The normal structure for ethyne(acetylene) is:

H - C = C - H

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H - C - C - H

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H - C H_{2} - C H_{2} - H

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H - C \equiv C - H

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H - C H - C H - H

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Solution

The correct option is D $H - C \equiv C - H$
The normal structure for ethyne (acetylene) is $H - C \equiv C - H$ .

A tripple bond is present between two carbon atoms. The tripple bond includes one sigma bond and two pi bonds. Both C atoms are sp hybridised and the molecule is linear. The molecular formula of acetylene is $C_{2} H_{2}$ .

Hence, the correct option is $D$

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C a | H H_{2} - C b | H H = \oplus C H_{2}

In the compound $^{1} C H_{2}$ = $^{2} C - H |^{3} C | H -^{4} H | C | H -^{5} C$ ≡ $^{6} C H$ , the hybridization of $C_{3}$ is:

H - C \equiv C^{-}, H_{2} C = - C H

^{-} C H_{3}

H C \equiv C - C H_{2} - C H_{2} = C H C H_{2} - | C = C H - H | C = O

C H_{3} - H

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

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

C_{6} H_{5} - H

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