Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy
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1 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 Course number: Ch120a Hours: 2-3pm Monday, Wednesday, Friday William A. Goddard, III, 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistant: Shane Flynn Recitation: Mon 3pm
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 E(eV) = /R(A) or E (kcal/mol) = /R(A) But covalent curve does not change until get large overlap the 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 Correcting for IP-EA at R= leads to a net bond of =4.6 ev experiment De = 4.23 ev Thus ionic character dominates E(eV) R(A) 3
4 GVB orbitals of NaCl Dipole moment = Debye Pure ionic Debye Thus q=0.79 e No overlap no bond R=6 A Overlap from Q transfer R=4.7 A Mostly Na + Cl - R=3.5 A Very Na + Cl - Re=2.4 A 4
5 electronegativity To provide a measure to estimate polarity in bonds, Linus Pauling developed a scale of electronegativity (χ) where the atom that gains charge is more electronegative and the one that loses is more electropositive He arbitrarily assigned χ=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 using atomic IP and EA (corrected for valence averaging and scaled by 5.2 to get similar numbers to Pauling: χ M = (IP+EA)/5.2 5
6 Electronegativity Based on M ++ 6
7 Comparison of Mulliken and Pauling electronegativities Mulliken Biggest flaw is the wrong value for H H is clearly much less electronegative than I 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 HS Fe 3+ is d 5 thus get HS S=5/2; LS=1/2 not important Fe 2+ is d 6 thus HS=2; LS=0, both important Ligand field splitting (Crystal field splitting) Negative neighbors at vertices of octahedron splits the d orbitals into t2g and eg irreducible representations of Td or Oh point group Five d orbitals t 2g [xy, xz, yz] e g [x 2 -y 2, 3z 2 -r 2 ] same energy atom octahedron tetrahedron e g [x 2 -y 2, 3z 2 -r 2 ] t 2g [xy, xz, yz] 12
13 More on HS and LS, octahedral site, Fe 3+ Fe 3+ is d 5 thus get HS S=5/2; LS=1/2 not important Weak field Strong field e g x 2 -y 2 3z 2 -r 2 e g x 2 -y 2 3z 2 -r 2 t 2g xy xz yz t 2g xy xz yz Exchange stabilization dominates, get high spin S=5/2 as for atom 5*4/2=10 exchange terms, ~220 kcal/mol Ligand interaction dominates, get low spin S=1/2 3+1 exchange terms ~88 kcal/mol 13
14 More on HS and LS, octahedral site, Fe 2+ Fe 2+ is d 6 thus HS=2; LS=0, both important Weak field Strong field e g x 2 -y 2 3z 2 -r 2 e g x 2 -y 2 3z 2 -r 2 t 2g xy xz yz Exchange stabilization dominates, get high spin S=2 as for atom t 2g xy xz yz Ligand interaction dominates, get low spin S=0 5*4/2=10 exchange terms, ~220 kcal/mol 3+3 exchange terms ~132 kcal/mol 14
15 Ionic radii, transition metals 15
16 Ionic radii Lanthanides and Actinide 16
17 Role of ionic sizes in determining crystal structures Assume that anions are large and packed so that they contact, then 2R A < L, L is distance between anions Assume anion and cation are in contact and calculate 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 17
18 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) B1 expect < (R C /R A ) < B2 or B1 (R C /R A ) > (R C /R A ) < neither 18
19 Sphalerite or Zincblende or B3 structure GaAs 19
20 Wurtzite or B4 structure 20
21 Radius ratios for 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 ) <
22 Structures for II-VI compounds B3 for 0.20 < (R C /R A ) < 0.55 B4 for 0.33 < (R C /R A ) < 0.53 B1 for 0.36 < (R C /R A ) <
23 CaF 2 or fluorite structure like B1, CsCl but with half the Cs missing Or Ca same positions as Ga for GaAs, but now have F at all tetrahedral sites Find for R C /R A >
24 Rutile (TiO 2 ) or Cassiterite (SnO 2 ) structure Related to NaCl with half the cations missing Find for R C /R A <
25 CaF 2 rutile CaF 2 rutile 25
26 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 /ν C where z C is the net charge on the cation and ν C is the coordination number Then z A = Σ i S I = Σ i z Ci /ν 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 26
27 α-quartz structure of SiO 2 Each Si bonds to 4 O, thus S=4/4=1 OSiO = each O must bond to 2 Si since 2*1=2 Si-O-Si = 155.x this is very open because the charge on O is very negative (expect 180 for NaONa, for HOH Helical chains single crystals optically active; α-quartz converts to β-quartz From wikipediaat 573 C rhombohedral (trigonal) hp9, P3121 No.152[10] 27
28 Example 2 of electrostatic balance: stishovite phase of SiO2 The stishovite phase of SiO 2 has six coordinate Si, S=4/6=2/3. Thus each O must have 3 Si neighbors:3*2/3=2 Rutile-like structure, with 6- coordinate Si; high pressure form Used to be the densest of the SiO2 polymorphs May be a new denser one From wikipedia tetragonal tp6, P42/mnm, No.136[17] 28
29 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 29
30 Corundum (α-al 2 O 3 ). Example 4 electrostatic balance Each Al 3+ is in a distorted octahedron, leading to S=3/6=1/2. Thus each O 2- must be coordinated to 4 Al 4*1/2=2 30
31 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) 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 = 4/6=2/3. How many Ti neighbors will each O have? 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 = 4/6=2/3. How many Ti neighbors will each O have? It cannot be 3 since 3*2/3=2 and there could be no bond from O to Ba. 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 = 4/6=2/3. How many Ti neighbors will each O have? It cannot be 3 since 3*2/3=2 and there could be no bond from O to Ba. It is likely not one since Ti does not make oxo bonds. 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 = 4/6=2/3. How many Ti neighbors will each O have? It cannot be 3 since 3*2/3=2 and there could be no bond from O to 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. 35
36 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 = 4/6=2/3. How many Ti neighbors will each O have? It cannot be 3 since 3*2/3=2 and there could be no bond from O to 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, ν Ba, leading to S Ba = 2/ν Ba, and how many Ba around each O, ν OBa. 36
37 Prediction of BaTiO3 structure : Ba coordination Since each O gets 2*2/3=4/3 electrons from the 2 Ti, we expect ν OBa * S Ba = 2/3, the missing charge for the O, we have only a few possibilities: ν Ba = 3 leading to S Ba = 2/ν Ba =2/3 leading to ν OBa = 1 ν Ba = 6 leading to S Ba = 2/ν Ba =1/3 leading to ν OBa = 2 ν Ba = 9 leading to S Ba = 2/ν Ba =2/9 leading to ν OBa = 3 ν Ba = 12 leading to S Ba = 2/ν Ba =1/6 leading to ν 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 Thus each of these Ba sees 4 O in the xy plane, 4 in the xz plane and 4 in the yz plane. 37
38 BaTiO3 structure (Perovskite) Ti Ba 38
39 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 µ(d) = µ(au) 39
40 Fractional ionic character of diatomic molecules Obtained from the experimental dipole moment in Debye, µ(d), and bond distance R(A) by δq = µ(au)/r(a 0 ) = C µ(d)/r(a) where C= Postive that head of column is negative 40
41 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) 41
42 Charge dependence of the energy (ev) of Cl atom E= Harmonic fit E=0 E= Cl + Cl Cl - Q=+1 Q=0 Q=-1 Get minimum at Q= Emin = = =
43 QEq parameters 43
44 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 44
45 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 45
46 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 46
47 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 λ = 0.5 Using R C =0.759a 0 47
48 QEq results for alkali halides 48
49 QEq for Ala-His-Ala Amber charges in parentheses 49
50 QEq for deoxy adenosine Amber charges in parentheses 50
51 QEq for polymers Nylon 66 PEEK 51
52 Stopped January 30 52
53 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 53
54 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. 54
55 <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 c a =1.01 ~ 1.06 a <100> tetragonal below Tc Domains separated by domain walls 55
56 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) 56
57 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) 57
58 Comparison to experiment Displacive small latent heat This agrees with experiment R O: T= 183K, S = 0.17±0.04 J/mol O T: T= 278K, S = 0.32±0.06 J/mol T C: T= 393K, S = 0.52±0.05 J/mol Diffuse xray scattering Expect some disorder, agrees with experiment Cubic Tetra. Ortho. Rhomb. 58
59 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., 2016 William 12, 1053 A. Goddard (1973) III, all rights reserved 59 59
60 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 Cubic I-43m = + + = 60
61 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 Raman Active Cubic Phase 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., 2016 William 12, 1053 A. Goddard (1973) III, all rights reserved 61 61
62 stopped 62
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