# Lecture 9 January 27, 2013 Ionic bonding and crystals

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Lecture 9 January 27, 2013 Ionic bonding and crystals
Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: Ch120a Hours: 2-3pm Monday, Wednesday, Friday William A. Goddard, III, 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants:Sijia Dong Samantha Johnson Ch120a-Goddard-L01

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) = eV minus EA(Cl) = eV = eV But as R is decreased the electrostatic energy drops as DE(eV) = /R(A) or DE (kcal/mol) = /R(A) But covalent curve does not change until get large overlap the 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 Correcting for IP-EA at R=∞ leads to a net bond of =4.6 eV experiment De = 4.23 eV Thus ionic character dominates E(eV) R(A)

GVB orbitals of NaCl No overlapno bond R=6 A
Dipole moment = Debye Pure ionic Debye Thus Dq=0.79 e Overlap from Q transfer R=4.7 A Mostly Na+ Cl- R=3.5 A Very Na+ Cl- Re=2.4 A

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 H H 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 Fitted to various crystals. Assumes O2- is 1.40A NaCl R= = 2.84, exper is 2.84 From R. D. Shannon, Acta Cryst. A32, 751 (1976)

HS Fe3+ is d5 thus get HSS=5/2; LS=1/2 not important Fe2+ 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 eg [x2-y2, 3z2-r2] t2g [xy, xz, yz] Five d orbitals same energy t2g [xy, xz, yz] eg [x2-y2, 3z2-r2] tetrahedron octahedron atom

More on HS and LS, octahedral site, Fe3+
Fe3+ is d5 thus get HSS=5/2; LS=1/2 not important Weak field Strong field x2-y2 3z2-r2 x2-y2 3z2-r2 eg eg xy xz yz xy xz yz t2g t2g 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+
Fe2+ is d6 thus HS=2; LS=0, both important Weak field Strong field x2-y2 3z2-r2 x2-y2 3z2-r2 eg eg xy xz yz xy xz yz t2g t2g 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

Role of ionic sizes in determining crystal structures
Assume 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 < (RC/RA ) < we expect B1 For (RC/RA ) > either is ok. For (RC/RA ) < must be some other structure

Radius Ratios of Alkali Halides and Noble metal halices
Rules work ok B1: 0.35 to 1.26 B2: 0.76 to 0.92 B1 expect < (RC/RA ) < 0.732 B2 or B1 (RC/RA ) > 0.732 (RC/RA ) < neither Based on R. W. G. Wyckoff, Crystal Structures, 2nd edition. Volume 1 (1963)

Sphalerite or Zincblende or B3 structure GaAs

Wurtzite or B4 structure

Thus B3,B4 should be the stable structures for
Radius ratios for 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) = (RC + RA) Thus RA < (RC + RA) or RC/RA > 0.225 Thus B3,B4 should be the stable structures for 0.225 < (RC/RA) <

Structures for II-VI compounds
B3 for 0.20 < (RC/RA) < 0.55 B4 for 0.33 < (RC/RA) < 0.53 B1 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

CaF2 rutile CaF2 rutile

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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
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
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 tetragonal tP6, P42/mnm, No.136[17] From wikipedia

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-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?

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, 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, 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, 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, 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
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)

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 = esu A Where m(D) = 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= Postive  that head of column is negative

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
Cl Cl- Cl+ Q=0 Q=-1 Q=+1 Harmonic fit Get minimum at Q=-0.887 Emin = = 8.291 = 9.352

QEq parameters

Interpretation of J, the hardness
Define an atomic radius as RA0 Re(A2) Bond distance of homonuclear diatomic H C N O Si S Li 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 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 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 Using RC=0.759a0 And l = 0.5

QEq results for alkali halides

Amber charges in parentheses
QEq for Ala-His-Ala Amber charges in parentheses

Amber charges in parentheses

QEq for polymers Nylon 66 PEEK

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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.

Phases of BaTiO3 Different phases of BaTiO3 c
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 Domains separated by domain walls above Tc <100> tetragonal below Tc O2- Ba2+/Pb2+ Ti4+ Ferroelectric crystals are non-polar (paraelectric) above its Curie temperature, but spontaneously polarized (ferroelectric) below the Curie temperature. Along with the spontaneous polarization, large distortion also occurs. As an example, BaTiO3, the most extensively investigated ferroelectric material, is cubic and non-polar above its Curie temperature, and is tetragonal and spontaneously polarized along <100> direction in room temperature. Upon further cooling, <110> polarized orthorhombic phase is formed at 278K, and <110> polarized rhomboredral phase round 183K. Typically, there is also a reduction of crystal symmetry along with transformation. The symmetry-related and crystallographically-equivalent variants can coexist as domains separated by domain walls.

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 Different phases of BaTiO3 Temperature 120oC 5oC -90oC <111> polarized rhombohedral <110> polarized orthorhombic <100> polarized tetragonal Non-polar cubic face edge vertex center 1960 Cochran Soft Mode Theory(Displacive Model)

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 1960 Cochran Soft Mode Theory(Displacive Model) Order-disorder 1966 Bersuker Eight Site Model 1968 Comes Order-Disorder Model (Diffuse X-ray Scattering)

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. Diffuse xray scattering Expect some disorder, agrees with experiment Ortho. Rhomb.

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 B. Ravel et al, Ferroelectrics, 206, 407 (1998) 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): (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
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. Model Inversion symmetry in Cubic Phase Raman Active Displacive Yes No FE/AFE B. Ravel et al, Ferroelectrics, 206, 407 (1998) A. M. Quittet et al, Solid State Comm., 12, 1053 (1973)

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