characteristics of soft matter (1) length scales between atomic and macroscopic (sometimes called...
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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