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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.

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Hi Jonas,

CRYSTAL does not implement point groups with infinite-order rotations, so \(D_{\infty h}\) cannot be used directly. For linear molecules like CO₂, the practical approach is to approximate the symmetry using a finite-order rotation group.

In this case, \(D_{4h}\) (i.e., 24) is a good option in CRYSTAL. It preserves the key degeneracies of linear molecules, including the doubly degenerate bending modes of CO₂. Using a lower-symmetry group like \(D_{2h}\) would artificially lift these degeneracies.

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Hi Orion,
Looking at your input, I think you want to plot the orbital projected electronic DOS (in CRYSTAL this is done with the DOSS keyword, PDOS in CRYSTAL normally refers to phonon DOS).

To obtain separate px py pz projections, the correct approach is:
set the first parameter of DOSS to 2 (because you want two separate projections).
After the main DOSS line, you must then add one line per projection, specifying which atomic orbitals (AOs) belong to each projection.
Each line has the form:

n AOs    index_1    index_2    ...    index_n

The AO indices correspond to the ones printed in the CRYSTAL output. To find them, search in the SCF output for "LOCAL ATOMIC FUNCTIONS BASIS SET", you will see a table with all the basis set, that will look like this:

*******************************************************************************
   ATOM   X(AU)   Y(AU)   Z(AU)  N. TYPE  EXPONENT  S COEF   P COEF   D/F/G COEF
*******************************************************************************
  1 O   -2.793  -4.838  -4.110
                                 1 S
                                        8.589E+03 1.895E-03 0.000E+00 0.000E+00
                                        1.297E+03 1.439E-02 0.000E+00 0.000E+00
                                        2.993E+02 7.073E-02 0.000E+00 0.000E+00
                                        8.738E+01 2.400E-01 0.000E+00 0.000E+00
                                        2.568E+01 5.948E-01 0.000E+00 0.000E+00
                                        3.740E+00 2.808E-01 0.000E+00 0.000E+00
                          2-     5 SP
                                        4.212E+01 1.139E-01 3.651E-02 0.000E+00
                                        9.628E+00 9.208E-01 2.372E-01 0.000E+00
                                        2.853E+00-3.274E-03 8.197E-01 0.000E+00
                          6-     9 SP
                                       9.057E-01 1.000E+00 1.000E+00 0.000E+00
                         10-    13 SP
                                        2.556E-01 1.000E+00 1.000E+00 0.000E+00
                         14-    18 D
                                        1.292E+00 0.000E+00 0.000E+00 1.000E+00

The indeces are the numbers befor each orbital shell (1 2-5 6-9 10-13 14-18).
The notation 2-5 means indices 2 through 5, because that shell contains 4 AOs.
From these blocks you must identify the p-type orbitals for each atom (Zn and Cl). The p shells will appear in groups of three basis functions (px, py, pz).

So, in your system:

Locate the Zn atom in this printed basis list
Identify the block corresponding to its p orbitals
Extract the AO indices for px, py, pz
Do the same for Cl

Then you define one projection per line. For example, structurally something like:

NEWK
12 12
1 0
DOSS
2 100 3 6 1 12 0
3 <list of Zn p AOs>
3 <list of Cl p AOs>
END

Keep in mind that, if your basis is double- or triple-Z, each p shell appears as a separate triplet of AOs (one triplet per Z). To correctly project all p you must include the corresponding AOs from every p-shell triplet for that atom.

Let me know if you manage to do this, or if you need further help.

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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!

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As a general suggestion, I would recommend adding SMEAR, and possibly enabling DIIS, which can also be helpful. However, when using SPINLOCK, be careful, you should only turn DIIS on after SPINLOCK is disabled, so make sure to use THREDIIS instead.

You can also try increasing FMIXING to around 90 or 95, and using slightly higher TOLINTEG values.

Additionally, increasing SHRINK to 2 2 could be helpful. If the system goes through a metallic state, this adjustment can improve the Fermi surface (computational cost should not increase significantly).

The input file should be similar to this: INPUT.d12

As an additional note, targeting a high-spin state with all Fe atoms having +5 spin may not be physically reasonable for this type of MOF. Such a configuration could lead to severe instability in the electronic structure, which often manifests as SCF convergence issues. It might be worth checking whether a lower or mixed-spin configuration is more appropriate for the system.

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Hi Wim,

Could you please provide the geometry of your system? An output file would also be very helpful.

In the meantime, please have a look at this thread.

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Hi,

Jefferson Maul has managed to generate this .cif file with 4 symmetry operators. As you suggest, there are probably more but this is what he could extract so far.

Beautiful system by the way: looks like a Christmas tree bauble!

Screenshot 2025-11-04 alle 08.17.24.png

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Hi,

job314 said in SCANMODE io error Read_int_1d:

So it turns out having fort.13 and fort.20 is not optional for restart.

Yes, the lack of these two files was the origin of the I/O error.

job314 said in SCANMODE io error Read_int_1d:

PS. I would also like to ask developers to allow uploading compressed files to this forum

Now also .zip, .tar, .tgz, .tar.gz files can be uploaded.

Cheers,

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hello
about VLAB I'll ask to the developer...
As for the MULTIWALL nanotubes.. there is another procedure you can follow, which use the keyword MULTIWALL. These are the steps, using as a test system a graphene sheet:
1) build a two-wall nanotube ---> see 2wall.jpg

Title
SLAB
77
2.47
1
6 0.33333333 0.6666666667 0.000000
NANOMULTI
2
WALL
1
ROLLINGV
6 6
WALL
2
ROLLINGV
12 12
NANOJMOL
ENDWALL
[Basis Set]

use the MULTIWALL.DAT file as a fort.34 and restart with the EXTERNAL keyword substituting one of the two wall with different atom(s). In this case, the inner wall becomes a nitrogen wall, see ---> ext2wall.jpg

Title
EXTERNAL
ATOMSUBS
24
1 7
2 7
3 7
4 7
5 7
6 7
7 7
8 7
9 7
10 7
11 7
12 7
13 7
14 7
15 7
16 7
17 7
18 7
19 7
20 7
21 7
22 7
23 7
24 7
END
[Basis Set]

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the ZnO@GaN were done exactly in this way... if you roll up a two-layers slab you end with a double wall nanotube

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