Lecture 9 January 30, Ionic bonding and crystals

Size: px

Start display at page:

Download "Lecture 9 January 30, Ionic bonding and crystals"

Dale Goodman
6 years ago
Views:

1 Lecture 9 January 30, 2013 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: Ch120a Hours: 2-3pm Monday, Wednesday, Friday William A. Goddard, III, wag@wag.caltech.edu 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants: Caitlin Scott <cescott@caltech.edu> Hai Xiao xiao@caltech.edu; Fan Liu <fliu@wag.caltech.edu> Ionic bonding and crystals Ch120a-1

2 Ionic bonding (chapter 9) Consider the covalent bond of Na to Cl. There Is very little contragradience, leading to an extremely weak bond. Alternatively, consider transferring the charge from Na to Cl to form Na + and Cl - 2

3 The ionic limit At R= the cost of forming Na + and Cl - is IP(Na) = ev minus EA(Cl) = ev = ev But as R is decreased the electrostatic energy drops as DE(eV) = /R(A) or DE (kcal/mol) = /R(A) Thus this ionic curve crosses the covalent curve at R=14.4/1.524=9.45 A Using the bond distance of NaCl=2.42A leads to a coulomb energy of 6.1eV leading to a bond of =4.6 ev The exper De = 4.23 ev Showing that ionic character dominates E(eV) R(A) 3

4 GVB orbitals of NaCl Dipole moment = Debye Pure ionic Debye Thus Dq=0.79 e 4

5 electronegativity To provide a measure to estimate polarity in bonds, Linus Pauling developed a scale of electronegativity ( ) where the atom that gains charge is more electronegative and the one that loses is more electropositive He arbitrarily assigned =4 for F, 3.5 for O, 3.0 for N, 2.5 for C, 2.0 for B, 1.5 for Be, and 1.0 for Li and then used various experiments to estimate other cases. Current values are on the next slide Mulliken formulated an alternative scale such that M = (IP+EA)/5.2 5

6 Electronegativity Based on M ++ 6

7 Comparison of Mulliken and Pauling electronegativities 7

8 Ionic crystals Starting with two NaCl monomer, it is downhill by 2.10 ev (at 0K) for form the dimer Because of repulsion between like charges the bond lengths, increase by 0.26A. A purely electrostatic calculation would have led to a bond energy of 1.68 ev Similarly, two dimers can combine to form a strongly bonded tetramer with a nearly cubic structure Continuing, combining 4x10 18 such dimers leads to a grain of salt in which each Na has 6 Cl neighbors and each Cl has 6 Na neighbors 8

9 The NaCl or B1 crystal All alkali halides have this structure except CsCl, CsBr, CsI (they have the B2 structure) 9

10 The CsCl or B2 crystal There is not yet a good understanding of the fundamental reasons why particular compound prefer particular structures. But for ionic crystals the consideration of ionic radii has proved useful 10

11 Ionic radii, main group Fitted to various crystals. Assumes O 2- is 1.40A NaCl R= = 2.84, exper is 2.84 From R. D. Shannon, Acta Cryst. A32, 751 (1976) 11

12 Ionic radii, transition metals 12

13 Ionic radii Lanthanides and Actinide 13

14 Role of ionic sizes in determining crystal structures Assume that the anions are large and packed so that they contact, so that 2R A < L, where L is the distance between anions Assume that the anion and cation are in contact. Calculate the smallest cation consistent with 2R A < L. R A +R C = L/ 2 > 2 R A R A +R C = ( 3)L/2 > ( 3) R A Thus R C /R A > Thus R C /R A > Thus for < (R C /R A ) < we expect B1 For (R C /R A ) > either is ok. For (R C /R A ) < must be some other structure 14

15 Radius Ratios of Alkali Halides and Noble metal halices Rules work ok B1: 0.35 to 1.26 B2: 0.76 to 0.92 Based on R. W. G. Wyckoff, Crystal Structures, 2 nd edition. Volume 1 (1963) 15

16 Wurtzite or B4 structure 16

17 Sphalerite or Zincblende or B3 structure GaAs 17

18 Radius rations B3, B4 The height of the tetrahedron is (2/3) 3 a where a is the side of the circumscribed cube The midpoint of the tetrahedron (also the midpoint of the cube) is (1/2) 3 a from the vertex. Hence (R C + R A )/L = (½) 3 a / 2 a = (3/8) = Thus 2R A < L = (8/3) (R C + R A ) = (R C + R A ) Thus R A < (R C + R A ) or R C /R A > Thus B3,B4 should be the stable structures for < (R C /R A ) <

19 Structures for II-VI compounds B3 for 0.20 < (R C /R A ) < 0.55 B1 for 0.36 < (R C /R A ) <

20 CaF 2 or fluorite structure Like GaAs but now have F at all tetrahedral sites Or like CsCl but with half the Cs missing Find for R C /R A >

21 Rutile (TiO 2 ) or Cassiterite (SnO 2 ) structure Related to NaCl with half the cations missing Find for R C /R A <

22 CaF 2 rutile CaF 2 rutile 22

23 Stopped L17, Feb 10 23

24 Electrostatic Balance Postulate For an ionic crystal the charges transferred from all cations must add up to the extra charges on all the anions. We can do this bond by bond, but in many systems the environments of the anions are all the same as are the environments of the cations. In this case the bond polarity (S) of each cation-anion pair is the same and we write S = z C /n C where z C is the net charge on the cation and n C is the coordination number Then z A = S i S I = S i z Ci /n i Example1 : SiO 2. in most phases each Si is in a tetrahedron of O 2- leading to S=4/4=1. Thus each O 2- must have just two Si neighbors 24

25 a-quartz structure of SiO 2 Each Si bonds to 4 O, OSiO = each O bonds to 2 Si Si-O-Si = 155.x Helical chains single crystals optically active; α-quartz converts to β-quartz at 573 C From wikipedia rhombohedral (trigonal) hp9, P3121 No.152[10] 25

26 Example 2 of electrostatic balance: stishovite phase of SiO2 The stishovite phase of SiO 2 has six coordinate Si, S=2/3. Thus each O must have 3 Si neighbors Rutile-like structure, with 6- coordinate Si; high pressure form densest of the SiO2 polymorphs From wikipedia tetragonal tp6, P42/mnm, No.136[17] 26

27 TiO 2, example 3 electrostatic balance Example 3: the rutile, anatase, and brookite phases of TiO 2 all have octahedral Ti. Thus S= 2/3 and each O must be coordinated to 3 Ti. top anatase phase TiO 2 front right 27

28 Corundum (a-al 2 O 3 ). Example 4 electrostatic balance Each Al 3+ is in a distorted octahedron, leading to S=1/2. Thus each O 2- must be coordinated to 4 Al 28

29 Olivine. Mg 2 SiO 4. example 5 electrostatic balance Each Si has four O 2- (S=1) and each Mg has six O 2- (S=1/3). Thus each O 2- must be coordinated to 1 Si and 3 Mg neighbors O = Blue atoms (closest packed) Si = magenta (4 coord) cap voids in zigzag chains of Mg Mg = yellow (6 coord) 29

30 Illustration, BaTiO 3 A number of important oxides have the perovskite structure (CaTiO 3 ) including BaTiO3, KNbO3, PbTiO3. Lets try to predict the structure without looking it up Based on the TiO 2 structures, we expect the Ti to be in an octahedron of O 2-, S TiO = 2/3. How many Ti neighbors will each O have? 30

31 Illustration, BaTiO 3 A number of important oxides have the perovskite structure (CaTiO 3 ) including BaTiO3, KNbO3, PbTiO3. Lets try to predict the structure without looking it up Based on the TiO 2 structures, we expect the Ti to be in an octahedron of O 2-, S TiO = 2/3. How many Ti neighbors will each O have? It cannot be 3 since there would be no place for the Ba. 31

32 Illustration, BaTiO 3 A number of important oxides have the perovskite structure (CaTiO 3 ) including BaTiO3, KNbO3, PbTiO3. Lets try to predict the structure without looking it up Based on the TiO 2 structures, we expect the Ti to be in an octahedron of O 2-, S TiO = 2/3. How many Ti neighbors will each O have? It cannot be 3 since there would be no place for the Ba. It is likely not one since Ti does not make oxo bonds. 32

33 Illustration, BaTiO 3 A number of important oxides have the perovskite structure (CaTiO 3 ) including BaTiO3, KNbO3, PbTiO3. Lets try to predict the structure without looking it up Based on the TiO 2 structures, we expect the Ti to be in an octahedron of O 2-, S TiO = 2/3. How many Ti neighbors will each O have? It cannot be 3 since there would be no place for the Ba. It is likely not one since Ti does not make oxo bonds. Thus we expect each O to have two Ti neighbors, probably at 180º. This accounts for 2*(2/3)= 4/3 charge. 33

34 Illustration, BaTiO 3 A number of important oxides have the perovskite structure (CaTiO 3 ) including BaTiO3, KNbO3, PbTiO3. Lets try to predict the structure without looking it up Based on the TiO 2 structures, we expect the Ti to be in an octahedron of O 2-, S TiO = 2/3. How many Ti neighbors will each O have? It cannot be 3 since there would be no place for the Ba. It is likely not one since Ti does not make oxo bonds. Thus we expect each O to have two Ti neighbors, probably at 180º. This accounts for 2*(2/3)= 4/3 charge. Now we must consider how many O are around each Ba, n Ba, leading to S Ba = 2/n Ba, and how many Ba around each O, n OBa. 34

35 Illustration, BaTiO 3 A number of important oxides have the perovskite structure (CaTiO 3 ) including BaTiO3, KNbO3, PbTiO3. Lets try to predict the structure without looking it up Based on the TiO 2 structures, we expect the Ti to be in an octahedron of O 2-, S TiO = 2/3. How many Ti neighbors will each O have? It cannot be 3 since there would be no place for the Ba. It is likely not one since Ti does not make oxo bonds. Thus we expect each O to have two Ti neighbors, probably at 180º. This accounts for 2*(2/3)= 4/3 charge. Now we must consider how many O are around each Ba, n Ba, leading to S Ba = 2/n Ba, and how many Ba around each O, n OBa. 35

36 Prediction of BaTiO3 structure : Ba coordination Since n OBa * S Ba = 2/3, the missing charge for the O, we have only a few possibilities: n Ba = 3 leading to S Ba = 2/n Ba =2/3 leading to n OBa = 1 n Ba = 6 leading to S Ba = 2/n Ba =1/3 leading to n OBa = 2 n Ba = 9 leading to S Ba = 2/n Ba =2/9 leading to n OBa = 3 n Ba = 12 leading to S Ba = 2/n Ba =1/6 leading to n OBa = 4 Each of these might lead to a possible structure. The last case is the correct one for BaTiO 3 as shown. Each O has a Ti in the +z and z directions plus four Ba forming a square in the xy plane The Each of these Ba sees 4 O in the xy plane, 4 in the xz plane and 4 in the yz plane. 36

37 BaTiO3 structure (Perovskite) 37

38 How estimate charges? We saw that even for a material as ionic as NaCl diatomic, the dipole moment a net charge of +0.8 e on the Na and -0.8 e on the Cl. We need a method to estimate such charges in order to calculate properties of materials. First a bit more about units. In QM calculations the unit of charge is the magnitude of the charge on an electron and the unit of length is the bohr (a 0 ) Thus QM calculations of dipole moment are in units of ea 0 which we refer to as au. However the international standard for quoting dipole moment is the Debye = esu A Where m(d) = m(au) 38

39 Fractional ionic character of diatomic molecules Obtained from the experimental dipole moment in Debye, m(d), and bond distance R(A) by dq = m(au)/r(a 0 ) = C m(d)/r(a) where C= Postive that head of column is negative 39

40 Charge Equilibration First consider how the energy of an atom depends on the net charge on the atom, E(Q) Including terms through 2 nd order leads to Charge Equilibration for Molecular Dynamics Simulations; A. K. Rappé and W. A. Goddard III; J. Phys. Chem. 95, 3358 (1991) (2) (3) 40

41 Charge dependence of the energy (ev) of an atom E= Harmonic fit E=0 E= Q=+1 Cl + Cl Cl - Q=0 Q=-1 Get minimum at Q= Emin = = =

42 QEq parameters 42

43 Interpretation of J, the hardness Define an atomic radius as R 0 A R e (A 2 ) Bond distance of H homonuclear C diatomic N O Si S Li Thus J is related to the coulomb energy of a charge the size of the atom 43

44 The total energy of a molecular complex Consider now a distribution of charges over the atoms of a complex: Q A, Q B, etc Letting J AB (R) = the Coulomb potential of unit charges on the atoms, we can write Taking the derivative with respect to charge leads to the chemical potential, which is a function of the charges or The definition of equilibrium is for all chemical potentials to be equal. This leads to 44

45 The QEq equations Adding to the N-1 conditions The condition that the total charged is fixed (say at 0) leads to the condition Leads to a set of N linear equations for the N variables Q A. AQ=X, where the NxN matrix A and the N dimensional vector A are known. This is solved for the N unknowns, Q. We place some conditions on this. The harmonic fit of charge to the energy of an atom is assumed to be valid only for filling the valence shell. Thus we restrict Q(Cl) to lie between +7 and -1 and Q(C) to be between +4 and -4 Similarly Q(H) is between +1 and -1 45

46 The QEq Coulomb potential law We need now to choose a form for J AB (R) A plausible form is J AB (R) = 14.4/R, which is valid when the charge distributions for atom A and B do not overlap Clearly this form as the problem that J AB (R) as R 0 In fact the overlap of the orbitals leads to shielding The plot shows the shielding for C atoms using various Slater orbitals And l = 0.5 Using R C =0.759a 0 46

47 QEq results for alkali halides 47

48 QEq for Ala-His-Ala Amber charges in parentheses 48

49 QEq for deoxy adenosine Amber charges in parentheses 49

50 QEq for polymers Nylon 66 PEEK 50

structure. Characteristic chemical formula of a perovskite ceramic: ABO 3, A atom has +2 charge. 12 coordinate at the corners of a cube.

51 Perovskites Perovskite (CaTiO3) first described in the 1830s by the geologist Gustav Rose, who named it after the famous Russian mineralogist Count Lev Aleksevich von Perovski crystal lattice appears cubic, but it is actually orthorhombic in symmetry due to a slight distortion of the structure. Characteristic chemical formula of a perovskite ceramic: ABO 3, A atom has +2 charge. 12 coordinate at the corners of a cube. B atom has +4 charge. Octahedron of O ions on the faces of that cube centered on a B ions at the center of the cube. Together A and B form an FCC structure 51

The stability of the perovskite structure depends on the relative ionic radii: Ferroelectrics if the cations are too small for close packing with the oxygens, they may

Since these ions carry electrical charges, such displacements can result in a net electric dipole moment (opposite charges separated by a small distance).

52 The stability of the perovskite structure depends on the relative ionic radii: Ferroelectrics if the cations are too small for close packing with the oxygens, they may displace slightly. Since these ions carry electrical charges, such displacements can result in a net electric dipole moment (opposite charges separated by a small distance). The material is said to be a ferroelectric by analogy with a ferromagnet which contains magnetic dipoles. At high temperature, the small green B-cations can "rattle around" in the larger holes between oxygen, maintaining cubic symmetry. A static displacement occurs when the structure is cooled below the transition temperature. 52

<100> polarized tetragonal Non-polar cubic

Different phases of BaTiO 3 Ba 2+ /Pb 2+ c

temperature Non-polar cubic above Tc a c a

53 <111> polarized rhombohedral <110> polarized orthorhombic Phases of BaTiO3 <100> polarized tetragonal Non-polar cubic -90 o C 5 o C 120 o C Temperature Different phases of BaTiO 3 Ba 2+ /Pb 2+ c Ti 4+ O 2- Six variants at room temperature Non-polar cubic above Tc a c a <100> tetragonal below Tc 1.01 ~ 1.06 Domains separated by domain walls 53

Nature of the phase transitions Displacive model Assume that the atoms prefer to distort toward a face or edge or vertex of the octahedron Increasing Temperature Different phases of BaTiO 3 <111>

54 Nature of the phase transitions Displacive model Assume that the atoms prefer to distort toward a face or edge or vertex of the octahedron Increasing Temperature Different phases of BaTiO 3 <111> polarized rhombohedral <110> polarized orthorhombic <100> polarized tetragonal Non-polar cubic -90 o C 5 o C 120 o C Temperature face edge vertex center 1960 Cochran Soft Mode Theory(Displacive Model) 54

55 Nature of the phase transitions Displacive model Assume that the atoms prefer to distort toward a face or edge or vertex of the octahedron Increasing Temperature 1960 Cochran Soft Mode Theory(Displacive Model) Order-disorder 1966 Bersuker Eight Site Model 1968 Comes Order-Disorder Model (Diffuse X-ray Scattering) 55

56 Comparison to experiment Displacive small latent heat This agrees with experiment R O: T= 183K, DS = 0.17±0.04 J/mol O T: T= 278K, DS = 0.32±0.06 J/mol T C: T= 393K, DS = 0.52±0.05 J/mol Diffuse xray scattering Expect some disorder, agrees with experiment Cubic Tetra. Ortho. Rhomb. 56

57 Problem displacive model: EXAFS & Raman observations (001) α d (111) EXAFS of Tetragonal Phase [1] Ti distorted from the center of oxygen octahedral in tetragonal phase. The angle between the displacement vector and (111) is α= Raman Spectroscopy of Cubic Phase [2] A strong Raman spectrum in cubic phase is found in experiments. But displacive model atoms at center of octahedron: no Raman 1. B. Ravel et al, Ferroelectrics, 206, 407 (1998) 2. A. M. Quittet et al, Solid copyright State Comm., 2011 William 12, 1053 A. Goddard (1973) III, all rights reserved 57 57

58 QM calculations The ferroelectric and cubic phases in BaTiO3 ferroelectrics are also antiferroelectric Zhang QS, Cagin T, Goddard WA Proc. Nat. Acad. Sci. USA, 103 (40): (2006) Even for the cubic phase, it is lower energy for the Ti to distort toward the face of each octahedron. How do we get cubic symmetry? Combine 8 cells together into a 2x2x2 new unit cell, each has displacement toward one of the 8 faces, but they alternate in the x, y, and z directions to get an overall cubic symmetry Ti atom distortions Microscopic Polarization Px Py Pz Macroscopic Polarization z Cubic I-43m = + + = 58

59 QM results explain EXAFS & Raman observations (001) α d (111) EXAFS of Tetragonal Phase [1] Ti distorted from the center of oxygen octahedral in tetragonal phase. The angle between the displacement vector and (111) is α= PQEq with FE/AFE model gives α=5.63 Raman Spectroscopy of Cubic Phase [2] A strong Raman spectrum in cubic phase is found in experiments. Model Inversion symmetry in Cubic Phase Raman Active Displacive Yes No FE/AFE No Yes 1. B. Ravel et al, Ferroelectrics, 206, 407 (1998) 2. A. M. Quittet et al, Solid copyright State Comm., 2011 William 12, 1053 A. Goddard (1973) III, all rights reserved 59 59

60 stopped 60

Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy

Lecture 20, November 11, 2016 Ionic bonding and crystals Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy