Lecture 9 January 30, 2013 Ionic bonding and crystals

© copyright 2011 William A. Goddard III, all rights reserved Ch120a-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]

316 Beckman Institute, x3093

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

California Institute of Technology

Teaching Assistants: Caitlin Scott <[email protected]>

Hai Xiao [email protected]; Fan Liu <[email protected]>

Lecture 9 January 30, 2013

Ionic bonding and crystals

Course number: Ch120a

Hours: 2-3pm Monday, Wednesday, Friday

mailto:[email protected]



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

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 eV

The exper De = 4.23 eV

Showing that ionic

character dominates


GVB orbitals

of NaCl

Dipole moment

= 9.001 Debye

Pure ionic

11.34 Debye

Thus Dq=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) < 0.55

B1 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

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

Stopped L17, Feb 10

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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 Si

Si-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

tetragonal

tP6, 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.











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.













Now we must consider how many O are around each Ba, nBa,

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

nOBa.


Prediction of BaTiO3 structure : Ba coordination

Since nOBa* SBa = 2/3, the missing charge for the O, we have

only a few possibilities:

nBa= 3 leading to SBa = 2/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=-1 Q=+1

Harmonic fit

= 8.291 = 9.352

Get minimum at Q=-0.887

Emin = -3.676


QEq parameters


Interpretation of J, the hardness

Define an atomic radius as

H 0.84 0.74

C 1.42 1.23

N 1.22 1.10

O 1.08 1.21

Si 2.20 2.35

S 1.60 1.63

Li 3.01 3.08

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 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

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


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

Temperature 120oC 5oC -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 cubic

above Tc <100> tetragonal

below 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

Temperature 120oC 5oC -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-disorder 1966 Bersuker Eight Site Model

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

Increasing

Temperature


Comparison to experiment

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

Cubic Tetra.

Ortho. Rhomb.

Diffuse xray scattering

Expect some disorder,

agrees with experiment


Problem displacive model: EXAFS & Raman observations

57

(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

59

(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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Lecture 9 January 30, 2013 Ionic bonding and crystals