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1 Knowledge of space groups and the implications of space group symmetry on the physical and chemical properties of solids are pivotal factors in all areas.

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Presentation on theme: "1 Knowledge of space groups and the implications of space group symmetry on the physical and chemical properties of solids are pivotal factors in all areas."— Presentation transcript:

1 1 Knowledge of space groups and the implications of space group symmetry on the physical and chemical properties of solids are pivotal factors in all areas of structural science. As we met to bring our ideas in teaching this subject to life, we both felt that teaching the concepts with repetitive textual and visual reinforcement, “early and often”, will provide a sound basis for students in this subject. We chose PowerPoint as the delivery vehicle for the tutorial, owing to the facility of combining narrative pedagogy with animations of the buildup of space groups, and the ease of making subtle changes and/or corrections. And, we realize that others may wish to alter the approach, e.g., by rewriting the text in another language, or altering the approach (….hopefully, the desired changes will be small!). The student may also wish to consider M. E. Kastner's Crystallographic Courseware, which provides an alternative approach to the teaching of Space Group Symmetry.M. E. Kastner's Crystallographic Courseware

2 2 We trust that the current and subsequent versions of our tutorial will become ein lebendiges Buch (a living book) to which all can contribute and enjoy, and from which all can learn.

3 3 The tutorial is based in part upon an approach to teaching space groups developed in a course in X-ray Crystallography at Brandeis University. However, through vigorous discussions, the ideas have been continuously reworked. Further thanks are due to Professors David Y. Curtin and Iain C. Paul, as well as Prof. Menahem Kaftory (The Technion) and Prof. Roland Boese (Universität Essen), without whose elegant teaching and presentation ideas this work would not have been possible.Professors David Y. Curtin and Iain C. PaulProf. Menahem Kaftory Prof. Roland Boese We wish to thank Dr. Michael J. Bennett, Henley-on-Thames, UK, who imparted vast knowledge, inspiration and technique to BMF during our mutual tenure in Prof. F. A. Cotton's group in the 1960's. What we have done here would not have even begun without the gems he provided in the past.Dr. Michael J. BennettProf. F. A. Cotton's

4 4 Acknowledgements We are particularly grateful to the National Science Foundation Summer Research Program in Solid State and Materials Chemistry, directed by Prof. Shiou-Jyh Hwu, Clemson University. The Program provides an opportunity for a faculty member and student to carry out research in a host laboratory; Prof. Jasinski and student Lisa Bennett worked on solid-state reactivity during Summer 2003 at Brandeis. For the faculty member to return to the program in subsequent years, a joint research program on educational aspects of solid-state chemistry is required. With the encouragement provided by the program requirements and additional NSF support through grants DMR and DMR to BMF, we designed and wrote this tutorial during Summer 2004 and 2005.Prof. Shiou-Jyh Hwu

5 5 Finally, we wish to thank Dr. Eugene Cheung (Cardiff University) and Professors Jen Swift (Georgetown) and Mike Ward (Minnesota) for reviewing the tutorial and providing oodles of helpful comments. Local comments and stimulating discussions at Brandeis University, from Professors Dan Oprian, Chris Miller and undergraduate students Jeremy Heyman and Stephen C. Wilson were of immeasurable help.Professors Jen SwiftMike WardDan OprianChris Miller We invite your comment and constructive criticism. The design of this tutorial will allow us to incorporate changes on a regular basis. Bruce M. Foxman Brandeis University Jerry P. Jasinski Keene State College September 2006

6 6 COLORS No doubt you have already noticed that we’re using underlined green text to indicate a hyperlinked page or pages. In order to minimize issues involving broken links, all links are included in the tutorial, and the exact link references may be found in the Credits section (Chapter 7) of this PowerPoint presentation. Blue is often used for emphasis. Red is usually employed when we are referring to symmetry elements or unit cell parameters a, b, c, , , . Other colors, used less often, are generally employed to emphasize particular points, or to improve the readability of a particular presentation.

7 7 Where should you be, in your knowledge of X-ray structure determination, or crystallography before beginning this tutorial? We’re starting with an assumed knowledge of point groups, unit cells, lattices and space group symmetry elements (there will be some minor review of these topics). In the future we'll be adding a special section on unit cells and lattices recently added to the tutorial. Finally, if you have around 3 minutes, listen to a humorous musical revue of Bravais lattices at : We recommend reviewing or understanding the equivalent of Chapters 1-4 in Sands, Introduction to Crystallography (Dover, 1975) or Sections 2.1 and in Stout and Jensen, X-Ray Structure Determination: A Practical Guide (Wiley, 1989).

8 8 Paul von Groth This tutorial is divided into sections by crystal class. There are 32 crystallographic point groups, or crystal classes. The 32 crystal classes correspond to the external shapes of crystals actually observed. The crystal class may be obtained from the space group symbol by “removing” the translations from the symbol…more about that later! In this section we consider the two triclinic space groups, P1 and P-1bar, which belong to crystal classes 1 and 1-bar, respectively.

9 9 In all of the tutorial presentations which show a unit cell diagram, we have followed the style of the International Tables for Crystallography, Volume A, Space Group Symmetry, which contains diagrams of the 230 space groups.International Tables for Crystallography, Volume A, Space Group Symmetry We will use the Hermann-Mauguin symmetry notation 1 throughout, and will be deriving each space group from the Hermann-Mauguin symbol. Our diagrams are “not exactly” identical to the diagrams in Volume A, but rather are a composite overlay of the unit cell diagram, symmetry elements, and molecules which reside in the Equivalent General Positions. 1 There will be an opportunity to learn more of the history of this notation on an upcoming slide.

10 10 In the triclinic system, a  b  c;       90º. Triclinic crystals either have only 1 symmetry (a 360° rotation, crystal class 1) or possess a center of inversion (crystal class ). For convenience, we’ll often write “1-bar” instead of. 1 Primitive Triclinic (click to rotate) 1

11 11 On each slide you will see the ac projection of a triclinic or monoclinic unit cell. The b axis, which points toward the viewer, is either inclined to the a and c axes (triclinic) or perpendicular to the page (monoclinic). In the orthorhombic case, we'll use an ab projection. Each atom or group of atoms will be displayed by an open circle. An open circle with a large comma inside will be used to indicate opposite chirality to the reference molecule. Each slide opens with the ac (orthorhombic, ab) projection and a reminder about the axial directions. Upon tapping the advance button, various events will occur, depending upon the space group under consideration. We’ll describe these precisely as we approach each example.

12 12 We need to define the coordinate system we'll be working in. Of course, we'll need to specify the positions of atoms or molecules within the unit cells under consideration. Let's imagine that a particular atom is located (in Å) at (X, Y, Z). A preferred method to specify the location of this atom would be to use a formalism that is independent of the size of the unit cell. To do this we use fractional coordinates (x, y, z), where x = X/a y = Y/b and z = Z/c Within the unit cell, values of x, y and z are thus constrained to decimal values between 0 and 1. And, for example, a location in a unit cell adjacent to the reference cell but displaced along a would thus be (1 + x, y, z).

13 13 What is our reference molecule or atom ? This is the first atom or group that appears on the screen as an open circle. It will always have the coordinates (x, y, z), and we'll draw an ac projection of the unit cell, with axis b coming out of the page. As you push the advance button, the operations of the group are applied …. in the following case, space group P1, we have only unit translations to apply, in sequence. Finally, we must specify the location of the atom or group along the third dimension (its "height" in/out of the screen). We’ll do this by placing the “prefix” of the y-coordinate next to the open circle. For the reference molecule, this will simply be a “+”, as it is located at (x, y, z)  +y. An indicator such as “½+” signifies “½+y”; “-” is just “-y”. Remember that x, y and z are fractional coordinates, and for molecules within the unit cell, each has a value between 0 and 1.

14 14 We’ll start by adding the translations, and recording the unique set of symmetry position(s) generated for a single atom. As the atoms are added, think about how many are actually within the unit cell; we’ll call this number Z. Beside the number of symmetry related atoms, we’ll list their positions. See if you can assign coordinates to each molecule added; some answers are given as you proceed. At a time after all the unique atoms within the cell have been generated we’ll draw a box around the “General Position”, the set within the unit cell equivalent by symmetry. The General Position thus contains a certain number of equivalent points per cell; the number is referred to as Z. We will also say that the multiplicity (the number) of the General Position is Z.

15 15 In the International tables for X-ray Crystallography, Volume A. section 2.11, the term "General Position" is defined as follows: A set of symmetrically equivalent points...is said to be in 'general position' if each of its points is left invariant only by the identity operation but by no other symmetry operation of the space group. Each space group has only one general position. We'll see, time and time again, that, in the general position, no symmetry requirements are imposed upon the molecule. Thus, general positions always have 'site-symmetry' 1 (Hermann-Mauguin notation) or alternatively C 1 (Schoenflies notation). More info about Hermann-Mauguin notation appears on slides 17, 25 and throughout the tutorial. We hope you will recall from your knowledge of point groups that the identity operation is a trivial operation of making no change, e.g., a 360° rotation.

16 16 The term "Special Position" (not yet encountered...) is defined as follows: A set of symmetrically equivalent points...is said to be in 'special position' if each of its points is mapped onto itself by at least one further symmetry operation of the space group. Later we'll see that a special position always corresponds to the location of a point group symmetry element (inversion  ), rotation (n = 2, 3, 4 or 6  rotations of 360/n degrees), reflection (m)). A molecule located on a special position is expected to possess the point group symmetry of the special position.

17 17 Our first example is space group No. 1, P1 – the symbol informs us that we have a primitive lattice, and only “1” symmetry. A number, n, as a symmetry element in Hermann- Mauguin notation (used throughout the tutorial) refers to a rotation by 360/n degrees. Thus 1 refers to a 360° rotation. This element is also called the identity in group theory, as it represents a trivial operation of making no change. We will consider the effect of symmetry on a molecule or group located at a “general position” in the unit cell, with coordinates (+x, +y, +z). Let’s try it now.

18 Z=1; a c

19 19 So, we’ve got one molecule per unit cell (Z = 1). Which two phrases describe the properties of handedness and centrosymmetry for P1? A space group that contains a center of symmetry is referred to as “Centrosymmetric”; if there is no center of symmetry, it is called “Non-centrosymmetric”. A space group that MUST contain pairs of enantiomers is referred to as “Non-enantiomorphous”. A space group that may contain molecules of one hand only is referred to as “Enantiomorphous”. Let’s consider some properties of space groups that we’d like to track as we take this tour…

20 Z=1; EnantiomorphousNon-centrosymmetric a c

21 Z=1; Reference atom An equivalent position Prefix Number of equivalent positions contained within one unit cell Now, let's review some of the jargon introduced earlier: In this area we'll list the set of equivalent points belonging to the general position Gridline for c/2 Gridline for a/2 Note that there are "halfway marks" on each unit cell, a visual guide: a c

22 22 A final point or reminder regarding P (primitive lattices): The primitive translation operators, expressed in fractional coordinates, are (1,0,0), (0,1,0), (0,0,1). These operations apply to any and all lattice points and may be applied once or many times. Thus, lattice points such as : (x,1+y,z)(x-1,y,z) (2+x,y-3,z+4) are all “symmetry-related” to the point (x,y,z). A common error made in the early stages of understanding is to suppose that, e.g., (x,1-y,z) is “the same as”, or is translation-related to (x,y-1,z). Note the difference! In general, adding or subtracting an integer to a fractional number is NOT the same as first taking the negative of that number, and then adding or subtracting integers!

23 23 Hmmm…I don’t see what you mean! Let me think about it. So…. I need to satisfy myself that (x, 1-y, z) and (x, y-1, z) are different, that is, they are not related by a simple translation. OK. I’ll choose a point ….uh….(0.1, 0.2, 0.3). I’ll calculate (x, y-1, z) = (0.1, -0.8, 0.3). This is clearly related to my “blue” point by the translation (0, 1, 0). Now, I’ll produce the point (x, 1-y, z) from my original “blue” point: (0.1, 0.8, 0.3). Ouch! This differs from the original by (0, 0.6, 0), and the brown point by (0, 1.6, 0)….it isn’t related to either point by a primitive translation! I see the key to it now: we can add positive or negative integers to a point, but a translation will NEVER change the sign of the coordinate. Sue N. Smart

24 24 OK! Now let’s take the same diagram, that is, the one we just finished, and add a center of symmetry at the origin. The symbol for this is a small circle. The other thing to look for is the appearance of other centers of symmetry as you add molecules: these are the ones that are generated by the interaction (i.e., the multiplication) of the operations of this group. After we add the center, where is the next molecule? Will it be in front of the screen (as for the (x, y, z) molecule), or behind? Think about its sign (the sign for the y-coordinate) before you push the button. Yes, its coordinates are (-x,-y,-z), and thus we place a “-” next to it. If the new molecule that is generated has the opposite chirality to its symmetry-related mate, we’ll put a comma inside to indicate that.

25 25 To reiterate, for each space group, as we did for P1, we’ll derive the group from the Hermann- Mauguin symbol. New figures will "pop up", and you should continually pause and consider the figure's HCE : Height, Chirality, and which new Elements have been generated. Carl Hermann Charles-Victor Mauguin Ah, oui. We will use the acronym HCE to remind ourselves of what to watch for and learn in each diagram: 1.H : ask yourself what the exact “height” (location normal to the screen) of the transformed figure is. 2.C : does the figure have the opposite or identical chirality to the prior group? 3.E : which new element(s) have just been generated? (Pronunciation Hint)

26 26 + Z=2;, -, Non-enantiomorphous, -, - Is this space group enantiomorphous or non-enantiomorphous? HCEHCE Symbol: = 1-bar = inversion center a c

27 27 Z=2; ++, -, - ++, -, - Is this space group centrosymmetric or non-centrosymmetric? Centrosymmetric a c

28 28 Z=2; (0, 0, 0) ; (½, 0, 0); (0, 0, ½); (½, 0, ½); (0, ½, 0); (½, ½, 0); (0, ½, ½); (½, ½, ½) Note that there are two molecules per unit cell, and that other centers of symmetry (at a/2, b/2, c/2 and combinations thereof) were generated as we added atoms or groups. It turns out that there are eight centers of symmetry: where are they? ++, -, - ++, -, - a c

29 29 Above is a perspective view of a unit cell in P1bar, with all centers of symmetry shown. The eight independent centers of symmetry (i. e., those NOT related by any symmetry operation) are shown in blue.

30 30 We say that the eight centers of symmetry are independent, since any single center of symmetry is unrelated to another by the operations of the group : Yes, of course, since we can always add a positive or negative integer (unit cell translation) to any point. Formally, the two have different locations, but they are identical upon unit translation along the a axis. It’s easy to see that the centers of symmetry are unrelated; plug any value into the above set of equivalent positions, and we get the same value back: try (0,0,0) – this is obvious. If we try (½,0,0) we generate (-½,0,0). Do these two points represent the same position?

31 31 Each of the eight centers of symmetry corresponds to a special position in. Note that each point will be mapped onto itself by an operation of the group (cf. slides 16 & 30). Special positions always correspond to a point group symmetry element, i. e. a rotation axis, reflection, inversion or rotary inversion axis. A special position always has reduced multiplicity compared to the general position. We say that the multiplicity (or number) of molecules placed in the general positions (Z) would be equal to two, while the multiplicity (the number) of molecules placed on each of the special positions would be equal to 1. As you should now realize, the general position has no symmetry requirements, while a group occupying a special position must have the corresponding symmetry.

32 32 Why is the multiplicity for the special positions in equal to 1? Recall that the equivalent positions for are: ),, ( zyx If we "run" any of the special position coordinates through these general positions, we only get one value in return: (0, 0, 0) gives identically (0, 0, 0), and, (½, ½, ½) gives (-½, -½, -½). Placing these two points within the same unit cell by unit translations along a, b and c renders them identical.

33 33 So. It is easy to see that P1-bar must contain racemic molecules….but could P1 contain a racemate? Of course. In P1, the left- and right- handed molecules are simply unrelated by any symmetry operation, while in P1-bar, they are related by the crystallographic inversion center. Kathleen Lonsdale Margaret C. Etter

34 34 We've completed the triclinic space groups, but there are a few final relevant points worth making from both experimental and pedagogic points of view. It is very useful for us to know something about the density of a crystal. If the measured and calculated densities agree, we may be confident that we know the stoichiometry of the crystal contents well. Alternatively, if the two values do not match, we can calculate the unit cell's molecular weight from the measured density, and often deduce the stoichiometry.

35 35 Can we actually define the density of a crystal in terms of the language & ideas we have used thus far? The mass, m, of one unit cell is just nM/N 0, where n = the number of molecules per unit cell, M = formula weight in grams, and N 0 = Avogadro's number. V is the unit cell volume. Normally we will use formula weight in grams and volume in cm 3.

36 36 Crystal density is conveniently measured by the neutral buoyancy technique. Imagine what would happen if we placed crystals of density 1.30 g-cm -3 in a solution of heptane,  = 0.68 g-cm -3 : Finally, if we adjusted the amounts of the two liquids such that the crystals neither sank nor floated, we could measure the weight of 10 mL of that solution, divide by 10, and get the crystal density! Now imagine what would happen if we used CCl 4,  = 1.59 :

37 37 Often, it is most useful to measure the density by the neutral buoyancy technique, and then calculate n, the number of molecules in the unit cell. Thus, a compound C 12 H 12 N 2 O 2 crystallizes in the triclinic system, and has formula weight , a density of g-cm -3 and a unit cell volume of Å 3. Let's calculate n =  VN o /M = 1.341(  )(6.02  )/(216.24) = 2.002! So, there's 2 molecules per unit cell…..can we tell, in general, whether the space group is P1 or P-1bar from such data? More pertinently: can we ever determine the space group from density information?

38 38 Bonjour, Professeur! I have a crystal…I have measured the density and determined the crystal system. It’s triclinic, and Z = 1. Therefore, it MUST belong to space group P1 !!!!!!!!!! Silly boy! No. You may be correct, but it just as well could be in P1-bar, with a molecule simply occupying a special position (the center of symmetry). You CANNOT obtain the space group from a density measurement! René-Just Haüy A. Étudiant

39 39 Ah, Professeur! I beg to differ! Now I have a different crystal. Again I have measured its density and determined the crystal system. It’s triclinic, and this time Z = 2. It is very clear that the space group is P1-bar !!! No. No. No. Again, you may be correct, but it just as well could be in P1, with two molecules each occupying a general position; the two are unrelated by symmetry. As I said, you CANNOT obtain the space group from a density measurement! René-Just Haüy A. Étudiant

40 40 End of Section 1, Introduction & Pointgroups 1 and 1-bar


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