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

Ch120a-Goddard-L01 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>

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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

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.

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 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 Aψa 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)]= The total energy is 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 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 y x x2-y2 destabilized by heme N lone pairs z2 destabilized by 5th ligand imidazole or 6th ligand CO

Summary 4 coord and 5 coord states

Out of plane motion of Fe – 4 coordinate

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

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 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

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 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

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 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

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? (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

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

Compare chemistry of column 10

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