# What is the hybridization of the central carbon in buta‐1,2‐diene?

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?

• 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.
– ron
Oct 5, 2014 at 12:39

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 $$\mathrm{sp}.$$
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°.