Lecture 15 February 8, 2010 Ionic bonding and oxide crystals. Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy. William A. Goddard, III, [email protected] 316 Beckman Institute, x3093
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Lecture 15 February 8, 2010
Ionic bonding and oxide crystals
Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy
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: WeiGuang Liu <[email protected]>
Ted Yu <[email protected]>
Monday Feb. 8, 2pm L14 TODAY(caught up)
Midterm was given out on Friday. Feb. 5, due on Wed. Feb. 10
It is five hour continuous take home with 0.5 hour break,
open notes for any material distributed in the course or on the course web site
but closed book otherwise
No collaboration
Friday Feb. 12, postpone lecture from 2pm to 3pm
Form 3 OO sigma bonds, but pp pairs overlap
Analog: cis HOOH bond is 51.17.6=43.5 kcal/mol. Get total bond of 3*43.5=130.5 which is 11.5 more stable than O2.
Correct for strain due to 60º bond angles = 26 kcal/mol from cyclopropane.
Expect ring O3 to be unstable with respect to O2 + O by ~14 kcal/mol
But if formed it might be rather stable with respect various chemical reactions.
Ab Initio Theoretical Results on the Stability of Cyclic Ozone L. B. Harding and W. A. Goddard III J. Chem. Phys. 67, 2377 (1977) CN 5599
VB view
MO view
Re=1.10A
R=1.50A
R=2.10A
Si2 has this configuration
2
1
0
1
2
1
2
4
3
4
4
2
2
From 19301962 the 3Pu was thought to be the ground state
Now 1Sg+ is ground state
2
2
MO configuration
Have two strong p bonds,
but sigma system looks just like Be2 which leads to a bond of ~ 1 kcal/mol
The lobe pair on each Be is activated to form the sigma bond. The net result is no net contribution to bond from sigma electrons. It is as if we started with HCCH and cut off the Hs
VB view
MO view
The VB wavefunctions for H2+
Φg= (хL + хR) and Φu= (хL  хR) lead to
eg = (hLL + 1/R) + t/(1+S) ≡ ecl + Egx
eu = (hLL + 1/R)  t/(1S) ≡ ecl + Eux
where t = (hLR  ShLL) is the VB interference or resonance energy and
ecl = (hLL + 1/R) is the classical energy
As shown here the t dominates the bonding and antibonding of these states
egx = t/(1+S) while eux = t/(1S)
Consider first very long R, where S~0
Then egx = t while eux = t
so that the bonding and antibonding effects are similar.
Now consider a distance R=2.5 bohr = 1.32 A near equilibrium
Here S= 0.4583
Where the 1S term in the denominator makes the u state 3 times as antibonding as the g state is bonding.
хL
хR
The above discussions show that the interference or exchange part of KE dominates the bonding, tKE=KELR –S KELL
This decrease in the KE due to overlapping orbitals is dominated by
tx = ½ [< (хL). ((хR)>  S[< (хL)2>
Dot product is large and negative in the shaded region between atoms, where the L and R orbitals have opposite slope (congragradience)
For H2, the classical energy is slightly attractive, but again the difference between bonding (g) and anti bonding (u) is essentially all due to the exchange term.
1Eg = Ecl + Egx
3Eu = Ecl + Eux
Ex/(1  S2)
Each energy is referenced to the value at R=∞, which is
1 for Ecl, Eu, Eg 0 for Exu and Exg
+Ex/(1 + S2)
where Ex = {(hab + hba) S + Kab –EclS2} = T1 + T2
Here T1 = {(hab + hba) S –(haa + hbb)S2} = 2St
Where t = (hab – Shaa) contains the 1e part
T2 ={Kab –S2Jab} contains the 2e part
Clearly the Ex is dominated by T1 and clearly T1 is dominated by the kinetic part, TKE.
T2
T1
Ex
Thus we can understand bonding by analyzing just the KE part if Ex
TKE
E(hartree)
Eu1x
Eg1x
R(bohr)
The one electron exchange for H2 leads to
Eg1x ~ +2St /(1 + S2)
Eu1x ~ 2St /(1  S2)
which can be compared to the H2+ case
egx ~ +t/(1 + S)
eux ~ t/(1  S)
For R=1.6bohr (near Re), S=0.7 Eg1x ~ 0.94t vs. egx ~ 0.67t
Eu1x ~ 2.75t vs. eux ~ 3.33t
For R=4 bohr, S=0.1
Eg1x ~ 0.20t vs. egx ~ 0.91t
Eu1x ~ 0.20t vs. eux ~ 1.11t
Consider a very small R with S=1. Then
Eg1x ~ 2t vs. egx ~ t/2
so that the 2e bond is twice as strong as the 1e bond but at long R, the 1e bond is stronger than the 2e bond
No bonding at the VB or MO level
Only simultaneous electron correlation (London attraction) or van der Waals attraction, C/R6
where s = Re(1/2)1/6
=0.89 Re
Ar2
s
Re
De
The weak binding in He2 and other noble gas dimers was explained in terms of QM by Fritz London in 1930
The idea is that even for a spherically symmetric atoms such as He the QM description will have instantaneous fluctuations in the electron positions that will lead to fluctuating dipole moments that average out to zero. The field due to a dipole falls off as 1/R3 , but since the average dipole is zero the first nonzero contribution is from 2nd order perturbation theory, which scales like
C/R6 (with higher order terms like 1/R8 and 1/R10)
The weak binding in He2 and other nobel gas dimers was explained in terms of QM by Fritz London in 1930
The idea is that even for a spherically symmetric atoms such as He the QM description will have instantaneous fluctuations in the electron positions that will lead to fluctuating dipole moments that average out to zero. The field due to a dipole falls off as 1/R3 , but since the average dipole is zero the first nonzero contribution is from 2nd order perturbation theory, which scales like
C/R6 (with higher order terms like 1/R8 and 1/R10)
Consequently it is common to fit the interaction potentials to functional forms with a long range 1/R6 attraction to account for London dispersion (usually referred to as van der Waals attraction) plus a short range repulsive term to account for short Range Pauli Repulsion)
MO view
ΨMO(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2
VB view
ΨVB(He2) = A[(La)(Lb)(Ra)(Rb)]= (L)2(R)2
Net BO=0
Pauli orthog of R to L repulsive
Substitute sg = R +Land sg = R L
Get ΨMO(He2) ≡ ΨMO(He2)
MO view
Ψ(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2
Two bonding and two antibonding BO= 0
Ψ(He2+) = A[(sga)(sgb)(sua)]= (sg)2(su) BO = ½
Get 2Su+ symmetry.
Bond energy and bond distance similar to H2+, also BO = ½

MO view
Ψ(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2
Two bonding and two antibonding BO= 0
Ψ(He2+) = A[(sga)(sgb)(sua)]= (sg)2(su) BO = ½
Get 2Su+ symmetry.
Bond energy and bond distance similar to H2+, also BO = ½
VB view
Substitute sg = R +Land sg = L R
Get ΨVB(He2) ≡ A[(La)(Lb)(Ra)]  A[(La)(Rb)(Ra)]
= (L)2(R)  (R)2(L)
2Sg+
(sg)1(su)2
2Su+
(sg)2(su)
BO=0.5

+
He2 Re=3.03A De=0.02 kcal/mol
No bond
H2 Re=0.74xA De=110.x kcal/mol
BO = 1.0
MO good for discuss spectroscopy,
VB good for discuss chemistry
H2+ Re=1.06x A De=60.x kcal/mol
BO = 0.5
Check H2 and H2+ numbers
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
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 6.11.5=4.6 eV
The exper De = 4.23 eV
Showing that ionic character dominates
E(eV)
R(A)
Dipole moment = 9.001 Debye
Pure ionic 11.34 Debye
Thus Dq=0.79 e
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 such that
M= (IP+EA)/5.2
Based on M++
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
All alkali halides have this structure except CsCl, CsBr, CsI
(they have the B2 structure)
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
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)
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
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)
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
B3 for 0.20 < (RC/RA) < 0.55
B1 for 0.36 < (RC/RA) < 0.96
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
Related to NaCl with half the cations missing
Find for RC/RA < 0.67
CaF2
rutile
CaF2
rutile
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 cationanion 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
Each Si bonds to 4 O,
OSiO = 109.5°
each O bonds to 2 Si
SiOSi = 155.x °
Helical chains
single crystals optically active;
αquartz converts to βquartz at 573 °C
rhombohedral (trigonal)hP9, P3121 No.152[10]
From wikipedia
The stishovite phase of SiO2 has six coordinate Si, S=2/3.
Thus each O must have 3 Si neighbors
Rutilelike structure, with 6coordinate Si;
high pressure form
densest of the SiO2 polymorphs
tetragonaltP6, P42/mnm, No.136[17]
From wikipedia
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
Each Al3+ is in a distorted octahedron, leading to S=1/2.
Thus each O2 must be coordinated to 4 Al
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)
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?
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.
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.
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.
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.
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.
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.
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 = 1010 esu A
Where m(D) = 2.5418 m(au)
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
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)
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
Define an atomic radius as
RA0
Re(A2)
Bond distance of homonuclear diatomic
H0.840.74
C1.421.23
N1.221.10
O1.081.21
Si2.202.35
S1.601.63
Li3.013.08
Thus J is related to the coulomb energy of a charge the size of the atom
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
Adding to the N1 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
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
Amber charges in parentheses
Amber charges in parentheses
Nylon 66
PEEK
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
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 Bcations 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.
<111> polarized rhombohedral
<110> polarized orthorhombic
<100> polarized tetragonal
Nonpolar cubic
Temperature
120oC
90oC
5oC
Different phases of BaTiO3
Ba2+/Pb2+
Ti4+
O2
c
Domains separated by domain walls
a
Nonpolar cubic
above Tc
Six variants at room temperature
<100> tetragonal
below Tc
<111> polarized rhombohedral
<110> polarized orthorhombic
<100> polarized tetragonal
Nonpolar 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
Displacive model
Assume that the atoms prefer to distort toward a face or edge or vertex of the octahedron
Increasing Temperature
Orderdisorder
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
d
(001)
α
(111)
79
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
The ferroelectric and cubic phases in BaTiO3 ferroelectrics are also antiferroelectric Zhang QS, Cagin T, Goddard WA Proc. Nat. Acad. Sci. USA, 103 (40): 1469514700 (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
d
(001)
α
(111)
81
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.
Ti atom distortions and polarizations determined from QM calculations. Ti distortions are shown in the FEAFE fundamental unit cells. Yellow and red strips represent individual TiO chains with positive and negative polarizations, respectively. Low temperature R phase has FE coupling in all three directions, leading to a polarization along <111> direction. It undergoes a series of FE to AFE transitions with increasing temperature, leading to a total polarization that switches from <111> to <011> to <001> and then vanishes.
Thermodynamic Functions
Transition Temperatures and Entropy Change FEAFE
Vibrations important to include
Proper description of Electrostatics is critical
Allow each atom to have two charges:
A fixed core charge(+4 for Ti)with a Gaussian shape
A variable shell charge with a Gaussian shape but subject to displacement and charge transfer
Electrostatic interactions between all charges, including the core and shell on same atom, includes Shielding as charges overlap
Allow Shell to move with respect to core, to describe atomic polarizability
Selfconsistent charge equilibration(QEq)
Four universal parameters for each element:
Get from QM
R
O
T
C
1. G. Shirane and A. Takeda, J. Phys. Soc. Jpn., 7(1):1, 1952
Common Alternative free energy from Vibrational states at 0K
We use 2PTVAC: free energy from
MD at 300K
Velocity AutoCorrelation Function
Velocity Spectrum
System Partition Function
Thermodynamic Functions: Energy, Entropy, Enthalpy, Free Energy
Free Energy (J/mol)
Temperature (K)
AFE coupling has higher energy and larger entropy than FE coupling.
Get a series of phase transitions with transition temperatures and entropies
Theory (based on low temperature structure)
233 K and 0.677 J/mol (R to O)
378 K and 0.592 J/mol (O to T)
778 K and 0.496 J/mol (T to C)
Experiment (actual structures at each T)
183 K and 0.17 J/mol (R to O)
278 K and 0.32 J/mol (O to T)
393 K and 0.52 J/mol (T to C)
Displacive
Orderdisorder
Develop model to explain all the following experiments (FEAFE)
Pm3m Phase
Frozen Phonon of BaTiO3 Pm3m phase
Brillouin Zone
15 Phonon Braches (labeled at T from X3):
TO(8) LO(4) TA(2)LA(1)
PROBLEM: Unstable TO phonons at BZ edge centers:M1(1), M2(1), M3(1)
P4mm (T) Phase
Amm2 (O) Phase
R3m (R) Phase
Unstable TO phonons:
M1(1), M2(1)
Unstable TO phonons:
M3(1)
NO UNSTABLE PHONONS
Cubic
Tetra.
Ortho.
Rhomb.
Diffuse X diffraction of BaTiO3 and KNbO3,
R. Comes et al, Acta Crystal. A., 26, 244, 1970
Photon K’
Phonon Q
Photon K
Cross Section
Scattering function
Dynamic structure factor
DebyeWaller factor
The partial differential cross sections (arbitrary unit) of Xray thermal scattering were calculated in the reciprocal plane with polarization vector along [001] for T, [110] for O and [111] for R. The AFE Soft phonon modes cause strong inelastic diffraction, leading to diffuse lines in the pattern (vertical and horizontal for C, vertical for T, horizontal for O, and none for R), in excellent agreement with experiment (25).
Agree with experiment?
96
+ + + + + + + + + + + + + + +
+ + + +     + + + +    
P
P
P
P
P
P
                
    + + ++     + + + +
+ + + + + + + + + + + + + + +
+ + + +     + + + +    
E=0
E
                
    + + ++     + + + +
CASE I
CASE II
CASE III
experimental
Polarized light optical micrographs of domain patterns in barium titanate (E. Burscu, 2001)
Charge and polarization distributions at the 90 degrees domain wall in barium titanate ferroelectric Zhang QS, Goddard WA Appl. Phys. Let., 89 (18): Art. No. 182903 (2006)
97
z
o
y
Type I
Type II
Type III
98
Ly
z
o
y
C
A
D
A
B
Wall center
Transition layer
Domain structure
A
A
B
B
C
C
D
D
99
Ly = 2048 Å =204.8 nm
Displacement dY
Zoom out
Displace away from domain wall
Displacement dZ
Displacement reduced near domain wall
Zoom out
99
z
o
y
Wall center: expansion, polarization switch, positively charged
Transition layer: contraction, polarization relaxed, negatively charged
Domain structure: constant lattice spacing, polarization and charge density
100
L = 2048 Å
Polarization P
Free charge ρf
100
z
A
C
o
y
B
D
A
B
C
D
101
L = 128 Å
Polarization P
Displacement dY
Displacement dZ
Free charge ρf
Wall center: expanded, polarization switches, positively charged
Transition layer: contracted, polarization relaxes, negatively charged
101
z
o
y
102
L= 8 Å
Displacement dZ
Polarization P
Wall center: polarization switch
102
z
o
y
Type I
Type II
Type III
103
Ly
z
L
o
y
Wall center
Transition Layer
Domain Structure
L=724 Å (N=128)
104
z
L
o
y
Wall center: Orthorhombic phase, Neutral
Transition Layer: Opposite charged
Domain Structure
L=724 Å (N=128)
Displacement dY
Displacement dZ
Free Charge Density
C is determined by the periodic boundary condition:
3D Poisson’s Equation
1D Poisson’s Equation
Solution
z
L
o
y
L=724 Å (N=128)
Polarization Charge Density
Free Charge Density
Electric Field
Electric Potential
180° domain wall
90° domain wall
108
0.1μm
Oxgen deficient dendrites in LiTaO3 (Bursill et al, Ferroelectrics, 70:191, 1986)