Problem Set 6 - Wolfgang Seidl


1. For the determination of a unit cell one has to choose reflexes of good shape with no tailing, shoulders or broadening (means they should be approximately gaussian) and they should be of high intensity. Both these requirements hopefully lead to a choice of reflexes with most exactly d-values. If enough peaks are available it is also senseful to choose reflexes at higher angles for a better zero error alignment. But it has to be taken into account that cells with similar unit cell axes often lead to nearly gaussian but slightly broader peaks, which contain more than one reflex. If one isn` t aware of that the choosing can yield no result at all, execpt perhaps a misshaped triclinic cell.
I choose the follwowing values:
 
 

2 q
Int.
2 q
Int.
8.07
7727
21.64
685
10.25
3634
23.45
626
14.82
1731
24.27
1793
15.05
557
26.57
5337
16.90
448
26.91
1425
17.15
1861
27.72
5418
18.07
740
29.81
1054
18.81
2587
30.06
729
19.36
1031
30.27
1193
20.63
680
30.39
1246

What I got were loads of different triclinic cells with adventureous axis lengths and angles and FOMs at maximum of 10 for perhaps 16 to 18 peaks. For this I varied the zero error from -0.05 to 0.05 in steps of 0.01.
Having encountered this problem I rewrited a former *.pks file of a STOE file with all the observed reflexes, expanded the usually 20 accounted reflexes to 35 and did a trial on the Visser algorithm. And surprise, surprise, I got a monoclinic cell with 19 lines indexed and a FOM of 72 with a = 23.7259, b = 6.1715, c = 22.0614 Å, b = 97.90 and a C centering. That looked quite possible. So I simulated the cell with your hkl generating software and indeed I got nearly all reflections and even some of the "shoulder" peaks. Then as shot into the dark I put in the obtained data in the refinement program and it yielded a refined cell of a = 23.7286, b = 6.1711, c = 22.0644 Å, b = 97.9013 and a zero error of -0.0135. Now knowing the "exact" error I restarted the cell finding process and tumbled over an astonishing result. I couldn´t find the cell at all!
Finally I tried the reflexes you wanted to be refined, added the rest of peaks were I was absolutely sure after the indexing that they were correct up to the necessary twenty and set the zero error to -0.012, -0.013, -0.014 and -0.015. The obtained cell was approximately the same monoclinic one as mentioned above, but with a FOM varying from 60,4 to 20.5 to 10.3 to 20.5 where all twenty used reflexes could be indexed. I looked up the course manual and found that you stated that a FOM must be at least over 10 to give a possible solution, but not a right one!
 
 

2 q
Int.
2 q
Int.
7.50
67
18.07
740
8.07
7727
18.27
185
10.25
3634
23.45
626
11.77
104
24.27
1793
14.82
1731
26.57
5337
15.05
557
26.91
1425
16.09
199
29.81
1054
16.64
2838
30.06
729
16.90
448
30.27
1193
17.15
1861
30.39
1246

Therefore I ask two questions:

1. How am I supposed to find the right zero error of -0.0135 when the data are only exact to 0.01 and the FOM changes so drastically from -0.012 to -0.014?

2. Comparing the simulated reflexes from the refined cell with the experimental data it becomes obvious that a few of the reflexes are slighly shifted, especially the intense ones. Didn´t you get more precise data?

I would appreciate that one could take a personal look into the diffractogramm by zooming in and look were the reflex exactly lies. If this is not possible then please choose a better example or cancel the exercise! My suggestion is to give a good and easier example with slightly more symmetry (orthorhombic cell with centering) to exercise peak exclusion and a harder one just to see how tedious it can be to find a suitable, well fitting cell!
By the way, in my opinion it is much too late for this exercise!

Now having let out the steam that built up in my head I can continue answering the questions!


2. The cell a = 23.7259, b = 6.1715, c = 22.0614 Å, b = 97.90 and C centering  is simulated and closely compared to the experimental data. It is found that there remain a few reflexes in the experimental data with unneglectable intensities that can´t be assigned. Hence I suspected that the centering might not be right and tried a primitive monoclinic cell with half the c axis as above. And indeed all measured peaks could then be clearly assigned even for the shoulders a match could be found. Therefore I was sure that the compound has a monoclinic cell with a = 23.7259, b = 6.1715, c = 11.0307 Å, b = 97.90


3. As asked I checked for glide planes and screw axes.

Glide planes have zonal reflection conditions of the type 0kl, h0l, or hk0 with one of the two indices being an even integer. Searching through the table I found for h0l a large number of reflexes with even h, so that h0l: h = 2n is fulfilled which is a glide plane perpendicular to b. Found indices are: -201, 201, -401, 401, -202, 202, -601, 601, -203, 203, -602, 602.
Having found a glide plane it should also be possible to find a twofold screw axis/mirror plane. The mirror plane perpendicular to b should create pairs of hkl and -hkl which it does (see indices above and -311, 311; -312, 312 and so on) and the screw axis has the extinction condition h00: h = 2n which is also fulfilled with the observed hkl = 200, 400, 600, 800. It has to be mentioned that this is not an original 21 screw axis as it comes from the glide plane.


4. The reflection conditions now allow two different space groups: Pa (7) or P 2/a (13). Space group 13 has an inversion center. As it is mentioned that it is a naturally-occuring organic material I would suggest that the space group P 2/a (13) is the right one, respectively P 2/c (13) with a and c axes exchanged. The organic material is very likely to be a pure enantiomorphic compound which looses its enantiomorphic information by addition of hydrochloric acid and becomes racemic. A racemic compound is very likely to have a center of inversion in the crystal as then a really dense packing is possible. From the reflection conditions it can´t be surely stated which of the two possible solutions is the right one.


5. After using the refinement program the table is filled with the necessary data:
 

2q
  h
  k
  l
 
2q
  h
  k
  l
 
2q
  h
  k
  l
  7.50
 2
0
0 
 
14.82
 1
 1
0 
 
16.90
-2 
0 
2 
  8.07
 0
0 
1 
 
15.05
 4
0 
0 
 
17.15
1 
1 
1 
10.25
 -2
0 
1 
 
16.09
 -4
0 
1 
 
18.07
4
0 
1 
11.77
 2
0 
1 
 
16.64
 -1
1 
1 
 
18.27
3 
1 
0 
  a (Å) b (Å) c (Å) a (°) b (°) g (°)
Value 23.72855  6.17111  11.03219  90.0000  97.9013  90.0000 
ESD 0.00822 0.00189  0.00453  0.0000  0.0157  0.0000 

The refined zero error value is -0.0135 ° 2q, which could be better. Usually the diffractometer should be calibrated over a large range to the lowest possible zero offset. Especially for higher angles a small zero offset allows very precise data which suit for automatic indexing and refinement.

Lowering of the ESD´s with this sample is difficult due to the fact that at high angles a lot of peak overlap occurs, which creates problems in assigning the right 2q values and reflexes are very close together. Another problem arises from the dataset istelf. Having the 2q data only as axact as 0.01 one could hardly expect to get better ESD´s than round about 0.01. Comparing to the refined values the ESD´s are for this measurement and the given data not too bad!