William A. Goddard, III, wag@wagltech 316 Beckman Institute, x3093

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 Ch120a-Goddard-

L01

1

Nature of the Chemical Bond with applications to catalysis, materials

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

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

Lecture 9 January 27, 2013Ionic bonding and crystals

Course number: Ch120aHours: 2-3pm Monday, Wednesday, Friday

Teaching Assistants:Sijia Dong <[email protected]>Samantha Johnson <[email protected]>

mailto:[email protected]

mailto:[email protected]

© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L25 2

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 limitAt 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)But covalent curve does not change until get large overlap

R(A)

E(eV)

Using the bond distance of NaCl=2.42A leads to a coulomb energy of 6.1eV Correcting for IP-EA at R=∞ leads to a net bond of 6.1-1.5=4.6 eVexperiment De = 4.23 eVThus ionic character dominates

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


GVB orbitals of NaCl

Dipole moment = 9.001 Debye

Pure ionic 11.34 Debye

Thus Dq=0.79 e

R=6 A

R=4.7 A

R=3.5 A

Re=2.4 A

No overlapno bond

Overlap from Q transfer

Mostly Na+ Cl-

Very Na+ Cl-


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 using atomic IP and EA (corrected for valence averaging and scaled by 5.2 to get similar numbers to Pauling: M= (IP+EA)/5.2


Electronegativity

Based on M++


Comparison of Mulliken and Pauling electronegativities

Mulliken Biggest flaw is the wrong value for HH is clearly much less electronegative than I


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

HS

Fe3+ is d5 thus get HSS=5/2; LS=1/2 not importantFe2+ is d6 thus HS=2; LS=0, both important

Ligand field splitting (Crystal field splitting)Negative neighbors at vertices of octahedron splits the d orbitals into t2g and eg irreducible representations of Td or Oh point group

atom octahedron tetrahedron

t2g [xy, xz, yz]

eg [x2-y2, 3z2-r2] t2g [xy, xz, yz]

eg [x2-y2, 3z2-r2] Five d orbitals same energy


More on HS and LS, octahedral site, Fe3+

Fe3+ is d5 thus get HSS=5/2; LS=1/2 not important

x2-y2

3z2-r2 eg

t2gxyxzyz

Weak fieldx2-y2

3z2-r2 eg

t2gxyxzyz

Strong field

Exchange stabilization dominates, get high spin S=5/2 as for atom

Ligand interaction dominates, get low spin S=1/2

5*4/2=10 exchange terms, ~220 kcal/mol

3+1 exchange terms ~88 kcal/mol


More on HS and LS, octahedral site, Fe2+

x2-y2

3z2-r2 eg

t2gxyxzyz

Weak fieldx2-y2

3z2-r2 eg

t2gxyxzyz

Strong field

Exchange stabilization dominates, get high spin S=2 as for atom

Ligand interaction dominates, get low spin S=0

5*4/2=10 exchange terms, ~220 kcal/mol

3+3 exchange terms ~132 kcal/mol

Fe2+ is d6 thus HS=2; LS=0, both important


Ionic radii, transition metals


Ionic radii Lanthanides and Actinide


Role of ionic sizes in determining crystal structuresAssume that anions are large and packed so that they contact, then 2RA < L, L is distance between anions

Assume anion and cation are in contact and calculate 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

B1 expect 0.414 < (RC/RA ) < 0.732

B2 or B1 (RC/RA ) > 0.732

(RC/RA ) < 0.414 neither


Sphalerite or Zincblende or B3 structure GaAs


Wurtzite or B4 structure


Radius ratios for B3, B4

The height of the tetrahedron is (2/3)√3 a where a is the side of the circumscribed cubeThe 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) = 0.612Thus 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) < 0.55B4 for 0.33 < (RC/RA) < 0.53B1 for 0.36 < (RC/RA) < 0.96


CaF2 or fluorite structure

like B1, CsCl but with half the Cs missing

Or Ca same positions as Ga for GaAs, but now have F at all tetrahedral sites

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


rutileCaF2

rutile

CaF2


Stopped L17, Feb 10


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 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

http://en.wikipedia.org/wiki/Rhombohedral

http://en.wikipedia.org/wiki/Silicon_dioxide


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]

http://en.wikipedia.org/wiki/Tetragonal

http://en.wikipedia.org/wiki/Silicon_dioxide


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 (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. 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, nBa, leading to SBa = 2/nBa, and how many Ba around each O, nOBa.







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


Prediction of BaTiO3 structure : Ba coordinationSince 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




Each of these might lead to a possible structure.

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

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

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


BaTiO3 structure (Perovskite)


How estimate charges?

We saw that even for a material as ionic as NaCl diatomic, the dipole moment a net charge of +0.8 e on the Na and -0.8 e on the Cl.

We need a method to estimate such charges in order to calculate properties of materials.

First a bit more about units.

In QM calculations the unit of charge is the magnitude of the charge on an electron and the unit of length is the bohr (a0)

Thus QM calculations of dipole moment are in units of ea0 which we refer to as au. However the international standard for quoting dipole moment is the Debye = 10-10 esu A

Where m(D) = 2.5418 m(au)


Fractional ionic character of diatomic molecules

Obtained from the experimental dipole moment in Debye, m(D), and bond distance R(A) by dq = m(au)/R(a0) = C m(D)/R(A) where 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 l = 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


Stopped January 30


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 Perovskicrystal 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

06.1~01.1ac

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


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

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, DS = 0.17±0.04 J/molO T: T= 278K, DS = 0.32±0.06 J/molT C: T= 393K, DS = 0.52±0.05 J/mol

Cubic Tetra.

Ortho. Rhomb.

Diffuse xray scatteringExpect some disorder, agrees with experiment


Problem displacive model: EXAFS & Raman observations

61

(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 calculationsThe 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

Tepe

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

63

(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


stopped

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William A. Goddard, III, wag@wagltech 316 Beckman Institute, x3093