C∗H2CH_{3} - CH_{2} - CH = \overset{*}{C}H_{2}

Answer

CH3−CH2−C≡C∗HCH_{3} - CH_{2} - C \equiv \overset{*}{C}H

Explanation

Solution

: The greater the s character of the hybrid orbitals, the greater is the electronegativity. Thus, a carbnon atom having an sp hybrid orbital with 50% s character is more electronegative than those possessing sp2sp^{2} or sp3sp^{3} hybridised orbitals.

The asterisk marked carbon is CH3−CH2−C≡CHCH_{3} - CH_{2} - C \equiv CH is sp hybridised hence, it is most electronegative.