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(D) To determine the hybridization of carbon atoms,we count the number of sigma bonds and lone pairs (if any) attached to each carbon atom.$(i)$ $CH_3

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$(ii)$

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(D) To determine the hybridization of carbon atoms,we count the number of sigma bonds and lone pairs (if any) attached to each carbon atom.$(i)$ $CH_3-CH_2-CH_2-CH_3$: All carbon atoms are bonded to four other atoms via single bonds,so all 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 contains both $sp^2$ and $sp^3$ hybridized carbons.$(iii)$ $CH_2=CH-CH=CH_2$: All carbon atoms are involved in double bonds,so all are $sp^2$ hybridized.$(iv)$ $HC\equiv CH$: Both carbon atoms are involved in a triple bond,so both are $sp$ hybridized.Therefore,only compound $(ii)$ has more than one kind of hybridization for carbon.

In which of the compounds given below is there more than one kind of hybridisation $(sp, sp^2, sp^3)$ for carbon?
$(i)$ $CH_3CH_2CH_2CH_3$
$(ii)$ $CH_3-CH=CH-CH_3$
$(iii)$ $CH_2=CH-CH=CH_2$
$(iv)$ $HC\equiv CH$

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In which of the compounds given below is there more than one kind of hybridisation $(sp, sp^2, sp^3)$ for carbon?$(i)$ $CH_3CH_2CH_2CH_3$$(ii)$ $CH_3-CH=CH-CH_3$$(iii)$ $CH_2=CH-CH=CH_2$$(iv)$ $HC\equiv CH$