Lecture 16 February 20 Transition metals, Pd and Pt

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1 Lecture 16 February 20 Transition metals, Pd and Pt 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, 316 Beckman Institute, x3093 Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Teaching Assistants: Ross Fu Fan Liu Ch120a-1

2 Last Time 2

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

4 Transition metals 4

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

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 Fe 2+ with a d 6 configuration Each N has a doubly occupied sp 2 s orbital pointing at it. 6

7 Energies of the 5 Fe 2+ d orbitals x 2 -y 2 z 2 =2z 2 -x 2 -y 2 yz xz xy 7

8 Exchange stabilizations 8

9 Skip energy stuff 9

10 Energy for 2 electron product wavefunction Consider the product wavefunction Ψ(1,2) = ψ a (1) ψ b (2) And the Hamiltonian H(1,2) = h(1) + h(2) +1/r /R In the details slides next, we derive E = < Ψ(1,2) H(1,2) Ψ(1,2)>/ <Ψ(1,2) Ψ(1,2)> E = h aa + h bb + J ab + 1/R where h aa =<a h a>, h bb = SKIP for now J ab <ψ a (1)ψ b (2) 1/r 12 ψ a (1)ψ b (2)>=ʃ [ψ a (1)] 2 [ψ b (1)] 2 /r 12 Represent the total Coulomb interaction between the electron density r a (1)= ψ a (1) 2 and r b (2)= ψ b (2) 2 Since the integrand r a (1) r b (2)/r 12 is positive for all positions of 1 and 2, the integral is positive, J ab > 0 Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 10

11 Details in deriving energy: normalization First, the normalization term is <Ψ(1,2) Ψ(1,2)>=<ψ a (1) ψ a (1)><ψ b (2) ψ b (2)> Which from now on we will write as SKIP for now <Ψ Ψ> = <ψ a ψ a ><ψ b ψ b > = 1 since the ψ i are normalized Here our convention is that a two-electron function such as <Ψ(1,2) Ψ(1,2)> is always over both electrons so we need not put in the (1,2) while one-electron functions such as <ψ a (1) ψ a (1)> or <ψ b (2) ψ b (2)> are assumed to be over just one electron and we ignore the labels 1 or 2 Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 11

12 Details of deriving energy: one electron terms Using H(1,2) = h(1) + h(2) +1/r /R We partition the energy E = <Ψ H Ψ> as E = <Ψ h(1) Ψ> + <Ψ h(2) Ψ> + <Ψ 1/R Ψ> + <Ψ 1/r 12 Ψ> Here <Ψ 1/R Ψ> = <Ψ Ψ>/R = 1/R since R is a constant <Ψ h(1) Ψ> = <ψ a (1)ψ b (2) h(1) ψ a (1)ψ b (2)> = = <ψ a (1) h(1) ψ a (1)><ψ b (2) ψ b (2)> = <a h a> = h aa Where h aa <a h a> <ψ a h ψ a > Similarly <Ψ h(2) Ψ> = <ψ a (1)ψ b (2) h(2) ψ a (1)ψ b (2)> = SKIP for now = <ψ a (1) ψ a (1)><ψ b (2) h(2) ψ b (2)> = <a a> = h bb The remaining term we denote as J ab <ψ a (1)ψ b (2) 1/r 12 ψ a (1)ψ b (2)> so that the total energy is E = h aa + h bb + J ab + 1/R Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 12

13 The energy for an antisymmetrized product, A ψ a ψ b The total energy is that of the product plus the exchange term which is negative with 4 parts SKIP for now E ex =-< ψ a ψ b h(1) ψ b ψ a >-< ψ a ψ b h(2) ψ b ψ a >-< ψ a ψ b 1/R ψ b ψ a > - < ψ a ψ b 1/r 12 ψ b ψ a > The first 3 terms lead to < ψ a h(1) ψ b ><ψ b ψ a >+ <ψ a ψ b ><ψ b h(2) ψ a >+ <ψ a ψ b ><ψ b ψ a >/R But <ψ b ψ a >=0 Thus all are zero Thus the only nonzero term is the 4 th term: -Kab=- < ψ a ψ b 1/r 12 ψ b ψ a > which is called the exchange energy (or the 2-electron exchange) since it arises from the exchange term due to the antisymmetrizer. Summarizing, the energy of the Aψ a ψ b wavefunction for H 2 is E = h aa + h bb + (J ab K ab ) + 1/R Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 13

14 The energy of the antisymmetrized wavefunction The total electron-electron repulsion part of the energy for any wavefunction Ψ(1,2) must be positive SKIP for now E ee = (d 3 r 1 )((d 3 r 2 ) Ψ(1,2) 2 /r 12 > 0 This follows since the integrand is positive for all positions of r 1 and r 2 then We derived that the energy of the A ψ a ψ b wavefunction is E = h aa + h bb + (J ab K ab ) + 1/R Where the E ee = (J ab K ab ) > 0 Since we have already established that J ab > 0 we can conclude that J ab > K ab > 0 Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 14

15 Separate the spinorbital into orbital and spin parts Since the Hamiltonian does not contain spin the spinorbitals can be factored into spatial and spin terms. For 2 electrons there are two possibilities: Both electrons have the same spin ψ a (1)ψ b (2)=[Φ a (1)a(1)][Φ b (2)a(2)]= [Φ a (1)Φ b (2)][a(1)a(2)] So that the antisymmetrized wavefunction is Aψ a (1)ψ b (2)= A[Φ a (1)Φ b (2)][a(1)a(2)]= =[Φ a (1)Φ b (2)- Φ b (1)Φ a (2)][a(1)a(2)] Also, similar results for both spins down Aψ a (1)ψ b (2)= A[Φ a (1)Φ b (2)][b(1)b(2)]= =[Φ a (1)Φ b (2)- Φ b (1)Φ a (2)][b(1)b(2)] Since <ψ a ψ b >= 0 = < Φ a Φ b ><a a> = < Φ a Φ b > We see that the spatial orbitals for same spin must be orthogonal Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 15

16 Energy for 2 electrons with same spin The total energy becomes E = h aa + h bb + (J ab K ab ) + 1/R where h aa <Φ a h Φ a > and h bb <Φ b h Φ b > where J ab = <Φ a (1)Φ b (2) 1/r 12 Φ a (1)Φ b (2)> SKIP for now We derived the exchange term for spin orbitals with same spin as follows K ab <ψ a (1)ψ b (2) 1/r 12 ψ b (1)ψ a (2)> `````= <Φ a (1)Φ b (2) 1/r 12 Φ b (1)Φ a (2)><a(1) a(1)><a(2) a(2)> K ab where K ab <Φ a (1)Φ b (2) 1/r 12 Φ b (1)Φ a (2)> Involves only spatial coordinates. Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 16

17 Energy for 2 electrons with opposite spin Now consider the exchange term for spin orbitals with opposite spin SKIP for now K ab <ψ a (1)ψ b (2) 1/r 12 ψ b (1)ψ a (2)> `````= <Φ a (1)Φ b (2) 1/r 12 Φ b (1)Φ a (2)><a(1) b(1)><b(2) a(2)> = 0 Since <a(1) b(1)> = 0. Thus the total energy is E ab = h aa + h bb + J ab + 1/R With no exchange term unless the spins are the same Since <ψ a ψ b >= 0 = < Φ a Φ b ><a b> There is no orthogonality condition of the spatial orbitals for opposite spin electrons In general < Φ a Φ b > =S, where the overlap S 0 Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 17

18 Summarizing: Energy for 2 electrons When the spinorbitals have the same spin, Aψ a (1)ψ b (2)= A[Φ a (1)Φ b (2)][a(1)a(2)] The total energy is E aa = h aa + h bb + (J ab K ab ) + 1/R SKIP for now But when the spinorbitals have the opposite spin, Aψ a (1)ψ b (2)= A[Φ a (1)Φ b (2)][a(1)b(2)]= The total energy is E ab = h aa + h bb + J ab + 1/R With no exchange term Thus exchange energies arise only for the case in which both electrons have the same spin Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 18

19 Consider further the case for spinorbtials with opposite spin Neither of these terms has the correct permutation symmetry separately for space or spin. But they can be combined [Φ a (1)Φ b (2)-Φ b (1)Φ a (2)][a(1)b(2)+b(1)a(2)]= A[Φ a (1)Φ b (2)][a(1)b(2)]-A[Φ b (1)Φ a (2)][a(1)b(2)] Which describes the Ms=0 component of the triplet state [Φ a (1)Φ b (2)+Φ b (1)Φ a (2)][a(1)b(2)-b(1)a(2)]= A[Φ a (1)Φ b (2)][a(1)b(2)]+A[Φ b (1)Φ a (2)][a(1)b(2)] Which describes the Ms=0 component of the singlet state Thus for the ab case, two Slater determinants must be combined to obtain the correct spin and space permutational symmetry Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 19

20 Consider further the case for spinorbtials with opposite spin The wavefunction [Φ a (1)Φ b (2)-Φ b (1)Φ a (2)][a(1)b(2)+b(1)a(2)] Leads directly to 3 E ab = h aa + h bb + (J ab K ab ) + 1/R Exactly the same as for [Φ a (1)Φ b (2)-Φ b (1)Φ a (2)][a(1)a(2)] [Φ a (1)Φ b (2)-Φ b (1)Φ a (2)][b(1)b(2)] These three states are collectively referred to as the triplet state and denoted as having spin S=1 The other combination leads to one state, referred to as the singlet state and denoted as having spin S=0 [Φ a (1)Φ b (2)+Φ b (1)Φ a (2)][a(1)b(2)-b(1)a(2)] We will analyze the energy for this wavefunction next. SKIP for now Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 20

21 Consider the energy of the singlet wavefunction [Φ a (1)Φ b (2)+Φ b (1)Φ a (2)][a(1)b(2)-b(1)a(2)] (ab+ba)(ab-ba) The next few slides show that SKIP for now 1 E = {(h aa + h bb + (h ab + h ba ) S 2 + J ab + K ab + (1+S 2 )/R}/(1 + S 2 ) Where the terms with S or Kab come for the exchange \ Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 21

22 energy of the singlet wavefunction - details [Φ a (1)Φ b (2)+Φ b (1)Φ a (2)][a(1)b(2)-b(1)a(2)] (ab+ba)(ab-ba) 1 E = numerator/ denominator Where numerator =<(ab+ba)(ab-ba) H (ab+ba)(ab-ba)> = =<(ab+ba) H (ab+ba)><(ab-ba) (ab-ba)> denominator = <(ab+ba)(ab-ba) (ab+ba)(ab-ba)> Since <(ab-ba) (ab-ba)>= 2 <ab (ab-ba)>= We obtain 2[<a a>-<a b>]=2 numerator =<(ab+ba) H (ab+ba)> = 2 <ab H (ab+ba)> denominator = <(ab+ba) (ab+ba)>=2 <ab (ab+ba)> Thus 1 E = <ab H (ab+ba)>/<ab (ab+ba)> SKIP for now Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 22

23 energy of the singlet wavefunction - details 1 E = <ab H (ab+ba)>/<ab (ab+ba)> SKIP for now Consider first the denominator <ab (ab+ba)> = <a a> + <a b> = 1 + S 2 Where S= <a b>= is the overlap The numerator becomes <ab (ab+ba)> = <a h a> + <a h b> + + <a a> + <a b> + + <ab 1/r 12 (ab+ba)> + (1 + S 2 )/R Thus the total energy is 1 E = {(h aa + h bb + (h ab + h ba ) S 2 + J ab + K ab + (1+S 2 )/R}/(1 + S 2 ) Ch120a-Goddard-L03 copyright William A. Goddard III, all rights reserved 23

24 Ferrous Fe II x 2 -y 2 destabilized by heme N lone pairs z 2 destabilized by 5 th ligand imidazole or 6 th ligand CO y x 24

25 Summary 4 coord and 5 coord states 25

26 Out of plane motion of Fe 4 coordinate 26

27 Add axial base N-N Nonbonded interactions push Fe out of plane is antibonding 27

28 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 6 th ligand 28

29 Bonding of O 2 with O to form ozone O 2 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 29

30 Bond O 2 to Mb Simple VB structures get S=1 or triplet state In fact MbO 2 is singlet Why? 30

31 change in exchange terms when Bond O 2 to Mb O 2 ps O 2 pp Assume perfect VB spin pairing Then get 4 cases Thus average K dd is ( )/4 = K dd 5*4/2 up spin down spin 7 K dd 4*3/2 + 2*1/2 7 K dd 4*3/2 + 2*1/2 6 K dd 3*2/2 + 3*2/2 31

32 Bonding O 2 to Mb Exchange loss on bonding O 2 32

33 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 = = + 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 O 2 does not gain the full VB strength, leading to bond of only 8kcal/mol instead of 33 33

34 Bond CO to Mb H 2 O and N 2 do not bond strongly enough to promote the Fe to an excited state, thus get S=2 34

35 compare bonding of CO and O2 to Mb 35

36 New material 36

2 Think of as bond from Tidz2 to H1s Covalent 2 electron CH

37 GVB orbitals for bonds to Ti Ti ds character, 1 elect H 1s character, 1 elect Covalent 2 electron TiH bond in Cl 2 TiH 2 Think of as bond from Tidz2 to H1s Covalent 2 electron CH bond in CH 4 Csp 3 character 1 elect H 1s character, 1 elect 37

38 Bonding at a transition metaal Bonding to a transition metals can be quite covalent. Examples: (Cl 2 )Ti(H 2 ), (Cl 2 )Ti(C 3 H 6 ), Cl 2 Ti=CH 2 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 TiCl 2 group has ~ same electronegativity as H or CH 3 The covalent bond can be thought of as Ti(dz2-4s) hybrid spin paired with H1s A{[(Tids)(H1s)+ (H1s)(Tids)](ab-ba)} 38

39 But TM-H bond can also be s-like Cl 2 TiH + Ti (4s) 2 (3d) 2 The 2 Cl pull off 2 e from Ti, leaving a d 1 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 d 5 s 1 configuration H bonds to 4s because of exchange stabilization of d 5 Mn-H bond character 0.07 Mnd+0.71Mnsp+1.20H 39

40 Bond angle at a transition metal For two p orbitals expect 90, HH nonbond repulsion increases it What angle do two d orbitals want H-Ti-H plane 76 Metallacycle plane 40

Best bond angle for 2 pure Metal bonds using d orbitals Assume that the first bond has pure d z2 or ds character to a ligand along the z axis Can we make a 2 nd bond, also of pure ds character

41 Best bond angle for 2 pure Metal bonds using d orbitals Assume that the first bond has pure d z2 or ds character to a ligand along the z axis Can we make a 2 nd 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 ), this is ½ the tetrahedral angle of (also the magic spinning angle for solid state NMR). 41

42 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 3 F state (7) and a 3 P state (3), with 3 F lower. This is because the electron repulsion between say a d xy and d x2-y2 is higher than between sasy d z2 and d xy. 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 42

43 How predict character of Transition metal bonds? 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 (3d) 2 (4s)(3d) 5 Now make bond to less electronegative ligands, H or CH 3 Use 4s if available, otherwise use d orbitals 43

71H ClMnH Mn (4s) 2 (3d) 5 The Cl pulls off 1 e from Mn, leaving a d 5 s 1

44 But TM-H bond can also be s-like Cl 2 TiH + Ti (4s) 2 (3d) 2 The 2 Cl pull off 2 e from Ti, leaving a d 1 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 d 5 s 1 configuration H bonds to 4s because of exchange stabilization of d 5 Mn-H bond character 0.07 Mnd+0.71Mnsp+1.20H 44

45 Example (Cl) 2 VH 3 + resonance configuration 45

46 Example ClMometallacycle butadiene 46

47 Example [Mn CH] 2+ 47

48 Summary: start with Mn + s 1 d 5 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 48

49 Summary: start with Mn + s 1 d 5 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 49

50 Compare chemistry of column 10 50

51 Ground state of group 10 column Pt: (5d) 9 (6s) 1 3 D ground state Pt: (5d) 10 (6s) 0 1 S excited state at 11.0 kcal/mol Pt: (5d) 8 (6s) 2 3 F excited state at 14.7 kcal/mol Ni: (5d) 8 (6s) 2 3 F ground state Ni: (5d) 9 (6s) 1 3 D excited state at 0.7 kcal/mol Ni: (5d) 10 (6s) 0 1 S excited state at 40.0 kcal/mol Pd: (5d) 10 (6s) 0 1 S ground state Pd: (5d) 9 (6s) 1 3 D excited state at 21.9 kcal/mol Pd: (5d) 8 (6s) 2 3 F excited state at 77.9 kcal/mol 51

52 Salient differences between Ni, Pd, Pt 2 nd row (Pd): 4d much more stable than 5s Pd d 10 ground state 3 rd row (Pt): 5d and 6s comparable stability Pt d 9 s 1 ground state 52

53 Ground state configurations for column 10 Ni Pd Pt 53

54 Next section Theoretical Studies of Oxidative Addition and Reductive Elimination: J. J. Low and W. A. Goddard III J. Am. Chem. Soc. 106, 6928 (1984) wag 190 Reductive Coupling of H-H, H-C, and C-C Bonds from Pd Complexes J. J. Low and W. A. Goddard III J. Am. Chem. Soc. 106, 8321 (1984) wag 191 Theoretical Studies of Oxidative Addition and Reductive Elimination. II. Reductive Coupling of H-H, H-C, and C-C Bonds from Pd and Pt Complexes J. J. Low and W. A. Goddard III Organometallics 5, 609 (1986) wag

55 Mysteries from experiments on oxidative addition and reductive elimination of CH and CC bonds on Pd and Pt Why are Pd and Pt so different 55

56 Mysteries from experiments on oxidative addition and reductive elimination of CH and CC bonds on Pd and Pt Why is CC coupling so much harder than CH coupling? 56

57 Step 1: examine GVB orbitals for (PH 3 ) 2 Pt(CH 3 ) 57

58 Analysis of GVB wavefunction 58

59 Alternative models for Pt centers 59

60 60

61 61

62 energetics Not agree with experiment 62

63 Possible explanation: kinetics 63

64 Consider reductive elimination of HH, CH and CC from Pd Conclusion: HH no barrier CH modest barrier CC large barrier 64

65 Consider oxidative addition of HH, CH, and CC to Pt Conclusion: HH no barrier CH modest barrier CC large barrier 65

66 Summary of barriers This explains why CC coupling not occur for Pt while CH and HHcoupling is fast But why? 66

67 How estimate the size of barriers (without calculations) 67

68 Examine HH coupling at transition state Can simultaneously get good overlap of H with Pd sd hybrid and with the other H Thus get resonance stabilization of TS low barrier 68

69 Examine CC coupling at transition state Can orient the CH 3 to obtain good overlap with Pd sd hybrid OR can orient the CH 3 to obtain get good overlap with the other CH 3 But CANNOT DO BOTH SIMULTANEOUSLY, thus do NOT get resonance stabilization of TS high barier 69

70 Examine CH coupling at transition state H can overlap both CH 3 and Pd sd hybrid simultaneously but CH 3 cannot thus get ~ ½ resonance stabilization of TS 70

71 Now we understand Pt chemistry But what about Pd? Why are Pt and Pd so dramatically different 71

72 stop 72

Lecture 11 January 30, Transition metals, Pd and Pt

Lecture 11 January 30, Transition metals, Pd and Pt Lecture 11 January 30, 2011 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy Course number: Ch120a Hours: