Why does greater orbital overlap mean a stronger bond?

According to valence bond theory, orbital overlap produces a bond. However, I don’t understand why having greater orbital overlap renders a bond stronger. It’s intuitive, I suppose, but I haven’t been able to find an actual explanation as to why that is. Are there quantum mechanical effects behind this?

• Since elctrons are waves, and waves with same phase sum. Greater overlap is like having regions of constructive interference, where the waves of the orbitals sum. More overlap, more constructive interference (in bonding orbitals anyways) – Scient Jun 29 '16 at 14:07

An intuitive explanation is given in the comment above. Better orbital overlap generally indicates more constructive, favorable interaction between the two atoms.

Let's consider a homonuclear diatomic bond (e.g. $\ce{H2}$). The qualitative understanding is similar for two different atoms, but the math from quantum mechanics is a bit easier with a homonuclear bond. So there's one bonding orbital ($\sigma$) and one anti bonding orbital ($\sigma^*$).

The energy of the orbitals will be:

$e_1 = \frac{H_{11} + H_{12}}{1 + S_{12}}$ $e_2 = \frac{H_{11} - H_{12}}{1 - S_{12}}$

You can put in some numbers and see what happens (e.g., as the overlap $S_{12}$ increases from 0 to 1, the bonding interaction becomes more and more energetically favorable).

N.B. There's a slight asymmetry - the anti bonding orbital is higher in energy than the bonding orbital is lower in energy.

• I’m not that advanced into quantum chemistry yet, so could you tell me where the formulas for the MO energies come from or direct me to a page where I can learn about them? I don’t know what your symbols mean. – lightweaver Jul 26 '16 at 4:44
• @lightweaver - sorry, just saw your comment. A simple model is Hückel, which assumes $S_{12}$ is zero, $H_{11}$ is a constant $\alpha$ and $H_{12}$ is a constant $\beta$. – Geoff Hutchison Jul 30 '16 at 1:56
• The different letters are simply integrals between the orbitals on atom 1 and atom 2. The equations come from the energies in the $\ce{H2}$ molecule. – Geoff Hutchison Jul 30 '16 at 1:57