I found your comments on generating Gaussian basis sets for pesky atoms like Iodine very helpful. It makes me suspect you would understand the answers to some long standing questions that have been in my mind.

Q1) I am interested in how charges move around in CH2I as a function of the vibrational excitation of the symmetric and asymmetric CH stretches. I run the calculation in C2v symmetry to find the equilibrium (planar) geometry and the vibrational normal modes (which I do in high precision with the keyword hpmodes). This generates x,y,z displacements of each atom to 5 sig figs, which I can then easily use to find the geometry of a vibrationally excited radical with say Q = 0.1. Why shouldn’t the sum of mass weighted displacements of the atoms for any vibrational mode add up to zero? (I am using the masses stated in the program, good to 8 digits)

Q2) If my reference geometry is such that the z axis is along the CI bond and the xz plane contains the HCH atoms, why is it that my normal mode displacements are sometimes not exactly symmetric (to 5 digits at least)?

Thanks!

David

2) Yes. always. ]]>

Thank you for your comments. In between coordinates for your QST3 or QST2 you have to add a line with a comment, typically indicating which structure belongs to products, reagents and the proposed TS. Maybe in your IRC you are missing the comment line or you lost a hydrogen atom somewhere along the way (that has happened to me before).

I hope this helps!

]]>I was going to reply that I don’t think that is a formal comparison since the number of electrons is different, but then again you may compare population changes in the C atom bearing the halogen and see how it changes. So, yes, I think you can do it and in fact you have given me an idea I needed for my own work. Thanks!

Have a nice day

]]>I hope this helps.

]]>Indeed we could label NBOs the way you say in your first paragraph; I haven’t seen that ever and maybe there is a reason but I don’t see any reason against that labeling system.

Regarding the equivalence of MOs and NBOs the answer is no. NBOs come from the re-orthonormalization done on the basis of the chosen basis set, hence MOs and NBOs are constructed diferently and should not be compared. However, they should give rise to the same electron density.

I hope this helps

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