Lecture 15 February 8, 2010 Ionic bonding and oxide crystals

1© copyright 2010 William A. Goddard III, all rights reservedCh120a-Goddard-L15

Nature of the Chemical Bond with applications to catalysis, materials

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

Lecture 15 February 8, 2010Ionic bonding and oxide crystals

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]>Ted Yu <[email protected]>


Course scheduleMonday Feb. 8, 2pm L14 TODAY(caught up)

Midterm was given out on Friday. Feb. 5, due on Wed. Feb. 10It 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 sitebut closed book otherwiseNo collaboration

Friday Feb. 12, postpone lecture from 2pm to 3pm


Last time


Ring ozone

Form 3 OO sigma bonds, but p pairs overlap Analog: cis HOOH bond is 51.1-7.6=43.5 kcal/mol. Get total bond of 3*43.5=130.5 which is 11.5 more stable than O2.Correct for strain due to 60º bond angles = 26 kcal/mol from cyclopropane. Expect ring O3 to be unstable with respect to O2 + O by ~14 kcal/molBut 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

Re=1.10A

R=1.50A

R=2.10A

VB view MO view


2

2

2

2

2

2

4

4

2

The configuration for C2

4

1

1

3

1

From 1930-1962 the 3u was thought to be the ground state

Now 1g+ is ground state

Si2 has this configuration

0


Ground state of C2

MO configuration

Have two strong 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+

Φg = (хL + хR) and Φu = (хL - хR) lead tog = (hLL + 1/R) + /(1+S) ≡ ecl + Eg

x

u = (hLL + 1/R) - /(1-S) ≡ ecl + Eux

where = (hLR - ShLL) is the VB interference or resonance energy and

cl = (hLL + 1/R) is the classical energy

As shown here the dominates the bonding and antibonding of these states


Analysis of classical and interference energies

egx = /(1+S) while eu

x = -/(1-S)

Consider first very long R, where S~0Then eg

x = while eux = -

so that the bonding and antibonding effects are similar.

Now consider a distance R=2.5 bohr = 1.32 A near equilibrium

Here S= 0.4583

= -0.0542 hartree leading to

egx = 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, KE=KELR –S KELL

This decrease in the KE due to overlapping orbitals is dominated by

Dot product is large and negative

in the shaded region between

atoms, where the L and R orbitals have

opposite slope (congragradience)

x = ½ [< (хL). ((хR)> - S [< (хL)2>

хL хR


The VB exchange energies for H2

1Eg = Ecl + Egx

3Eu = Ecl + Eux

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.

Each energy is referenced to the value at

R=∞, which is

-1 for Ecl, Eu, Eg 0 for Ex

u and Exg

+Ex/(1 + S2)

-Ex/(1 - S2)


Analysis of the VB exchange energy, Ex

where Ex = {(hab + hba) S + Kab –EclS2} = TT

Here T{(hab + hba) S –(haa + hbb)S2} = 2S

Where = (hab – Shaa) contains the 1e part

T{Kab –S2Jab} contains the 2e part

Clearly the Ex is dominated by T and clearly T is dominated by the kinetic part, T. E

x

T2

T1

TThus we can understand bonding by analyzing just the KE part if Ex


Analysis of the exchange energies

The one electron exchange for H2 leads to

Eg1x ~ +2S /(1 + S2)

Eu1x ~ -2S /(1 - S2)

which can be compared to the H2

+ case

egx ~ +/(1 + S)

eux ~ -/(1 - S) Eu

1x

Eg1x

R(bohr)

E(hartree)

Consider a very small R with S=1. Then

Eg1x ~ 2vs. eg

x ~

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

For R=1.6bohr (near Re), S=0.7 Eg

1x ~ 0.94vs. egx ~

Eu1x ~ -2.75vs. eu

x ~ For R=4 bohr, S=0.1

Eg1x ~ 0.20vs. eg

x ~

Eu1x ~ -0.20vs. eu

x ~


Noble gas dimers

Ar2

Re

De

LJ 12-6 Force FieldE=A/R12 –B/R6

= De[12 – 26]De[12 – 6]= R/Re= R/where = Re(1/2)1/6

=0.89 Re

No bonding at the VB or MO level

Only simultaneous electron correlation (London attraction) or van der Waals attraction, -C/R6


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

VB view

MO view

ΨMO(He2) = A[(g)(g)(u)(u)]= (g)2(u)2

ΨVB(He2) = A[(L)(L)(R)(R)]= (L)2(R)2

Substitute g = R + Land g = R - L

Get ΨMO(He2) ≡ ΨMO(He2)

Net BO=0

Pauli orthog of R to L repulsive


Remove an electron from He2 to get He2+

Ψ(He2) = A[(g)(g)(u)(u)]= (g)2(u)2

Two bonding and two antibonding BO= 0

Ψ(He2+) = A[(g)(g)(u)]= (g)2(u) BO = ½

Get 2u+ symmetry.

Bond energy and bond distance similar to H2+, also BO = ½

MO view


Remove an electron from He2 to get He2+

Ψ(He2) = A[(g)(g)(u)(u)]= (g)2(u)2

Two bonding and two antibonding BO= 0

Ψ(He2+) = A[(g)(g)(u)]= (g)2(u) BO = ½

Get 2u+ symmetry.

Bond energy and bond distance similar to H2+, also BO = ½

VB view

MO view

Substitute g = R + Land g = L - R

Get ΨVB(He2) ≡ A[(L)(L)(R)] - A[(L)(R)(R)]

= (L)2(R) - (R)2(L)

-


He2+

He2 Re=3.03A De=0.02 kcal/molNo bond

2u+

2g+

(g)2(u)

(g)1(u)2

-

+

BO=0.5

H2 Re=0.74xA De=110.x kcal/molBO = 1.0H2

+ Re=1.06x A De=60.x kcal/molBO = 0.5

Check H2 and H2+ numbers

MO good for discuss spectroscopy,

VB good for discuss chemistry


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 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

To provide a measure to estimate polarity in bonds, Linus Pauling developed a scale of electronegativity () where the atom that gains charge is more electronegative and the one that loses is more electropositive

He arbitrarily assigned

=4 for F, 3.5 for O, 3.0 for N, 2.5 for C, 2.0 for B, 1.5 for Be, and 1.0 for Li

and then used various experiments to estimate other cases . Current values are on the next slide

Mulliken formulated an alternative scale such that

M= (IP+EA)/5.2


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

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


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 < 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


Wurtzite or B4 structure


Sphalerite or Zincblende or B3 structure GaAs


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)


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.








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.







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.


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


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.


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




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

79

(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

81

(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


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.

96© copyright 2010 William A. Goddard III, all rights reservedCh120a-Goddard-L15 96

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

97

•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

98

)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

99

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

100© copyright 2010 William A. Goddard III, all rights reservedCh120a-Goddard-L15 100

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

101

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

102

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

103

)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

104

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)

108

•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


stop

Lecture 15 February 8, 2010 Ionic bonding and oxide crystals

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Transcript of Lecture 15 February 8, 2010 Ionic bonding and oxide crystals