Characteristics of Soft Matter 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.

Interaction Potentials 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 H O H O H H d- d+ d- d+ d+ d+ 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: 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 + -

O-C-O Examples of Nonpolar Molecules: u = 0 CO2 H 109º C C CH4 methane CCl4 Cl C 109º Have rotational and mirror symmetry

Have lost some rotational and mirror symmetry! Examples of Polar Molecules CHCl3 CH3Cl Cl H C C H Cl H Cl H Cl Have lost some rotational and mirror symmetry!

C=O u = 0.11 D N H H O S O O Dipole moments Bond moments N-H 1.31 D + - Vector addition of bond moments is used to find u for molecules. N H u = 1.47 D - + N-H 1.31 D H O - + u = 1.85 D O-H 1.51 D V. High! F-H 1.94 D S O O u = 1.62 D + - What is S-O bond moment? Find from vector addition.

Charge-Dipole Interactions + q Q - W(r) = -Cr -6 r 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 charges: Ions can induce ordering and alignment of polar molecules. Why? W(r) decreases as q decreases to 0.

Dipole-Dipole Interactions + + q1 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 q1 q2 r 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. W(r) = -Cr -3

w(r) (J) W(r) = -Cr -3 kT at 300 K w(r) (J) kT at 300 K At a typical spacing of 0.4 nm, w(r) is about 1 kT. Hence, thermal energy is able to disrupt the alignment. Side-by-side q1 = q2 = 90° -10-19 End-to-end q1 = q2 = 0 W(r) = -Cr -3 -2 x10-19 0.4 r (nm)

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: Interaction energy depends inversely on T, and because of constant motion, there is no angular dependence: W(r) = -Cr -6

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: Units of polarisability:

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 _ _ + + In an electric field Initial state

Simple Illustration of e- Polarisability Without a field: With a field: Fext Fint Force on the electron due to the field: Attractive, Coulombic force on the electron from nucleus: At equilibrium, the forces balance:

Simple Illustration of e- Polarisability Substituting expressions for the forces: Solving for the induced dipole moment: So we obtain an expression for the polarisability: Electronic polarisability is proportional to the size of the molecule!

Units of Electronic Polarisability ao is often expressed as: Units of volume

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 10.5 Smallest Units: (4o)10-30 m3 =1.11 x 10-40 C2m2J-1 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+ dispersed in CCl4 (non-polar). Field from the Ca2+ ion: From the literature, we find for CCl4: Affected by the permittivity of CCl4: e = 2.2 We find when r = 2 nm: u = 3.82 x 10-31 Cm Thus, an electron with charge e is shifted by: Å

Polarisability of Polar Molecules In a liquid, molecules are continuously rotating and turning, so the time-averaged dipole moment for a polar molecule in the liquid state is 0. An external electric field can partially align dipoles: + -  Let q represent the angle between the dipole moment of a molecule and an external E-field direction. The induced dipole moment is: The spatially-averaged value of <cos2q> = 1/3 As u = aE, we can define an orientational polarisability. The molecule still has electronic polarisability, so the total polarisability, a, is given as: 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. 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 - + - + r

Origin of the London or Dispersive Energy + - r The field produced by the instantaneous dipole is: u1  u2 So the induced dipole moment in the neighbour is: We can now calculate the interaction energy between the two dipoles (using equations for permanent dipoles): 

Origin of the London or Dispersive Energy + - r This result: compares favourably with the London result (1937) that was derived from a quantum-mechanical approach: hn is the ionisation energy, i.e. the energy to remove an electron from the molecule

London or Dispersive Energy The London result is of the form: where C is called the London constant: 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 E-field. The Clausius-Mossotti equation relates the dielectric constant e of a molecule with a volume v to a: •At the frequency of visible light, however, only the electronic polarisability, eo, is active. • At these frequencies, the Lorenz-Lorentz equation relates the refractive index (n2 = e) to ao:

Measuring Polarisability The van der Waals’ gas law can be written (with V = molar volume) as: The constant, a, is directly related to the London constant, C: 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.

Type of Interaction Interaction Energy, w(r) Summary Type of Interaction Interaction Energy, w(r) In vacuum: e=1 Charge-charge Coulombic Dipole-charge Dipole-dipole Keesom Charge-nonpolar Dipole-nonpolar Debye Nonpolar-nonpolar Dispersive

Comparison of the Dependence of Interaction Potentials on r Coulombic n = 2 n = 6 n = 3 van der Waals 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: • 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: with r = number of molecules per unit volume  s -3, where s is the molecular diameter. So, with n = 6: For one mole, Esubstance = (1/2)NAE Esubstance = sum of heats of melting + vaporisation. Predictions agree well with experiment! 1/2 to avoid double counting!

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 s, then the shortest centre-to-centre distance will likewise be s. • Thus the boiling point is approximately:

Comparison of Theory and Experiment Evaluated at close contact where r = s. Note that ao and C increase with s. C can be found experimentally from deviations from the ideal gas law:

Additivity of Interactions Molecule Mol. Wt. u (D) TB(°C) C-C H Dispersive only Ethane: CH3CH3 30 0 -89 C=O H Keesom + dispersive Formaldehyde: HCHO 30 2.3 -21 C-O-H H H-bonding + Keesom + dispersive Methanol: CH3OH 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 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. (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.) r q ze

Micellisation Immiscibility Association of molecules Protein folding Adhesion in water De-wetting “Froth flotation” Coagulation