Transcript Lecture 2 - Polarisability

In the previous lecture:
Characteristics of Soft Matter
• (1) Length scales between atomic and
macroscopic
• (2) The importance of thermal fluctuations
and Brownian motion
• (3) Tendency to self-assemble into
hierarchical structures (i.e. ordered on large
size scales)
• (4) Short-range forces and interfaces are
important.
In the previous lecture:
Interaction Potentials
• 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; active between ANY two molecules.
W(r) = -Cr -6
• Covalent bonds: usually the strongest type of
bond; directional forces - not described by a
simple potential.
Polarisability and van der
Waals’ Interactions
3SCMP
26 January, 2006
Lecture 2
See Israelachvili’s Intermolecular and Surface Forces, Ch. 4, 5 & 6
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.
Polarity of Molecules
• All attractions between molecules are electrostatic in
origin.
• A 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:
+
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.
Charge-Dipole Interactions

u
Q
-
r
+
q
W(r) = -Cr -6
• 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 charges:
Q1Q2
w (r ) =
4 or
• Ions can induce ordering and alignment of polar molecules.
• Why? 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.
Fixed-dipole Interactions

u1
q1

u2
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.
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.
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
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
Electronic polarisability is proportional to the size of the
molecule!
Units of Electronic Polarisability
ao is often expressed 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
(4o)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?
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 negligible 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.


- u1 +
a2
- u1 +
- u2 +
r
• The two dipoles then interact.
Origin of the London or Dispersive Energy
-

u1
+
-

u2
+
r
The field produced by the instantaneous dipole is:


u1
u1
2 1/ 2
E=
(
1
+
3
cos
q
)
q
3
4 o r
u2
So the induced dipole moment in the neighbour is:

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 equations for permanent
dipoles):

 a ou1
u1(
2
 
3)
4 o r
a ou1
u1u2
w (r ) =
=
3 f (q1,q 2 ,f ) =
3
4 o r
4 o r
( 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, o, 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
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=
2C
3
3
We can thus use the C-M, L-L and v.d.W. equations to
find values for ao and a.
Measuring Polarisability
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.
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
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:
4Cr
E=
(n 3) n _ 3
• with r = number of molecules per unit volume   -3,
where  is the molecular diameter. So, with n = 6:
4Cr 4C
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 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 4C
~
2 3 6
w (r )
TB =
3 k
2
Evaluated at close
contact where r = .
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.)
Micellisation
Immiscibility
Association of
molecules
Adhesion
in water
Coagulation
Protein
folding
De-wetting
“Froth
flotation”