3D Symmetry _2 (Two weeks)

3D lattice: Reading crystal7.pdf Building the 3D lattices by adding another translation vector to existing 2D lattices Oblique (symmetry 1) + triclinic General Triclinic Primitive

Oblique (symmetry 2) + projection 4 choices:

Double cell side centered Double cell side centered

Double cell body centered Some people use based centered, some use body centered. monoclinic

Rectangular (symmetry m) + 90o 90o already exist! Rectangular (symmetry g) + : the same. cm + ?

Rectangular (symmetry 2mm) + P2mm P2mg p2gg Orthorhombic primitive

Double cell side centered Orthorhombic base-centered Orthorhombic base-centered Double cell side centered

Orthorhombic body-centered rectangular

Centered Rectangular (symmetry 2mm) + C2mm the same

Face centered orthorhombic

Square (symmetry 4, 4mm) + P4 P4mm p4gm Tetragonal primitive

Tetragonal Body centered Tetragonal

p3 p3m1 p31m Hexagonal (symmetry 3, 3m) + not in this category Hexagonal primitive not in this category Why? Rhombohedral

Hexagonal primitive Rhombohedral triple cell

p6 p6mm p31m Hexagonal (symmetry 3m, 6, 6mm) + can only located at positions: Hexagonal primitive p31m Hexagonal & 6 related can only fit 3P!

11 lattice types already cubic (isometric) Special case of orthorhombic (222) with a = b = c Primitive (P) Body centered (I) Face centered (F) Base center (C) Tetragonal (I)? Cubic a = b  c [100]/[010]/[001] [111] Tetragonal (P)

Another way to look as cubic: Consider an orthorhombic and requesting the diagonal direction to be 3 fold rotation symmetry P222  P23 Primitive I222  I23 Body centered F222  F23 Face centered C222  I23

Bingo! 14 Bravais lattices!

Lattice type - compatibility with - point group reading crystal9.pdf. Crystal Class Bravais Lattices Point Groups Triclinic P (1P) 1, 1 Monoclinic P (2P), C(2I) 2, m, 2/m Orthorhombic P(222P), C(222C) F(222F), I(222I) 222, mm2, 2/m 2/m 2/m Rhombohedral P (3P), 3R 3, 3 , 32, 3m, 3 2/m Hexagonal P (3P) 6, 6 , 6/m, 622, 6mm, 6 m2, 6/m 2/m 2/m Tetragonal P (4P), I (4I) 4, 4 , 4/m, 422, 4mm, 4 2m, 4/m 2/m 2/m Isometric (Cubic) P (23P), F(23F), I (23I) 23, 2/m 3 , 432, 4 3m, 4/m 3 2/m

http://www.theory.nipne.ro/~dragos/Solid/Bravais_table.jpg  = P = I = T P

= P = I = B = T P http://users.aber.ac.uk/ruw/teach/334/bravais.php

Next, we can put the point groups to the compatible lattices, just like the cases in 2D space group. 3D Lattices (14) + 3D point groups  3D Space group There are also new type of symmetry shows up in 3D space group, like glide appears in 2D space (plane) group!

The naming (Herman-Mauguin space group symbol) is the same as previously mentioned in 2D plane group! The first letter identifies the type of lattice: P: Primitive; I: Body centered; F: Face centered C: C-centered; B: B-centered, A: A-centered The next three symbols denote symmetry elements in certain directions depending on the crystal system. (See next page)

Monoclinic a  b = 90o; c  b = 90o. b axis is chosen to correspond to a 2-fold axis of rotational symmetry axis or to be perpendicular to a mirror symmetry plane. Convention for assigning the other axes is c < a. a  c is obtuse (between 90º and 180º). Orthorhombic The standard convention is that c < a < b.  Once you define the cell following the convention  A, B, C centered

Hexagonal/ Rhombohedral Crystal System Symmetry Direction Primary Secondary Tertiary Triclinic None   Monoclinic [010] Orthorhombic [100] [001] Tetragonal [100]/[010] [110] Hexagonal/ Rhombohedral [120]/[1 0] Cubic [100]/[010]/ [001] [111]

Consider 2P Monoclinic + 2 /2 P2 /2

How about 2I Monoclinic + 2 There is a lattice point in the cell centered!

z (1) (3) (2) z z +1/2 (3) (1) (2) New type of operation In general Screw axis 21 2

Specifying

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

42 n1 n2 ……... nm-2 nm-1 No chirality

3 31 32 2 21 4 41 42 43

6 61 62 63 64 65

62

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

A: 21 21 P + 21 = P21 21 21 I + 2 = I2 or I + 21 = I21 A A: 2  E: 21 Same, only shifted E A: 21  E: 2 I2 = I21

Hexagonal lattice (P and R) with 3, 31, 32. Case P first! All translations in P have component on c of 0 or unity! A B C B C B and C: same point; B and C: equivalent point; Having

P3 P31 P32

All translations of R has component on c of 1/3 or 2/3! Case R! A E D 2/3 All translations of R has component on c of 1/3 or 2/3! E D 1/3 Screw at Designation of Space group A  D’ E’ 3 31 32 c/3 2c/3 2/3 c/3 2c/3 c 2/3 2c/3 c 4c/3 31 32 3 32 3 31 R3 R31 R32 R3 = = Hexagonal lattice (P, R) + 3, 31, 32  P3, P31, P32, R3.

Square lattice P with 4, 41, 42, 43. The translation of P have component on c of 0 or unity! A C B B B C C A 4 41 42 43  c/4 c/2 3c/4 B /2 0 /2 c/4 /2 c/2 /2 3c/4 B  0  c/2  c  3c/2 B 4 41 42 43 B 2 21 P4 P41 P42 P43

P4 P41 P42 P43

Homework: Discuss the cases of I4, I41, I42, I43.

How to obtain Herman-Mauguin space group symbol by reading the diagram of symmetry elements? First, know the Graphical symbols used for symmetry elements in one, two and three dimensions! International Tables for Crystallography (2006). Vol. A, Chapter 1.4, pp. 7–11. http://www.kristall.uni-frankfurt.de/media/exercises/Symbols-for-symmetryelements-ITC-Vol.A2.pdf

Symmetry planes normal to the plane of projection Graphical symbol Translation Symbol Reflection plane None   m Glide plane 1/2 along line   a, b, or c 1/2 normal to plane Double glide plane 1/2 along line & 1/2 normal to plane (2 glide vectors)   e Diagonal glide plane 1/2 along line, 1/2 normal to plane (1 glide vector)   n Diamond glide plane 1/4 along line & 1/4 normal to plane   d

Symmetry planes parallel to plane of projection Graphical symbol Translation Symbol Reflection plane None   m Glide plane 1/2 along arrow   a, b, or c Double glide plane 1/2 along either arrow   e Diagonal glide plane 1/2 along the arrow   n Diamond glide plane 1/8 or 3/8 along the arrows   d 3/8 1/8 The presence of a d-glide plane automatically implies a centered lattice!

Symmetry Element Graphical Symbol Translation Symbol Identity None   1 2-fold ⊥ page   2 2-fold in page 2 sub 1 ⊥ page 1/2   21 2 sub 1 in page 3-fold   3 3 sub 1 1/3   31 3 sub 2 2/3   32 4-fold   4 4 sub 1 1/4   41 4 sub 2   42 4 sub 3 3/4   43 6-fold   6 6 sub 1 1/6   61 6 sub 2   62 6 sub 3   63

Symmetry Element Graphical Symbol Translation Symbol 6 sub 4 2/3   64 6 sub 5 5/6   65 Inversion None   1 3 bar   3 4 bar   4 6 bar   6 = 3/m 2-fold and inversion   2/m 2 sub 1 and inversion   21/m 4-fold and inversion   4/m 4 sub 2 and inversion   42/m 6-fold and inversion   6/m 6 sub 3 and inversion   63/m

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

From the point group mmm  orthorhombic For orthorhombic: primary direction is (100), secondary direction is (010), and tertiary is (001). lattice for orthorhombic: C Short symbol No. 17 orthorhombic that can be derived

Principles for judging crystal system by space group 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)

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). http://chemistry.osu.edu/~woodward/ch754/sym_itc.htm

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

Example: Generating a Crystal Structure http://chemistry.osu.edu/~woodward/ch754/sym_itc.htm Description of crystal structure of Sr2AlTaO6 Space Group = Fm 3 m; a= 7.80 Å Atomic Positions Atom x y z Sr 0.25 Al 0.0 Ta 0.5 O

From the space group tables http://www.cryst.ehu.es/cgi-bin/cryst/programs/nph-wp-list?gnum=225 32 f 3m xxx, -x-xx, -xx-x, x-x-x, xx-x, -x-x-x, x-xx, -xxx 24 e 4mm x00, -x00, 0x0, 0-x0,00x, 00-x d mmm 0 ¼ ¼, 0 ¾ ¼, ¼ 0 ¼, ¼ 0 ¾, ¼ ¼ 0, ¾ ¼ 0 8 c 4 3m ¼ ¼ ¼ , ¼ ¼ ¾ 4 b m 3 m ½ ½ ½ a 000

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) Sr (000) (½½0) (½0½) (0½½) 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)

Ta (000) (½½0) (½0½) (0½½) 4b: ½ ½ ½  (½½½) (00½) (0½0) (½00) (000) (½½0) (½0½) (0½½) O x00 24e: ¼ 0 0  (¼00) (¾½0) (¾0½) (¼½½) -x00 ¾ 0 0  (¾00) (¼½0) (¼0½) (¾½½) 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)

Bond distances: Al ion is octahedrally coordinated by six O Al-O distance d = 7.80 Å  0.25−0 2 + 0−0 2 + 0−0 2 = 1.95 Å Ta ion is octahedrally coordinated by six O Ta-O distance d = 7.80 Å  0.25−0.5 2 + 0.5−0.5 2 + 0.5−0.5 2 = 1.95 Å Sr ion is surrounded by 12 O Sr-O distance: d = 2.76 Å

Determining a Crystal Structure from Symmetry & Composition Example: Consider the following information: Stoichiometry = SrTiO3 Space Group = Pm 3 m a = 3.90 Å Density = 5.1 g/cm3

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10-8 cm)3 = 5.93  10-23 cm3 (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

 number of molecules per unit cell : 1 SrTiO3.  (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) 6 e 4mm x00, -x00, 0x0, 0-x0,00x, 00-x 3 d 4/mmm ½ 0 0, 0 ½ 0, 0 0 ½ c 0 ½ ½ , ½ 0 ½ , ½ ½ 0 1 b m 3 m ½ ½ ½ a 000 http://www.cryst.ehu.es/cgi-bin/cryst/programs/nph-wp-list?gnum=221

Calculate the Ti-O bond distances: 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) Atom x y z Sr 0.5 Ti O