Lecture 15, February 20, 2015 Transition metals – … (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

© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 Ch125a-Goddard-

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Elements of Quantum Chemistry with Applications to Chemical Bonding and

Properties of Molecules and Solids

William A. Goddard, III, [email protected] Beckman Institute, x3093

Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics,

California Institute of Technology

Special Instructor: Julius Su <[email protected]>Teaching Assistants: Hai Xiao <[email protected]>

Mark Fornace <[email protected]>

Lecture 15, February 20, 2015Transition metals – Heme-Fe

Course number: Ch125a; Room 115 BIHours: 11-11:50am Monday, Wednesday, Friday

© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 2

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


Ground states of neutral atoms

Sc (4s)2(3d)1Ti (4s)2(3d)2V (4s)2(3d)3Cr (4s)1(3d)5Mn (4s)2(3d)5Fe (4s)2(3d)6Co (4s)2(3d)7Ni (4s)2(3d)8Cu (4s)1(3d)10

Sc++ (3d)1Ti ++ (3d)2V ++ (3d)3 Cr ++ (3d)4Mn ++ (3d)5Fe ++ (3d)6 Co ++ (3d)7 Ni ++ (3d)8 Cu++ (3d)10


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 () each with one heme subunit

Each subunit resembles myoglobin (Mb) which has one heme

Hb

Mb


The action is at the heme or Fe-Porphyrin molecule

Essentially all action occurs at the heme, which is basically anFe-Porphyrin molecule

The rest of the Mb serves mainly to provide a hydrophobic envirornment at the Fe and to protect the heme


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 orbital pointing at it.


Energies of the 5 Fe2+ d orbitals

x2-y2

z2=2z2-x2-y2

xy

xz

yz


Exchange stabilizations

© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15

Skip energy stuff

9


Consider the product wavefunctionΨ(1,2) = ψa(1) ψb(2) And the HamiltonianH(1,2) = h(1) + h(2) +1/r12 + 1/RIn the details slides next, we deriveE = < 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 a(1)=| ψa(1)|2 and b(2)=| ψb(2)|2

Since the integrand a(1) b(2)/r12 is positive for all positions of 1 and 2, the integral is positive, Jab > 0

Energy for 2 electron product wavefunction

SKIP for now


Details in deriving energy: normalization

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 normalizedHere 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

SKIP for now


Using H(1,2) = h(1) + h(2) +1/r12 + 1/RWe partition the energy E = <Ψ| H|Ψ> asE = <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 isE = haa + hbb + Jab + 1/R

Details of deriving energy: one electron terms

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 partsEex=-< 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>=0Thus all are zeroThus 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.

Summarizing, the energy of the Aψaψb wavefunction for H2 isE = haa + hbb + (Jab –Kab) + 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 positiveEee =∫ (d3r1)((d3r2)|Ψ(1,2)|2/r12 > 0This follows since the integrand is positive for all positions of r1and r2 thenWe derived that the energy of the A ψa ψb wavefunction isE = haa + hbb + (Jab –Kab) + 1/R Where the Eee = (Jab –Kab) > 0Since we have already established that Jab > 0 we can conclude thatJab > Kab > 0

SKIP for now


Separate the spinorbital into orbital and spin parts

Since the Hamiltonian does not contain spin the spinorbitals canbe factored into spatial and spin terms. For 2 electrons there are two possibilities:Both electrons have the same spinψa(1)ψb(2)=[Φa(1)(1)][Φb(2)(2)]= [Φa(1)Φb(2)][(1)(2)]So that the antisymmetrized wavefunction isAψa(1)ψb(2)= A[Φa(1)Φb(2)][(1)(2)]==[Φa(1)Φb(2)- Φb(1)Φa(2)][(1)(2)]Also, similar results for both spins downAψa(1)ψb(2)= A[Φa(1)Φb(2)][(1)(2)]==[Φa(1)Φb(2)- Φb(1)Φa(2)][(1)(2)]

Since <ψa|ψb>= 0 = < Φa| Φb><|> = < Φa| Φb>We see that the spatial orbitals for same spin must be orthogonal


Energy for 2 electrons with same spinThe total energy becomesE = 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)>

We derived the exchange term for spin orbitals with same spin asfollowsKab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)> `````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><(1)|(1)><(2)|(2)>

≡ Kabwhere Kab ≡ <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)>Involves only spatial coordinates.

SKIP for now


Now consider the exchange term for spin orbitals with opposite spinKab ≡ <ψa(1)ψb(2) |1/r12 |ψb(1)ψa(2)> `````= <Φa(1)Φb(2) |1/r12 |Φb(1)Φa(2)><(1)|(1)><(2)|(2)>

= 0Since <(1)|(1)> = 0.

Energy for 2 electrons with opposite spin

Thus the total energy isE = haa + hbb + Jab + 1/R With no exchange term unless the spins are the same

Since <ψa|ψb>= 0 = < Φa| Φb><|> There is no orthogonality condition of the spatial orbitals for opposite spin electronsIn general < Φa| Φb> =S, where the overlap S ≠ 0

SKIP for now


Summarizing: Energy for 2 electronsWhen the spinorbitals have the same spin, Aψa(1)ψb(2)= A[Φa(1)Φb(2)][(1)(2)]The total energy isE = haa + hbb + (Jab –Kab) + 1/R

But when the spinorbitals have the opposite spin, Aψa(1)ψb(2)= A[Φa(1)Φb(2)][(1)(2)]=The total energy isE = haa + hbb + Jab + 1/R With no exchange term

Thus exchange energies arise only for the case in which both electrons have the same spin

SKIP for now


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

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

Thus for the 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)][(1)(2)+(1)(2)] Leads directly to 3E = haa + hbb + (Jab –Kab) + 1/R Exactly the same as for [Φa(1)Φb(2)-Φb(1)Φa(2)][(1)(2)] [Φa(1)Φb(2)-Φb(1)Φa(2)][(1)(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[Φa(1)Φb(2)+Φb(1)Φa(2)][(1)(2)-(1)(2)]We will analyze the energy for this wavefunction next.

SKIP for now


Consider the energy of the singlet wavefunction

[Φa(1)Φb(2)+Φb(1)Φa(2)][(1)(2)-(1)(2)] ≡ (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)][(1)(2)-(1)(2)] ≡ (ab+ba)(-)1E = numerator/ denominator Where numerator =<(ab+ba)(-)|H|(ab+ba)(-)> =

=<(ab+ba)|H|(ab+ba)><(-)|(-)>denominator = <(ab+ba)(-)|(ab+ba)(-)>Since <(-)|(-)>= 2 <|(-)>=

2[<><<>]=2We obtainnumerator =<(ab+ba)|H|(ab+ba)> = 2 <ab|H|(ab+ba)> denominator = <(ab+ba)|(ab+ba)>=2 <ab|(ab+ba)>Thus 1E = <ab|H|(ab+ba)>/<ab|(ab+ba)>

SKIP for now


energy of the singlet wavefunction - details

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)

SKIP for now


Ferrous FeII

x

y

z2 destabilized by 5th ligand imidazole

or 6th ligand CO

x2-y2 destabilized by heme N lone pairs


Four coordinate Fe-Heme – High

spin case, S=2 or q

The 5th axial ligand will destabilize q2 since dz2 is doubly occupiedA pi acceptor would stabilize q1 wrt q2Bonding O2 to 5 coordinate will stabilize q3 wrt q1Future discuss only q1 and denote as q

Doubly occupied zx (or yz) but xy case just 9 kcal/mol higher.Presumably get some delocalization of dorbitals with porbitals of porphyrin


Four coordinate Fe-Heme – Intermediate spin, S=1 or tDoubly occupied xy and zx (or yz) but two doubly occupied dorbitals just 3 kcal/mol higher.


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


Summary 4 coord states


Out of plane motion of Fe – 4 coordinate


Add axial base

N-N Nonbonded interactions push Fe out of plane

is antibonding


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

Free atom to 4 coord to 5 coord


Bond CO to Mb

Strongly bound CO 0 electrons in dz2, stabilize S=0 or s state

H2O (and N2) do not bond strongly enough to promote the Fe to an excited state, thus get S=2 or q state

Never get S=1 or t state

Mb-H2O Mb-CO


Bonding of O2 with O to form ozone

O2 has available a porbital for a bond to a p orbital of the O atom

And the 3 electron system for a bond to a p orbital of the O atom


Bond O2 to Mb

Simple VB structures

get S=1 or triplet state

In fact MbO2 is S=0 singlet

Why?

Bond to s Bond to t Bond to q

This state should also bind to O2


change in exchange terms when Bond O2 to Mb

O2p

O2p

10 Kdd

5*4/2

7 Kdd

4*3/2 +

2*1/2

6 Kdd

3*2/2 +

3*2/2

7 Kdd

4*3/2 +

2*1/2

Assume perfect VB spin pairing. get 4 cases

up spin

down spinThus average Kdd is (10+7+7+6)/4 =7.5

Assume perfect VB spin pairing

Then get 4 cases


Bonding O2 to Mb

Exchange loss on bonding O2

S=2 S=1S=1

S=0


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 H = -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


compare bonding of CO and O2 to Mb

Documents

Lecture 15, February 20, 2015 Transition metals – … (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