© copyright 2011 william a. goddard iii, all rights reservedch120a-goddard-l25 ch120a- goddard- l01...
TRANSCRIPT
© 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, [email protected] 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: Ch120aHours: 2-3pm Monday, Wednesday, Friday
Teaching Assistants:Sijia Dong <[email protected]>Samantha Johnson <[email protected]>
© 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-
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 3
The ionic limitAt 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)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 6.1-1.5=4.6 eVexperiment De = 4.23 eVThus ionic character dominates
the ionic curve crosses the covalent curve at R=14.4/1.524=9.45 A
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GVB orbitals of NaCl
Dipole moment = 9.001 Debye
Pure ionic 11.34 Debye
Thus Dq=0.79 e
R=6 A
R=4.7 A
R=3.5 A
Re=2.4 A
No overlapno bond
Overlap from Q transfer
Mostly Na+ Cl-
Very Na+ Cl-
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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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 6
Electronegativity
Based on M++
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Comparison of Mulliken and Pauling electronegativities
Mulliken Biggest flaw is the wrong value for HH is clearly much less electronegative than I
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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 has 6 Na neighbors
© 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)
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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
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Ionic radii, main group
From R. D. Shannon, Acta Cryst. A32, 751 (1976)
Fitted to various crystals. Assumes O2- is 1.40A
NaCl R=1.02+1.81 = 2.84, exper is 2.84
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Ionic radii, transition metals
HS
Fe3+ is d5 thus get HSS=5/2; LS=1/2 not importantFe2+ 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
atom octahedron tetrahedron
t2g [xy, xz, yz]
eg [x2-y2, 3z2-r2] t2g [xy, xz, yz]
eg [x2-y2, 3z2-r2] Five d orbitals same energy
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 13
More on HS and LS, octahedral site, Fe3+
Fe3+ is d5 thus get HSS=5/2; LS=1/2 not important
x2-y2
3z2-r2 eg
t2g
xyxzyz
Weak field
x2-y2
3z2-r2 eg
t2g
xyxzyz
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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 14
More on HS and LS, octahedral site, Fe2+
x2-y2
3z2-r2 eg
t2g
xyxzyz
Weak field
x2-y2
3z2-r2 eg
t2g
xyxzyz
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
Fe2+ is d6 thus HS=2; LS=0, both important
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 15
Ionic radii, transition metals
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Ionic radii Lanthanides and Actinide
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Role of ionic sizes in determining crystal structuresAssume 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
Thus RC/RA > 0.414
RA+RC = (√3)L/2 > (√3) RA
Thus RC/RA > 0.732
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
© 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, 2nd
edition. Volume 1 (1963)
Rules work ok
B1: 0.35 to 1.26
B2: 0.76 to 0.92
B1 expect 0.414 < (RC/RA ) < 0.732
B2 or B1 (RC/RA ) > 0.732
(RC/RA ) < 0.414 neither
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Sphalerite or Zincblende or B3 structure GaAs
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Wurtzite or B4 structure
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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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 22
Structures for II-VI compounds
B3 for 0.20 < (RC/RA) < 0.55B4 for 0.33 < (RC/RA) < 0.53B1 for 0.36 < (RC/RA) < 0.96
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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
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Rutile (TiO2) or Cassiterite (SnO2) structure
Related to NaCl with half the cations missing
Find for RC/RA < 0.67
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rutile
CaF2
rutile
CaF2
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Stopped L17, Feb 10
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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
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 28
a-quartz structure of SiO2
Helical chains single crystals optically active; α-quartz converts to β-quartz at 573 °C
rhombohedral (trigonal)hP9, P3121 No.152[10]
Each Si bonds to 4 O, OSiO = 109.5°each O bonds to 2 SiSi-O-Si = 155.x °
From wikipedia
© 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 SiO2 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
tetragonaltP6, P42/mnm, No.136[17]
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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
front right
anatase phase TiO2
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 31
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
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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)
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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?
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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.
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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.
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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.
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 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.
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 38
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.
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 39
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.
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 40
BaTiO3 structure (Perovskite)
© 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 (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)
© 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, 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
© 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 2nd order leads to
(2) (3)
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 44
Charge dependence of the energy (eV) of an atom
E=0
E=-3.615
E=12.967
Cl Cl-Cl+
Q=0 Q=-1Q=+1
Harmonic fit
= 8.291 = 9.352
Get minimum at Q=-0.887Emin = -3.676
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 45
QEq parameters
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 46
Interpretation of J, the hardness
Define an atomic radius as
H 0.84 0.74C 1.42 1.23N 1.22 1.10O 1.08 1.21Si 2.20 2.35S 1.60 1.63Li 3.01 3.08
RA0 Re(A2) Bond distance of
homonuclear diatomic
Thus J is related to the coulomb energy of a charge the size of the atom
© 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: QA, QB, etc
Letting JAB(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
© 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 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
© 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 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 overlapClearly this form as the problem that JAB(R) ∞ as R 0In 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 RC=0.759a0
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 50
QEq results for alkali halides
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QEq for Ala-His-Ala
Amber charges in
parentheses
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QEq for deoxy adenosine
Amber charges in
parentheses
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QEq for polymers
Nylon 66
PEEK
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Stopped January 30
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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
© 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.
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 57
c
a
Temperature120oC5oC-90oC
<111> polarized rhombohedral
<110> polarized orthorhombic
<100> polarized tetragonal
Non-polar cubic
Different phases of BaTiO3
Six variants at room temperature
06.1~01.1a
c
Domains separated by domain walls
Non-polar cubicabove Tc
<100> tetragonalbelow Tc
O2-
Ba2+/Pb2+
Ti4+
Phases of BaTiO3
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 58
Nature of the phase transitions
1960 Cochran Soft 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
Temperature120oC5oC-90oC
<111> polarized rhombohedral
<110> polarized orthorhombic
<100> polarized tetragonal
Non-polar cubic
Different phases of BaTiO3
face edge vertex center
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 59
Nature of the phase transitions
1960 Cochran Soft Mode Theory(Displacive Model)
Displacive model
Assume that the atoms prefer to distort toward a face or edge or vertex of the octahedron
Order-disorder1966 Bersuker Eight Site Model
1968 Comes Order-Disorder Model (Diffuse X-ray Scattering)
Increasing Temperature
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 60
Comparison to experiment
Displacive small latent heatThis agrees with experimentR O: T= 183K, DS = 0.17±0.04 J/molO T: T= 278K, DS = 0.32±0.06 J/molT C: T= 393K, DS = 0.52±0.05 J/mol
Cubic Tetra.
Ortho. Rhomb.
Diffuse xray scatteringExpect some disorder, agrees with experiment
© 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)
© 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): 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
Te
pe
ratu
re
x
CubicI-43m
TetragonalI4cm
RhombohedralR3m
OrthorhombicPmn21
y
z
o
FE AFE/
FE AFE/
FE AFE/
Px Py Pz
+ +
+ +
+ +
+ +
=
=
=
=
MacroscopicPolarization
Ti atom distortions
=
=
=
=
Microscopic Polarization
© 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)
Model Inversion symmetry in Cubic Phase
Raman Active
Displacive Yes No
FE/AFE No Yes
© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 64
stopped