Presentation on theme: "1 We are now ready to move to the orthorhombic system. There are 59 orthorhombic space groups. Orthorhombic crystals belong to crystal classes 222, mm2."— Presentation transcript:
1 We are now ready to move to the orthorhombic system. There are 59 orthorhombic space groups. Orthorhombic crystals belong to crystal classes 222, mm2 or mmm. In this section we’ll consider many (but not yet all) of the orthorhombic space groups in crystal class 222, namely, space groups P222, P222 1, P2 1 2 1 2, P2 1 2 1 2 1, and I222. Later versions of the tutorial will complete the 222 groups, and move on to the remaining orthorhombic groups. Note that all space groups which belong to crystal class 222 will be enantiomorphous and noncentrosymmetric.
2 In the orthorhombic system, a b c; = = = 90º. We can thus have three possible crystal classes or point groups: 222, mm2 or mmm. 1 http://phycomp.technion.ac.il/~sshaharr/intro.html http://phycomp.technion.ac.il/~sshaharr/intro.html Primitive Orthorhombic 1 End-Centered Orthorhombic (A- or B- or C- ) 1 Body-Centered Orthorhombic (I) 1 Face-Centered Orthorhombic (F) 1 (Click on any green label above to rotate)
3 Martin J. Buerger In order to derive the primitive orthorhombic space groups in crystal class 222, we need to consider how a pair of 2 (or a pair of 2 1 axes) parallel to the crystallographic a and b axes may be combined. Once we've figured out how to approach this, we'll jump in, do the derivations, and see what is generated along the crystallographic c direction. This will require your careful attention, but the results are very useful and very interesting!
4 Hmmm….getting started should be fairly straightforward: a pair of twofold axes may either intersect or they may not. Similarly, a pair of 2 1 screw axes may intersect or they may not. Let's see what happens when we try a derivation of an orthorhombic space group beginning with a pair of intersecting 2's. Hey, I'm really getting into this! A. Student
5 b a Z=4; + + - - + + - - + + - - + + - - Now for the test: do you know what this group is called? HCEHCE Height Chirality Elements
7 That worked, and we got space group P222. What will we get if we have two nonintersecting 2's? I'll try it: first one will go along a, while I'll put the second one || b, but at c = ¼ so that it doesn't intersect the first twofold axis. Right now I'm not certain what I'll get, but it cannot be the same group! Here goes….. Will B. Learning
8 Z=4; HCEHCE + ½ - ½ + - + ½ - ½ + - + ½ - ½ + - + ½ - ½ + - So…. do you know what this group is called? b a
9 Z=4; HCEHCE + ½ - ½ + - + ½ - ½ + - + ½ - ½ + - + ½ - ½ + - b a
10 So….when we have two intersecting 2's, we derive P222, and when the axes do not intersect, we obtain P222 1. Now what we need to do is to try this for a pair of 2 1 axes: the drill should now be familiar! First we'll see what we get with a pair of intersecting 2 1 's, and then with a pair of non- intersecting 2 1 's. The new group will likely be P2 1 2 1 X, where X = a new generated element. Before we begin to look at that, we'll add a shortcut to our derivations: we've noted several times before that the combinations of symmetry elements and translations lead to additional elements that appear every half-unit cell; this is a very general observation. Look back at the one we just did:
11 Ja! + ½ - ½ + - + ½ - ½ + - + ½ - ½ + - + ½ - ½ + - Remember that, at the end of our derivation of P222 1, we saw the axes in red appear. These are not related to the black axes by translation but are generated! The result is that all elements appear every half-unit cell. b a
12 After that "interlude", we need to remind ourselves where we were! A few slides back, we said that our next step would be to attempt the derivation for a pair of 2 1 axes: the drill should now be familiar! First we'll see what we get with a pair of intersecting 2 1 's, and then with a pair of non-intersecting 2 1 's. The new group will likely be P2 1 2 1 X, where X = a new generated element.
13 HCEHCE - + - - ++ + - + b a
14 HCEHCE - + - - ++ + - + We seem to have produced which space group? Oui! It is P2 1 2 1 2. But why is the "No. 18" in quotes? b a
15 Yes! We seem to have derived space group P2 1 2 1 2, given the appearance of the diagram, but our diagram has a different origin from that in the International Tables for Crystallography, Volume A. Thus we need to examine the conventions for choice of origin, as specified on page 21 of that volume.
16 From the International tables for Crystallography, Volume A, page 21, we read: (i) All centrosymmetric space groups are described with an inversion centre as origin. A second description is given if a space group contains points of high site symmetry that do not coincide with a centre of symmetry… (ii) For non-centrosymmetric space groups, the origin is at a point of highest site symmetry... If there is no site symmetry higher than 1, except for the cases listed below under (iii), the origin is placed on a screw axis, or a glide plane, or at the intersection of several such symmetry elements… (iii) In space group P 2 1 2 1 2 1 the origin is chosen in such a way that it is surrounded symmetrically by three pairs of 2 1 axes… Let us look at our putative P2 1 2 1 2 diagram again:
17 HCEHCE - + - - ++ + - + Do you see? If point group symmetry takes precedence, and the highest site symmetry in P2 1 2 1 2 is on the 2, we must then re-derive this group, using an origin shift of (¼, ¼, 0), so that the 2 will be generated along (0, 0, z). b a
18 Z=4; HCEHCE - - ++ + + + + + + b a
19 J. Monteath Robertson Now we'll derive P2 1 2 1 2 1 - recall that the convention is to place the three axes at locations equally displaced from the origin. And…we've been doing our derivations by starting with two axes || a and b, respectively. So, let's begin by putting the first 2 1, || a, at b = ¼, and the second 2 1, || b, at c = ¼. If all goes as we now may have learned to expect, we should generate a third 2 1, || c, at a = ¼.
20 Z=4; HCEHCE + ½-½- ½ + ½ - ½ + + + + - b a
21 Essentially, we've carried out the permutations and combinations for derivations of space groups starting with pairs of 2 or 2 1 axes. We did not look at combining a 2 and a 2 1 axis, but……in effect we've done that job too! How? Our results were the space groups P222, P222 1, P2 1 2 1 2 and P2 1 2 1 2 1. From observing these groups, it should be clear that combining a 2 and a 2 1 axis will produce either an additional 2 or an additional 2 1, depending upon whether the 2 and 2 1 intersect or not. The result would be a non-standard setting of one of the four above groups, e.g., P2 1 22 1 or P2 1 22, etc. The standard setting could be readily produced by a simple transformation of axes. For example, if we say that P222 1 has axes abc, then P2 1 22 will correspond to an axial transformation with axial settings cab relative to the standard group P222 1.
22 However, there is an obvious combination that we did not try: namely, to force three 2 1 axes to intersect! To test this case, we shall start with 2 1 axes along the crystallographic a, b and c directions; then we will perform the derivation of the “unknown” space group as before. After the Figure is complete, it should be possible to identify the space group as one of the 9 orthorhombic space groups belonging to crystal class 222.
23 HCEHCE ++ ½-½- ½-½- - - ½+ - - + ++ Remember the convention that the origin will be at highest site symmetry? It must be… Hmm…I see three intersecting 2's at (¼, ¼, ¼). Ooo… and I-centering! b a
24 I 222 ++ ½-½- ½-½- - - ½+ - - + ++ Remember the convention that the origin will be at highest site symmetry? It must be… Hmm…I see three intersecting 2's at (¼, ¼, ¼). Ooo… and I-centering! b a
25 Note that there is a 222 symmetry site in this "version" of I222. If we look in the International Tables, Volume A, we can find I222, but not this diagram. To derive the diagram in the Tables, we need to place the three 2's at the origin, and add I-centering operations…and, of course, then we should get the conventional, standard diagram, as well as three intersecting 2 1 's displaced by (¼, ¼, ¼). The take-home message is twofold, no pun intended! First, if we force three 2 1 's to intersect, we obtain a non-primitive lattice. Second, the non-primitive lattice contains pairs of parallel 2's and 2 1 's along a, b and c. Let's rederive I222 starting with I-centering and all the 2's and 2 1 's:
26 Z=8; + - - + + - - + + - - + + - - HCEHCE + ½ - ½ + ½ - b a