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?
A) \[{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {}^*{\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_3}\]
B) \[{\text{C}}{{\text{H}}_3} - {}^*{\text{CH}} = {\text{CH}} - {\text{C}}{{\text{H}}_3}\]
C) \[{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{C}}{}-^*{\text{C}} \equiv {\text{H}}\]
D) \[{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{CH}} = {}^*{\text{C}}{{\text{H}}_2}\]

Answer 1

Hint: Electronegativity of carbon atoms depends on their state of hybridisation. Determine the hybridisation of each carbon atom marked with an asterisk and then decide the most electronegative carbon atom.

Complete step by step answer:
Consider the compound (A) \[{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {}^*{\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_3}\],
The carbon atom marked with an asterisk in compound (A) forms four bond pairs. Thus, the carbon atom is $s{p^3}$ hybridised.
Percentage s-character in $s{p^3}$ hybridised orbital $ = \dfrac{1}{4} = 0.25 \times 100\% = 25\% $.
Thus, the percentage s-character of the carbon atom marked with asterisk in compound (A) is $25\% $.
Consider the compound (B) \[{\text{C}}{{\text{H}}_3} - {}^*{\text{CH}} = {\text{CH}} - {\text{C}}{{\text{H}}_3}\],
The carbon atom marked with an asterisk in compound (B) forms three bond pairs. Thus, the carbon atom is $s{p^2}$ hybridised.
Percentage s-character in $s{p^2}$ hybridised orbital $ = \dfrac{1}{3} = 0.33 \times 100\% = 33\% $.
Thus, the percentage s-character of the carbon atom marked with asterisk in compound (B) is $33\% $.
Consider the compound (C) \[{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{C}}{}-^*{\text{C}} \equiv {\text{H}}\],
The carbon atom marked with an asterisk in compound (C) forms two bond pairs. Thus, the carbon atom is sp hybridised.
Percentage s-character in sp hybridised orbital $ = \dfrac{1}{2} = 0.5 \times 100\% = 50\% $.
Thus, the percentage s-character of the carbon atom marked with asterisk in compound (C) is $50\% $.
Consider the compound (D) \[{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{CH}} = {}^*{\text{C}}{{\text{H}}_2}\],
The carbon atom marked with an asterisk in compound (D) forms three bond pairs. Thus, the carbon atom is $s{p^2}$ hybridised.
Percentage s-character in $s{p^2}$ hybridised orbital $ = \dfrac{1}{3} = 0.33 \times 100\% = 33\% $.
Thus, the percentage s-character of the carbon atom marked with asterisk in compound (D) is $33\% $.
We know that more the s-character, more is the electronegativity of the carbon atom.
The carbon atom with more s-character is the carbon atom marked with an asterisk in compound (C).
Thus, in compound (C) \[{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{C}}{}-^*{\text{C}} \equiv {\text{H}}\] The carbon marked with an asterisk is most electronegative.

Thus, the correct option is (C) \[{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{C}}{}-^*{\text{C}} \equiv {\text{H}}\].

Note: More the s-character, more is the electronegativity of the carbon atom. Thus, the sp hybridised carbon atom is most electronegative because the contribution of s and p orbitals is 50 %.