Geometry Editing

Poking further, since the last error is that of MPI and not of CRYSTAL, I launched the same input with NEWK 8 8 as a single-core process (with Pproperties), and this time the error was different:

85-C( 3 2 4) 86-C( 4 2 4) 87-C( 5 2 4) 88-C( 3 3 4) 89-C( 4 3 4) 90-R( 4 4 4) ERROR **** PROJVR **** NULL COMPONENT 0.222045E-15 0.100000E-07

for

ROTREF ATOMS 10 # this is Manganese 0 0 0 # I also tried this with different cell indices 12 # P5 0 0 0 8 # P2 0 0 0

and then

85-C( 3 2 4) 86-C( 4 2 4) 87-C( 5 2 4) 88-C( 3 3 4) 89-C( 4 3 4) 90-R( 4 4 4) ERROR **** RHOLSK **** BASIS SET LINEARLY DEPENDENT

for the following input:

ROTREF ATOMS 10 1 0 0 12 1 0 1 8 1 1 0

As well as for other indices

dmitoli Thank you, it was a formatting problem with my notebook, now the issue has been solved.

Best regards
Wang

If you fear that a reduced symmetry might influence the thermodynamic results, you can simply run the calculation without imposing any symmetry at all. For a 3-atom system like CO₂ the computational cost is so minimal that you can run both symmetry-free and \(D_{4h}\) calculations. Comparing the two results should give you an indication of whether the imposed symmetry has any impact on the quantities you are interested in.

Dear Jonas,

I tried using pymatgen to extract the point-symmetry information from your .xyz file (see the Python script below):

from pymatgen.core import Molecule from pymatgen.symmetry import analyzer bigstructure = Molecule.from_file("yourfile.xyz") PGstructure = analyzer.PointGroupAnalyzer(bigstructure) sym_mol = PGstructure.get_equivalent_atoms() print(sym_mol["eq_sets"])

This returns a Python data structure containing the symmetry-irreducible sets of atoms (only 6 for this system, which is indeed of Ih point group!).
When preparing the CRYSTAL input, be careful with the orientation of your asymmetric unit. In my case, for example, I had to change the sign of the x and y coordinates to make the symmetry consistent with CRYSTAL’s conventions.
Icosahedral point groups are available in CRYSTAL (Ih is point group number 47 in CRYSTAL), so the input fort this molecular cage reads:

Symm. structure MOLECULE 47 6 8 2.605032231 -11.914271806 11.762689798 6 4.344538598 15.664236912 4.366798884 6 3.427479683 14.428996321 8.326818862 6 -8.906580884 1.370529673 14.411150640 1 2.632960331 14.962480562 7.832453804 5 -8.776255231 -2.916256114 14.200279302 COORPRT TESTGEOM END

aerba Christmas is already in the air indeed!

Hi Giu, it worked!

Good! The clean way is:

CRYSTAL
1 0 0
P 21/A

I guess the extra "1"s you put are safe as they correspond to the identity operator.

GiacomoAmbrogio

Dear Ambrogio,

Thank you so much! the problem has been perfectly solved.

All the best,

wang

Thank you Giacomo Ambrogio.
I am greatful to the CRYSTAL code forum
This will help alot.

Best Wishes