William A. Goddard, III, wag@wag.caltech.edu Lecture 17 February 12, 2010 Hydrocarbons Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy 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: Wei-Guang Liu <wgliu@wag.caltech.edu> Ted Yu <tedhyu@wag.caltech.edu>

Course schedule All lectures on schedule Friday Feb. 12, 2pm L17 Monday Feb. 15, caltech holiday Wednesday Feb. 17, 2pm L18 Friday Feb. 19, 2pm L19

Last time

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 7 (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): 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

QM results explain EXAFS & Raman observations 9 (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)

Ti atom distortions and polarizations determined from QM calculations Ti atom distortions and polarizations determined from QM calculations. Ti distortions are shown in the FE-AFE fundamental unit cells. Yellow and red strips represent individual Ti-O chains with positive and negative polarizations, respectively. Low temperature R phase has FE coupling in all three directions, leading to a polarization along <111> direction. It undergoes a series of FE to AFE transitions with increasing temperature, leading to a total polarization that switches from <111> to <011> to <001> and then vanishes.

Phase Transition at 0 GPa Thermodynamic Functions Transition Temperatures and Entropy Change FE-AFE Phase Eo (kJ/mol) ZPE Eo+ZPE R 22.78106 O 0.06508 22.73829 0.02231 T 0.13068 22.70065 0.05023 C 0.19308 22.66848 0.08050 Vibrations important to include

Mystery: Origin of Oxygen Vacancy Trees! Oxgen deficient dendrites in LiTaO3 (Bursill et al, Ferroelectrics, 70:191, 1986)

Oxygen Vacancy Structure (Vz) P P P Ti O 4.41Å 2.12Å 1.85Å 1.84Å 2.10Å Ti O 2.12Å 1.93Å O O Ti 1 domain No defect defect leads to domain wall 1.93Å O 2.12Å O O Remove Oz Ti 1.93Å O 2.12Å O O Ti 1.93Å O 2.12Å O O Ti P Leads to Ferroelectric Fatigue

Divacancy in the x-y plane V1 is a fixed Vx oxygen vacancy. V2 is a neighboring oxygen vancancy of type Vx or Vy. Interaction energy in eV.. Vacancy Interaction Short range attraction due to charge redistribution. Anisotropic: vacancy pair prefers to break two parallel chains (due to coherent local relaxation) Ti O O O O Ti Ti O O O y O Ti Ti O z O O z

Vacancy Clusters z Vx cluster in y-z plane: y Best 1D Best branch 2D 0.1μm z Vx cluster in y-z plane: y 0.335eV 0.360 eV 0.456 eV 0.636 eV 0.669 eV 0.650 eV 1.878 eV Best 1D Best branch 2D Dendritic Bad Prefer 1-D structure If get branch then grow linearly from branch get dendritic structure n-type conductivity, leads to breakdown

Woodward-Hoffmann rules orbital symmetry rules Frontier Orbital rules Roald Hoffmann Certain cycloadditions occur but not others 2s+2s 2s+4s 4s+4s

Woodward-Hoffmann rules orbital symmetry rules Frontier Orbital rules Certain cyclizations occur but not others conrotatory disrotatory disrotatory conrotatory

2+2 cycloaddition – Orbital correlation diagram GS Allowed Forbidden ES

WH rules – 2 + 4 Ground State Allowed

WH rules – 2 + 4 Excited State Forbidden A S S

Summary WH rules cycloaddition 2n + 2m n+m odd: Thermal allowed Photochemical forbidden n+m even: Thermal forbidden Photochemical allowed n=1, m=1: ethene + ethene n=1, m=2: ethene + butadience (Diels-Alder)

S WH rules – cyclization-GS A A A S A A S Forbidden Allowed A S S A S A S S Rotation, C2 Reflection, s

Summary WH rules cyclization n odd: thermal disrotatory Photochemical conrotatory n even: Thermal conrotatory Photochemical disrotatory n=2  butadiene n=3  hexatriene

GVB view reactions Reactant HD+T H D T During reaction, bonding orbital on D stays on D, Bonding orbital on H keeps its overlap with the orbital on D but delocalizes over H and T in the TS and localizes on T in the product. Thus highly overlapping bond for whole reaction Nonbonding Orbital on free T of reactant becomes partially antibonding in TS and localizes on free H of product, but it changes sign Product H+DT

GVB view reactions Reactant HD+T H D T Bond pair keeps high overlap while flipping from reactant to product Transition state nonbond orbital keeps orthogonal, hence changes sign Product H+DT H D T

GVB analysis of cyclization (4 e case) Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 4 VB orbitals: A,B,C,D reactant φB φA φB φA φB φC 2 3 4 φD 1 φC φA φD φC 2 3 φD 4 1 φB φA 2 3 Now ask how the CH2 groups 1 and 4 must rotate so that C and D retain positive overlap. Clearly 4n is conrotatory φC φD 1 4

Apply GVB model to 2 + 2 φB φA φA φB φC φD φD φB φD φA φC φC 4 VB orbitals:A,B,C,D reactant Transition state: ignore C φB φA φA φB φC φD φD φB φD Nodal plane 4 VB orbitals product φA φC \ φC

Transition state for 2 + 2 φB φA φD φC 4 3 1 2 2 1 3 4 Transition state: ignore C Orbitals A on 1 and B on 2 keep high overlap as the bond moves from 12 to 23 with B staying on 2 and A moving from 1 to 3 φB 2 1 φA Orbital D must move from 3 to 1 but must remain orthogonal to the AB bond. Thus it gets a nodal plane The overlap of D and C goes from positive in reactant to negative in product, hence going through 0. thus break CD bond. 3 4 φD Nodal plane φC Reaction Forbidden

GVB model fast analysis 2 + 2 4 VB orbitals:A,B,C,D reactant Move A from 1 to 3 keeping overlap with B Simultaneously D moves from 3 to 1 but must change sign since must remain orthogonal to A and B 2 1 φA φB φC φD 3 4 \ φB φA φD φC C and D start with positive overlap and end with negative overlap. Thus break bond  forbidden

Next examine 2+4

1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 GVB 2+4 φA φB φA φB φC φD 1 2 4 3 φF φE 5 6 φD φC 2 3 1 4 6 5 φE φF 1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1

2. Move EF bond; C changes phase again as it moves from 1 to 5 GVB 2+4 2. Move EF bond; C changes phase again as it moves from 1 to 5 φA φB φD φC 2 3 1 4 φA φB φD 2 3 6 5 1 4 φE φF φE φC 3. Now examine overlap of D with C. It is positive. Thus can retain bond CD as AB and EF migrate 6 5 φF Reaction Allowed

GVB 2+4 φA φB φC φB φD φC φD φA φA φB φD φE φF φE φF φE φC φF 2. Move EF bond; C changes phase again as it moves from 1 to 5 φA φB φA φB φC φD 1 2 4 3 φF φE 5 6 φD φC 2 3 1 4 φA φB φD 2 3 6 5 1 4 φE φF 3. Examine final overlap of D with C. It is positive. Thus can retain bond CD as AB and EF migrate φE 1. Move AB bond; Ignore D; C changes phase as it moves from 3 to 1 φC 6 5 φF Reaction Allowed

New material

Benzene and Resonance

Resonance

Benzene wavefunction

Allyl Radical

Allyl wavefunctions It is about 12 kcal/mol

Graphene

graphene Graphene: CC=1.4210A Bond order = 4/3 Benzene: CC=1.40 BO=3/2 Ethylene: CC=1.34 BO = 2 CCC=120° Unit cell has 2 carbon atoms Bands: 2pp orbitals per cell 2 bands 1x1 Unit cell This is referred to as graphene

Graphene band structure 1x1 Unit cell Unit cell has 2 carbon atoms Bands: 2pp orbitals per cell 2 bands of states each with N states where N is the number of unit cells 2 p electrons per cell  2N electrons for N unit cells The lowest N MOs are doubly occupied, leaving N empty orbitals. The filled 1st band touches the empty 2nd band at the Fermi energy 2nd band Get semi metal 1st band

Graphite Stack graphene layers as ABABAB Can also get ABCABC Rhombohedral AAAA stacking much higher in energy Distance between layers = 3.3545A CC bond = 1.421 Only weak London dispersion attraction between layers De = 1.0 kcal/mol C Easy to slide layers, good lubricant

energetics

Cn What is the structure of C3?

Cn

Energetics Cn Note extra stability of odd Cn by 33 kcal/mol, this is because odd Cn has an empty px orbital at one terminus and an empty py on the other, allowing stabilization of both p systems

Stability of odd Cn

Bond energies and thermochemical calculations

Bond energies and thermochemical calculations

Heats of Formation

Heats of Formation

Heats of Formation

Heats of Formation

Bond energies

Bond energies

Bond energies Both secondary

Average bond energies

Average bond energies

Average bond energies of little use in predicting mechanism Real bond energies Average bond energies of little use in predicting mechanism

Group values

Group functions of propane

Examples of using group values

Group values

Strain

Strain energy cyclopropane from Group values

Strain energy c-C3H6 using real bond energies

Stained GVB orbitals of cyclopropane

Benson Strain energies

Resonance in thermochemical Calculations

Resonance in thermochemical Calculations

Resonance energy butadiene

Allyl radical

Benzene resonance

Benzene resonance

Benzene resonance

Benzene resonance

Benzene resonance

Transition metals Aufbau (4s,3d) Sc---Cu (5s,4d) Y-- Ag (6s,5d) (La or Lu), Ce-Au

Transition metals

Ground states of neutral atoms Sc (4s)2(3d)1 Ti (4s)2(3d)2 V (4s)2(3d)3 Cr (4s)1(3d)5 Mn (4s)2(3d)5 Fe (4s)2(3d)6 Co (4s)2(3d)7 Ni (4s)2(3d)8 Cu (4s)1(3d)10 Sc++ (3d)1 Ti ++ (3d)2 V ++ (3d)3 Cr ++ (3d)4 Mn ++ (3d)5 Fe ++ (3d)6 Co ++ (3d)7 Ni ++ (3d)8 Cu++ (3d)10

GVB orbitals for bonds to Ti Ti ds character, 1 elect H 1s character, 1 elect Covalent 2 electron TiH bond in Cl2TiH2 Think of as bond from Tidz2 to H1s Csp3 character 1 elect H 1s character, 1 elect Covalent 2 electron CH bond in CH4

Bonding at a transition metaal Bonding to a transition metals can be quite covalent. Examples: (Cl2)Ti(H2), (Cl2)Ti(C3H6), Cl2Ti=CH2 Here the two bonds to Cl remove ~ 1 to 2 electrons from the Ti, making is very unwilling to transfer more charge, certainly not to C or H (it would be the same for a Cp (cyclopentadienyl ligand) Thus TiCl2 group has ~ same electronegativity as H or CH3 The covalent bond can be thought of as Ti(dz2-4s) hybrid spin paired with H1s A{[(Tids)(H1s)+ (H1s)(Tids)](ab-ba)}

But TM-H bond can also be s-like Cl2TiH+ Ti (4s)2(3d)2 The 2 Cl pull off 2 e from Ti, leaving a d1 configuration Ti-H bond character 1.07 Tid+0.22Tisp+0.71H ClMnH Mn (4s)2(3d)5 The Cl pulls off 1 e from Mn, leaving a d5s1 configuration H bonds to 4s because of exchange stabilization of d5 Mn-H bond character 0.07 Mnd+0.71Mnsp+1.20H

Bond angle at a transition metal For two p orbitals expect 90°, HH nonbond repulsion increases it H-Ti-H plane What angle do two d orbitals want 76° Metallacycle plane

Best bond angle for 2 pure Metal bonds using d orbitals Assume that the first bond has pure dz2 or ds character to a ligand along the z axis Can we make a 2nd bond, also of pure ds character (rotationally symmetric about the z axis) to a ligand along some other axis, call it z. For pure p systems, this leads to  = 90° For pure d systems, this leads to  = 54.7° (or 125.3°), this is ½ the tetrahedral angle of 109.7 (also the magic spinning angle for solid state NMR).

Best bond angle for 2 pure Metal bonds using d orbitals Problem: two electrons in atomic d orbitals with same spin lead to 5*4/2 = 10 states, which partition into a 3F state (7) and a 3P state (3), with 3F lower. This is because the electron repulsion between say a dxy and dx2-y2 is higher than between sasy dz2 and dxy. Best is ds with dd because the electrons are farthest apart This favors  = 90°, but the bond to the dd orbital is not as good Thus expect something between 53.7 and 90° Seems that ~76° is often best

stop