1 / 64

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

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.

sibley
Download Presentation

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

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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Ch120a-Goddard-L01 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, wag@wag.caltech.edu 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants:Sijia Dong <sdong@caltech.edu> Samantha Johnson <sjohnson@wag.caltech.edu>

  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-

  3. 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) 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 6.1-1.5=4.6 eV experiment De = 4.23 eV Thus ionic character dominates E(eV) R(A)

  4. GVB orbitals of NaCl No overlapno bond R=6 A Dipole moment = 9.001 Debye Pure ionic 11.34 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

  5. 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

  6. Electronegativity Based on M++

  7. Comparison of Mulliken and Pauling electronegativities Mulliken Biggest flaw is the wrong value for H H is clearly much less electronegative than I

  8. 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

  9. The NaCl or B1 crystal All alkali halides have this structure except CsCl, CsBr, CsI (they have the B2 structure)

  10. 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

  11. 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)

  12. Ionic radii, transition metals HS Fe3+ is d5 thus get HSS=5/2; LS=1/2 not important Fe2+is d6thus 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 t2g [xy, xz, yz] eg [x2-y2, 3z2-r2] eg [x2-y2, 3z2-r2] Five d orbitals same energy t2g [xy, xz, yz] tetrahedron octahedron atom

  13. 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

  14. 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

  15. Ionic radii, transition metals

  16. Ionic radii Lanthanides and Actinide

  17. 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 = (√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

  18. 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) B1 expect 0.414 < (RC/RA ) < 0.732 B2 or B1 (RC/RA ) > 0.732 (RC/RA) < 0.414 neither

  19. Sphalerite or Zincblende or B3 structure GaAs

  20. Wurtzite or B4 structure

  21. 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) = 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

  22. 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

  23. 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

  24. Rutile (TiO2) or Cassiterite (SnO2) structure Related to NaCl with half the cations missing Find for RC/RA < 0.67

  25. CaF2 rutile CaF2 rutile

  26. Stopped L17, Feb 10

  27. 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

  28. 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

  29. 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

  30. 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

  31. 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

  32. 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)

  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?

  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.

  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.

  36. 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.

  37. 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.

  38. 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.

  39. 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.

  40. BaTiO3 structure (Perovskite)

  41. 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)

  42. 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

  43. 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)

  44. 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

  45. QEq parameters

  46. Interpretation of J, the hardness Define an atomic radius as RA0 Re(A2) Bond distance of homonuclear diatomic 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 Thus J is related to the coulomb energy of a charge the size of the atom

  47. 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

  48. 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

  49. 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

  50. QEq results for alkali halides

More Related