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(D) To identify compounds containing both $sp$ and $sp^2$ hybridized carbons,we analyze each structure:$(I)$ $N \equiv C-CH_2-COOH$: The carbon in the

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$I$,$III$ and $V$

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(D) To identify compounds containing both $sp$ and $sp^2$ hybridized carbons,we analyze each structure:$(I)$ $N \equiv C-CH_2-COOH$: The carbon in the cyano group $(-CN)$ is $sp$ hybridized,and the carbonyl carbon in $-COOH$ is $sp^2$ hybridized. Thus,it contains both.$(II)$ $HC \equiv C-C \equiv CH$: All carbons are $sp$ hybridized.$(III)$ $H_2C=C=CH_2$: The terminal carbons are $sp^2$ hybridized,and the central carbon is $sp$ hybridized. Thus,it contains both.$(IV)$ Benzoic acid $(C_6H_5COOH)$: The benzene ring carbons and the carbonyl carbon are $sp^2$ hybridized. No $sp$ hybridized carbon is present.$(V)$ Benzonitrile $(C_6H_5CN)$: The benzene ring carbons are $sp^2$ hybridized,and the cyano carbon is $sp$ hybridized. Thus,it contains both.$(VI)$ Benzamide $(C_6H_5CONH_2)$: All carbons are $sp^2$ hybridized.Therefore,compounds $(I)$,$(III)$,and $(V)$ contain both $sp$ and $sp^2$ hybridized carbons.

From the following compounds,the ones which contain both $sp$ and $sp^2$ hybridized carbons are:

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