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, 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> Ch120a-Goddard-L01

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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

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 sp2 s orbital pointing at it.

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

Exchange stabilizations

Skip energy stuff

Energy for 2 electron product wavefunction Consider the product wavefunction Ψ(1,2) = ψa(1) ψb(2) And the Hamiltonian H(1,2) = h(1) + h(2) +1/r12 + 1/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 = haa + hbb + Jab + 1/R where haa =<a|h|a>, hbb =<b|h|b> Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)>=ʃ [ψa(1)]2 [ψb(1)]2/r12 Represent the total Coulomb interaction between the electron density ra(1)=| ψa(1)|2 and rb(2)=| ψb(2)|2 Since the integrand ra(1) rb(2)/r12 is positive for all positions of 1 and 2, the integral is positive, Jab > 0 SKIP for now

Details in deriving energy: normalization SKIP for now First, the normalization term is <Ψ(1,2)|Ψ(1,2)>=<ψa(1)|ψa(1)><ψb(2) ψb(2)> Which from now on we will write as <Ψ|Ψ> = <ψa|ψa><ψb|ψb> = 1 since the ψi 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 <ψa(1)|ψa(1)> or <ψb(2) ψb(2)> are assumed to be over just one electron and we ignore the labels 1 or 2

Details of deriving energy: one electron terms Using H(1,2) = h(1) + h(2) +1/r12 + 1/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/r12|Ψ> Here <Ψ|1/R|Ψ> = <Ψ|Ψ>/R = 1/R since R is a constant <Ψ|h(1)|Ψ> = <ψa(1)ψb(2) |h(1)|ψa(1)ψb(2)> = = <ψa(1)|h(1)|ψa(1)><ψb(2)|ψb(2)> = <a|h|a><b|b> = ≡ haa Where haa≡ <a|h|a> ≡ <ψa|h|ψa> Similarly <Ψ|h(2)|Ψ> = <ψa(1)ψb(2) |h(2)|ψa(1)ψb(2)> = = <ψa(1)|ψa(1)><ψb(2)|h(2)|ψb(2)> = <a|a><b|h|b> = ≡ hbb The remaining term we denote as Jab ≡ <ψa(1)ψb(2) |1/r12 |ψa(1)ψb(2)> so that the total energy is E = haa + hbb + Jab + 1/R SKIP for now

The energy for an antisymmetrized product, A ψaψb The total energy is that of the product plus the exchange term which is negative with 4 parts Eex=-< ψaψb|h(1)|ψb ψa >-< ψaψb|h(2)|ψb ψa >-< ψaψb|1/R|ψb ψa > - < ψaψb|1/r12|ψb ψa > The first 3 terms lead to < ψa|h(1)|ψb><ψbψa >+ <ψa|ψb><ψb|h(2)|ψa >+ <ψa|ψb><ψb|ψa>/R But <ψb|ψa>=0 Thus all are zero Thus the only nonzero term is the 4th term: -Kab=- < ψaψb|1/r12|ψb ψa > which is called the exchange energy (or the 2-electron exchange) since it arises from the exchange term due to the antisymmetrizer. SKIP for now Summarizing, the energy of the Aψaψb wavefunction for H2 is E = haa + hbb + (Jab –Kab) + 1/R

The energy of the antisymmetrized wavefunction The total electron-electron repulsion part of the energy for any wavefunction Ψ(1,2) must be positive Eee =∫ (d3r1)((d3r2)|Ψ(1,2)|2/r12 > 0 This follows since the integrand is positive for all positions of r1 and r2 then We derived that the energy of the A ψa ψb wavefunction is E = haa + hbb + (Jab –Kab) + 1/R Where the Eee = (Jab –Kab) > 0 Since we have already established that Jab > 0 we can conclude that Jab > Kab > 0 SKIP for now

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 ψa(1)ψb(2)=[Φa(1)a(1)][Φb(2)a(2)]= [Φa(1)Φb(2)][a(1)a(2)] So that the antisymmetrized wavefunction is Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)]= =[Φa(1)Φb(2)- Φb(1)Φa(2)][a(1)a(2)] Also, similar results for both spins down Aψa(1)ψb(2)= A[Φa(1)Φb(2)][b(1)b(2)]= =[Φa(1)Φb(2)- Φb(1)Φa(2)][b(1)b(2)] Since <ψa|ψb>= 0 = < Φa| Φb><a|a> = < Φa| Φb> We see that the spatial orbitals for same spin must be orthogonal

Energy for 2 electrons with same spin The total energy becomes E = haa + hbb + (Jab –Kab) + 1/R where haa ≡ <Φa|h|Φa> and hbb ≡ <Φb|h|Φb> where Jab= <Φa(1)Φb(2) |1/r12 |Φa(1)Φb(2)> SKIP for now We derived the exchange term for spin orbitals with same spin as follows Kab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)> `````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><a(1)|a(1)><a(2)|a(2)> ≡ Kab where Kab ≡ <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)> Involves only spatial coordinates.

Energy for 2 electrons with opposite spin Now consider the exchange term for spin orbitals with opposite spin Kab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)> `````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><a(1)|b(1)><b(2)|a(2)> = 0 Since <a(1)|b(1)> = 0. SKIP for now Thus the total energy is Eab = haa + hbb + Jab + 1/R With no exchange term unless the spins are the same Since <ψa|ψb>= 0 = < Φa| Φb><a|b> There is no orthogonality condition of the spatial orbitals for opposite spin electrons In general < Φa| Φb> =S, where the overlap S ≠ 0

Summarizing: Energy for 2 electrons When the spinorbitals have the same spin, Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)a(2)] The total energy is Eaa = haa + hbb + (Jab –Kab) + 1/R But when the spinorbitals have the opposite spin, Aψa(1)ψb(2)= A[Φa(1)Φb(2)][a(1)b(2)]= Eab = haa + hbb + Jab + 1/R With no exchange term SKIP for now Thus exchange energies arise only for the case in which both electrons have the same spin

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 [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)]= A[Φa(1)Φb(2)][a(1)b(2)]-A[Φb(1)Φa(2)][a(1)b(2)] Which describes the Ms=0 component of the triplet state [Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)]= A[Φa(1)Φb(2)][a(1)b(2)]+A[Φb(1)Φa(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

Consider further the case for spinorbtials with opposite spin The wavefunction [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)b(2)+b(1)a(2)] Leads directly to 3Eab = haa + hbb + (Jab –Kab) + 1/R Exactly the same as for [Φa(1)Φb(2)-Φb(1)Φa(2)][a(1)a(2)] [Φa(1)Φb(2)-Φb(1)Φa(2)][b(1)b(2)] These three states are collectively referred to as the triplet state and denoted as having spin S=1 SKIP for now The other combination leads to one state, referred to as the singlet state and denoted as having spin S=0 [Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)] We will analyze the energy for this wavefunction next.

Consider the energy of the singlet wavefunction [Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)] ≡ (ab+ba)(ab-ba) The next few slides show that 1E = {(haa + hbb + (hab + hba) S2 + Jab + Kab + (1+S2)/R}/(1 + S2) Where the terms with S or Kab come for the exchange \ SKIP for now

energy of the singlet wavefunction - details [Φa(1)Φb(2)+Φb(1)Φa(2)][a(1)b(2)-b(1)a(2)] ≡ (ab+ba)(ab-ba) 1E = 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)>= 2[<a|a><b|b>-<a|b><b|a>]=2 We obtain numerator =<(ab+ba)|H|(ab+ba)> = 2 <ab|H|(ab+ba)> denominator = <(ab+ba)|(ab+ba)>=2 <ab|(ab+ba)> SKIP for now Thus 1E = <ab|H|(ab+ba)>/<ab|(ab+ba)>

energy of the singlet wavefunction - details SKIP for now 1E = <ab|H|(ab+ba)>/<ab|(ab+ba)> Consider first the denominator <ab|(ab+ba)> = <a|a><b|b> + <a|b><b|a> = 1 + S2 Where S= <a|b>=<b|a> is the overlap The numerator becomes <ab|(ab+ba)> = <a|h|a><b|b> + <a|h|b><b|a> + + <a|a><b|h|b> + <a|b><b|h|a> + + <ab|1/r12|(ab+ba)> + (1 + S2)/R Thus the total energy is 1E = {(haa + hbb + (hab + hba) S2 + Jab + Kab + (1+S2)/R}/(1 + S2)

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

Summary 4 coord and 5 coord states

Out of plane motion of Fe – 4 coordinate

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

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

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

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

change in exchange terms when Bond O2 to Mb O2ps O2pp 10 Kdd 5*4/2 7 Kdd 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

Exchange loss on bonding O2 Bonding O2 to Mb Exchange loss on bonding O2

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

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

compare bonding of CO and O2 to Mb

New material

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

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 (4s)(3d)5 (3d)2 Now make bond to less electronegative ligands, H or CH3 Use 4s if available, otherwise use d orbitals

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

Example (Cl)2VH3 + resonance configuration

Example ClMo-metallacycle butadiene

Example [Mn≡CH]2+

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

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

Compare chemistry of column 10

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

Salient differences between Ni, Pd, Pt 2nd row (Pd): 4d much more stable than 5s  Pd d10 ground state 3rd row (Pt): 5d and 6s comparable stability  Pt d9s1 ground state

Ground state configurations for column 10 Ni Pd Pt

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 206

Why are Pd and Pt so different 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

Why is CC coupling so much harder than CH coupling? 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?

Step 1: examine GVB orbitals for (PH3)2Pt(CH3)

Analysis of GVB wavefunction

Alternative models for Pt centers

Not agree with experiment energetics Not agree with experiment

Possible explanation: kinetics

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

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

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

How estimate the size of barriers (without calculations)

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

Examine CC coupling at transition state Can orient the CH3 to obtain good overlap with Pd sd hybrid OR can orient the CH3 to obtain get good overlap with the other CH3 But CANNOT DO BOTH SIMULTANEOUSLY, thus do NOT get resonance stabilization of TS  high barier

Examine CH coupling at transition state H can overlap both CH3 and Pd sd hybrid simultaneously but CH3 cannot thus get ~ ½ resonance stabilization of TS

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

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