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

Lecture 15 February 15, 2013 Transition metals. 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.

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

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  1. Lecture 15 February 15, 2013 Transition metals Ch120a-Goddard-L01 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: Ross Fu <fur@caltech.edu>; Fan Liu <fliu@wag.caltech.edu>

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

  3. Ground states of neutral atoms

  4. Hemoglobin Blood has 5 billion erythrocytes/ml Each erythrocyte contains 280 million hemoglobin (Hb) molecules Each Hb has MW=64500 Dalton (diameter ~ 60A) Four subunits (a1, a2, b1, b2) each with one heme subunit Each subunit resembles myoglobin (Mb) which has one heme Hb Mb

  5. The action is at the heme or Fe-Porphyrin molecule Essentially all action occurs at the heme, which is basically an Fe-Porphyrin molecule The rest of the Mb serves mainly to provide a hydrophobic envirornment at the Fe and to protect the heme

  6. The heme group The net charge of the Fe-heme is zero. The VB structure shown is one of several, all of which lead to two neutral N and two negative N. Thus we consider that the Fe is Fe2+ with a d6 configuration Each N has a doubly occupied sp2s orbital pointing at it.

  7. Energies of the 5 Fe2+ d orbitals x2-y2 z2=2z2-x2-y2 yz xz xy

  8. Exchange stabilizations

  9. Ferrous FeII y x x2-y2 destabilized by heme N lone pairs z2 destabilized by 5th ligand imidazole or 6th ligand CO

  10. Four coordinate Fe-Heme – High spin case, S=2 or q The 5th axial ligand will destabilize q2 since dz2 is doubly occupied A pi acceptor would stabilize q1 wrt q2 Bonding O2 to 5 coordinate will stabilize q3 wrt q1 Future discuss only q1 and denote as q

  11. Four coordinate Fe-Heme – Intermediate spin, S=1 or t

  12. Four coordinate Fe-Heme – Low spin case, S=0 or s

  13. Out of plane motion of Fe – 4 coordinate

  14. Add axial base N-N Nonbonded interactions push Fe out of plane is antibonding

  15. Summary 4 coord and 5 coord states

  16. Free atom to 4 coord to 5 coord Net effect due to five N ligands is to squish the q, t, and s states by a factor of 3 This makes all three available as possible ground states depending on the 6th ligand

  17. Bonding of O2 with O to form ozone O2 has available a ps orbital for a s bond to a ps orbital of the O atom And the 3 electron p system for a p bond to a pp orbital of the O atom

  18. Bond O2 to Mb Simple VB structures  get S=1 or triplet state In fact MbO2 is singlet Why?

  19. change in exchange terms when Bond O2 to Mb O2ps O2pp 10 Kdd 7 Kdd 5*4/2 4*3/2 + 2*1/2 Assume perfect VB spin pairing Then get 4 cases 7 Kdd 6 Kdd up spin 4*3/2 + 2*1/2 3*2/2 + 3*2/2 Thus average Kdd is (10+7+7+6)/4 =7.5 down spin

  20. Bonding O2 to Mb Exchange loss on bonding O2

  21. Modified exchange energy for q state But expected t binding to be 2*22 = 44 kcal/mol stronger than q What happened? Binding to q would have DH = -33 + 44 = + 11 kcal/mol Instead the q state retains the high spin pairing so that there is no exchange loss, but now the coupling of Fe to O2 does not gain the full VB strength, leading to bond of only 8kcal/mol instead of 33

  22. Bond CO to Mb H2O and N2 do not bond strongly enough to promote the Fe to an excited state, thus get S=2

  23. compare bonding of CO and O2 to Mb

  24. GVB orbitals for bonds to Ti H 1s character, 1 elect Ti ds 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

  25. 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)}

  26. 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

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

  28. 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).

  29. 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

  30. How predict character of Transition metal bonds? (4s)(3d)5 (3d)2 Start with ground state atomic configuration Ti (4s)2(3d)2 or Mn (4s)2(3d)5 Consider that bonds to electronegative ligands (eg Cl or Cp) take electrons from 4s easiest to ionize, also better overlap with Cl or Cp, also leads to less reduction in dd exchange Now make bond to less electronegative ligands, H or CH3 Use 4s if available, otherwise use d orbitals

  31. 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

  32. Example (Cl)2VH3 + resonance configuration

  33. Example ClMo-metallacycle butadiene

  34. Example [Mn≡CH]2+

  35. Summary: start with Mn+ s1d5 dy2 s bond to H1s dx2-x2 non bonding dyz p bond to CH dxz p bond to CH dxy non bonding 4sp hybrid s bond to CH

  36. Summary: start with Mn+ s1d5 dy2 s bond to H1s dx2-x2 non bonding dyz p bond to CH dxz p bond to CH dxy non bonding 4sp hybrid s bond to CH

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