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When is there hybridization and when is there no hybridization? Do all structures have hybridization or only some specific structures have it and why?

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    $\begingroup$ Hybridization is there when you are thinking about it, and vanishes when you stop thinking about it. Hybridization is not a physical phenomenon, really. $\endgroup$ – Ivan Neretin Dec 29 '15 at 13:22
  • $\begingroup$ What do you mean? I know that hybridization is a model / theory, when considering this model, should I assume that all compounds have hybridization or only some? $\endgroup$ – bean Dec 29 '15 at 15:19
  • $\begingroup$ @bean you could've attached a few examples, like $\ce{NH3}$ and $\ce{PH3}$, think about what will happen when bond angles become ~90° (ie: P-H). $\endgroup$ – Sujith Sizon Dec 29 '15 at 16:14
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As Ivan already mentioned, hybridisation is not a physical phenomenon. Physically, the most accepted theory at the moment is that you have nuclei and the electrons roam freely in between them, forming standing waves (orbitals). These standing waves generally spread across the entire molecule unless it is a standing wave of a core electron.

However, it is non-trivial for us chemists to come up with a proper description of these molecular orbitals. Hence, we use a certain set of assumptions. For example, the LCAO approach for molecular orbitals used in some calculations assumes a set of hydrogen-like orbitals on every atom and linear combines them to create the most stable set of molecular orbitals. Since that is still rather hard to do, chemists have also derived a pen-and-paper only approach that does this on a per-atom basis (called hybridisation) and then checks what kind of orbitals would form which bonds — so that’s two steps of linear combinations following each other.

But all of this is only our thought experiment to better describe the physical picture of a molecule. Nature does not need hybridisation or linear combinations. Nature just ‘does it’ and comes up with a final solution. So whenever hybridisation ‘is involved’, it is always man thinking about how to look at the molecule.


Once you have taken that for granted, the question may remain:

But when do I need to take hybridisation into account for orbital considerations?

There’s a set of simple rules:

  • Elements of the third and higher periods will remain mostly unhybridised unless they require hybridisation for any specific issue ($\ce{SiH4}$ cannot be explained without $\mathrm{sp^3}$ hybridisation on silicon).

  • Halogens and noble gases will never be hybridised unless they are hepta-/ octacoordinated.

  • Generally assume a hybridisation for all elements of the second period. Use the one that best explains the geometry. For oxygen, assume at least one unhybridised p-orbital.

  • For a nitrogen with three single bonds, remember that the hybridisation will always switch between $\mathrm{sp^3}$ and $\mathrm{sp^2 + p}$. This is the nitrogen inversion.

  • Hydrogen needs no hybridisation.

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    $\begingroup$ Nice Jan! I like to use the hybridization "language" to try and describe physical reality. For me it is a simple and convenient language. A quick comment on your rule involving nitrogen - can't nitrogen also be sp hybridized as in a nitrile? $\endgroup$ – ron Dec 29 '15 at 19:21
  • $\begingroup$ @ron Yes, and plain sp² in imides, azobenzene and related; I was thinking only about $\ce{NR3}$ cases where there are three single bonds. $\endgroup$ – Jan Dec 29 '15 at 19:54

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