# How to treat delocalized pi bonds in DFT

For a structure such benzene, the pi bonds between the carbon atoms are said to be de-localized. Therefore the electrons are expected to be exchanged in between individual C-C bonds.

When dealing with DFT calculations of such a structure, should there be any additional considerations to ensure the accuracy of the results?

• Several article you can find using ELF to define character of bond, here is one of them which can be useful ch.imperial.ac.uk/rzepa/blog/?p=1903 Dec 4, 2021 at 18:10
• I don't think any other considerations are required in most cases. The usual quantum chemistry codes that work with Hartree-Fock or DFT wavefunctions usually handle delocalized pi-bonds accurately. On the other hand, running a valence bond type calculation would require some careful tuning of the settings. This is because the usual Hartree-Fock type calculations consider all orbitals are delocalized (in the sense that they can have contributions from all atomic orbitals), and does not attempt to localize them. Dec 5, 2021 at 12:20
• @SRMaiti Your comment could likely be converted to a reasonable answer.
– Tyberius
Dec 6, 2021 at 18:57
• @SRMaiti do you think you might have time some day to follow-through with the suggestion by Tyberius? I think it would be nice to clear this from the unanswered queue if possible! Jan 20, 2022 at 22:32
• Let's give some time for SRMaiti to write an answer, and if you believe someone with more experience in the field is required for additional references, those can always be added in a second answer :) Jan 27, 2022 at 2:02

The most common form of the self-consistent field (SCF) calculations, which are used in DFT and Hartree-Fock, assumes that each molecular orbital is a linear combination of all the atomic orbitals (aka basis function), and that the molecular orbitals are orthogonal to each other. $$\psi^\mathrm{MO}_\mathrm{i} = \sum_\mathrm{p} c_\mathrm{pi} \chi^\mathrm{AO}_\mathrm{p}$$

So you start with $$n$$ atomic orbitals, and by taking linear combinations, you get $$n$$ orthogonal MO's. These molecular orbitals are called canonical orbitals. (Assuming there is no linear dependency in the AO basis set)

As the SCF procedure assumes from the start that each MO can have contributions from all of the AO's, the concept of delocalization is built into the procedure by default.

Therefore, when you do a DFT calculation on a molecule like benzene, no other considerations are required. If there is any delocalization, it will be treated by the method.

For example, have a look at the HOMO-1 orbital of butadiene (which is supposed to the lowest $$\pi$$-orbital, which all of the 2p-orbitals of carbon overlapping constructively) from a B3LYP/6-31+G(2d,p) calculation: It is quite clear that the delocalization is accounted for.

As an aside, I want to add that even if you start with localized orbitals, the canonical wavefunction can be obtained by taking linear combination of the various different ways you can localize the orbitals. This is something I learnt in my quantum chemistry lectures, and the math works out, but it is difficult to have an intuitive sense of why this works. So for example you can write the structure of benzene in two Kekule forms, and it turns out that the canonical wavefunction of benzene is simply a linear combination of both - benzene exists as a superposition of the Kekule-type localized wavefunctions.

There can only be one set of canonical MO's that satisfy the orthonormality condition, but there can be multiple way to localize the MO's. There are actually some exotic quantum chemistry methods (such as ALMO), which actually start with localized wavefunctions (determined for specified fragments of the system), and then builds up the total wavefunction from those fragmented localized MO's. This works quite efficiently for large systems such as water clusters, because there is no delocalization, so electron-electron interaction only happens locally.

• +1 and welcome to the 5000 points club! Jan 28, 2022 at 22:20
• "So you start with n atomic orbitals, and by taking linear combinations, you get n orthogonal MO's." Only if the basis set has no linear dependencies, which is the case only in very small (minimal) basis sets. Feb 1, 2022 at 23:17
• @SusiLehtola I thought QM softwares usually remove linear dependencies before SCF starts? I wasn't thinking about linear dependency when writing the answer, but I guess you are right. Feb 2, 2022 at 12:46
• @SRMaiti exactly, which is why the number of MOs is generally smaller than the number of AOs. Feb 2, 2022 at 17:36