1 / 95

William A. Goddard, III, wag@kaist.ac.kr

Lecture 18, November 5, 2009. Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy. Course number: KAIST EEWS 80.502 Room E11-101 Hours: 0900-1030 Tuesday and Thursday.

cooper-jarvis
Download Presentation

William A. Goddard, III, wag@kaist.ac.kr

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. Lecture 18, November 5, 2009 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: KAIST EEWS 80.502 Room E11-101 Hours: 0900-1030 Tuesday and Thursday William A. Goddard, III, wag@kaist.ac.kr WCU Professor at EEWS-KAIST and Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Senior Assistant: Dr. Hyungjun Kim: linus16@kaist.ac.kr Manager of Center for Materials Simulation and Design (CMSD) Teaching Assistant: Ms. Ga In Lee: leeandgain@kaist.ac.kr Special assistant: Tod Pascal:tpascal@wag.caltech.edu EEWS-90.502-Goddard-L15

  2. Schedule changes TODAY Nov. 5, Thursday, 9am, L18, as scheduled Nov. 9-13 wag lecturing in Stockholm, Sweden; no lectures, Nov. 17, Tuesday, 9am, L19, as scheduled Nov. 18, Wednesday, 1pm, L20, additional lecture room 101 Nov. 19, Thursday, 9am, L21, as scheduled Nov. 24, Tuesday, 9am, L22, as scheduled Nov. 26, Thursday, 9am, L23, as scheduled Dec. 1, Tuesday, 9am, L24, as scheduled Dec. 2, Wednesday, 3pm, L25, additional lecture, room 101 Dec. 3, Thursday, 9am, L26, as scheduled Dec. 7-10 wag lecturing Seattle and Pasadena; no lectures, Dec. 11, Friday, 2pm, L27, additional lecture, room 101 EEWS-90.502-Goddard-L15

  3. Last time EEWS-90.502-Goddard-L15

  4. The configuration for C2 Si2 has this configuration 2 1 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 EEWS-90.502-Goddard-L15

  5. 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 EEWS-90.502-Goddard-L15

  6. Low-lying states of C2 EEWS-90.502-Goddard-L15

  7. EEWS-90.502-Goddard-L15

  8. London Dispersion The universal attractive term postulated by van der Waals was explained in terms of QM by Fritz London in 1930 The idea is that even for spherically symmetric atoms such as He, Ne, Ar, Kr, Xe, Rn 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 froms with a long range 1/R6 attraction to account for London dispersion (usually refered to as van der Waals attraction) plus a short range repulsive term to acount for short Range Pauli Repulsion) EEWS-90.502-Goddard-L15

  9. Noble gas dimers • LJ 12-6 • 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 EEWS-90.502-Goddard-L15

  10. Remove an electron from He2 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 EEWS-90.502-Goddard-L15

  11. 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 a coulomb energy of 6.1eV leacing to a bond of6.1-1.5=4.6 eV The exper De = 4.23 eV Showing that ionic character dominates E(eV) R(A) EEWS-90.502-Goddard-L15

  12. GVB orbitals of NaCL Dipole moment = 9.001 Debye Pure ionic 11.34 Debye Thus Dq=0.79 e EEWS-90.502-Goddard-L15

  13. 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 arbitrary assigned χ=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 χM= (IP+EA)/5.2 EEWS-90.502-Goddard-L15

  14. EEWS-90.502-Goddard-L15

  15. The NaCl or B1 crystal All alkali halides have this structure except CsCl, CsBr, Cs I (they have the B2 structure) EEWS-90.502-Goddard-L15

  16. 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 EEWS-90.502-Goddard-L15

  17. Ionic radii, main group Fitted to various crystals. Assumes O2- is 1.40A From R. D. Shannon, Acta Cryst. A32, 751 (1976) EEWS-90.502-Goddard-L15

  18. Ionic radii, transition metals EEWS-90.502-Goddard-L15

  19. Role of ionic sizes Assume that the anions are large and packed so that they contact, so that 2RA < L, where L is the distance between then 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 ) < 0.732 must be some other structure EEWS-90.502-Goddard-L15

  20. Radiius 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) EEWS-90.502-Goddard-L15

  21. Wurtzite or B4 structure EEWS-90.502-Goddard-L15

  22. Sphalerite or Zincblende or B3 structure EEWS-90.502-Goddard-L15

  23. 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 EEWS-90.502-Goddard-L15

  24. Structures for II-VI compounds B3 for 0.20 < (RC/RA) < 0.55 B1 for 0.36 < (RC/RA) < 0.96 EEWS-90.502-Goddard-L15

  25. 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 EEWS-90.502-Goddard-L15

  26. Rutile (TiO2) or Cassiterite (SnO2) structure Find for RC/RA < 0.67 Related to NaCl with half the cations missing EEWS-90.502-Goddard-L15

  27. CaF2 rutile CaF2 rutile EEWS-90.502-Goddard-L15

  28. 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 enviroments 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=1. Thus each O2- must have just two Si neighbors EEWS-90.502-Goddard-L15

  29. Some old some New material EEWS-90.502-Goddard-L15

  30. More examples electrostatic balance Example 2. The stishovite phase of SiO2 has six coordinate Si, leading to S=2/3. Thus each O must have 3 Si neighbors Example 3: the rutile, anatase, and brookhite phases of TiO2 all have octahedral T. Thus S= 2/3 and each O must be coordinated to 3 Ti. Example 4. Corundum (a-Al2O3). Each Al3+ is in a distorted octahedron, leading to S=1/2. Thus each O2- must be coordinated to 4 Al Example 5. Olivine. Mg2SiO4. 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 EEWS-90.502-Goddard-L15

  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 TiiO2 structures , we expect the Ti to be in an octahedron of O2-, STiO = 2/3. The question is 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. The Ba must provide the other 2/3. 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. Since nOBa* SBa = 2/3, the missing charge for the O, we have only a few possibilities: EEWS-90.502-Goddard-L15

  32. Prediction of BaTiO3 structure 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. EEWS-90.502-Goddard-L15

  33. BaTiO3 structure (Perovskite) EEWS-90.502-Goddard-L15

  34. New material EEWS-90.502-Goddard-L15

  35. 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) EEWS-90.502-Goddard-L15

  36. 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 EEWS-90.502-Goddard-L15

  37. 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) EEWS-90.502-Goddard-L15

  38. 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 EEWS-90.502-Goddard-L15

  39. QEq parameters EEWS-90.502-Goddard-L15

  40. 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 EEWS-90.502-Goddard-L15

  41. 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 EEWS-90.502-Goddard-L15

  42. 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. 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 for C to be between +4 and -4 Similarly Q(H) is between +1 and -1 EEWS-90.502-Goddard-L15

  43. 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 EEWS-90.502-Goddard-L15

  44. QEq results for alkali halides EEWS-90.502-Goddard-L15

  45. QEq for Ala-His-Ala Amber charges in parentheses EEWS-90.502-Goddard-L15

  46. QEq for deoxy adenosine Amber charges in parentheses EEWS-90.502-Goddard-L15

  47. QEq for polymers Nylon 66 PEEK EEWS-90.502-Goddard-L15

  48. 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

  49. 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 ferro-electric by analogy with a ferro-magnet 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. We have illustrated a displacement along the z-axis, resulting in tetragonal symmetry (z remains a 4-fold symmetry axis), but at still lower temperatures the symmetry can be lowered further by additional displacements along the x- and y-axes. We have a dynamic 3D-drawing of this ferro-electric transition.

  50. Phases 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

More Related