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3D Symmetry _2 (Two weeks)

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

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Oblique (symmetry 2) + projection 4 choices:

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Double cell side centered Double cell side centered

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

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

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

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

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rectangular Orthorhombic body-centered

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Centered Rectangular (symmetry 2mm) + C2mm the same

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Face centered orthorhombic

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

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Tetragonal Body centered Tetragonal

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Hexagonal (symmetry 3, 3m) + p31mp3m1 Hexagonal primitive Rhombohedral p3p3 not in this category Why?

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Hexagonal primitive Rhombohedral triple cell

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Hexagonal (symmetry 3m, 6, 6mm) + can only located at positions: p6p6 p6mm Hexagonal primitive p31m Hexagonal & 6 related can only fit 3P!

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

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

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Bingo! 14 Bravais lattices!

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Crystal ClassBravais LatticesPoint Groups TriclinicP (1P) MonoclinicP (2P), C(2I)2, m, 2/m Orthorhombic P(222P), C(222C) F(222F), I(222I) 222, mm2, 2/m 2/m 2/m RhombohedralP (3P), 3R HexagonalP (3P) TetragonalP (4P), I (4I) Isometric (Cubic) P (23P), F(23F), I (23I) Lattice type - compatibility with - point group reading crystal9.pdf.crystal9.pdf

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nipne.ro/~dragos/S olid/Bravais_table. jpg = P= I = T P

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= P = I = B = T P c.uk/ruw/teach/33 4/bravais.php

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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!

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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)

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Monoclinic a b = 90 o ; c b = 90 o. 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

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Crystal System Symmetry Direction PrimarySecondaryTertiary TriclinicNone Monoclinic[010] Orthorhombic[100][010][001] Tetragonal[001][100]/[010][110] Hexagonal/ Rhombohedral [001][100]/[010][120]/[1 0] Cubic [100]/[010]/ [001] [111][110]

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Monoclinic + 2 P2P2 p2p2 2D Consider 2P Monoclinic + 2 /2 P2P2

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

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z z z +1/2 New type of operation 2 Screw axis (1) (2) (3) (1) (2) (3) 2121 In general

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Specifying

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For a 3-fold screw axis: fold screw axis:

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4242 n1n1 n2n2 ……...n m-2 n m-1 No chirality

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6262

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

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P = P2 1 A: 2 1 I + 2 = I2 or I = I2 1 E A A: 2 E: 2 1 A: 2 1 E: 2 Same, only shifted I2 = I

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Hexagonal lattice (P and R) with 3, 3 1, 3 2. Case P first! A All translations in P have component on c of 0 or unity! B B C C B and C : same point; B and C: equivalent point; Having

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P3P3 P31P31 P32P32

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All translations of R has component on c of 1/3 or 2/3! A 1/3 2/3 D D E E A Screw at D ’ E ’ Designation of Space group c/3 2c/3 2 /3 c/3 2c/3 c 2c/3 c 4c/3 2 / R3R31R32R3R31R32 R3R3R3R3 = = Hexagonal lattice (P, R) + 3, 3 1, 3 2 P3, P3 1, P3 2, R3. Case R!

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The translation of P have component on c of 0 or unity! A A A Square lattice P with 4, 4 1, 4 2, 4 3. B B B C C C 0 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 B B P4P41P42P43 P4P41P42P43

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P4P4 P41P41 P42P42 P43P43

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Homework: Discuss the cases of I4, I4 1, I4 2, I4 3.

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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. frankfurt.de/media/exercises/Symbols-for- symmetryelements-ITC-Vol.A2.pdf

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Symmetry planes normal to the plane of projection Symmetry planeGraphical symbolTranslationSymbol Reflection planeNone m Glide plane1/2 along line a, b, or c Glide plane 1/2 normal to plane a, b, or c 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

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1/8 Symmetry planeGraphical symbolTranslationSymbol Reflection planeNone m Glide plane1/2 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 Symmetry planes parallel to plane of projection The presence of a d-glide plane automatically implies a centered lattice!

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Symmetry Element Graphical SymbolTranslationSymbol IdentityNone 1 2-fold ⊥ page None 2 2-fold in pageNone 2 2 sub 1 ⊥ page 1/ sub 1 in page1/ foldNone 3 3 sub 11/ sub 22/ foldNone 4 4 sub 11/ sub 21/ sub 33/ foldNone 6 6 sub 11/ sub 21/ sub 31/2 6 3

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Symmetry Element Graphical SymbolTranslationSymbol 6 sub 42/ sub 55/6 6 5 InversionNone 1 3 barNone 3 4 barNone 4 6 barNone 6 = 3/m 2-fold and inversion None 2/m 2 sub 1 and inversion None 2 1 /m 4-fold and inversion None 4/m 4 sub 2 and inversion None 4 2 /m 6-fold and inversion None 6/m 6 sub 3 and inversion None 6 3 /m

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mb-glide n-glide c-glide || b a a m n || a b b c c c c || c

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

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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), 4 1, 4 2 or 4 3 (i.e. P , I4/m, P4/mcc) Hexagonal – The primary symmetry symbol will always be a 6, (-6), 6 1, 6 2, 6 3, 6 4 or 6 5 (i.e. P6mm, P6 3 /mcm) Trigonal – The primary symmetry symbol will always be a 3, (-3) 3 1 or 3 2 (i.e P31m, R3, R3c, P312) Principles for judging crystal system by space group

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Orthorhombic – All three symbols following the lattice descriptor will be either mirror planes, glide planes, 2-fold rotation or screw axes (i.e. Pnma, Cmc2 1, 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, P2 1 /n) Triclinic – The lattice descriptor will be followed by either a 1 or a (-1).

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

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Example: Generating a Crystal Structure htm Description of crystal structure of Sr 2 AlTaO 6 Atomxyz Sr0.25 Al0.0 Ta0.5 O

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From the space group tables bin/cryst/programs/nph-wp-list?gnum=225 32f3mxxx, -x-xx, -xx-x, x-x-x, xx-x, -x-x-x, x-xx, -xxx 24e4mmx00, -x00, 0x0, 0-x0,00x, 00-x 24dmmm0 ¼ ¼, 0 ¾ ¼, ¼ 0 ¼, ¼ 0 ¾, ¼ ¼ 0, ¾ ¼ 0 8c¼ ¼ ¼, ¼ ¼ ¾ 4b½ ½ ½ 4a000

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Sr 8c; Al 4a; Ta 4b; O 24e 40 atoms in the unit cell stoichiometry Sr 8 Al 4 Ta 4 O 24 Sr 2 AlTaO 6 F: face centered (000) (½ ½ 0) (½ 0 ½) (0 ½ ½) 8c: ¼ ¼ ¼ (¼¼¼) (¾¾¼) (¾¼¾) (¼¾¾) ¼ ¼ ¾ (¼¼¾) (¾¾¾) (¾¼¼) (¼¾¼) ¾ + ½ = 5/4 =¼ Sr Al 4a: (000) (½ ½ 0) (½ 0 ½) (0 ½ ½) (000) (½½0) (½0½) (0½½)

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Ta 4b: ½ ½ ½ (½½½) (00½) (0½0) (½00) O 24e: ¼ 0 0 (¼00) (¾½0) (¾0½) (¼½½) (000) (½½0) (½0½) (0½½) ¾ 0 0 (¾00) (¼½0) (¼0½) (¾½½) x00 -x00 0 ¼ 0 (0¼0) (½¾0) (½¼½) (½¾½) 0x00x0 0-x0 0 ¾ 0 (0¾0) (½¼0) (½¾½) (0¼½) 0 0 ¼ (00¼) (½½¼) (½0¾) (0½¾) 00x 00-x 0 0 ¾ (00¾) (½½¾) (½0¼) (0½0¼)

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Sr ion is surrounded by 12 O Sr-O distance: d = 2.76 Å

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Determining a Crystal Structure from Symmetry & Composition

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First step: calculate the number of formula units per unit cell : Formula Weight SrTiO 3 = (16.00) = g/mol (M) Unit Cell Volume = (3.90 cm) 3 = 5.93 cm 3 (V) (5.1 g/cm 3 ) (5.93 cm 3 ) : weight in a unit cell ( g/mole) / (6.022 /mol) : weight of one molecule of SrTiO 3

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number of molecules per unit cell : 1 SrTiO 3. (5.1 g/cm 3 ) (5.93 cm 3 )/ ( g/mole/6.022 /mol) = e4mmx00, -x00, 0x0, 0-x0,00x, 00-x 3d4/mmm½ 0 0, 0 ½ 0, 0 0 ½ 3c4/mmm0 ½ ½, ½ 0 ½, ½ ½ 0 1b½ ½ ½ 1a000 From the space group tables (only part of it) list?gnum=221

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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 3c) = 2.76 Å (0 ½ ½) d 3d) = 1.95 Å (½ 0 0, Better) Atomxyz Sr0.5 Ti000 O0.500

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