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1 CH3. Intro to Solids Lattice geometries Common structures Lattice energies Born-Haber model Thermodynamic effects Electronic structure.

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Presentation on theme: "1 CH3. Intro to Solids Lattice geometries Common structures Lattice energies Born-Haber model Thermodynamic effects Electronic structure."— Presentation transcript:

1 1 CH3. Intro to Solids Lattice geometries Common structures Lattice energies Born-Haber model Thermodynamic effects Electronic structure

2 2 A Stacked 2D hexagonal arrays BC

3 3 Packing efficiency It can be easily shown that all close-packed arrays have a packing efficiency (V occ /V tot ) of 0.74 This is the highest possible value for same-sized spheres, though this is hard to prove “…And suppose…that there were one form, which we will call ice-nine - a crystal as hard as this desk - with a melting point of, let us say, one- hundred degrees Fahrenheit, or, better still, … one-hundred-and-thirty degrees.” Kurt Vonnegut, Jr. Cat’s Cradle

4 4 Close-packing of polymer microspheres

5 5 hcp vs ccp Also close-packed: (ABAC) n (ABCB) n Not close packed: (AAB) n (ABA) n Why not ? (ACB) n (AB) n hcp(ABC) n ccp

6 6 Unit cells for hcp and fcc Hexagonal cell = hcp Cubic cell ccp = fcc Unit cells, replicated and translated, will generate the full lattice

7 7 Generating lattices lattice point CNDistance from origin (a units) (½,½,0) (1,0,0) (½,½,1) (1,1,0) ( 3 / 2,½,0) Etc…

8 8 Oh and Td sites in ccp r Oh : a = 2r s + 2r Oh a / √ 2 = 2r s r Oh / r s = spheres / cell 4 Oh sites / cell 8 Td sites / cell fcc lattice showing some Oh and Td sites

9 9 Ionic radii are related to coordination number

10 10 Element Structures at STP (ABCB) n

11 11 Ti phase transitions RT → 882°C hcp 882 → 1667° bcc 1667 → 3285° liquid 3285 → gas

12 12 Classes of Alloys (a)Substitutional (b)Interstitial (c)intermetallic

13 13 Some alloys Alloy Composition Cu, Niany Cu and Ni are ccp, r(Cu) = 1.28, r(Ni) = 1.25 Å Cast iron Fe, C (2+ %), Mn, Si r(Fe) = 1.26, r(C) = 0.77 Stainless Steels Fe, Cr, Ni, C … Brass CuZn (  ) = bcc r(Zn) = 1.37, hcp substitutional interstitial intermetallic

14 14 A few stainless steels Chemical Composition % (Max unless noted) StainlessCMnPSSiCrNiMoN

15 15 Zintl phases KGe

16 16 NaCl (rocksalt) Cl Na B C A a b c –fcc anion array with all Oh sites filled by cations –the stoichiometry is 1:1 (AB compound) –CN = 6,6 –Look down the body diagonal to see 2D hex arrays in the sequence (AcBaCb) n –The sequence shows coordination, for example the c layer in AcB Oh coordination

17 17 CaC 2 Tetragonal distortion of rocksalt structure (a = b ≠ c) Complex anion also decreases (lowers) symmetry

18 18 Other fcc anion arrays

19 19 Antifluorite / Fluorite Antifluorite is an fcc anion array with cations filling all Td sites 8 Td sites / unit cell and 4 spheres, so this must be an A 2 B-type salt. Stacking sequence is (AabBbcCca) n CN = 4,8. Anion coordination is cubic. Fluorite structure reverses cation and anion positions. An example is the mineral fluorite CaF 2

20 20 Sphalerite (ZnS) fcc anion array with cations filling ½ Td sites Td sites are filled as shown Look down body diagonal of the cube to see the sequence (AaBbCc) n… If all atoms were C, this is diamond structure. B C A a b c

21 21 Sphalerite

22 22 Semiconductor lattices based on diamond / sphalerite Group 14: C, Si, Ge,  Sn, SiC 3-5 structures: cubic-BN, AlN, AlP, GaAs, InP, InAs, InSb, GaP,… 2-6 structures: BeS, ZnS, ZnSe, CdS, CdSe, HgS… 1-7 structures: CuCl, AgI

23 23 Structure Maps more covalent more ionic incr. radius, polarizability

24 24 Lattices with hcp anion arrays

25 25 NiAs hcp anion array with cations filling all Oh sites cation layers all eclipsing one another stacking sequence is (AcBc) n CN = 6,6 AcB and BcA gives Oh cation coordination, but cBc and cAc gives trigonal prismatic (D 3h ) anion coordination

26 26 CdI 2 hcp anion array with cations filling ½ Oh sites in alternating layers Similar to NiAs, but leave out every other cation layer stacking sequence is (AcB) n CN = (6, 3) anisotropic structure, strong bonding within AcB layers, weak bonding between layers the layers are made from edge-sharing CdI 6 octahedra

27 27 LiTiS 2 (AcBc’) n Ti S Li

28 28 LDH structures Mg(OH) 2 (brucite) Mg x Al 1-x (OH) 2. An

29 29 Rutile (TiO 2 ) hcp anion array with cations filling ½ Oh sites in alternating rows the filled cation rows are staggered CN = 6, 3 the filled rows form chains of edge- sharing octahedra. These chains are not connected within one layer, but are connected by the row of octahedra in the layers above and below. Lattice symmetry is tetragonal due to the arrangement of cations.

30 30 Rutile TiO 2-x and SiO 2

31 31 Wurtzite (ZnS) hcp anion array with cations filling ½ Td sites Stacking sequence = (AaBb) n CN = 4, 4 wurtzite and sphalerite are closely related structures, except that the basic arrays are hcp and ccp, respectively. Many compounds can be formed in either structure type: ZnS, has two common allotropes, sphalerite and wurtzite

32 32 ReO 3 Re is Oh, each O is shared between 2 Re, so there are ½ * 6 = 3 O per Re, overall stoichiometry is thus ReO 3 Neither ion forms a close-packed array. The oxygens fill 3/4 of the positions for fcc (compare with NaCl structure). The structure has ReO 6 octahedra sharing all vertices.

33 33 Perovskite (CaTiO 3 ) Similar to ReO 3, with a cation (CN = 12) at the unit cell center. Simple perovskites have an ABX 3 stoichiometry. A cations and X anions, combined, form a close-packed array, with B cations filling 1/4 of the Oh sites. An ordered AA’BX 3 perovskite

34 34 Superconducting copper oxides Many superconducting copper oxides have structures based on the perovskite lattice. An example is: YBa 2 Cu 3 O 7. In this structure, the perovskite lattice has ordered layers of Y and Ba cations. The idealized stoichiometry has 9 oxygens, the anion vacancies are located mainly in the Y plane, leading to a tetragonal distortion and anisotropic (layered) character.YBa 2 Cu 3 O 7

35 35 Charged spheres For 2 spherical ions in contact, the electrostatic interaction energy is: E el = (e 2 / 4   0 ) (Z A Z B / d) e = e - charge = x C  0 = vac. permittivity = x C 2 J -1 m -1 Z A = charge on ion A Z B = charge on ion B d = separation of ion centers Assumes a uniform charge distribution (unpolarizable ions). With softer ions, higher order terms (d -2, d -3,...) can be included.

36 36 Consider an infinite linear chain of alternating cations and anions with charges +e or –e The electrostatic terms are: E el = (e 2 /4  0 )(Z A Z B / d) [2(1) - 2(1/2) + 2(1/3) - 2(1/4) +…] = (e 2 /4  0 )(Z A Z B /d) (2 ln2) Infinite linear chains

37 37 Madelung constants Generalizing the equation for 3D ionic solids, we have: E el = (e 2 / 4   0 ) * (Z A Z B / d) * A where A is called the Madelung constant and is determined by the lattice geometry

38 38 Madelung constants Some values for A and A / n: lattice A CNstoich A / n CsCl1.763(8,8)AB0.882 NaCl1.748(6,6)AB0.874 sphalerite1.638(4,4)AB0.819 wurtzite1.641(4,4)AB0.821 fluorite2.519(8,4) AB rutile2.408(6,3) AB

39 39 Born-Meyer model Electrostatic forces are net attractive, so d → 0 (the lattice collapse to a point) without a repulsive term Add a pseudo hard-shell repulsion: C‘ e -d/d* where C' and d* are scaling factors ( d * has been empirically fit as Å) V rep mimics a step function for hard sphere compression (0 where d > hard sphere radius, very large where d < radius)

40 40 Born-Meyer eqn The total interaction energy, E: E = E el + E repulsive = (e 2 / 4  0 )(NAZ A Z B /d) + NC'e -d/d* Since E has a single minimum d, set dE/dd = 0 and solve for C‘: E = -  H L = (e 2 /4  0 ) (NAZ A Z B /d 0 ) (1 - d * /d 0 ) (Born-Meyer equation) Note sign conventions !!!

41 41 Further refinements E el ’ include higher order terms E vdw NC’’r -6 instantaneous polarization E ZPE Nh   lattice vibrations For NaCl: E total = E el ’ + E rep + E vdw + E ZPE kJ/mol

42 42 Kapustinskii approximation: The ratio A/n is approximately constant, where n is the number of ions per formula unit (n is 2 for an AB - type salt, 3 for an AB 2 or A 2 B - type salt,...) Substitute the average value into the B-M eqn, combine constants, to get the Kapustinskii equation:   H L = kJÅ/mol (nZ A Z B / d 0 ) (1 - d * /d 0 ) with d 0 in Å

43 43 Kapustinskii eqn Using the average A / n value decreases the accuracy of calculated E’s. Use only when lattice structure is unknown.  H L (Z A,Z B,n,d 0 ). The first 3 of these parameters are given from in the formula unit, the only other required info is d 0. d 0 can be estimated for unknown structures by summing tabulated cation and anion radii. The ionic radii depend on both charge and CN.

44 44 Example: Use the Kapustiskii eqn to estimate  H L for MgCl 2 1.Z A = +2, Z B = -1, n = 3 2.r(Mg 2+ ) CN 8 = 1.03 Å r(Cl - ) CN 6 = 1.67 Å 3.d 0 ≈ r + + r - ≈ 2.7 Å  H L (Kap calc) = 2350 kJ/mol  H L (best calc) = 2326  H L (B-H value) = 2526

45 45 Unit cell volume relation Note that d*/d 0 is a small term for most salts, so (1 - d*/d 0 ) ≈ 1, Then for a series of salts with the same ionic charges and formula units:  H L ≈ 1 / d 0 For cubic structures:  H L ≈ 1 / V 1/3 where V is the unit cell volume

46 46  H L vs V -1/3 for cubic lattices V 1/3 is proportional to lattice E for cubic structures. V is easily obtained by powder diffraction.

47 47 Born – Haber cycle  H f {KCl(s)} =  H sub (K) + I(K) + ½ D 0 (Cl 2 ) – E a (Cl) -  H L All enthalpies are measurable except  H L Solve to get  H L (B-H)  H L EaEa  H sub ½ D 0 I  H f  H f {KCl(s)} =  H {K(s) + ½Cl 2 (g) → KCl(s)}

48 48 Is MgCl 3 stable ?   H L is from the Kapustinskii eqn, using d 0 from MgCl 2  The large positive  H f means it is not stable.  I(3) is very large, there are no known stable compounds containing Mg 3+. Energies required to remove core electrons are not compensated by other energy terms.  H f =  H at,Mg + 3/2 D 0 (Cl 2 ) + I(1) Mg + I(2) Mg + I(3) Mg - 3 E a (Cl) -  H L = /2 (240) (350) ≈ kJ/mol

49 49 Entropic contributions  G =  H - T  S Example: Mg(s) + Cl 2 (g) → MgCl 2 (s)  S sign is usually obvious from phase changes.  S is negative (unfavorable) here due to conversion of gaseous reactant into solid product. Using tabulated values for molar entropies:  S 0 rxn =  S 0 (MgCl 2 (s)) -  S 0 (Mg(s)) -  S 0 (Cl 2 (g)) = = -166 J/Kmol -T  S at 300 K ≈ + 50 ; at 600 K ≈ +100 kJ/mol Compare with  H f {MgCl 2 (s)} = -640 kJ/mol  S term is usually a corrective term at moderate temperatures. At high T it can dominate.

50 50 Thermochemical Radii What are the radii of polyatomic ions ? (Ex: CO 3 2-, SO 4 2-, PF 6 -, B(C 6 H 6 ) -, N(Et) 4 + ) If  H L is known from B-H cycle, use B-M or Kap eqn to determine d 0. If one ion is not complex, the complex ion “radius” can be calculated from: d 0 = r cation + r anion Tabulated thermochemical radii are averages from several salts containing the complex ion. This method can be especially useful when for ions with unknown structure, or low symmetry.

51 51 Thermochemical Radii Example:  H L (BH) for Cs 2 SO 4 is 1658 kJ/mol Use the Kap eqn:  H L = 1658 = 1210(6/d 0 )( /d 0 ) solve for d 0 = 4.00 Å Look up r (Cs + ) = 1.67 Å r (SO 4 2- ) ≈ = 2.33 Å The tabulated value is 2.30 Å (an avg for several salts)

52 52 Predictive applications O 2 (g) + PtF 6 (l) → O 2 PtF 6 (s) Neil Bartlett (1960); side-reaction in preparing PtF 6 E a (PtF 6 ) = 787 kJ/mol. Compare E a (F) = 328  I(Xe) ≈ I(O 2 ), so Xe + PtF 6 - (s) may be stable if  H L is similar. Bartlett reported the first noble gas compound in O 2 (g) → O 2 + (g) + e kJ/mol e - + PtF 6 (g) → PtF 6 - (g) O 2 + (g) + PtF 6 - (g) → O 2 PtF 6 (s) - 470* O 2 (g) + PtF 6 (g) → O 2 PtF 6 (s) ≈ - 93 * Estimated from the Kap eqn

53 53 Some consequences of  H L Ion exchange / displacement Thermal / redox stabilities Solubilities

54 54 Exchange / Displacement Large ion salt + small ion salt is better than two salts with large and small ions combined. Example: Salt  H L sum CsF750 NaI kJ/mol CsI620 NaF This can help predict some reactions like displacements, ion exchange, thermal stability.

55 55 Thermal stability of metal carbonates An important industrial reaction involves the thermolysis of metal carbonates to form metal oxides according to: MCO 3 (s) → MO (s) + CO 2 (g)  G must be negative for the reaction to proceed. At the lowest reaction temp:  G = 0 and T min =  H /  S  S is positive because gas is liberated. As T increases,  G becomes more negative (i.e. the reaction becomes more favorable).  S depends mainly on  S 0 {CO 2 (g)} and is almost independent of M.

56 56 Thermal stability of metal carbonates MCO 3 (s) → MO (s) + CO 2 (g) T min almost directly proportional to  H.  H L favors formation of the oxide (smaller anion) for smaller cations. So T min for carbonates should increase with cation size.

57 57 Solubility MX (s) --> M + (aq) + X - (aq)  S is positive, so a negative  H is not always required for a spontaneous rxn. But  H is usually related to solubility. Use a B-H analysis to evaluate the energy terms that contribute to dissolution: MX(s) → M + (g) + X - (g)  H lat M + (g) + n L → ML' n + (aq)  H solv, M X - (g) + m L → XL' m - (aq)  H solv, X L' n + L' n → (n + m) L  H L-L MX(s) → M + (aq) + X - (aq)  H solution, MX Driving force for dissolution is ion solvation, but this must compensate for the loss of lattice enthalpy. LiClO 4 and LiSO 3 CF 3 deliquesce (absorb water from air and dissolve) due to dominance of  H solv

58 58 Solubility The energy balance favors solvation for large-small ion combinations, salts of ions with similar sizes are often less soluble.

59 59 Solubility  H L terms dominate when ions have higher charges; these salts are usually less soluble. Some aqueous solubilities at 25°C:  H solution solubility salt (kJ/mol)(g /100 g H 2 O) LiF LiCl LiI MgF MgO

60 60 Orbitals and Bands

61 61 Band and DOS diagrams

62 62  vs T

63 63 Intrinsic Semiconductors  = n q   = conductivity n = carrier density q = carrier charge  = carrier mobility P = electron population ≈ e -(Eg)/2kT EgEg C5.5 eV Si1.1 eV Ge0.7 eV GaAs1.4 eV

64 64 Bandgap vs 

65 65 Arrhenius relation ln  1 / T Slope = -E g /2k Arrhenius relation:    e -Eg/2kT

66 66 Extrinsic Semiconductors n-type p-type p-type example: B-doped Si n-type example: P-doped Si


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