Monoclinic + 2

1. Monoclinic + 2 2D P2 p2 c b a

2. z (1) (2) z z +1/2 (3) New type of operation 2 Screw axis

3. Specifying

4. For a 3-fold screw axis: 3 31 32 P31 P32 4-fold screw axis: 43 41 41 42 43

5. 42 n1 n2 ……... nm-2 nm-1 No chirality

6. 2 21 3 31 32 4 41 42 43

7. 6 61 62 63 64 65

8. 62

9. Example to combine lattice with screw symmetry D A A: 2-fold + translation (to arise at B, C, or D) B C Rotation symmetry of B, C, and D is the same as A. A: 2 P + 2 = P2

10. A: 21 P + 21 = P21 I + 2 = I2 or I + 21 = I21 A A: 2  E: 21 E The same A: 21 E: 2 I2 = I21

11. Space groups isogonal with point group 3: Hexagonal lattice (P and R) with 3, 31, 32. A All translation of P have component on c of 0 or unity! B C B C

12. P3 P31 P32

13. A The translation of R have component on c of 0 or unity! E 2/3 D E D 1/3 Screw at Designation of Space group A  D’E’ 3 31 32 0 c/3 2c/3 2/3 2/3 2/3 c/3 2c/3 c 2/3 2/3 2/3 2c/3 c 4c/3 31 32 3 32 3 31 R3 R31 R32 R3 R3 = =

15. Glide plane also exists for 3D space group with more possibility Symmetry planes normal to the plane of projection

16. Symmetry planes parallel to plane of projection 3/8 1/8 The presence of a d-glide plane automatically implies a centered lattice!

19. http://img.chem.ucl.ac.uk/sgp/mainmenu.htm http://xrayweb.chem.ou.edu/notes/symmetry.html In a recent version of the International Tables for Crystallography, Vol. A, the double glide called the e-glide is described.14 The e glide occurs only in centered cells and is defined by one plane with two perpendicular glide vectors related by a centering operation. This type of glide was proposed by the third Nomenclature Report of the IUCr.15 Take a look at the notation for space group (non-trivia). There are convention (see the reason for them for a few examples). Then, we would like to pass around representation of some none-trivia space groups.

20. c-glide b n-glide ||c 21 c || a n 2 21 2 b-glide m m c || b a

21. Maximum length Intermedium length Conventions: Magnitudes of Translation determine a, b, and c in orthorhombic: A-, B-, C- base centered What symmetry go first?

23. lattice for orthorhombic: C Short symbol No. 17 orthorhombic that can be derived

24. http://chemistry.osu.edu/~woodward/ch754/sym_itc.htm • Cubic – The secondary symmetry symbol will always be either 3 or –3 (i.e. Ia3, Pm3m, Fd3m) • Tetragonal – The primary symmetry symbol will always be either 4, (-4), 41, 42 or 43 (i.e. P41212, I4/m, P4/mcc) • Hexagonal – The primary symmetry symbol will always be a 6, (-6), 61, 62, 63, 64 or 65 (i.e. P6mm, P63/mcm) • Trigonal – The primary symmetry symbol will always be a 3, (-3) 31 or 32 (i.e P31m, R3, R3c, P312)

25. Orthorhombic– All three symbols following the lattice descriptor will be either mirror planes, glide planes, 2-fold rotation or screw axes (i.e. Pnma, Cmc21, Pnc2) • Monoclinic – The lattice descriptor will be followed by either a single mirror plane, glide plane, 2-fold rotation or screw axis or an axis/plane symbol (i.e. Cc, P2, P21/n) • Triclinic – The lattice descriptor will be followed by either a 1 or a (-1).

26. What can we do with the space group information contained in the International Tables? 1. Generating a Crystal Structure from its Crystallographic Description 2. Determining a Crystal Structure from Symmetry & Composition

27. Example: Generating a Crystal Structure http://chemistry.osu.edu/~woodward/ch754/sym_itc.htm Description of crystal structure of Sr2AlTaO6 Space Group = Fmm; a= 7.80 ÅAtomic Positions

28. From the space group tables http://www.cryst.ehu.es/cgi-bin/cryst/programs/nph-wp-list?gnum=225

29. Sr 8c; Al 4a; Ta 4b; O 24e 40 atoms in the unit cell stoichiometry Sr8Al4Ta4O24Sr2AlTaO6 F: face centered  (000) (½ ½ 0) (½ 0 ½)(0 ½ ½) (000) (½½0) (½0½)(0½½) Sr 8c: ¼ ¼ ¼  (¼¼¼) (¾¾¼) (¾¼¾)(¼¾¾) ¼ ¼ ¾  (¼¼¾) (¾¾¾) (¾¼¼)(¼¾¼) Al ¾ + ½ = 5/4 =¼ 4a: 0 0 0  (000) (½ ½ 0) (½ 0 ½)(0 ½ ½)

30. (000) (½½0) (½0½)(0½½) Ta 4b: ½½½ (½½½) (00½) (0½0)(½00) (000) (½½0) (½0½)(0½½) O x00 24e: ¼ 0 0 (¼00) (¾½0) (¾0½)(¼½½) ¾ 0 0 (¾00) (¼½0) (¼0½)(¾½½) -x00 0x0 0 ¼ 0  (0¼0) (½¾0) (½¼½)(½¾½) 0-x0 0 ¾ 0  (0¾0) (½¼0) (½¾½)(0¼½) 00x 0 0 ¼ (00¼) (½½¼) (½0¾)(0½¾) 00-x 0 0 ¾ (00¾) (½½¾) (½0¼)(0½0¼)

31. Bond distances: Al ion is octahedrally coordinated by six O Al-O distance d = 7.80 Å = 1.95 Å Ta ion is octahedrally coordinated by six O Ta-O distance d = 7.80 Å = 1.95 Å Srion is surrounded by 12 O Sr-O distance: d = 2.76 Å

32. Determining a Crystal Structure from Symmetry & Composition Example: Consider the following information: Stoichiometry = SrTiO3Space Group = Pmma = 3.90 ÅDensity = 5.1 g/cm3

33. First step: calculate the number of formula units per unit cell : Formula Weight SrTiO3 = 87.62 + 47.87 + 3(16.00) = 183.49 g/mol (M) Unit Cell Volume = (3.9010-8cm)3 = 5.93 10-23cm3 (V) (5.1 g/cm3)(5.93 10-23 cm3):weight in a unit cell (183.49 g/mole) /(6.0221023/mol) : weight of one molecule of SrTiO3

34.  (5.1 g/cm3)(5.93 10-23 cm3)/ (183.49 g/mole/6.022 1023/mol) = 0.99  number of molecules per unit cell : 1 SrTiO3. From the space group tables (only part of it) http://www.cryst.ehu.es/cgi-bin/cryst/programs/nph-wp-list?gnum=221

35. Sr: 1a or 1b; Ti: 1a or 1b  Sr 1a Ti 1b or vice verse O:3c or 3d Evaluation of 3c or 3d: Calculate the Ti-O bond distances: d (O @ 3c) = 2.76 Å(0 ½ ½) D (O @ 3d) =1.95 Å (½ 0 0, Better)