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:
 
 

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!
 
 

Comment: It is important to choose the peaks with the largest d spacings as the peaks of first choice. If you take the first 20 given, you won't succeed. (It wouldn't be much of an exercise if it was that simple?) However, if you now leave in the low-angle weak peaks, but leave out the higher-angle weak peak at 20.84°, it indexes first time. (You don't need to change the diffractometer zero to make this work OK.) As it happens, the peak you leave out is genuine, but this behaviour is quite typical of indexing programs...

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?

Reply: Indexing programs should be able to cope with a systematic zero error of less than 1 step width (assuming the step is small), and they should be able to cope with random errors of around 0.01°, so the precision of the data should be sufficient.

This particular problem is an example of zone dominance mentioned in the "pitfall" section on indexing. This does make this assignment harder than the rest, but it also allows us to distinguish between students that are excellent, good, or simply average in contrast to some of the other assignments.

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?

Reply: The data was actually collected by a new PhD student in the group. More precise data could have been obtained using a silicon wafer to avoid sample penetration completely. This would have given more symmetric peaks and a zero error even closer to zero.

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!

Additional comment: It may be that this particular example is a pathological case with respect to zone indexing. I used it because it illustrates certain points with respect to space-group determination that I think are useful as regards teaching. It was also real data. I could generated an example with less problems, but then I would have only 2q data and no diffraction pattern to show (unless I create that too).

I accept your points, and the exercise will be improved by giving much stronger hints with regard to the first step. If this fails, then I will have to find (or create) better data. Having said this, at least you have seen how frustrating indexing can be, though this was not one of the aims of the exercise.

Having the assignment closer to the relevant course sections would be ideal. I will be investigating what is possible in practice here. Many thanks for the feedback.

3 Marks

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

Really excellent: well-spotted regarding the doubled c axis obtained in Q1! (You were unlucky to get this, but it did result from your choice of peaks - the FOM and density of reflections should also tell you that the initial cell was too large, unless of course it was centred.)
5 Marks

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.

Comment: Excellent with regard to searching for glide planes, except that you should never consider even the zonal reflections 0kl and hk0 for monoclinic (y axis unique).
4 Marks

After finding the glide, the only other symmetry element that may be present is a 2 or 2(1) axis, for which you need to look at 0k0 and not h00.
0 Marks

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.

Comment: Excellent. Given the error in Q3 above, you get full marks for the deductions made here. You were very close to the correct answer in any case. (See model answer.)
5 Marks

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

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!

Comment: In the table of cell parameters, you should give the precision appropriate to the esd, i.e. 3 decimal places for cell length and 2 decimal places for cell angle beta. Cell angles that are precisely 90° by symmetry should be given as such (i.e. no decimal place and no esd value at all since 0.0000 could imply that the error is simply less than 0.0000, which is different to no error.

See model answer for comments on zero error and high angle data.
6 Marks