Transcript Slide 1

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: Wei-Guang Liu <[email protected]>
Ted Yu <[email protected]>
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
1
Course schedule
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
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
2
Last time
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
3
Ring ozone
Form 3 OO sigma bonds, but pp pairs overlap
Analog: cis HOOH bond is 51.1-7.6=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
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
4
GVB orbitals of N2
VB view
MO view
Re=1.10A
R=1.50A
R=2.10A
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
5
The configuration for C2
Si2 has this configuration
1
1
2
4
4
0
1
2
4
3
2
2
2
From 1930-1962 the 3Pu was
thought to be the ground
state
2
2
1S + is ground state
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
Now
6
Ground state of C2
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
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
7
Low-lying states of C2
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
8
VB view
MO view
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
9
The VB interference or resonance energy for H2+
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/(1-S) ≡ 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
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
10
Analysis of classical and interference energies
egx = t/(1+S) while eux = -t/(1-S)
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
t= -0.0542 hartree leading to
egx = -0.0372 hartree while
eux = + 0.10470 hartree
ecl = 0.00472 hartree
Where the 1-S term in the denominator
makes the u state 3 times as
antibonding
as the g ©state
is2010
bonding.
Ch120a-Goddard-L15
copyright
William A. Goddard III, all rights reserved
11
Contragradience
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
хL
large and negative
in the shaded
region between
atoms, where the L
and R orbitals have
opposite slope
(congragradience)
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
хR
12
The VB exchange energies for H2
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.
-Ex/(1 - S2)
+Ex/(1 + S2)
Ch120a-Goddard-L15
1E
g
3E
u
= Ecl + Egx
= Ecl + Eux
Each energy is
referenced to
the value at
R=∞, which is
-1 for Ecl, Eu, Eg
0 for Exu and Exg
© copyright 2010 William A. Goddard III, all rights reserved
13
Analysis of the VB exchange energy, Ex
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.
Thus we can
understand
bonding by
analyzing just the
KE part if Ex
Ch120a-Goddard-L15
T2
T1
Ex
TKE
© copyright 2010 William A. Goddard III, all rights reserved
14
Analysis of the exchange energies
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
E(hartree)
Eu1x
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
1x
E
g
but at long R, the 1e bond is
R(bohr)
15
stronger
than the 2e bond
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
Noble gas dimers
No bonding at the VB or MO level
Only simultaneous electron
correlation (London attraction) or
van der Waals attraction, -C/R6
s
Ar2
Re
De
Ch120a-Goddard-L15
LJ 12-6 Force Field
E=A/R12 –B/R6
= De[r-12 – 2r-6]
= 4 De[t-12 – t-6]
r= R/Re
t= R/s
where s = Re(1/2)1/6
=0.89 Re
© copyright 2010 William A. Goddard III, all rights reserved
16
London Dispersion
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)
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
17
London Dispersion
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)
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
18
Some New and old material
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
19
MO and VB view of He dimer, He2
MO view
ΨMO(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2
Net BO=0
VB view
ΨVB(He2) = A[(La)(Lb)(Ra)(Rb)]= (L)2(R)2
Substitute sg = R + L and sg = R - L
Get ΨMO(He2) ≡ ΨMO(He2)
Ch120a-Goddard-L15
Pauli  orthog of R
to L  repulsive
© copyright 2010 William A. Goddard III, all rights reserved
20
Remove an electron from He2 to get 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
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
21
Remove an electron from He2 to get 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
VB view
Substitute sg = R + L and sg = L - R
Get ΨVB(He2) ≡ A[(La)(Lb)(Ra)] - A[(La)(Rb)(Ra)]
= (L)2(R) - (R)2(L)
-
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
22
He2+
+
2S +
g
(sg)1(su)2
2S +
u
(sg
)2(s
u)
BO=0.5
MO good for discuss spectroscopy,
VB good for discuss chemistry
Check H2 and H2+ numbers
Ch120a-Goddard-L15
He2 Re=3.03A
De=0.02 kcal/mol
No bond
H2 Re=0.74xA
De=110.x kcal/mol
BO = 1.0
H2+ Re=1.06x A
De=60.x kcal/mol
BO = 0.5
© copyright 2010 William A. Goddard III, all rights reserved
23
New material
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
24
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
25
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)
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
Ch120a-Goddard-L15
E(eV)
© copyright 2010 William A. Goddard III, all rights reserved
R(A)
26
GVB orbitals
of NaCl
Dipole moment
= 9.001 Debye
Pure ionic
11.34 Debye
Thus
Dq=0.79 e
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
27
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 such that
M= (IP+EA)/5.2
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
28
Electronegativity
Based on M++
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
29
Comparison of Mulliken and Pauling electronegativities
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
30
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-L15
6 Na neighbors © copyright 2010 William A. Goddard III, all rights reserved
31
The NaCl or B1 crystal
All alkali halides
have this
structure except
CsCl, CsBr, CsI
(they have the B2
structure)
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
32
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
33
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-L15
copyrightA32,
2010 William
A. Goddard III, all rights reserved
34
Ionic radii, transition metals
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
35
Ionic radii Lanthanides and Actinide
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
36
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 = 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-L15
(RC/RA ) < 0.414 must
be2010
some
structure
© copyright
William other
A. Goddard
III, all rights reserved
37
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)
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
38
Wurtzite or B4 structure
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
39
Sphalerite or Zincblende or B3 structure GaAs
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
40
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
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
41
Structures for II-VI compounds
B3 for 0.20 < (RC/RA) < 0.55
B1 for 0.36 < (RC/RA) < 0.96
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
42
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
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
43
Rutile (TiO2) or Cassiterite (SnO2) structure
Related to NaCl
with half the
cations missing
Find for RC/RA < 0.67
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
44
CaF2
rutile
CaF2
rutile
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
45
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
46
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-L15
rhombohedral
(trigonal)
hP9, P3121
No.152[10]
© copyright 2010 William A. Goddard III, all rights reserved
47
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-L15
tetragonal
tP6, P42/mnm,
No.136[17]
© copyright 2010 William A. Goddard III, all rights reserved
48
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-L15
right
© copyright 2010 William A. Goddard III, all rights reserved
49
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
50
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
51
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
52
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
53
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
54
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
55
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
56
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
57
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
58
BaTiO3 structure (Perovskite)
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
59
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
60
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 
61
© copyright 2010 William A. Goddard III, all rights reserved
thatCh120a-Goddard-L15
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
(3)
62
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-L15
ClQ=-1
= 8.291
Get minimum at Q=-0.887
Emin = -3.676
= 9.352
© copyright 2010 William A. Goddard III, all rights reserved
63
QEq parameters
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
64
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
65
Ch120a-Goddard-L15
© copyright 2010 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 2010 William A. Goddard III, all rights reserved
Ch120a-Goddard-L15
66
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
67
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-L15
Using RC=0.759a0
© copyright 2010 William A. Goddard III, all rights reserved
68
QEq results for alkali halides
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
69
QEq for Ala-His-Ala
Amber
charges in
parentheses
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
70
QEq for deoxy adenosine
Amber
charges in
parentheses
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
71
QEq for polymers
Nylon 66
PEEK
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
72
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
73
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
74
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
75
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
76
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
77
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-L15
© copyright 2010 William A. Goddard III, all rights reserved
Rhomb.
78
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-L15
© copyright
2010 William
A. Goddard
79
79
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-L15 z
=
Pz
Py
Px
+
Macroscopic
Polarization
+
© copyright 2010 William A. Goddard III, all rights reserved
=
80
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-L15
© copyright
2010 William
A. Goddard
81
81
Ti atom distortions and polarizations determined from QM calculations. Ti distortions are shown in
the FE-AFE fundamental unit cells. Yellow and red strips represent individual Ti-O 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.
Microscopic Polarization
Ti atom
distortions
Cubic
I-43m
=
+
z
o
x
Pz
Py
Px
Macroscopic
Polarization
=
+
FE / AFE
y
Tetragonal
I4cm
+
=
+
=
FE / AFE
=
Teperature
Orthorhombic
Pmn21
+
=
+
+
=
FE / AFE
Rhombohedral
R3m
Ch120a-Goddard-L15
+
=
© copyright 2010 William A. Goddard III, all rights reserved
82
Phase Transition at 0 GPa
Thermodynamic Functions
ZPE =
Transition Temperatures and
Entropy Change FE-AFE
1
 (q, v)

2 q ,v
  (q, v) 
1

E = Eo +   (q, v) coth
2 q ,v
 2 k BT 
Phas
e

  (q, v)  
 
F = Eo + k BT  ln 2 sinh

q ,v
 2 k BT  

R
0
22.78106 0
O
0.06508
22.73829 0.02231
T
0.13068
22.70065 0.05023
C
0.19308
22.66848 0.08050
  (q, v) 




(
q
,
v
)
coth

q ,v
 2 k BT 

  (q, v)  
 
- k B  ln 2 sinh

q ,v
 2 k BT  

S=
1
2T
Eo
(kJ/mol)
ZPE
(kJ/mol)
Eo+ZPE
(kJ/mol)
Vibrations important to include
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
83
Polarizable QEq
Proper description of Electrostatics is critical E = E Coulomb + E vdW
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
Self-consistent charge equilibration (QEq)

 c 2
ic 3 2 c
c
r (r ) = ( p ) Qi exp(-i  | r - ri | )
 s 2
is 3 2
s 
s
s
ri (r ) = ( p ) Qi exp(-i  | r - ri | )
c
i
Four universal parameters for each element:
Get from QM
io , J io , Ric , Ris & qic
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
84
Validation
Phase
Properties
EXP
QMd
P-QEq
Cubic
(Pm3m)
a=b=c (A)
B(GPa)
εo
4.012a
4.007
167.64
4.0002
159
4.83
a=b(A)
c(A)
Pz(uC/cm2)
B(GPa)
3.99c
4.03c
15 to
26b
3.9759
4.1722
3.9997
4.0469
17.15
135
a=b(A)
c(A)
γ(degree)
Px=Py(uC/cm2)
B(Gpa)
4.02c
3.98c
89.82c
15 to
31b
4.0791
3.9703
89.61
a=b=c(A)
α=β=γ(degree)
Px=Py=Pz(uC/cm2)
B(GPa)
4.00c
89.84c
14 to
33b
4.0421
89.77
Tetra.
(P4mm)
Ortho.
(Amm2)
Rhomb.
(R3m)
6.05e
98.60
97.54
97.54
4.0363
3.9988
89.42
14.66
120
4.0286
89.56
12.97
120
a.
H. F. Kay and P. Vousden, Philosophical Magazine 40, 1019 (1949)
b.
H. F. Kay and P. Vousden, Philosophical Magazine 40, 1019 (1949) ;W. J. Merz, Phys. Rev. 76, 1221 (1949); W. J.
Merz, Phys. Rev. 91, 513 (1955); H. H. Wieder, Phys. Rev. 99,1161 (1955)
c.
G.H. Kwei, A. C. Lawson, S. J. L. Billinge, and S.-W. Cheong, J. Phys. Chem. 97,2368
Ch120a-Goddard-L15
© copyright
William
A. Society
Goddard
III, all rights
d.
M. Uludogan, T. Cagin, and W.
A. Goddard, 2010
Materials
Research
Proceedings
(2002),reserved
vol. 718, p. D10.11.
85
QM Phase Transitions at 0 GPa, FE-AFE
R
Experiment [1]
Transition
O
T C
This Study
T(K)
ΔS (J/mol)
T(K)
ΔS (J/mol)
R to O
183
0.17±0.04
228
0.132
O to T
278
0.32±0.06
280
0.138
T to C
393
0.52±0.05
301
0.145
1. G. Shirane and A. Takeda,
J. Phys.2010
Soc.William
Jpn., 7(1):1,
1952 III, all rights reserved
Ch120a-Goddard-L15
© copyright
A. Goddard
86
Free energies for Phase Transitions
Common Alternative free energy
from Vibrational states at 0K
We use 2PT-VAC: free energy from
MD at 300K
Velocity Auto-Correlation Function
C vv = V (0)  V (t )  c
=
 dV (0)  V (t ) r ()
U ({ri , i = 1...3 N }) = U o ( ri o , i = 1...3 N )
1 3 N  2U
+ 
2 i , j =1 ri r j
v
Dri Dr j
Rio , r jo
Velocity Spectrum
~
C vv (v) =


-
dte2pivt C vv (t )
3N
~
S (v) = 2 b  m j C vv (v)
j =1

System Partition Function
Q=
 dvS (v) ln Q(v)
0
Thermodynamic Functions: Energy, Entropy, Enthalpy, Free Energy
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
87
Free energies predicted for BaTiO3 FE-AFE phase structures.
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)
Free Energy (J/mol)
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)
Ch120a-Goddard-L15
Temperature (K)
© copyright 2010 William A. Goddard III, all rights reserved
88
Nature of the phase transitions
Displacive
1960
Cochran
Soft Mode Theory(Displacive Model)
Order-disorder
1966
Bersuker
Eight Site Model
1968
Comes
Order-Disorder Model (Diffuse X-ray Scattering)
Develop model to explain all the following experiments (FE-AFE)
EXP
Displacive
Order-Disorder
FE-AFE (new)
Small Latent Heat
Yes
No
Yes
Diffuse X-ray
diffraction
Yes
Yes
Yes
Distorted structure in No
EXAFS
Yes
Yes
Intense Raman in
Cubic Phase
Yes
Yes
Ch120a-Goddard-L15
No
© copyright 2010 William A. Goddard III, all rights reserved
89
Space Group & Phonon DOS
Phase
Displacive Model
FE/AFE Model (This Study)
Symmetry 1
atoms
Symmetry 2
atoms
C
Pm3m
5
I-43m
40
T
P4mm
5
I4cm
40
O
Amm2
5
Pmn21
10
R
R3m
5
R3m
5
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
90
Frozen Phonon Structure-Pm3m(C) Phase - Displacive
Pm3m Phase
Frozen Phonon of BaTiO3 Pm3m phase
Brillouin Zone
Ch120a-Goddard-L15
Γ
(0,0,0)
X1
(1/2, 0, 0)
X2
(0, 1/2, 0)
X3
(0, 0, 1/2)
M1
(0,1/2,1/2)
M2
(1/2,0,1/2)
M3
(1/2,1/2,0)
R
(1/2,1/2,1/2)
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)
© copyright 2010 William A. Goddard III, all rights reserved
91
Frozen Phonon Structure – Displacive model
P4mm (T) Phase
Unstable TO phonons:
M1(1), M2(1)
Ch120a-Goddard-L15
Amm2 (O) Phase
Unstable TO phonons:
M3(1)
R3m (R) Phase
NO UNSTABLE
PHONONS
© copyright 2010 William A. Goddard III, all rights reserved
92
Next Challenge: Explain X-Ray Diffuse Scattering
Cubic
Tetra.
Ortho.
Rhomb.
Diffuse X diffraction of BaTiO3 and KNbO3,
R. Comes et al, Acta Crystal. A., 26, 244, 1970
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
93
X-Ray Diffuse Scattering
Photon K’
Phonon Q
Photon K
Cross Section
Scattering function
Dynamic structure factor
Debye-Waller factor
Ch120a-Goddard-L15
s 1
K'
=N
S1 (Q)

K
1
(n(Q, v) + )
2  F (Q, v) 2
S1 (Q) = 
1
 (Q, v)
v
F1 (Q, v) = 
i


fi
exp- Wi Q  + iQ  ri  Q  e*i Q, v 
Mi
1

2
 n(q, v) +   Q  ei (q, v) 

2

Wi (Q) =

2MN m q ,v
 (q, v)
© copyright 2010 William A. Goddard III, all rights reserved
94
Diffuse X-ray diffraction predicted for the BaTiO3 FE-AFE
phases.
Qx
Qx
-4
-3
-2
-1
0
1
2
4
-5
5
5
3
3
2
2
1
1
Qz
4
4
0
-4
-3
-2
-1
-1
-2
-2
-3
-3
-4
-4
-5
-5
-4
-3
-2
-1
0
1
2
3
4
5
3
4
5
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
T (350K)
Qx
-5
5
0
0
-1
Qx
1
2
3
4
5
-5
5
4
3
3
2
2
1
1
Qz
4
0
-4
-3
-2
-1
0
1
2
0
-1
-1
-2
-2
-3
-3
-4
-4
-5
-5
O (250K)
Ch120a-Goddard-L15
3
C (450K)
Qz
The partial differential
cross sections (arbitrary
unit) of X-ray 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).
Qz
-5
5
R (150K)
© copyright 2010 William A. Goddard III, all rights reserved
95
Summary Phase Structures and Transitions
•Phonon structures
•FE/AFE transition
Agree with experiment?
EXP
Displacive Order-Disorder
FE/AFE(This Study)
Small Latent Heat
Yes
No
Yes
Diffuse X-ray
diffraction
Yes
Yes
Yes
Distorted structure
in EXAFS
No
Yes
Yes
Intense Raman in
Cubic Phase
No
Yes
Yes
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
96
96
Domain Walls Tetragonal Phase of BaTiO3
Consider 3 cases
experimental
Polarized light
optical
micrographs of
domain patterns in
barium titanate (E.
Burscu, 2001)
CASE I
CASE II
+++++++++++++++
CASE III
++++
----
++++
P
P
P
P
P
-----------------
+++++++++++++++
E=0
----
P
- - - - +++ + - - - ++++
----
++++
++++
----
E
-----------------
- - - - +++ + - - - -
++++
•Open-circuit
•Short-circuit
•Open-circuit
•Surface charge not neutralized
Charge
and
•Surface charge neutralized •Surface charge not neutralized •Domain stucture
cs
Epolarization
= E el + E vdw
E cs = E el + E vdw
E cs = E el + E vdw
distributions at the
 
97
Ch120a-Goddard-L15
© copyright 2010 William-A.PGoddard
97
 E III, all rights reserved+ E dw + E surface
180° Domain Wall of BaTiO3 – Energy vs length
(001)
z
(001)
o
Ly
y
Type I
Type II
Type III
Ch120a-Goddard-L15
Type I
L>64a(256Å)
Type II
4a(16Å)<L<32a(128Å)
Type III
L=2a(8Å)
© copyright 2010 William A. Goddard III, all rights reserved
98
98
180° Domain Wall –
Type I, developed
Displacement dY
(001)
(001)
Ly = 2048 Å =204.8 nm
z
C
o
Zoom out
A
A
D
B
A B C
Displacement dZ
Displacement
reduced near
domain wall
A B C
Ch120a-Goddard-L15
Displace away
from domain
wall
D
Zoom out
Wall center
y
Transition layer
D
Domain structure
© copyright 2010 William A. Goddard III, all rights reserved
99
99
180° Domain Wall –
Type I, developed
(001)
z
L = 2048 Å
Polarization P
(001)
Free charge ρf
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
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
100
100
180° Domain Wall – Type
II, underdeveloped
(001)
(001)
z
L = 128 Å
A
C
Displacement dY
B
A
B
Displacement dZ
Free charge ρf
D
C
o
y
Polarization P
D
Wall center: expanded, polarization switches,
positively charged
Transition layer: contracted, polarization relaxes,
negatively charged
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
101
101
180° Domain Wall – Type
III, antiferroelectric
(001)
(001)
z
L= 8 Å
o
Polarization P
Displacement dZ
y
Wall center: polarization switch
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
102
102
180° Domain Wall of BaTiO3 – Energy vs length
(001)
z
(001)
o
Ly
y
Type I
Type II
Type III
Ch120a-Goddard-L15
Type I
L>64a(256Å)
Type II
4a(16Å)<L<32a(128Å)
Type III
L=2a(8Å)
© copyright 2010 William A. Goddard III, all rights reserved
103
103
90° Domain Wall of BaTiO3
(010)
(0 0 1)
L
z
2  2N  2 2
o
y
L=724 Å (N=128)
•Wall energy is 0.68 erg/cm2
•Stable only for L362 Å
(N64)
Wall center
Transition Layer
Domain Structure
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
104
104
(010)
90° Domain Wall of BaTiO3
(0 0 1)
L
L=724 Å (N=128)
Displacement dY
Displacement dZ
z
o
y
Free Charge Density
Wall center: Orthorhombic phase, Neutral
Transition Layer: Opposite charged
Domain Structure
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
105
90° Wall – Connection to Continuum Model
3-D Poisson’s Equation
r
 2

U
=

eo

r = r p + r f

 r p = -  P

1-D Poisson’s Equation
 d 2U
r
=
 2
eo
 dy

r = r p + r f

 r = - dPy
 p
dy
y 
1 y
Solution U ( y) =   Py  d -   r f  dd  + c  y
0 

eo  o
C is determined by the periodic boundary condition: U (0) = U ( 2L)
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
106
90° Domain Wall of BaTiO3
(010)
L=724 Å (N=128)
Polarization Charge Density
Electric Field
Ch120a-Goddard-L15
(0 0 1)
L
Free Charge Density
z
o
y
Electric Potential
© copyright 2010 William A. Goddard III, all rights reserved
107
Summary III (Domain Walls)
180° domain wall
•Three types – developed, underdeveloped and AFE
•Polarization switches abruptly across the wall
•Slightly charged symmetrically
90° domain wall
•Only stable for L36 nm
•Three layers – Center, Transition & Domain
•Center layer is like orthorhombic phase
•Strong charged – Bipolar structure – Point Defects and Carrier
injection
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
108
108
Mystery: Origin of Oxygen Vacancy Trees!
0.1μm
Oxgen deficient dendrites in LiTaO3 (Bursill et al,
Ferroelectrics, 70:191, 1986)
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
109
stop
Ch120a-Goddard-L15
© copyright 2010 William A. Goddard III, all rights reserved
110