Characteristics of Soft Matter(1) Length scales between atomic and macroscopic

(sometimes called mesoscopic) are relevant.(2) Weak, short-range forces and interfaces are important.(3) The importance of thermal fluctuations and Brownian

motion

(4) Tendency to self-assemble into hierarchical structures (i.e. ordered on multiple size scales beyond the molecular)

In the previous lecture:

PH3-SM (PHY3032)

Soft Matter PhysicsLecture 2:

Polarisability and van der Waals’ Interactions:

Why are neutral molecules attractive to each other?

11 October, 2011

See Israelachvili’s Intermolecular and Surface Forces, Ch. 4, 5 & 6

What are the forces that operate over short distances to hold condensed matter together?

Nitrogen condensed in the liquid phase.

What are the forces that operate over short distances to cause adhesion?

http://www.cchem.berkeley.edu/rmgrp/about_gecko.jpg

Interaction Potentials

• For two atoms/molecules/segments separated by a distance of r, the interaction energy can be described by an attractive potential energy: watt(r) = - Cr -n = -C/r n, where C and n are constants.

• There is also repulsion because of the Pauli exclusion principle: electrons cannot occupy the same energy levels.

• Treat atoms/molecules like hard spheres with a diameter, s. Use a simple repulsive potential:

wrep(r) = +(s/r)

• The interaction potential w(r) = watt + wrep

r

s

“Hard-Sphere” Interaction Potential

+

w(r)

-

Attractive potential

r

watt(r) = -C/rn

+

w(r)

-

Repulsive potential

rswrep(r) = (s/r)

r

s

Hard-Sphere Interaction Potentials

+

w(r)

-Total potential:

r

w(r) = watt + wrep

s

Minimum of potential = equilibrium spacing in a solid = s

The force, F, acting between atoms (molecules) with this interaction energy is:

drdw

F

where a negative force is attractive. As r ∞, F 0.

Interaction Potentials

• Gravity: all atoms/molecules have a mass!• Coulomb: applies to ions and charged molecules; same

equations as in electrostatics• van der Waals: classification of interactions that applies

to non-polar and to polar molecules (i.e. without or with a uniform distribution of charge). IMPORTANT in soft matter!

• How can we describe their potentials?

Gravity: n = 1

r

m1m2

r

mGmrmGmrw 211

21)(

G = 6.67 x 10-11 Nm2kg-1

When molecules are in contact, w(r) is typically ~ 10-52 J

Negligible interaction energy – much weaker than thermal energy (kT)!

Coulombic Interactions: n = 1

r

Q1Q2 rQQ

rwo4

21=)(

• With Q1 = z1e, where e is the charge on the electron, and z1 is an integer value.

• eo is the permittivity of free space and e is the relative permittivity of the medium between ions (can be vacuum with e = 1 or can be a gas or liquid with e > 1).

• The interaction potential is additive in ionic crystals.

• When molecules are in close contact, w(r) is typically ~ 10-18 J, corresponding to about 200 to 300 kT at room temp.

The sign of w depends on whether charges are alike or opposite.

van der Waals Interactions: n = 6

r

a1a2

6)(

r

Crw

u2 u1

• Interaction energy (and the constant, C) depends on the dipole moment (u) of the molecules and their polarisability (a).

• When molecules are in close contact, w(r) is typically ~ 10-21 to 10-20 J, corresponding to about 0.2 to 2 kT at room temp., i.e. of a comparable magnitude to thermal energy!

• v.d.W. interaction energy is much weaker than covalent bond strengths.

Covalent Bond Energies

From Israelachvili, Intermolecular and Surface Forces

1 kJ mol-1 = 0.4 kT per molecule at 300 K

(Homework: Show why this is true.)

Therefore, a C=C bond has a strength of 240 kT at this temp.!

Hydrogen bonding

• 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. It cannot make a covalent network.• The H’s proton is unshielded by electrons and makes an

electropositive end (d+) to the bond: ionic character in a covalent bond.

• Hydrogen 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.

HO

HH

HO

d+

d+

d+d+

d-d-

• When w(r) is a minimum, dw/dr = 0.• Solve for r to find equilibrium spacing for a solid, where r = re.• (Confirm minimum by checking curvature from 2nd derivative.)• The force between two molecules is F = -dw/dr• Thus, F = 0 when r = re.• If r < re, the external F is compressive (+ve); If r > re, the external

F is tensile (-ve).• When dF/dr = d2w/dr2 =0, the attractive F is at its maximum.

Significance of Interaction Potentials

re = equilibrium spacing-

+

r

How much energy is required to remove a molecule from the condensed phase?

Q: Does a central molecule interact with ALL the others?

nrCrw =)(

Applies to pairs

L

s = molecular spacing when molecules are in contact

r = density = number of molec./volume

Individual molecules

•

s

Total Interaction Energy, E

Interaction energy for a pair: w(r) = -Cr -n

Volume of thin shell:

Number of molecules at a distance, r :

Total interaction energy between a central molecule and all others in the system (from s to L), E:

drrv 24

)4()( 2drrrN

Lr

rnrn

CE

3-

1

)3-(

4 3-3

)(1)3(

4 nn Ln

CE

But L >> s ! When can we neglect the term?

r -n+2=r -(n-2) System L

nrCrNrwE

24)()(

Conclusions about E

• There are three cases:• (1) When n<3, then the exponent is negative. As s <<L, then

(s/L)n-3>>1 and is thus significant.• In this case, E varies with the size of the system, L! (This result

applies to gravitational potential (n= 1) in a solar system.) • (2) But when n>3, (s/L)n-3<<1 and can be neglected. Then E is

independent of system size, L. • When n>3, a central molecule is not attracted strongly by ALL

others - just its closer neighbours!

[ ]3

33 )3(

4≈)(1

)3(

4n

nn n

CLn

C

E=

33

)(1)3(

4

n

n Ln

CE

The Third Case: n = 33)( Crrw

drrrN 24)(

lnln44

LCrCdr

ELr

r

s will be very small (typically 10-9 m), but lns is not negligible. L cannot be neglected in most cases.

Which values of n apply to various molecular interaction potentials? Are they >, < or = 3?

http://www.chem1.com/acad/webtext/atoms/atpt-4.html

Electron Probability Distributions

s orbitals in the H atom H orbitals with n = 3

• Symmetric distribution of electrons in atomic orbitals

• Position of the electrons cannot be known with certainty.

Polarity of Molecules• All interaction potentials (and forces) between molecules are

electrostatic in origin: even when the molecules have no net charge.• In a non-polar molecule the centre of electronic (-ve) charge coincides

with the centre of nuclear (+ve) charge.• But, a charge-neutral molecule is polar when its electronic charge

distribution is not symmetric about its nuclear (+ve charged) centre.

O nucleus 8+

O nucleus 8+

--

O2 is non-polar CO is polar

O nucleus 8+

C nucleus 6+ --

Centre of +ve charge

Centre of -ve charge

Dipole Moments

A “convenient” (and conventional) unit for polarity is called a Debye (D):

1 D = 3.336 x 10-30 Cm

qu =

The polarity of a molecule is described by its dipole moment, u, given as:

when charges of +q and - q are separated by a distance .

If 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, then we have that u = 1.602 x 10-29 Cm.

+ -

Examples of Nonpolar Molecules: u = 0

CO2 O=C=O

CCl4

ClC

Cl

ClCl

109º

methane

Have rotational and mirror symmetry

120

Top view

C

H

HH

H109º

Tetrahedral co-ordination

CH4C

H

H

H

H

Tetrahedron

Examples of Polar Molecules

CH3Cl CHCl3

Cmxu 301024.6

Cmxu 301054.3

ClC

H

ClCl

C

Cl

HH

H

Have lost some rotational and mirror symmetry!

Unequal sharing of electrons between two unlike atoms leads to polarity in the bond CH polarity ≠CCl polarity.

Dipole moments

N

H HH u = 1.47 D

-

+

H

HO

- +

u = 1.85 D

Bond moments

N-H 1.31 D

O-H 1.51 D

F-H 1.94 D

What is the S=O bond moment?

Find from vector addition knowing O-S-O bond angle.

V. High!

Vector addition of bond moments is used to find u for molecules.

SO Ou = 1.62 D

+

-

H H

Given that the 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?

q/2

O

1.51 D

uH2O = 2 cos(q/2)uOH = 2 cos (52.25 °) x 1.51 D = 1.85 D

Vector Addition of Bond Moments

Charge-Dipole Interactions

• There is an electrostatic (i.e. Coulombic) interaction between a charged molecule (an ion) and a static polar molecule.

• The interaction potential can be compared to the Coulomb potential for two point charges (Q1 and Q2):

• Ions can induce ordering and alignment of polar molecules.• Why? Equilibrium state when w(r) is minimum. w(r) decreases (becomes

more negative) as q increases to 0 degrees.

24

cos)(

r

Qurw

o

r

QQrw

o4)( 21

Qqu

r

+

-w(r) = -Cr -2

Interactions between Fixed Dipoles

• There are Coulombic interactions between the +ve and -ve charges associated with each dipole.

• The interaction energy, w(r), depends on the relative orientation of the dipoles:

• 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.

q1

q21u 2u

f

]cossinsincoscos2[4

)( 2121321

r

uurw

o

r

Note: W(r) = -Cr -3

-

+

-

+

w(r)

(J)

r (nm)

At a typical spacing of 0.4 nm, w(r) is about 1 kT. Hence, thermal energy is able to disrupt the alignment.

nm10.=

nm10.=

-10-19

-2 x10-19

0

0.4

kT at 300 K

Dqu 1=||=||

End-to-end

Side-by-side

W(r) = -Cr -3

q1 = q2 = 0

q1 = q2 = 90°

From Israelachvili, Intermol. & Surf. Forces, p. 59

End-to-end alignment lowers the energy more than side-by-side. But, small values of r cannot be achieved.

Freely-Rotating Dipoles• In liquids and gases, dipoles do not have a fixed position and

orientation on a lattice, but instead constantly “tumble” about.• Freely-rotating dipoles occur when the thermal energy is greater

than the fixed dipole interaction energy:

• The interaction energy depends inversely on T, and because of constant motion, there is no angular dependence:

321

4 r

uukT

o>

62

22

21

)4(3)(

kTr

uurw

o

Note: w(r) = -Cr -6

This interaction energy is called the Keesom energy.

Polarisability• All molecules can have a dipole induced by an external

electromagnetic field, • The strength of the induced dipole moment, |uind|, is

determined by the polarisability, a, of the molecule:

E

uind

=

Units of polarisability: J

mCNmC

CNCm

CmJCm 222

===

E

Polarisability of Nonpolar Molecules• An electric field will shift the electron cloud of a molecule.

• The extent of polarisation is determined by its electronic polarisability, ao.

_E

_

Initial state In an electric field

+ +

Simple Bohr Model of e- Polarisability

eEu oind

==

Force on the electron due to the field: EeFext

=

Attractive Coulombic force on the electron from nucleus:

32

2

2

2

int 4=

4=sin

4=

)(=

R

ueRR

e

R

edR

RdwF

o

ind

oo

At equilibrium, the forces balance:int= FFext

Without a field: With a field:

Fext

Fint

int= FFext

eEu oind

==

34 R

ueEe

o

ind

=Substituting expressions

for the forces:

Solving for the induced dipole moment: ERu oind

34=

So we obtain an expression for the polarisability:34 Roo =

From this crude argument, we predict that electronic polarisability is proportional to the size of a molecule!

Simple Bohr Model of e- Polarisability

Units of Electronic Polarisability

3112

122

mmJC

JmC

Units of volume

Polarisability is often reported as:o

o

4

e0 is the permittivity of free space (vacuum)

Electronic Polarisabilities

He 0.20

H2O 1.45

O2 1.60

CO 1.95

NH3 2.3

CO2 2.6

Xe 4.0

CHCl3 8.2

CCl4 10.5Largest

Smallest

Unitsao/(4o): 10-30 m3

Numerically equivalent to ao in units of 1.11 x 10-40 C2m2J-1

ao/(4o) (10-30 m3)

Example: Polarisation Induced by an Ion’s Field

Consider Ca2+ dispersed in CCl4 (non-polar).

What is the induced dipole moment in CCl4 at a distance of r = 2 nm?

- +

By how much is the electron cloud of the CCl4 shifted?

From Israelachvili, Intermol.& Surf. Forces, p. 72

Ca2+

CCl4

Example: Polarisation Induced by an IonCa2+ dispersed in CCl4 (non-polar). Eu oind

=

Affected by the permittivity of CCl4: e = 2.2

2

24 r

eu

o

oind

=

330105.104

mxo

o From the literature, we

find for CCl4:

24

2

r

eE

o=

Field from the Ca2+ ion:

We find at a “close contact” of r = 2 nm:

uind = 3.82 x 10-31 Cm

Thus, an electron with charge e is shifted by:

02.01038.2 12 mxe

u Å

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.

• An instantaneous dipole, resulting from fluctuations in the electronic distribution, creates an electric field that can polarise a neighbouring molecule.

• The two dipoles then interact.

1 2

2- +1u

+ +- - 2u

1u

r

Origin of the London or Dispersive Energy

+ +- - 2u

1u

r

2123

1 314

/)cos+(= r

uE

o

u1

u2

The field produced by the instantaneous dipole is:

)(===

fr

uEuu

o

ooind 3

12 4

So the induced dipole moment in the neighbour is:

62

21

3

31

1

21321

444

4 ru

rr

uu

fr

uurw

o

o

o

o

o

o )(

)(),,()(

We can now calculate the interaction energy between the two dipoles (using the equation for fixed, permanent dipoles - slide 27):

Instantaneous dipole

Induced dipole

Origin of the London or Dispersive Energy

+ +- - 2u

1u

r

62

21

)4()(

r

urw

o

o

We recall the approximation that 300 4 R

So we see thatR

R0

02

4

We note that u can be approximated as eR, where R is the atomic radius.

62

22

)4()(

r

Rerw

o

o

Substituting for R2, we find:R

e

rrw

o

o

0

2

62

2

4)4()(

R

In the Bohr model, then angular momentum of the electron is quantised as mvr = nh/2p and the electron energy is quantised as:

R

Zeh

0

2

4

Origin of the London or Dispersive Energy

when it is orbiting with a frequency of .

Substituting in the right-hand side of this equation (Z = 1), we see:

62

2

0

2

62

2

)4(4)4()(

r

h

R

e

rrw

o

o

o

o

This result compares favourably with London’s result (1937) that was derived from a quantum-mechanical approach using perturbations in the Schrödinger equation:

62

2

)4(4

3)(

r

hrw

o

o

h n is the ionisation energy, i.e. the energy to remove an electron from the molecule

London or Dispersive Energy

62

2

)4(4

3)(

r

hrw

o

o

The London result is of the form: 6

)(r

Crw

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.

where C is called the London constant:

2

2

)4(4

3

o

o hC

Must consider the pair interaction energies of all “near” neighbours.

SummaryType of Interaction Interaction Energy, w(r)

Charge-charge rQQ

o421 Coulombic

Nonpolar-nonpolar 62

2

443

r

hrw

o

o

)(_=)(

Dispersive

Charge-nonpolar 42

2

42 rQ

o )(_

Dipole-charge24 r

Qu

ocos_

42

22

46 kTruQ

o )(_

Dipole-dipole

62

22

21

43 kTruu

o )(_

Keesom

321

22

21

4 rfuu

o ),,(_

Dipole-nonpolar

62

2

4 ru

o )(_

Debye

62

22

4231

ru

o )()cos+(_

In vacuum: e=1

van der Waals’ Interactions

• Refers collectively to all interactions between polar or nonpolar molecules that vary as r -6.

• Includes the Keesom, Debye and dispersive interactions.

• Values of interaction energy are usually only a few kT (at RT), and hence are considered weak.

Comparison of the Dependence of Interaction Potentials on r

Not a comparison of the magnitudes of the energies!

n = 1

n = 2

n = 3n = 6

Coulombic

van der Waals

Fixed dipole-dipole

Charge-fixed dipole

Interaction energy 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 on Slide 43:

42

22

46 kTruQ

o )(_

• 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).

+Q

Polarisability of Polar MoleculesIn a liquid, molecules are continuously rotating and turning, so the time-averaged dipole moment for a polar molecule in the liquid state is 0.

Let q represent the angle between the dipole moment of a molecule and an external E-field direction.

The spatially-averaged value of <cos2q> = 1/3

The induced dipole moment is: 22

cos=kT

Euuind

An external electric field can partially align dipoles:

E +

-

The molecule still has electronic polarisability, so the total polarisability, a, is given as:

kTu

o 3

2

+= Debye-Langevin equation

kTu

orient 3

2

=As u = aE, we can define an orientational polarisability.

Measuring Polarisability

• Polarisability is dependent on the frequency of the E-field. • The Clausius-Mossotti equation relates the dielectric constant

(permittivity) e of a molecule having a volume v to a:

43

21

4v

o

)(

43

21

4 2

2 vnn

o

o )(

• 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, n (n2 = e) to ao:

So we see that measurements of the refractive index can be used to find the electronic polarisability.

Frequency Dependence of Polarisability

From Israelachvili, Intermol. Surf. Forces, p. 99

it.wikipedia.org/wiki/Legge_di_Van_der_Waals

PV diagram for CO2

RTnbVV

aP ))(( 2

Non-polar gases condense into liquids because of the dispersive (London) attractive energy.

Van der Waals Gas Equation:

P

V

Measuring Polarisability• The van der Waals’ gas law can be written (with V = molar

volume) as:

RTnbVV

aP ))(( 2

33

2

C

a

The constant, a, is directly related to the London constant, C:

where s 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

From Israelachvili, Intermol.& Surf. Forces

Polarisability determined from van der Waals gas (a) and u measurements.

Polarisability determined from dielectric/index measurements.

<

<

<

High f

Low f

Problem Set 11. 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

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 12.25 14.45A12 6.20 9.11 12.13

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.

(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.)

3. Show that 1 kJ mole-1 = 0.4 kT per molecule at 300 K.

,2)(6

6

12

12

rA

rAru

rq

ze