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I'm trying to calculate the adsorption energies of various adsorbates on Fe2O3 surfaces (although, with there being many possible surfaces for Fe2O3 it is quite complex, but that's for another time...). I was wondering should I converge my adsorption energies versus the number of unit cells or add layers of atoms such that I am adding fractions of unit cells to the bottom of my slab? If its the case of unit cells, the Fe2O3 unit cell as a slab is already reasonably large, but doubling or even tripling it leads to it becoming unusably expensive for the resources I have available. Alternatively, adding atoms is less expensive but here we are breaking the stoichiometry of the unit cell.

What would you advise?

Here is my unit cell with

  • A/B lengths = 5.035
  • C length = 13.750.

Image of unit cell of Fe2O3

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    $\begingroup$ I dont have time to write full answer at the moment, but asymmetrical cells can be used if they are accounted for correctly, but in general it is very good for the cell to have inversion symmetry. It makes your analysis much easier. How large is the Fe2O3 surface unit cell? Can you include a picture to give us some context. $\endgroup$
    – Tristan Maxson
    Oct 16, 2020 at 15:59
  • $\begingroup$ Sure thing, the unit cell is here imgur.com/a/L3Cq9hs, the unit cell in terms of a,b,c is 5.03530, 5.03530, 13.74950. $\endgroup$
    – Charlie A
    Oct 19, 2020 at 14:54
  • $\begingroup$ Added it to the main question. I look at materials project and find the lowest energy cell to be smaller. Are you sure this cell doesnt either reduce down? This may not help when you go to actually make a surface though. materialsproject.org/materials/mp-19770 $\endgroup$
    – Tristan Maxson
    Oct 19, 2020 at 14:58
  • $\begingroup$ I grabbed this as a cif file off the ICSD. If I use a smaller starting unit cell this could certainly help converging the surface based on number of layers $\endgroup$
    – Charlie A
    Oct 21, 2020 at 14:24
  • $\begingroup$ What softwares are you using? I will try to help you get to an answer but there are lots of variables here. $\endgroup$
    – Tristan Maxson
    Oct 21, 2020 at 14:48

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