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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:

Similar Questions

Consider a positively charged $C_2H_3$ species in which the positively charged carbon is $sp$-hybridized,the uncharged carbon is $sp^2$-hybridized and an empty $p$-orbital is perpendicular to the $\pi$ system. What is the best description of this cation?

Number of $sp$ hybridised carbon atoms in the given compound is:
$CH_2 = CH - C \equiv C - CH_2 - CN$

Which of the following bonds is most acidic?

Amongst the given options,the compound$(s)$ in which all the atoms are in one plane in all the possible conformations (if any),is (are):
$(A)$ $CH_2=CH-CH=CH_2$
$(B)$ $HC\equiv C-CH=CH_2$
$(C)$ $H_2C=C=O$
$(D)$ $H_2C=C=CH_2$

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

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