Download Dipole moments and Review

Chemistry 431
Lecture 28
Introduction to electrostatics
Charge and dipole moment
Polarizability
van der Waal’s forces
Review of transition dipole
NC State University
Review of Electrostatics
The Coulombic force on charge j due to charge i is:
q iq j
1
Fj =
r ij
2
4πε 0 r ij
The Coulombic force is additive. The combined force is
a superposition. The force on charge k due to a number
of charges with the index j is:
q j qk
1
Fk =
r jk
Σ
2
4πε 0 j ≠ k r jk
The constant ε0 is the permittivity of vacuum. In MKS units
the value is ε0 = 8.854 x 10-12 C2 N-1 m-2. In the cgs-esu
unit system the permittivity of free space is 1/4π and the
constant 1/4πε0 does not appear in the Coulomb force.
Electric Field
The electric field is is the force per unit charge. The most
precise statement is that it is the force per unit charge in
the limit that the charge is infinitesimally small:
∂Fj
Ej =
∂q
When applied to the Coulomb force the electric field
becomes:
r jk
1
Ek =
qj 2
Σ
4πε 0 j ≠ k r jk
Electrostatic Potential
The electric field is the negative gradient of the scalar
potential:
E = – ∇φ
The potential at a distance r from a charge is:
q
φ= –
4πε 0r
The electric field represents the force per unit charge.
The potential is the work per unit charge.
W12 = φ(q2 – q1)
In MKS units the potential has units of V where 1 V = 1 J/C.
Definition of a dipole moment
A dipole is defined as a charge displaced through a distance.
It is a vector quantity, i.e. it has direction:
μ = q(d2 – d1)
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 is
Expectation value of the dipole operator er.
μ=e
Ψ r Ψdr
all space
Potential and Field due to a Dipole
The potential due to a dipole is:
μ⋅r
φ(r) =
4πε 0r 3
The assumption in this equation is that the distance between
the charge and dipole, r, is large relative to the separation of
charges in the dipole, d, r >> d.
The electric field due to a dipole is:
3(μ ⋅ r)r μ
1
E=
– 3
5
4πε 0
r
r
Using the expression W = -μ.E we can calculate the interaction
energy of two dipoles.
W=
1 μ 1 ⋅μ 2 – 3(μ 1 ⋅ r)(μ 2 ⋅ r)
4πε 0 r 3
r5
Example: Effect of Dipole Orientation
Consider two dipoles, which have the orientations below that
we can call aligned
and head-to-tail
Aligned:
μ1. μ2 = μ2 , μ1. r = μ r , μ2. r = μ r , W = -2μ2/4πε0r3
Head-to-tail: μ1. μ2 = -μ2 , μ1. r = 0 , , μ2. r = 0 , W = -μ2/4πε0r3
Interaction of electric moments
with the electric field
The interaction of a collection of charges subjected to an
electric field is given by:
W = qφ – μ⋅E + ...
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:
F =Σ qiE(r i)
i
or a torque:
T =Σ r i × qiE(r i)
i
on a collection of charges.
Polarizability
In the presence of an externally applied electric field the
eipole moment of the molecule can also be expressed
as an expansion in terms of moments:
μ = μ 0 – α⋅E + 1 β :EE + ...
2
The leading term in this expansion is the permanent dipole
moment, μ0. The polarizability is a tensor whose components
can be described as a follows:
∂μ x
α xy =
∂Ey
0
Where the 0 subscript refers to the fact that the derivative is
Evaluated at zero field. The β tensor is called the hyperpolarizability and is third ranked tensor.
Polarizability as second rank tensor
The dipole moment components each can depend on
as many as three different polarizability components as
described by the matrix:
α xx α xy α xz
μx
μ y = α yx α yy α yz
μz
α zx α zy α zz
Ex
Ey
Ez
If a molecule has a center of symmetry (e.g. CCl4) then
The polarizability is a scalar (i.e. the induced dipole moment
Is always in the direction of the applied field). However, for
non-centrosymmetric molecules components can be induced
in other directions. The directions are often determined by
the directions of chemical bonds, which may not be aligned
with the field. This is the significance of the tensor.
Properties of the polarizability tensor
Like the quadrupole moment, the polarizability can be
made diagonal in the principle axes of the molecule.
In the laboratory frame of reference the polarizability
depends on the orientation of the molecule.
The average polarizability is independent of orientation.
It is given by the Trace, which is written Tr α.
Tr α = 1 α xx + α yy + α zz
3
The polarizability increases with the number of electrons in
The molecule or with the volume of the charge distribution.
Classically, for a molecule of radius a, α = 4πε0a3.
Van der Waal’s Forces
1/r6 Interactions
1. Keesom - permanent dipole/permanent dipole
2 2
μ
2
1μ 2
–
3 4πε 0 2kTr 6
2. Debye - permanent dipole/induced dipole
–
α 0,1μ 22 + α 0,2μ 21
2 6
4πε 0 r
3. London - induced dipole/induced dipole
α 0,1α 0,2
ν 1ν 2
3h
–
2 ν 1 + ν 2 4πε 0 2r 6
Dipolar Interactions
The field around a dipole can be resolved into two
components as shown in the Figure. The
components are:
E||
E^
2μ 1 cos θ
E || = –
4πε 0 r 3
μ 1 sin θ
E⊥ = –
4πε 0 r 3
The total field is:
μ1 1
2
E=–
1
+
3cos
θ
3
4πε 0 r
m1
The Debye Term
A permanent dipole on molecule 1 will induce a
dipole moment on molecule 2.
μ 2 = α 2E
The total energy of the second dipole is:
Φ2 = – 1 α 2 E 2
2
Substituting for E and averaging over all orientations
yields:
2
α 2μ 2
Φ2 = –
2 6
4πε 0 r
A similar equation can be derived for F1 to yield the
Debye equation.
The Keesom Term
The Keesom term arises from the interaction of two
permanent dipoles. Here we consider the Debye
term for the polarizability of a polar solvent.
NA
μ2
P=
α0 +
3ε 0
3kT
Using a similar reasoning applied to the Debye term
we can substitute in m2/3kT for a to obtain the
Keesom term.
2 2
μ
1μ 2
2
–
3 4πε 0 2kTr 6
London Interactions
No permanent dipole is required for London forces
to apply. The London equation for the attraction
between two particles represents a quantum
mechanical effect. The derivation uses a harmonic
oscillator model. Consider a dipole-dipole interaction:
2e 1 e
2μ 2
±
3 = ±
4πε 0r
4πε 0r 3
since the definition of a dipole is:
μ1 = e
1
2
Harmonic oscillator model
Consider the electrons in a material as a harmonic
oscillator. The nuclei represent the restoring force.
The potential energy is given by:
K
Φ=
2
in which:
2
1
2
e
K=α
0
Combining these various contributions we have:
ΦT = K
2
2
1
+
2
2
2e 1 e
±
4πε 0r
3
2
Harmonic oscillator model
The energies of the harmonic oscillator are:
E = n 1 + 1/2 hν 1 + n 2 + 1/2 hν 2
in which:
2α 0
1–
, ν2 = ν
3
4πε 0r
ν1 = ν
2α 0
1+
4πε 0r 3
An illustration of the two induced dipoles for the
London interaction is shown below:
-
+
-
+
l1
l2
r
Harmonic stabilization energy
Taking the lowest energy harmonic oscillator state:
E = h ν1 + ν2
2
Two independent oscillators in their ground state
have energy:
E = h 2ν
2
The difference is the contribution of dispersion
forces to the interaction energy:
Φ = h ν 1 + ν 2 – 2ν
2
The London term
Plugging in the frequencies obtained above and
solving yields:
hνα 20
Φ=–
2 6
2 4πε 0 r
for identical molecules or:
α 0,1α 0,2
ν 1ν 2
3h
Φ=–
2 ν 1 + ν 2 4πε 0 2r 6
two different types of molecule 1 and 2.
The van der Waal’s parameter β
The van der Waal’s potential is the sum of the three
terms derived:
Φ=–
1
4πε 0
2
4
2μ
β 11
3
1
1
2
2
2α 0,1μ 1 +
+ hνα 0,1 6 = – 6
3kT 4
r
r
In this case it is derived for a pair of identical
molecules of type 1. Thus, the parameter β is an
interaction parameter for molecules of type 1 that
includes Keeson, Debye and London terms.
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: α-helix
Electrostatic Interactions
Coulomb’s Law: V = q1q2/εr
Example of a hydrogen bond
-N-H…..O=CExample of a Salt Bridge
Main Chain
Lysine
Main Chain
Glutamate
Hydrogen bonding in water
Hydrophobic interactions
Contributions to ΔG
-
0
+
-TΔS
Conformational
Entropy
ΔH
Internal
Interactions
-TΔS
Hydrophobic
Effect
Net:
ΔG
Folding
Review of transitions
The Fermi Golden Rule can be used to
calculate many types of transitions
Transition
1. Optical transitions
2. NMR transitions
3. Electron transfer
4. Energy transfer
5. Atom transfer
6. Internal conversion
7. Intersystem crossing
H(t) dependence
Electric field
Magnetic field
Non-adiabaticity
Dipole-dipole
Non-adiabaticity
Non-adiabaticity
Spin-orbit coupling
Optical electromagnetic radiation permits
transitions among electronic states
Η t = – μ⋅E t
where μ is the dipole operator and the dot
represents the dot product. If the dipole μ is
aligned with the electric vector E(t) then
H(t) = - μE(t). If they are perpendicular then
H(t) = 0.
μ = er
where e is the charge on an electron and r is
the distance.
The time-dependent perturbation has the
form of an time-varying electric field
E t = E 0cos ωt
where ω 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 [-μ.E(t)]2 is μ2E02/6.
Absorption of visible or ultraviolet
radiation leads to electronic transitions
σ∗
Polarization
of
Radiation
s
s
σ
Absorption of visible or ultraviolet
radiation leads to electronic transitions
Transition
moment
σ∗
s
s
The change in
nodal structure
also implies a
change in orbital
angular momentum.
σ
The interaction of electromagnetic
radiation with a transition moment
The electromagnetic wave has an angular
momentum of 1. Therefore, an atom or
molecule must have a change of 1 in its
orbital angular momentum to conserve this
quantity. This can be seen for hydrogen atom:
Electric vector
of radiation
l=0
l=1
The Fermi Golden Rule for optical
electronic transitions
π e Ψ1|q|Ψ2
k 12 =
6h
2
2
E
2
0
δ ω – ω12
The rate constant is proportional to the square
of the matrix element e< Ψ1|q| Ψ2> times a
delta function. The delta function is an energy
matching function:
δ(ω - ω12) = 1 if ω = ω12
δ(ω - ω12) = 0 if ω ≠ ω12.
A propagating wave of electromagnetic radiation
of wavelength l has an oscillating electric dipole, E
(magnetic dipole not shown)
λ
E
The oscillating electric dipole, E, can induce an oscillating
dipole in a molecule as the radiation
passes through the sample
λ
E
The oscillating electric dipole, E, can induce an oscillating
dipole in a molecule as the radiation passes through the
sample
l
v=1
R
O
R
∆E = hc/λ
v=0
R
O
R
The type of induced oscillating dipole depends on λ.
If λ corresponds to a vibrational energy gap, then radiation
will be absorbed, and a molecular vibrational transition will
result
The oscillating electric dipole, E, can induce an oscillating
dipole in a molecule as the radiation passes through the
sample
λ
LUMO
∆E = hc/λ
HOMO
If λ corresponds to a electronic energy gap, then radiation
will be absorbed, and an electron will be promoted to an
unfilled MO
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.,
aμ b = ∫ψ μψ b dτ ≠ 0
*
a
∞
where μ = μx + μy + μz is the electric dipole moment
operator which transforms in the same manner as the porbitals
ψa⊗μ⊗ψb= ψa⊗(μx + μy + μz)⊗ψb, must contain the totally
symmetric irrep
…or put another way,
ψa⊗ψb must transform as any one of μx, μy, μz
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
C2v
A2
B1
A2 ⊗ B1
E
1
1
1
C2
1
-1
-1
σv
-1
1
-1
σ'v
-1
-1
1
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 = A
•g 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.
Electric vector
of radiation
l=0
l=1
For a hydrogen atom, we can view the electromagnetic
radiation as mixing the 1s and 2p orbitals transiently.
Is the orbital transition dyz → px electric dipole
allowed in C2v symmetry?
μ dyz
⎛ b1 ⎞
⎛ a1 ⎞
⎛ b2 ⎞
⎜ ⎟
⎜ ⎟
⎜ ⎟
b1 ⊗ ⎜ b2 ⎟ ⊗ b2 = ⎜ a2 ⎟ ⊗ b2 = ⎜ b1 ⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝ a1 ⎠
⎝ b1 ⎠
⎝ a2 ⎠
px
None of the three components contains the a1
representation, 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
How about an a1→b2 orbital transition?
μ
⎛ b1 ⎞
⎛ a2 ⎞
⎛ a2 ⎞
⎜ ⎟
⎜ ⎟
⎜ ⎟
b 2 ⊗ ⎜ b2 ⎟ ⊗ a1 = ⎜ a1 ⎟ ⊗ a1 = ⎜ a 1 ⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝ a1 ⎠
⎝ b2 ⎠
⎝ b2 ⎠
Since my makes the transition allowed, the transition is
said to be "y-allowed" or "y-polarized"
Remember the shortcut: a1⊗b2 = b2 which transforms as μy
Problem Indicate whether each of the following metal localized transitions is electric
dipole allowed in PtCl42-.
(b) dyz → dz2 (c) dx2-y2 → px,py (d) pz → s
(a) dxy → pz
Example:
Myoglobin/Hemoglobin
Heme spectroscopy
Transition moment
Franck-Condon active
Vibronic coupling
Myoglobin Structure
Globular α-helical protein
G helix
F helix
A helix
Heme
B helix
E helix
H helix
The iron in heme is the binding site
for oxygen and other diatomics
Heme is iron protoporphyrin IX. Fe is
found in Fe2+ and Fe3+
oxidation states.
Diatomics bind to Fe2+.
Examples, CO, NO, O2.
O2 is the physiologically
relevant ligand, but it can
oxidize iron and it is difficult
to study directly.
N
O
|||
C
N
Fe
N
N
O
O-
O
O-
The porphine ring is an aromatic ring
that has a fourfold symmetry axis
The ring and metal can be
considered separately.
The ring has been succesfully
modeled using the Gouterman
four orbital model.
In globins the iron is Fe(II)
and can be either high spin
or low spin.
MbCO ------ low spin
Deoxy Mb - high spin
N
N
N
N
The four orbital model is used to represent
the highest occupied and lowest unoccupied
molecular orbitals of porphyrins
The two highest occupied
orbitals (a1u,a2u) are nearly
equal in energy. The eg
orbitals are equal in energy.
Transitions occur from:
a1u→ eg and a2u → eg.
a2u π
∗
eg π
M1
a1u π
The transitions from ground state π orbitals
a1u and a2u to excited state π* orbitals eg
can mix by configuration interaction
Both excited state configurations
are Eu so they can mix.
Two electronic transitions
are observed. One is very
strong (B or Soret) and the
M1
other is weak (Q).
The transition moments are:
B band Rs0 = M1 + M2
a2u π
0
Q band rs = M1 - M2 ≈ 0
∗
eg π
M2
a1u π
Porphine orbitals
eg
a1u
eg
a2u
Four orbital model
of metalloporphyrin spectra
There are four excited state configurations possible in
D4h symmetry. These are denoted B (strong) and Q(weak).
0
y
|B 〉 =
1
2
a 2ue gy + a 1ue gx
0
y
|Q 〉 =
1
2
a 2ue gy – a 1ue gx
0
x
|B 〉 =
1
2
a 2ue gx + a 1ue gy
|Q 〉 =
1
2
a 2ue gx – a 1ue gy
0
x
The transition moment for absorption
The absorption probability amplitude for a
Franck-Condon active transition is:
e<G|σ|B>
EB – EG – hω – iΓ B
Here i and f represent individual vibrational
levels in each electronic manifold. The
polarization can be σ = x, y, or z.
Remember that the ground state is totally
symmetric (filled shell molecules). Here it
is A1g.
Character table for D4h point group
E 2C4(z)
C2 2C'2 2C''2 i
2S4 σh 2σv 2σd
linears,
rotations
quadratic
The1 absorption
forx2a+y2, z2
1
1
1 probability
1 1 1 1amplitude
1
transition
1 Franck-Condon
1
1 -1
-1 active
1 1
1 -1 -1 Ris:
z
A1g 1
A2g
B1g 1 -1
1
1
-1
1
-1
1
1
-1
x2-y2
B2g 1 -1
1
-1
1
1
-1
1
-1
1
xy
Eg
-2
0
0
2
0
-2 0
0
A1u 1 1
1
1
1
-1 -1
-1 -1
-1
A2u 1 1
1
-1
-1
-1 -1
-1 1
1
B1u 1 -1
1
1
-1
-1 1
-1 -1
1
B2u 1 -1
1
-1
1
-1 1
-1 1
-1
Eu
-2
0
0
-2 0
2
0
2 0
2 0
0
(Rx, Ry)
z
(x, y)
(xz, yz)
Character table for D4h point group
E 2C4(z)
C2 2C'2 2C''2 i
2S4 σh 2σv 2σd
linears,
rotations
quadratic
The1 absorption
forx2a+y2, z2
1
1
1 probability
1 1 1 1amplitude
1
transition
1 Franck-Condon
1
1 -1
-1 active
1 1
1 -1 -1 Ris:
z
A1g 1
A2g
B1g 1 -1
1
1
-1
1
-1
1
1
-1
x2-y2
B2g 1 -1
1
-1
1
1
-1
1
-1
1
xy
Eg
-2
0
0
2
0
-2 0
0
A1u 1 1
1
1
1
-1 -1
-1 -1
-1
A2u 1 1
1
-1
-1
-1 -1
-1 1
1
B1u 1 -1
1
1
-1
-1 1
-1 -1
1
B2u 1 -1
1
-1
1
-1 1
-1 1
-1
Eu
-2
0
0
-2 0
2
0
2 0
2 0
0
(Rx, Ry)
z
(x, y)
(xz, yz)
Character table for D4h point group
E 2C4(z)
C2 2C'2 2C''2 i
2S4 σh 2σv 2σd
linears,
rotations
quadratic
The1 absorption
forx2a+y2, z2
1
1
1 probability
1 1 1 1amplitude
1
transition
1 Franck-Condon
1
1 -1
-1 active
1 1
1 -1 -1 Ris:
z
A1g 1
A2g
B1g 1 -1
1
1
-1
1
-1
1
1
-1
x2-y2
B2g 1 -1
1
-1
1
1
-1
1
-1
1
xy
Eg
-2
0
0
2
0
-2 0
0
A1u 1 1
1
1
1
-1 -1
-1 -1
-1
A2u 1 1
1
-1
-1
-1 -1
-1 1
1
B1u 1 -1
1
1
-1
-1 1
-1 -1
1
B2u 1 -1
1
-1
1
-1 1
-1 1
-1
Eu
-2
0
0
-2 0
2
0
2 0
2 0
0
(Rx, Ry)
z
(x, y)
(xz, yz)
Character table for D4h point group
E 2C4(z)
C2 2C'2 2C''2 i
2S4 σh 2σv 2σd
linears,
rotations
quadratic
The1 absorption
forx2a+y2, z2
1
1
1 probability
1 1 1 1amplitude
1
transition
1 Franck-Condon
1
1 -1
-1 active
1 1
1 -1 -1 Ris:
z
A1g 1
A2g
B1g 1 -1
1
1
-1
1
-1
1
1
-1
x2-y2
B2g 1 -1
1
-1
1
1
-1
1
-1
1
xy
Eg
-2
0
0
2
0
-2 0
0
A1u 1 1
1
1
1
-1 -1
-1 -1
-1
A2u 1 1
1
-1
-1
-1 -1
-1 1
1
B1u 1 -1
1
1
-1
-1 1
-1 -1
1
B2u 1 -1
1
-1
1
-1 1
-1 1
-1
Eu
-2
0
0
-2 0
2
0
2 0
2 0
0
(Rx, Ry)
z
(x, y)
(xz, yz)
Mixing of the excited state
configurations
There are two transitions that both have Eu symmetry.
Thus, they can add constructively and destructively.
0
y
|B 〉 =
1
2
a 2ue gy + a 1ue gx
Constructive (allowed)
0
y
|Q 〉 =
1
2
a 2ue gy – a 1ue gx
Destructive (forbidden)
|Bx0 〉 =
1
2
a 2ue gx + a 1ue gy
Constructive (allowed)
|Q x0 〉 =
1
2
a 2ue gx – a 1ue gy
Destructive (forbidden)
Vibrational modes that couple the
states are determined by the
direct product
Γcoupling = Eu Eu which we can determine from the
character table.
E 2C4(z)
C2 2C'2 2C''2 i
2S4 σh 2σv 2σd
Eu
2 0
-2
0
0
-2 0
2
0
0
Γc
4 0
4
0
0
4
4
0
0
0
linears,
rotations
quadratic
(x, y)
This reducible representation can be decomposed into
four irreps: A1g + A2g + B1g + B2g
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.
-5
N
N
5
Δm=9
Δm=1
-4
4
-3
3
-2
-1
2
N
N
1
imφ
Φ=
e
2π
1
m=0
MCD spectra
∂f ν
Δε ν = A 1 –
∂ν
Δε m ν
εν
=
A1μ B
D0
C0
+ B0 +
f ν
k BT
∂f ν
–
∂ν
f
A1
ΔL z = 2
D0
Franzen, JPC Accepted
MbCO MCD spectra follow the PM
B
Q
The spectra are A-term MCD as shown by the derivatives of the
absorption spectrum (red). The (Q MCD) = 9 x (B MCD). Franzen, JPC Accepted
Deoxy MCD spectra are anomalous
B
C-term 4 time larger than MbCO!
Q
A-term but with vibronic structure
MCD spectra: Vibronic coupling
in the Perimeter Model
Franzen, JPC Accepted
MCD spectra: Vibronic coupling
in the Perimeter Model
Metal
Porphyrin
Vibronic Distortions