How does the MO-LCAO theory explain the direction of valence bonds?

As all the bells sound, the common explanation of geometry of a molecule (consider the water molecule) is due to overlaping electron density.


The valence bond theory of the water molecule describes the two O-H bonds as resulting from the overlap of the H 1s orbitals with the two half-filled 2p orbitals of the oxygen atom. Since the two 2p orbitals are at right angles to one another, valence bond theory predicts a bent geometry for the water molecule with a bond angle of 90°. To do this, draw the following picture:
[![enter image description here](https://i.stack.imgur.com/LJPRe.png)](https://i.stack.imgur.com/LJPRe.png)


The opening of the angle to a value greater than the predicted one of 90° can be accounted for in terms of a lessening of the repulsion between the hydrogen nuclei.


Ok.


But what do we have in MO-LCAO (aka HF) theory?


Methods of quantum chemical calculations give us the following picture of orbitals


[![enter image description here](https://i.stack.imgur.com/jcaZk.png)](https://i.stack.imgur.com/jcaZk.png)


For me personally, the picture of molecular orbitals in the second case does not at all agree with the picture of atomic orbitals in the first case. How can you identify the first in the second picture? Or in other words, how does the MO-LCAO (aka HF) theory explain the direction of valence bonds?




The relevant MO analysis for understanding the effect of bond angle on energy is a "Walsh diagram," which for compounds of the type $\ce{AH2}$ looks like this:


[![SH2 Walsh diagram](https://i.stack.imgur.com/oPskqm.png)](https://i.stack.imgur.com/oPskqm.png)


Source is [this post](https://chemistry.stackexchange.com/questions/42911/constructing-a-walsh-diagram-of-ah2) on ChemSE which copies the image from the well known textbook Orbital Interactions in Chemistry by Albright, Burdett and Whangbo.


What it shows is how the energies of each MO are affected as the bond angle decreases from 180 on the left to 90 on the right, in this case for the specific molecule $\ce{SH2}$. The diagram for $\ce{H2O}$ would look qualitatively similar but differ quantitatively such that the minimum total energy would fall at the observed bond angle for that molecule.


Things to notice:


1. s-p mixing between orbitals with "a" symmetry creates electron density on the oxygen that is similar to an sp hybrid orbital on the z axis
2. Decreasing the H-A-H bond angle from 180 reduces the overlap of the hydrogen atomic orbitals with the central atom $p\_x$ orbital, which increases the energy of the $1b\_2$ MO, but it creates overlap with the $p\_z$ orbital (with which there is zero overlap at 180), stabilizing the $2a\_1$ MO. It is largely the balancing of these two effects that determines the optimum bond angle.
3. The large effects seen on the anti-bonding orbitals are irrelevant as long as they are not occupied.


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