Unit 5.1 - The Space group P2(1)/c and the Asymmetric Unit

Unit 5.1 of our course The Fascination of Crystals and Symmetry
Additonal resources at: crystalsymmetr...
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[@olliebudd8098]

You sorted my degree out mate cheers

[@moonfirerankeight4918]

Hello mister Frank. I watched most of your videos and I would like to say thank you because it helped me understand crystallography a lot better.

[@FrankHoffmann1000]

Hello Moonfire! Thank you very much! I am glad that this course was helpful for you!

[@3mro49]

Hi Frank. I was wondering if you have work sheets related to the topic to examine ourselves! Thanks for your explanation.

[@FrankHoffmann1000]

No, sorry, I do not have work sheets. Best wishes! Frank

[@priyapandey286]

Very interesting, your videos prove to be a great help for understanding the concepts. Can I find any structure determination and refinement tutorial of yours? The way you explain makes everything easy and understable. Thank you for all these videos.

[@FrankHoffmann1000]

Thanks for your kind comment! Crystal structure analysis might be a topic for future videos. But I have no concrete time plan, yet.

[@narminsalimova7334]

In the lecture it is said that for space group P/c, it is located at the height one quarter, but why? Shouldn't it be 1/2 instead of 1/4? Because in the previous lectures, you draw a table and mentioned that if order is 2 and symbol is like in the picture now for 21, then t=1/2. Can U explain please, I can't figure out where this 1/4 came from ?

[@FrankHoffmann1000]

Dear Narmin, first, the space group is still P2(1)/c - the space group P/c was not considered here or in any other units (by the way, a space group P/c doesn't exist, however Pc (No. 7) exists). Second, please do not mix up glide planes with screw axes. If I talk about order it can only refer to the order of either a rotational axis or a screw axis. Indeed the screw axis 2(1) has the order two because it involves a rotation by 180°. Third, the t = 1/2 is a) the translational part of a 2(1) screw axis and also that of a glide plane c. But, fourth, then you should not mix up such translational componenets of symmetry operations with any _locations_ of symmetry elements. The 1/4 is the _location_ of the glide plane c along the height in the unit cell, here along the b direction. Okey - now - why it is located at 1/4? First of all, there is another glice plane c at the height 3/4. Second, it should be clear that such mirrors or glide planes cannot be located at an arbitrary place within the unit cell, say at 5/17. Why? In short: Because a multiple application of any symmetry element should generate a position that must match with the location in, for instance, the neighboring cell, so by a symmetry-related location that is generated by translations of whole unit cells along one of the vectors of the lattice. best Frank

[@azice6034]

Great series. Wishing I knew about it this summer. It’s absolutely spectacular. I have a couple of questions, first of, how do you know that P2(1)/c is actually P12(1)/c1? That is, how do you know that there are 1’s in the a and c viewing directions and that the 2(1)/c is in the c direction specifically? Couldn’t you have something like P2(1)/c11 or P112(1)/c where it is in the a or c viewing directions?

[@FrankHoffmann1000]

It is abolutely possible to have something like P2(1)/c11 or P112(1)/c. It depends only on your choice regarding the choice of the axis in which the symmetry appears. Note that in the monoclinic crystal system symmetry is present in only _one_ direction. Usually you choose the axes in such the way that this symmetry direction is the b-direction, menaing: it is purely a convention. In Eastern Europe is is not uncommon to choose the c-direction. The a direction is usually not chosen.

[@krypticcrystallographer1369]

Super helpful. Would love to see you Zoom in to have a talk at my University. We have a Crystallographer, Dr Charles Lake, if you know of him.

[@FrankHoffmann1000]

Thanks for your nice comment - I don't know Charles Lake but that may change in the future.

[@Pterodactylandi]

Hello Frank, I have an exam in the next two days and I was wondering how it is possible to find the crystal system given a space group? Is the only way to do it with this book you’re talking about? I know you mention going from space groups to crystal classes in another video but then how do you take it back that last step? Thanks.

[@FrankHoffmann1000]

Dear Andrea, no, you don't need this book. Well, for most of the space groups it is relatively easy to derive the crystal system. Look at the list of space groups, a good source being the wikipedia entry: en.wikipedia.org/wiki/List_of_space_groups For the triclinic system you can just memorize that there is either no symmetry ("1") or maximum an center of inversion ("-1"). Tetragonal, simple: every space group that has as second symbol (after the symbol for the kind of centering) something with four-fold rotational symmetry belongs to this crystal system. Trigonal, again simple: every space group that has as second symbol (after the symbol for the kind of centering) something with three-fold rotational symmetry belongs to this crystal system. Hexagonal, once again simple: every space group that has as second symbol (after the symbol for the kind of centering) something with three-fold rotational symmetry belongs to this crystal system. Monoclinic: every space group that has symmetry in only one direction and in which this symmetry is restricted to something two-fold and/or mirror symmetry belongs to this crystal system. So, the orthorhombic and cubic system are left. These are sometimes not so easily to classify. For instance, intuively there is a trend to assign he space group Fddd to the cubic crystal system. However, we additionally need something with a three-fold symmetry in the cubic system, because the characteristic of the cubic system is that there are 4 three-fold axes of rotation. Therefore, if you have excluded all the other crystal systems and only the orthorhomic or cubic system are left, look for a "3" or "-3" and you are done! best! Frank

[@ayeshaasghar3940]

Hi, Frank Is it possible a racemate crystalize in P21/c and it's asymmetric unit reduced to 1 molecule. If racemic mixture present asymmetric unit should contain both RS conformation.

[@FrankHoffmann1000]

Hi Ayesha - Yes, it is possibe that the asymmetric unit contain only one molecule (enantiomer); the other will then be generated through symmetry (specifically the glide plane), meaning that the asymmetric unit does not have to contain both enantiomers.

[@neopets25]

Hello again Frank! Thanks again for your amazing videos. I have two questions if you would allow: 1. When you construct the diagram at 6:24, you only used the center of inversion and the 2-1 screw axis. What happened to the glide plane c? (Is this why there are two screw axes along c?) If I chose to construct the diagram using 2-1 screw axis and glide plane c (without using the center of inversion) would I have arrived at the same result? 2. What is the significance of the specified height of the mirror plane (1/4 along b direction)? I don't see it used anywhere in the diagrams. Thank you so much!

[@FrankHoffmann1000]

1) Yes, you would arrive at the same result. This again makes clear that the so-called generators (these are the symmetry elements that generate all possible positions in the cell) do not necessarily include all symmetry elements that are present. This is, because the combination of certain symmetry elements automatically generate other symmetry elements. In this case: 2(1) and a center of inversion generate a c-glide plane, 2(1) and a c-glide plane generate a center of inversion. 2) The significance would beome clearer if you choose to generate the diagram with 2(1) and c, for instance. In order to carry out the reflection part of the c-glide plane properly, you need, of course, this specification. All symmetry elements have a certain position, but in the 2D projection used here (along the b direction) you don't see, where this mirror component is - therefore this 1/4.

[@neopets25]

@@FrankHoffmann1000 Thank you again Frank!! Your channel has been so much help to students like me learning by ourselves. After your channel, I'd like to learn structure solutions, refining crystal data, and use of SHELXL software for analysis. Do you have any recommendation on books I can buy/read? Thanks again so much!

[@FrankHoffmann1000]

I can definitely recommend ‚ Crystal Structure Refinement: A Crystallographer's Guide to SHELXL‘ by Peter Müller!

[@segovemoc4776]

I checked this website img.chem.ucl.ac.uk/sgp/medium/sgp.htm and found out that there are not shown the general positions for space groups belonging to cubic crystal system in the diagrams (e.g. img.chem.ucl.ac.uk/sgp/medium/195az1.htm). I assume it is because the fractional numbers generated by 3-fold axes under angle 54,7 would be to crazy do depict. On the other hand the written general positions (list of x,y,z coordinates) do not show these numbers. Could you please explain why it is so?

[@FrankHoffmann1000]

Sorry, Sego, your comment was marked as Spam by KZbin, probably because one of the links doesn't work. Well, yes, probably they left out the general positions in the diagram because it looks almost chaotic... Concerning your question with the fractional numbers: why should this be the case? Remember that x,y, and z are placeholders and they can have any(!) value. So, a genral positions could also be (0.27896, 0.395557, 0.439126). If this is the first coordinate, 11 other additional coordinates would be generated, namely x.-y,-z = 0.27896, -0.39557, -0.439126 and so forth. best regards Frank

[@loopgarcia3862]

Hi Frank. Thank you for the video. So for P21/c the asymmetric unit is one-quarter of the unit cell. How can you determine this value for other groups? Can you read it directly from the international tables or you have to deduce it from the diagrams, which looks rather indirect?

[@FrankHoffmann1000]

Hi Albert, you can read it directly from the International Tables. There is a specific entry "Asymmetric unit" which specifies it in the form of ranges of the fractional coordinates, for instance 0 < x < 1, 0 < y < 1/4, 0 < z < 1. But note that the choice of the asymmetric unit is in many cases not uniquely defined, and there are more choices. The explanatory part inside the ITA states: "In cases where the asymmetric unit is not uniquely determined by symmetry, its choice may depend on the purpose of its application."

[@loopgarcia3862]

@@FrankHoffmann1000 Thank you very much for answering the questions, Frank. And thanks for all these educational videos! I have two final questions on this topic: 1) Can you relate the number of general positions (GPs) to the number of asymmetric units in the cell? I.e P21/c as shown in the video has 4 GPs and 4 asymmetric units in the cell. I was wondering if this is always the case, as it would be an easy way to determine the number of asymmetric units in the cell. 2) . I thought that the number of GPs should be the same as the number in the first entry in the table for Wyckoff positions (identity). This is the case seen in this video (P21/C 4 GPs and multiplity = 4) and in the unit 4.7 at 3:30 for Pmm2 (first entry in the column for Wyckoff positions = 4 and 4 GPs). However, in the international tables for Space group I/4m (number 87) the number of general positions do not correspond with the first entry in the table for Wyckoff positions, that shows a multiplicity of 16 even though there are 8 GPs What is the reason for this?.In this case, I think the unit cell contains 8 asymmetric units, is this correct? thank you very much for all your help.

[@FrankHoffmann1000]

@@loopgarcia3862 You're welcome. From a chemical point of view, it is somewhat unusual to speak of the _number_ of asymmetric units, because the AU is often referred to the _one_, smallest chemical unit which gives the whole structure by applying all symmetry operations. But it is of course perfectly allowed, although it is more common to speak then about the number of fundamental domains, if the AU is defined as a fraction of space. In fact, the number of AU correlates with the number of GPs: they are identical. This is also the case with space group 87 - there are 16 GPs and 16 fundamental domains, except that only the first 8 GPs are listed.... For centerings, the procedure is always to split the coordinates into subsets, which are then listed above the coordinates. For the body centering there are the two coordinate sets (0,0,0)+ and (1/2, 1/2, 1/2)+... Note that a body centering is equivalent to a translation operation x+1/2, y+1/2, z+1/2. For the positions now only the coordinates for the set (0,0,0)+ are specified... In particular with the face centerings one saves thereby much paperwork...

[@ManjuVV]

Thank you very much for your excellent information sir, kindly let me know where can i get in detail explanation of non standard settings of monoclinic unit cell.

[@FrankHoffmann1000]

In Unit 5.2... kzbin.info/www/bejne/roPIg62hoqqjg9E

[@natesh999]

Why is the viewing direction in the order a,b, c? For Monoclinic systems, in the previous lecture you mentioned that the first direction is always along 'b'

[@FrankHoffmann1000]

Dear Natesh, well, we can put this in two ways: First, we can say that the viewing directions for the triclinic, monoclinic and orthorhombic crystal system are along a, b, and c. Then we have to say, well, for the triclinic system there is maximum a center of inversion, which has no direction, for the monoclinic system there is always symmetry only along one direction, which is the b direction by convention, and for the orthorhombic system along all these three directions certain symmetry elements are present. Second, we can say there is no viewing direction for the triclinic system, there is only one (and thereby the first direction is simultaneously the last) direction for the monoclinic system (the b direction) and for the orthorhombic system they are a, b, and c. okey? Frank

[@natesh999]

Frank Hoffmann Thank you, yes that does help! :)

[@webbxu2319]

Hello, sir, first thank you for your helpful videos. I have a question about the point at 6:52. I'm confused that inversion will not change the chirality, thus the other gernerated general points will not have the coma, which seems to contradict to the diagram.

[@FrankHoffmann1000]

Hi Webb, they important thing here is that both kind of points/circles can be a chiral object! And if so, each inversion converts the motif into a motif with opposite handedness. The inversion of a point with no comma turns it into a point with comma and vice versa; this is the way the International Tables handle this. The International Tables could have also expressed it in a way that the inverted motif will become an inverted comma but they decided to express image and mirror-image with the pair "no comma and comma". best Frank

[@mm7966]

Hello sir, thank you for your great video. I have a question at 6:24. I do we know if we can apply the center of inversion there?

[@FrankHoffmann1000]

Hello, I am not sure if I understand your question completely. The question is not, if the center of inversion can be applied - it has to be applied always, because it is present there. Or do you mean the order of the operations? The order can be chosen completely arbitrarily for the sake of the reconstuction of the general position diagram.

[@dhoonygo]

It's a great explanation! So can I draw the diagram when given some space group notation? I think in that notation, there are no information about location of generators in diagram. for example in P2 (1)/c, location of glide plane c which is 1/4 isn't given in notation. also relative distance between inversion center and 2 (1) screw axis isn't given in notation. I mean, how can I put generator on proper location on certain diagram?

[@FrankHoffmann1000]

Dear Donghoon, no, usually it is not possible to draw the general position diagram, if you only know the pure space group notation. Maybe for some of the very simple space groups, and maybe, if you are an absolutely experienced freak, for some slightly more complicated ones, but in general this is not possible. This is the reason why the International Tables exist :-) There are, of course, some logical relations and some derivations of the positions of certain symmetry elements, for instance the location of the inversion center, but I prefer to look into our 'Bible' :-) best! Frank

[@dhoonygo]

Thank you so much ! it really helps me.

[@joelaw999]

Hi Frank! I think centers of inversion are not of the same height in this example, right? Then how to explain the diagram at 9:50 where two pairs of objects share the same center of inversion? I think there should be two centers of different heights

[@FrankHoffmann1000]

Hi Joe, the centers of inversion which you see here on the plane of projection do have the same height; however there are more in cell, namely further nine at b = 0.5 and another nine at b = 1, but these are just the translated ones from b = 0. In principle: Why shouldn't this be possible that two pairs of objects share the same center of inversion? If you look all 4 circles (indicating general positions) then you will see that it is indeed the case that you can compart these 4 circles into two pairs that are each symmetry-related to each other. ok? best Frank

[@joelaw999]

Frank Hoffmann Yeah I got it! So the two pairs I mentioned actually are about 2 different centers, one at b=0 and the other at b=0.5 as u said!

[@renatosilva9795]

Hello Dr. Hoffmann. Thank you again for the videos. I have a question. How can i determine the space group from a X-ray diffraction pattern? Do you have a suggestion for other video or book? I have a sample i know the elements but i would like to know the space group. Cheers!

[@FrankHoffmann1000]

If you carry out an X-ray diffraction experiment, you usually don't stop at determining the space group; instead you probably proceed with the solution and refinement of the crystal structure. But you are right that this is usually the first thing to do, and the determination of the space group is based on the inspection of the reflections that are "missing", that are systematically absent or extinct. All symmetry elements with a translational component (centerings, screw axes, glide planes) cause extinctions. And through the inspection which classes of reflections are extinct, you can derive these symmetry elements, hence the space group. My book recommendation: global.oup.com/ukhe/product/x-ray-crystallography-9780198700975?cc=de&lang=en& It's amazingly concise and a very good introduction! best Frank

[@renatosilva9795]

@@FrankHoffmann1000 Thank you very much! I will read the book. Is there any softwere which can used to help in this task?

[@FrankHoffmann1000]

@@renatosilva9795 Yes, a great application in that regard is XrayView 5.0. You can download it here: www.phillipslab.org/downloads

[@renatosilva9795]

@@FrankHoffmann1000 Thanks a lot! I am learning a lot with your videos.

[@nathalycalel2047]

Thank You for the explanation You know why no protein has ever been found to crystallize in this space group?

[@FrankHoffmann1000]

Yes, because proteins are enantiopure, chiral objects and chiral objects can crystallize only in those space groups that have rotations and translations (but not any kind of mirror components) as symmetry elements.

[@TheMachinist97]

Thank you very much. I have a question, how can we determine the centering-type ( A,B,C,I,F) just from the space group diagram?

[@FrankHoffmann1000]

Uhhh, that is definitely a tough task and I do not know any recipe for that. One reason might be that it is hard to memorize all symmetry rules. For instance: Compare P2 with C2 - would you expect that we have a 2(1) screw axis in the space group C2? I think you get a kind of 'feeling' for that, if you are used to stare at space group diagrams but that's all.

[@TheMachinist97]

@@FrankHoffmann1000 Thank you very much professor... I noticed that for orthorhombic and monoclinic, the symmetry operators can indicate the centering type, e.g, for C-centeted there exists ( x+1/2, y+1/2, z) , for I-centered exists ( x+1/2, y+1/2, z+1/2), And for F-centeted exists ( x+1/2, y+1/2, z), ( x+1/2, y, z+1/2), ( x, y+1/2, z+1/2)...I don't know if this observation is also valid for the remaining crystal systems, I didn't check it on them to be honest.. Anyhow, much appreciation , the lecture is so informative and helpful..

[@vipulkumarsharma3159]

Hello sir, first thank you for these video lectures ,these are very helpful to understand the crystal structure.but sir i have a question that if we know only the notation of space group then how can we draw the top left standard projection which is i think very important .

[@FrankHoffmann1000]

Hi Vipul, I am not sure, if I understand your question. The locations of the symmetry elements within the unit cell of a particular spaced group are fixed. They are completely independent of the concrete atomistic structure of your crystal. The only thing that you have to take into consideration is the choice of the origin. Sometimes there are two possibilities, but not in P21/c. The convention here is that the origin is at the center of inversion. best wishes Frank

[@vipulkumarsharma3159]

@@FrankHoffmann1000 sir, i would like to know that if we have p21/c symmetry group and according to the knowledge, the top left projection is drawn which is drawn in every point group according to their notation respectively . so my question is that on what basis it is drawn? or what should i read to understand it because i read many books but i don't find any such relation or theory threw which we can draw the left most projection . just as in p21/c a center of inversion and a 2-fold screw axis is drawn at fix position so how it is drawn ? please suggest something . thank you

[@FrankHoffmann1000]

Sorry, but I think I do not understand your question. What does "the top left projection is drawn which is drawn in every point group according to their notation respectively" mean? You can choose of course another projection, in which you look along the a- or c-direction instead. In the International Tables for Crystallography you will find all these three projections. The top left is the standard projection because it is oriented in sich a way that you look along the b direction, by convention the only direction in which symmetry elements appear in this particular space group P21/c What do you mean with "every point group"? We are dealing with space groups not points groups. Furthermore we are dealing here only with _one_ space group, P21/c. There is no 'theory' in a strict sense behind a particular space group. Space group P21/c simply means that you have these symmetry elements at the given locations as drawn in the diagram, nothing more, but also nothing less. Don't know, if this helps.

[@vipulkumarsharma3159]

+Frank Hoffmann sorry sir point group is my mistake. but my question is that can we draw the projection if we only have h-m notation for any space group? if yes then what will be the first step please explain.

[@FrankHoffmann1000]

I think, I am slowly beginning to understand your question. Well, usually this is only possible for the most simple space groups. We can begin with P1, which means that we have no symmetry except pure translations - trivial case. In P-1 we have only a center of inversion - theoretical considerations lead to the conclusion that this center of inversion has to be located exactly in the center of the cell. In Pm (space group No. 6) we have one mirror plane. Similar considerations lead to the conclusion that it has to be located at b = 0, 0.5, and 1. The guideline is always that the symmetry operation within the cell have to be compatible with the translational symmetry of the underlying lattice. You can't do this 'on the fly'. It took several years for Fedorov and Schoenflies to deduce these 230 types of space groups (framework of symmetry elements). Therefore, all space groups are listed together with the locations of their symmetry elements in the International Tables.

[@nourel-houdadegachi3673]

Why did we apply the rotation and reflection on the y axis? Is the selection of the axis random or always we choose the y axis? Because the change of axis is followed by a change in coordinates Please answer me Thanks in advance

[@FrankHoffmann1000]

The space group P2(1)/c belongs to the monoclinic crystal system and in the monoclinic crystal system we have symmetry only in one direction, and this is by convention the _b_ direction or _y_ axis. The full symbol for P2(1)/c would be P12(1)/c1, no symmetry in the _a_ direction, a two-one screw rotation and a glide plane c in the _b_ direction and again no symmtry in the _c_ direction. Concerning the viewing direction and space group symbols, please refer to unit 4.5: kzbin.info/www/bejne/hJK1m3qcppZsfac

[@nourel-houdadegachi3673]

Well, I understand, thanks, the axis is related to the type of crystal system that belongs to the space group

[@kishorebehera1]

Sir How to search space group of compounds in international union of crystallography website?and what is the website link to search this things?

[@FrankHoffmann1000]

Structural databases are not hosted by the IUCr. There are two main resources for structural data: 1. The Cambridge Structural Database (CSD). The CSD contains structural data derived from x-ray and neutron diffraction studies for over one million small-molecule (metal-)organic and (metal-)organic framework crystal structures. You can access basic search functions by the web interface WebCSD: www.ccdc.cam.ac.uk/structures/ Additionally, WebCSD now also covers data from the Inorganic Crystal Structure Database (ICSD) maintained by the FIZ Karlsruhe. For more advanced search functions you need a subscription. 2. The Crystallography Open Database (COD). It is an open-access collection of crystal structures of organic, inorganic, metal-organics compounds and minerals, excluding biopolymers. www.crystallography.net/cod/

[@kishorebehera1]

@@FrankHoffmann1000 Thank you sir

[@ivanbombana7282]

Good morning. Why must I write 3m and not 3mmm?

[@FrankHoffmann1000]

Good afternoon! How does your question belong tho this video? Did you already watch unit 2.9, 3.5 and 3.6? In short: because there are only two (viewing) directions in the trigonal crystal system. Please make sure that you first understand the systematic how point (and space) groups are derived: You look at certain directions and write down the symmetry elements that you find along or perpendicular to these directions. These directions are different for the different crystal systems. best regards Frank

[@nourel-houdadegachi3673]

why location of glide plane c which is 1/4 isn't given in notation.

[@FrankHoffmann1000]

Do you mean in the space group notation? Well, this is not only true for the glide plane c, but for all other symmetry elements, too. The location of the 21-screw axis is also not given. I think, It would be a very confusing notation system if you include the location for every symmetry element.

[@nourel-houdadegachi3673]

Okay ,Thank you

[@KULDEEPSINGH-hl6oe]

Hi sir, can you provide me cif file of Al2Mn2O7 monoclinic phase P2/m space group or the atomic positions o elements. I am struggling with refinement. Thank you

[@FrankHoffmann1000]

The only paper that I can find concerning this compound is that of F. Khachnaoui et al. but the authors do not provide atomic positions although they claimed that they carried out a Rietveld refinement. You should contact the authors of this paper in order to request the coordinates.

[@KULDEEPSINGH-hl6oe]

@@FrankHoffmann1000 Sir the email they provide doesn't work.

[@FrankHoffmann1000]

@@KULDEEPSINGH-hl6oe Interesting and also a bit obscure! I have no further idea, sorry.

[@KULDEEPSINGH-hl6oe]

@@FrankHoffmann1000 Thanks a lot sir, if you find ever please reply to me.

[@FrankHoffmann1000]

@@KULDEEPSINGH-hl6oe I will - I contacted two of the authors via ResearchGate - let's see, if and what they will answer...

[@adityabhardwaj2440]

How centre of inversion came in picture , we just need to do screw and glide operation .

[@FrankHoffmann1000]

Dear Aditya, you are correct: To generate all equivalent positions in this space group we do not need this center of inversion. However, does this mean that these centers of inversion are not present? No, they are present. They are automatically present if and because the 2(1) screw axis and the glide plane perpendicular to this axis are present. The space group symbol specifies only the so called "generators", i.e. those symmetry operations/elements which are necessary to generate all (general) positions, but this set is not complete in the sense that there maybe not other symmetry elements. Indeed, it is very often the case that the presence of two symmetry elements generate a third one. Think for instance at two perpendicular mirror planes: Then automatically there is also a two-fold axis of rotation present (along the cutting edge of these two planes). ok? best Frank