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## Lecture 26 March 07, 2011 BaTiO 3 and Hypervalent

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**Lecture 26 March 07, 2011**BaTiO3 and Hypervalent Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy 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: Wei-Guang Liu <wgliu@wag.caltech.edu> Caitlin Scott <cescott@caltech.edu>**The ionic limit**At R=∞ the cost of forming Na+ and Cl- is 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) 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 6.1-1.5=4.6 eV The exper De = 4.23 eV Showing that ionic character dominates E(eV) R(A)**GVB orbitals of NaCl**Dipole moment = 9.001 Debye Pure ionic 11.34 Debye Thus Dq=0.79 e**Electronegativity**Based on M++**The NaCl or B1 crystal**All alkali halides have this structure except CsCl, CsBr, CsI (they have the B2 structure)**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**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. A32, 751 (1976)**Role of ionic sizes in determining crystal structures**Assume that the anions are large and packed so that they contact, so that 2RA < L, where L is the distance between anions Assume that the anion and cation are in contact. Calculate the smallest cation consistent with 2RA < L. RA+RC = (√3)L/2 > (√3) RA Thus RC/RA > 0.732 RA+RC = L/√2 > √2 RA Thus RC/RA > 0.414 Thus for 0.414 < (RC/RA ) < 0.732 we expect B1 For (RC/RA ) > 0.732 either is ok. For (RC/RA ) < 0.414 must be some other structure**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, 2nd edition. Volume 1 (1963)**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 (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**Structures for II-VI compounds**B3 for 0.20 < (RC/RA) < 0.55 B1 for 0.36 < (RC/RA) < 0.96**CaF2 or fluorite structure**Like GaAs but now have F at all tetrahedral sites Or like CsCl but with half the Cs missing Find for RC/RA > 0.71**Rutile (TiO2) or Cassiterite (SnO2) structure**Related to NaCl with half the cations missing Find for RC/RA < 0.67**CaF2**rutile CaF2 rutile**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**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 rhombohedral (trigonal)hP9, P3121 No.152[10] From wikipedia**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 6-coordinate Si; high pressure form densest of the SiO2 polymorphs tetragonaltP6, P42/mnm, No.136[17] From wikipedia**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 right front**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**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)**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**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.**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.**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)**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 that 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; • K. Rappé and W. A. Goddard III; J. Phys. Chem. 95, 3358 (1991) (2) (3)**Charge dependence of the energy (eV) of an atom**E=12.967 E=0 E=-3.615 Cl+ Cl Cl- Q=+1 Q=0 Q=-1 Harmonic fit Get minimum at Q=-0.887 Emin = -3.676 = 8.291 = 9.352**Interpretation of J, the hardness**Define an atomic radius as RA0 Re(A2) Bond distance of homonuclear diatomic H 0.84 0.74 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 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**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**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 Using RC=0.759a0 And l = 0.5**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.**Phases of BaTiO3**<111> polarized rhombohedral <110> polarized orthorhombic <100> polarized tetragonal Non-polar cubic Temperature 120oC -90oC 5oC Different phases of BaTiO3 Ba2+/Pb2+ Ti4+ O2- c Domains separated by domain walls a Non-polar cubic above Tc Six variants at room temperature <100> tetragonal below Tc**Nature of the phase transitions**<111> polarized rhombohedral <110> polarized orthorhombic <100> polarized tetragonal Non-polar cubic Temperature 120oC -90oC 5oC 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 face edge vertex center**MEMS Actuator performance parameters:**Actuation strain Work per unit volume Frequency Goal: Obtain cyclic high actuations by 90o domain switching in ferroelectrics Design thin film micro devices for large actuations Ferroelectric Actuators 10 8 s h a p e m e m o r y a l l o y 90o domain switching 10 7 s o l i d - l i q u i d f a t i g u e d S M A p n e u m a t i c t h e r m o - 10 6 P Z T 10 5 e l e c t r o m a g n e t i c ( E M ) Work per volume (J/m3) m u s c l e 10 4 e l e c t r o s t a t i c ( E S ) E M 1 0 3 E S Z n O m i c r o b u b b l e 1 0 2 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 1 2 3 4 5 6 7 C y c l i n g F r e q u e n c y ( H z ) Characteristics of common actuator materials Tetragonal perovskites: 1% (BaTiO3), 6.5% (PbTiO3)) P. Krulevitch et al, MEMS 5 (1996) 270-282**Bulk Ferroelectric Actuation**s s V 0 V s s Strains, BT~1%, PT~5.5% • Apply constant stress and cyclic voltage • Measure strain and charge • In-situ polarized domain observation US Patent # 6,437, 586 (2002) Eric Burcsu, 2001**Ferroelectric Model MEMS Actuator**[100] [010] • BaTiO3-PbTiO3 (Barium Titanate (BT)-Lead Titanate (PT) • Perovskite pseudo-single crystals (biaxially textured thin films) MEMS TestBed**Application: Ferroelectric Actuators Must understand role of**domain walls in mediate switching 1.0 E 2 Experiments in BaTiO3 Strain (%) Domain walls lower the energy barrierby enabling nucleation and growth 0 1 -10,000 0 10,000 90° domain wall Electric field (V/cm) Switching gives large strain, … but energy barrier is extremely high! Essential questions: Are domain walls mobile? Do they damage the material? In polycrystals? In thin films? Use MD with ReaxFF**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 Order-disorder**Comparison to experiment**Cubic Tetra. Ortho. Rhomb. 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**Problem displacive model: EXAFS & Raman observations**d (001) α (111) 49 • 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 • B. Ravel et al, Ferroelectrics, 206, 407 (1998) • A. M. Quittet et al, Solid State Comm., 12, 1053 (1973)**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