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Before $\ce{sp^3}$ hybridisation, does the $\ce{C}$ -atom get excited to $\ce{1s^2~2s^1~2p^3}$ state, as it happens before $\ce{sp^2}$ hybridisation? What is the energy of the $\ce{sp^3}$ hybridised orbital?

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  • $\begingroup$ No it does not just like it does not in the other case. Hybridization is mumbo-jumbo that makes absolutely wrong assumptions, but somehow get correct result in some cases. $\endgroup$ – permeakra May 27 '14 at 20:55
  • $\begingroup$ @permeakra hybridization "gets the answer right" because it is applied once the geometry is known. I find it useless for prediction. $\endgroup$ – Eric Brown May 31 '14 at 15:17
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To follow up on Swedish Architect's answer (and Martin's comment):

The energy of an orbital is a poorly defined concept, but what can be measured is the energy it takes to excite an electron from a molecular orbital (or an atomic orbital in some cases-- but that is not relevant for hybridization). In the case of an sp3 hybridized carbon one may infer something about the 'energy' of the orbital from the bond dissociation energy of of the bond formed by such an orbital (as previously mentioned this is a fairly imprecise use of language, but the concept does have some instructional value).

This reference (http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociationenergy.pdf) provides some of these. In particular on p4.46 of that reference in the chart, CH3CH3 (sp3) is listed at 410 kJ/mol, while CH2CH2 (sp2) is listed at 427 kJ/mol, and HCCH (sp) is listed at 523 kJ/mol. this is the amount of energy required to abstract a hydrogen atom from that systems, which is a reflect of how 'strong' the bond is, which is a reflection of how low in energy the 'orbital' is. In essence, the higher the bond dissociation energy, the lower the energy of the corresponding orbital is. Since all of the molecules in question here from combinations of the hydrogen orbital with the carbon orbital, as a crude approximation, you can assign a relative energy.

Please be advised, this analysis really only has meaning for trying to teach the concept. The actual molecular orbital includes contributions from all atoms in a molecule, and the fact that the ethane has more hydrogens than ethene or ethyne is perturbing this analysis slightly.

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sp3 hybridisation is what we call the formation of hybrid atomic orbitals, by the combination of an s orbital and 3 p orbitals on the carbon atom. I think you understand this. However, like in a dative bond, even though one of p orbitals is empty before the carbon becomes hybridized, it can still combine with the other atomic orbitals, to create new hybrid atomic orbitals. So no, the atom doesn't have to get excited to 1s2 2s1 2p3 before hybridization.

Another way of thinking about sp3 hybridization, a more advanced view, is that they are merely a nice simple way of describing the bonding in atoms. If we were to use the molecular orbital approach we would have to keep going back to our computers to work out the MO's. This isn't really needed, unless we want exact energies and a detailed description of the bonding. In the case of sp3 hybridization, say in methane, the carbon s orbital interacts with ALL of the hydrogen orbitals to form a MO. Then the carbon p-orbitals interact with the hydrogen 1s orbitals. Here the case is that several AO are interacting to form one MO. Then the sp3 model can just be said to be a simplification of a more complex interaction of AOs. This hasn't directly answered your question, yet. The carbon atom doesn't have to be excited to the 1s2 2s1 2p3 state to interact like described above.

Your next question. The energy of the sp3 hybridised orbital is not one, but 4 orbitals. The energy cannot really be found because they don't really exist. After they bond, however, the energy changes based on the bond, as they form a new MO, so it may be more useful to know the energy of each MO in the individual case.

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    $\begingroup$ Hybridisation is not an interaction, it is only a linear combination of functions. It can therefore only be a description of a bonding situation and will always be a result of a geometric arrangement. As an hybrid orbital is only a mathematical concept, it does not have a physical meaning and therefore it does not have an energy that may be assigned to it. $\endgroup$ – Martin - マーチン May 27 '14 at 10:00
  • $\begingroup$ @Martin, yeah I think I badly worded it in the top paragraph, so I've edited it. I also said in my original answer that hybrid orbitals don't have energies, because they don't really exist. $\endgroup$ – Swedish Architect May 27 '14 at 13:23

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