Structure, Bonding and Properties

Transcription

1 Structure, Bonding and Properties Atomic Arrangements In gases there is no order In liquids there is short range order In solids there is long range range The order is determined by the type of atomic bonds Lattices A grid like pattern Composed of unit cells Unit cells are stacked together endlessly to form the lattice (with no empty spaces between cells) Scanning Tunneling Microscope Image of Iron in the (110) plane

2 Structure Subatomic level Electronic structure of individual atoms that defines interaction among atoms (interatomic bonding). Atomic level Arrangement of atoms in materials (for the same atoms can have different properties, e.g. two forms of carbon: graphite and diamond) Microscopic structure Arrangement of small grains of material that can be identified by microscopy. Macroscopic structure Structural elements that may be viewed with the naked eye. Monarch butterfly ~ 0.1 m

3 Amorphous Solids: The atoms are not orderly arranged in 3 D. Some can have order only in two dimensions such as the layered materials (clays, graphite, MoS2 ). While there is no long range order in the amorphous materials, certain bond distances are maintained and some short range order can be achieved. Crystalline Materials: Atoms are orderly arranged in 3 D for long distances. Crystalline solids can be classified as single crystals or monocrystals and polycrystals. Polycrystals exhibit a 2 D defect known as grain boundaries.

4 2-D lattice

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6 Lattice: A 3 dimensional system of points that designate the positions of the components (atoms, ions, or molecules) that make up the substance. Unit Cell: The smallest repeating unit of the lattice. The lattice is generated by repeating the unit cell in all three dimensions

7 Crystal Systems Crystallographers have shown that only seven different types of unit cells are necessary to create all point lattice Cubic a= b = c ; α = β = γ = 90 Tetragonal a= b c ; α = β = γ = 90 Rhombohedral a= b = c ; α = β = γ 90 Hexagonal a= b c ; α = β = 90, γ =120 Orthorhombic a b c ; α = β = γ = 90 Monoclinic a b c ; α = γ = 90 β Triclinic a b c ; α γ β 90 The basis vectors a, b and c define the unit cell; their magnitudes a, b and c respectively, are the lattice parameters of the unit cell. The angles b^c, c^a and a^b, are conventionally labelled α, β and γ respectively.

8 Bravais Lattices Many of the seven crystal systems have variations of the basic unit cell. August Bravais ( ) showed that 14 standards unit cells could describe all possible lattice networks. The number of ways in which points can be arranged regularly in 3 D, such that the stacking of unit cells fills space, is not limitless.

9 Certain unit cells are compatible with body centering, face centering or side centering. For example the orthorhombic unit cells can be: C F A Bravais lattice is a lattice in which every lattice point has exactly the same environment.

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11 Symmetry Although the properties of a crystal can be anisotropic, there may be different directions along which they are identical. These directions are said to be equivalent and the crystal is said to possess symmetry. For example, that a particular edge of a cube cannot be distinguished from any other is a measure of its symmetry; an orthorhombic parallelepiped has lower symmetry, since its edges can be distinguished by length. otranslation orotation oreflection (Mirror) oglide oscrew

12 Translation: Operation required as definition of unit cell. Rotation: and 6 Fold Rotation Axis corresponding to angles of rotation of 360, 180, 120, 90 and 60 degrees.

13 Although objects themselves may appear to have 5 fold, 7 fold, 8 fold, or higher fold rotation axes, these are not possible in crystals. The reason is that the external shape of a crystal is based on a geometric arrangement of atoms. Note that if we try to combine objects with 5 foldand 8 fold apparent symmetry, that we cannot combine them in such a way that they completely fill space, as illustrated below.

14 Mirror Symmetry A mirror symmetry operation is an imaginary operation that can be performed to reproduce an object. The operation is done by imagining that you cut the object in half, then place a mirror next to one of the halves of the object along the cut. If the reflection in the mirror reproduces the other half of the object, then the object is said to have mirror symmetry. The plane of the mirror is an element of symmetry referred to as a mirror plane, and is symbolized with the letter m.

15 Center of Symmetry Another operation that can be performed is inversion through a point. Rotoinversion Combinations of rotation with a center of symmetry perform the symmetry operation of rotoinversion. 2 fold Rotoinversion The operation of 2 fold rotoinversion involves first rotating the object by 180o then inverting it through an inversion center. This operation is equivalent to having a mirror plane perpendicular to the 2 fold rotoinversion axis.

16 3 fold Rotoinversion This involves rotating the object by 120o (360/3 = 120), and inverting through a center. A cube is good example of an object that possesses 3 fold rotoinversion axes. A 3 fold rotoinversion axis is denoted as Thus, this crystal has the following symmetry elements: 1 4 fold rotation axis (A4) 4 2 fold rotation axes (A2), 2 cutting the faces & 2 cutting the edges. 5 mirror planes (m), 2 cutting across the faces, 2 cutting through the edges, and one cutting horizontally through the center. Note also that there is a center of symmetry (i). The symmetry content of this crystal is thus: i, 1A4, 4A2, 5m Later you will see that this belongs to crystal class 4/m2/m2/m.

17 Certain Bravais lattice types are compatible with some symmetry operations: 14 Bravais Lattices + Compatible Symmetry Elements 32 Crystal Symmetry Classes

18 Inversion Mirror Rotation Rotations Mirrors Improper rotations 14 Bravais Lattices + Compatible Symmetry Elements 32 Crystal Symmetry Classes

19 Two Translational Symmetry Elements 32 Crystal Symmetry Classes + Translational Symmetry Operations 230 Space Groups

20 All combinations of point symmetry elements are not possible A three fold axis can not just have one two fold axis perpendicular to it. In three dimensions the existence of two perpendicular two folds implies the existence of a third perpendicular two fold The allowed combinations of point symmetry elements are called point groups Point symmetry elements compatible with 3D translations

21 Only 32 point groups are consistent with periodicity in 3D. The 32 point groups Schönflies and Hermann Maugin symbols for crystallographic point groups

22 Combining symmetry elements For three dimensions 32 point groups 14 Bravais lattices but only 230 space groups For two dimensions 5 lattices 10 point groups but only 17 plane groups

23 Examples: Two fold Rotation.. Mirror Plane

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25 Space groups are numbered (1-230) from the lowest to the highest symmetry. First letter of space group notation indicates the type of unit cell: P = primitive I = Body-centered F = Face-centered C = side-centered Other symbols indicate symmetry operations: m = mirror plane 3 = three-fold rotation axis 2 1 = two-fold screw axis c = glide axis

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27 Polymorphism or Allotropy Many elements or compounds exist in more than one crystalline form under different conditions of temperature and pressure. This phenomenon is termed polymorphism and if the material is an elemental solid is called allotropy. Example: Iron (Fe Z = 26) liquid above 1539 C. δ-iron (BCC) between 1394 and 1539 C. γ-iron (FCC) between 912 and 1394 C. α-iron (BCC) between -273 and 912 C. 912 o C 1400 o C 1539 o C α iron γ iron δ iron liquid iron BCC FCC BCC

28 Another example of allotropy is carbon. Pure, solid carbon occurs in three crystalline forms diamond, graphite; and large, hollow fullerenes. Two kinds of fullerenes are shown here: buckminsterfullerene (buckyball) and carbon nanotube.

29 x Crystallographic Planes and Directions Atom Positions in Cubic Unit Cells A cube of lattice parameter a is considered to have a side equal to unity. Only the atoms with coordinates x, y and z greater than or equal to zero and less than unity belong to that specific cell. z 0,0,1 0,1,1 0 x, y, z < 1 1,0,1 1,1,1 ½, ½, ½ 1,0,0 0,0,0 1,1,0 0,1,0 y

30 Directions in The Unit Cell For cubic crystals the crystallographic directions indices are the vector components of the direction resolved along each of the coordinate axes and reduced to the smallest integer. z Example direction A 1,0,1 1,0,0 0,0,1 0,1,1 0,0,0 ½, ½, ½ A 1,1,1 1,1,0 0,1,0 y a) Two points origin coordinates 0,0,0 and final position coordinates 1,1,0 b) 1,1,0-0,0,0 = 1,1,0 c) No fractions to clear d) Direction [110] x

31 z 0,0,1 Example direction B a) Two points origin coordinates 1,1,1 and final position coordinates 0,0,0 b) 0,0,0-1,1,1 = -1,-1,-1 c) No fractions to clear _ d) Direction [111] x C B 0,0,0 1,1,1 ½, 1, 0 y Example direction C a) Two points origin coordinates ½,1,0 and final position coordinates 0,0,1 b) 0,0,1 - ½,1,0 = -½,-1,1 c) There are fractions to clear. Multiply times 2. 2( -½,-1,1) = -1,-2,2 d) Direction [ 12 2]

32 Notes About the Use of Miller Indices for Directions A direction and its negative are not identical; [100] is not equal to [bar100]. They represent the same line but opposite directions.. direction and its multiple are identical: [100] is the same direction as [200]. We just forgot to reduce to lowest integers. Certain groups of directions are equivalent; they have their particular indices primarily because of the way we construct the coordinates. For example, a [100] direction is equivalent to the [010] direction if we re-define the co-ordinates system. We may refer to groups of equivalent directions as directions of the same family. The special brackets < > are used to indicate this collection of directions. Example: The family of directions <100> consists of six equivalent directions < 100 > [100],[010],[001],[010],[001],[100]

33 Miller Indices for Crystallographic planes in Cubic Cells Planes in unit cells are also defined by three integer numbers, called the Miller indices and written (hkl). Miller s indices can be used as a shorthand notation to identify crystallographic directions (earlier) AND planes. Procedure for determining Miller Indices locate the origin identify the points at which the plane intercepts the x, y and z coordinates as fractions of unit cell length. If the plane passes through the origin, the origin of the co-ordinate system must be moved! take reciprocals of these intercepts clear fractions but do not reduce to lowest integers enclose the resulting numbers in parentheses (h k l). Again, the negative numbers should be written with a bar over the number.

34 z A y Example: Miller indices for plane A a) Locate the origin of coordinate. b) Find the intercepts x = 1, y = 1, z = 1 c) Find the inverse 1/x=1, 1/y=1, 1/z=1 d) No fractions to clear e) (1 1 1) x

35 More Miller Indices - Examples c c c 1/5 a b a 2/3 b a b c c a b a b c b a Notes About the Use of Miller Indices for Planes A plane and its negative are parallel and identical. Planes and its multiple are parallel planes: (100) is parallel to the plane (200) and the distance between (200) planes is half of the distance between (100) planes.

36 Certain groups of planes are equivalent (same atom distribution); they have their particular indices primarily because of the way we construct the co-ordinates. For example, a (100) planes is equivalent to the (010) planes. We may refer to groups of equivalent planes as planes of the same family. The special brackets { } are used to indicate this collection of planes. In cubic systems the direction of miller indices [h k l] is normal o perpendicular to the (h k l) plane. in cubic systems, the distance d between planes (h k l ) is given by the formula constant. Example: d = h 2 + a k 2 + l where a is the lattice The family of planes {100} consists of three equivalent planes (100), (010) and (001) 2

37 A family of crystal planes contains all those planes are crystallographically equivalent. Planes have the same atomic packing density a family is designated by indices that are enclosed by braces. - {111}: (111), (111), (111), (111), (111), (111), (111), (111)

38 Single Crystal Polycrystalline materials Anisotropy and isotropy

39 Two Types of Indices in the Hexagonal System a 3 c a 1 a 2 a 1,a 2,and c are independent, a 3 is not! a 3 = - (a 1 + a 2 ) Miller: (hkl) (same as before) Miller-Bravais: (hkil) i = - (h+k) c (001) = (0001) - - (110) = (1100) c a 3 a 1 a (110) = (1100) a 3 a 1 a 2 (100) = (1010) -

40 Structures of Metallic Elements H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Primitive Cubic Body Centered Cubic Cubic close packing (Face centered cubic) Hexagonal close packing

41 Structure BCC FCC HCP Metal Lattice Constant a, nm c, nm Atomic Radius, nm Chromium Iron Molybdenum Potassium Sodium Tungsten Aluminum Copper Gold Nickel Silver Zinc Magnesium Cobalt Titanium

42 SIMPLE CUBIC STRUCTURE (SC) Rare due to poor packing (only Po has this structure) Close-packed directions are cube edges. Coordination # = 6 (# nearest neighbors) Number of atoms per unit cell= 1 atom

43 Atomic Packing Factor (APF) APF = Volume of atoms in unit cell* Volume of unit cell *assume hard spheres APF for a simple cubic structure = 0.52 a close-packed directions contains 8 x 1/8 = 1 atom/unit cell R=0.5a atoms unit cell APF = 1 volume 4 atom 3 π (0.5a)3 a 3 volume unit cell 6

44 Body Centered Cubic (BCC) Close packed directions are cube diagonals. --Note: All atoms are identical; the center atom is shaded differently only for ease of viewing. atoms unit cell APF = 2 a 3 Coordination # = 8 Unit cell contains: x 1/8 = 2 atoms/unit cell Close-packed directions: length = 4R = 3 a 4 3 π ( 3a/4)3 volume unit cell R volume atom APF for a BCC = 0.68 a

45 Face Centered Cubic (FCC) Close packed directions are face diagonals. --Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing. atoms unit cell APF = 4 a 3 Coordination # = 12 Close-packed directions: length = 4R = 2 a Unit cell contains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell 4 3 π ( 2a/4)3 volume unit cell volume atom a APF for a FCC = 0.74

46 Hexagonal Close-Packed (HCP) The APF and coordination number of the HCP structure is the same as the FCC structure, that is, 0.74 and 12 respectively. An isolated HCP unit cell has a total of 6 atoms per unit cell. 2 atoms shared by two cells = 1 atom per cell 12 atoms shared by six cells = 2 atoms per cell 3 atoms

47 Close-Packed Structures Both the HCP and FCC crystal structures are close-packed structure. Consider the atoms as spheres: Place one layer of atoms (2 Dimensional solid). Layer A Place the next layer on top of the first. Layer B. Note that there are two possible positions for a proper stacking of layer B.

48 The third layer (Layer C) can be placed in also teo different positions to obtain a proper stack. (1)exactly above of atoms of Layer A (HCP) or (2)displaced A B A : hexagonal close packed A B C : cubic close packed

49 120 A B C A A B C : cubic close pack A B A : hexagonal close pack 90 A B A

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51 Packing of Non-Identical Spheres The holes between the atoms of a crystal, called interstices, can house smaller atoms without appreciable distortion of the host. Many compounds of two or more elements have a structure, which can be described by the smaller atoms/ions filling the interstices between the larger atoms/ions. Different structures arise from the different numbers and sizes of the interstices in the fcc, hcp, bcc and simple cubic structures. The way the interstices are distributed is also important. Two important interstices are the tetragonal and the octahedral interstices in close packed structures

52 Tetrahedral/Octahedral The tetragonal interstice, surrounded by four atoms. The octahedral interstice, surrounded by six atoms. The six atoms surround (or coordinate) the interstice in the shape of an octahedron. (Also consider as three atoms below and three atoms above.)

53 FCC Interstitials FCC Octahedral In the fcc structure, consider the interstitial site shown. Six host atoms surround it. These six atoms surround (or coordinate) the interstitial site in the shape of an octahedron. There is one octahedral site at the centre of the FCC cell (½,½,½) and one on each of the twelve cell edges (½,0,0). A total of 13 octahedral sites. Calculate the octahedral void radius as a fraction of the parent atom radius in a FCC structure. r r void atom = 0.414

54 FCC Tetrahedron In the fcc structure, consider the interstitial site shown. Four atoms surround it. These four atoms surround the interstitial site in the shape of a tetrahedron. There are eight tetrahedral sites in the FCC unit cell located at (¼,¼,¼). Calculate the tetrahedral void radius as a fraction of the parent atom radius in a FCC structure. r r void atom = 0.225

55 HCP Interstitials HCP Octahedral In the hcp structure, consider the interstitial site shown. Six host atoms surround it. These six atoms surround the interstitial site in the shape of an octahedron. There are six octahedral sites. HCP Tetragonal In the hcp structure, consider the interstitial site shown. Four atoms surround it. These four atoms surround the interstitial site in the shape of a tetrahedron. Total of 8 tetrahedral sites.

56 BCC Interstitials Note: the bcc structure is not close packed. In the bcc structure the octahedron and tetrahedron are not regular, they do not have edges of equal lengths. BCC Octahedral In the bcc structure, consider the interstitial site shown. Six host atoms surround it. These six atoms surround the interstitial site in the shape of an octahedron. There is one octahedral site on each of the six BCC cell faces (½,½,0) and one on each of the twelve cell edges (½,0,0). Total of 18 sites.

57 BCC Tetrahedral In the bcc structure, consider the interstitial site shown. Four atoms surround it. These four atoms surround the interstitial site in the shape of a tetrahedron. There are four tetrahedral sites on each of the six BCC cell faces (½,¼,0). Total of 24 sites. Using this diagram calculate the octahedral void radius as a fraction of the parent atom radius in a BCC crystal r r void atom = 0.155

58 Using this diagram calculate the tetrahedral void radius as a fraction of the parent atom radius in a BCC crystal r r void atom = 0.291

59 Interstitial sites Locations between the normal atoms or ions in a crystal into which another - usually different - atom or ion is placed. o Cubic site - An interstitial position that has a coordination number of eight. An atom or ion in the cubic site touches eight other atoms or ions. o Octahedral site - An interstitial position that has a coordination number of six. An atom or ion in the octahedral site touches six other atoms or ions. o Tetrahedral site - An interstitial position that has a coordination number of four. An atom or ion in the tetrahedral site touches four other atoms or ions.

60 Crystals having filled Interstitial Sites Octahedral, Oh, Sites Tetrahedral, Th, Sites FCC Lattice has: 3 [=12(¼)] Oh sites at edge centers + 1 Oh site at body center FCC Lattice has: 8 Th sites at ¼, ¼, ¼ positions Interstitial sites are important because we can derive more structures from these basic FCC, BCC, HCP structures by partially or completely different sets of these sites

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62 Interstitial coordinates It is easy to identify the atomic coordinates of the interstitials in the fcc and bcc structures. Use the unit cell diagrams below to help identify the interstitial positions in the hcp structure.

63 Density Calculations Since the entire crystal can be generated by the repetition of the unit cell, the density of a crystalline material, ρ = the density of the unit cell = (atoms in the unit cell, n ) (mass of an atom, M) / (the volume of the cell, V c ) Atoms in the unit cell, n = 2 (BCC); 4 (FCC); 6 (HCP) Mass of an atom, M = Atomic weight, A, in amu (or g/mol) is given in the periodic table. To translate mass from amu to grams we have to divide the atomic weight in amu by the Avogadro number N A = atoms/mol The volume of the cell, V c = a 3 (FCC and BCC) a = 2R 2 (FCC); a = 4R/ 3 (BCC) where R is the atomic radius.

64 Density Calculation n: number of atoms/unit cell na A: atomic weight ρ = V V C N C : volume of the unit cell A N A : Avogadro s number (6.023x10 23 atoms/mole) Example Calculate the density of copper. R Cu =0.128nm, Crystal structure: FCC, A Cu = 63.5 g/mole n = 4 atoms/cell, V C = a = ( 2R 2) = 16 2R ρ = [16 2( (4)(63.5) 8 ) ] = 8.89g / cm 8.94 g/cm 3 in the literature 3

65 Example Rhodium has an atomic radius of nm (1.345A) and a density of 12.41g.cm -3. Determine whether it has a BCC or FCC crystal structure. Rh (A = g/mol) Solution V c n ρ = na V N C A If rhodium is BCC then n = 2 and a a n 3 3 a = n A = ρn x(0.1345nm) = 2 A If rhodium is FCC then n = 4 and a n: number of atoms/unit cell A: atomic weight V C : volume of the unit cell = 12.41g. cm 3 3 a x(0.1345nm) = = nm n 4 Rhodium has a FCC structure g. mol x10 atoms. mole 3 3 = (4r = nm 3 = ( 4 r 3 ) 3 ) = r = r N A : Avogadro s number (6.023x10 23 atoms/mole) = x cm 3 = nm 3

66 Linear And Planar Atomic Densities L l = 2a a = 4R 3 Crystallographic direction A B Linear atomic density = 2R/L l = Planar atomic density: L l = 2π R 2 /(Area A D E B )

67 Structure of Ceramics Ceramics keramikos - burnt stuff in Greek - desirable properties of ceramics are normally achieved through a high temperature heat treatment process (firing). Usually a compound between metallic and nonmetallic elements Always composed of more than one element (e.g., Al 2 O 3, NaCl, SiC, SiO 2 ) Bonds are partially or totally ionic, can have combination of ionic and covalent bonding (electronegativity) Generally hard, brittle and electrical and thermal insulators Can be optically opaque, semi-transparent, or transparent Traditional ceramics based on clay (china, bricks, tiles, porcelain), glasses. New ceramics for electronic, computer, aerospace industries.

68 Crystal Structures in Ceramics with predominantly ionic bonding Crystal structure is defined by The electric charge: The crystal must remain electrically neutral. Charge balance dictates chemical formula (Ca 2+ and F - form CaF 2 ). Relative size of the cation and anion. The ratio of the atomic radii (r cation /r anion ) dictates the atomic arrangement. Stable structures have cation/anion contact. r r Cation Anion

69 Coordination Number: the number of anions nearest neighbors for a cation. As the ratio gets larger (that is as r cation /r anion ~ 1) the coordination number gets larger and larger. Holes in sphere packing Triangular Tetrahedral Octahedral

70 Calculating minimum radius ratio for a triangle: B O O B 1 2 AO = 1 AB 2 AO r r r a c a A AB ra + r c = r a r a + r c = cos30 = = C = cosα ( α = 30 ) AO = 1 AB 2 AO r r r a c a AB ra + r c = r a r a A + r c = cos45 = = 2 2 C = cosα ( α = 45 ) o for an octahedral hole

71 C.N. = 2 r C /r A < C.N. = < r C /r A < C.N. = < r C /r A < C.N. = < r C /r A < C.N. = < r C /r A < 1.0

72 Ionic (and other) structures may be derived from the occupation of interstitial sites in close-packed arrangements.

73 Comparison between structures with filled octahedral and tetrahedral holes o/t fcc(ccp) hcp all oct. NaCl NiAs all tetr. CaF 2 (ReB 2 ) o/t (all) (Li 3 Bi) (Na 3 As) ½t sphalerite (ZnS) wurtzite (ZnS) (½ o CdCl 2 CdI 2 ) Location and number of tetrahedral holes in a fcc (ccp) unit cell -Z = 4(number of atoms in the unit cell) -N = 8(number of tetrahedral holes in the unit cell)

74 Crystals having filled Interstitial Sites Octahedral, Oh, Sites Ionic Crystals prefer the NaCl Structure: NaCl structure has Na+ ions at all 4 octahedral sites Large interatomic distance LiH, MgO, MnO, AgBr, PbS, KCl, KBr Na+ ions Cl- ions

75 Crystals having filled Interstitial Sites Tetrahedral, Th, Sites Both the diamond cubic structure And the Zinc sulfide structures have 4 tetrahedral sites occupied and 4 tetrahedral sited empty. Zn atoms Covalently Bonded Crystals Prefer this Structure Shorter Interatomic Distances than ionic Group IV Crystals (C, Si, Ge, Sn) Group III--Group V Crystals (AlP, GaP, GaAs, AlAs, InSb) Zn, Cd Group VI Crystals (ZnS, ZnSe, CdS) Cu, Ag Group VII Crystals (AgI, CuCl, CuF) S atoms

76 The "zinc blende" lattice is face centered cubic (fcc) with two atoms in the base at (0,0,0) and (¼, ¼, ¼).

77 AX Type Crystal Structures Rock Salt Structure (NaCl) NaCl structure: r C = r Na = nm, r A = r Cl = nm r C /r A = 0.56 Coordination = 6 Cl NaCl, MgO, LiF, FeO, CoO Na Cesium Chloride Structure (CsCl) CsCl Structure: r C = r Cs = nm, r A = r Cl = nm r C /r A = 0.94 Coordination = 8 Cl Cs Is this a body centered cubic structure?

78 Zinc Blende Structure (ZnS) radius ratio = Coordination = 4 S Zn ZnS, ZnTe, SiC have this crystal structure A m X p -Type Crystal Structures If the charges on the cations and anions are not the same, a compound can exist with the chemical formula A m X p, where m and/or p 1. An example would be AX 2, for which a common crystal structure is found in fluorite (CaF 2 ).

79 CaF 2 Fluorite The lattice is face centered cubic (fcc) with three atoms in the base, one kind (the cations) at (0,0,0), and the other two (anions of the same kind) at (¼, ¼, ¼), and (¼, ¾, ¼).

80 Fluorite CaF 2 Fluorite (CaF 2 ): r C = r Ca = nm, r A = r F = nm r C /r A = 0.75 From the table for stable geometries we see that C.N. = 8 Other compounds that have this crystal structure include UO 2, PuO 2, and ThO 2. A m B n X p -Type Crystal Structures It is also possible for ceramic compounds to have more than one type of cation; for two types of cations (represented by A and B), their chemical formula may be designated as A m B n X p. Barium titanate (BaTiO 3 ), having both Ba 2+ and Ti 4+ cations, falls into this classification. This material has a perovskite crystal structure and rather interesting electromechanical properties

81 CaTiO 3 -dielectric BaTiO 3 - ferroelectric Pb(Mg 1/3 Nb 2/3 )O 3 - relaxor ferroelectric Pb(Zr 1-x Ti x )O 3 - piezoelectric (Ba 1-x La x )TiO 3 semiconductor Perovskite - an Inorganic Chameleon ABX 3 - three compositional variables, A, B and X (Y 1/3 Ba 2/3 )CuO 3 -x - superconductor Na x WO 3 - mixed conductor; electrochromic SrCeO 3 - H - protonic conductor RECoO 3-x - mixed conductor (Li 0.5-3x La 0.5+x )TiO 3 - lithium ion conductor LaMnO 3-x - Giant magnetoresistance

82 The lattice is essentially cubic primitive, but may be distorted to some extent and then becomes orthorhombic or worse. It is also known as the BaTiO 3 or CaTiO 3 lattice and has three different atoms in the base. In the example it would be Ba at (0,0,0), O at (½, ½,,0) and Ti at (½, ½, ½). A particular interesting perovskite (at high pressures) is MgSiO 3. It is assumed to form the bulk of the mantle of the earth, so it is the most abundant stuff on this planet, neglecting its Fe/Ni core. The mechanical properties (including the movement of dislocations) of this (and related) minerals are essential for geotectonics forming the continents, making and quenching volcanoes, earthquakes

83 The perovskite structure CaTiO 3 -TiO 6 octahedra -CaO 12 cuboctahedra (Ca 2+ and O 2- form a cubic close packing) preferred basis structure of piezoelectric, ferroelectric and superconducting materials

84 Perovskite Structure ABO 3 e.g. KNbO 3 SrTiO 3 LaMnO 3 SrTiO 3 cubic, a = 3.91 Å In SrTiO 3, Ti-O = a/2 = Å Sr-O = a 2/2 = Å CN of A=12, CN of B=6 OR

85 The fractional coordinates for cubic perovskite are: A = (½, ½, ½) A = ( 0, 0, 0) B = (0, 0, 0) OR B = (½, ½, ½) X = (½,0,0) (0,½,0) (0,0,½) X = (½,½,0) (½,0,½) (0,½ ½) Draw one of these as a projection.

86 In SrTiO 3, Ti-O ~ 1.95 Å a typical bond length for Ti-O; stable as a cubic structure larger In BaTiO 3, Ti-O is stretched, > 2.0 Å Too long for a stable structure. Ti displaces off its central position towards one oxygen square pyramidal coordination

87 This creates a net dipole moment : Displacement by 5-10% Ti-O bond length Random dipole orientations Aligned dipole orientations paraelectric ferroelectric Under an applied electric field, dipole orientations can be reversed, i.e. the structure is polarisable Dipoles tend to be frozen in at room temperature; as increase temperature, thermal vibrations increase the polarisability

88 Cubic (Pm3m) T > 393 K Ti-O O Distances (Å)( Tetragonal (P4mm) 273 K < T < 393 K Ti-O O Distances (Å)( 1.83, , , 2.21 Toward a corner Orthorhombic (Amm2) 183 K < T < 273 K Ti-O O Distances (Å)( , , , Toward an edge BaTiO 3 Phase Transitions In the cubic structure BaTiO 3 is paraelectric.. That is to say that the orientations of the ionic displacements are not ordered and dynamic. Below 393 K BaTiO 3 becomes ferroelectric and the displacement of the Ti 4+ ions progressively displace upon cooling. Rhombohedral (R3m) 183 K < T < 273 K Ti-O O Distances (Å)( , 1.88, Toward a face See Kwei et al. J. Phys. Chem. 97, 2368 (1993),

89 Density Calculations in Ceramic Structures n ' ( ) ρ = C A V A c + N n : number of formula units in unit cell (all ions that are included in the chemical formula of the compound = formula unit) ΣA C : sum of atomic weights of cations in the formula unit ΣA A : sum of atomic weights of anions in the formula unit V c : volume of the unit cell N A : Avogadro s number, 6.023x10 23 (formula units)/mol A A

90 Example: NaCl n = 4 in FCC lattice ΣA C = A Na = g/mol ΣA A = A Cl = g/mol r Na =0.102x10-7 r Cl =0.181x10-7 cm V c = a 3 = (2r Na +2r Cl ) 3 V c = ( ) 3 cm 3 ( ) 4 3 = 2. 14g ρ = cm [ ]

91 Structure, Bonding and Properties BaTiO 3 : Ferroelectric (T C ~ 130 C, ε r > 1000) Ba 2+ ion stretches the octahedra (Ti-O dist. ~ 2.00Å), this lowers energy of CB (LUMO) and stabilizes SOJT dist. SrTiO 3 : Insulator, Normal dielectric (ε r ~ x) Sr 2+ ion is a good fit (Ti-O dist. ~ 1.95Å), this compound is close to a ferroelectric instability and is called a quantum paraelectric. PbTiO 3 : Ferroelectric (T C ~ 490 C) Displacements of both Ti 4+ and Pb 2+ (6s 2 6p 0 cation) stabilize ferroelectricity BaSnO 3 : Insulator, Normal dielectric (ε r ~ x) Main group Sn 4+ has no low lying t 2g orbitals and no tendency toward SOJT dist. KNbO 3 : Ferroelectric (T C ~ x) Behavior is very similar to BaTiO 3 KTaO 3 : Insulator, Normal dielectric (ε r ~ x) Ta 5d orbitals are more electropositive and have a larger spatial extent than Nb 4d orbitals (greater spatial overlap with O 2p), both effects raise the energy of the t 2g LUMO, diminishing the driving force for a SOJT

92 Transformations Many physical properties depend on the crystallographic directions and the anisotropy of a material is best described by tensors. (a)tensors can be used to describe physical properties (b) Symmetry effects on physical properties can be described by how the tensor transforms under a symmetry operation (c) the magnitude of a property in any arbitrary direction can be evaluated by transforming the tensor. (d) It can be used to draw a geometric representation of the property. (e) It provides a way of averaging the properties over a certain direction. (f) It relates properties of single crystals with polycrystals

93 Tensor: A specific type of matrix representation that can relate the directionality of either a material property (property tensors conductivity, elasticity) or a condition/state (condition tensors stress, strain). Tensor of zero rank: scalar quantity (density, temperature). Tensor of first rank: vector quantity (force, electric field, flux of atoms). Tensor of second rank: relates two vector quantities (flux of atoms with concentration gradient). Tensor third rank: relates vector with a second rank tensor (electric field with strain in a piezoelectric material) Tensor Fourth rank: relates two second rank tensors (relates strain and stress Elasticity)

94 The key to understanding property or condition tensors is to recognize that tensors can be specified with reference to some coordinate system which is usually defined in 3 D space by orthogonal axes that obey a right hand rule. Rotation Matrix and Euler Angles: Rotation Matrix and Euler Angles: Several schemes can be used to produce a rotation matrix. The three Euler angles are given as three counterclockwise rotations: (a)a rotation about a z axis, defined as φ1 (b)a rotation about the new x axis, defined as Φ (c)a rotation about the second z position, defined as φ2

95 The rotation matrix a is given by the matrix multiplication of the rotation matrices of each individual rotations: ] ] Φ Φ Φ Φ Φ + Φ Φ Φ + Φ = Φ Φ Φ Φ = = Φ cos cos sin sin sin cos sin cos cos cos sin sin cos sin cos cos sin sin sin sin cos cos sin cos sin sin cos cos cos cos sin 0 sin cos cos sin 0 sin cos cos sin 0 sin cos φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ φ a a a a a

96 Mathematically, the transformation converts a set of orthogonal axes (X 1, Y 1, Z 1 ) into another (X 2, Y 2, Z 2 ). The two set of axes are related to one another by nine direction cosines (a 11, a 12, a 13, a 21, a 22, a 23, a 31, a 32, a 33 ). The first subscript refers to the new axis.