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State the hybridisation of asterisked carbon in $\ce{CH3-CH=\overset{\ast}{C}=CH2}.$

I am not sure between $\mathrm{sp^2}$ and $\mathrm{sp}$ hybridization. Because of the presence of double bonds I think it should be $\mathrm{sp^2},$ but because of the two π-bonds I think $\mathrm{sp}$.

Which one is correct?

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  • $\begingroup$ The molecule you've drawn belongs to a class of molecules called "allenes." Here are some diagrams of the molecule's structure and bonding that may be helpful. $\endgroup$
    – ron
    Oct 5, 2014 at 12:39

2 Answers 2

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You must be clear that the hybridization is decided by the number of σ-electrons, so for $\ce{CH4}$ the number of σ-electrons is four and so its hybridization is $\mathrm{sp^3}.$

In case of $\ce{CH3-CH=\overset{\ast}{C}=CH2}$ the $\ce{\overset{\ast}{C}}$ is having two σ- and two π-bonds, so its hybridization is $\mathrm{sp}.$

While deciding hybridization we need to only care about bonds, and that too only the σ-bonds. Remember that possible hybridization for $\ce{C}$ are $\mathrm{sp^3},$ $\mathrm{sp^2}$ and $\mathrm{sp}.$

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As the asterisked carbon has two neighbors, the hybridization of this asterisked carbon is $\mathrm{sp},$ i.e. the angle $\ce{−CH=\overset{\ast}{C}=CH2}$ is 180°.

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