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Lecture 9 January 27, 2013 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, [email protected] 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants:Sijia Dong <[email protected]> Samantha Johnson <[email protected]> Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved Ch120a1 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 ClCh120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 2 The ionic limit At R=∞ the cost of forming Na+ and Clis IP(Na) = 5.139 eV minus EA(Cl) = 3.615 eV = 1.524 eV But as R is decreased the electrostatic energy drops as DE(eV) = 14.4/R(A) or DE (kcal/mol) = -332.06/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 6.1-1.5=4.6 eV experiment De = 4.23 eV Thus ionic character dominates Ch120a-Goddard-L25 E(eV) © copyright 2011 William A. Goddard III, all rights reserved R(A) 3 GVB orbitals of NaCl No overlapno bond R=6 A Dipole moment = 9.001 Debye Overlap from Q transfer Pure ionic 11.34 Debye Thus Dq=0.79 e R=4.7 A Mostly Na+ ClR=3.5 A Very Na+ Cl- Re=2.4 A Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 4 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 using atomic IP and EA (corrected for valence averaging and scaled by 5.2 to get similar numbers to Pauling: M= (IP+EA)/5.2 Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 5 Electronegativity Based on M++ Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 6 Comparison of Mulliken and Pauling electronegativities Mulliken Biggest flaw is the wrong value for H H is clearly much less electronegative than I Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 7 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 4x1018 such dimers leads to a grain of salt in which each Na has 6 Cl neighbors and each Cl hasCh120a-Goddard-L25 6 Na neighbors © copyright 2011 William A. Goddard III, all rights reserved 8 The NaCl or B1 crystal All alkali halides have this structure except CsCl, CsBr, CsI (they have the B2 structure) Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 9 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 Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 10 Ionic radii, main group Fitted to various crystals. Assumes O2- is 1.40A NaCl R=1.02+1.81 = 2.84, exper is 2.84 From R. D. Shannon, Acta©Cryst. 751 (1976) Ch120a-Goddard-L25 copyrightA32, 2011 William A. Goddard III, all rights reserved 11 Ionic radii, transition metals HS Fe3+ is d5 thus get HSS=5/2; LS=1/2 not important Fe2+ is d6 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 same energy atom Ch120a-Goddard-L25 eg [x2-y2, 3z2-r2] t2g [xy, xz, yz] t2g [xy, xz, yz] eg [x2-y2, 3z2-r2] octahedron tetrahedron © copyright 2011 William A. Goddard III, all rights reserved 12 More on HS and LS, octahedral site, Fe3+ Fe3+ is d5 thus get HSS=5/2; LS=1/2 not important Weak field eg t2g Strong field x2-y2 3z2-r2 xy xz yz t2g Exchange stabilization dominates, get high spin S=5/2 as for atom 5*4/2=10 exchange terms, ~220 kcal/mol Ch120a-Goddard-L25 eg x2-y2 3z2-r2 xy xz yz Ligand interaction dominates, get low spin S=1/2 3+1 exchange terms ~88 kcal/mol © copyright 2011 William A. Goddard III, all rights reserved 13 More on HS and LS, octahedral site, Fe2+ Fe2+ is d6 thus HS=2; LS=0, both important Weak field eg t2g Strong field x2-y2 3z2-r2 xy xz yz t2g Exchange stabilization dominates, get high spin S=2 as for atom 5*4/2=10 exchange terms, ~220 kcal/mol Ch120a-Goddard-L25 eg x2-y2 3z2-r2 xy xz yz Ligand interaction dominates, get low spin S=0 3+3 exchange terms ~132 kcal/mol © copyright 2011 William A. Goddard III, all rights reserved 14 Ionic radii, transition metals Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 15 Ionic radii Lanthanides and Actinide Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 16 Role of ionic sizes in determining crystal structures Assume that anions are large and packed so that they contact, then 2RA < L, L is distance between anions Assume anion and cation are in contact and calculate smallest cation consistent with 2RA < L. RA+RC = L/√2 > √2 RA RA+RC = (√3)L/2 > (√3) RA Thus RC/RA > 0.414 Thus RC/RA > 0.732 Thus for 0.414 < (RC/RA ) < 0.732 we expect B1 For (RC/RA ) > 0.732 either is ok. ForCh120a-Goddard-L25 (RC/RA ) < 0.414 must be2011 some other structure © copyright William A. Goddard III, all rights reserved 17 Radius Ratios of Alkali Halides and Noble metal halices Rules work ok B1: 0.35 to 1.26 B2: 0.76 to 0.92 B1 expecton 0.414 < (RC/RA ) < 0.732 Based R. W. G.B1 Wyckoff, B2 or (RC/RA ) > 0.732 Crystal (RStructures, neither C/RA ) < 0.414 2nd edition. Volume 1 (1963) Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 18 Sphalerite or Zincblende or B3 structure GaAs Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 19 Wurtzite or B4 structure Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 20 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 (RC + RA)/L = (½) √3 a / √2 a = √(3/8) = 0.612 Thus 2RA < L = √(8/3) (RC + RA) = 1.633 (RC + RA) Thus 1.225 RA < (RC + RA) or RC/RA > 0.225 Thus B3,B4 should be the stable structures for 0.225 < (RC/RA) < 0. 414 Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 21 Structures for II-VI compounds B3 for 0.20 < (RC/RA) < 0.55 B4 for 0.33 < (RC/RA) < 0.53 B1 for 0.36 < (RC/RA) < 0.96 Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 22 CaF2 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 RC/RA > 0.71 Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 23 Rutile (TiO2) or Cassiterite (SnO2) structure Related to NaCl with half the cations missing Find for RC/RA < 0.67 Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 24 CaF2 rutile CaF2 rutile Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 25 Stopped L17, Feb 10 Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 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 = zC/nC where zC is the net charge on the cation and nC is the coordination number Then zA = Si SI = Si zCi /ni Example1 : SiO2. in most phases each Si is in a tetrahedron of O2- leading to S=4/4=1. Thus each O2- must have just two Si neighbors Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 27 a-quartz structure of SiO2 Each Si bonds to 4 O, OSiO = 109.5° 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 Ch120a-Goddard-L25 rhombohedral (trigonal) hP9, P3121 No.152[10] © copyright 2011 William A. Goddard III, all rights reserved 28 Example 2 of electrostatic balance: stishovite phase of SiO2 The stishovite phase of SiO2 has six coordinate Si, S=2/3. Thus each O must have 3 Si neighbors Rutile-like structure, with 6coordinate Si; high pressure form densest of the SiO2 polymorphs From wikipedia Ch120a-Goddard-L25 tetragonal tP6, P42/mnm, No.136[17] © copyright 2011 William A. Goddard III, all rights reserved 29 TiO2, example 3 electrostatic balance Example 3: the rutile, anatase, and brookite phases of TiO2 all have octahedral Ti. Thus S= 2/3 and each O must be coordinated to 3 Ti. top anatase phase TiO2 front Ch120a-Goddard-L25 right © copyright 2011 William A. Goddard III, all rights reserved 30 Corundum (a-Al2O3). Example 4 electrostatic balance Each Al3+ is in a distorted octahedron, leading to S=1/2. Thus each O2- must be coordinated to 4 Al Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 31 Olivine. Mg2SiO4. example 5 electrostatic balance Each Si has four O2- (S=1) and each Mg has six O2- (S=1/3). Thus each O2- 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) Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 32 Illustration, BaTiO3 A number of important oxides have the perovskite structure (CaTiO3) including BaTiO3, KNbO3, PbTiO3. Lets try to predict the structure without looking it up Based on the TiO2 structures , we expect the Ti to be in an octahedron of O2-, STiO = 2/3. How many Ti neighbors will each O have? Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 33 Illustration, BaTiO3 A number of important oxides have the perovskite structure (CaTiO3) including BaTiO3, KNbO3, PbTiO3. Lets try to predict the structure without looking it up Based on the TiO2 structures , we expect the Ti to be in an octahedron of O2-, STiO = 2/3. How many Ti neighbors will each O have? It cannot be 3 since there would be no place for the Ba. Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 34 Illustration, BaTiO3 A number of important oxides have the perovskite structure (CaTiO3) including BaTiO3, KNbO3, PbTiO3. Lets try to predict the structure without looking it up Based on the TiO2 structures , we expect the Ti to be in an octahedron of O2-, STiO = 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. Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 35 Illustration, BaTiO3 A number of important oxides have the perovskite structure (CaTiO3) including BaTiO3, KNbO3, PbTiO3. Lets try to predict the structure without looking it up Based on the TiO2 structures , we expect the Ti to be in an octahedron of O2-, STiO = 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. Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 36 Illustration, BaTiO3 A number of important oxides have the perovskite structure (CaTiO3) including BaTiO3, KNbO3, PbTiO3. Lets try to predict the structure without looking it up Based on the TiO2 structures , we expect the Ti to be in an octahedron of O2-, STiO = 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, nBa, leading to SBa = 2/nBa, and how many Ba around each O, nOBa. Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 37 Illustration, BaTiO3 A number of important oxides have the perovskite structure (CaTiO3) including BaTiO3, KNbO3, PbTiO3. Lets try to predict the structure without looking it up Based on the TiO2 structures , we expect the Ti to be in an octahedron of O2-, STiO = 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, nBa, leading to SBa = 2/nBa, and how many Ba around each O, nOBa. Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 38 Prediction of BaTiO3 structure : Ba coordination Since nOBa* SBa = 2/3, the missing charge for the O, we have only a few possibilities: nBa= 3 leading to SBa = 2/nBa=2/3 leading to nOBa = 1 nBa= 6 leading to SBa = 2/nBa=1/3 leading to nOBa = 2 nBa= 9 leading to SBa = 2/nBa=2/9 leading to nOBa = 3 nBa= 12 leading to SBa = 2/nBa=1/6 leading to nOBa = 4 Each of these might lead to a possible structure. The last case is the correct one for BaTiO3 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. Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 39 BaTiO3 structure (Perovskite) Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 40 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 (a0) Thus QM calculations of dipole moment are in units of ea0 which we refer to as au. However the international standard for quoting dipole moment is the Debye = 10-10 esu A Where m(D) = 2.5418 m(au) Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 41 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(a0) = C m(D)/R(A) where C=0.743470. Postive 42 © copyright 2011 William A. Goddard III, all rights reserved thatCh120a-Goddard-L25 head of column is negative Charge Equilibration First consider how the energy of an atom depends on the net charge on the atom, E(Q) Including terms through 2nd order leads to Charge Equilibration for Molecular Dynamics Simulations; A. K. Rappé and W. A. Goddard III; J. Phys. Chem. 95, 3358 (1991) (2) Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved (3) 43 Charge dependence of the energy (eV) of an atom E=12.967 Harmonic fit E=0 E=-3.615 Cl+ Q=+1 Cl Q=0 Ch120a-Goddard-L25 ClQ=-1 = 8.291 Get minimum at Q=-0.887 Emin = -3.676 = 9.352 © copyright 2011 William A. Goddard III, all rights reserved 44 QEq parameters Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 45 Interpretation of J, the hardness Define an atomic radius as RA0 Re(A2) Bond distance of homonuclear H 0.84 0.74 diatomic C 1.42 1.23 N 1.22 1.10 O 1.08 1.21 Si 2.20 2.35 S 1.60 1.63 Li 3.01 3.08 Thus J is related to the coulomb energy of a charge the size of the 46 Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved atom The total energy of a molecular complex Consider now a distribution of charges over the atoms of a complex: QA, QB, etc Letting JAB(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 © copyright 2011 William A. Goddard III, all rights reserved Ch120a-Goddard-L25 47 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 QA. 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 Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 48 The QEq Coulomb potential law We need now to choose a form for JAB(R) A plausible form is JAB(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 JAB(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 Ch120a-Goddard-L25 Using RC=0.759a0 © copyright 2011 William A. Goddard III, all rights reserved 49 QEq results for alkali halides Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 50 QEq for Ala-His-Ala Amber charges in parentheses Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 51 QEq for deoxy adenosine Amber charges in parentheses Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 52 QEq for polymers Nylon 66 PEEK Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 53 Stopped January 30 Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 54 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: ABO3, 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 Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 55 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. Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 56 Phases of BaTiO3 <111> polarized rhombohedral <110> polarized orthorhombic -90oC <100> polarized tetragonal 120oC 5oC Non-polar cubic Temperature Different phases of BaTiO3 Ba2+/Pb2+ c Ti4+ O2- a Non-polar cubic above Tc Six variants at room temperature c 1.01 ~ 1.06 a <100> tetragonal below Tc Domains separated by domain walls Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 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 Different phases of BaTiO3 <111> polarized rhombohedral <110> polarized orthorhombic -90oC face <100> polarized tetragonal 120oC 5oC edge Non-polar cubic vertex Temperature center 1960 Cochran Soft Mode Theory(Displacive Model) Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 58 Nature of the phase transitions Displacive model Assume that the atoms prefer to distort toward a face or edge or vertex of the octahedron 1960 Cochran Increasing Temperature Soft Mode Theory(Displacive Model) Order-disorder 1966 Bersuker Eight Site Model 1968 Comes Order-Disorder Model (Diffuse X-ray Scattering) Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 59 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. Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved Rhomb. 60 Problem displacive model: EXAFS & Raman observations d (001) EXAFS of Tetragonal Phase[1] •Ti distorted from the center of oxygen octahedral in tetragonal phase. α (111) •The angle between the displacement vector and (111) is α= 11.7°. 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 State Comm., 12, 1053 (1973) III, all rights reserved Ch120a-Goddard-L25 © copyright 2011 William A. Goddard 61 61 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): 14695-14700 (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 Microscopic Polarization Ti atom distortions Cubic I-43m Ch120a-Goddard-L25 z = Pz Py Px + Macroscopic Polarization + © copyright 2011 William A. Goddard III, all rights reserved = 62 QM results explain EXAFS & Raman observations d (001) EXAFS of Tetragonal Phase[1] •Ti distorted from the center of oxygen octahedral in tetragonal phase. α (111) •The angle between the displacement vector and (111) is α= 11.7°. 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. 1. Model Inversion symmetry in Cubic Phase Raman Active Displacive Yes No FE/AFE No Yes B. Ravel et al, Ferroelectrics, 206, 407 (1998) 2. A. M. Quittet et al, Solid State Comm., 12, 1053 (1973) III, all rights reserved Ch120a-Goddard-L25 © copyright 2011 William A. Goddard 63 63 stopped Ch120a-Goddard-L25 © copyright 2011 William A. Goddard III, all rights reserved 64