Lecture
This presentation is the property of its rightful owner.
Sponsored Links
1 / 36

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


  • 91 Views
  • Uploaded on
  • Presentation posted in: General

Lecture 15 February 15, 2013 Transition metals. 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.

Download Presentation

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

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


William a goddard iii wag wag caltech edu 316 beckman institute x3093

Lecture 15 February 15, 2013

Transition metals

Ch120a-Goddard-L01

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, [email protected]

316 Beckman Institute, x3093

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

Teaching Assistants: Ross Fu <[email protected]>;

Fan Liu <[email protected]>


Transition metals

Transition metals

(4s,3d) Sc---Cu

(5s,4d) Y-- Ag

(6s,5d) (La or Lu), Ce-Au


Ground states of neutral atoms

Ground states of neutral atoms


Hemoglobin

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 (a1, a2, b1, b2) 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

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 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 fe 2 d orbitals

Energies of the 5 Fe2+ d orbitals

x2-y2

z2=2z2-x2-y2

yz

xz

xy


Exchange stabilizations

Exchange stabilizations


Ferrous fe ii

Ferrous FeII

y

x

x2-y2 destabilized by heme N lone pairs

z2 destabilized by 5th ligand imidazole or 6th ligand CO


Four coordinate fe heme high spin case s 2 or q

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


Four coordinate fe heme intermediate spin s 1 or t

Four coordinate Fe-Heme – Intermediate spin, S=1 or t


Four coordinate fe heme low spin case s 0 or s

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


Out of plane motion of fe 4 coordinate

Out of plane motion of Fe – 4 coordinate


Add axial base

Add axial base

N-N Nonbonded interactions push Fe out of plane

is antibonding


Summary 4 coord and 5 coord states

Summary 4 coord and 5 coord states


Free atom to 4 coord to 5 coord

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 o 2 with o to form ozone

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 o 2 to mb

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 o 2 to mb

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 o 2 to mb

Bonding O2 to Mb

Exchange loss on bonding O2


Modified exchange energy for q state

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

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

compare bonding of CO and O2 to Mb


Gvb orbitals for bonds to ti

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 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

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

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

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 orbitals1

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

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 like1

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 2 vh 3

Example (Cl)2VH3

+ resonance configuration


Example clmo metallacycle butadiene

Example ClMo-metallacycle butadiene


Example mn ch 2

Example [Mn≡CH]2+


William a goddard iii wag wag caltech edu 316 beckman institute x3093

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


William a goddard iii wag wag caltech edu 316 beckman institute x3093

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


  • Login