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How does one obtain the value of dispersion coefficient $C_6$ for elements that come after Xe. I am going through the original paper [1]. I can see that: $$\tag{1}C^a_6 = 0.05N(I^a \alpha^a),$$ where $\alpha^a$ is the dipole polarization and $I^a$ is ionization potential. This does not seem that complicated but in the subsequent paragraphs they tend to complicate things. Where they are talking about if the element in the compounds vary a lot from free elements then we need to derive coefficients from molecular properties and there is quite a lot of other things too.

Is there any simpler way to go about doing this?

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2 Answers 2

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There's many ways to calculate a $C_6$ value, for which it doesn't matter whether the element is heavier or lighter than Xe.

One of the simplest equations for $C_6$ is in the abstract of a 1931 paper by Slater and Kirkwood (often called the Slater-Kirkwood approximation):

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Therefore we have (in the notation of Brandt in 1955, and probably others between 1931-1955):

$$\tag{1} C_6 = -\frac{3}{4}\left(a_0N\alpha^3\right)^{1/2}e^2, $$

with $a_0$ being the Bohr radius, $N$ being the number of electrons in the molecular valence shell, $\alpha$ being the average dipole polarizability, and $e$ being the charge of an electron.

For a heteronuclear system another simple formula is given here:

$$\tag{2} C_6^{AB} = -\frac{3}{2}\frac{I_AI_B}{I_A+I_B}\alpha_A\alpha_B, $$

with $I_A$ and $I_B$ being the first ionization energies for the systems $A$ and $B$, and $\alpha_A$ and $\alpha_B$ being again their average dipole polarizabilities.

You are correct that other formulas exist, for example, I found at least two in this paper:

enter image description here

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I tried to turn the question What are the different ways of calculating dispersion constants? into a list of all the known methods to approximate $C_6$, but I wanted it to be in the form of with each user explaining just one method (otherwise you might get 10 answers from me, which I think would look weird).

Beyond the element Xe, relativistic effects will become quite important when attempting to calculate the dipole polarizability $\alpha$ or the ionization energy $I$ using ab initio methods. Another famous relativistic effect was described in this 1948 paper by Casimir and Polder, in which they showed that for very large distances the $C_6$ actually becomes a $C_7$.

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  • $\begingroup$ Thanks for your answer, this will be really helpful. Could you please also refer me to some resource which shows how to calculate dipole polarizability and ionization energy using DFT. I am actually super new to dispersion correction so I am not sure what I am doing at the moment $\endgroup$ Jun 9 at 20:28
  • $\begingroup$ @ParmeetSinghEP066 I'm glad my answer will be helpful to you. I recommend that you ask a separate question if you would like to calculate the dipole polarizability (and that you don't restrict it to just DFT, since there's better ways to get dipole polarizabilities). As for ionazation energy, have you thought about this yourself a little bit? $\endgroup$ Jun 9 at 20:35
  • $\begingroup$ Ionization energy is defined as the energy required to create an ion from a neutral atom. Now I am a bit confused why do I have to run a dft calculation for this? I am assuming this ionization energy will somehow depend upon my system. I will also create another post for this $\endgroup$ Jun 9 at 21:20
  • $\begingroup$ It's you that mentioned DFT. I don't now the reason for that. If you're dealing with a diatomic molecule, then the $I_A$ and $I_B$ are the ionization energies of the two atoms in the diatomic, and therefore you can find them online (without doing a DFT calculation). The more details you provide, the better people can help you. $\endgroup$ Jun 9 at 21:32
  • $\begingroup$ Sorry, my bad @NikeDattani! I actually have a big double perovskite system so would the value for my Ionization energy and dipole polarizability depend upon my system or would they be same as that I could find for individual elements in some experimental research papers ?? $\endgroup$ Jun 9 at 21:39
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You already have a potentially good answer from Nike Dattani, however I might frame this as an entirely different question. Unless you have a good reason to determine you own dispersion constants / corrections, can you just use an existing dispersion correction such as DFT-D4 (rather than D2). Dispersion should work with any element on the periodic table, as of this commit.

You can interface this onto any calculator using the ASE-D4 calculator if you need to. This might solve your issue from a practical standpoint, even if this isn't your original question.

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