Member since May 2018 · 7 posts
Group memberships: Members
Subject: How to constrain atoms part of a rigid body to special positions?
Hello,

I am dealing with the Rietveld refinement of a metal-organic framework which has cubic symmetry (Fm-3m) and contains tetrafluoroterephtalic acid as a linker.
The linker is defined by five crystallographically independent atoms (named O2, C1, C2, C3 and F), three of which should sit in special positions: O2 (x, 0, z), C1 (x, 0, x) and C2 (x, 0, x).

I have set up the following rigid group:

rigid
z_matrix  C2
z_matrix  C3    C2    1.39
z_matrix  F      C3    1.35        C2    120
z_matrix  C1    C2    1.48        C3    120    F  0
z_matrix  O2    C1    1.30        C2    120    C3  = torsion;

prm torsion  43.27588` max 90 min 0

prm Linkrot1 -179.98276`_LIMIT_MIN_-180 max 180 min -180
prm Linkrot2 -180.00000`_LIMIT_MIN_-180 max 180 min -180
prm Linkrot3  14.92436` max 180 min -180

With such a rigid group, the position of C2 is constrained to be the following: Linktrans1, 0, Linktrans1

The problem arises when I try to constrain the y coordinate for both O2 and C1 to be 0 and the x and z coordinates for C1 to have the same value.
I have tried to create and refine parameters, but these seem to conflict with the rigid body.

Is there a way of getting around this issue?

Thanks,
Marco
User title: John Evans
Member since Aug 2009 · 190 posts
Dear Marco,

I haven't looked at your problem properly.  However in general when you have rigid bodies on special positions you're not free to refine all the rotation angles or translations.  e.g. if you had an octahedron with it's centre at a 4/m site (4 along z) then you couldn't refine xrot or yrot as it would move atoms off the mirror plane.  Is your problem related to that?

John
AlanCoelho #3
Member since Aug 2011 · 256 posts
Group memberships: Members
Marco

You can check num_posns to see of the symmetry has been broken by the rigid body. Note, the program will generate equivalent positions depending on what is refined. Thus he following is not a special position:

site x@ 0 y @ 0 z @ 0 ...

If the symmetry is broken then occ_merge may be able to help where a site just about to split will still have the correct overall occupancy.

alan
Member since May 2018 · 7 posts
Group memberships: Members
Dear John and Alan,

I have been trying to get my head around the issue building upon your suggestions, but so far I haven't managed to find a convenient and reliable way to refine my crystal structure.

Regarding Alan's comment, my initial question was exactly about the issue that the coordinates of the atoms belonging to a rigid group are automatically refined when the rigid group undergoes translations and rotations. As far as I understood, there isn't much that I can do to play with them at the site level and try to get them to stick to special positions by e.g. constraining x and z to have the same value.

Following John's suggestion, I have tried to play around with rotation angles and figure out whether I could fix them to some straightforward values or find some relationship between them that would allow to reduce the degrees of freedom, but with little success.

Besides the difficulty of forcing the atoms belonging to the rigid group to obey the restrictions imposed by symmetry, the refinement is made even more tricky by the fact that the aromatic ring and the carboxylic groups in my linker do not lie on the same plane, so that the carboxylic groups lie on a mirror plane and the atoms belonging to the aromatic ring are displaced over two positions, above and below the mirror plane.
I have tried to use rigid groups containing either 5, 4 or 3 atoms, but in each case I end up having issues that prevent me from getting a nice solution.
Member since May 2018 · 7 posts
Group memberships: Members
Hi,

I finally managed to get an acceptable refinement.
Going through the details without having visual support is not very helpful, but if anyone reads this and is interested in knowing more, I can share the .inp file and answer any question (or at least try to answer).

Marco
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