[Rutherford]

Can sp mixing be rationalized this way:
In second period elements with more than a half filled p orbital the energy pattern of the MOs is regular.
In second period elements with less than or half filled p orbital, as the s and p atomic orbitals have similar energies, so the formed σ_g MO would have similar energy as σ_u, but because of electron cloud repulsion it has a big increase in energy, hence it is above the π orbitals?

[Mitch]

It isn't an electrostatic repulsion issue. All orbitals of the same symmetry will mix, but the extent of mixing depends on how close the energies of the orbitals are.

See J. Chem. Ed. 68(9), 737. (1991)

[Rutherford]

Okay, but why would the mixing cause a rise in energy for the sigma orbital?

[Rutherford]

Could anyone explain this?

[Corribus]

A starting point is because when you take linear combinations of two eigenfunctions of the Hamiltonian, both the (normalized) additive and subtractive combinations are also solutions to the Schrodinger equation.

A less rigorous but also less mathematical explanation is that the sign/phase of each wavefunction is arbitrary, so when you add two, they can either add positively or negatively. This is very easy to see in the case of p-orbitals that are aligned along the same axis. If we designate the phase alignment of a single p-orbital as {+·-}, where the · is the node at the nucleus, then it is arbitrarily either {+·-} or {-·+}. This makes no difference - the orbital has the same energy in free space. When two are brought together, though, they can either be {+·-}{+·-} or {+·-}{-·+} or {-·+}{-·+} or {-·+}{+·-}. Scenarios 1 and 3 are equivalent and result in destructive interference and 2 and 4 are equivalent and result in constructive interference. This is the basis for the energy differences, because 1/3 and 2/4 are no longer equivalent states of mixing. In the latter case, the wavefunctions destructively interfere with each other between the nuclei, interpreted as the electrons being located most of the time at the opposite sides of the putative molecule. This leaves the two positively charged nuclei with no negative counterbalance between them, so they repel. In energetic terms, there is an unfavorable Coulombic interaction between the two positive nuclei. This is the repulsive "antibonding" state. In the 1/3 scenario, the wavefunctions constructively interact between the nuclei, so most electron density is in this space. The positive Coulombic interactions between the two nuclei and the two electrons is more favorable in this case, which leads to the energetically stabilized "bonding state".

There's no discussion of electron exchange in this simplified discussion - and that's a big part of the energy stabilization/destabilization - but at least it should help you see why the combination of two orbitals leads to both a stabilized and destabilized (relative to isolated orbitals) delocalized molecular orbitals.

(Mixing of multiple orbitals or orbitals with different angular momenta is more complex but essentially the same principles are at work.)

[Rutherford]

I know that but why would s-p mixing cause the rise in energy for the sigma orbital? I don't understand how s-p mixing works.

[Corribus]

It works no differently - the orbitals add constructively and destructively (or both, if there is no net interaction).

For an S orbital interacting with a P orbital oriented so that one of the lobes of the P orbital is pointing toward the S orbital, you have again 4 possibilities:

{+}{+·-}
{+}{-·+}
{-}{+·-}
{-}{-·+}

1 and 4 are identical, and interact constructively and 2 and 3 are identical and interact destructively.
Note that if the P orbital is oriented such that its axis is perpendicular to the vector between the two orbitals, there is no net interaction because part of the interaction is constructive and part is destructive. This is where molecular symmetry comes into play.

[Rutherford]

I can't imagine this. What would the σ_2p orbital look like when it's formed from both p-p overlap and s-p mixing? I should certainly be different than a normal σ orbital obtained from 2*p orbitals.

Why wouldn't s-p mixing cause decrease in energy of the σ_2p orbital and increase of energy in σ_2s orbital?