# Calculating nearest neighbor distances

For cubic materials, there are equations in which to calculate nearest neighbor (NN), second nearest neighbor, etc. Can this be done with tetragonal crystal structures? I want to calculate NN, 2NN, and 3NN of $\ce{TiO2}$ rutile with a tetragonal crystal structure but am unsure how to do it. Any suggestions/help would be appreciated!

• There is no equation for cubic materials. What you probably have in mind is the formula for the simple cubic structure having one atom per unit cell. Rutile is very much not like that. True, it is tetragonal, but to say that is the same as to say nothing. You will need atomic coordinates, symmetry elements, and on top of that, some textbook on crystallography. Then (and not before) you'll have your first distances, etc. – Ivan Neretin Oct 10 '17 at 15:22
• @IvanNeretin Could I just use coordination analysis to create the radial distribution function and take the first peak for 1NN, second peak for 2NN, etc? – Jackson Hart Oct 10 '17 at 15:27
• It depends on what do you mean by coordination analysis. – Ivan Neretin Oct 10 '17 at 15:29
• @IvanNeretin If I create the RDF of my material, it is sufficient to use that to calculate the NN distances right? – Jackson Hart Oct 10 '17 at 15:30
• @IvanNeretin So I took your advance and looked at the crystal. I see that 1.94 and 1.98 peaks correspond to O-Ti, 2.5 peak is O-O, 2.7 peak is also O-O, 2.9 peak is Ti-Ti. I do not have a lot of experience using NN. How would you go about choosing what cutoff distance to use? – Jackson Hart Oct 10 '17 at 16:15

First, you can obtain CIF-file from COD, then load it with Olex2 (free, available on Windows, Linux, MacOS) and execute command envi <r>, which will print a list of the atoms about special position within a sphere of radius $r$. For example, envi 3 ($r = \pu{3 Å}$) yields in the following assigned pairs:

    O    SYMM  Ti    Ti    Ti    O     O     O     O     O     O     O     O     O     O     O
Ti  1.95 1_554 -     -     -     -     -     -     -     -     -     -     -     -     -     -
Ti  1.95 I     98.8  -     -     -     -     -     -     -     -     -     -     -     -     -
Ti  1.98 2_544 130.6 130.6 -     -     -     -     -     -     -     -     -     -     -     -
O   2.54 4_454 49.4  49.4  180.0 -     -     -     -     -     -     -     -     -     -     -
O   2.78 1_544 93.4  150.2 44.5  135.5 -     -     -     -     -     -     -     -     -     -
O   2.78 1_545 150.2 93.4  44.5  135.5 64.4  -     -     -     -     -     -     -     -     -
O   2.78 3_655 150.2 93.4  44.5  135.5 89.1  54.3  -     -     -     -     -     -     -     -
O   2.78 3_654 93.4  150.2 44.5  135.5 54.3  89.1  64.4  -     -     -     -     -     -     -
O   2.78 1_554 45.5  96.2  117.2 62.8  111.5 159.1 106.5 73.5  -     -     -     -     -     -
O   2.78 I     96.2  45.5  117.2 62.8  159.1 111.5 73.5  106.5 64.4  -     -     -     -     -
O   2.78 3_554 45.5  96.2  117.2 62.8  73.5  106.5 159.1 111.5 90.9  125.7 -     -     -     -
O   2.78 3_555 96.2  45.5  117.2 62.8  106.5 73.5  111.5 159.1 125.7 90.9  64.4  -     -     -
O   2.96 2_543 40.6  139.4 90.0  90.0  57.8  122.2 122.2 57.8  57.8  122.2 57.8  122.2 -     -
O   2.96 2_545 139.4 40.6  90.0  90.0  122.2 57.8  57.8  122.2 122.2 57.8  122.2 57.8  180.0 -


But this is only based on our guess for allowed $r$ value and is rather speculative. Answering your secondary question from the comments – "How would you go about choosing what cutoff distance to use?" – I would suggest to use Voronoi-Dirichlet (VD) polyhedra set [1] for every crystallographically inequivalent atom (e.g. for Ti and O pair in this case). You can do it with the ToposPro (Free, available on Windows), using Dirichlet subroutine.

### Titanium:

Central atom:Ti1 0.500 0.500 0.500 Rsd:1.220
Atom:1.949 < r < 3.487  <r>=2.570   Top: 1.621 < R < 1.785  <R>=1.742
CN=6:0:4 NV=16 V=7.607/20.812 S=23.299 Cpac=0.509 Ccov=3.131
G3=0.083992444
Face distribution: {3/4 6/6 }
Vertex distribution: {3/16 }

N  Atom   x      y      z     Dist.  SAng.
1  O  1  0.3048  0.3048  1.0000  1.94857  16.77659
2  O  1  0.6952  0.6952  1.0000  1.94857  16.77659
3  O  1  0.3048  0.3048  0.0000  1.94857  16.77659
4  O  1  0.6952  0.6952  0.0000  1.94857  16.77659
5  O  1  0.1952  0.8048  0.5000  1.98000  16.44522
6  O  1  0.8048  0.1952  0.5000  1.98000  16.44522
*7  O  1 -0.1952  0.1952  0.5000  3.48704   0.00080
*8  O  1  0.8048  1.1952  0.5000  3.48704   0.00080
*9  O  1  0.1952 -0.1952  0.5000  3.48704   0.00080
*10  O  1  1.1952  0.8048  0.5000  3.48704   0.00080


VD-polyhedra for Ti atom with NN denoted with the bonds:

### Oxygen:

Central atom:O1 0.305 0.305 0.000 Rsd:1.413
D(CP):0.050  ( 0.2971 0.2971 -0.0000 )
D(VDP):0.069  ( 0.2941 0.2941 -0.0000 )
Atom:1.949 < r < 3.487  <r>=2.788   Top: 1.621 < R < 1.820  <R>=1.747
CN=6:8:4 NV=32 V=11.805/52.029 S=28.544 Cpac=0.328 Ccov=2.138
G3=0.085774387
Face distribution: {3/4 6/14 }
Vertex distribution: {3/32 }

N  Atom   x      y      z     Dist.  SAng.
1  Ti 1  0.5000  0.5000 -0.5000  1.94857  16.77659
2  Ti 1  0.5000  0.5000  0.5000  1.94857  16.77659
3  Ti 1  0.0000  0.0000  0.0000  1.98000  16.44522
4  O  1  0.6952  0.6952  0.0000  2.53648   3.12774
#5  O  1  0.8048  0.1952 -0.5000  2.77800   4.78494
#6  O  1  0.1952 -0.1952 -0.5000  2.77800   4.78494
#7  O  1  0.8048  0.1952  0.5000  2.77800   4.78494
#8  O  1  0.1952  0.8048 -0.5000  2.77800   4.78494
#9  O  1 -0.1952  0.1952 -0.5000  2.77800   4.78494
#10  O  1  0.1952  0.8048  0.5000  2.77800   4.78494
#11  O  1  0.1952 -0.1952  0.5000  2.77800   4.78494
#12  O  1 -0.1952  0.1952  0.5000  2.77800   4.78494
*13  O  1  0.3048  0.3048 -1.0000  2.95870   1.71934
*14  O  1  0.3048  0.3048  1.0000  2.95870   1.71934
15  O  1  0.6952 -0.3048  0.0000  3.32530   2.57704
16  O  1 -0.3048  0.6952  0.0000  3.32530   2.57704
*17  Ti 1  1.0000  0.0000  0.0000  3.48704   0.00080
*18  Ti 1  0.0000  1.0000  0.0000  3.48704   0.00080


VD-polyhedra for O atom with NN denoted with the bonds:

Briefly, we are looking for interactions that are not labeled with # or *; to sum it up, there are the following sets of NN:

• 6 NN for Ti: 4×O@1.94857 Å, 2×O@1.98000 Å;
• 6 NN for O: 2×Ti@1.94857 Å, 1×Ti@1.98000 Å, 1×O@2.53648 Å, 2×O@3.32530 Å.

### Reference

1. Blatov, V. A.; Shevchenko, A. P.; Serenzhkin, V. N. Acta Cryst A, 1995, 51 (6), 909–916 DOI: 10.1107/S0108767395006799.
• I really appreciate this detailed answer of calculating the different NN distances. Is there a general consensus on which one of these to use for cluster analysis of a given property? Each value will produce vastly different results. – Jackson Hart Oct 10 '17 at 16:52
• Also In this case, the 2nd nearest neighbor for Titanium would be 1.98, and for Oxygen would be 2.53 correct? – Jackson Hart Oct 10 '17 at 16:58
• I always relied on VD-approach for this matter. It provides algorithmic, descriptive and unambiguous result. – andselisk Oct 10 '17 at 16:58
• @JacksonHart Oh, my bad, I posted wrong number in conclusion for O (already fixed, listing was correct). It is 1.98 both for Ti and O. – andselisk Oct 10 '17 at 17:01
• I am not sure that VD would work in my case. First, I have to use WS analysis to calculate vacancies, and then I want to determine the size of clustering of vacancies. So I need to load in the MD dump file, perform WS analysis, and then specify the cutoff distance between vacancies, which would form a void. – Jackson Hart Oct 10 '17 at 17:01