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Sometimes carbon forms sp3 orbital and sometimes sp2 and sp. Also, carbon chains are more stable with sp3 orbitals. So, why and how does carbon hybridize itself to other orbitals? My question is more about stability. It's not about how it is significant to organic chemistry. For example, Methane which has 'sp3' orbitals is stable because of its sigma bonds. But, ethene, which has a pi bond is not stable and is more reactive than methane. So, why does carbon form this 'pi bond' while on the other hand has a better option of a sigma bond.

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marked as duplicate by Wildcat, M.A.R., Jan, Todd Minehardt, Martin - マーチン Oct 23 '15 at 3:27

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    $\begingroup$ Hybridization is just a mathematical construct. It has no physical meaning. $\endgroup$ – bon Oct 22 '15 at 15:23
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    $\begingroup$ Re your edit: No, ethene is as stable as methane (or ethane) is. In fact, they are both thermodynamically unstable towards combustion into $\ce{CO2}$ and $\ce{H2O}$. Their ‘stability’ is actually only kinetical inertness. $\endgroup$ – Jan Oct 23 '15 at 13:44
  • $\begingroup$ On the other hand, ethene cannot form a set of two sigma bonds unless it resorted to diborane-like banana bonds. But a set of sigma and pi is so much more favourable than two banana bonds (whatever Chuck Boldwyn might say). $\endgroup$ – Jan Oct 23 '15 at 13:45
  • $\begingroup$ I hesitate to reopen this question. It is not a strict duplicate anymore, but to me it is unclear what you are actually asking. Hybridisation has absolutely nothing to do with stability, like bon already said. It is also not very good manners to remove the formatting that other caring members of this community have introduced. $\endgroup$ – Martin - マーチン Oct 24 '15 at 15:19
  • $\begingroup$ @Jan The formulation of two banana-bonds is equal to sigma and pi bond. The big but here is, that a banana bond is no sigma bond, so there are not two sigma bonds versus one sigma and one pi bond. True is: ethene cannot form two sigma bonds. $\endgroup$ – Martin - マーチン Oct 24 '15 at 15:21
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Hybridization is part of the language chemists use to speak to one another. When someone says that a carbon atom is $\ce{sp^2}$ hybridized, it's a shorthand way of telling others

  • that there is a p-orbital on that carbon
  • that the angles around the carbon are roughly 120°
  • that the bonding orbitals (the $\ce{s^1p^2}$ orbitals) have 1 part in 3, or 33%, s-character and 2 parts in 3, or 67%, p-character
  • that the bonds around that carbon are likely to be a bit shorter than the bonds around an $\ce{sp^3}$ hybridized carbon (because the $\ce{sp^2}$ bonds have more s-character and are therefore closer to the nucleus, e.g. shorter).

So when someone says that a carbon atom is $\ce{sp^2}$ hybridized, they're really saying a lot more.

Using hybridization to describe an atom is similar to using VSEPR or MO theory to describe a molecule. Hybridization is just another way to try and describe reality; just like the other methods, often it works well, but sometimes it doesn't.

Also keep in mind that there is a continuum of hybridizations available to a molecule all the way from pure s to pure p with everything else in between.

$\ce{s^1p^0...s^1p^1...s^1p^2...s^1p^{2.2}...s^1p^3...s^1p^4...s^0p^1}$

Of course, we normally drop the "1" and the "0" superscripts and write those hybridizations as

$\ce{s...sp...sp^2...sp^{2.2}...sp^3...sp^4...p}$

As to how carbon (or any other element) "decides" which hybrization to use, it adopts the hybridization that produces the overall lowest energy for the molecule. Let's take the case of a carbocation. Suppose we start with an $\ce{sp^3}$ hybridized carbon in a molecule like $\ce{R_3C-I}$ and subject it to conditions where it ionizes to an $\ce{sp^2}$ hybridized carbocation and $\ce{I-}$.

enter image description here

(image source)

Your question is basically, why did the central carbon rehybridize from $\ce{sp^3}$ to $\ce{sp^2}$? The answer is, the ionized molecule (the carbocation) will reoptimize its geometry (rehybridize) in order to attain the lowest energy possible. In this example note that the carbocation p-orbital is empty - it contains no electrons. So why keep that orbital $\ce{sp^3}$ hybridized, why waste a part of a valuable, low-energy s-orbital in a bond with no electrons in it to stabilize? Instead, why not remove the s-character from the $\ce{sp^3}$ bond making it a pure p-orbital and mix that s-character into the other 3 orbitals (or bonds) to convert them from $\ce{sp^3}$ to $\ce{sp^2}$ thereby lowering the energies of the electrons that are in those 3 bonds. In fact, this rehybridization, or geometry reoptimization, is exactly what does occur in this reaction in order to stabilize the electrons as much as possible and produce a molecule (or ion) with the lowest possible energy.

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