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

1. Lecture 17, November 4, 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 TODAYWednesday, 1pm, L17, additional lecture, room 101 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. States arrising 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 EEWS-90.502-Goddard-L15

5. 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- EEWS-90.502-Goddard-L15

6. The states of O2 molecule Moss and Goddard JCP 63, 3623 (1975) (pu)3(pg)3 (pu)4(pg)2 EEWS-90.502-Goddard-L15

7. Role of O2 in atmosphere Moss and Goddard JCP 63, 3623 (1975) Strong Get 3P + 1D O atom Weak Get 3P + 3P O atom EEWS-90.502-Goddard-L15

8. 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 EEWS-90.502-Goddard-L15 Temperature (K)

9. ionosphere night Heaviside-Kennelly layer Reflects radio waves to allow long distance communications D layer electrons (daytime) EEWS-90.502-Goddard-L15

10. 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) EEWS-90.502-Goddard-L15

11. 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 EEWS-90.502-Goddard-L15

12. VB description of O2 + + - + Must have resonance of two VB configurations EEWS-90.502-Goddard-L15

13. GVB orbitals at Re Problem: one VB configuration not enough + EEWS-90.502-Goddard-L15

14. 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 EEWS-90.502-Goddard-L15

15. 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 EEWS-90.502-Goddard-L15

16. 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 EEWS-90.502-Goddard-L15

17. 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 EEWS-90.502-Goddard-L15

18. 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) EEWS-90.502-Goddard-L15

19. Pi GVB orbitals ozone Some delocalization of central Opp pair Increased overlap between L and R Opp due to central pair EEWS-90.502-Goddard-L15

20. 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 EEWS-90.502-Goddard-L15

21. Photochemical smog High temperature combustion: N2 + O2  2NO 2NO + O2  2NO2 Auto exhaust  NO NO2 + hn NO + O O + O2 + M  O3 + M O3 + NO  NO2 + O2 Get equilibrium Add in hydrocarbons NO2 + O2 + HC + hn  MeC=O)-OO-NO2 peroxyacetylnitrate EEWS-90.502-Goddard-L15

22. new EEWS-90.502-Goddard-L15

23. 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 EEWS-90.502-Goddard-L15

24. Ring ozone Form 3 OO sigma bonds, but pp paires 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 EEWS-90.502-Goddard-L15

25. 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 EEWS-90.502-Goddard-L15

26. GVB orbitals of N2 Re=1.10A R=1.50A R=2.10A EEWS-90.502-Goddard-L15

27. Hartree Fock Orbitals N2 EEWS-90.502-Goddard-L15

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29. The configuration for C2 2 1 1 2 1 2 4 3 4 4 2 2 2 2 EEWS-90.502-Goddard-L15

30. 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

31. 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

32. C2, Si2, EEWS-90.502-Goddard-L15

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

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37. Van der Waals interactions For an ideal gas the equation of state is given by pV =nRT where p = pressure; V = volume of the container n = number of moles; R = gas constant = NAkB NA = Avogadro constant; kB = Boltzmann constant Van der Waals equation of state (1873) [p + n2a/V2)[V - nb] = nRT Where a is related to attractions between the particles, (reducing the pressure) And b is related to a reduced available volume (due to finite size of particles) EEWS-90.502-Goddard-L15

38. 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

39. 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

40. 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

41. Ionic bonding (chapter 9) Consider the covalent bond of Na to Cl. There Is very liitle contragradience, leading to an extremely weak bond. Alternativly, consider transfering the charge from Na to Cl to form Na+ and Cl- EEWS-90.502-Goddard-L15

42. 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

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

44. 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

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47. 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 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 neigbors EEWS-90.502-Goddard-L15

48. 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

49. 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

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