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© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 Ch120a- Goddard- L01 1 Nature of the Chemical Bond with applications to.

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1 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 Ch120a- Goddard- L01 1 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy William A. Goddard, III, 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Lecture 9 January 27, 2013 Ionic bonding and crystals Course number: Ch120a Hours: 2-3pm Monday, Wednesday, Friday Teaching Assistants:Sijia Dong Samantha Johnson

2 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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 -

3 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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 R(A) E(eV) 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 the ionic curve crosses the covalent curve at R=14.4/1.524=9.45 A

4 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 4 GVB orbitals of NaCl Dipole moment = Debye Pure ionic  Debye Thus  q=0.79 e R=6 A R=4.7 A R=3.5 A Re=2.4 A No overlap  no bond Overlap from Q transfer Mostly Na + Cl - Very Na + Cl -

5 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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 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

6 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 6 Electronegativity Based on M ++

7 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 7 Comparison of Mulliken and Pauling electronegativities Mulliken Biggest flaw is the wrong value for H H is clearly much less electronegative than I

8 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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

9 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 9 The NaCl or B1 crystal All alkali halides have this structure except CsCl, CsBr, CsI (they have the B2 structure)

10 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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

11 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 11 Ionic radii, main group From R. D. Shannon, Acta Cryst. A32, 751 (1976) Fitted to various crystals. Assumes O 2- is 1.40A NaCl R= = 2.84, exper is 2.84

12 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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 atom octahedron tetrahedron t 2g [xy, xz, yz] e g [x 2 -y 2, 3z 2 -r 2 ] t 2g [xy, xz, yz] e g [x 2 -y 2, 3z 2 -r 2 ] Five d orbitals same energy

13 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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 x 2 -y 2 3z 2 -r 2 egeg t 2g xy xz yz Weak field x 2 -y 2 3z 2 -r 2 egeg t 2g xy xz yz Strong field Exchange stabilization dominates, get high spin S=5/2 as for atom Ligand interaction dominates, get low spin S=1/2 5*4/2=10 exchange terms, ~220 kcal/mol 3+1 exchange terms ~88 kcal/mol

14 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 14 More on HS and LS, octahedral site, Fe 2+ x 2 -y 2 3z 2 -r 2 egeg t 2g xy xz yz Weak field x 2 -y 2 3z 2 -r 2 egeg t 2g xy xz yz Strong field Exchange stabilization dominates, get high spin S=2 as for atom Ligand interaction dominates, get low spin S=0 5*4/2=10 exchange terms, ~220 kcal/mol 3+3 exchange terms ~132 kcal/mol Fe 2+ is d 6 thus HS=2; LS=0, both important

15 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 15 Ionic radii, transition metals

16 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 16 Ionic radii Lanthanides and Actinide

17 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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 Thus R C /R A > R A +R C = (√3)L/2 > (√3) 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

18 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 18 Radius Ratios of Alkali Halides and Noble metal halices Based on R. W. G. Wyckoff, Crystal Structures, 2 nd edition. Volume 1 (1963) Rules work ok B1: 0.35 to 1.26 B2: 0.76 to 0.92 B1 expect < (R C /R A ) < B2 or B1 (R C /R A ) > (R C /R A ) < neither

19 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 19 Sphalerite or Zincblende or B3 structure GaAs

20 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 20 Wurtzite or B4 structure

21 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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 Thus B3,B4 should be the stable structures for < (R C /R A ) <

22 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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 ) < 0.96

23 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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 > 0.71

24 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 24 Rutile (TiO 2 ) or Cassiterite (SnO 2 ) structure Related to NaCl with half the cations missing Find for R C /R A < 0.67

25 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 25 rutile CaF 2 rutile CaF 2

26 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 Stopped L17, Feb 10 26

27 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 27 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 / C where z C is the net charge on the cation and C is the coordination number Then z A =  i S I =  i z Ci / 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

28 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 28  -quartz structure of SiO 2 Helical chains single crystals optically active; α-quartz converts to β-quartz at 573 °C rhombohedral rhombohedral (trigonal) hP9, P3121 No.152[10][10] Each Si bonds to 4 O, OSiO = 109.5° each O bonds to 2 Si Si-O-Si = 155.x ° From wikipedia

29 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 29 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 From wikipedia Rutile-like structure, with 6- coordinate Si; high pressure form densest of the SiO2 polymorphs tetragonal tetragonal tP6, P42/mnm, No.136[17][17]

30 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 30 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 front right anatase phase TiO 2

31 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 31 Corundum (  -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

32 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 32 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)

33 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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?

34 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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.

35 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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.

36 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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 = 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.

37 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 37 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, Ba, leading to S Ba = 2/ Ba, and how many Ba around each O, OBa.

38 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 38 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, Ba, leading to S Ba = 2/ Ba, and how many Ba around each O, OBa.

39 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 39 Prediction of BaTiO3 structure : Ba coordination Since OBa * S Ba = 2/3, the missing charge for the O, we have only a few possibilities: Ba = 3 leading to S Ba = 2/ Ba =2/3 leading to OBa = 1 Ba = 6 leading to S Ba = 2/ Ba =1/3 leading to OBa = 2 Ba = 9 leading to S Ba = 2/ Ba =2/9 leading to OBa = 3 Ba = 12 leading to S Ba = 2/ Ba =1/6 leading to 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.

40 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 40 BaTiO3 structure (Perovskite)

41 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 41 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  (D) =  (au)

42 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 42 Fractional ionic character of diatomic molecules Obtained from the experimental dipole moment in Debye,  (D), and bond distance R(A) by  q =  (au)/R(a 0 ) = C  (D)/R(A) where C= Postive  that head of column is negative

43 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 43 Charge Equilibration Charge Equilibration for Molecular Dynamics Simulations; A.K. Rappé and W. A. Goddard III; J. Phys. Chem. 95, 3358 (1991) 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 (2)(3)

44 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 44 Charge dependence of the energy (eV) of an atom E=0 E= E= ClCl - Cl + Q=0Q=-1Q=+1 Harmonic fit = 8.291= Get minimum at Q= Emin =

45 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 45 QEq parameters

46 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 46 Interpretation of J, the hardness Define an atomic radius as H C N O Si S Li RA0RA0 R e (A 2 ) Bond distance of homonuclear diatomic Thus J is related to the coulomb energy of a charge the size of the atom

47 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 47 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 or Taking the derivative with respect to charge leads to the chemical potential, which is a function of the charges The definition of equilibrium is for all chemical potentials to be equal. This leads to

48 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 48 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

49 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 49 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 = 0.5 Using R C =0.759a 0

50 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 50 QEq results for alkali halides

51 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 51 QEq for Ala-His-Ala Amber charges in parentheses

52 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 52 QEq for deoxy adenosine Amber charges in parentheses

53 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 53 QEq for polymers Nylon 66 PEEK

54 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 Stopped January 30 54

55 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 55 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

56 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 56 Ferroelectrics The stability of the perovskite structure depends on the relative ionic radii: 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.

57 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 57 c a Temperature 120 o C 5oC5oC-90 o C polarized rhombohedral polarized orthorhombic polarized tetragonal Non-polar cubic Different phases of BaTiO 3 Six variants at room temperature Domains separated by domain walls Non-polar cubic above Tc tetragonal below Tc O 2- Ba 2+ /Pb 2 + Ti 4+ Phases of BaTiO3

58 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 58 Nature of the phase transitions 1960CochranSoft Mode Theory(Displacive Model) Displacive model Assume that the atoms prefer to distort toward a face or edge or vertex of the octahedron Increasing Temperature Temperature 120 o C 5oC5oC-90 o C polarized rhombohedral polarized orthorhombic polarized tetragonal Non-polar cubic Different phases of BaTiO 3 faceedgevertexcenter

59 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 59 Nature of the phase transitions 1960CochranSoft Mode Theory(Displacive Model) Displacive model Assume that the atoms prefer to distort toward a face or edge or vertex of the octahedron Order-disorder 1966BersukerEight Site Model 1968ComesOrder-Disorder Model (Diffuse X-ray Scattering) Increasing Temperature

60 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 60 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 Cubic Tetra. Ortho. Rhomb. Diffuse xray scattering Expect some disorder, agrees with experiment

61 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 61 Problem displacive model: EXAFS & Raman observations 61 (001) (111) d α 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= 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)

62 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 62 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

63 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 63 QM results explain EXAFS & Raman observations 63 (001) (111) d α 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= 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.B. Ravel et al, Ferroelectrics, 206, 407 (1998) 2.A. M. Quittet et al, Solid State Comm., 12, 1053 (1973) ModelInversion symmetry in Cubic Phase Raman Active DisplaciveYesNo FE/AFENoYes

64 © copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 64 stopped


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