(B) Electronegativity of a carbon atom depends on its $s$-character in the hybrid orbital. Greater the $s$-character,higher is the electronegativity.

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(B) Electronegativity of a carbon atom depends on its $s$-character in the hybrid orbital. Greater the $s$-character,higher is the electronegativity. $sp$ hybridization has $50\% \ s$-character,$sp^2$ has $33.3\% \ s$-character,and $sp^3$ has $25\% \ s$-character. Therefore,$sp$-hybridized carbon is the most electronegative.

Class 11 Chemistry 1 Organic Chemistry: Classification and Nomenclature of Organic Compounds Bonding and hybridisation in organic compounds

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Which of the following carbon atoms is the most electronegative?

A
$sp^3$-hybridized carbon
B
$sp$-hybridized carbon
C
$sp^2$-hybridized carbon
D
Always the same regardless of hybridization

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Chapter

1 Organic Chemistry: Classification and Nomenclature of Organic Compounds

Subtopic

Bonding and hybridisation in organic compounds

In the given structure,the number of $sp$ and $sp^{2}$ hybridized carbon atoms present respectively are:

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Explain how the electronegativity of carbon atoms is related to their state of hybridization in an organic compound.

Medium

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In the compound $CH_2=CH-CH_2-C \equiv CH$,what is the hybridization type of the $C_2-C_3$ bond?

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In the molecule $CH_2=C=CH-CH_3$ (where the atoms are numbered $1, 2, 3,$ and $4$ respectively),the hybridization of carbon $1, 2, 3,$ and $4$ are:

MediumJEE MAIN 2021

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The compounds containing $sp$-hybridised carbon atom are:
$(I)$ $H_3C-N(CH_3)-CHO$
$(II)$ Pyridine
$(III)$ $H_3C-CN$
$(IV)$ $H_2C=C=CH-CH_3$

MediumKVPY 2016

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Which of the following carbon atoms is the most electronegative?

Similar Questions

In the given structure,the number of $sp$ and $sp^{2}$ hybridized carbon atoms present respectively are:

Explain how the electronegativity of carbon atoms is related to their state of hybridization in an organic compound.

In the compound $CH_2=CH-CH_2-C \equiv CH$,what is the hybridization type of the $C_2-C_3$ bond?

In the molecule $CH_2=C=CH-CH_3$ (where the atoms are numbered $1, 2, 3,$ and $4$ respectively),the hybridization of carbon $1, 2, 3,$ and $4$ are:

The compounds containing $sp$-hybridised carbon atom are:$(I)$ $H_3C-N(CH_3)-CHO$$(II)$ Pyridine$(III)$ $H_3C-CN$$(IV)$ $H_2C=C=CH-CH_3$

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The compounds containing $sp$-hybridised carbon atom are:
$(I)$ $H_3C-N(CH_3)-CHO$
$(II)$ Pyridine
$(III)$ $H_3C-CN$
$(IV)$ $H_2C=C=CH-CH_3$