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Lecture 14 February 5, 2010

Lecture 14 February 5, 2010. 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

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Lecture 14 February 5, 2010

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  1. Lecture 14 February 5, 2010 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>

  2. Course schedule Friday Feb. 5, 2pm L14 TODAY(caught up) Midterm given out on Friday. Feb. 5, due on Wed. Feb. 10 It will be five hour take home with 30 min. break, open notes

  3. Last time

  4. Separated atom limit MO notation Separated atoms notation

  5. Separated atoms limit Note that in each case we get one bonding combination (no new nodal plane) and one antibonding combination (new nodal plane, red lines)

  6. At large R 2ps better bonding than 2pp In earlier lectures we considered the strength of one-electron bonds where we found that Since the overlap of ps orbitals is obviously higher than pp We expect that bonding antibonding

  7. Summarizing united atom limit Note for 3d, the splitting is 3ds < 3dp < 3dd Same argument as for 2p

  8. Correlation diagram for Carbon row homonuclear diatomics C2 N2 O2 F2 United atom limit O2+ separated atom limit N2+

  9. Homonuclear Diatomics Molecules – the valence bond view Consider bonding two Ne atoms together Clearly there will be repulsive interactions as the doubly occupied orbitals on the left and right overlap, leading to repulsive interactions and no bonding. In fact as we will consider later, there is a weak attractive interaction scaling as -C/R6, that leads to a bond of 0.05 kcal/mol, but we ignore such weak interactions here The symmetry of this state is 1Sg+

  10. Halogen dimers Next consider bonding of two F atoms. Each F has 3 possible configurations (It is a 2P state) leading to 9 possible configurations for F2. Of these only one leads to strong chemical binding This also leads to a 1Sg+ state. Spectroscopic properties are listed below . Note that the bond energy decreases for Cl2 to Br2 to I2, but increases from F2 to Cl2. we will get back to this later.

  11. Di-oxygen or O2 molecule Next consider bonding of two O atoms. Each O has 3 possible configurations (It is a 3P state) leading to 9 possible configurations for O2. Of these one leads to directly to a double bond This suggests that the ground state of O2 is a singlet state. At first this seemed plausible, but by the late 1920’s Mulliken established experimentally that the ground state of O2 is actually a triplet state, which he had predicted on the basis of molecular orbitial (MO) theory. This was a fatal blow to VB theory, bringing MO theory to the fore, so we will consider next how Mulliken was able to figure this out in the 1920’s without the aid of computers.

  12. O2 MO configuration (1pg)2 2 For O2 the ordering of the MOs Is unambiguous 4 2 2 2 Next consider states of (1pg)2 2 2

  13. States based on (p)2 D- S- S+ D+ Have 4 spatial combinations Which we combine as where x and y denote px and py φ1, φ2 denote the angle about the axis and F is independent of φ1, φ2 Rotating about the axis by an angle g, these states transform as

  14. States arising from (p)2 Adding spin we get MO theory explains the triplet ground state and low lying singlets O2 Energy (eV) 1.636 (p)2 0.982 0.0 Ground state

  15. Using the correleation diagram Choices for N2 In order to use the correlation diagram to predict the states of diatomic molecules, we need to have some idea of what effective R to use (actually it is the effective overlap with large R small S and small R large S). Mulliken’s original analysis [Rev. Mod. Phys. 4, 48 (1932)] was roughly as follows. 1. N2 was known to be nondegenerate and very strongly bound with no low-lying excited states 2 4 2 2 4 2 4 2 2 2 2

  16. N2 MO configurations This is compatible with several orderings of the MOs Largest R 2 4 2 4 2 2 2 4 2 2 Smallest R 2

  17. N2+ But the 13 electron molecules BeF, BO, CO+, CN, N2+ Have a ground state with 2S symmetry and a low lying 2S sate. In between these two 2S states is a 2P state with spin orbital splitting that implies a p3 configuration This implies that Is the ground configuration for N2 and that the low lying states of N2+ are This agrees with the observed spectra

  18. Correlation diagram for Carbon row homonuclear diatomics C2 N2 O2 F2 United atom limit O2+ separated atom limit N2+

  19. 1s and 2s cases A A B B

  20. Bond Anti BO 1 2 2.5 3 2.5 2 1 0

  21. More about O2

  22. First excited configuration (1pu)3 (1pu)3 (1pg)3 (1pg)3 (1pg)2 Ground configuration excited configuration 1Su+ 1Du Only dipole allowed transition from 3Sg- 3Su- 3Su+ 1Su- 3Du Strong transitions (dipole allowed) DS=0 (spin) Sg Su or Pu but S-  S-

  23. The states of O2 molecule Moss and Goddard JCP 63, 3623 (1975) (pu)3(pg)3 (pu)4(pg)2

  24. Role of O2 in atmosphere Moss and Goddard JCP 63, 3623 (1975) Strong Get 3P + 1D O atom Weak Get 3P + 3P O atom

  25. Implications UV light > 6 eV (l < 1240/6 = 207 nm) can dissociate O2 by excitation of 3Su+ which dissociates to two O atom in 3P state UV light > ~7.2 eV can dissociate O2 by excitation of 3Su- which dissociates to one O atom in 3P state and one in 1D (maximum is at ~8.6 eV, Schumann-Runge bands) Net result is dissociation of O2 into O atoms

  26. Regions of the atmosphere mesosphere O + hn O+ + e- Heats from light O2 + hn O + O 100 stratopause altitude (km) O + O2 O3 stratosphere 50 O3 + hn O + O2 Heats from light 30 20 tropopause 10 troposphere Heated from earth 200 300 Temperature (K)

  27. ionosphere night Heaviside-Kennelly layer Reflects radio waves to allow long distance communications D layer day

  28. nightglow At night the O atoms created during the day can recombine to form O2 The fastest rates are into the Herzberg states, 3Su+ 1Su- 3Du Get emission at ~2.4 eV, 500 nm Called the nightglow (~ 90 km)

  29. Problem with MO description: dissociation 3Sg- state: [(pgx)(pgy)+ (pgy) (pgx)] As R∞ (pgx)  (xL – xR) and (pgy)  (yL – yR) Get equal amounts of {xL yL and xR yR} and {xLyR and xR yL} Ionic: [(O-)(O+)+ (O+)(O-)] covalent: (O)(O) But actually it should dissociate to neutral atoms

  30. Back to valence bond (and GVB) Four ways to combine two 3P states of O to form a s bond bad Closed shell Open shell Looks good because make p bond as in ethene, BUT have overlapping doubly occupied orbitals  antibonding Each doubly occupied orbital overlaps a singly occupied orbital, not so repulsive

  31. Analysis of open shell configurations Each can be used to form a singlet state or a triplet state, e.g. Singlet: A{(xL)2(yR)2[(yL)(xR) + (xR)(yL)](ab-ba)} Triplet: A{(xL)2(yR)2[(yL)(xR) - (xR)(yL)](ab+ba)} and aa, bb Since (yL) and (xR) are orthogonal, high spin is best (no chance of two electrons at same point) as usual

  32. VB description of O2 + + - + Must have resonance of two VB configurations

  33. Bond H to O2 Bring H toward px on Left O Overlap doubly occupied (pxL)2 thus repulsive Overlap singly occupied (pxL)2 thus bonding Get HOO bond angle ~ 90º S=1/2 (doublet) Antisymmetric with respect to plane: A” irreducible representation (Cs group) Bond weakened by ~ 51 kcal/mol due to loss in O2 resonance 2A” state

  34. Bond 2nd H to HO2 to form hydrogen peroxide Bring H toward py on right O Expect new HOO bond angle ~ 90º Expect HOOH dihedral ~90º Indeed H-S-S-H: HSS = 91.3º and HSSH= 90.6º But H-H overlap leads to steric effects for HOOH, net result: HOO opens up to ~94.8º HOOH angle  111.5º trans structure, 180º only 1.2 kcal/mol higher

  35. Rotational barriers 7.6 kcal/mol Cis barrier HOOH 1.19 kcal/mol Trans barrier HSSH: 5.02 kcal/mol trans barrier 7.54 kcal/mol cis barrier

  36. Compare bond energies (kcal/mol) O23Sg- 119.0 50.8 67.9 HO-O 68.2 H-O2 51.5 17.1 HO-OH 51.1 HOO-H 85.2 Interpretation: OO s bond = 51.1 kcal/mol OO p bond = 119.0-51.1=67.9 kcal/mol (resonance) Bonding H to O2 loses 50.8 kcal/mol of resonance Bonding H to HO2 loses the other 17.1 kcal/mol of resonance Intrinsic H-O bond is 85.2 + 17.1 =102.3 compare CH3O-H: HO bond is 105.1

  37. Bond O2 to O to form ozone Require two OO s bonds get States with 4, 5, and 6 pp electrons Ground state is 4p case Get S=0,1 but 0 better Goddard et al Acc. Chem. Res. 6, 368 (1973)

  38. sigma GVB orbitals ozone

  39. Pi GVB orbitals ozone Some delocalization of central Opp pair Increased overlap between L and R Opp due to central pair

  40. Bond O2 to O to form ozone lose O-O p resonance, 51 kcal/mol New O-O s bond, 51 kcal/mol Gain O-Op resonance,<17 kcal/mol,assume 2/3 New singlet coupling of pL and pR orbitals Total splitting ~ 1 eV = 23 kcal/mol, assume ½ stabilizes singlet and ½ destabilizes triplet Expect bond for singlet of 11 + 12 = 23 kcal/mol, exper = 25 Expect triplet state to be bound by 11-12 = -1 kcal/mol, probably between +2 and -2

  41. Alternative view of bonding in ozone Start here with 1-3 diradical Transfer electron from central doubly occupied pp pair to the R singly occupied pp. Now can form a p bond the L singly occupied pp. Hard to estimate strength of bond

  42. New material

  43. Ring ozone Form 3 OO sigma bonds, but pp pairs overlap Analog: cis HOOH bond is 51.1-7.6=43.5 kcal/mol. Get total bond of 3*43.5=130.5 which is 11.5 more stable than O2. Correct for strain due to 60º bond angles = 26 kcal/mol from cyclopropane. Expect ring O3 to be unstable with respect to O2 + O by ~14 kcal/mol, But if formed it might be rather stable with respect various chemical reactions. Ab Initio Theoretical Results on the Stability of Cyclic Ozone L. B. Harding and W. A. Goddard III J. Chem. Phys. 67, 2377 (1977) CN 5599

  44. Photochemical smog High temperature combustion: N2 + O2  2NO Thus Auto exhaust  NO 2 NO + O2 2 NO2 NO2 + hn NO + O O + O2 + M  O3 + M O3 + NO  NO2 + O2 Get equilibrium Add in hydrocarbons NO2 + O2 + HC + hn  Me(C=O)-OO-NO2 peroxyacetylnitrate

  45. More on N2 The elements N, P, As, Sb, and Bi all have an (ns)2(np)3 configuration, leading to a triple bond Adding in the (ns) pairs, we show the wavefunction as This is the VB description of N2, P2, etc. The optimum orbitals of N2 are shown on the next slide. The MO description of N2 is Which we can draw as

  46. GVB orbitals of N2 Re=1.10A R=1.50A R=2.10A

  47. Hartree Fock Orbitals N2

  48. The configuration for C2 2 1 1 2 1 2 4 3 4 4 2 2 2 2

  49. 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

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