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(A) To determine the hybridization of carbon atoms,we look at the number of sigma bonds and lone pairs (if any) attached to each carbon atom:$(i) CH_3

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$(ii) 	ext{ and } (iii)$

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(A) To determine the hybridization of carbon atoms,we look at the number of sigma bonds and lone pairs (if any) attached to each carbon atom:$(i) CH_3CH_2CH_2CH_3$: All carbon atoms are $sp^3$ hybridized.$(ii) CH_3-CH=CH-CH_3$: The terminal carbons are $sp^3$ hybridized,while the carbons involved in the double bond are $sp^2$ hybridized. Thus,it has more than one type of hybridization.$(iii) CH_2=CH-C\equiv CH$: The terminal $CH_2$ is $sp^2$,the middle $CH$ is $sp^2$,and the carbons in the triple bond are $sp$ hybridized. Thus,it has more than one type of hybridization.$(iv) HC\equiv CH$: Both carbon atoms are $sp$ hybridized. Only one type of hybridization is present.Therefore,compounds $(ii)$ and $(iii)$ contain more than one type of hybridization.

Among the following compounds,which have more than one type of hybridization for carbon atoms?
$(i) CH_3CH_2CH_2CH_3$
$(ii) CH_3-CH=CH-CH_3$
$(iii) CH_2=CH-C\equiv CH$
$(iv) HC\equiv CH$

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Among the following compounds,which have more than one type of hybridization for carbon atoms?$(i) CH_3CH_2CH_2CH_3$$(ii) CH_3-CH=CH-CH_3$$(iii) CH_2=CH-C\equiv CH$$(iv) HC\equiv CH$