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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

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

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?

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  • $\begingroup$ You calculate the MO diagram for all possible directions (or a sufficiently large number of directions), and then find out which direction minimises the energy: en.wikipedia.org/wiki/Energy_minimization $\endgroup$
    – orthocresol
    Commented Feb 10, 2022 at 20:59
  • $\begingroup$ @orthocresol Yes, let's say I determined the minimum energy and the molecular orbital that corresponds to it. By the form of the orbital, I do not see any overlap, as in the VB- theory. What then is the point of VB-theory? $\endgroup$
    – Sergio
    Commented Feb 10, 2022 at 21:04
  • $\begingroup$ I completely disagree with the described way in the VB model. It's valence bond for a reason, it accounts for bonding domains naturally, so, if all you start with four domains about oxygen, idealised as sp³, and go from there. Modern calculations have shown that this approach does not represent the mathematically ideal solution. Are the lone pairs in water equivalent? $\endgroup$
    – Martin - マーチン ♦
    Commented Feb 10, 2022 at 21:28
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    $\begingroup$ As I understand it, and correct me if I'm wrong, you seek to 'reconcile' the MOT and VBT pictures. I would contend that there's no reason that MOT should explain VBT (although people have tried; I'm not the right person to summarise this, though, as it's way outside my domain of knowledge). The theories of bonding exist not to rationalise one another; rather, their aim is to explain experimentally measurable things such as geometries, chemical reactivity, or various spectroscopies. In this case, both MOT and VBT correctly explain the H–O–H bond angle. Beyond that, though, all bets are off. $\endgroup$
    – orthocresol
    Commented Feb 10, 2022 at 22:37
  • $\begingroup$ @orthocresol Yes, I seek to 'reconcile' the MOT and VBT pictures. Even if in VBT these pictures at least make it possible to visualize the direction of the connection, then in MOT it is generally not clear what the pictures of orbitals mean and why they are drawn. $\endgroup$
    – Sergio
    Commented Feb 11, 2022 at 8:38

1 Answer 1

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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

Source is this post 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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  • $\begingroup$ +1 - my immediate thought was "that's exactly what I'd say" and then I saw you linked to my previous answer. :-) $\endgroup$
    – Geoff Hutchison
    Commented Feb 11, 2022 at 16:39

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