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Electronegativity and Hybridisation

Electronegati...

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?

C H_{3} - C H_{2} - * C H_{2} - C H_{3}

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C H_{3} - * C H = C H - C H_{3}

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C H_{3} - C H_{2} - C * C \equiv H

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C H_{3} - C H_{2} - C H = * C H_{2}

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Solution

The correct option is A $C H_{3} - C H_{2} - C * C \equiv H$
Greater the $s$ character of the hybrid orbitals, greater is the electronegativity. Thus, a carbon atom having $s p$ hybrid orbital with $50 %$ s character is more electronegative than those possessing $s p^{2} o r s p^{3}$ hybridised orbitals.
The asterisk marked carbon in $C H_{3} - C H_{2} - C \equiv C H$ is sp hybridised, hence, it is most electronegative.

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Similar questions

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H | C H_{3} - C | C H_{3} - C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{2} - C H_{3}

C H_{3} - C H_{3}, C H_{2} = C H_{2}

C H \equiv C H

What are the hybridization states of each carbon atom(s) in the following molecules ?

(i) $C H_{3} * C H_{2} C H_{2} C H_{3}$

(ii) $C H_{2} = C^{*} H - C H = C H_{2}$

(iii) $C H_{3} * C H = C H - C H_{3}$

(iv) $H - * C \equiv C - H$

S_{N} 2

K I

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H ∣ C H_{3} - C H_{2} - C l

(I) (I I)

C H_{3} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H_{2} - C H_{2} - B r

(I) (I I)

C H_{3} - C H ∣ C H_{3} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H_{3} ∣ C ∣ C H_{3} - C H_{2} C l

(I) (I I)

C H_{3} - C H_{2} - C H_{2} - B r ∣ C H - C H_{3} a n d C H_{3} - C H_{3} ∣ C H - C H_{2} - B r ∣ C H - C H_{3}

(I) (I I)

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