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State the hybridisation of asterisked carbon in $\ce{CH3-CH=C^{✪}=CH2}$

I am not sure between $sp^2$ and $sp$ hybridization. Because of double bond I think it should be $sp^2$ but because of 2 $\pi$ bonds iI think $sp$.

So which one is correct?

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  • $\begingroup$ You are correct that a carbon with two double bonds is not $sp^2$. I'd suggest thinking about the geometry around that atom. What shape will it have to handle both double bonds? $\endgroup$ – Geoff Hutchison Oct 5 '14 at 4:24
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    $\begingroup$ To add to what Geoff Hutchison wrote, an atom with similar geometry and hybridization that you have probably encountered previously is the carbon in CO2. $\endgroup$ – Jason Patterson Oct 5 '14 at 4:52
  • $\begingroup$ @hey 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 '14 at 12:39
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    $\begingroup$ The hybridization is the same as carbon dioxide. $\endgroup$ – LDC3 Oct 5 '14 at 14:25
  • $\begingroup$ @GeoffHutchison Sorry it's linear. $\endgroup$ – Freddy Oct 5 '14 at 15:54
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You must be clear that the hybridization is decided by the number of sigma electrons So for $\ce{CH4}$ the number of sigma electrons is 4 and so its hybridization is $sp^3$.

In your case: $\ce{CH3-CH=C^{✪}=CH2}$ the $*$ carbon is having 2 sigma and 2 pi bonds so its hybridization is $sp$.

Remember that possible hybridization for $\ce{C}$ are $sp^3$,$sp^2$ and $sp$.

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  • $\begingroup$ So, while deciding hybridization we need to only care about bonds, Thanks. $\endgroup$ – Freddy Oct 10 '14 at 12:54
  • $\begingroup$ @Freddy Yes, and that too only the sigma bonds. $\endgroup$ – Singh Oct 10 '14 at 12:55
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As the asterisked carbon has two neighbors, the hybridization of this asterisked carbon is sp . i.e. the angle −CH=C✪=CH2 is 180 degrees.

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