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© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 Ch125a-Goddard-
1
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 3
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 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 () each with one heme subunit
Each subunit resembles myoglobin (Mb) which has one heme
Hb
Mb
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 5
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 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 sp2 orbital pointing at it.
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 7
Energies of the 5 Fe2+ d orbitals
x2-y2
z2=2z2-x2-y2
xy
xz
yz
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 8
Exchange stabilizations
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 10
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 11
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 12
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
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© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 13
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 14
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 15
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 16
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 17
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
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© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 18
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 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[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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 20
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 21
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 22
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 23
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 24
Ferrous FeII
x
y
z2 destabilized by 5th ligand imidazole
or 6th ligand CO
x2-y2 destabilized by heme N lone pairs
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 25
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 26
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.
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 27
Four coordinate Fe-Heme – Low spin case, S=0 or s
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 28
Summary 4 coord states
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 29
Out of plane motion of Fe – 4 coordinate
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 30
Add axial base
N-N Nonbonded interactions push Fe out of plane
is antibonding
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 31
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 32
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 33
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 34
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 35
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
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 36
Bonding O2 to Mb
Exchange loss on bonding O2
S=2 S=1S=1
S=0
© copyright 2015 William A. Goddard III, all rights reservedCh125-Goddard-L15 37
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