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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?
$(i)$ $CH_3CH_2CH_2CH_3$
$(ii)$ $CH_3-CH=CH-CH_3$
$(iii)$ $CH_2=CH-CH=CH_2$
$(iv)$ $HC\equiv CH$

Similar Questions

The hybridisation of the negatively charged carbon atoms in $(1)$ and $(2)$ are:
$(1)$ $CH_3^-$
$(2)$ $H_2C^- - CHO - CH_3$

What are the hybridisation states of each carbon atom in the following compounds?
$CH_2=C=O, \quad CH_3-CH=CH_2, \quad (CH_3)_2CO, \quad CH_2=CH-CN, \quad C_6H_6$

Determine the hybridization of the $C_3$ and $C_6$ carbon atoms in the given structure,respectively. (Numbering starts from the left end of the longest chain containing the principal functional group).

Which of the following represents the given mode of hybridisation $sp^2-sp^2-sp-sp$ from left to right?

The base strength order of $H_3C-CH_2^-$ $(i)$,$H_2C=CH^-$ $(ii)$,and $HC \equiv C^-$ $(iii)$ is:

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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)$ $HC\equiv CH$