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I would like to know what is the level of energy of a hybrid orbital?

For instance, lets consider the $\ce{N2}$ molecule. According to its geometry, we know that there is an orbital 2p and 2s that are going to form 2 sp orbitals.
In the molecular orbital energy diagram, where should we place the these 2 sp orbitals? Would they be in the middle of a 2p and a 2s? And which p orbital do we take to take the middle ($p_x$,$p_y$,$p_z$, $\pi$ or $\sigma$?) Will the sp* be higher or lower that the $2p_x$ and $2p_y$?

Jan
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Bbruyne
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  • Hi and welcome to chemistry.stackexchange.com. Since you already took the tour, all that remains is to point you towards the [help], where any unclear aspects of how the site works will be explained =) – Jan Jun 16 '16 at 12:09
  • Note that molecular orbital theory does not require hybrid orbitals. – Ben Norris Jun 16 '16 at 13:06
  • Right, I did not know that molecular orbital theory does not require the use of hybrid orbitals, cheers. – Bbruyne Jun 16 '16 at 13:33
  • @Bbruyne If you decide to use hybrid orbitals then, yes, your guess is correct: Since an $\mathrm{sp}$ orbital is an equal mixture of an $\mathrm{s}$ and a $\mathrm{p}$ orbital it would be energetically in the middle between those orbitals. For the mixing you would usually use the $\mathrm{p}$ orbital that points in the direction of the bonding axis which conventially is taken to be the $\mathrm{p}_{z}$ orbital. – Philipp Jun 16 '16 at 14:11
  • @Bbruyne What do you mean by $\mathrm{sp}^{}$? There is no such thing as an anti-bonding $\mathrm{sp}$ orbital if that is what you mean. Mixing orbitals to form hybrid orbitals is conceptually totally different from letting, say, two $\mathrm{s}$ orbitals interact to form a $\sigma$ and a $\sigma^{}$ orbital. – Philipp Jun 16 '16 at 14:15
  • I assume that each sp orbital of the two N will interact to form a σ and a σ* orbital. – Bbruyne Jun 16 '16 at 14:49
  • @Bbruyne Yes, the two $\mathrm{sp}$ orbitals pointing towards each other would form form a $\sigma$ and a $\sigma^{}$ orbital. The $\mathrm{sp}$ orbitals pointing away from each other would interact as well but much weaker due to the low orbital overlap. Thus the splitting between the resulting $\sigma$ and $\sigma^{}$ orbitals would be quite small and so those $\mathrm{sp}$ orbitals could essentially be considered as non-bonding orbitals (holding the $\ce{N2}$ lone pairs). – Philipp Jun 16 '16 at 15:10

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When we have an sp hybrid orbital, it is usually made of the s orbital and the p orbital that points in the bounding axis ($p_z$). It's energy will be the mean of the energy of the initial orbitals.

In the example of $\ce{N2}$, it is essential to bear in mind that each sp won't form an $\sigma$ and an $\sigma^*$ orbital. Only the one pair of the sp orbitals, the one that overlap the most, will do. The other pair will contribute to the $\ce{N2}$ lone pairs because there is nearly no overlapping. The 2 others bonds that form the triple bonds are made with the 2 left orbitals $p_x$ and $p_y$.

By the way, it is not necessary to use hybrid orbitals for the molecular orbital theory.

Credits to Philipp and Ben Norris.

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Bbruyne
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  • "Its energy will be the mean of the energy of the initial orbitals" How does this align with the answer posted here? https://chemistry.stackexchange.com/questions/41097/why-does-the-mixing-of-sigma-2s-and-2p-orbitals-lower-the-energy-of-the-sigma-2s?rq=1 – Blaise Dec 15 '18 at 16:22