Lecture 16, November 9, 2015: Transition metals – …+ (3d)10 Sc+ (4s)1(3d)1 Ti+ (4s) (3d) V+...

© copyright 2015 William A. Goddard III, all rights reserved Ch125-Goddard-L15 Ch125a-Ch120a-

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Ch 125a: Elements of Quantum Chemistry with Applications to Chemical Bonding and Properties of

Molecules and Solids

William A. Goddard, III, [email protected] 316 Beckman Institute, x3093

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

California Institute of Technology

Teaching Assistants: Ch125a Sijia Dong <[email protected]> Ch120a Kurtis Carsch < [email protected] >

Lecture 16, November 9, 2015: Transition metals – Heme-Fe

Room 115 BI Hours: 11-11:50am Monday, Wednesday, Friday

Ch 120a:Nature of the Chemical bond

© copyright 2015 William A. Goddard III, all rights reserved Ch125-Goddard-L15

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


Ground states of atoms Occupation depends on Charge

Sc (4s)2(3d) 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

Sc+ (4s)1(3d)1

Ti+ (4s)1(3d)2

V+ (4s)0(3d)3

Cr+ (4s)0(3d)5

Mn+ (4s)1(3d)5

Fe+ (4s)1(3d)6

Co+ (4s)0(3d)7

Ni+ (4s)0(3d)8

Cu+ (4s)0(3d)10

Neutral (4s)2(3d)n-2 +1 (4s)1(3d)n-2 +2 (4s)0(3d)n-2


Ground state of M+ metals Mostly s1dn-1

Exceptions: 1st row: V--Cr, Co--Cu 2nd row: Nb--Mo, Ru--Ag 3rd row: La, Pt--Au

Strong stabilization of

d5 and d10

1st and 3rd row similar s vs. d 2nd row: very strong d stabilization


Exchange energies: compare d5 to s1d4

Get 5*4/2=10 exchange terms 4 Ksd + 6 Kdd Exchange Stabilization ~ 138 (Cr); 127 (Mo); 132.2 (W)

Ksd Kdd Mn+ 4.8 19.8 Tc+ 8.3 15.3 Re+ 11.9 14.1

kcal/mol

A[(d1α)(d2α)(d3α)(d4α)(sα)]

Cr+, Mo+, and W+: s1d4

For high spin (S=5/2) Cr+, Mo+, and W+: s0d5

For high spin (S=5/2) A[(d1α)(d2α)(d3α)(d4α)(d5α)]

Get 5*4/2=10 exchange terms 0 Ksd + 10 Kdd Exchange Stabilization ~ 198 (Cr); 153 (Mo); 141 (W)

Hund’s rule (high spin best): compare high spin S=5/2 (10 Kdd) and S=3/2 (6 Kdd) High spin wins by 4Kdd ~ 79.2 Cr, 61.2 Mo, 56.4W

Exchange stabilization of s0d5 over s1d4

60 Cr; 26 Mo; 9 W (kcal/mol)


Bond energies MH+

Cr

Mo

Re

Ag

Cu

Au

A{(d1α)(d2α)(d3α)(d4α)(sdbα)[(sdb)H+H(sdb)](αβ−βα)} sdb is α half the time and β half the time

Form bond to H, must lose half the exchange stabilization for the orbital bonded to the H


Change in atomic Exchange energies upon bonding

Get 6*5/2=15 exchange terms 5Ksd + 10 Kdd Responsible for Hund’s rule Ksd Kdd Mn+ 4.8 19.8 Tc+ 8.3 15.3 Re+ 11.9 14.1

kcal/mol

Form bond to H, must lose half the exchange stabilization for the orbital bonded to the H

A[(d1α)(d2α)(d3α)(d4α)(d5α)(sα)]

Mn+: s1d5

For high spin (S=3)

A{(d1α)(d2α)(d3α)(d4α)(sdbα)[(sdb)H+H(sdb)](αβ−βα)} sdb is α half the time and β half the time


Size of atomic orbitals, M+

Valence s similar for all three rows, 5s biggest

Big decrease 6s from La(Z=57) to Hf(Z=72)

Valence d very small for 3d


Charge transfer in MH+ bonds electropositive

Electronegative

Ru,Rh,Pd

Pt,Au,Hg

1st row all electropositive

2nd row: Ru,Rh,Pd

electronegative 3rd row:

Pt, Au, Hg electronegative


Can we have concerted 2s + 2s reactions for transition metals?

2s + 2s forbidden for organics

X

Cl2Ti Cl2Ti Cl2Ti ? ? 2s + 2s forbidden for organometallics?

Cl2Ti Cl2Ti Cl2Ti Me

Me

Me

Me

Me

Me


Physics behind Woodward-Hoffmann Rules

For a reaction to be allowed, the number of bonds must be conserved. Consider H2 + D2

2 bonds TS ? bonds 2 bonds

Bonding 2 elect

nonbonding 1 elect

antibonding 0 elect

Have 3 electrons, 3 MO’s

Have 1 bond. Next consider 4th atom, can we get 2 bonds?

To be allowed must have 2 bonds at TS How assess number of bonds at the TS. What do the dots mean? Consider first the fragment


Now consider a TM case: Cl2TiH+ + D2

Orbitals of reactants

GVB orbitals of TiH bond for Cl2TiH+

GVB orbitals of D2


GVB orbitals at the TS for Cl2TiH+ + D2 Cl2TiD+ + HD


GVB orbitals for the Cl2TiD+ + HD product Note get phase change for both orbitals


Follow the D2 bond as it

evolves into the HD bond


Follow the TiH bond as it

evolves into the TiD bond


Is Cl2TiH+ + D2 Cl2TiD+ + HD allowed?

Bonding 2 elect

nonbonding 1 elect

antibonding 0 elect

when add Ti 4th atom, can we get 2 bonds?

Answer: YES. The two orbitals can have high overlap at the TS orthogonal in the TS, thus the reaction is allowed

Now referred to a sigma metathesis instead of 2s + 2s

Now the bonding orbital on Ti is d-like. Thus at TS have


Barriers small, thus allowed

Increased d character in bond smaller barrier

2s + 2s Reactions at Transition Metals: I. The Reactions of D2 with Cl2TiH+, Cl2TiH, and Cl2ScH M. L. Steigerwald and W. A. Goddard III; J. Am. Chem. Soc. 106, 308 (1984)

Now referred to a sigma metathesis instead of 2s + 2s


Are all MH reactions with D2 allowed? No

Example: ClMn-H + D2

Here the Mn-Cl bond is very polar

Mn(4s-4pz) lobe orbital with Cl:3pz

This leaves the Mn: (3d)5(4s+4pz), S=3 state to bond to the H But spin pairing to a d orbital would lose

4*Kdd/2+Ksd/2= (40+2.5) = 42.5 kcal/mol

whereas bonding to the (4s+4pz) orbital loses

5*Ksd/2 = 12.5 kcal/mol

As a result the H bonds to (4s+4pz), leaving a high spin d5.

Now the exchange reaction is forbidden


Thus ClMn-H bond is sp-like 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

This cannot overlap the antisymmetric orbital delocalized over the three H atoms in the TS As a result at the Transition state the MnH bond has the character of H3

- with both electrons on the H3.

This leads to a high barrier, ~45 kcal/mol


Show reaction for ClMnH + D2

Show example reactions


Symmetry considerations see Ch226a-1967-chapII

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The octahedral Point Group, O. The group of all proper rotations (no reflections or inversions) leaving the cube invariant (page 88)

Symmetry transformations are in

the same class if they do the same thing (can be transformed into each other by a symmetry

transformation). Classes for

group O


Irreducible Representation page 24

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Representation: a set of functions {φi : i=1,,d} that are transformed into each other by the various symmetries of a group {Rk: k=1…g} where g is the order of the group Irreducible Representation: the minimum sets of functions that must transform into each other under the symmetry transformations These states are necessarily degenerate Thus if H φi = εi φi Then RkH φi = H [Rkφi]=Rk (εi φi)= εi [ Rk φi] H [Rkφi]=εi [ Rk φi] Thus [Rkφi] is an eigenfunction of H with the same eigenvalue


Transformations of d orbitals page 95

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Consider the five d orbitals of an atom and the O point group The 3 functions xy, yz, and xz form an irreducible representation Here Rk takes each function into itself (3x3 diagonal matrix with trace 3 C4z takes xy to –xy; yz into –xz and xz into +yz leading to a 3x3 matrix with trace -1 C4z

2 takes xy to +xy; yz into –yz and xz into -xz leading to a 3x3 matrix with trace +1 C3 about the 3 fold diagonal takes xy into yz, yz into zx and zx into xy leading to a 3x3 matrix with trace 0 C2 about the 2 fold diagonal in the [011] direction takes yz to yz but xz to yz and yz to –zx leading to a 3x3 matrix with trace 0

summarizing


Character Table for O, page 90

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The traces of the transformation matrices for the irreducible representations of O

T2

Similarly is E


Splitting of d atomic levels in octahedral environment

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Example: (FeF6)4- in an octahedral environment, Oh Have Fe2+ d6 for the atom this has high spin S=2 with one doubly occupied d orbital and the others all up spin But with 6 F- at the corners of an octahedron the eg orbitals are destabilized. For small destabilization (weak field) the high spin wins (t2g)4(eg)2 with S=2 but for very strong fields the low spin wins (t2g)6(eg)0 with S=0 However intermediate spin (t2g)5(eg)1 with S=1 never wins

Oh include inversion

2 fold degenerate

3 fold degenerate

Eg

T2g

Convention: many-electron states upper case, orbitals lower case


Fe2+ in octahedral field

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yz xy

x2-y2 2z2-x2-y2

atom

high spin S=2

Intermediate spin S=1

Low spin S=0

t2g

e2g

xz

∆

∆ << Kdd ∆ ~ Kdd ∆ >> Kdd

For S=2 and S=0 the total wavefunction symmetry is octahedral (nondegenerate) For S=1 the molecule distorts (Jahn-Teller coupling)


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 (α1, α2, β1, β2) 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 an Fe-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 σ orbital pointing at it.

y

y x


The symmetry group of the heme is D4h

33

D4h add inversion

xy

xz, yz

x2-y2 2z2 -x2-y2

xz, yz

xy

x2-y2

2z2 -x2-y2

atom

heme

high spin S=2

Int. spin S=1

Low spin S=0

t2g

e2g


Energies of the 5 Fe2+ d orbitals

x2-y2

z2=2z2-x2-y2

xy

xz

yz


Exchange stabilizations


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 occupied A pi acceptor would stabilize q1 wrt q2 Bonding O2 to 5 coordinate will stabilize q3 wrt q1 Future 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 dπ orbitals with pπ orbitals of porphyrin


Four coordinate Fe-Heme – Intermediate spin, S=1 or t Doubly occupied xy and zx (or yz) but two doubly occupied dπ orbitals 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 pσ orbital for a σ bond to a pσ orbital of the O atom

And the 3 electron π system for a π bond to a pπ 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

O2pσ

O2pπ

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 spin Thus 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=1 S=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

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Lecture 16, November 9, 2015: Transition metals – …+ (3d)10 Sc+ (4s)1(3d)1 Ti+ (4s) (3d) V+...

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