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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 $2p_y$?

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  • $\begingroup$ Hi and welcome to chemistry.stackexchange.com. Since you already took the tour, all that remains is to point you towards the help center, where any unclear aspects of how the site works will be explained =) $\endgroup$ – Jan Jun 16 '16 at 12:09
  • $\begingroup$ Note that molecular orbital theory does not require hybrid orbitals. $\endgroup$ – Ben Norris Jun 16 '16 at 13:06
  • $\begingroup$ Right, I did not know that molecular orbital theory does not require the use of hybrid orbitals, cheers. $\endgroup$ – Bbruyne Jun 16 '16 at 13:33
  • $\begingroup$ @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. $\endgroup$ – Philipp Jun 16 '16 at 14:11
  • $\begingroup$ @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. $\endgroup$ – Philipp Jun 16 '16 at 14:15
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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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