# How many electrons and orbitals should I include in a CASSCF calculation?

I'm interested in running CASSCF calculations to get a dissociation curve for a number of systems of the form: atom-atom, atom-diatomic or diatomic-diatomic (e.g. Li+ ... O-, K+ ... O2, or N(3-) ... Ar). The two interacting species can be charged, but usually would not involve any radicals.

Is there a way to get the right number of electrons and number of orbitals to use in CASSCF from just knowing what atoms or diatomic species are involved?

P.S. I'm using Orca. The manual says to run an MP2 calculation and then "occupation numbers between 1.98 and 0.02 should be in the active space"; but also it seems like there is trial and error after that.

• This necessarily depends on your desired precision and computational budget. If it is not affordable to include all orbitals with MP2 natural orbital occupations in (0.02,1.98), it is not uncommon to include, say, only those orbitals with occupations in (0.1,1.9). Likewise, if you are aiming at very high accuracy, then you may probably want to include even more orbitals. Oct 25, 2022 at 7:37

This "rule of thumb" that you mentioned, in which the arbitrary limits of 1.98 and 0.02 are chosen for the occupation numbers, is likely meant for systems that are too large for the above strategy. For the systems that you described as examples in your post, it doesn't seem necessary to resort to such an arbitrary choice. For $$\ce{Li+-O-}$$ there's only 2 possible electrons coming from Li+, and at most 9 electrons coming from O-. If you were to include the 1s and 2s orbitals for Li+ and the 1s,2s,2px,2py,2pz orbitals for O-, then you have only 7 spatial orbitals in total. A CAS(11e,7o)SCF calculation is usually no problem on a computer from this century (it was fun to use the word "century" like this, but "this decade" or even "the last two decades" wouldn't be enough!). Next you can add the three p-orbitals for Li+ and the next s- and p-orbitals for O-, which would take you to a CAS(11e,14o) calculation, which is also easy and probably accurate enough for almost all purposes! Similar analysis will lead to the same conclusion for the other examples that you gave in your question.