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Title: Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy


1
Nature of the Chemical Bond with applications to
catalysis, materials science, nanotechnology,
surface science, bioinorganic chemistry, and
energy
Lecture 15 February 8, 2010 Ionic bonding and
oxide crystals
  • William A. Goddard, III, wag_at_wag.caltech.edu
  • 316 Beckman Institute, x3093
  • Charles and Mary Ferkel Professor of Chemistry,
    Materials Science, and Applied Physics,
    California Institute of Technology

Teaching Assistants Wei-Guang Liu
ltwgliu_at_wag.caltech.edugt Ted Yu lttedhyu_at_wag.caltech
.edugt
2
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
3
Last time
4
Ring ozone
Form 3 OO sigma bonds, but pp pairs overlap
Analog cis HOOH bond is 51.1-7.643.5 kcal/mol.
Get total bond of 343.5130.5 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
5
GVB orbitals of N2
VB view
MO view
Re1.10A
R1.50A
R2.10A
6
The configuration for C2
Si2 has this configuration
2
1
1
0
2
2
1
4
3
4
4
2
2
2
From 1930-1962 the 3Pu was thought to be the
ground state Now 1Sg is ground state
2
7
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
8
Low-lying states of C2
9
VB view
MO view
10
The VB interference or resonance energy for H2
The VB wavefunctions for H2 Fg (?L ?R) and
Fu (?L - ?R) lead to
eg (hLL 1/R) t/(1S) 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
11
Analysis of classical and interference energies
egx t/(1S) while eux -t/(1-S)
Consider first very long R, where S0 Then egx
t while eux -t so that the bonding and
antibonding effects are similar.
Now consider a distance R2.5 bohr 1.32 A near
equilibrium Here S 0.4583
  • -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 is
    bonding.

12
Contragradience
The above discussions show that the interference
or exchange part of KE dominates the bonding,
tKEKELR S KELL This decrease in the KE due to
overlapping orbitals is dominated by
tx ½ lt (??L). ((??R)gt - S lt (??L)2gt
Dot product is large and negative in the shaded
region between atoms, where the L and R orbitals
have opposite slope (congragradience)
13
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.
1Eg Ecl Egx 3Eu Ecl Eux
-Ex/(1 - S2)
Each energy is referenced to the value at R8,
which is -1 for Ecl, Eu, Eg 0 for Exu and Exg
Ex/(1 S2)
14
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.
T2
T1
Ex
Thus we can understand bonding by analyzing just
the KE part if Ex
TKE
15
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 R1.6bohr (near Re), S0.7 Eg1x 0.94t vs.
egx 0.67t Eu1x -2.75t vs. eux -3.33t
For R4 bohr, S0.1 Eg1x 0.20t vs. egx
0.91t Eu1x -0.20t vs. eux -1.11t
Consider a very small R with S1. 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
16
Noble gas dimers
No bonding at the VB or MO level Only
simultaneous electron correlation (London
attraction) or van der Waals attraction, -C/R6
  • LJ 12-6 Force Field
  • EA/R12 B/R6
  • Der-12 2r-6
  • 4 Det-12 t-6
  • R/Re
  • R/s
  • where s Re(1/2)1/6
  • 0.89 Re

Ar2
s
Re
De
17
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)
18
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)
19
Some New and old material
20
MO and VB view of He dimer, He2
Net BO0
Pauli ? orthog of ?R to ?L ? repulsive
Substitute sg ?R ?L and sg ?R - ?L Get
?MO(He2) ?MO(He2)
21
Remove an electron from He2 to get He2
MO view
?(He2) A(sga)(sgb)(sua)(sub) (sg)2(su)2 Two
bonding and two antibonding ? BO 0 ?(He2)
A(sga)(sgb)(sua) (sg)2(su) ? BO ½ Get 2Su
symmetry. Bond energy and bond distance similar
to H2, also BO ½
22
Remove an electron from He2 to get He2
MO view
?(He2) A(sga)(sgb)(sua)(sub) (sg)2(su)2 Two
bonding and two antibonding ? BO 0 ?(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
?VB(He2) A(?La)(?Lb)(?Ra) -
A(?La)(?Rb)(?Ra) (?L)2(?R) - (?R)2(?L)
23
He2
He2 Re3.03A De0.02 kcal/mol No bond
H2 Re0.74xA De110.x kcal/mol BO 1.0
MO good for discuss spectroscopy, VB good for
discuss chemistry
H2 Re1.06x A De60.x kcal/mol BO 0.5
Check H2 and H2 numbers
24
New material
25
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-
26
The ionic limit
At R8 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 R14.4/1.5249.45 A
Using the bond distance of NaCl2.42A leads to a
coulomb energy of 6.1eV leading to a bond of
6.1-1.54.6 eV The exper De 4.23 eV Showing
that ionic character dominates
E(eV)
R(A)
27
GVB orbitals of NaCl
Dipole moment 9.001 Debye Pure ionic? 11.34
Debye Thus Dq0.79 e
28
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 (IPEA)/5.2
29
Electronegativity
Based on M
30
Comparison of Mulliken and Pauling
electronegativities
31
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
32
The NaCl or B1 crystal
All alkali halides have this structure except
CsCl, CsBr, CsI (they have the B2 structure)
33
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
34
Ionic radii, main group
Fitted to various crystals. Assumes O2- is 1.40A
NaCl R1.021.81 2.84, exper is 2.84
From R. D. Shannon, Acta Cryst. A32, 751 (1976)
35
Ionic radii, transition metals
36
Ionic radii Lanthanides and Actinide
37
Role of ionic sizes in determining crystal
structures
Assume that the anions are large and packed so
that they contact, so that 2RA lt L, where L is
the distance between anions Assume that the anion
and cation are in contact. Calculate the smallest
cation consistent with 2RA lt L.
RARC (v3)L/2 gt (v3) RA Thus RC/RA gt 0.732
RARC L/v2 gt v2 RA Thus RC/RA gt 0.414
Thus for 0.414 lt (RC/RA ) lt 0.732 we expect B1
For (RC/RA ) gt 0.732 either is ok. For (RC/RA )
lt 0.414 must be some other structure
38
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)
39
Wurtzite or B4 structure
40
Sphalerite or Zincblende or B3 structure GaAs
41
Radius rations B3, B4
The height of the tetrahedron is (2/3)v3 a where
a is the side of the circumscribed cube The
midpoint of the tetrahedron (also the midpoint of
the cube) is (1/2)v3 a from the vertex. Hence
(RC RA)/L (½) v3 a / v2 a v(3/8)
0.612 Thus 2RA lt L v(8/3) (RC RA) 1.633 (RC
RA) Thus 1.225 RA lt (RC RA) or RC/RA gt 0.225
Thus B3,B4 should be the stable structures for
0.225 lt (RC/RA) lt 0. 414
42
Structures for II-VI compounds
B3 for 0.20 lt (RC/RA) lt 0.55 B1 for 0.36 lt
(RC/RA) lt 0.96
43
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 gt 0.71
44
Rutile (TiO2) or Cassiterite (SnO2) structure
Related to NaCl with half the cations missing
Find for RC/RA lt 0.67
45
CaF2
rutile
CaF2
rutile
46
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 S4/41. Thus
each O2- must have just two Si neighbors
47
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
a-quartz converts to ß-quartz at 573 C
rhombohedral (trigonal)hP9, P3121 No.15210
From wikipedia
48
Example 2 of electrostatic balance stishovite
phase of SiO2
The stishovite phase of SiO2 has six coordinate
Si, ? S2/3. Thus each O must have 3 Si neighbors
Rutile-like structure, with 6-coordinate Si
high pressure form densest of the SiO2
polymorphs
tetragonaltP6, P42/mnm, No.13617
From wikipedia
49
TiO2, example 3 electrostatic balance
Example 3 the rutile, anatase, and brookite
phases of TiO2 all have octahedral Ti. Thus S
2/3 and each O must be coordinated to 3 Ti.
top
anatase phase TiO2
right
front
50
Corundum (a-Al2O3). Example 4 electrostatic
balance
Each Al3 is in a distorted octahedron, leading
to S1/2. Thus each O2- must be coordinated to 4
Al
51
Olivine. Mg2SiO4. example 5 electrostatic balance
Each Si has four O2- (S1) and each Mg has six
O2- (S1/3). 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)
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?
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.
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.
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.
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.
57
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.
58
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/nBa2/3 leading to nOBa
1 nBa 6 leading to SBa 2/nBa1/3 leading to
nOBa 2 nBa 9 leading to SBa 2/nBa2/9
leading to nOBa 3 nBa 12 leading to SBa
2/nBa1/6 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.
59
BaTiO3 structure (Perovskite)
60
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)
61
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 C0.743470.
Postive ? that head of column is negative
62
Charge Equilibration
First consider how the energy of an atom depends
on the net charge on the atom, E(Q)
Including terms through 2nd order leads to
  • Charge Equilibration for Molecular Dynamics
    Simulations
  • K. Rappé and W. A. Goddard III J. Phys. Chem.
    95, 3358 (1991)

(2)
(3)
63
Charge dependence of the energy (eV) of an atom
Harmonic fit
Get minimum at Q-0.887 Emin -3.676
8.291
9.352
64
QEq parameters
65
Interpretation of J, the hardness
Define an atomic radius as
RA0
Re(A2)
Bond distance of homonuclear diatomic
H 0.84 0.74 C 1.42 1.23 N 1.22 1.10 O 1.08 1.21 Si
2.20 2.35 S 1.60 1.63 Li 3.01 3.08
Thus J is related to the coulomb energy of a
charge the size of the atom
66
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
67
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. AQX, 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
68
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) ? 8 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 RC0.759a0
And l 0.5
69
QEq results for alkali halides
70
QEq for Ala-His-Ala
Amber charges in parentheses
71
QEq for deoxy adenosine
Amber charges in parentheses
72
QEq for polymers
Nylon 66
PEEK
73
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
74
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.
75
Phases of BaTiO3
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
Increasing Temperature
Different phases of BaTiO3
face
edge
vertex
center
1960 Cochran Soft Mode Theory(Displacive Model)
77
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
1960 Cochran Soft Mode Theory(Displacive Model)
Order-disorder
1966 Bersuker Eight Site Model
1968 Comes Order-Disorder Model (Diffuse X-ray Scattering)
78
Comparison to experiment
Displacive ? small latent heat This agrees with
experiment R ? O T 183K, DS 0.170.04 J/mol O
? T T 278K, DS 0.320.06 J/mol T ? C T
393K, DS 0.520.05 J/mol
Diffuse xray scattering Expect some disorder,
agrees with experiment
79
Problem displacive model EXAFS Raman
observations
  • EXAFS of Tetragonal Phase1
  • Ti distorted from the center of oxygen octahedral
    in tetragonal phase.
  • The angle between the displacement vector and
    (111) is a 11.7.

Raman Spectroscopy of Cubic Phase2 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)

80
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
81
QM results explain EXAFS Raman observations
  • EXAFS of Tetragonal Phase1
  • Ti distorted from the center of oxygen octahedral
    in tetragonal phase.
  • The angle between the displacement vector and
    (111) is a 11.7.

PQEq with FE/AFE model gives a5.63
Raman Spectroscopy of Cubic Phase2 A strong
Raman spectrum in cubic phase is found in
experiments.
Model Inversion symmetry in Cubic Phase Raman Active
Displacive Yes No
FE/AFE No Yes
  1. B. Ravel et al, Ferroelectrics, 206, 407 (1998)
  2. A. M. Quittet et al, Solid State Comm., 12, 1053
    (1973)

82
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 lt111gt direction. It undergoes
a series of FE to AFE transitions with increasing
temperature, leading to a total polarization that
switches from lt111gt to lt011gt to lt001gt and then
vanishes.
83
Phase Transition at 0 GPa
Thermodynamic Functions
Transition Temperatures and Entropy Change FE-AFE
Phase Eo (kJ/mol) ZPE (kJ/mol) EoZPE (kJ/mol)
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
Vibrations important to include
84
Polarizable QEq
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 Self-consistent charge
equilibration (QEq)
Four universal parameters for each element Get
from QM
85
Validation
Phase Properties EXP QMd P-QEq
Cubic (Pm3m) abc (A) B(GPa) eo 4.012a 6.05e 4.007 167.64 4.0002 159 4.83
Tetra. (P4mm) ab(A) c(A) Pz(uC/cm2) B(GPa) 3.99c 4.03c 15 to 26b 3.9759 4.1722 98.60 3.9997 4.0469 17.15 135
Ortho. (Amm2) ab(A) c(A) ?(degree) PxPy(uC/cm2) B(Gpa) 4.02c 3.98c 89.82c 15 to 31b 4.0791 3.9703 89.61 97.54 4.0363 3.9988 89.42 14.66 120
Rhomb. (R3m) abc(A) aß?(degree) PxPyPz(uC/cm2) B(GPa) 4.00c 89.84c 14 to 33b 4.0421 89.77 97.54 4.0286 89.56 12.97 120
  1. H. F. Kay and P. Vousden, Philosophical Magazine
    40, 1019 (1949)
  2. 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)
  3. G.H. Kwei, A. C. Lawson, S. J. L. Billinge, and
    S.-W. Cheong, J. Phys. Chem. 97,2368
  4. M. Uludogan, T. Cagin, and W. A. Goddard,
    Materials Research Society Proceedings (2002),
    vol. 718, p. D10.11.

86
QM Phase Transitions at 0 GPa, FE-AFE
Transition Experiment 1 Experiment 1 This Study This Study
Transition T(K) ?S (J/mol) T(K) ?S (J/mol)
R to O 183 0.170.04 228 0.132
O to T 278 0.320.06 280 0.138
T to C 393 0.520.05 301 0.145
1. G. Shirane and A. Takeda, J. Phys. Soc. Jpn.,
7(1)1, 1952
87
Free energies for Phase Transitions
88
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) 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)
89
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 EXAFS No Yes Yes
Intense Raman in Cubic Phase No Yes Yes
90
Space Group Phonon DOS
Phase Displacive Model Displacive Model FE/AFE Model (This Study) 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
91
Frozen Phonon Structure-Pm3m(C) Phase - Displacive
Frozen Phonon of BaTiO3 Pm3m phase
Pm3m Phase
Brillouin Zone
G (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)
92
Frozen Phonon Structure Displacive model
P4mm (T) Phase
Amm2 (O) Phase
R3m (R) Phase
Unstable TO phonons M1(1), M2(1)
Unstable TO phonons M3(1)
NO UNSTABLE PHONONS
93
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
94
X-Ray Diffuse Scattering
Photon K
Phonon Q
Photon K
Cross Section
Scattering function
Dynamic structure factor
Debye-Waller factor
95
Diffuse X-ray diffraction predicted for the
BaTiO3 FE-AFE phases.
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).
96
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
96
97
Domain Walls Tetragonal Phase of BaTiO3 Consider
3 cases
CASE I
CASE II
CASE III
experimental
Polarized light optical micrographs of domain
patterns in barium titanate (E. Burscu, 2001)
  • Open-circuit
  • Surface charge not neutralized
  • Domain stucture
  • Short-circuit
  • Surface charge neutralized
  • Open-circuit
  • Surface charge not neutralized

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
98
180 Domain Wall of BaTiO3 Energy vs length
Ly
Type I Lgt64a(256Å)
Type II 4a(16Å)ltLlt32a(128Å)
Type III L2a(8Å)
99
180 Domain Wall Type I, developed
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
100
180 Domain Wall Type I, developed
L 2048 Å
Polarization P
Free charge ?f
100
101
180 Domain Wall Type II, underdeveloped
L 128 Å
Polarization P
Displacement dY
Displacement dZ
Free charge ?f
Wall center expanded, polarization switches,
positively charged
Transition layer contracted, polarization
relaxes, negatively charged
101
102
180 Domain Wall Type III, antiferroelectric
L 8 Å
Displacement dZ
Polarization P
Wall center polarization switch
102
103
180 Domain Wall of BaTiO3 Energy vs length
Ly
Type I Lgt64a(256Å)
Type II 4a(16Å)ltLlt32a(128Å)
Type III L2a(8Å)
104
90 Domain Wall of BaTiO3
L724 Å (N128)
  • Wall energy is 0.68 erg/cm2
  • Stable only for L?362 Å (N?64)

104
105
90 Domain Wall of BaTiO3
L724 Å (N128)
Displacement dY
Displacement dZ
Free Charge Density
106
90 Wall Connection to Continuum Model
1-D Poissons Equation
3-D Poissons Equation
Solution
107
90 Domain Wall of BaTiO3
L724 Å (N128)
Polarization Charge Density
Free Charge Density
Electric Field
Electric Potential
108
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

108
109
Mystery Origin of Oxygen Vacancy Trees!
Oxgen deficient dendrites in LiTaO3 (Bursill et
al, Ferroelectrics, 70191, 1986)
110
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