Chemistry 431Lecture 28

Introduction to electrostaticsCharge and dipole moment

Polarizabilityvan der Waals forces

Review of transition dipole

NC State University

Review of ElectrostaticsThe Coulombic force on charge j due to charge i is:

The Coulombic force is additive. The combined force isa superposition. The force on charge k due to a numberof charges with the index j is:

The constant 0 is the permittivity of vacuum. In MKS unitsthe value is 0 = 8.854 x 10-12 C2 N-1 m-2. In the cgs-esuunit system the permittivity of free space is 1/4 and theconstant 1/40 does not appear in the Coulomb force.

Fj = 140qiqjr ij2

r ij

Fk = 140qjqkr jk2

j k

r jk

Electric FieldThe electric field is is the force per unit charge. The mostprecise statement is that it is the force per unit charge inthe limit that the charge is infinitesimally small:

When applied to the Coulomb force the electric field becomes:

E j =Fjq

Ek = 140qj

r jkr jk2

j k

Electrostatic Potential

The electric field is the negative gradient of the scalarpotential:

The potential at a distance r from a charge is:

The electric field represents the force per unit charge.The potential is the work per unit charge.

In MKS units the potential has units of V where 1 V = 1 J/C.

E =

= q40r

W12 = (q2 q1)

Definition of a dipole momentA dipole is defined as a charge displaced through a distance.It is a vector quantity, i.e. it has direction:

The units of dipole are Cm as well as Debye.1 Debye = 3.33 x 10-30 Cm

One can also use units electron-Angstroms.4.8 Debye = 1 eA

The quantum mechanical definition of a dipole moment isExpectation value of the dipole operator er.

= q(d2 d1)

= e r drall space

Potential and Field due to a DipoleThe potential due to a dipole is:

The assumption in this equation is that the distance betweenthe charge and dipole, r, is large relative to the separation ofcharges in the dipole, d, r >> d.The electric field due to a dipole is:

Using the expression W = -.E we can calculate the interactionenergy of two dipoles.

(r) = r40r 3

E = 1403( r)r

r 5

r 3

W = 1401 2

r 3 3(1 r)(2 r)

r 5

Example: Effect of Dipole OrientationConsider two dipoles, which have the orientations below thatwe can call aligned

and head-to-tail

Aligned: 1. 2 = 2 , 1. r = r , 2. r = r , W = -22/40r3Head-to-tail: 1. 2 = -2 , 1. r = 0 , , 2. r = 0 , W = -2/40r3

Interaction of electric moments with the electric field

The interaction of a collection of charges subjected to anelectric field is given by:

The picture is that of a charge interacting with the potential,the dipole interacting with a field, etc.An electric field can exert a force:

or a torque:

on a collection of charges.

W = q E + ...

F = qiE(r i)i

T = r i qiE(r i)i

PolarizabilityIn the presence of an externally applied electric field the eipole moment of the molecule can also be expressedas an expansion in terms of moments:

The leading term in this expansion is the permanent dipolemoment, 0. The polarizability is a tensor whose componentscan be described as a follows:

Where the 0 subscript refers to the fact that the derivative isEvaluated at zero field. The tensor is called the hyper-polarizability and is third ranked tensor.

= 0 E + 12 :EE + ...

xy =xEy 0

Polarizability as second rank tensorThe dipole moment components each can depend on as many as three different polarizability components asdescribed by the matrix:

If a molecule has a center of symmetry (e.g. CCl4) then The polarizability is a scalar (i.e. the induced dipole momentIs always in the direction of the applied field). However, fornon-centrosymmetric molecules components can be inducedin other directions. The directions are often determined bythe directions of chemical bonds, which may not be alignedwith the field. This is the significance of the tensor.

xyz

=xx xy xzyx yy yzzx zy zz

ExEyEz

Properties of the polarizability tensorLike the quadrupole moment, the polarizability can bemade diagonal in the principle axes of the molecule.

In the laboratory frame of reference the polarizabilitydepends on the orientation of the molecule.

The average polarizability is independent of orientation.It is given by the Trace, which is written Tr .

The polarizability increases with the number of electrons inThe molecule or with the volume of the charge distribution.

Classically, for a molecule of radius a, = 40a3.

Tr = 13 xx + yy + zz

Van der Waals Forces

1/r6 Interactions1. Keesom - permanent dipole/permanent dipole

2. Debye - permanent dipole/induced dipole

3. London - induced dipole/induced dipole

0,122 + 0,212

402r6

231222

402kTr6

3h212

1 + 2

0,10,240

2r6

Dipolar InteractionsThe field around a dipole can be resolved into twocomponents as shown in the Figure. The components are:

E || = 2140

cos r3

E = 1

40sin

r3

m1

E||

E^The total field is:

E = 1401r3 1 + 3cos

2

The Debye TermA permanent dipole on molecule 1 will induce a dipole moment on molecule 2.

The total energy of the second dipole is:

Substituting for E and averaging over all orientationsyields:

A similar equation can be derived for F1 to yield the Debye equation.

2 = 2E

2 = 122E2

2 = 222

402r6

The Keesom TermThe Keesom term arises from the interaction of twopermanent dipoles. Here we consider the Debyeterm for the polarizability of a polar solvent.

Using a similar reasoning applied to the Debye term we can substitute in m2/3kT for a to obtain the Keesom term.

P = NA300 +

23kT

231222

402kTr6

London InteractionsNo permanent dipole is required for London forcesto apply. The London equation for the attractionbetween two particles represents a quantum mechanical effect. The derivation uses a harmonic oscillator model. Consider a dipole-dipole interaction:

since the definition of a dipole is:

22

40r3=

2 e 1 e 240r3

1 = e 1

Harmonic oscillator modelConsider the electrons in a material as a harmonicoscillator. The nuclei represent the restoring force.The potential energy is given by:

in which:

Combining these various contributions we have:

= K 12

2

K = e2

0

T = K2 12 + 2

2 2 e 1 e 2

40r3

Harmonic oscillator modelThe energies of the harmonic oscillator are:

in which:

An illustration of the two induced dipoles for the London interaction is shown below:

E = n1 + 1/2 h1 + n2 + 1/2 h2

1 = 1 20

40r3, 2 = 1 +

2040r3

+ -l1

+ -l2

r

Harmonic stabilization energyTaking the lowest energy harmonic oscillator state:

Two independent oscillators in their ground state have energy:

The difference is the contribution of dispersion forces to the interaction energy:

E = h2 1 + 2

E = h2 2

= h2 1 + 2 2

The London termPlugging in the frequencies obtained above and solving yields:

for identical molecules or:

two different types of molecule 1 and 2.

= h02

2 402r6

= 3h212

1 + 2

0,10,240

2r6

The van der Waals parameter The van der Waals potential is the sum of the three terms derived:

In this case it is derived for a pair of identical molecules of type 1. Thus, the parameter is an interaction parameter for molecules of type 1 thatincludes Keeson, Debye and London terms.

= 140

2 20,112 +2143kT +

34h0,1

2 1r6 =

11r6

Protein folding energetics

Non-covalent forces in proteins

What holds them together? Hydrogen bonds Salt-bridges Dipole-dipole interactions Hydrophobic effect Van der Waals forces

What pulls them apart? Conformational Entropy

Dipole-Dipole Interactions

Dipoles often line up in this manner.Example: -helix

Electrostatic InteractionsCoulombs Law: V = q1q2/r

Example of a Salt Bridge

Example of a hydrogen bond -N-H..O=C-

Main Chain Main Chain

Lysine Glutamate

Hydrogen bonding in water

Hydrophobic interactions

Contributions to G

-TS

-TS

H

G

InternalInteractions

ConformationalEntropy

HydrophobicEffect

Net:

Folding

- 0 +

Review of transitions

The Fermi Golden Rule can be used to calculate many types of transitions

Transition H(t) dependence1. Optical transitions Electric field2. NMR transitions Magnetic field3. Electron transfer Non-adiabaticity4. Energy transfer Dipole-dipole5. Atom transfer Non-adiabaticity6. Internal conversion Non-adiabaticity7. Intersystem crossing Spin-orbit coupling

Optical electromagnetic radiation permits transitions among electronic states

t = E twhere is the dipole operator and the dotrepresents the dot product. If the dipole isaligned with the electric vector E(t) then H(t) = - E(t). If they are perpendicular thenH(t) = 0.

= erwhere e is the charge on an electron and r isthe distance.

The time-dependent perturbation has the form of an time-varying electric field

E t = E0cos twhere is the angular frequency. The electric field oscillation drives a polarization in an atom or molecule. A polarization is a coherent oscillation between two electronic states. The symmetry of the states must be correct in order for the polarization to be created.The orientation average and time average over the square of the field is [-.E(t)]2 is 2E02/6.

Absorption of visible or ultraviolet radiation leads to electronic transitions

s s

Polarizationof

Radiation

Absorption of visible or ultraviolet radiation leads to electronic transitions

s s

Transitionmoment

The change innodal structurealso implies a change in orbitalangular momentum.

The interaction of electromagnetic radiation with a transition moment

The electromagnetic wave has an angular momentum of 1. Therefore, an atom ormolecule must have a change of 1 in itsorbital angular momentum to conserve thisquantity. This can be seen for hydrogen atom:

l = 0l = 1

Electric vectorof radiation

The Fermi Golden Rule for optical electronic transitions

k 12 = e 1|q|2

2E0

2

6h2 12The rate constant is proportional to the squareof the matrix element e< 1|q| 2> times a delta function. The delta function is an energymatching function:( - 12) = 1 if = 12( - 12) = 0 if 12.

A propagating wave of electromagnetic radiationof wavelength l has an oscillating electric dipole, E

(magnetic dipole not shown)

E

The oscillating electric dipole, E, can induce an oscillating dipole in a molecule as the radiation

passes through the sample

E

The oscillating electric dipole, E, can induce an oscillating dipole in a molecule as the radiation passes through the

sample

l

The type of induced oscillating dipole depends on .If corresponds to a vibrational energy gap, then radiation will be absorbed, and a molecular vibrational transition will result

v=0

v=1

E = hc/

OR

R

OR

R

The oscillating electric dipole, E, can induce an oscillating dipole in a molecule as the radiation passes through the

sample

If corresponds to a electronic energy gap, then radiation will be absorbed, and an electron will be promoted to an unfilled MO

HOMO

LUMO

E = hc/

The absorption of light by molecules is is subject to several selection rules. From a group theory perspective, the basis of these selection rules is that the transition between two states a and b is electric dipole allowed if the electric dipole moment matrix element is non-zero, i.e.,

a b = a*

b d 0

where = x + y + z is the electric dipole moment operator which transforms in the same manner as the p-

orbitalsab= a(x + y + z)b, must contain the totally

symmetric irrepor put another way,

ab must transform as any one of x, y, z

Direct Products: The representation of the product of two representations is given by the product of the

characters of the two representations.

Verify that under C2v symmetry A2 B1 = B2

1-1-11A2 B1

-11-11B1

-1-111A2

'vvC2EC2v

As can be seen above, the characters of A2 B1 are those of the B2 irrep.

Verify that A2 B2 = B1, B2 B1= A2

Also verify that the product of any non degenerate representation with itself is totally symmetric and the product of any representation with the totally symmetric representation yields the original representation

Note that,

A x B = B; while A x A = B x B = Ag x u = u; while g x g = u x u =g.

Light can be depicted as mutually orthogonal oscillating electric and magnetic dipoles. In infrared and electronic absorption spectroscopies, light is said to be absorbed when the oscillating electric field component of light induces an electric dipole in a molecule.

For a hydrogen atom, we can view the electromagneticradiation as mixing the 1s and 2p orbitals transiently.

l = 0l = 1

Electric vectorof radiation

Is the orbital transition dyz px electric dipole allowed in C2v symmetry?

b1 b1b2a1

b2 =a1a2b1

b2 =b2b1a2

None of the three components contains the a1representation, so this transition is electric dipole

forbidden

A transition between two non-degenerate states will be allowed only if the direct product of the two state

symmetries is the same irrep as one of the components of the electric dipole

px dyz

How about an a1b2 orbital transition?

Problem Indicate whether each of the following metal localized transitions is electric dipole allowed in PtCl42-.

(a) dxy pz (b) dyz dz2 (c) dx2-y2 px,py (d) pz s

b2 b1b2a1

a1 =a2a1b2

a1 =a2a1b2

Since my makes the transition allowed, the transition is said to be "y-allowed" or "y-polarized"

Remember the shortcut: a1b2 = b2 which transforms as y

Example: Myoglobin/Hemoglobin

Heme spectroscopy Transition moment

Franck-Condon activeVibronic coupling

Myoglobin StructureGlobular -helical protein

Heme

A helixF helix

B helixB helix

E helix

G helix

H helix

The iron in heme is the binding site for oxygen and other diatomics

Heme is iron proto-porphyrin IX. Fe isfound in Fe2+ and Fe3+oxidation states.Diatomics bind to Fe2+.Examples, CO, NO, O2.O2 is the physiologicallyrelevant ligand, but it canoxidize iron and it is difficultto study directly.

N N

NN

Fe

O O-O O-

O|||C

The porphine ring is an aromatic ring that has a fourfold symmetry axis

The ring and metal can beconsidered separately.The ring has been succesfullymodeled using the Goutermanfour orbital model.In globins the iron is Fe(II)and can be either high spin or low spin.MbCO ------ low spinDeoxy Mb - high spin

N N

NN

The four orbital model is used to represent the highest occupied and lowest unoccupied

molecular orbitals of porphyrins

eg

a1u a2u

The two highest occupiedorbitals (a1u,a2u) are nearly equal in energy. The egorbitals are equal in energy.Transitions occur from:a1u eg and a2u eg.

M1

The transitions from ground state orbitalsa1u and a2u to excited state * orbitals eg

can mix by configuration interaction

eg

a1u a2u

Both excited state configurationsare Eu so they can mix.Two electronic transitionsare observed. One is verystrong (B or Soret) and the other is weak (Q).The transition moments are:B band Rs0 = M1 + M2Q band rs0 = M1 - M2 0

M1 M2

Porphine orbitals

eg eg

a1u a2u

Four orbital model of metalloporphyrin spectra

|By0 = 12 a2uegy + a1uegx

|Qy0 = 12 a2uegy a1uegx

|Bx0 = 12 a2uegx + a1uegy

|Qx0 = 12 a2uegx a1uegy

There are four excited state configurations possible inD4h symmetry. These are denoted B (strong) and Q(weak).

The transition moment for absorption

eEB EG h iB

The absorption probability amplitude for a Franck-Condon active transition is:

Here i and f represent individual vibrationallevels in each electronic manifold. The polarization can be = x, y, or z.Remember that the ground state is totallysymmetric (filled shell molecules). Here itis A1g.

The absorption probability amplitude for a Franck-Condon active transition is:

Character table for D4h point group

(x, y)0020-200-202Eu

-11-11-11-11-11B2u

1-1-11-1-111-11B1u

z11-1-1-1-1-1111A2u

-1-1-1-1-111111A1u

(xz, yz)(Rx, Ry)00-20200-202Eg

xy1-11-111-11-11B2g

x2-y2-111-11-111-11B1g

Rz-1-1111-1-1111A2g

x2+y2, z21111111111A1g

quadraticlinears,rotations2d2vh2S4i2C''22C'2C22C4(z)E

The absorption probability amplitude for a Franck-Condon active transition is:

Character table for D4h point group

(x, y)0020-200-202Eu

-11-11-11-11-11B2u

1-1-11-1-111-11B1u

z11-1-1-1-1-1111A2u

-1-1-1-1-111111A1u

(xz, yz)(Rx, Ry)00-20200-202Eg

xy1-11-111-11-11B2g

x2-y2-111-11-111-11B1g

Rz-1-1111-1-1111A2g

x2+y2, z21111111111A1g

quadraticlinears,rotations2d2vh2S4i2C''22C'2C22C4(z)E

The absorption probability amplitude for a Franck-Condon active transition is:

Character table for D4h point group

(x, y)0020-200-202Eu

-11-11-11-11-11B2u

1-1-11-1-111-11B1u

z11-1-1-1-1-1111A2u

-1-1-1-1-111111A1u

(xz, yz)(Rx, Ry)00-20200-202Eg

xy1-11-111-11-11B2g

x2-y2-111-11-111-11B1g

Rz-1-1111-1-1111A2g

x2+y2, z21111111111A1g

quadraticlinears,rotations2d2vh2S4i2C''22C'2C22C4(z)E

The absorption probability amplitude for a Franck-Condon active transition is:

Character table for D4h point group

(x, y)0020-200-202Eu

-11-11-11-11-11B2u

1-1-11-1-111-11B1u

z11-1-1-1-1-1111A2u

-1-1-1-1-111111A1u

(xz, yz)(Rx, Ry)00-20200-202Eg

xy1-11-111-11-11B2g

x2-y2-111-11-111-11B1g

Rz-1-1111-1-1111A2g

x2+y2, z21111111111A1g

quadraticlinears,rotations2d2vh2S4i2C''22C'2C22C4(z)E

Mixing of the excited state configurations

There are two transitions that both have Eu symmetry. Thus, they can add constructively and destructively.

|By0 = 12 a2uegy + a1uegx

|Qy0 = 12 a2uegy a1uegx

|Bx0 = 12 a2uegx + a1uegy

|Qx0 = 12 a2uegx a1uegy

Constructive (allowed)

Destructive (forbidden)

Constructive (allowed)

Destructive (forbidden)

Vibrational modes that couple the states are determined by the

direct productcoupling = Eu Eu which we can determine from the character table.

This reducible representation can be decomposed intofour irreps: A1g + A2g + B1g + B2g

0040400404c

(x, y)0020-200-202Eu

quadraticlinears,rotations2d2vh2S4i2C''22C'2C22C4(z)E

Magnetic Circular Dichroism

The Perimeter Model

The porphine ring has D4h symmetry.The aromatic ring has 18 electrons.The p system approximates circular electron path.

N N

NN

m = 01

2

3

4

5

-1-2

-3

-4

-5

= 12

eim

m=1m=9

MCD spectra

Franzen, JPC Accepted

= A1 f

+ B0 +C0k BT

f

m

= A1B

D0

f

f

Lz = 2A1D0

MbCO MCD spectra follow the PM

The spectra are A-term MCD as shown by the derivatives of the absorption spectrum (red). The (Q MCD) = 9 x (B MCD).

B Q

Franzen, JPC Accepted

Deoxy MCD spectra are anomalous

C-term 4 time larger than MbCO!

B Q

A-term but with vibronic structure

MCD spectra: Vibronic couplingin the Perimeter Model

Franzen, JPC Accepted

MCD spectra: Vibronic couplingin the Perimeter Model

Metal Porphyrin Vibronic Distortions