Lecture 15 February 8, 2010
This presentation is the property of its rightful owner.
Sponsored Links
1 / 110

Lecture 15 February 8, 2010 Ionic bonding and oxide crystals PowerPoint PPT Presentation


  • 87 Views
  • Uploaded on
  • Presentation posted in: General

Lecture 15 February 8, 2010 Ionic bonding and oxide crystals. Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy. William A. Goddard, III, [email protected] 316 Beckman Institute, x3093

Download Presentation

Lecture 15 February 8, 2010 Ionic bonding and oxide crystals

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Lecture 15 february 8 2010 ionic bonding and oxide crystals

Lecture 15 February 8, 2010

Ionic bonding and oxide crystals

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

William A. Goddard, III, [email protected]

316 Beckman Institute, x3093

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

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

Ted Yu <[email protected]>


Course schedule

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

Last time


Ring ozone

Ring ozone

Form 3 OO sigma bonds, but pp pairs overlap

Analog: cis HOOH bond is 51.1-7.6=43.5 kcal/mol. Get total bond of 3*43.5=130.5 which is 11.5 more stable than O2.

Correct for strain due to 60º bond angles = 26 kcal/mol from cyclopropane.

Expect ring O3 to be unstable with respect to O2 + O by ~14 kcal/mol

But if formed it might be rather stable with respect various chemical reactions.

Ab Initio Theoretical Results on the Stability of Cyclic Ozone L. B. Harding and W. A. Goddard III J. Chem. Phys. 67, 2377 (1977) CN 5599


Gvb orbitals of n2

GVB orbitals of N2

VB view

MO view

Re=1.10A

R=1.50A

R=2.10A


The configuration for c2

The configuration for C2

Si2 has this configuration

2

1

0

1

2

1

2

4

3

4

4

2

2

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

Now 1Sg+ is ground state

2

2


Ground state of c 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

Low-lying states of C2


Lecture 15 february 8 2010 ionic bonding and oxide crystals

VB view

MO view


The vb interference or resonance energy for h 2

The VB interference or resonance energy for H2+

The VB wavefunctions for H2+

Φg= (хL + хR) and Φu= (хL - хR) lead to

eg = (hLL + 1/R) + t/(1+S) ≡ ecl + Egx

eu = (hLL + 1/R) - t/(1-S) ≡ ecl + Eux

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

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

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


Analysis of classical and interference energies

Analysis of classical and interference energies

egx = t/(1+S) while eux = -t/(1-S)

Consider first very long R, where S~0

Then egx = t while eux = -t

so that the bonding and antibonding effects are similar.

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

Here S= 0.4583

  • = -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

Contragradience

хL

хR

The above discussions show that the interference or exchange part of KE dominates the bonding, tKE=KELR –S KELL

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

tx = ½ [< (хL). ((хR)> - S[< (хL)2>

Dot product is large and negative in the shaded region between atoms, where the L and R orbitals have opposite slope (congragradience)


The vb exchange energies for h 2

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 R=∞, which is

-1 for Ecl, Eu, Eg 0 for Exu and Exg

+Ex/(1 + S2)


Analysis of the vb exchange energy e x

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

Analysis of the exchange energies

E(hartree)

Eu1x

Eg1x

R(bohr)

The one electron exchange for H2 leads to

Eg1x ~ +2St /(1 + S2)

Eu1x ~ -2St /(1 - S2)

which can be compared to the H2+ case

egx ~ +t/(1 + S)

eux ~ -t/(1 - S)

For R=1.6bohr (near Re), S=0.7 Eg1x ~ 0.94t vs. egx ~ 0.67t

Eu1x ~ -2.75t vs. eux ~ -3.33t

For R=4 bohr, S=0.1

Eg1x ~ 0.20t vs. egx ~ 0.91t

Eu1x ~ -0.20t vs. eux ~ -1.11t

Consider a very small R with S=1. Then

Eg1x ~ 2t vs. egx ~ t/2

so that the 2e bond is twice as strong as the 1e bond but at long R, the 1e bond is stronger than the 2e bond


Noble gas dimers

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

  • E=A/R12 –B/R6

  • = De[r-12 – 2r-6]

  • = 4 De[t-12 – t-6]

  • = R/Re

  • = R/s

    where s = Re(1/2)1/6

    =0.89 Re

Ar2

s

Re

De


London dispersion

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 dispersion1

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

Some New and old material


Mo and vb view of he dimer he 2

MO and VB view of He dimer, He2

MO view

ΨMO(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2

VB view

ΨVB(He2) = A[(La)(Lb)(Ra)(Rb)]= (L)2(R)2

Net BO=0

Pauli  orthog of R to L  repulsive

Substitute sg = R +Land sg = R -L

Get ΨMO(He2) ≡ ΨMO(He2)


Remove an electron from he 2 to get he 2

Remove an electron from He2 to getHe2+

MO view

Ψ(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2

Two bonding and two antibonding  BO= 0

Ψ(He2+) = A[(sga)(sgb)(sua)]= (sg)2(su)  BO = ½

Get 2Su+ symmetry.

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


Remove an electron from he 2 to get he 21

Remove an electron from He2 to getHe2+

-

MO view

Ψ(He2) = A[(sga)(sgb)(sua)(sub)]= (sg)2(su)2

Two bonding and two antibonding  BO= 0

Ψ(He2+) = A[(sga)(sgb)(sua)]= (sg)2(su)  BO = ½

Get 2Su+ symmetry.

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

VB view

Substitute sg = R +Land sg = L -R

Get ΨVB(He2) ≡ A[(La)(Lb)(Ra)] - A[(La)(Rb)(Ra)]

= (L)2(R) - (R)2(L)


Lecture 15 february 8 2010 ionic bonding and oxide crystals

He2+

2Sg+

(sg)1(su)2

2Su+

(sg)2(su)

BO=0.5

-

+

He2 Re=3.03A De=0.02 kcal/mol

No bond

H2 Re=0.74xA De=110.x kcal/mol

BO = 1.0

MO good for discuss spectroscopy,

VB good for discuss chemistry

H2+ Re=1.06x A De=60.x kcal/mol

BO = 0.5

Check H2 and H2+ numbers


New material

New material


Ionic bonding chapter 9

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

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 DE(eV) = - 14.4/R(A) or DE (kcal/mol) = -332.06/R(A)

Thus this ionic curve crosses the covalent curve at R=14.4/1.524=9.45 A

Using the bond distance of NaCl=2.42A leads to a coulomb energy of 6.1eV leading to a bond of 6.1-1.5=4.6 eV

The exper De = 4.23 eV

Showing that ionic character dominates

E(eV)

R(A)


Gvb orbitals of nacl

GVB orbitals of NaCl

Dipole moment = 9.001 Debye

Pure ionic 11.34 Debye

Thus Dq=0.79 e


Electronegativity

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


Electronegativity1

Electronegativity

Based on M++


Comparison of mulliken and pauling electronegativities

Comparison of Mulliken and Pauling electronegativities


Ionic crystals

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

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

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

Ionic radii, main group

Fitted to various crystals. Assumes O2- is 1.40A

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

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


Ionic radii transition metals

Ionic radii, transition metals


Ionic radii lanthanides and actinide

Ionic radii Lanthanides and Actinide


Role of ionic sizes in determining crystal structures

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 = (√3)L/2 > (√3) RA

Thus RC/RA > 0.732

RA+RC = L/√2 > √2 RA

Thus RC/RA > 0.414

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

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

Wurtzite or B4 structure


Sphalerite or zincblende or b3 structure gaas

Sphalerite or Zincblende or B3 structure GaAs


Radius rations b3 b4

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

Structures for II-VI compounds

B3 for 0.20 < (RC/RA) < 0.55

B1 for 0.36 < (RC/RA) < 0.96


Caf 2 or fluorite structure

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 tio 2 or cassiterite sno 2 structure

Rutile (TiO2) or Cassiterite (SnO2) structure

Related to NaCl with half the cations missing

Find for RC/RA < 0.67


Lecture 15 february 8 2010 ionic bonding and oxide crystals

CaF2

rutile

CaF2

rutile


Electrostatic balance postulate

Electrostatic Balance Postulate

For an ionic crystal the charges transferred from all cations must add up to the extra charges on all the anions.

We can do this bond by bond, but in many systems the environments of the anions are all the same as are the environments of the cations. In this case the bond polarity (S) of each cation-anion pair is the same and we write

S = zC/nC where zC is the net charge on the cation and nC is the coordination number

Then zA = Si SI = Si zCi /ni

Example1 : SiO2. in most phases each Si is in a tetrahedron of O2- leading to S=4/4=1.

Thus each O2- must have just two Si neighbors


A quartz structure of sio 2

a-quartz structure of SiO2

Each Si bonds to 4 O,

OSiO = 109.5°

each O bonds to 2 Si

Si-O-Si = 155.x °

Helical chains

single crystals optically active;

α-quartz converts to β-quartz at 573 °C

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

From wikipedia


Example 2 of electrostatic balance stishovite phase of sio2

Example 2 of electrostatic balance: stishovite phase of SiO2

The stishovite phase of SiO2 has six coordinate Si,  S=2/3.

Thus each O must have 3 Si neighbors

Rutile-like structure, with 6-coordinate Si;

high pressure form

densest of the SiO2 polymorphs

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

From wikipedia


Tio 2 example 3 electrostatic balance

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 al 2 o 3 example 4 electrostatic balance

Corundum (a-Al2O3). Example 4 electrostatic balance

Each Al3+ is in a distorted octahedron, leading to S=1/2.

Thus each O2- must be coordinated to 4 Al


Olivine mg 2 sio 4 example 5 electrostatic balance

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 batio 3

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 batio 31

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 batio 32

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 batio 33

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 batio 34

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 batio 35

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

Prediction of BaTiO3 structure : Ba coordination

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

nBa= 3 leading to SBa = 2/nBa=2/3 leading to nOBa = 1

nBa= 6 leading to SBa = 2/nBa=1/3 leading to nOBa = 2

nBa= 9 leading to SBa = 2/nBa=2/9 leading to nOBa = 3

nBa= 12 leading to SBa = 2/nBa=1/6 leading to nOBa = 4

Each of these might lead to a possible structure.

The last case is the correct one for BaTiO3 as shown.

Each O has a Ti in the +z and –z directions plus four Ba forming a square in the xy plane

The Each of these Ba sees 4 O in the xy plane, 4 in the xz plane and 4 in the yz plane.


Batio3 structure perovskite

BaTiO3 structure (Perovskite)


How estimate charges

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

Fractional ionic character of diatomic molecules

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


Charge equilibration

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

Charge dependence of the energy (eV) of an atom

E=12.967

E=0

E=-3.615

Cl+

Cl

Cl-

Q=+1

Q=0

Q=-1

Harmonic fit

Get minimum at Q=-0.887

Emin = -3.676

= 8.291

= 9.352


Qeq parameters

QEq parameters


Interpretation of j the hardness

Interpretation of J, the hardness

Define an atomic radius as

RA0

Re(A2)

Bond distance of homonuclear diatomic

H0.840.74

C1.421.23

N1.221.10

O1.081.21

Si2.202.35

S1.601.63

Li3.013.08

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


The total energy of a molecular complex

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

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

The QEq Coulomb potential law

We need now to choose a form for JAB(R)

A plausible form is JAB(R) = 14.4/R, which is valid when the charge distributions for atom A and B do not overlap

Clearly this form as the problem that JAB(R)  ∞ as R 0

In fact the overlap of the orbitals leads to shielding

The plot shows the shielding for C atoms using various Slater orbitals

Using RC=0.759a0

And l = 0.5


Qeq results for alkali halides

QEq results for alkali halides


Qeq for ala his ala

QEq for Ala-His-Ala

Amber charges in parentheses


Qeq for deoxy adenosine

QEq for deoxy adenosine

Amber charges in parentheses


Qeq for polymers

QEq for polymers

Nylon 66

PEEK


Perovskites

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

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

Phases of BaTiO3

<111> polarized rhombohedral

<110> polarized orthorhombic

<100> polarized tetragonal

Non-polar cubic

Temperature

120oC

-90oC

5oC

Different phases of BaTiO3

Ba2+/Pb2+

Ti4+

O2-

c

Domains separated by domain walls

a

Non-polar cubic

above Tc

Six variants at room temperature

<100> tetragonal

below Tc


Nature of the phase transitions

Nature of the phase transitions

<111> polarized rhombohedral

<110> polarized orthorhombic

<100> polarized tetragonal

Non-polar cubic

Temperature

120oC

-90oC

5oC

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


Nature of the phase transitions1

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

Order-disorder


Comparison to experiment

Comparison to experiment

Cubic

Tetra.

Ortho.

Rhomb.

Displacive  small latent heat

This agrees with experiment

R  O: T= 183K, DS = 0.17±0.04 J/mol

O  T: T= 278K, DS = 0.32±0.06 J/mol

T  C: T= 393K, DS = 0.52±0.05 J/mol

Diffuse xray scattering

Expect some disorder, agrees with experiment


Problem displacive model exafs raman observations

Problem displacive model: EXAFS & Raman observations

d

(001)

α

(111)

79

  • 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

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

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


Qm calculations

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

QM results explain EXAFS & Raman observations

d

(001)

α

(111)

81

  • 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.

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

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


Lecture 15 february 8 2010 ionic bonding and oxide crystals

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.


Phase transition at 0 gpa

Phase Transition at 0 GPa

Thermodynamic Functions

Transition Temperatures and Entropy Change FE-AFE

Vibrations important to include


Polarizable qeq

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

Validation

  • 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

QM Phase Transitions at 0 GPa, FE-AFE

R

O

T

C

1. G. Shirane and A. Takeda, J. Phys. Soc. Jpn., 7(1):1, 1952


Free energies for phase transitions

Free energies for Phase Transitions

Common Alternative free energy from Vibrational states at 0K

We use 2PT-VAC: free energy from

MD at 300K

Velocity Auto-Correlation Function

Velocity Spectrum

System Partition Function

Thermodynamic Functions: Energy, Entropy, Enthalpy, Free Energy


Free energies predicted for batio3 fe afe phase structures

Free energies predicted for BaTiO3 FE-AFE phase structures.

Free Energy (J/mol)

Temperature (K)

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 transitions2

Nature of the phase transitions

Displacive

Order-disorder

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


Space group phonon dos

Space Group & Phonon DOS


Frozen phonon structure p m3m c phase displacive

Frozen Phonon Structure-Pm3m(C) Phase - Displacive

Pm3m Phase

Frozen Phonon of BaTiO3 Pm3m phase

Brillouin Zone

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

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

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

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

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

Summary Phase Structures and Transitions

  • Phonon structures

  • FE/AFE transition

Agree with experiment?

96


Domain walls tetragonal phase of batio 3 consider 3 cases

Domain Walls Tetragonal Phase of BaTiO3Consider 3 cases

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

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

P

P

P

P

P

P

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

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

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

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

E=0

E

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

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

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 batio 3 energy vs length

180° Domain Wall of BaTiO3 – Energy vs length

z

o

y

Type I

Type II

Type III

98

Ly


180 domain wall type i developed

180° Domain Wall – Type I, developed

z

o

y

C

A

D

A

B

Wall center

Transition layer

Domain structure

A

A

B

B

C

C

D

D

99

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 developed1

180° Domain Wall – Type I, developed

z

o

y

Wall center: expansion, polarization switch, positively charged

Transition layer: contraction, polarization relaxed, negatively charged

Domain structure: constant lattice spacing, polarization and charge density

100

L = 2048 Å

Polarization P

Free charge ρf

100


180 domain wall type ii underdeveloped

180° Domain Wall – Type II, underdeveloped

z

A

C

o

y

B

D

A

B

C

D

101

L = 128 Å

Polarization P

Displacement dY

Displacement dZ

Free charge ρf

Wall center: expanded, polarization switches, positively charged

Transition layer: contracted, polarization relaxes, negatively charged

101


180 domain wall type iii antiferroelectric

180° Domain Wall – Type III, antiferroelectric

z

o

y

102

L= 8 Å

Displacement dZ

Polarization P

Wall center: polarization switch

102


180 domain wall of batio 3 energy vs length1

180° Domain Wall of BaTiO3 – Energy vs length

z

o

y

Type I

Type II

Type III

103

Ly


90 domain wall of batio 3

90° Domain Wall of BaTiO3

z

L

o

y

Wall center

Transition Layer

Domain Structure

L=724 Å (N=128)

  • Wall energy is 0.68 erg/cm2

  • Stable only for L362 Å (N64)

104


90 domain wall of batio 31

90° Domain Wall of BaTiO3

z

L

o

y

Wall center: Orthorhombic phase, Neutral

Transition Layer: Opposite charged

Domain Structure

L=724 Å (N=128)

Displacement dY

Displacement dZ

Free Charge Density


90 wall connection to continuum model

90° Wall – Connection to Continuum Model

C is determined by the periodic boundary condition:

3-D Poisson’s Equation

1-D Poisson’s Equation

Solution


90 domain wall of batio 32

90° Domain Wall of BaTiO3

z

L

o

y

L=724 Å (N=128)

Polarization Charge Density

Free Charge Density

Electric Field

Electric Potential


Summary iii domain walls

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

Mystery: Origin of Oxygen Vacancy Trees!

0.1μm

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


Lecture 15 february 8 2010 ionic bonding and oxide crystals

stop


  • Login