lecture 26 march 07, 2011 batio 3 and hypervalent

1© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L26

Nature of the Chemical Bond with applications to catalysis, materials

science, nanotechnology, surface science, bioinorganic chemistry, and energy

Lecture 26 March 07, 2011

BaTiO3 and Hypervalent

William A. Goddard, III, [email protected] Beckman Institute, x3093

Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics,

California Institute of Technology

Teaching Assistants: Wei-Guang Liu <[email protected]>Caitlin Scott <[email protected]>


Last time


The ionic limit

At R=∞ the cost of forming Na+ and Cl-

is IP(Na) = 5.139 eV minus EA(Cl) = 3.615 eV = 1.524 eV But as R is decreased the electrostatic energy drops as E(eV) = - 14.4/R(A) or E (kcal/mol) = -332.06/R(A)Thus this ionic curve crosses the covalent curve at R=14.4/1.524=9.45 A

R(A)

E(eV)

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 eVThe exper De = 4.23 eVShowing that ionic character dominates


GVB orbitals of NaCl

Dipole moment = 9.001 Debye

Pure ionic 11.34 Debye

Thus q=0.79 e


Electronegativity

Based on M++


Comparison of Mulliken and Pauling electronegativities


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

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

Fitted to various crystals. Assumes O2- is 1.40A

NaCl R=1.02+1.81 = 2.84, exper is 2.84


Ionic radii, transition metals


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

Thus RC/RA > 0.414

RA+RC = (√3)L/2 > (√3) RA

Thus RC/RA > 0.732

Thus for 0.414 < (RC/RA ) < 0.732 we expect B1

For (RC/RA ) > 0.732 either is ok.

For (RC/RA ) < 0.414 must be some other structure


Radius Ratios of Alkali Halides and Noble metal halices

Based on R. W. G. Wyckoff,

Crystal Structures, 2nd

edition. Volume 1 (1963)

Rules work ok

B1: 0.35 to 1.26

B2: 0.76 to 0.92


Sphalerite or Zincblende or B3 structure GaAs


Wurtzite or B4 structure


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


Structures for II-VI compounds

B3 for 0.20 < (RC/RA) < 0.55B1 for 0.36 < (RC/RA) < 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 > 0.71


Rutile (TiO2) or Cassiterite (SnO2) structure

Related to NaCl with half the cations missing

Find for RC/RA < 0.67


rutile

CaF2

rutile

CaF2


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/C where zC is the net charge on the cation and C is the coordination number

Then zA = i SI = i zCi /i

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


-quartz structure of SiO2

Helical chains single crystals optically active; α-quartz converts to β-quartz at 573 °C

rhombohedral (trigonal)hP9, P3121 No.152[10]

Each Si bonds to 4 O, OSiO = 109.5°each O bonds to 2 SiSi-O-Si = 155.x °

From wikipedia


Example 2 of electrostatic balance: stishovite phase of SiO2

The stishovite phase of SiO2 has six coordinate Si, S=2/3. Thus each O must have 3 Si neighbors

From wikipedia

Rutile-like structure, with 6-coordinate Si;

high pressure form

densest of the SiO2 polymorphs

tetragonaltP6, P42/mnm, No.136[17]


TiO2, example 3 electrostatic balance

Example 3: the rutile, anatase, and brookite phases of TiO2 all have octahedral Ti. Thus S= 2/3 and each O must be coordinated to 3 Ti.

top

front right

anatase phase TiO2


Corundum (-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


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)


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


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, Ba, leading to SBa = 2/Ba, and how many Ba around each O, OBa.


Prediction of BaTiO3 structure : Ba coordination

Since OBa* SBa = 2/3, the missing charge for the O, we have only a few possibilities:

Ba= 3 leading to SBa = 2/Ba=2/3 leading to OBa = 1




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 (D) = 2.5418 (au)


Fractional ionic character of diatomic molecules

Obtained from the experimental dipole moment in Debye, (D), and bond distance R(A) by q = (au)/R(a0) = C (D)/R(A) where C=0.743470. Postive that head of column is negative


Charge Equilibration

Charge Equilibration for Molecular Dynamics Simulations;

A. K. Rappé and W. A. Goddard III; J. Phys. Chem. 95, 3358 (1991)

First consider how the energy of an atom depends on the net charge on the atom, E(Q)

Including terms through 2nd order leads to

(2) (3)


Charge dependence of the energy (eV) of an atom

E=0

E=-3.615

E=12.967

Cl Cl-Cl+

Q=0 Q=-1Q=+1

Harmonic fit

= 8.291 = 9.352

Get minimum at Q=-0.887Emin = -3.676


QEq parameters


Interpretation of J, the hardness

Define an atomic radius as

H 0.84 0.74C 1.42 1.23N 1.22 1.10O 1.08 1.21Si 2.20 2.35S 1.60 1.63Li 3.01 3.08

RA0 Re(A2) Bond distance of

homonuclear diatomic

Thus J is related to the coulomb energy of a charge the size of the atom


The total energy of a molecular complex

Consider now a distribution of charges over the atoms of a complex: QA, QB, etc

Letting JAB(R) = the Coulomb potential of unit charges on the atoms, we can write

or

Taking the derivative with respect to charge leads to the chemical potential, which is a function of the charges

The definition of equilibrium is for all chemical potentials to be equal. This leads to


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


The QEq Coulomb potential law

We need now to choose a form for JAB(R) A plausible form is JAB(R) = 14.4/R, which is valid when the charge distributions for atom A and B do not overlapClearly this form as the problem that JAB(R) ∞ as R 0In fact the overlap of the orbitals leads to shielding The plot shows the shielding for C atoms using various Slater orbitals

And = 0.5 Using RC=0.759a0


QEq results for alkali halides


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.


c

a

Temperature120oC5oC-90oC

<111> polarized rhombohedral

<110> polarized orthorhombic

<100> polarized tetragonal

Non-polar cubic

Different phases of BaTiO3

Six variants at room temperature

06.1~01.1a

c

Domains separated by domain walls

Non-polar cubicabove Tc

<100> tetragonalbelow Tc

O2-

Ba2+/Pb2+

Ti4+

Phases of BaTiO3


Nature of the phase transitions

1960 Cochran Soft Mode Theory(Displacive Model)

Displacive model

Assume that the atoms prefer to distort toward a face or edge or vertex of the octahedron

Increasing Temperature

Temperature120oC5oC-90oC

<111> polarized rhombohedral

<110> polarized orthorhombic

<100> polarized tetragonal

Non-polar cubic

Different phases of BaTiO3

face edge vertex center


Ferroelectric Actuators

• MEMS Actuator performance parameters:– Actuation strain– Work per unit volume– Frequency

• Goal:– Obtain cyclic high

actuations by 90o domain switching in ferroelectrics

– Design thin film micro devices for large actuations

Characteristics of common actuator materials

100 101 102 103 104 105 106 107

102

103

104

105

106

107

108

microbubble ZnO

muscle

solid-liquid

thermo-pneumatic

PZT

Cycling Frequency (Hz)

shape memory alloy

fatigued SM A

electromagnetic (E M)

electrostatic (ES)E M

ES

Work

per

volu

me (

J/m

3)

90o domain switching

Tetragonal perovskites:

1% (BaTiO3), 6.5% (PbTiO3))

P. Krulevitch et al, MEMS 5 (1996) 270-282


Bulk Ferroelectric Actuation

– Apply constant stress and cyclic voltage– Measure strain and charge– In-situ polarized domain observation 0 V

V

US Patent # 6,437, 586 (2002)

Eric Burcsu, 2001

Strains, BT~1%, PT~5.5%


Ferroelectric Model MEMS Actuator

•BaTiO3-PbTiO3 (Barium Titanate (BT)-Lead Titanate (PT)

•Perovskite pseudo-single crystals (biaxially textured thin films)

MEMS Test Bed

[010] [100]


Application: Ferroelectric Actuators Must understand role of domain walls in mediate switching

Switching gives large strain,

… but energy barrier is extremely high!

E

90° domain wall

Domain walls lower the energy barrierby enabling nucleation and growth

Essential questions: Are domain walls mobile? Do they damage the material?In polycrystals? In thin films?

Experiments in BaTiO3

1

2

0 10,000-10,000

0

1.0

Electric field (V/cm)S

trai

n (%

)

Use MD with ReaxFF




Displacive model

Assume that the atoms prefer to distort toward a face or edge or vertex of the octahedron

Order-disorder1966 Bersuker Eight Site Model

1968 Comes Order-Disorder Model (Diffuse X-ray Scattering)

Increasing Temperature


Comparison to experiment

Displacive small latent heatThis agrees with experimentR O: T= 183K, S = 0.17±0.04 J/molO T: T= 278K, S = 0.32±0.06 J/molT C: T= 393K, S = 0.52±0.05 J/mol

Cubic Tetra.

Ortho. Rhomb.

Diffuse xray scatteringExpect some disorder, agrees with experiment


Problem displacive model: EXAFS & Raman observations

49

(001)

(111)

d

α

EXAFS of Tetragonal Phase[1]

•Ti distorted from the center of oxygen octahedral in tetragonal phase.

•The angle between the displacement vector and (111) is α= 11.7°.

Raman Spectroscopy of Cubic Phase[2]

A strong Raman spectrum in cubic phase is found in experiments. But displacive model atoms at center of octahedron: no Raman

1. B. Ravel et al, Ferroelectrics, 206, 407 (1998)

2. A. M. Quittet et al, Solid State Comm., 12, 1053 (1973)


QM calculations

The ferroelectric and cubic phases in BaTiO3 ferroelectrics are also antiferroelectric Zhang QS, Cagin T, Goddard WA Proc. Nat. Acad. Sci. USA, 103 (40): 14695-14700 (2006)

Even for the cubic phase, it is lower energy for the Ti to distort toward the face of each octahedron.

How do we get cubic symmetry?

Combine 8 cells together into a 2x2x2 new unit cell, each has displacement toward one of the 8 faces, but they alternate in the x, y, and z directions to get an overall cubic symmetry

Te

pe

ratu

re

x

CubicI-43m

TetragonalI4cm

RhombohedralR3m

OrthorhombicPmn21

y

z

o

FE AFE/

FE AFE/

FE AFE/

Px Py Pz

+ +

+ +

+ +

+ +

=

=

=

=

MacroscopicPolarization

Ti atom distortions

=

=

=

=

Microscopic Polarization


QM results explain EXAFS & Raman observations

51

(001)

(111)

d

α

EXAFS of Tetragonal Phase[1]

•Ti distorted from the center of oxygen octahedral in tetragonal phase.

•The angle between the displacement vector and (111) is α= 11.7°.

PQEq with FE/AFE model gives α=5.63°

Raman Spectroscopy of Cubic Phase[2]

A strong Raman spectrum in cubic phase is found in experiments.

1. B. Ravel et al, Ferroelectrics, 206, 407 (1998)

2. A. M. Quittet et al, Solid State Comm., 12, 1053 (1973)

Model Inversion symmetry in Cubic Phase

Raman Active

Displacive Yes No

FE/AFE No Yes


New material


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.

Te

pe

ratu

re

x

CubicI-43m

TetragonalI4cm

RhombohedralR3m

OrthorhombicPmn21

y

z

o

FE AFE/

FE AFE/

FE AFE/

Px Py Pz

+ +

+ +

+ +

+ +

=

=

=

=

MacroscopicPolarization

Ti atom distortions

=

=

=

=

Microscopic Polarization


Phase Transition at 0 GPa

v BB

v B

v BBo

v Bo

v

Tk

vk

Tk

vv

TS

Tk

vTkEF

Tk

vvEE

vZPE

,

,

,

,

,

2

),(sinh2ln

2

),(coth),(

2

1

2

),(sinh2ln

2

),(coth),(

2

1

),(2

1

q

q

q

q

q

q

qq

q

qq

q

Thermodynamic Functions Transition Temperatures and Entropy Change FE-AFE

Phase

Eo

(kJ/mol)

ZPE

(kJ/mol)

Eo+ZPE

(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


Four universal parameters for each element:Get from QM

Polarizable QEq

)||exp()()(

)||exp()()(

2

2

23

23

si

si

si

si

ci

ci

ci

ci

rrQr

rrQrsi

ci

Allow each atom to have two charges:A fixed core charge (+4 for Ti) with a Gaussian shapeA variable shell charge with a Gaussian shape but subject to displacement and charge transferElectrostatic interactions between all charges, including the core and shell on same atom, includes Shielding as charges overlapAllow Shell to move with respect to core, to describe atomic polarizabilitySelf-consistent charge equilibration (QEq)

ci

si

ci

oi

oi qRRJ &,,,

Proper description of Electrostatics is critical vdWCoulomb EEE


Validation

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

d. M. Uludogan, T. Cagin, and W. A. Goddard, Materials Research Society Proceedings (2002), vol. 718, p. D10.11.

Phase Properties EXP QMd P-QEq

Cubic(Pm3m)

a=b=c (A)B(GPa)εo

4.012a

6.05e

4.007167.64

4.00021594.83

Tetra.(P4mm)

a=b(A)c(A)Pz(uC/cm2)B(GPa)

3.99c

4.03c

15 to 26b

3.97594.1722

98.60

3.99974.046917.15135

Ortho.(Amm2)

a=b(A)c(A) γ(degree)Px=Py(uC/cm2)B(Gpa)

4.02c

3.98c

89.82c

15 to 31b

4.07913.970389.61

97.54

4.03633.998889.4214.66120

Rhomb.(R3m)

a=b=c(A)α=β=γ(degree)Px=Py=Pz(uC/cm2)B(GPa)

4.00c

89.84c

14 to 33b

4.042189.77

97.54

4.028689.5612.97120


QM Phase Transitions at 0 GPa, FE-AFE

Transition Experiment [1] 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. Soc. Jpn., 7(1):1, 1952

R O T C


Free energies for Phase Transitions

v

cvv

tVVd

tVVC

)()()0(

)()0(

Velocity Auto-Correlation Function

N

jvvj

vvivt

vv

vCmvS

tCdtevC

3

1

2

)(~

2)(

)()(~

Velocity Spectrum

ji

rR

N

ji ji

oioi

rrrr

U

NirUNirU

oj

oi

,

3

1,

2

2

1

)3...1,(})3...1,({

System Partition Function

0

)(ln)( vQvdvSQ

Thermodynamic Functions: Energy, Entropy, Enthalpy, Free Energy

We use 2PT-VAC: free energy from MD at 300K

Common Alternative free energy from Vibrational states at 0K


AFE coupling has higher energy and larger entropy than FE coupling.

Get a series of phase transitions with transition temperatures and entropies

Free energies predicted for BaTiO3 FE-AFE phase structures.

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)

Free Energy (J/mol)

Temperature (K)




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

Develop model to explain all the following experiments (FE-AFE)

Displacive

Order-disorder1966 Bersuker Eight Site Model

1968 Comes Order-Disorder Model (Diffuse X-ray Scattering)


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


Frozen Phonon Structure-Pm3m(C) Phase - Displacive

Brillouin Zone

Frozen Phonon of BaTiO3 Pm3m phasePm3m Phase

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)

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)


Frozen Phonon Structure – Displacive model

Unstable TO phonons:

M1(1), M2(1)

Unstable TO phonons:

M3(1)

P4mm (T) Phase Amm2 (O) Phase R3m (R) Phase

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

v

i

mi

iiii

i

i

v

v

vevn

MNW

viWM

fvF

vFv

vnS

SK

KN

,

2

*1

1

1

'1

),(

),(21

),(

2)(

,exp),(

),(1

2

),(

)21

),(()(

)(

q q

qQqQ

QeQrQQQ

QQ

QQ

Q

Cross Section

Scattering function

Dynamic structure factor

Debye-Waller factor

Photon K’


C (450K)

- 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5

Q x

-5

-4

-3

-2

-1

0

1

2

3

4

5

Qz

T (350K)

O (250K) R (150K)

- 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5

Q x

-5

-4

-3

-2

-1

0

1

2

3

4

5

Qz

- 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5

Q x

-5

-4

-3

-2

-1

0

1

2

3

4

5

Qz

- 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5

Q x

-5

-4

-3

-2

-1

0

1

2

3

4

5

Qz

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

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).

Diffuse X-ray diffraction predicted for the BaTiO3 FE-AFE phases.


Diffuse X diffraction of BaTiO3 and KNbO3,

R. Comes et al, Acta Crystal. A., 26, 244, 1970

FE-AFE Explains X-Ray Diffuse Scattering

Cubic Tetra.

Ortho. Rhomb.

Experimental

(100) (010)

Strong Strong

Cubic Phase

(001) Diffraction Zone

(100) (001)

Weak Strong

Tetra. Phase


Ortho. Phase


(100) (001)

Strong Weak

Rhomb. Phase


(100) (010)

Very weak Very weak

experimental

68© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L26 68

Summary Phase Structures and Transitions

•Phonon structures

•FE/AFE transition

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

Agree with experiment?


experimental

Domain Walls Tetragonal Phase of BaTiO3 Consider 3 cases

69

•Short-circuit •Surface charge neutralized

vdwelcs EEE

P P

+ + + + + + + + + + + + + + +

- - - - - - - - - - - - - - - - -

E=0 E

+ + + + + + + + + + + + + + +

- - - - - - - - - - - - - - - - -

+ + + + - - - - + + + + - - - -

- - - - + + + + - - - - + + + +

P

P

P

P

+ + + + - - - - + + + + - - - -

- - - - + + + + - - - - + + + +

•Open-circuit •Surface charge not neutralized

•Open-circuit •Surface charge not neutralized•Domain stucture

CASE I CASE II CASE III

EP

EEE vdwelcs

surfacedw

vdwelcs

EE

EEE

Polarized light optical

micrographs of domain patterns in barium titanate (E.

Burscu, 2001)

Charge and polarization distributions at the 90 degrees domain wall in barium titanate ferroelectric Zhang QS, Goddard WA Appl. Phys. Let., 89 (18): Art. No. 182903 (2006)


180° Domain Wall of BaTiO3 – Energy vs length

y

z

o

70

)001( )100(

Ly

Type I

Type II

Type III

Type I L>64a(256Å)

Type II 4a(16Å)<L<32a(128Å)

Type III L=2a(8Å)


180° Domain Wall – Type I, developed

71

Displacement dY

Displacement dZ

Wall center Transition layer Domain structure

C

AA

B

D

A B C D

A B C D

Ly = 2048 Å =204.8 nm

Zoom out

Zoom out

y

z

o

)001( )100(

Displace away from domain

wall

Displacement reduced near domain wall

72© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L26 72

Polarization P Free charge ρf

L = 2048 Å

Wall center: expansion, polarization switch, positively chargedTransition layer: contraction, polarization relaxed, negatively chargedDomain structure: constant lattice spacing, polarization and charge density

y

z

o

)001( )100(180° Domain Wall – Type I, developed


180° Domain Wall – Type II, underdeveloped

73

Displacement dY Displacement dZ Polarization P

A B C D

Wall center: expanded, polarization switches, positively charged

Transition layer: contracted, polarization relaxes, negatively charged

A C

B D Free charge ρf

L = 128 Å

)001( )100(

y

z

o


180° Domain Wall – Type III, antiferroelectric

74

Displacement dZ Polarization P

Wall center: polarization switch

L= 8 Å

)001( )100(

y

z

o


180° Domain Wall of BaTiO3 – Energy vs length

y

z

o

75

)001( )100(

Ly

Type I

Type II

Type III

Type I L>64a(256Å)

Type II 4a(16Å)<L<32a(128Å)

Type III L=2a(8Å)


90° Domain Wall of BaTiO3

76

z

yo2222 N

Wall center

Transition Layer

Domain Structure

•Wall energy is 0.68 erg/cm2

•Stable only for L362 Å (N64)

L=724 Å (N=128)

)010()001(

L



Wall center: Orthorhombic phase, Neutral

Transition Layer: Opposite charged

Domain Structure

Displacement dY Displacement dZ Free Charge Density

)010()001(

L z

yoL=724 Å (N=128)


90° Wall – Connection to Continuum Model

dy

dP

dy

Ud

yp

fp

o

2

2

1-D Poisson’s Equation

C is determined by the periodic boundary condition: )2()0( LUU

Solution ycdddPyUy

o

y

fyo

0

1)(

3-D Poisson’s Equation

Pp

fp

o

U

2



Polarization Charge Density Free Charge Density

Electric Field Electric Potential

)010()001(

L z

yo

L=724 Å (N=128)


Summary III (Domain Walls)

80

•Three types – developed, underdeveloped and AFE

•Polarization switches abruptly across the wall

•Slightly charged symmetrically

•Only stable for L36 nm

•Three layers – Center, Transition & Domain

•Center layer is like orthorhombic phase

•Strong charged – Bipolar structure – Point Defects and Carrier injection

180° domain wall

90° domain wall


Mystery: Origin of Oxygen Vacancy Trees!

Oxgen deficient dendrites in LiTaO3 (Bursill et al, Ferroelectrics, 70:191, 1986)

0.1μm


c

a

Vz

Vy

Vx

Aging Effects and Oxygen Vacancies

Problems•Fatigue – decrease of ferroelectric polarization upon continuous large signal cycling•Retention loss – decrease of remnant polarization with time•Imprint – preference of one polarization state over the other.•Aging – preference to relax to its pre-poled state

Three types of oxygen vacancies in BaTiO3 tetragonal phase:

Vx, Vy & Vz

Pz


Oxygen Vacancy Structure (Vz)

Ti

Ti

Ti

Ti

O

O

O

O O

O O

O O

O O

2.12Å

2.12Å

2.12Å

1.93Å

1.93Å

1.93Å

Ti

Ti

Ti

Ti

O

O

O O

O O

O O

O O

4.41Å

2.12Å

1.85Å

1.84Å

2.10Å

Remove Oz

Ti

Ti

Ti

Ti

O

O

O

O O

O O

O O

O O

2.12Å

2.12Å

2.12Å

1.93Å

1.93Å

1.93Å

P P P

PLeads to Ferroelectric Fatigue

1 domain

No defect

defect leads to domain

wall


Single Oxygen Vacancy

Vy(0eV) Vx(0eV)

TSxz(1.020eV)

TSxz(0.011eV)

TSxy(0.960eV)

Tk

Dq

Tk

EaD

B

B

o

*

2

)exp(2

Diffusivity

Mobility


Divacancy in the x-y plane•V1 is a fixed Vx oxygen vacancy.

•V2 is a neighboring oxygen vancancy of type Vx or Vy.

•Interaction energy in eV..

1. Short range attraction due to charge redistribution.

2. Anisotropic: vacancy pair prefers to break two parallel chains (due to coherent local relaxation)

Vacancy Interaction

Ti

Ti

Ti

Ti

OO

O O

OO

O

O O

OTi

Ti

Ti

Ti

OO

O O

OO

O

O O

O

O

O

z

y

z


Vacancy Clusters

•Prefer 1-D structure•If get branch then grow linearly from branch•get dendritic structure•n-type conductivity, leads to breakdown

Vx cluster in y-z plane:

0.335eV 0.360 eV 0.456 eV 0.636 eV 0.669 eV 0.650 eV 1.878 eV

y

z

1D 2DDendriticBest Best

branch

0.1μm

Bad


Summary Oxygen Vacancy

•Vacancies trap domain boundary– Polarization Fatigue

•Single Vacancy energy and transition barrier rates

• Di-vacany interactions: lead to short range ordering

•Vacancy Cluster: Prefer 1-D over 2-D structures that favor Dielectric Breakdown


Hysterisis Loop of BaTiO3 at 300K, 25GHz by MD

88

Apply Dz at f=25GHz (T=40ps).T=300K.

Monitor Pz vs. Dz.

o

PDE

Get Pz vs. Ez.Ec = 0.05 V/A at f=25 GHz.

Dz

(V/A)

Time (ps)

Applied Field (25 GHz)

Applied Field (V/A)

Polarization (C/cm2)

VDP

VPP

EEEOo

vdwel

3

2

Dipole Correction

Electric Displacement Correction

Ec

Pr


O Vacancy Jump When Applying Strain

89

X-direction strain induces x-site O vacancies (i.e., neighboring Ti’s in x direction) to y or z-sites.

x

z

y

x

z

y

o

O atom

O vacancy site


Effect of O Vacancy on the Hystersis Loop

90

•Introducing O Vacancy reduces both Pr & Ec.

•O Vacancy jumps when domain wall sweeps.

Perfect Crystal without O vacancy

Crystal without 1 O vacancy.O Vacancy jumps when domain wall sweeps.

Supercell: 2x32x2Total Atoms: 640/639

Can look at bipolar case where switch domains from x to y

Ec

Pr


Summary Ferroelectrics

1. The P-QEq first-principles self-consistent polarizable charge equilibration force field explains FE properties of BaTiO3

2. BaTiO3 phases have the FE/AFE ordering. Explains phase structures and transitions

3. Characterized 90º and 180º domain walls: Get layered structures with spatial charges

4. The Oxygen vacancy leads to linearly ordered structures dendritic patterns. Should dominate ferroelectric fatigue and dielectric breakdown

92

Hypervalent compounds

It was quite a surprize to most chemists in 1962 when Neil Bartlett reported the formation of a compound involving Xe-F bonds.

But this was quickly followed by the synthesis of XeF4 (from Xe and F2 at high temperature and XeF2 in 1962 and later XeF6.Indeed Pauling had predicted in 1933 that XeF6 would be stable, but no one tried to make it.

Later compounds such as ClF3 and ClF5 were synthesized

These compounds violate simple octet rules and are call hypervalent

93

Noble gas dimers

Recall that there is no chemical bonding in He2, Ne2 etc

This is explained in VB theory as due to repulsive Pauli repulsion from the overlap of doubly occupied orbitals

(g)2(u)2

It is explained in MO theory as due to filled bonding and antibonding orbitals

94

Noble gas dimer positive ions

On the other hand the positive ions are strongly bound

This is explained in MO theory as due to one less antibonding electron than bonding, leading to a three electron bond for He2

+ of 55 kcal/mol, the same strength as the one electron bond of H2

+ (g)2(u)1

-

The VB explanation is less straightforward. We consider that there are two equivalent VB structures neither of which leads to much bonding, but superimposing them leads to resonance stabilization

Using (g) = L+R and (u)=L-R

Leads to (with negative sign

95

Re-examine the bonding of HeH

Why not describe HeH as (g)2(u)1 where (g) = L+R and (u)=L-RWould this lead to bonding?The answer is no, as easily seen with the VB form, where the right structure is 23.6-0.7=23.9 eV above the left. Thus the energy for the (g)2(u)1 state would be +12.0 – 2.5 = 9.5 eV unbound at R=∞Adding in ionic stabilization lowers the energy by 14.4/2.0 = 7.2 eV (overestimate because of shielding) , still unbound by 2.3 eV

-

He H He+H-

IP=+24.6 eV EA = 0.7 eV

96

Examine the bonding of XeF

Xe Xe+

The energy to form Xe+ F- can be estimated from

Consider the energy to form the charge transfer complex

Using IP(Xe)=12.13eV, EA(F)=3.40eV, and R(IF)=1.98 A,

we get E(Xe+ F-) = 1.45eV (unbound)

Thus there is no covalent bond for XeF, which has a weak bond of ~ 0.1 eV and a long bond

97

Examine the bonding in XeF2

The energy to form Xe+F- is +1.45 eVNow consider, the impact of putting a 2nd F on the back side of the Xe+ Xe+

Since Xe+ has a singly occupied pz orbital pointing directly at this 2nd F, we can now form a covalent bond to itHow strong would the bond be?Probably the same as for IF, which is 2.88 eV.Thus we expect F--Xe+F- to have a bond strength of ~2.88 – 1.45 = 1.43 eV!Of course for FXeF we can also form an equivalent bond for F-Xe+--F. Thus we get a resonance, which we estimate below

We will denote this 3 center – 4 electron charge transfer bond as

FXeF

98

Estimate stability of XeF2 (eV)

XeF2 is stable with respect to the free atoms by 2.7 eV

Bond energy F2 is 1.6 eV.

Thus stability of XeF2 with respect to Xe + F2 is 1.1 eV

1.3

2.7

Energy form F Xe+ F- at R=∞

F-Xe+ covalent bond length (from IF)

Energy form F Xe+ F- at R=Re

F-Xe+ covalent bond energy (from IF)

Net bond strength of F--Xe+ F-

Resonance due to F- Xe+--F

Net bond strength of XeF2

99

Stability of gas of XeF2

The XeF2 molecule is stable by 1.1 eV with respect to Xe + F2

But to assess whether one could make and store XeF2, say in a bottle, we have to consider other modes of decomposition.

The most likely might be that light or surfaces might generate F atoms, which could then decompose XeF2 by the chain reaction

XeF2 + F {XeF + F2} Xe + F2 + F

Since the bond energy of F2 is 1.6 eV, this reaction is endothermic by 2.7-1.6 = 1.1 eV, suggesting the XeF2 is relatively stable.

Indeed XeF2 is used with F2 to synthesize XeF4 and XeF6.

100

The VB analysis indicates that the stability for XeF4 relative to XeF2 should be ~ 2.7 eV, but maybe a bit weaker due to the increased IP of the Xe due to the first hypervalent bond and because of some possible F---F steric interactions.

There is a report that the bond energy is 6 eV, which seems too high, compared to our estimate of 5.4 eV.

XeF4

Putting 2 additional F to overlap the Xe py pair leads to the square planar structure, which allows 3 center – 4 electron charge transfer bonds in both the x and y directions.

101

XeF6

Since XeF4 still has a pz pair, we can form a third hypervalent bond in this direction to obtain an octahedral XeF6 molecule.

Indeed XeF6 is stable with this structure

Pauling in 1933 suggested that XeF6 would be stabile, 30 years in advance of the experiments.

He also suggested that XeF8 is stable.

However this prediction is wrong

Here we expect a stability a little less than 8.1 eV.

102

Estimated stability of other Nobel gas fluorides (eV)

Using the same method as for XeF2, we can estimate the binding energies for the other Noble metals.

KrF2 is predicted to be stable by 0.7 eV, which makes it susceptible to decomposition by F radicals

1.3 1.3 1.3 1.3 1.3 1.3

2.71.0 3.9-5.3-2.9 -0.1

RnF2 is quite stable, by 3.6 eV, but I do not know if it has been observed

103

XeCl2

Since

EA(Cl)=3.615 eV

R(XeCl+)=2.32A

De(XeCl+)=2.15eV,

We estimate that XeCl2 is stable by 1.14 eV with respect to Xe + Cl2.

However since the bond energy of Cl2 is 2.48 eV, the energy of the chain decomposition process is exothermic by 2.48-1.14=1.34 eV, suggesting at most a small barrier

Thus XeCl2 would be difficult to observe

104

Halogen Fluorides, ClFn

The IP of ClF is 12.66 eV comparable to the IP of 12.13 for Xe.

This suggests that the px and py pairs of Cl could be used to form hypervalent bonds leading to ClF3 and ClF5.

We estimate that ClF3 is stable by 2.8 eV.

Stability of ClF3 relative to ClF + 2F

Indeed the experiment energy for ClF3 ClF + 2F is 2.6 eV, quite similar to

XeF2. Thus ClF3 is endothermic

by 2.6 -1.6 = 1.0 eV

105

Geometry of ClF3

106

ClHF2

We estimate that Is stable to ClH + 2F by 2.7 eV

This is stable with respect to ClH + F2 by 1.1 ev

But D(HF) = 5.87 eV, D(HCl)=4.43 eV, D(ClF) = 2.62 eV

Thus F2ClH ClF + HF is exothermic by 1.4 eV

F2ClH has not been observed

107

ClF5

108

BrFn and IFn

109

SFn

110

SF6

The VB rationalization for octahedral SF6 would be to assume that S is promoted from (3s)2(3p)4 to (3s)0(3p)6 which would lead to 3 hypervalent bonds in the x, y, and z directions.

With an “empty” 3s orbital, the EA for SF6 would be very high

111

PFn

The VB view is that the PF3 was distorted into a planar geometry, leading the 3s lone pair to become a 3pz pair, which can then form a hypervalent bond to two additional F atoms to form PF5

112

Donor-acceptor bonds to oxygen

113

Ozone, O3

The simple VB description of ozone is, where the terminal p electrons are not doing much

This is analogous to the s system in the covalent description of XeF2.

Thus we can look at the p system of ozone as hypervalent, leading to charge transfer to form

114

Diazomethane

leading to


Origin of reactivity in the hypervalentreagent o-iodoxybenzoic acid (IBX)

Hypervalent O-I-O linear bond

Application of hypervalent concepts

Enhancing 2-iodoxybenzoic acid reactivity by exploiting a hypervalent twist Su JT, Goddard WA; J. Am. Chem. Soc., 127 (41): 14146-14147 (2005)


Hypervalent iodine assumes many metallic personalities

IOH

O O

O

I

OAc

OAc

I

OH

OTs

I

O

Oxidations

Radicalcyclizations

CC bondformation

Electrophilicalkene activation

CrO3/H2SO4

Pd(OAc)2

SnBu3Cl

HgCl2

this remarkable chemistry of iodine can be understood in terms of hypervalent concepts

Martin, J. C. organo-nonmetallic chemistry – Science 1983 221(4610):509-514

Hypervalent I alternative


stop

lecture 26 march 07, 2011 batio 3 and hypervalent

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