I recently learned that there is such thing as a hybridization in chemistry, but I don't really get it.

For example, in $\ce{C2H2}$ (Ethene), they say the there are 3 electrons in three ${sp^2}$ and one electron in a normal $p$ orbital. How do we decide what orbitals get hybridized and what doesn't?


closed as too broad by Mithoron, A.K., Todd Minehardt, airhuff, aventurin Oct 9 '18 at 5:14

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    $\begingroup$ Welcome to ChemSE. Please keep in mind that the concept of hybridization you mention here applies on atomic, and eventually on molecular orbitals; and not on electrons. $\endgroup$ – Buttonwood Oct 8 '18 at 21:15

We do not decide either which. Hybridization is an observational theory - it is a theory crafted to fit observation, it is not made from the bottom up. That doesn't make it any less valuable (well... maybe it does...) but your question becomes rather easy to answer. We decide which orbitals become hybridized based on how we measure the molecule to be. We see the symmetry.

Hybridization is just a model. But then again so are most things we learn in chemistry. I would say you are not missing out on anything if you just skip learning it and go directly to molecular orbital theory. Your teacher might argue though, do what (s)he says.


You have overall 4 valence electrons in the carbon atom. They can be involved either in sigma-bonds or in pi-bonds. All atomic orbitals involved in bonding do so as molecular orbitals (remember LCAO as mathematical approach).

When a carbon is involved in single bonds only, its ${s}$- and three ${p}$-orbitals are ${sp^3}$-hybridized. When a carbon is involved in a double bond (alkenes, carbonyls etc.), three of four orbitals are ${sp^2}$-hybridized, because the double bond consists of one $\pi$-bond and one $\sigma$-bond. When a carbon is involved in a triple bond (alkynes, nitriles etc.) two of four orbitals are ${sp}$-hybridized, because the triple bond consists of two $\pi$-bonds and one $\sigma$-bond.

  • $\begingroup$ Welcome SimonE to ChemSE. I edited your answer mainly because the concept of hybridization applies to orbitals, rather than to electrons. $\endgroup$ – Buttonwood Oct 8 '18 at 21:16

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