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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

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

Last time

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

GVB orbitals of N2

VB view

MO view

Re1.10A

R1.50A

R2.10A

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

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

Low-lying states of C2

VB view

MO view

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

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.

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)

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)

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

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

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

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)

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)

Some New and old material

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)

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

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)

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

New material

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-

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)

GVB orbitals of NaCl

Dipole moment 9.001 Debye Pure ionic? 11.34

Debye Thus Dq0.79 e

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

Electronegativity

Based on M

Comparison of Mulliken and Pauling

electronegativities

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

The NaCl or B1 crystal

All alkali halides have this structure except

CsCl, CsBr, CsI (they have the B2 structure)

The CsCl or B2 crystal

There is not yet a good understanding of the

fundamental reasons why particular compound

prefer particular structures. But for ionic

crystals the consideration of ionic radii has

proved useful

Ionic radii, main group

Fitted to various crystals. Assumes O2- is 1.40A

NaCl R1.021.81 2.84, exper is 2.84

From R. D. Shannon, Acta Cryst. A32, 751 (1976)

Ionic radii, transition metals

Ionic radii Lanthanides and Actinide

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

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)

Wurtzite or B4 structure

Sphalerite or Zincblende or B3 structure GaAs

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

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

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

Rutile (TiO2) or Cassiterite (SnO2) structure

Related to NaCl with half the cations missing

Find for RC/RA lt 0.67

CaF2

rutile

CaF2

rutile

Electrostatic Balance Postulate

For an ionic crystal the charges transferred from

all cations must add up to the extra charges on

all the anions. We can do this bond by bond, but

in many systems the environments of the anions

are all the same as are the environments of the

cations. In this case the bond polarity (S) of

each cation-anion pair is the same and we write S

zC/nC where zC is the net charge on the cation

and nC is the coordination number Then zA Si

SI Si zCi /ni

Example1 SiO2. in most phases each Si is in a

tetrahedron of O2- leading to S4/41. Thus

each O2- must have just two Si neighbors

a-quartz structure of SiO2

Each Si bonds to 4 O, OSiO 109.5 each O bonds

to 2 Si Si-O-Si 155.x

Helical chains single crystals optically active

a-quartz converts to ß-quartz at 573 C

rhombohedral (trigonal)hP9, P3121 No.15210

From wikipedia

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

TiO2, example 3 electrostatic balance

Example 3 the rutile, anatase, and brookite

phases of TiO2 all have octahedral Ti. Thus S

2/3 and each O must be coordinated to 3 Ti.

top

anatase phase TiO2

right

front

Corundum (a-Al2O3). Example 4 electrostatic

balance

Each Al3 is in a distorted octahedron, leading

to S1/2. Thus each O2- must be coordinated to 4

Al

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)

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?

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.

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.

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.

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.

Illustration, BaTiO3

A number of important oxides have the perovskite

structure (CaTiO3) including BaTiO3, KNbO3,

PbTiO3. Lets try to predict the structure

without looking it up Based on the TiO2

structures , we expect the Ti to be in an

octahedron of O2-, STiO 2/3. How many Ti

neighbors will each O have? It cannot be 3 since

there would be no place for the Ba. It is likely

not one since Ti does not make oxo bonds. Thus

we expect each O to have two Ti neighbors,

probably at 180º. This accounts for 2(2/3) 4/3

charge. Now we must consider how many O are

around each Ba, nBa, leading to SBa 2/nBa, and

how many Ba around each O, nOBa.

Prediction of BaTiO3 structure Ba coordination

Since nOBa SBa 2/3, the missing charge for the

O, we have only a few possibilities nBa 3

leading to SBa 2/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.

BaTiO3 structure (Perovskite)

How estimate charges?

We saw that even for a material as ionic as NaCl

diatomic, the dipole moment ? a net charge of

0.8 e on the Na and -0.8 e on the Cl. We need a

method to estimate such charges in order to

calculate properties of materials. First a bit

more about units. In QM calculations the unit of

charge is the magnitude of the charge on an

electron and the unit of length is the bohr

(a0) Thus QM calculations of dipole moment are in

units of ea0 which we refer to as au. However the

international standard for quoting dipole moment

is the Debye 10-10 esu A Where m(D) 2.5418

m(au)

Fractional ionic character of diatomic molecules

Obtained from the experimental dipole moment in

Debye, m(D), and bond distance R(A) by dq

m(au)/R(a0) C m(D)/R(A) where C0.743470.

Postive ? that head of column is negative

Charge Equilibration

First consider how the energy of an atom depends

on the net charge on the atom, E(Q)

Including terms through 2nd order leads to

- Charge Equilibration for Molecular Dynamics

Simulations - K. Rappé and W. A. Goddard III J. Phys. Chem.

95, 3358 (1991)

(2)

(3)

Charge dependence of the energy (eV) of an atom

Harmonic fit

Get minimum at Q-0.887 Emin -3.676

8.291

9.352

QEq parameters

Interpretation of J, the hardness

Define an atomic radius as

RA0

Re(A2)

Bond distance of homonuclear diatomic

H 0.84 0.74 C 1.42 1.23 N 1.22 1.10 O 1.08 1.21 Si

2.20 2.35 S 1.60 1.63 Li 3.01 3.08

Thus J is related to the coulomb energy of a

charge the size of the atom

The total energy of a molecular complex

Consider now a distribution of charges over the

atoms of a complex QA, QB, etc Letting JAB(R)

the Coulomb potential of unit charges on the

atoms, we can write

Taking the derivative with respect to charge

leads to the chemical potential, which is a

function of the charges

or

The definition of equilibrium is for all chemical

potentials to be equal. This leads to

The QEq equations

Adding to the N-1 conditions

The condition that the total charged is fixed

(say at 0) leads to the condition

Leads to a set of N linear equations for the N

variables QA. 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

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

QEq results for alkali halides

QEq for Ala-His-Ala

Amber charges in parentheses

QEq for deoxy adenosine

Amber charges in parentheses

QEq for polymers

Nylon 66

PEEK

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

Ferroelectrics

The stability of the perovskite structure

depends on the relative ionic radii if the

cations are too small for close packing with the

oxygens, they may displace slightly. Since these

ions carry electrical charges, such displacements

can result in a net electric dipole moment

(opposite charges separated by a small distance).

The material is said to be a ferroelectric by

analogy with a ferromagnet which contains

magnetic dipoles. At high temperature, the small

green B-cations can "rattle around" in the larger

holes between oxygen, maintaining cubic symmetry.

A static displacement occurs when the structure

is cooled below the transition temperature.

Phases of BaTiO3

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)

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)

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

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

- B. Ravel et al, Ferroelectrics, 206, 407 (1998)
- A. M. Quittet et al, Solid State Comm., 12, 1053

(1973)

QM calculations

The ferroelectric and cubic phases in BaTiO3

ferroelectrics are also antiferroelectric Zhang

QS, Cagin T, Goddard WA Proc. Nat. Acad. Sci.

USA, 103 (40) 14695-14700 (2006)

Even for the cubic phase, it is lower energy for

the Ti to distort toward the face of each

octahedron. How do we get cubic symmetry? Combine

8 cells together into a 2x2x2 new unit cell, each

has displacement toward one of the 8 faces, but

they alternate in the x, y, and z directions to

get an overall cubic symmetry

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

- B. Ravel et al, Ferroelectrics, 206, 407 (1998)
- A. M. Quittet et al, Solid State Comm., 12, 1053

(1973)

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.

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

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

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

- H. F. Kay and P. Vousden, Philosophical Magazine

40, 1019 (1949) - 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) - G.H. Kwei, A. C. Lawson, S. J. L. Billinge, and

S.-W. Cheong, J. Phys. Chem. 97,2368 - M. Uludogan, T. Cagin, and W. A. Goddard,

Materials Research Society Proceedings (2002),

vol. 718, p. D10.11.

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

Free energies for Phase Transitions

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)

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

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

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)

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

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

X-Ray Diffuse Scattering

Photon K

Phonon Q

Photon K

Cross Section

Scattering function

Dynamic structure factor

Debye-Waller factor

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).

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

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

180 Domain Wall of BaTiO3 Energy vs length

Ly

Type I Lgt64a(256Å)

Type II 4a(16Å)ltLlt32a(128Å)

Type III L2a(8Å)

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

180 Domain Wall Type I, developed

L 2048 Å

Polarization P

Free charge ?f

100

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

180 Domain Wall Type III, antiferroelectric

L 8 Å

Displacement dZ

Polarization P

Wall center polarization switch

102

180 Domain Wall of BaTiO3 Energy vs length

Ly

Type I Lgt64a(256Å)

Type II 4a(16Å)ltLlt32a(128Å)

Type III L2a(8Å)

90 Domain Wall of BaTiO3

L724 Å (N128)

- Wall energy is 0.68 erg/cm2
- Stable only for L?362 Å (N?64)

104

90 Domain Wall of BaTiO3

L724 Å (N128)

Displacement dY

Displacement dZ

Free Charge Density

90 Wall Connection to Continuum Model

1-D Poissons Equation

3-D Poissons Equation

Solution

90 Domain Wall of BaTiO3

L724 Å (N128)

Polarization Charge Density

Free Charge Density

Electric Field

Electric Potential

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

Mystery Origin of Oxygen Vacancy Trees!

Oxgen deficient dendrites in LiTaO3 (Bursill et

al, Ferroelectrics, 70191, 1986)

stop