1、SOLID STATE,Crystals Structures 1. Unit Cell,Crystal structure basics unit cells symmetry lattices,Some important crystal structures and properties close packed structures octahedral and tetrahedral holes basic structures ferroelectricity,Objectives,By the end of this section you should: be able to

2、identify a unit cell in a symmetrical pattern know that there are 7 possible unit cell shapes be able to define cubic, tetragonal, orthorhombic and hexagonal unit cell shapes,Why Solids?,most elements solid at room temperature atoms in fixed position “simple” case - crystalline solid Crystal Structu

3、re,Why study crystal structures? description of solid comparison with other similar materials - classification correlation with physical properties,Early ideas,Crystals are solid - but solids are not necessarily crystalline Crystals have symmetry (Kepler) and long range order Spheres and small shape

4、s can be packed to produces regular shapes (Hooke, Hauy),?,Group discussion,Kepler wondered why snowflakes have 6 corners, never 5 or 7. By considering the packing of polygons in 2 dimensions, demonstrate why pentagons and heptagons shouldnt occur.,Empty space not allowed,Definitions1. The unit cell

5、,“The smallest repeat unit of a crystal structure, in 3D, which shows the full symmetry of the structure”,The unit cell is a box with: 3 sides - a, b, c 3 angles - , , , Seven unit cell shapes,Cubica=b=c=90 Tetragonala=bc=90 Orthorhombicabc=90 Monoclinicabc =90, 90 Triclinicabc 90 Hexagonala=bc=90,

6、=120 Rhombohedrala=b=c=90,2D example - rocksalt (sodium chloride, NaCl),We define lattice points ; these are points with identical environments,Choice of origin is arbitrary - lattice points need not be atoms - but unit cell size should always be the same.,This is also a unit cell - it doesnt matter

7、 if you start from Na or Cl,- or if you dont start from an atom,This is NOT a unit cell even though they are all the same - empty space is not allowed!,In 2D, this IS a unit cellIn 3D, it is NOT,All M.C. Escher works (c) Cordon Art-Baarn-the Netherlands.All rights reserved.,Summary,Unit cells must l

8、ink up - cannot have gaps between adjacent cells All unit cells must be identical Unit cells must show the full symmetry of the structure next section,Crystals Structures 2: Symmetry,By the end of this section you should: be able to recognize rotational symmetry and mirror planes know about centres

9、of symmetry be able to identify the basic symmetry elements in cubic, tetragonal and orthorhombic shapes understand centring and recognize face-centred, body-centred and primitive unit cells. Know some simple structures (Fe, Cu, NaCl, CsCl),Symmetry“Something possesses symmetry if it looks the same

10、from 1 orientation”,Rotational symmetry Can rotate by 120 about the C-Cl bond and the molecule looks identical - the H atoms are indistinguishable,This is called a rotation axis - in particular, a three fold rotation axis, as rotate by 120 (= 360/3) to reach an identical configuration,In general: n-

11、fold rotation axis = rotation by (360/n),We talk about the symmetry operation (rotation) about a symmetry element (rotation axis),? Think of examples for n=2,3,4,5,6,Mirror Plane Symmetry “Arises when one half of an object is the mirror image of the other half”,This molecule has two mirror planes,On

12、e is horizontal, in the plane of the paper - bisects the H-C-H bonds Other is vertical, perpendicular to the plane of the paper and bisects the Cl-C-Cl bonds,Centre of Symmetry“present if you can draw a straight line from any point, through the centre, to an equal distance the other side, and arrive

13、 at an identical point” (phew!),Centre of symmetry at S,No centre of symmetry,All M.C. Escher works (c) Cordon Art-Baarn-the Netherlands.All rights reserved.,Unit cell symmetries - cubic,4 fold rotation axes (passing through pairs of opposite face centres, parallel to cell axes) TOTAL = 3,Unit cell

14、symmetries - cubic,4 fold rotation axes TOTAL = 3,3-fold rotation axes (passing through cube body diagonals) TOTAL = 4,Unit cell symmetries - cubic,4 fold rotation axes TOTAL = 3,3-fold rotation axes TOTAL = 4,2-fold rotation axes (passing through diagonal edge centres) TOTAL = 6,Mirror planes - cub

15、ic,3 equivalent planes in a cube,6 equivalent planes in a cube,Tetragonal Unit Cella = b c ; = = = 90,c a, b,elongated / squashed cube,Reduction in symmetry,CubicTetragonal Three 4-axesOne 4-axis Two 2-axes Four 3-axesNo 3-axes Six 2-axesTwo 2-axes Nine mirrorsFive mirrors See tutorial,Example,CaC2

16、- has a rocksalt-like structure but with non-spherical carbides,2-,C,C,Carbide ions are aligned parallel to c c a,b tetragonal symmetry,Cubic Unit Cell,a=b=c, =90,a,c,b,Many examples of cubic unit cells: e.g. NaCl, CsCl, ZnS, CaF2, BaTiO3,All have different arrangements of atoms within the cell. So

17、to describe a crystal structure we need to know: the unit cell shape and dimensions the atomic coordinates inside the cell,Primitive and Centred Lattices,Copper metal is face-centred cubic Identical atoms at corners and at face centres Lattice type F also Ag, Au, Al, Ni.,Primitive and Centred Lattic

18、es,-Iron is body-centred cubic Identical atoms at corners and body centre (nothing at face centres) Lattice type I Also Nb, Ta, Ba, Mo.,Primitive and Centred Lattices,Caesium Chloride (CsCl) is primitive cubic Different atoms at corners and body centre. NOT body centred, therefore. Lattice type P Al

19、so CuZn, CsBr, LiAg,Primitive and Centred Lattices,Sodium Chloride (NaCl) - Na is much smaller than Cs Face Centred Cubic Rocksalt structure Lattice type F Also NaFl, KBr, MgO.,Another type of centring,Side centred unit cell,Notation: A-centred if atom in bc plane B-centred if atom in ac plane C-cen

20、tred if atom in ab plane,Unit cell contents Counting the number of atoms within the unit cell,Many atoms are shared between unit cells,Atoms Shared Between:Each atom counts: corner8 cells1/8 face centre2 cells1/2 body centre1 cell1 edge centre4 cells1/4,lattice typecell contents P1 =8 x 1/8 I2 =(8 x

21、 1/8) + (1 x 1) F4 =(8 x 1/8) + (6 x 1/2) C ?,e.g. NaCl Na at corners: (8 1/8) = 1 Na at face centres (6 1/2) = 3 Cl at edge centres (12 1/4) = 3 Cl at body centre = 1 Unit cell contents are 4(Na+Cl-),2 =(8 x 1/8) + (2 x 1/2),Summary,Crystals have symmetry Each unit cell shape has its own essential

22、symmetry In addition to the basic primitive lattice, centred lattices also exist. Examples are body centred (I) and face centred (F),Crystals Structures 3: Miller Index,By the end of this section you should: understand the concept of planes in crystals know that planes are identified by their Miller

23、 Index and their separation, d be able to calculate Miller Indices for planes know the d-spacing equation for orthogonal crystals be able to use the d-spacing equation,Lattice Planes and Miller Indices,Imagine representing a crystal structure on a grid (lattice) which is a 3D array of points (lattic

24、e points). Can imagine dividing the grid into sets of “planes” in different orientations,All planes in a set are identical The planes are “imaginary” The perpendicular distance between pairs of adjacent planes is the d-spacing Need to label planes to be able to identify them,Find intercepts on a,b,c

25、: 1/4, 2/3, 1/2 Take reciprocals 4, 3/2, 2 Multiply up to integers: (8 3 4) if necessary,Exercise - What is the Miller index of the plane below?,Find intercepts on a,b,c: 1/2, 1, 1/2 Take reciprocals 2, 1, 2 Multiply up to integers: (2, 1, 2),Plane perpendicular to y cuts at , 1, (0 1 0) plane,Gener

26、al label is (h k l) which intersects at a/h, b/k, c/l (hkl) is the MILLER INDEX of that plane.,This diagonal cuts at 1, 1, (1 1 0) plane,NB an index 0 means that the plane is parallel to that axis,Using the same set of axes draw the planes with the following Miller indices:,(0 0 1),(1 1 1),d-spacing

27、 formula,For orthogonal crystal systems (i.e. =90) :-,For cubic crystals (special case of orthogonal) a=b=c :-,e.g. for(1 0 0)d = a (2 0 0)d = a/2 (1 1 0)d = a/2 etc.,A tetragonal crystal has a=4.7 , c=3.4 . Calculate the separation of the: (1 0 0) (0 0 1) (1 1 1) planes,A cubic crystal has a=5.2 (=

28、0.52nm). Calculate the d-spacing of the (1 1 0) plane,4.7 3.4 2.4 ,Summary,We can imagine planes within a crystal Each set of planes is uniquely identified by its Miller index (h k l) We can calculate the separation, d, for each set of planes (h k l),Crystals Structures 4: Closed Packed Structures,B

29、y the end of this section you should: understand the concept of close packing know the difference between hexagonal and cubic close packing know the different types of interstitial sites in a close packed structure recognise and demonstrate that cubic close packing is equivalent to a face centred cu

30、bic unit cell,Inorganic Crystal Structures,All crystal structures may be described in terms of the unit cell and atomic coordinates of the contents Many inorganic structures may be described as arrays of space filling polyhedra - tetrahedra, octahedra, etc. Many structures - ionic, metallic, covalen

31、t - may be described as close packed structures,Close packed structures - metals,Most efficient way of packing equal sized spheres. In 2D, have close packed layers,Coordination number (CN) = 6. This is the maximum possible for 2D packing.,Can stack close packed (c.p.) to give 3D structures.,Two main

32、 stacking sequences:,If we start with one cp layer, two possible ways of adding a second layer (can have one or other, but not a mixture) :,Two main stacking sequences:,If we start with one cp layer, two possible ways of adding a second layer (can have one or other, but not a mixture) :,Lets assume

33、the second layer is B (red). What about the third layer? Two possibilities: (1) Can have A position again (blue). This leads to the regular sequence ABABABA. Hexagonal close packing (hcp) (2) Can have layer in C position, followed by the same repeat, to give ABCABCABC Cubic close packing (ccp),Hexag

34、onal close packed,Cubic close packed,No matter what type of packing, the coordination number of each equal size sphere is always 12,We will see that other coordination numbers are possible for non-equal size spheres,These will be studied further in the lab The reasons why a particular metal prefers

35、a particular structure are still not well understood,Metals usually have one of three structure types: ccp (=fcc, see next slide), hcp or body centred cubic (bcc),Close packed ionic structures,Ionic structures - cations (+ne) and anions (-ne) In many ionic structures, the anions, which are larger th

36、an the cations, form a c.p. array and the cations occupy interstitial holes within this anion array. Two main types of interstitial site: TETRAHEDRAL : CN = 4 OCTAHEDRAL :CN = 6,Tetrahedral T+,Tetrahedral T-,Octahedral O,ccp = fcc ?,Build up ccp layers (ABC packing),Add construction lines - can see

37、fcc unit cell,c.p layers are oriented perpendicular to the body diagonal of the cube -,Hexagonal close packed structures (hcp),hcp,bcc,Summary,Close packing occurs in a variety of metals We can envisage layers, positions denoted by A, B and C, so that hexagonal close packing is represented by ABABA

38、and cubic close packing is represented by ABCABCA ccp is equivalent to face-centred cubic Small ions can occupy interstitial sites in a close packed structure - both tetrahedral (4) and octahedral (6) sites exist,Crystals Structures-5: Interstitial Sites,By the end of this section you should: know h

39、ow atom positions are denoted by fractional coordinates be able to calculate bond lengths for octahedral and tetrahedral sites in a cube be able to calculate the size of interstitial sites in a cube,Objectives,By the end of this section you should: know how atom positions are denoted by fractional c

40、oordinates be able to calculate bond lengths for octahedral and tetrahedral sites in a cube be able to calculate the size of interstitial sites in a cube,Fractional coordinates,Used to locate atoms within unit cell,0, 0, 0 , , 0 , 0, 0, , ,Note: atoms are in contact along face diagonals (close packe

41、d),1. 2. 3. 4.,Octahedral Sites,Coordinate , , Distance = a/2,Coordinate 0, , 0 =1, , 0 Distance = a/2,In a face centred cubic anion array, cation octahedral sites at: , 0 0, 0 0, 0 0 ,Tetrahedral sites,Relation of a tetrahedron to a cube:,i.e. a cube with alternate corners missing and the tetrahedr

42、al site at the body centre,Can divide the f.c.c. unit cell into 8 minicubes by bisecting each edge; in the centre of each minicube is a tetrahedral site,So 8 tetrahedral sites in a fcc,Bond lengths important dimensions in a cube,Face diagonal, fd (fd) = (a2 + a2) = a 2,Body diagonal, bd (bd) = (2a2

43、+ a2) = a 3,Octahedral: half cell edge, a/2 Tetrahedral: quarter of body diagonal, 1/4 of 3a Anion-anion: half face diagonal, 1/2 of 2a,Bond lengths:,Sizes of interstitials,fcc / ccp,Spheres are in contact along face diagonals octahedral site, bond distance = a/2 radius of octahedral site = (a/2) -

44、r tetrahedral site, bond distance = a3/4 radius of tetrahedral site = (a3/4) - r,Summaryf.c.c./c.c.p anions,4 anions per unit cell at:000000 4 octahedral sites at:000000 4 tetrahedral T+ sites at: 4 tetrahedral T- sites at:,A variety of different structures form by occupying T+ T- and O sites to dif

45、fering amounts: they can be empty, part full or full. We will look at some of these later. Can also vary the anion stacking sequence - ccp or hcp,Summary,By understanding the basic geometry of a cube and use of Pythagoras theorem, we can calculate the octahedral bond length (a/2) and tetrahedral bon

46、d length (3a/4) in a fcc structure As a consequence, we can calculate the radius of the octahedral interstice= (a/2) - r and of the tetrahedral interstice = (a3/4) - r, where r is the radius of the packing ion.,Crystals Structures 6: Some Examples,By the end of this section you should: be able to dr

47、aw a simple crystal structure projection appreciate that a variety of important crystal structures can be described by close-packing be able to compare and contrast similar structures,Crystal Structure Projections,Another way of describing structures Make drawings of a structure projected down one a

48、xis onto a unit cell face,b,a,ORIGIN,Example 1 - Rocksalt,Example 2 - Zinc Blende (Sphalerite),Example 3 - Fluorite structure,Descriptions of Structures,With ccp anion array: Rock salt, NaClO occupied Zinc Blende, ZnST+ (or T-) occupied * Antifluorite, Na2OT+ and T- occupied With hcp anion array: Wu

49、rtzite, ZnS T+ (or T-) occupied With ccp cation array: Fluorite, ZrO2T+ and T- occupied * major structure in semiconductor/microelectronics industry,Rocksalt structure Most alkali halides (LiF, KCl, etc.) Many divalent metal chalcogenides (CaO, BaS, etc.) Some carbides, nitrides (TiC, LaN, etc.) Sph

50、alerite or Zinc Blende Structure Many chalcogenides (ZnSe, HgS, etc.) Many pnictides (BN, GaAs, etc.) Some halides (CuX) Diamond (C, Si) Major structure in the semiconductor/microelectronics industry Fluorite/Antifluorite structure Some divalent halides (CaF2) Some tetravalent oxides (PbO2) Some mon

51、ovalent chalcogenides (Li2O, Na2S),Sphalerite (ZnS) vs Diamond structure,Ball and stick shows us the 4-fold coordination in both structures,Looking at tetrahedra in the structure helps us see the “diamond shape”,Fluorite structure,Can be thought of as a 3D array of alternating empty and occupied cub

52、es,Cadmium Chloride, CdCl2 structure,ccp Cl ions, half of O occupied by Cd, T+ T- sites empty. Layered structure,Layers of Cd ions sandwiched between chloride layers. Adjacent sandwiches are linked by Cl-Cl van der Waals bonds.,In the closely related cadmium iodide structure (CdI2), layers are simil

53、ar but with hcp (ABABAB) packing of iodide ions. These structures are common in divalent transition metal halides (generally not fluorides, which are too ionic and adopt alternative structures) Bonding must have considerable covalent character - if we consider coordination of Cl - three nearest neig

54、hbour Cd ions to one side, with twelve next-nearest-neighbour Cl ions. The anti-CdCl2 structure occurs in Cs2O,Nickel Arsenide (NiAs) structure,h.c.p. analogue of rocksalt structure h.c.p. As with octahedral Ni,c pointing towards us,c pointing upwards,In the c-direction, the Ni-Ni distance is rather

55、 short. Overlap of 3d orbitals gives rise to metallic bonding. The NiAs structure is a common structure in metallic compounds made from (a) transition metals with (b) heavy p-block elements such as As, Sb, Bi, S, Se.,Coordination of As is also 6 but as a trigonal prism:,Summary (see lab/web pages),S

56、ummary of AX structures, wurtzite ZnSCN = 4 sphalerite,NaCl, NiAsCN = 6,CsClCN = 8,General trend is to get higher coordination numbers with larger (heavier) cations. This is seen also with AX2 structures,Summary of AX2 structures,SiO2, BeF2silica structureCN = 4 : 2,TiO2, MgF2rutile structureCN = 6

57、: 3,CdCl2, CdI2layer structureCN = 6 : 3,Note: rutile structure will be introduced in the lab,PbO2, CaF2fluorite structureCN = 8 : 4,Compare:1) Be, Mg, Ca fluorides 2) Si, Ti, Pb dioxides,Crystals Structures 7: Understanding,By the end of this section you should: understand the concept of the radius

58、 of an atom or ion know the trends in ionic radius with coordination number, oxidation state, group know about the radius ratio and be able to calculate this for octahedral and 8-fold coordination,Ionic radii and bond distances,Ionic radii cannot be “measured” - estimated from trends in known struct

59、ures (reference - Shannon, Acta Cryst. (1976) A32 751),Oxide ion: r0 taken as 1.26 ,Refs: Krug et al. Zeit. Phys Chem. Frankfurt 4 36 (1955) Krebs, Fundamentals of Inorganic Crystal Chemistry, (1968),Notes: Ion radii for given element increase with CN,Notes: Ion radii for given element increase with CN Ion radii for given element decreas