Energy level of hybrid orbitals in the molecular orbital energy diagram

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 $2p_y$?

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• 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

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 $p_y$.