[molpro-user] basis set definition

[Gerald Knizia]

On Mon, 2012-02-06 at 02:13 -0800, Vera Cathrine wrote:
> The problem is that I couldn’t understand from the the manual how to
> define my specified basis set for metal in input file (see the
> following)

Unfortunately, the syntax you chose does not work (but maybe something
like that should). The way this can be specified is this:

basis = {
  default,6-31g++(d,p)
  !
  ! MANGANESE (14s,11p,6d,3f) -> [8s,6p,4d,1f]
  ! MANGANESE (14s,11p,6d,3f)->[8s,6p,4d,1f]
  s, MN , 243694., 35995.0, 8223.56, 2353.12,
780.965,288.519,115.701,49.1175, 16.0885, 6.70430,
1.80517, .703011, .106385, .039616
  c, 1.6, .00029, .00225, .01152, .04559, .14039, .31472
  c, 7.8, .40974, .21079
  c, 9.9, 1
[...]
}

Note the "default" line.

> I use 6-31g++(d,p) for other atoms.

You might want to consider using cc-pVnZ or def2-SVP/TZVPP/QZVPP basis
sets, which are better optimized (the former for correlated
calculations, the latter for both DFT/HF and correlated calculations).
Molpro can deal with the so called "general contractions", so using such
sets is not slower than the Pople-type sets (in fact, for the same
quality these are faster because they have spherical instead of
cartesian components).

> I have switched off the symmetry, my system has charge +2 and spin
> multiplicity of 6 (high spin complex). Can I define my wave function
> in this open shell system uniquely just by supplying charge and
> multiplicity without supplying the symmetry and occ and open
> variables?

This will often work even if the wave function is not uniquely defined.
In this case Molpro guesses initial occupations. In the molecule
genuinely has no symmetry (like this), degeneracies do usually not occur
so the default "wf,charge=xx,spin=xx" should do the trick.
-- 
Gerald Knizia