# Physical reason for why delocalisation leads to stability?

Why does delocalization of electrons generally make compounds more stable (e.g. in carboxylate anions, where the lone pair on the negatively charged oxygen is delocalised into the C=O π* orbital)? Is there any proper explanation involving quantum mechanics?

• Electrons are always delocalised across the entire molecule according to quantum mechanics. It is just that the Lewis depiction of localised bonds is so much simpler to teach that we use it, and then we need resonance to overcome its weaknesses. – Jan Jun 3 '16 at 18:08

A simple QM explanation I can think of would invoke Hückel theory. It would go this way:

Suppose you have a two-state system (analogous to having two orbitals) with eigenvalues of the Hamiltonian $$\lambda_1$$ and $$\lambda_2$$. These correspond to the orbital energies. So just for argument's sake, let's say that

$$\begin{equation} \mathbf{H} = \begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix} \end{equation}$$

then obviously you have $$\lambda_1 = 1, \lambda_2 = 2$$. Let's say this system has two electrons. They would both go into the orbital with the lower energy, i.e. $$\lambda_1$$. In the context of organic chemistry, this would typically mean that the first orbital would be bonding in nature (hence filled), and the second orbital would be antibonding (hence empty), although that's not always the case.

Now, you perturb the system by making the non-diagonal elements of the Hamiltonian negative (analogous to allowing the orbitals to overlap, i.e. allowing delocalisation of electron density between the two). Just for argument's sake, let's change $$\mathbf{H}_{12}$$ and $$\mathbf{H}_{21}$$ to both be $$-0.1$$. If you recalculate the eigenvalues you get

$$\lambda_1 = 0.99, \lambda_2 = 2.01$$

The new eigenvectors will be linear combinations of the old eigenvectors – one in-phase and one out-of-phase combination. The "bonding" or in-phase combination of the two orbitals will have decreased in energy, whereas the out-of-phase combination of the two orbitals will have an increased energy. If you put both electrons into the bonding orbital, you get a net stabilisation. In general, interactions between filled and unfilled orbitals are stabilising in nature, since both electrons occupy a lower-energy bonding orbital.

On the other hand, interactions between two filled orbitals are destabilising, because you would have four electrons and you would have to fill both of the new orbitals. If you add up the two new eigenvalues, the total energy is greater than it was before when $$\mathbf{H}$$ was unperturbed.

This is not evident in my example above. The reason is that I made the assumption that $$\langle i | j \rangle = \mathbf{S}_{ij} = \delta_{ij}$$. In extended Huckel theory you would solve the generalised eigenvalue equation, $$\mathbf{Hc} = E\mathbf{Sc}$$. However with my assumption, $$\mathbf{S}$$ is simply the identity matrix and this reduces to $$\mathbf{Hc} = E\mathbf{c}$$. A theorem from linear algebra states that the sum of the eigenvalues of $$\mathbf{H}$$ is simply equal to the trace of $$\mathbf{H}$$. Therefore, as long as we assume $$\mathbf{S}_{ij} = \delta_{ij}$$, merely changing the off-diagonal values will not change the sum of the orbital energies.

In general, the basis vectors are not orthogonal. If you solved the full equation, $$\mathbf{Hc} = E\mathbf{Sc}$$ then you would find that the sum of the new eigenvalues is greater than the sum of the old, unperturbed eigenvalues.

Obviously this is ignoring other important aspects, e.g. electron-electron repulsions (that is a problem intrinsic to Huckel theory), but it's a start.

• Great explanation and easily understood! Thanks! – Jason Jun 3 '16 at 18:13
• Even simpler: Particle in a box. The longer the box, the lower the energy. Delocalization equals a longer box. However, this trite model is not quite as good as orthocresol's explanation. – Lighthart Jun 3 '16 at 21:19
• Actually, is it always true that delocalisation leads to stability? Consider the case of various antiaromatic compounds. In particular, let us look at cyclobutadiene. It exists more stably in its "rectangular form" with localised double bonds, rather than with electronic conjugation across the whole molecule. – Tan Yong Boon Feb 4 '19 at 4:25
• @TanYongBoon Would one describe an antiaromatic system as well delocalized? In cyclobutadiene, two bonds have constructive pi intereference, while two other bonds have destructive pi interference. I'm not an expert in it, but perhaps the instability is due in part precisely to less delocalization being available. – Blaise Oct 21 '19 at 7:39
• @Blaise I would like to disagree. In the "fully-antiaromatic" cyclobutadiene, the 4 $\ce {C-C}$ bond lengths tends towards being equivalent, favouring maximum delocalisation. The degree of delocalisation cannot be predicted by just looking at the MOs in the MO diagram. I think your description is also slightly incorrect. In the MOs of cyclobutadiene, two are nonbonding, one is bonding while one is antibonding. However, the degree of delocalisation cannot be inferred from the diagram. – Tan Yong Boon Oct 21 '19 at 8:29