TiO2 Solution

Structure Solution – Inorganics

Learning outcomes: We’ll use the information from indexing and Pawley fitting to try and solve the structure of TiO₂. A fairly trivial example but illustrates the methodology.

Information: Indexing/Pawley refinements give a=4.59468, c=2.95884 and space group P42nm. The refined zero point was 0.06548. We should remember that P42/mnm has the same extinction conditions. The input file for structure solution could be generated directly from the file used for Pawley fitting. Here we’ll create it from scratch.

1. Save the files above in your working directory.

2. Go through the menu for “Simple Rietveld Refinement”. N.B. if you’re already run a Rietveld refinement on TiO₂in this directory in an earlier tutorial you should either delete or rename the file called d5_05005_riet_01.inp. If not jedit will not let you create a new version.

3. Click on “Select Data File” and navigate to find d5_05005.raw.

4. Click on Instrument/Corrections and select “Durham_d5000_scint” and “Refine zero point”. You could change the zero point to 0.06548 and fix it by changing “zero” to “!zero”.

5. As we don’t know the structure click on “Structure – no cif”. Click on the first 3 icons and enter information as prompted. Use P42nm for the space group.

6. Click on “Lattice parameters” and “Tetrgonal”. Type in the cell parameters of a=4.59468 and c=2.95884 over the “#” symbols. You might want to delete the “@” symbols to stop the cell refining during structure solution.

7. Click on “site with beq” twice to get two atomic sites. Make the first one a Ti at (0, 0, 0) and the second one O at (0, 0, 0). At the end of each line type “rand_xyz 5”. This moves atoms randomly around the unit cell. The 5 scales how much they move. The atomic sites should look like:

site Ti x @ 0 y @ 0 z @ 0 occ Ti 1 beq @ 1 rand_xyz 5
site O x @ 0 y @ 0 z @ 0 occ O 1 beq @ 1 rand_xyz 5

8. We’re going to take a very simple approach of moving atoms randomly through the cell. We need to do something about special positions – i.e. what happens to multiplicities when an atom moves from a general x,y,z site towards a special position. Here we’ll take care of this by adding the lines below after the atomic coordinates:

occ_merge Ti* occ_merge_radius .6
occ_merge O* occ_merge_radius .6

9. Add the phrase “view_structure” at the bottom of the file to watch the structure during annealing.

10. Go to the top of the input file and in the section labelled “general information about the refinement” add a line saying “continue_after_convergence”.

11. “Save/Send this file to topas” then run in topas. In case you’ve made a mistake there’s a sample input file linked here.

12. You should be able to see a series of Rietveld refinements as atoms are moved randomly around the cell. Stop the refinement after a couple of minutes. You should get an Rwp around 15%. Stop the refinement and then run it again using the “step” button in topas. The structure you see should look recognisably that of rutile.

13. Examine the coordinates in the results file. I got:

site Ti x @ 0.50042` y @ 1.49394` z @ 0.68035` occ Ti 0.2610 beq @ 0.4297` rand_xyz 5
site O x @ 0.82094` y @ -0.79166` z @ -1.31724` occ O 0.5668 beq @ 0.5171` rand_xyz 5

14. It’s clear that both Ti and O are refining close to special positions. i.e. Ti is close to 0.5 0.5 z and O is close to x, x, z. Try the solution again with this symmetry imposed. i.e. change the atomic lines to those below and remove the “occ_merge” sections. Change occupancies accordingly. You might find it useful to put num_posns # at the end of the site line which tells you the multiplicity of the site:

site Ti x 0.50000 y 0.50000 z @ 2.54439` occ Ti 1 beq @ 0.5070` rand_xyz 5
site O x x_o 1.30629` y =x_o;:1.30629` z @ -2.94530` occ O 1 beq @ -0.1060` rand_xyz 5
' occ_merge Ti* occ_merge_radius .6
' occ_merge O* occ_merge_radius .6
view_structure

15. You should get essentially the same Rwp.

16. The true symmetry of rutile is actually P42/mnm – you should be able to see this from the refined coordinates and/or by viewing the structure in Topas (you should see a mirror plane perpendicular to c). Compare your structure solution with the earlier Rietveld refinements. Try solving the structure in this space group.

17. Add the phrase “Decompose(0.005)” to your input file immediately after the line containing the zero point information. Run the refinement again to compare the speed. This line converts the powder pattern to a “single crystal like” pattern in which you only retain peak heights not the full profile. Comment this line out to see the whole fit.

18. Try adding the phrase “Auto_T(10)” to the top of the input file. You can now delete the rand_xyz 5 lines from the sample coordinates. This applies a “default” set of simulated annealing shifts/temperature regimes. The scaling used is a scheme that Alan has found to be optimal for a variety of different systems. Some details are in the manual.

19. There are lots of structure solution examples in the “TOPAS-Academic, INP format” menu under “Structure Solution”.