Crystal Structures Metals tend to have a close-packing (c. p.) structure. In 2D – 6 circles fit around one circle In 3D – 12 spheres fit around one sphere (3 above and 3 below) ABCABC… – cubic closest packing (ccp) ABABAB… – hexagonal closest packing (hcp) AAAA… – coordination number is 8 in 3D ( ), so this packing structure is not close-packing. ccp = fcc (face-centered cubic) – must turn on diagonal to see the planes unit cell – smallest structural unit primitive or simple cubic ⅛(8) = 1 atom per unit cell 52% packing efficiency

hexagonal unit cell closest packing of first and second layers layer a layer b layer c hexagonal closest packing cubic closest packing abab… (74%) abcabc… (74%) expanded side views face-centered unit cell

body-centered cubic 2 atoms per unit cell 68% packing efficiency face-centered cubic ⅛(8) + ½(6) = 4 atoms per unit cell 74% packing efficiency 1/8 atom at 8 corners 1 atom at center 1/8 atom at 8 corners 1/2 atom at 6 faces Au, Ag and Cu have a fcc packing structure, and are very ductile ( 延展性的 ) and very malleable ( 可鍛的 ). Mg is hexagonal close-packed (hcp) and show some ductility. W has a body centered cubic packing structure and is brittle.

fcc structure has 4 sets of c. p. planes and are ductile and malleable (easy to slide). hcp structure has only one set of c. p. plane and are less malleable. Miller Indices – 3 numbers which define a lattice plane (hkl) For a cubic unit cell, the (100) set of planes cuts x at 1a, where a is the unit cell edge length, cuts y at ∞ (parallel), and cuts z at ∞ (parallel). Miller indices – reciprocal of where it cuts axes 1/1, 1/∞, 1/∞  (100) ½a, 1/∞, 1/∞  (200) _ Negative numbers are allowed. 1 (called one bar) indicates a negative number. Miller indices for planes – (hkl)for sets of equivalent planes such as (100), (010), and (001) – {hkl}for crystal faces/surfaces – {hkl} for growth direction – [hkl]For equivalent sets of directions, e.g. cubic [100], [010] etc., can write the general symbol.

Bravais Lattices

d-spacing formula (see p. 447)

Indices of Directions For cubic systems, an [hkl] direction is always perpendicular to the (hkl) plane of the same indices.

Miller Indices of Crystal Faces Truncated Rhombic Dodecahedral Cu 2 O Nanocages J. Am. Chem. Soc. 2008, 130, High-Resolution TEM Image of A Corner- and Edge- Truncated Cu 2 O Nanocube Adv. Funct. Mater. 2007, 17, 3773

Atomic Arrangements of Atoms on Different Planes of a fcc Metal Top and side views of the most stable structures of thiophenol on Au {111}, {110}, and {100} surfaces. BE denotes binding energy. Inorg. Chem. 2011, 50, 8106

Area Density of Gold Atoms on Different Planes The area density of gold atoms on the {111}, {100}, and {110} planes of gold are obtained by using the smallest 2-dimensional repeating areas as the unit cells. Calculations of the area density (r A ) of the gold atoms are given below. Here Au–Au bond length is Å, and Au roll-to-roll distance is Å. Knowing the particle size, one can calculate the total surface atoms on a particle.

Crystal Structure and Different Planes of Cu 2 O The Cu atoms arrange in a fcc sublattice, the O atoms in a bcc sublattice. The unit cell contains 4 Cu atoms and 2 O atoms.fccbcc Cu atoms occupy half of the tetrahedral sites. Crystal structures of Cu 2 O oriented to show the (a) (100), (b) (111), and (c, d) (110) planes. Surface Cu atoms on the {110} and {111} faces are shown with yellow circles.

Crystal structure models of cuprite Cu 2 O showing the (a, b) {100}, (c, d) {111}, and (e, f) {110} surfaces. Oxygen atoms are shown in red, while copper atoms are shown in white. Two different viewing angles have been presented for each surface to more clearly show the number of surface atoms. For panel b, the uppermost layer of oxygen atoms have been removed to clearly show the number of surface copper atoms. Panel d shows the {111} planes rotated 63º with respect to the model shown in panel c. The blue triangle in panel d encloses the same area as that shown in panel c. The cubic unit cell parameter is a.

Ionic Structures – anionic lattices with cations occupying interstices (anions are larger) Structure CN of Anions CN of Cations NaCl66 CsCl88 ZnS (cubic, zinc blende)44 ZnS (hexagonal, wurtzite)44 CaF 2 (fluorite)48 TiO 2 (rutile, see p. 50)36 eutactic – cations are too large for holes, anions are not in contact with each other. e.g. NaCl LiCl would be regular c. p. Three types of holes: (a) Td holes, CN = 4(b) Oh holes, CN = 6 (c) cubic holes, CN = 8 In closest packing structures, there are twice as many Td sites as Oh sites. 8 Td sites and 4 Oh sites: (¼)12 at edges + 1 in the center eutactic 1:1 compounds 1:2 compounds

NaCl – Cl – ions form fcc lattice and Na + ions occupy all Oh sites Cl – = ⅛(8) + ½(6) = 4Na + = ¼(12) + 1(1) = 44 formula units for NaCl CsCl – Cl – ions form simple cubic lattice with Cs + ions in all cubic sites. CaF 2 (fluorite) – Ca 2+ ions form fcc lattice with F – ions in all the Td sites Ca ÅF – Å K 2 S and Na 2 O form an antifluorite structure. O 2– ions form fcc lattice with Na + in all Td sites. 4 formula units per cell. ZnS (cubic) – S 2– ions form fcc lattice with Zn 2+ ions in ½ Td sites Wurtzite ZnO

Ti Ca

ZnS (hexagonal) – hcp S 2– framework with Zn 2+ ions in ½ Td sites S 2– ions form ABAB… structure Diamond – ZnS cubic structure8 atoms in unit cell of diamond NiAs (see p. 42 & 43) has Ni ions in Oh sites and As ions in trigonal prismatic sites. CdI 2 – hcp I ions with Cd in ½ Oh sites. Van der Walls forces hold I-I together. Can remove layers with Scotch tape. (see p. 51)

If remove ½ cations from rock salt structure, one gets CdCl 2 structure. CdCl 2 – cubic closest packed Cl – ions with Cd 2+ in ½ Oh sites. It is layered and soft. Can remove layers with Scotch tape. (see p. 53) TiO 2 (rutile) has a hcp structure with half of the Oh sites filled. (see p. 48 & 49) AnionInterstitial SitesStructure OhT + T – 1.ccp1NaCl 2.ccp1ZnS (zinc blende) 3.ccp11antifluorite (e.g. Na 2 O) 4.ccp ½CdCl 2 5.hcp1NiAs 6.hcp1ZnS (wurtzite) 7.hcp ½CdI 2 8.hcp ½TiO 2 (rutile)

nickel arsenide structures TiO 2 rutile structure

The CdI 2 structure: (a) the basal plane of the hexagonal unit cell, with two possible choices of origin (b) the layer stacking sequence. The CdCl 2 structure

Radius Ratio Rule Consider minimum size of Oh hole. Want r M /r x ratio. (2r x ) 2 + (2r x ) 2 = (2r x + 2r M ) 2 cos 45 º = 2r x /(2r x + 2r M ) = 1/√2 face diagonal = √2r body diagonal = √3r  r 2 + (√2r) 2 = (body diagonal) 2 2r x + 2r M = 2√2r x 2r M = (2√2 – 2)r x r M = (√2 – 1)r x r M /r x = minimum size of a Oh hole. If the ratio is larger than this number, the cations would be in Oh holes. What is the minimum size of a cubic hole? body diagonal = 2(r x + r M ) = √3r, where r = cell edge = 2r x 2(r x + r M ) = 2√3r x r M = (√3 – 1)r x r M /r x = √3 – 1 = r

CN Radius Ratio Structure 20–0.155linear –0.255trigonal –0.414Td –0.732Oh –1.0cubic With a radius ratios of 1.2, cations form the lattice structure and anions will be in holes. Ti 4+ /O 2–  0.75 Å/1.22 Å = 0.61 Ti 4+ ions want Oh sites (CN = 6) Twice as many O 2–, therefore O 2– ions need CN = 3 TiO 2 has a hcp O 2– structure with Ti 4+ in ½ of the Oh sites. Space Filling Polyhedra – anions at vertices - Td and Oh structures that share corners, edges, and/or faces (many examples in Crystal Growth and Design and Chemistry of Materials) NaCl

- emphasize on CN of cations - good for framework connectivity, topology and finding voids NaCl – infinite edge-sharing Oh Edge-sharing Oh leaves Td voids. Antifluorite (Na 2 O) – infinite edge-sharing Td Edge-sharing Td leaves Oh voids. Cubic ZnS – infinite corner-sharing Td (ZnS 4 units) Hexagonal ZnS – infinite corner-sharing ZnS 4 Td Na 2 O cubic ZnS hexagonal ZnS

Linking Polyhedra corner-sharing Oh edge-sharing Oh edge-sharing Td face-sharing Oh M–M distance = 1.16 MX corner-sharing Td M–M distance = 2 MX face-sharing Td M–M distance = 0.67 MX This is not allowed because M–M distance is less than the MX distance.

Factors Affecting Structures 1.Ionic interactions 2.Covalent interactions 3.Non-bonding electrons – crystal field effects (high spin and low spin transition metals) Pauling’s Structural Rules 1.Anions form polyhedra around cations 2.Sum of ionic radii give cation-anion distance 3.Use radius ratio to determine CN of cation 4.Charge balance 5.Polyhedra sharing edges and faces have decreased stability due to cation – cation repulsion Radius Ratio just right cation too largecation too small 6 coordinated LiClOK. eutactic structure NaClmust decrease CN

Silicates – crystal structures composed of linking SiO 4 4– tetrahedra units Si:O ratio# of bridging O# of non-bridging O structure 1:404isolated SiO 4 4– Td 1:3.513Si 2 O 7 6– dimer 1:322(SiO 3 ) n 2n– chains or Si 3 O 9 6– rings 1:2.531(Si 2 O 5 ) n 2n– infinite sheets 1:240SiO 2 3D framework silica, quartz SiO 3 Si 2 O 5 Si 4 O 11 Si 6 O 17

(a–e) Layers of SiO 4 tetrahedra sharing 3 vertices. In (a) and (b) the small black circles represent Si atoms and the open circles O atoms. Oxygen atoms lying above Si atoms are drawn more heavily. The O atoms are omitted from the nets (c)–(e).

Plan of the structure of  -quartz. Small black circles represent Si atoms. The oxygen atoms lie at different heights above the plane of the paper. The arrangement of the Si atoms in (a)  -quartz and (b)  -quartz. Beryl, Be 3 Al 2 Si 6 O 18

Examples of silicates: beryl ( 綠柱玉 ) – Be 3 Al 2 Si 6 O 18 Si:O = 1:3 chains or rings zircon ( 镐石 ) – ZrSiO 4 isolated SiO 4 4– ~ 1.5 ÅZr 4+ ~ 0.98 Å Na 2 Si 3 O 7 – Si:O = 1:2.33between sheets and 3D network actually infinite double sheet silicate mica ( 雲母 ) or muscovite ( 白雲母 ) – KAl 2 (OH) 2 (Si 3 Al)O 10 Al substitute some Si sites (Si + Al):O = 1:2.5 infinite sheets Spinels ( 尖晶石 ) – AB 2 O 4 structure Spinels form ccp lattice. MgAl 2 O 4 is a normal spinel ([A] tet [B 2 ] oct O 4 ) Mg 2+ in Td sites (⅛ occupied) Al 3+ in Oh sites (½ occupied)  4 Oh sites per unit cell O 2– – cubic closest packed muscovite

Zircon crystal structure

Spinel structure AB 2 O 4. The structure can be thought of as 8 octants of alternating AO 4 tetrahedra and B 4 O 4 cubes as shown in the left-hand diagram; the 4 O ions have the same orientation in all 8 octants and so build up into a fcc lattice of 32 ions which coordinate A tetrahedrally and B octahedrally. The 4 A octants contain 4 A ions and the 4 B octants contain 16 B ions. The unit cell is completed by an encompassing fcc of A ions ( ● ) as shown in the right-hand diagram; this is shared with adjacent unit cells and comprises the remaining 4 A ions in the complete unit cell A 8 B 16 O 32. The location of two of the B 4 O 4 cubes is shown for orientation.

Structural model of Ge 1 Sb 2 Te 4 in the spinel phase (Ge = red, Sb = blue, Te = yellow) shown in a supercell containing 56 atoms. Nature Materials 2006, 5, 56–62.

Inverse spinel ([B] tet [A, B] oct O 4 ) – example: NiFe 2 O 4 Ni 2+ in Oh sites (¼ occupied) Fe 3+ in Oh sites (¼ occupied) Fe 3+ in Td sites (⅛ occupied) Perovskite with the general formula ABX 3 example: SrTiO 3 Structure of tetragonal BaTiO 3 projected onto the ac plane. BaTiO 3 is ferroelectric.