Lecture 2 - Polarisability
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Transcript Lecture 2 - Polarisability
In the previous lecture:
Characteristics of Soft Matter
• (1) Length scales between atomic and macroscopic
(sometimes called mesoscopic)
• (2) The importance of thermal fluctuations and
Brownian motion
• (3) Tendency to self-assemble into hierarchical
structures (i.e. ordered on multiple size scales beyond
the molecular)
• (4) Short-range forces and interfaces are important.
In the previous lecture:
Interaction Potentials: w = -Cr -n
• If n<3, molecules interact with all others in the system of
size, L. If n<3, molecules interact only with the nearer
neighbours.
• Gravity: negligible at the molecular level. W(r) = -Cr -1
• Coulombic: relevant for salts, ionic liquids and charged
molecules.
W(r) = -Cr -1
• van der Waals’ Interaction: usually quite weak; causes
attraction between ANY two molecules. W(r) = -Cr -6
• Covalent bonds: usually the strongest type of bond;
directional forces - not described by a simple potential.
• Hydrogen bonding: stronger than van der Waals
bonds; charge attracting resulting from unshielded proton
in H.
Lecture 2:
Polarisability and van der Waals’
Interactions:
Why are neutral molecules attractive to each other?
3SMS
23 January, 2007
See Israelachvili’s Intermolecular and Surface Forces, Ch. 4, 5 & 6
Polarity of Molecules
• The intermolecular force is found from F = - dw/dr
• All interaction potentials (and forces) between molecules are
electrostatic in origin.
• A neutral molecule is polar when its electronic charge distribution
is not symmetric about its nuclear (+ve charged) centre.
• In a non-polar molecule the centre of electronic (-ve) charge
does not coincide with the centre of nuclear (+ve) charge.
_
+
_
+
Dipole Moments
The polarity of a molecule is described by its dipole
moment, u, given as:
u = q
where charges of +q and -q are separated by a distance
Typically, q is the charge on the electron:1.602 x10-19 C
and the magnitude of is on the order of 1Å= 10-10 m,
giving u = 1.602 x 10-29 Cm.
A “convenient” (and conventional) unit for polarity is
called a Debye (D):
+
1 D = 3.336 x 10-30 Cm
.
Examples of Nonpolar Molecules: u = 0
CO2
O-C-O
H
H
C
H
CH4
methane
109ºC
H
H
H
H
H
Cl
CCl4
Have rotational and mirror symmetry
109ºC
Cl
Cl
Cl
Examples of Polar Molecules
CH3Cl
CHCl3
Cl
H
C
C
H
H
H
u = 6.24 x10 _ 30 Cm
Cl
Cl
Cl
u = 3.54 x10 _ 30 Cm
Have lost some rotational and mirror symmetry!
Dipole moments
+
-
C=O
+
H
Vector addition of bond
moments is used to find u
for molecules.
u = 0.11 D
N
-
H
u = 1.47 D
-
N-H
1.31 D
O-H
1.51 D
F-H
1.94 D
H
H
O
Bond moments
H
u = 1.85 D
V. High!
+
u = 1.62 D
S
O
+
O-
What is S-O bond moment?
Find from vector addition knowing
O-S-O bond angle.
Vector Addition of Bond Moments
Given that H-O-H bond angle is 104.5° and that the bond
moment of OH is 1.51 D, what is the dipole moment of
water?
O
1.51 D
q/2
H
H
uH2O = 2 cos(q/2)uOH = 2 cos (52.25 °) x 1.51 D = 1.85 D
Charge-Dipole Interactions
u
Q
-
r
+
q
W(r) = -Cr -2
• There is an electrostatic (i.e. Coulombic) interaction between
a charged molecule (an ion) and a static polar molecule.
Qu cos q
w (r ) = _
4 o r 2
• The interaction potential
can be compared to the Coulomb potential for two point
charges (Q1 and Q2):
Q1Q2
w (r ) =
4 or
• Ions can induce ordering and alignment of polar molecules.
• Why? Equilibrium state when W(r) is minimum. W(r) decreases as q
decreases to 0.
Dipole-Dipole Interactions
u1
-
+
+
q1
-
u2
q2
• There are Coulombic interactions between the +ve
and -ve charges associated with each dipole.
• In liquids, thermal energy causes continuous motion,
i.e. tumbling, of dipoles in relation to each other.
• In solids, dipoles are usually fixed on a lattice with a
certain orientation, described by q1 and q2.
Fixed-dipole Interactions
u1
-
u2
+
q1
-
+
q2
f
r
• The interaction energy, w(r), depends on the relative
orientation of the dipoles:
u1u2
w (r ) = _
3 [ 2 cosq 1cos q 2 _ sinq1 sinq 2 cos f ]
4 o r
• Molecular size influences the minimum possible r.
• For a given spacing r, the end-to-end alignment has a
lower w, but usually this alignment requires a larger r
compared to side-by-side (parallel) alignment.
Note: W(r) = -Cr -3
0
w(r)
kT at 300 K
= 0.1nm
(J)
= 0.1nm
Side-by-side
q1 = q2 = 90°
-10-19
| u |= q | |= 1D
At a typical spacing
of 0.4 nm, w(r) is
about 1 kT. Hence,
thermal energy is
able to disrupt the
alignment.
From Israelachvili,
Intermol.& Surf.
Forces, p. 59
End-to-end
q1 = q2 = 0
W(r) = -Cr -3
-2 x10-19
r (nm)
0.4
Freely-Rotating Dipoles
• In some cases, dipoles do not have a fixed position
and orientation on a lattice but constantly move about.
• This occurs when thermal energy is greater than the
fixed dipole interaction energy:
u1u2
kT >
4 o r 3
• Interaction energy depends inversely on T, and
because of constant motion, there is no angular
dependence:
u12u2 2
w (r ) = _
3( 4 o )2 kTr 6
Note: W(r) = -Cr -6
Polarisability
• All molecules can have a dipole induced by
an external electromagnetic field, E
• The strength of the induced dipole moment,
|uind|, is determined by the polarisability, a, of
the molecule:
uind
a=
E
Units of
polarisability:
Cm Cm C 2m C 2m 2
=
=
=
J
N
N
J
C
Cm
Polarisability of Nonpolar Molecules
• An electric field will shift the electron cloud of
a molecule.
_
+
Initial state
E
_
+
In an electric field
• The extent of polarisation is determined by its
electronic polarisability, ao.
Simple Illustration of e- Polarisability
Without a field:
With a field:
Fext
Fint
uind = a o E = e
Fext = eE
Force on the electron due to the field:
Attractive Coulombic force on the electron from nucleus:
2
2
euind
dw (R )
e
e
Fint =
=
sinq =
=
2
2
dR
4 o R
4 o R R 4 o R 3
At equilibrium, the forces balance:
Fext = Fint
Simple Illustration of e- Polarisability
uind = a o E = e
Fext = Fint
Substituting expressions
for the forces:
euind
eE =
4 o R 3
3
Solving for the induced dipole moment: uind = 4 o R E
So we obtain an expression for the
polarisability: a = 4 R 3
o
o
From this crude argument, we predict that electronic polarisability is
proportional to the size of the molecule!
Units of Electronic Polarisability
ao is often reported as:
2
ao
4 o
2 _1
C m J
2 _1
C J m
_1
=m
3
Units of volume
Electronic Polarisabilities
He
0.20
Smallest
H2O
1.45
O2
1.60
CO
1.95
(4o)10-30 m3
NH3
2.3
=1.11 x 10-40 C2m2J-1
CO2
2.6
Xe
4.0
CHCl3
8.2
CCl4
10.5
Units:
Largest
Example: Polarisation induced by an ion
Ca2+ dispersed in CCl4 (non-polar).
-
+
What is the induced dipole moment in CCl4 at a distance of 2 nm?
By how much is the electron cloud of the CCl4 shifted?
From Israelachvili, Intermol.& Surf. Forces, p. 72
Example: Polarisation Induced by an Ion
Ca2+
uind = a o E
dispersed in CCl4 (non-polar).
Field from the
Ca2+ ion:
From the literature,
we find for CCl4:
E=
2e
4 o r 2
ao
= 10.5x10 _ 30 m 3
4 o
Affected by the
permittivity of
CCl4: = 2.2
a o 2e
uind =
4 o r 2
We find when r = 2 nm:
u = 3.82 x 10-31 Cm
Thus, an electron with
charge e is shifted by:
u
= = 2.38x10 _ 12 m = 0.02 Å
e
Polarisability of Polar Molecules
In a liquid, molecules are continuously rotating and turning, so the timeaveraged dipole moment for a polar molecule in the liquid state is 0.
An external electric field can partially align dipoles:
E
+
-
q
Let q represent the angle between the dipole moment of a
molecule and an external E-field direction. 2
u E
The induced dipole moment is:
u =
cos 2 q
ind
kT
The spatially-averaged value of <cos2q> = 1/3
As u = aE, we can define an orientational polarisability.
a orient
The molecule still has electronic polarisability, so the total
polarisability, a, is given as:
2
u
a = ao +
3kT
u2
=
3kT
Debye-Langevin
equation
Origin of the London or Dispersive Energy
• The dispersive energy is quantum-mechanical in origin,
but we can treat it with electrostatics.
• Applies to all molecules, but is insignificant in charged or
polar molecules.
a1
a2
• An instantaneous dipole, resulting from fluctuations in
the electronic distribution, creates an electric field that can
polarise a neighbouring molecule.
- u1 +
a2
- u1 +
- u2 +
r
• The two dipoles then interact.
Origin of the London or Dispersive Energy
-
Instantaneous
dipole
u1
+
-
u2
+
Induced dipole
r
The field produced by the instantaneous dipole is:
E=
u1
4 o r
2 1/ 2
(
1
+
3
cos
q)
3
u1
q
So the induced dipole moment in the
neighbour is:
u2
a ou1
uind = u2 = a o E =
3 f (q )
4 o r
We can now calculate the interaction energy between the two
dipoles (using the equation for permanent dipoles
- slide 12):
w (r ) =
u1u2
4 o r
3 f (q1,q 2 ,f ) =
a ou1
u1(
3)
4 o r
4 o r
3
=
2
a ou1
( 4 o )2 r 6
Origin of the London or Dispersive Energy
-
u1
+
-
u2
+
r
2
a ou1
This result:
w (r ) =
( 4 o )2 r 6
compares favourably with the London result (1937) that
was derived from a quantum-mechanical approach:
3 a o 2 h
w (r ) =
4 ( 4 o )2 r 6
h is the ionisation energy,
i.e. the energy to remove an
electron from the molecule
London or Dispersive Energy
3 a o 2 h
w (r ) =
4 ( 4 o )2 r 6
The London result is of the form:
w (r ) =
C
r
6
where C is called the London constant:
3 a o 2 h
C=
4 ( 4 o )2
In simple liquids and solids consisting of non-polar
molecules, such as N2 or O2, the dispersive energy is solely
responsible for the cohesion of the condensed phase.
Must consider the pair interaction energies of all “near”
neighbours.
Measuring Polarisability
• Polarisability is dependent on the frequency of the Efield.
• The Clausius-Mossotti equation relates the dielectric
constant of a molecule with a volume v to a:
a
_ 1 3v
=(
)
4 o
+ 2 4
•At the frequency of visible light, however, only the
electronic polarisability, ao, is active.
• At these frequencies, the Lorenz-Lorentz equation
relates the refractive index (n2 = ) to ao:
ao
n 2 _ 1 3v
=( 2
)
4 o
n + 2 4
Frequency dependence of polarisability
From Israelachvili, Intermol. Surf. Forces, p. 99
PV diagram for CO2
(P + a
V
2 )(V
it.wikipedia.org/wiki/Legge_di_Van_der_Waals
_ b ) = RT
Measuring Polarisability
• The van der Waals’ gas law can be written (with V = molar
volume) as:
(P + a
V
2 )(V
_ b ) = RT
The constant, a, is directly related to the London
constant, C:
a=
2C
3
3
where is the molecular diameter (closest molecular
spacing). We can thus use the C-M, L-L and v.d.W.
equations to find values for ao and a.
Measuring Polarisability
Summary
Type of Interaction
Charge-charge
In vacuum: =1
Interaction Energy, w(r)
Q1Q2
Coulombic
4 o r
_ Qu cos q
Dipole-charge
_ Q 2u 2
4 o r 2
6( 4 o )2 kTr 4
_ u12u2 2f (q1,q 2 ,f )
Dipole-dipole
_ u12u2 2
Charge-nonpolar
Dipole-nonpolar
_ Q 2a
3( 4 o )2 kTr 6
Keesom
2( 4 o )2 r 4
_ u 2a (1 + 3 cos 2 q )
_ u 2a
Nonpolar-nonpolar
4 o r 3
( 4 o )2 r 6
Dispersive
2( 4 o )2 r 6
Debye
3 a o 2 h
w (r ) = _
4 ( 4 o )2 r 6
van der Waals’ Interactions
• Refers to all interactions between polar
or nonpolar molecules, varying as r -6.
• Includes Keesom, Debye and dispersive
interactions.
• Values of interaction energy are usually
only a few kT.
Comparison of the Dependence
of Interaction Potentials on r
n = 1 Coulombic
n=2
n=6
van der
Waals
n=3
Dipole-dipole
Not a comparison of the magnitudes of the energies!
Interaction between ions and
polar molecules
• Interactions involving charged molecules (e.g. ions)
tend to be stronger than polar-polar interactions.
• For freely-rotating dipoles with a moment of u
interacting with molecules with a charge of Q we saw:
_ Q 2u 2
6( 4 o )2 kTr 4
• One result of this interaction energy is the condensation of
water (u = 1.85 D) caused by the presence of ions in the
atmosphere.
• During a thunderstorm, ions are created that nucleate rain
drops in thunderclouds (ionic nucleation).
Cohesive Energy
• Def’n.: Energy required to separate all molecules in
the condensed phase or energy holding molecules in
the condensed phase.
• In Lecture 1, we found that for a single molecule, and
with n>3:
4Cr
E=
(n 3) n _ 3
• with r = number of molecules per unit volume -3,
where is the molecular diameter. So, with n = 6:
4Cr 4C
E=
3
1/2 to avoid double
3
3 6
• For one mole, E
= (1/2)N E counting!
substance
A
• Esubstance = sum of heats of melting + vaporisation.
• Predictions agree well with experiment!
Boiling Point
• At the boiling point, TB, for a liquid, the thermal energy
of a monoatomic molecule, 3/2 kTB, will exactly equal the
energy of attraction between molecules.
• Of course, the strongest attraction will be between the
“nearest neighbours”, rather than pairs of molecules that
are farther away.
• The interaction energy for van der Waals’ interactions is
of the form, w(r) = -Cr -6. If molecules have a diameter of
, then the shortest centre-to-centre distance will
likewise be .
• Thus the boiling point is approximately:
w ( )
TB =
3 k
2
Comparison of Theory and Experiment
Note that ao and C increase with .
Emole
C can be found experimentally from
deviations from the ideal gas law:
(P + a 2 )(V _ b ) = RT
V
N A 4C
~
2 3 6
w (r )
TB =
3 k
2
Evaluated at close
contact where r = .
Cohesive energy = energy required to separate molecules
to a large distance (solid gas)
Additivity of Interactions
Molecule
Mol. Wt.
H
H
C-C
H
u (D)
H
H
H
Ethane: CH3CH3
30
TB(°C)
Dispersive only
0
-89
H
C=O
H
Formaldehyde: HCHO
Keesom + dispersive
30
2.3
-21
H
H
C-O-H
H
Methanol: CH3OH
H-bonding + Keesom + dispersive
32
1.7
64
Problem Set 1
1. Noble gases (e.g. Ar and Xe) condense to form crystals at very low temperatures. As the atoms do not undergo any
chemical bonding, the crystals are held together by the London dispersion energy only. All noble gases form crystals having
the face-centred cubic (FCC) structure and not the body-centred cubic (BCC) or simple cubic (SC) structures. Explain why
the FCC structure is the most favourable in terms of energy, realising that the internal energy will be a minimum at the
equilibrium spacing for a particular structure. Assume that the pairs have an interaction energy, u(r), described as
u(r ) = 2 A12
12
A6
r
r
6
,
where r is the centre-to-centre spacing between atoms. The so-called "lattice sums", An, are given below for each of the three
cubic lattices.
SC
BCC
FCC
A6
8.40
14.45
12.25
A12
6.20
12.13
9.11
Then derive an expression for the maximum force required to move a pair of Ar atoms from their point of contact to an infinite
separation.
2. (i) Starting with an expression for the Coulomb energy, derive an expression for the interaction energy between an ion of
charge ze and a polar molecule at a distance of r from the ion. The dipole moment is tilted by an angle q with relation to r, as
shown below.
ze
r
q
(ii) Evaluate your expression for a Mg2+ ion (radius of 0.065 nm) dissolved in water (radius of 0.14 nm) when the water
dipole is oriented normal to the ion and when the water and ion are at the point of contact. Express your answer in units of
kT. Is it a significant value? (The dipole moment of water is 1.85 Debye.)
Interaction between ions and
polar molecules
• Interactions involving charged molecules (e.g. ions)
tend to be stronger than polar-polar interactions.
• For freely-rotating dipoles with a moment of u
interacting with molecules with a charge of Q we saw:
_ Q 2u 2
6( 4 o )2 kTr 4
• One result of this interaction energy is the condensation of
water (u = 1.85 D) caused by the presence of ions in the
atmosphere.
• During a thunderstorm, ions are created that nucleate rain
drops in thunderclouds (ionic nucleation).
Hydrogen bonding
d-
O
H
d+
d-
Hd+
H O
d+
Hd+
• In a covalent bond, an electron is shared between two
atoms.
• Hydrogen possesses only one electron and so it can
covalently bond with only ONE other atom.
• The proton is unshielded and makes an electropositive
end to the bond: ionic character.
• Bond energies are usually stronger than v.d.W., typically
25-100 kT.
• The interaction potential is difficult to describe but goes
roughly as r-2, and it is somewhat directional.
• H-bonding can lead to weak structuring in water.
Hydrophobic Interactions
A water “cage”
around another
molecule
• “Foreign” molecules in water can increase the local
ordering - which decreases the entropy. Thus their
presence is unfavourable.
• Less ordering of the water is required if two or more of
the foreign molecules cluster together: a type of
attractive interaction.
• Hydrophobic interactions can promote self-assembly.
Hydrophobic Interactions
• The decrease in entropy (associated with the ordering of molecules)
makes it unfavourable to mix water with “hydrophobic molecules”.
• For example, when mixing n-butane with water:
DG = DH - TDS = -4.3 +28.7 = +24.5 kJ mol-1.
Unfavourable (+ve DG) because of the decrease in entropy!
• This value of DG is consistent with a “surface area” of n-butane of 1
nm2 and g 40 mJ m-2 for the water/butane interface; an increase in
DG = gDA is needed to create a new interface!
• Although hydrophobic means “water-fearing”, there is an attractive van
der Waals’ force (as discussed later in this lecture) between water and
other molecules - there is not a repulsion! Water is more strongly
attracted to itself, because of H bonding, however, in comparison to
hydrophobic molecules.
Micellisation
Immiscibility
Association of
molecules
Adhesion
in water
Coagulation
Protein
folding
De-wetting
“Froth
flotation”