In the molecule HC^{1} equiv C^{2} - CH^{3} = | vedclass

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(A) The electronegativity of a carbon atom depends on its hybridization state. The electronegativity order is $sp > sp^{2} > sp^{3}$. Carbon $1$ $(C^{

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(A) The electronegativity of a carbon atom depends on its hybridization state. The electronegativity order is $sp > sp^{2} > sp^{3}$. Carbon $1$ $(C^{1})$ is $sp$ hybridized. Carbon $2$ $(C^{2})$ is $sp$ hybridized. Carbon $3$ $(C^{3})$ is $sp^{2}$ hybridized. Carbon $4$ $(C^{4})$ is $sp^{2}$ hybridized. Carbon $5$ $(C^{5})$ is $sp^{3}$ hybridized. Since $sp^{3}$ hybridized carbon has the lowest s-character $(25\%)$,it exhibits the lowest electronegativity. Therefore,the fifth carbon atom $(C^{5})$ has the lowest electronegativity.

In the molecule $HC^{1} \equiv C^{2} - CH^{3} = CH^{4} - CH_{3}^{5}$,which carbon atom exhibits the lowest electronegativity?

Similar Questions

Which of the following does not contain an $sp^2$ hybridized carbon?

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

In the following molecule,the two carbon atoms marked by asterisk $()$ possess the following type of hybridized orbitals: $CH_3-C^ \equiv C^*-CH_3$

Which of the following compounds do not have $sp^3$ carbon atom$(s)$?
$I$. Acetone
$II$. Acetic acid
$III$. Buta-$1,3$-diene
$IV$. Propyne
$V$. Naphthalene

What is the hybridization state of carbon atoms $1, 2, 3,$ and $4$ in the given reaction?

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