# MO Diagram “greatest stabilisation when orbitals are close in energy”

I was studying MO diagrams in Chemistry and we were given a diagram explaining this: I understood the diagram furthest to the left describing the situation where the Electrophile (E) and Nucleophile (Nu) result in the largest energy gain when both are equal in energy.

However, the lecturer later said that when there is a large difference in energy between E and Nu (far right), the energy gain is negligible. He didn't go further to explaining why this was true.

I tried looking on the internet for an answer but was led to s-p mixing and content I am not familiar with nor find relevant to the question.

I'm confused here. What explains this phenomena? Or, is it just an effect observed without explanation - like a Postulate?

## 2 Answers

Fundamentally it's mathematics. The energies of the MOs are the eigenvalues of the matrix $$\begin{pmatrix}\alpha_1 & \beta \\ \beta & \alpha_2 \end{pmatrix},$$ where $$(\alpha_1,\alpha_2)$$ are the energies of the two constituent orbitals and $$\beta$$ is (loosely speaking) the overlap between them.

For the same size of $$\beta$$, if $$\alpha_1 \approx \alpha_2$$, then the eigenvalues can differ quite a lot from $$(\alpha_1,\alpha_2)$$. On the other hand, if $$\alpha_1$$ is very different from $$\alpha_2$$, then the eigenvalues will be closer to $$(\alpha_1,\alpha_2)$$.

As an illustration consider the matrix

$$\begin{pmatrix}1.1 & 0.5 \\ 0.5 & 1 \end{pmatrix}$$

This has eigenvalues of $$1.55$$ and $$0.55$$, which are relatively distant from the "original energies" of $$1.1$$ and $$1$$. On the other hand, the matrix

$$\begin{pmatrix}3 & 0.5 \\ 0.5 & 1 \end{pmatrix}$$

has eigenvalues $$3.12$$ and $$0.88$$, which are closer to $$3$$ and $$1$$.

Taking a more abstract perspective, the eigenvalues of $$\begin{pmatrix}\alpha_1 & \beta \\ \beta & \alpha_2 \end{pmatrix}$$ are

$$\lambda_\pm = \frac{\alpha_1 + \alpha_2}{2} \pm \frac{\sqrt{(\alpha_1 - \alpha_2)^2 + 4\beta^2}}{2}$$

and you can play around with this expression to gain some insight. For example, in the limit where $$(\alpha_1 - \alpha_2)^2 \gg 4\beta^2$$ (corresponding to a large energy difference between the original interacting orbitals), the eigenvalues reduce to

$$\lambda_\pm \to \frac{\alpha_1 + \alpha_2}{2} \pm \frac{\alpha_1 - \alpha_2}{2}$$

which are simply $$\alpha_1$$ and $$\alpha_2$$.

• Perhaps this is overly complicated. The math is clear for advanced level chemistry, but if @BobSmith doesn't understand what happens when orbitals mix, maybe he needs to first grasp the concept underneath the math. He will probably be lost wondering why you do the calculations that you do. – chemicalromance Feb 4 '19 at 23:57
• These formulae do not seem to show that the antibonding MO is more antibonding than the bonding MO is bonding. I suppose it is because you are not working with the rigorous physics here and you are just using simpler mathematics to illustrate the point that is relevant to this question? – Tan Yong Boon Feb 4 '19 at 23:57
• @chemicalromance I agree with your point, but I make no apologies for writing what I have written. The nature of SE allows multiple people to write multiple answers and the asker may accept anything they deem to be most useful. This may not be directly helpful to OP, but it may be useful to somebody else with a different background. Unfortunately, I don't really have the time or desire to explain Hückel theory from the ground up. The point is: if you think you have a more helpful approach, please post it (indeed, I see you already have!). – orthocresol Feb 5 '19 at 0:13
• @TanYongBoon In essence, yes, it is a simplification. It is explained here in greater detail chemistry.stackexchange.com/a/52160/16683 – orthocresol Feb 5 '19 at 0:15
• @orthocresol So the reason is that you are employing simple Huckel theory which does not take into account the overlap integral S? – Tan Yong Boon Feb 5 '19 at 0:21

You can think (in a gross simplification) of an orbital as an orbit, a circular path where electrons go to feel the attraction of the nucleus. If the attraction is big, then the orbit is small and the energy is low. If the attraction is low, the orbit is larger and the energy rises.

Now, when two atomic orbitals (AO from now on) mix, the electrons switch from feeling the attraction of just one nucleus to being attracted by two. That is, they "jump" to new orbits called molecular orbitals (MO from now on), one bonding (lower energy) and one antibonding (higher energy). The bond will be stronger when the overall energy of all the electrons is lower.

When two AOs mix:

• If both AOs have the same energy, then they go from the same initial state to the same final state and the electrons are shared equally between the atoms. You can see that in the leftmost diagram.

• If one AO is lower in energy (Nu in this example), then the attraction from the nucleus of this atom will be greater even when combined to form MOs; the electrons in the bonding MO will be shared but they will "spend more time" closer to the Nu atom. Thus we draw the bonding MO slightly closer to the Nu AO, and the antibonding MO closer to the E AO, as evidenced by the diagram in the middle. As the difference in energy increases, we move the MOs even closer in energy to the AOs because the Nu atom attracts the electrons even more strongly, like the diagram to the right.

Now, in the special case of a nucleophile bonding with an electrophile, the electrophile's AO is empty while the nucleophile's is full (again an oversimplification1). As all the electrons are there, the initial state has an overall energy determined by the energy of the nucleophile's AO.

When the bond forms, all the electrons occupy the only bonding MO. Then, the final state has an energy determined by the energy of the bonding MO.

Hence, the bond strength is determined by the overall difference in the energy of the electrons in the initial and final states, which in this diagram1 is directly proportional to the energy difference between the Nu AO and the bonding MO (confusingly labeled "energy gain" in the graphic, even though the overall energy is lower). So, for equal energies of the starting AOs, this so-called "energy gain" is biggest.

1 Often there are more than one AO and more electrons involved.

2 This varies wildly from molecule to molecule, because not every atom has the same electronic configuration.