Inorganic materials chemistry and functional materials

Transcription

1 Chemical bonding Inorganic materials chemistry and functional materials Helmer Fjellvåg and Anja Olafsen Sjåstad Lectures at CUTN spring 2016

2 CRYSTALLOGRAPHY - SYMMETRY

3 Symmetry NATURE IS BEAUTIFUL The chief forms of beauty in nature are order, symmetry and definiteness. Aristotle, 350 BC The scientist does not study nature because it is useful. He studies it because he takes pleasure in it, and he takes pleasure in it because it is beautiful. Henry Poincaré, 1908

4 Symmetry

5 Symmetry MOLECULAR SYMMETRY Molecules can be classified in terms of their symmetry What types of symmetry can we find in a molecule? Rotational axes, mirror planes, inversion centers

6 Symmetry MOLECULAR SYMMETRY

7 Symmetry ROTATION Described with symmetry element C n, where the angle of rotation is 360 /n. If more than one rotational axis is present, we identify the principal axis as the one with the highest symmetry; the highest n. Ex: A square planar molecule with four equal substituents has a principal C 4 -axis.

8 Symmetry ROTATION

9 Symmetry MIRROR PLANE Mirror planes are denoted with a small sigma; σ. We can identify three types of mirror planes: Planes perpendicular to the principal axis: σ h. Planes containing the principal axis: σ v. Planes containing the principal axis, but bisecting the angle between two adjacent 2- fold axes: σ d.

10 Symmetry MIRROR PLANE

11 Symmetry INVERSION CENTRE If all points in a molecule can be reflected through the centre and this produces an indistinguishable configuration, it contains a centre of inversion. Denoted i.

12 Symmetry IMPROPER ROTATION If a rotation about an n-fold axis followed by a reflection through a plane perpendicular to this axis produces an indistinguishable configuration, we call this an improper rotation, denoted S n.

13 Symmetry IDENTITY For mathematical reasons, a last operator is needed, an operator that leaves the molecule unchanged. All molecules have the identity operation. We denote this operator: E.

14 Symmetry SUCCESSIVE OPERATORS With all operators defined, we can permute a molecule with successive symmetry operations. Successive operations can be added together, and can correspond to other operations. Ex. Rotating 180 two times around a C 2 -axis will bring the molecule back to the original state. C 2 x C 2 = C 2 2 = E.

15 Symmetry GROUP THEORY A mathematical group is a set of elements that are related through given rules. The product of any two elements in the group must be an element of the group. There must be an identity element; an element that commute with all other elements and leave them unchanged. The associative law of multiplication must hold. All elements must have a reciprocal element which is also an element of the group. The product of an element and its reciprocal element must always be the identity element.

16 Symmetry MATHEMATICAL GROUPS An example of a mathematical group is the set of all integers Z, with standard addition as the operator. This group is denoted (Z,+). Adding an integer to another integer will always give another integer. Adding 0 to any element will give that element, so 0 is the identity element. a+(b+c) = (a+b)+c for all integers a,b,c. The reciprocal element for all elements is it s negative counterpart, as a + (-a) = 0.

17 Symmetry MATHEMATICAL GROUPS As an example of something similar that is not a group is the set of all integers Z, with standard multiplication as the operator, (Z, ). Multiplying an integer with another integer will give another integer. Multiplying any element with 1 will give that element, so 1 is the identity element. a(bc) = (ab)c for all integers a,b,c. There does not exist any integers that can be multiplied with another integer to produce 1.

18 Symmetry GROUP THEORY IN CHEMISTRY Why do you need this? The set of symmetry operations that can be applied to a molecule form a mathematical group! An indefinite number of such point groups exist. Out of these only 32 groups are relevant for crystalline solids These 32 are termed: Crystallographic point groups

19 Symmetry MOLECULES versus SOLIDS - I Inorganic molecules: rather small in size Interactions between molecules ar weak: - Dispersion forces (induced dipols; dipol dipol; hydrogen bonds) Molecular solids Crystalline solids with strong bonding in 1, 2 or 3 dimensions (chains, layers, network) - Strong ionic, covalent, or metallic bonding

20 Symmetry MOLECULES versus SOLIDS - II Crystalline solids with strong bonding in 1, 2 or 3 dimensions (chains, layers, network) - Strong ionic, covalent, or metallic bonding Periodic structures; unit cells repeating in 3D - Metric of unit cell - Mathematical lattice for the structure - Symmetry operations compatible with a lattice - Crystal system, Bravais lattice, crystallographic point group, space group Local symmetry for a given atom in the structure

21 Symmetry Symmetry The Unit Cell smallest repeating unit that shows full symmetry of the crystal structure. By stacking the «boxes» (unit cells) in three dimensions, a complete description of the obtained structure is obtained. Unit cell - Has a metric - Contains atoms - Shows various symmetries 3D stacking of unit cells crystallite

22 Symmetry Symmetry A unit cell in 3D is defined by three pairs of parallel planes Definition of relations between angles (,, ) and axes (a, b, c) in unit cell. : angle between a and b Unit Cell : angle between b and c : angle between a and c

23 Symmetry Symmetry Units for unit cell dimensions and atomic/ionic radii Ångström unit is frequently used: The Ångström unit is in the range of the size of atoms and molecules, and microscopic biological structures. It is useful for discussing chemical bonding and arrangement of atoms in crystals. Å is not an SI-unit 1 Å = 0.1 nm = 100 pm = m

24 Symmetry Symmetry What unit cell is correct for the 2D lattice in (a)? 1) Smallest unit that shows full lattice symmetry 2) Choice of origin optional (inversion center and/or atoms on cell corners, edges)

25 Symmetry Symmetry 7 crystal classes/systems and their 7 primitive unit cells RHOMBOHEDRAL R a = b c = = 90 o = 120 o a b c = = 90 o 90 o

26 Symmetry Symmetry Seven crystal systems Hierarchy implies: Higher crystal system contains at least one symmetry element that the lower system does not contain. Hierarchy of crystal systems Hexagonal Cubic Tetragonal Orthorombic Trigonal Monoclinic Triclinic

27 Symmetry Symmetry 7 crystal systems and their 7 primitive Bravais lattices Consider now JUST mathematical points. If one such point is placed at origin for the seven different types of unit cells shown for seven crystal systems, we then have unit cells for seven Mathematical lattices; Bravais lattices What is a space lattice/lattice? Mathematically repeating pattern of points. REQUIREMENT: All lattice points have identical surroundings a b c = = 90 o 90 o This is exactly the cases for our 7 lattices These contain 1 mathematical point (8 x 1/8) per unit cell Are termed primitive Bravais lattices P

28 Symmetry Symmetry Counting of points of the unit cell A point inside one unit cell 1 point per unit cell A face point is shared between 2 cells 1 / 2 point per unit cell An edge-point is shared between 4 cells 1 / 4 point per unit cell A corner-pont is shared between 8 cells 1 / 8 point per unit cell

29 Symmetry Symmetry Question: are there more ways to arrange mathematical points in space and obtain a repeating lattice with identical surroundings at each lattice point? P = Primitive 1 lattice point I = Body centered 2 lattice points Lattices with more than one point in the unit cell C = Side centered 2 lattice points F = Face centered 4 lattice points Centering is on the side normal to the c-axis Other options: A, B centering

30 Symmetry Symmetry 7 crystal systems 4 lattice types P,F,I,C 14 Bravais Lattices a b c = = 90 o 90 o Ma All lattice points have identical surroundings. A unit cell in 3D is defined by three pairs of parallel planes

31 Symmetry Symmetry Exercise 7 (p. 14) Draw a tetragonal primitive (P) lattice. Place an extra lattice point in ½, ½, 0; i.e. make a C centring of the lattice. Show that a tetragonal C lattice does not exist as such, i.e., that it is equivalent to another tetragonal P lattice. Tetragonal - P a = b c = = = 90 o

32 Symmetry Symmetry -Values of coordinates (x, y, z) are always between 0 and 1 with respect to the unit cell; i.e. we use fractional coordinates. - The unit cell is repeating; the structure is periodic any integer multipla of the unit vectors can be added/substracted to get identical points in other unit cells c Positioning of atoms atomic coordinates A = (0.5,0.5,0.5) B = (0.5,1.5,0.5) ( A = (0.5, 1.5-1, 0.5)) C = (0.5,2.5,0.5) ( 0.5, 2.5-2, 0.5) c (1,0,0) (0,0,0) a b b a Can add and subtract whole numbers at will. The unit cell vectors (axes) are basis for these translations

33 Symmetry Symmetry From mathematical Bravais lattice to crystal structure 2D unit cell Mathematical lattice Filling the unit cell with atoms (motif) -The unit cell defines the mathematical room (Bravais lattice) - The room then has to be filled with its basis/motif (basis = non-equivalent atoms = atoms with different surroundings) - «Crystal structure = lattice + basis/motif»

34 Symmetry Example Symmetry 7 (p. 14) Identify lattice and basis for the CsCl type structure Strategy: 1) See that the spheres in corners are different from the sphere in centre (i.e. disconnected) 2) Add several unit cells, to find the repeating unit that applies for both Cs + and Cl. 3) Recognize that the repeating unit for both Cs + and Cl are identical; i.e. same Bravais lattice CsCl structure: Bravais lattice + basis cubic-primitive + basis [Cs (0,0,0) and Cl (½, ½, ½)]

35 Symmetry Symmetry Differences between CsCl- and bcc structure Cl Cs + CsCl type structure = cubic primitive lattice (P) + basis Basis = Cs (0,0,0) and Cl (½, ½, ½) bcc (W) = cubic body centred lattice (I) + basis Basis = W (0,0,0) Tungsten (W) is I-centered (bcc) since W atoms in (0,0,0) and (½, ½, ½) have identical surroundings.

36 Symmetry Symmetry Unit cell; Graphite Two-dimesnional sheets; graphene sheets; are stacked ABAB... along the c-axis. Strong bonding within sheets; van der Waals forces between sheets Graphene sheets Graphene sheets Graphene sheets Hexagonal, P ALL atoms listed

37 Symmetry Symmetry Symmetry operations Assembly of symmetry operations Point groups Lattice +Translation Space groups Two systems for nomenclature (areas of use) we learn both now: 1. Schönflies (spectroscopy) 2. Hermann-Mauguin (crystallography)

38 Point groups POINT GROUPS There are indefinite number of point groups There are 32 crystallographic point groups We will first use Schönflies notation to describe some molecules We will use Hermann Maugin notation to put point group symmetry into the frame of periodic structures You should be able to assign the correct point group to ANY molecular object

39 Point groups

40 Point groups POINT GROUP DECISION TREE Makes point group assignment relatively straightforward Follow the tree along the correct branches, and you ll find your point group Practice at some everyday examples

41 Point groups AMERICAN FOOTBALL

42 Point groups PYRAMID

43 Point groups BOOMERANG

44 Point groups POCl 3

45 S 8 Point groups

46 Point groups BF 3

47 Point groups Allene

48 Point groups THE CHARACTER TABLE Every point group has a corresponding character table This is a set of irreducible representations that span the group Let us see what this means

49 Point groups THE WATER MOLECULE Symmetry operations? C 2 rotation axis Two vertical mirror planes Identity operations Has C 2v symmetry

50 Point groups THE WATER MOLECULE The hydrogen atoms are labeled for clarity and a coordinate system is added.

51 Point groups THE WATER MOLECULE As the result of symmetry operations, all atoms are shifted to new coordinates In such a 3D-system, mapping a point to a new position can be described fully by a 3x3 matrix

52 Point groups THE WATER MOLECULE The character of a matrix is the sum of the numbers on the diagonal from upper left to lower right

53 Point groups THE WATER MOLECULE The set of character forms a representation, in this case a reducible representation This can be further reduced, first by block diagonalization

54 Point groups THE WATER MOLECULE By doing this, we diagonalize the x, y and z coordinates and they become independent of each other:

55 Point groups THE WATER MOLECULE We are closing in on the character table, which is the complete set of irreducible operations for the group Mathematics sets some important constraints on how to form such a complete set 1. The number of symmetry operations in the group is the order of the group.

56 Point groups THE WATER MOLECULE 2. Symmetry operations are arranged in class, all operations in one class have identical transformation matrix characters 3. The number of irreducible representations equals the number of classes 4. The sum of the squares of the characters under the identity operation equals the order of the group

57 Point groups THE WATER MOLECULE 5. For any irreducible representation, the sum of the squares of the characters multiplied with the number of operations in the class equals the order of the group 6. Irreducible operations are orthogonal to each other 7. All groups include a totally symmetric representation, with characters of 1 for all operations

58 Point groups THE WATER MOLECULE By using the three existing representations, we can now make the required fourth representation and write up the complete character table for the point group C 2v

59 Point groups CHARACTER TABLES All point groups have their own character table Why do we need this? Symmetry is an important property of a system that can be used to deduce physical properties and response to certain characterization techniques. The character table summarizes this symmetry in a simple way.

60 Point groups CHARACTER TABLES and spectroscopy IR: Change in dipole moment during vibration means IR-activity. How is this described in the character table? Raman: Change in polarizability tensor during vibration means Raman-acivity. How is this described in the character table?

61 Point groups; rotation Symmetry operations 1. Rotation axis, n-fold (rotation by 360 o /n) 4-fold rotation axis = 360 o /n = 360 o /4 = 90 o Crystals (periodic lattices ) only display rotational symmetries of 2, 3, 4 and 6 Molecules can in addition have n = 5, 7,

62 Point groups; identity 2. Identity, i (rotation axis n = 1; 360 o /1) (x, y, z) (x, y, z) (i.e. the object is transferred into itself) (Required operation according to group theory; and for us the identity operation has no other practical use.)

63 Point groups; mirror planes 3. Mirror plane, m horizontal Use the symbol (, ) to indicate the mirror image vertical A mirror plane transfers a right handed object to a left handed object

64 Point groups; inversion 4. Inversion, 1 (x, y, z) ( x, y, z) Inversion symmetry transfers a right handed object to a left handed object The object with inversion symmetry is said to be centrosymmetric

65 Point groups; rotation-inversion 5. Roto-inversion axis, n Rotation-inversion axis Rotation 360 o /n + inversion 4 Methane CH 4 Is a combined symmetry operation: rotation 360 o /n followed directly by an inversion

66 Point groups - notation Point symmetry elements according to Hermann-Mauguin notation

67 EXAMPLE - EXERCISE Which symmetry operations can you identify for the following objects: H 2 O NH 3 CO 2 CH 4 d xy orbital Exercise 10 (p. 16) Hint: Visit Symmetry@Otterbein;

68 (Hermann-Mauguin notation) Point groups - symbols 32 crystallographic point groups (classified according to their crystal system) Note: Usually the point group symbol starts with the highest or characteristic symmetry (n) (see p. 18: not fully consistent)

69 Point groups - symbols Exercise 13 (p. 19) Which symmetry elements can you identify in the following point group symbols: 1, 3, 6/m, 4/m? 1 Inversion (centrosymmetry) Triclinic 3 1 three-fold rotation axis Trigonal 6/m 1 six-fold rotation axis + mirror plane normal to 6-fold rot.axis 4/m 1 fourfold roto-inversion axis + mirror plane normal to 4 rot-inv. axis Hexagonal Tetragonal

70 Point groups - symbols Rules for the first, second and third symbol (point group) Fictive point group: j k l Symbol nb Symbol 1: Symmetry direction 1 (Characteristic symmetry element of the crystal system) Symbol 2: Symmetry direction 2 Symbol 3: Symmetry direction 3 (for mirror planes: normal vector)

71 Construction of stereographic projections - I Surround the object/crystal by a sphere Project the object down on a xy-equatorial plane The projected point is determined by the intersection of a connection line from the point of interest to the pole of the opposite side. The two halves of the spheres are distinguished by assigning + and, and/or by the use filled and open symbols Pole + xy-plane Pole

72 Point groups stereographic projections Construction of stereographic projections - II A thin circle limits the stereograms in the paper plane, representing the intersection of the xy-plane with the sphere Symmetry elements are drawn using relevant graphical symbols (see Table) The highest rotation axis is always chosen to be perpendicular to the projection plane Vertical mirror planes - drawn as thick lines Horizontal mirror plane drawn as a thick line around the circle periphery (in xy-plane)

73 Exercise: Draw the point groups mmm and 4mm in stereographic projections. b a

74 EXAMPLE - EXERCISE Exercise: Are there more symmetry elements in the projection drawn for mmm? Yes, there are three 2-fold rotation axes All three are normal to one of the mirror planes The full point group symbol is therefore 2/m 2/m 2/m

75 General and special positions General and special positions - A general position is not situated on any symmetry element. - A special position is situated on one or more symmetry elements. If a point is situated on a symmetry element, the latter operations will transfer the point just onto itself.

76 General and special positions for 2/m For convenience we select symmetry direction 1 to be II to c General position Special position (x, y, z) generates 4 equivalent points (x, y, 0) generates 2 equivalent points Special position Special position (0, 0, z) generates 2 equivalent points (0, 0, 0) generates 1 equivalent point

77 Symmetry operations with translation Glide planes and screw axes Going from point symmetry to symmetry for solids, two new symmetry elements appear Screw axis (n m ) = rotation (n) + translation (m/n) Glide plane (a, b, c, n, d) = mirroring + ½ translation along an axis (or more axes) Symmetry planes Symbol Translation Axial glide a a/2 b b/2 c c/2 Face diagonal glide n (a+b)/2, (b+c)/2, (a+c)/2 (a+b+c)/2 for cubic and tetragonal only Diamond glide d (a±b)/4, (b±c)/4, (a±c)/4 (Body diagonal) (a±b±c)/4 for cubic and tetragonal only

78 Symmetry screw axis Screw axis 2 1 Two-fold rotation (360 o /2 = 180 o ) + Translation (1/2) parallel to rotation axis Comparison with 2-fold rotation a 0

79 Symmetry glide planes Glide plane Example: a-glide Mirroring + ½ glide along a Comparison with just mirroring a 0

80 Space groups 230 space-groups Space group symbol: Xefg Used to describe periodic objects solids Combination of 32 point groups with the 14 Bravais lattices, glide plans and screw axes Lattice: P (R) F, I A, B, C Symmetry operations without translation: Inversion -1,1 Rotation n Mirror m Rotation-inversion n Characteristic symmetry elements (dependent on crystal system) Symmorphic space groups (73 groups) Symmetry operations with translation: Screw-axis n m, 2 1,6 3, etc. Glideplane a,b,c,n,d Non-symmorphic space groups (157 groups)

81 Space groups Information from space group on point group Why? E.g. be able to find number of general positions for a structure To identify the corresponding point group of a space group one must: 1) Remove symbol for the Bravais lattice (but note whether it is primitive or centered) 2) Exchange symbols for symmetry elements with translation with corresponding symbols for symmetry elements without translation n m n a,b,c,d or n m for a screw axis (go to rotation axis) for glide plane (go to a mirror plane) Examples: P6 3 /mmc 6 3 /mmc 6/mmm Pnma nma mmm

82 Space groups How to read International tables of crystallography See hand-outs, or page 31 in Compendium

83 EXAMPLE - EXERCISE Draw a 2-dimensional orthogonal unit cell with cell edges a = 3.80 Å and b = 3.90 Å. Exercise 1 (p. 5) Place the following atoms in the unit cell: Cu(1) (0,0); O(1) (0,0.5); O(2) (0.5,0) Calculate Cu-Cu distances and Cu-O distances. How do the calculated Cu-O distances fit your expectations of ionic/covalent bonding? a b Cu-Cu: a 2 + b 2 = (Cu-Cu) 2 Cu-Cu = ( ) = 4.72 Å Cu-O1 = 1/2b = 1.95 Å Cu-O2 = 1/2a = 1.90 Å Cu(1) O(1) O(2) From Shannon: Cu(II) CN = Å Cu(II) CN = Å O( II) (CN = 6) 1.26 Å Cu-O = 1.97 Å Cu-O = 2.13 Å

84 EXAMPLE - EXERCISE

85 EXAMPLE - EXERCISE How to draw a structure: CuO Given the information: a = 465 pm, b = 341 pm, c = 511 pm, = 99.5 Space group C2/c Cu in 4c O in 4e y = Question: What can you conclude directly from the given information? Crystal system: a b c, = = 90, 90 ( hence monoclinic) Bravais-lattice: C monoclinic, side centered (in ab plane) two lattice points Corresponding crystallographic point group: 2/m (4 points) Maximum general points for the space group is therefore 2 x 4 = 8 Number of formula units (CuO) per unit cell = 4 (4c and 4e positions are occupied)

86 From International Tables of Crystallography EXAMPLE - EXERCISE

87 EXAMPLE - EXERCISE C2/c O Cu

88 EXAMPLE - EXERCISE a = 465 pm, b = 341 pm, c = 511 pm, = 99.5 Space group C2/c Cu in 4c O in 4e y = O 1. (0.00, 0.416, 0.25) 3. (0.00, 0.584, 0.75) 2. (0.50, 0.916, 0.25) 4. (0.50, 0.084, 0.75) 1. (0.25, 0.25, 0.00) 3. (0.75, 0.25, 0.50) Cu 2. (0.75, 0.75, 0.00) 4. (0.25, 0.75, 0.50)

89 EXAMPLE - EXERCISE O 1. (0.00, 0.416, 0.25) 3. (0.00, 0.584, 0.75) 2. (0.50, 0.916, 0.25) 4. (0.50, 0.084, 0.75) Draw -angle in the paper plane Project along short axis a = 465 pm = 99.9 o Cu 1. (0.25, 0.25, 0.00) 3. (0.75, 0.25, 0.50) 2. (0.75, 0.75, 0.00) 4. (0.25, 0.75, 0.50) a = 465 pm, b = 341 pm, c = 511 pm, = 99.5 is the angle between a and c axis b-axis normal to ac plane

90 EXAMPLE - EXERCISE a = 465 pm = 99.5 o 0.25 O Cu 1. (0.00, 0.416, 0.25) 3. (0.00, 0.584, 0.75) 2. (0.50, 0.916, 0.25) 4. (0.50, 0.084, 0.75) 1. (0.25, 0.25, 0.00) 3. (0.75, 0.25, 0.50) 2. (0.75, 0.75, 0.00) 4. (0.25, 0.75, 0.50) a = 465 pm, b = 341 pm, c = 511 pm, = 99.5

91 EXAMPLE - EXERCISE What is the coordination number (CN) of Cu and O in this crystal structure? Do we know some other compounds that could be similar to CuO? NiO: NaCl-type structure; Ni in the center of octahedra MnO CoO Could CuO take a similar structure? Space group and crystal system do indicate some issues. Why? Cu(II): 3d 9 Jahn-Teller ion expect deformation of octahedron

92 EXAMPLE - EXERCISE What is the CN of Cu in this crystal structure? a = 465 pm O Cu = 99.5 o a = 465 pm, b = 341 pm, c = 511 pm, = What is expected Cu-O bond? Ionic radius Cu(II)? Ionic radius (O-II)? What is around Cu within a radius of approx. 200 pm?

93 EXAMPLE - EXERCISE a c

94 EXAMPLE - EXERCISE a c

95 EXAMPLE - EXERCISE a Coordinates along b-axis given (normal to paper) c Change color of atoms with shortest Cu O bonds)

96 EXAMPLE - EXERCISE c a

97 EXAMPLE - EXERCISE a Cu(II) is 6-coordinated (4+2) in a deformed octahedron c Cu(II) 3d 9 Jahn-Teller ion Remember: Work 3D when determining coordination number

98 EXAMPLE - EXERCISE What is the CN of O in this crystal structure? a = 465 pm = 99.9 o O Cu Refresh Ionic compound M m X x M X m x = CN( M ) CN( X ) x m

99 Crystallographic planes and directions Crystallographic planes and directions Crystallographic planes and directions are given relative to the coordinate system used to define the unit cell We use Miller indexes to define parallel planes (h k l) We use [u v w] to give directions in a crystal

100 Crystallographic planes and directions Crystal planes - Miller indices ( h k l ) c a Recipe: 1. Identify the plane that is adjacent to an equivalent plane that passes through origin 2. Find intersection of this plane with unit cell edges (a, b, c) 3. The reciprocals of these values are the Miller indices of the planes (1/ 0) b (001) a h b c 1 h k l 1/ 1/ 1/

101 Crystallographic planes and directions Crystal planes - Miller indices ( h k l )

102 Crystallographic planes and directions Crystal planes - Miller indices ( h k l )

103 Crystallographic planes and directions Crystal planes - Miller indices ( h k l )

104 Crystallographic planes and directions Crystal planes Miller indices ( h k l ) parallell planes

105 (2,0,2) Crystallographic planes and directions Directions [101] and [202] c [112] (1,1,2) 1) Directions are given as [uvw] 2) Directions in crystals are defined from origo: [100] II to a-axis [010] II to b-axis [001] II to c-axis (1,0,1) (½, ½,1) 3) Parallel directions have same index b a [½ 0 ½] = [101] = [202] = n[101]

106 Crystallographic planes and directions Crystal planes and crystal directions more on notation A given plane (h k l) A set of equivalent planes {h k l} Cubic system (1 0 0), (0 1 0), (0 0 1) {1 0 0} A given direction A set of equivalent directions [uvw] <uvw> The equivalent planes and directions are a result of the symmetry of the system [-111] c [111] e.g. fcc <111> [111] [111] [111] [111] a [111] [111] [111] [111] b

107 Supplementary information Collection of 32 point groups Supplementary information

108 Supplementary information Triclinic Monoclinic

109 Supplementary information Orthorombic Tetragonal

110 Supplementary information Trigonal

111 Supplementary information Hexagonal

112 Supplementary information Cubic

113 Solid solutions What is a solid solution? Water Alcohol Mixing on the molecular scale Cu Ni Cu-Ni alloy Mixing on the atomic scale in a crystalline solid

114 Solid solutions What types of solid solutions exist? Substitutional solid solution Interstitial solid solution Aliovalent substitutions defect generating solid solution Cu-Ni alloy Mixing on the atomic scale in a crystalline solid

115 Solid solutions Metals: substitutional solid solution Hume-Rothery substitutional solubility rules 1. The crystal structure of each element of the involved pair is the same 2. The atomic sizes of the involved atoms do not differ more than 15% 3. The elements do not differ greatly in electronegativity 4. Elements should have same valence

116 Solid solutions Solid solutions and defects Intrinsic defects associated with stoichiometric and pure crystals Extrinsic defects associated with substitutants or impurities (0.1 1 %) What about substitutants > 1%??? Solid solution Substitutional solid solution Aliovalent (heterovalent) substitution Interstitial solid solution

117 Solid solutions Requirements: for substitutional solid solution to form (for ionic/polar covalent compounds) 1) The ions must be of same charge 2) The ions must be similar in size. (15-20 % difference acceptable) 3) The crystal structures of the end members must be isostructural for complete solid solubility 4) Partial solid solubility is possible for non-isostructural end members Mg 2 SiO 4 (Mg in octahedra) - Zn 2 SiO 4 (Zn in tetrahedra) 5) The involved atoms must have preference for the same type of sites Cr 3+ only in octahedral sites, Al 3+ both octahedral and tetrahedral sites LiCrO 2 - LiCr 1-x Al x O 2 - LiAlO 2

118 Solid solutions Substitutional solid solution ionic compounds Al 2 O 3 corundum Al 3+ covalent radius 1.18 Å (Al 2-x Cr x )O 3 corundum Statistic distribution of Al(III) and Cr(III) atoms Cr 2 O 3 corundum Cr 3+ covalent radius 1.18 Å

119 Solid solutions Metals substitutional solid solution Effect of temperature High temperature Low temperature Disordered Au 0.5 Cu 0.5 bcc Ordered Au 0.5 Cu 0.5 primitive cubic Hume-Rothery substitutional solubility rules 1. Crystal structure of each element in the pair is the same 2. Atomic sizes of the atoms do not differ more than 15% 3. The elements do not differ greatly in electronegativity 4. Elements should have same valence

120 Sphere packings - holes Closest packing of spheres in 3D hcp Where to put the third layer? Option -1 Layer -A Layer-B Layer-A Layer-B Degree of filling = 74 % Each sphere have 12 nearest neighbors CN = 12 Hexagonal close-packed (hcp) (heksagonal tetteste kulepakking) ABABAB.

121 Sphere packings - holes Closest packing in 3D ccp Where to put the third layer? Option 2 Layer-A Layer-B Layer-C Layer-A Degree of filling = 74 % Each sphere have 12 nearest neighbors CN = 12 Cubic close-packed (ccp/fcc); (kubisk tetteste kulepakking) ABCABCABC..

122 Solid Solid solutions Voids/interstitial holes interstital solid solutions Example: ccp(fcc), hcp, bcc metals/alloys For hcp and ccp: Octahedral hole A B Possible to fill with in smaller atom: r M /r X 0.41 Tetrahedral hole Space for small Interstitial atoms r M /r X 0.22

123 Sphere packings - holes Close packing and holes in hcp and ccp 74% of the volume is filled by the spheres 26% voids Octahedral hole (1 per packing sphere) Tetrahedral holes (2 per packing sphere) T + and T hcp: Trigonal bipyramidal hole (5 spheres generating hole) (can not be filled at the same time as the tetrahedral holes) 125

124 Solid solutions Metals interstitial solid solutions Small atoms can enter into the smaller holes in the metal structure and create phases as: Interstitial carbides steel industry Stoichiometric carbides Fe 3 C cementite Interstitial and stoichiometric hydrides Hydrogen storage For interstitial solid solutions, the Hume-Rothery rules are: 1. Solute atoms must be smaller than the interstitial sites in the solvent lattice. 2. The solute and solvent should have similar electronegativities.

125 Solid solutions Aliovalent substitution in ionic compounds AO x Defect formation; charge compensation Too much positive charge Substitution by higher valence cations (B) into AO x Cation vacancies Interstitial anions Too much negative charge Substitution by lower valence cations (C) Anion vacancies Interstitial cations Side note: Similar patterns can be made for anion substitution; but not included here as anion substitution occurs to less extent in solid solutions

126 Density The unit cell: Contains Z number of formula units of the compound in question Note: the cations and anions can take more than one Wückoff position each! When checking/calculating/drawing: be aware of Wückoff site multiplicity The unit cell volume and the mass of the atoms in the unit cell can be calculated exact. NOTE: you anticipate that the atomic arrangement is perfect i.e. that there are no defects (interstitials; vacancies) X ray = m V unitcell = Formula weight number of units / cell Unitcell volume N A The calculated density is sometimes called «X-ray density»

127 Density Comparison of experimental (pycnometric) calculated; X-ray density x-ray > exp. Why??? Experimental density is burdened with errors: - Poor wetting of the particles/material - Inner porosity in the material not accessible by the liquid Calculated density is not reflecting the fully true picture - Major amounts of defects have not been accounted for

128 Applications of symmetry REPETITION Let s quickly take a look at the NH 3 -molecule What are the symmetry elements? What is the point group? How do we create the character table?

129 Applications of symmetry REPETITION Block diagonalization, this time x and y are coupled:

130 Applications of symmetry REPETITION Note the three classes of elements. There are two C 3 s, both C 3 and C 32, but the transformation matrix is the same so they belong to the same class. There are three vertical mirror planes, but they are all invariant, so they also belong to the same class.

131 Applications of symmetry REPETITION How do we find the last representation? Orthogonality and the fact that the character must be symmetric under E.

132 Applications of symmetry APPLICATIONS Let s take it a step further and look at vibrational spectroscopy of the N 2 O 4 -molecule. Self test: What is the point group?

133 Applications of symmetry N 2 O 4 You might have learned that the number of vibrational modes of a molecule is 3N 6 where N is the number of atoms. Is there a direct logic to this equation? -3 because of translation -3 because of rotation

134 Applications of symmetry N 2 O 4 12 vibrational modes should be possible for this molecule Point group: D 2h In fact, we can now use the character table to determine the symmetry of all 18 motions of this molecule and assign them to translation, rotation or vibration

135 Applications of symmetry N 2 O 4 Let s first label all the atoms: Now we need to see what happens to the axes under the different symmetry operations

136 Applications of symmetry N 2 O 4 We use characters and say that: If an atom moves, the character is 0. If an atom is stationary and the axis direction is unchanged, the character is 1. If an atom is stationary and the axis directions is reversed, the character is -1.

137 Applications of symmetry N 2 O 4 Now sum all characters in each class to make a reducible representation.

138 Applications of symmetry N 2 O 4 Now comes the tricky part. We have to reduce this by using the character table and the following rule:

139 Applications of symmetry N 2 O 4 Well, the order is 8. So we should have:

140 Applications of symmetry N 2 O 4 Now we use the character table to remove translations and rotations:

141 Applications of symmetry N 2 O 4 By looking at the character table we can easily imagine how some of these vibrations might look. Take the the three A g s for example, they have to be symmetric:

142 Applications of symmetry N 2 O 4 Can we deduce if the vibrations are IR active? The dipole moment must change, and only representations with x, y or z symmetry do this.

143 Applications of symmetry N 2 O 4 Can we deduce if the vibrations are IR active? The dipole moment must change, and only representations with x, y or z symmetry does this.

144 Applications of symmetry N 2 O 4 Can we deduce if the vibrations are Raman active? The polarizability of the molecule must change, and only representations that transform like products of x, y and z do this.

145 Applications of symmetry N 2 O 4 Can we deduce if the vibrations are Raman active? The polarizability of the molecule must change, and only representations that transform like products of x, y and z do this.

146 Crystallography space groups A two-fold rotation axis parallel with the z-axis is described as 2[001]

147 Crystallography space groups 2/m is identical to the inversion operation (-1): A b-axial glide plane with (y,z) reflection (mirror) is described as:

148 Crystallography space groups Diagonal glide plane, perpendicular to the c-axis: -

149 Crystallography space groups

150 Crystallography space groups

151 Crystallography space groups

152 Crystallography space groups Space group Pmm2: symmetry operations: m[100]; m[010]; 2[001] All these have ORIGIN as a common point (see figure right) P orthorhombic Bravais lattice; crystallographic space group: mm2

153 Crystallography space groups How many points are generated (to be drawn in left figure)? Answer: = (number of equivalent points of crystallographic point group) x (identical points of the Bravail lattice) mm2 gives 4 identical points (cf. stereographic projection) Bravais lattice is primitive Hence: nb of points = 4 x 1 = 4

154 Crystallography space groups Triclinic crystal system space group P-1; primitive Bravais lattice crystallographic point group -1

155 Crystallography space groups Additional symmetry exists; in addition to the inversion center in origin:

156 Crystallography space groups Monoclinic crystal system: a b c AND one angle deviating from 90 o, Which one? STANDARD setting: C-centering (if it does exist; and not use A or B centering) Consequence: b-axis is the so-called unique axis; i.e. perpendiacular to a- and c-axes; and 90 o.

157 Crystallography space groups According to standard notation, ab-projection is drawn for the space group Then practical to draw the deviating angle in the plane of projection. Then c-axis is unique axis; and any centering (if existing) is chosen as B Symmetry (from the space group symbol) B Bravais lattice; 2 equi.points Crystallographic space group: 2 Expects: 2 x 2 = 4 pointsto the drawn

158 Crystallography space groups Stepwise approach: Sketching: points sketching symmetry operations b 1 a O+ O+ Add 2[001] 2 O+ O+ O+ O+ Add B lattice 3 O+ O+½ O+ O+½

159 Crystallography space groups O+ O+ 3 O+ O+½ O+ O+½ Operate on the new point with the 2-fold rotation axis: O+ O+ 4 O+ O+½ O+½ O+½ O+½ O+ The unit cell contains now all 4 expected points Now continue with analysis of symmetry elements being present

160 Crystallography space groups O+ O+ 4 O+ O+½ O+½ O+½ O+½ O+ We have now the following points: (1) x,y,z; (2) x,-y,z; (3) x+½; y,z; (4) ½-x,-y,z How are the points (1) and (4) connected? We can add coordinates for more atoms in the sketch; e.g. (5) at x,1-y,z and ask: How is (1) and (5) connected by symmetry? And so on... In this way we identify all symmetry for this space gorup, and become able to draw the complete figure of symmetry operations

161 Crystallography space groups Example: How are the points z,y,z, and x+1/2, -y, z+1/2 related? We then write this in terms of the Seitz operator, and split the matrices: We realize quickly that there is a two-fold rotation axis: From the first comumn matrix we realize that this is actually a 2 1 [001] axis. What about the second column matrix? This tells you where this symmetry operation is located in the unit cell!! Lacoated at half the coordinate values: i.e. at 0.5 x (1/2 0 0) = (1/4, 0 0)

162 Crystallography space groups Two settings of the non-centrosymmetric space group Bb or Cc Respectively, with c-unique and b-unique axis

163 Crystallography space groups

164 Crystallography space groups We will now consider space group Pnma in detail and compare our findings/procedure with what tabulated in the International Tables for Crystallography; which is the following figures:

165 Crystallography space groups Pnma What can we read out of the space group symbol? Primitive Bravais lattice; 1 lattice point (0,0,0) Corresponding crystallographic point group: mmm A point group in the orthorhombic crystal system 8 general points for mmm Full symbol 2/m 2/m 2/m Altogether 8 point symmetry operations I, -1, m[100], m[010], m[001], 2[100], 2[010], 2[001], Expected 8 x 1 = 8 points in the unti cell for Pnma Pnma: Essential operators are: n[100] and m[010] and a[001] This is our starting point now...

166 Crystallography space groups In mathematical for, these tre essential symmetry operators are: They have one point in common: the origin of the unit cell

167 Crystallography space groups Points Ø+1/2 a[001] m[010] Symmetry n[100] Ø+ O+ Ø- Coordinates for the transformed points are: (1) x,y,z (2) -x, y+½, z+½ n[100] (3) x, -y, z m[010] (4) y+½, y, -z a[001] We have now located four out of the eight target positions

168 Crystallography space groups The next steps are now to adopt combinations of the three symmetry operators (5) By operating m[010] on point (2) n m (6) By operating a[010] on point (2) n a (7) By operating a[010] on point (5) n m a (8) By operating a[010] on point (3) m a (5): m[010] (-x,y+½,z+½) = (-x, ½-y, ½+z) (6): a[001](-x,y+½,z+½) = (½-x, ½+y, ½-z) Points (3) (5) O+1/2 Ø+1/2 (2) Ø+ O+ (1) a[001] m[010] Symmetry n[100] O-1/2 (6) Ø- (4)

169 Crystallography space groups The next steps are now to adopt combinations of the three symmetry operators (5) By operating m[010] on point (2) n m (6) By operating a[010] on point (2) n a (7) By operating a[010] on point (5) n m a (8) By operating a[010] on point (3) m a (7): a[010] (-x,y-½,z+½) = (½-x, ½-y, ½-z) (8): a[001](x,y,z) = (½+x, -y, -z) Points (3) (5) O+1/2 Ø+1/2 (2) Ø+ O+ (1) a[001] m[010] Symmetry n[100] (8) (7) Ø-1/2 O-1/2 (6) O- Ø- (4)

170 Crystallography space groups Points m[010] Symmetry (3) (5) O+1/2 Ø+1/2 (2) Ø+ O+ (1) a[001] n[100] (8) (7) Ø-1/2 O-1/2 (6) O- Ø- (4) The next step is now to identify all other symmetry operations and draw them into the upper right figure. We do this by again asking what type of symmetry connects various points (beyound what considered already) How is (1) and (5) connected? How is (1) and (6) connected? How is (1) and (7) connected? How is (1) and (8) connected?

171 Crystallography space groups (1) x,y,z (2) -x, y+½, z+½ (3) x, -y, z (4) y+½, y, -z (5) (-x, ½-y, ½+z) (6) (½-x, ½+y, ½-z (7) (½-x, ½-y, ½-z) (8) (½+x, -y, -z) How is (1) and (5) connected? How is (1) and (6) connected? How is (1) and (7) connected? How is (1) and (8) connected? 1 5: we see that there is a 2[001] transforming x, y into x, -y: There is a translational component of ½ along c; hence a 2 1 [001] axis. Which is located at (0,1/4,0) 1 6: we see that there is a 2[010] transforming x, z into x, -z: There is a translational component of ½ along b; hence a 2 1 [010] axis. Which is located at (1/4,0,1/4) 1 7: we see that there is inversion center transforming x,y,z into x,-y,-z: Which is located at (1/4,1/4,1/4) 1 8: we see that there is a 2[100] transforming y, z into y, -z: There is a translational component of ½ along a; hence a 2 1 [100] axis. Which is located at (0,0,0)

172 Crystallography space groups 2 1 [001] axis at (0,1/4,0) 2 1 [010] axis at (1/4,0,1/4) Inversion at (1/4,1/4,1/4) 2 1 [100] axis at (0,0,0) We will now draw these into the figure o Points m[010] Symmetry (3) (5) O+1/2 Ø+1/2 (2) Ø+ O+ (1) (8) (7) Ø-1/2 O-1/2 (6) O- Ø- (4) a[001] 1/4 o 1/4 n[100] The final step will be to complete the figure (let symmetry act on what is drawn)

173 Crystallography space groups Note: we found inversion symmetry at (1/4,1/4,1/4) A centrosymmetric spacegroup should have its origin at a point of inversion. Hence, we now need to do a coordinate shift for having inversion at origin. Thereafter, our figure will (when fully completed) be identical to that of the International Tables of Crystallography. We will discuss these representations in light of other information in the International Tables of Crystallography.

174 Crystallography phase transitions Example on phase transition; NiAs to MnP-type NiAs-type structure Hexagonal As: in hcp Ni: octahedral holes a = b c = = 90 o = 120 o Unit cells drawn as projection on the ab-plane Adding atoms at a later stage

175 Crystallography phase transitions * * * * * * Compendium West

176 Sphere packings - holes Closest packing of spheres in 3D hcp Where to put the third layer? Option -1 Layer -A Layer-B Layer-A Layer-B Degree of filling = 74 % Each sphere have 12 nearest neighbors CN = 12 Hexagonal close-packed (hcp) (heksagonal tetteste kulepakking) ABABAB.

177 Crystallography phase transitions Example on phase transition; NiAs to MnP-type b a NiAs-type structure Hexagonal As: in hcp Ni: oct.holes a = b c = = 90 o = 120 o Unit cells drawn as projection on the ab-plane Adding atoms at a later stage Space group P6 3 /mmc Primitive Bravais lattice One point (0,0,0) 6/mmm

178 Crystallography space groups The orthorhombic MnP-type structure is structurally very related to the hexagonal NiAs-type structure A continuous, displacive, 2.order phase transition may occur for a given compound, as function of pressure or temperature (or substitution) We will explore this in detail, starting with defining relationship between the two unit cells of these structures

179 Crystallography phase transitions Example on phase transition; NiAs to MnP-type b h Vectors b o a h c o NiAs-type structure Hexagonal As: in hcp Ni: oct.holes a = b c = = 90 o = 120 o MnP Space group Pnma Primitive Bravais lattice One point (0,0,0) mmm Space group P6 3 /mmc Primitive Bravais lattice One point (0,0,0) 6/mmm

180 Crystallography phase transitions Example on phase transition; NiAs to MnP-type b h Vectors b o c o a h NOTE: The orthorhombic cell is A-centered as long as c o /b o = 3 = a o = c h b o = a h c o = 2b h + a h In case of an orthorhombic structure, the requirement on a = b (for the hexagonal structure) is no longer active. If this length requirement is still in place, we have so far just descirbed the hexagonal structure in an orthorhombic setting. We call this unit cell for an orthohexagonal unit cell. Then c o /b o = 3 = A consequence of the hexagonal orthorhombic phase transition is then the possibility of c o /b o (but c o /b o = 3 is not a good order parameter for the transition)

181 Crystallography phase transitions Example on phase transition; NiAs to MnP-type b h Vectors b o c o a h Vector relations; describing orthorhombic unit cell in terms of the hexagonal one: a o = c h b o = a h c o = 2b h + a h In a matrix representation we write (note the way of writing!):

182 Crystallography phase transitions Example on phase transition; NiAs to MnP-type b h Vectors b o c o a h And then the opposite relationship: Vector relations; describing hexagonal unit cell in terms of the orthorhombic one: a h = b o b h = -1/2b o + 1/2c o c h = a o In a matrix representation we write (note the way of writing!):

183 Crystallography phase transitions Example on phase transition; NiAs to MnP-type

184 Crystallography space groups

185 Crystallography space groups See additional information for completeness

186 Crystallography space groups

187 Crystallography space groups

188 Crystallography space groups

189 Crystallography space groups

190 Crystallography space groups