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CHE-1H26: Elements of Chemical Physics Part IV: Electrical Properties

Dr Yimin Chao Room 1.45 Email: [email protected] http://www.uea.ac.uk/~qwn07jsu

1 Outline

1. THE ELECTRIC – Revision 2. AND POLARISABILITY 3. PERMANENT AND INDUCED ELECTRIC MOMENTS IN 4. MOLAR POLARISABILITY, AND OPTICAL ACTIVITY 5. INTERMOLECULAR AND MULTIPOLE INTERACTIONS

2 I. THE

Electric – Law :

The force intensity between two static charged particles q1 and q2

1 q q q1 magnitude F = 1 2 F πε 2 4 0 r F r q2

F ε = × −12 2 −1 −2 units: -Newton (N) 0 .8 854 10 C N m

q -Coulomb (C) constant -Meter (m) r 1 − = .8 988 ×10 9 C 2 Nm 2 πε 4 0

3 direction: depending on the sign of the charge

F(r o/2)

F(r o) F(2r o) -F(2r o) -F(r o)

-F(r o/2)

4 Notes: 1. The electric force can describe the interactions between or any charged particles

2.______Some typical values of forces. ______Type of Force Intensity Electrical force between two charges of 1C each placed 1 m apart 8.98 × 10 9 N Electrical Force between an and a placed 1mm apart 2.309 ×10 −23 N Compare with the Gravitational force between the electron and the proton separated the same distance 1.016 ×10 −61 N i. Is it correct not to take into account the gravitational force between these quantities? ii. Here we have a table with the mass and charge of the elementary particles that compose an ______Particle Mass m (kg) Charge q (C)

−31 −19 Electron me= 9.1091 ×10 qelectron ≡ -e = -1.602177 ×10 C −27 Proton mp= 1.6725 ×10 qproton ≡ +e

−27 mn= 1.6748 ×10 qneutron = 0

5 F ← Vectorial Representation 1 2 q1

1 q1q2 F1←2 = r r rˆ πε − 2 12 4 0 r| 1 r2 | r r 1 q q 1 12 q = − r 1 r2 rˆ 2 πε − 2 21 4 0 r| 1 r2 | O r2

= −F 2←1 Vectorial configuration for two opposite charges

r r Where r − r rˆ = r1 r2 12 − unit vector r| 1 r 2 |

Note: r r r r − = − | r1 r 2 || r2 r1|

6 Electric Fields:

Electric Fields E is defined as the force that a positive test charge would feel in a particular due to the presence of certain number of charges.

F0 E0 = q 0

7 Electric field of a point charge

r 1 qq F = 0 rˆ 0 πε 2 4 0 r

The electric field produced at point P by an isolated point charge q at S:

r 1 q E = rˆ πε 2 4 0 r

8 Electric field produced by point charge s

If there are several charges present in the system, then the electric field generated is: the sum of overall the individual electric field generated by each particle

r r 1  q q q  E (r) = r 1r rˆ + r 2r rˆ + ... + r Nr rˆ total πε  − 2 1 − 2 2 − 2 N  4 0 | r r1 | | r r2 | | r rN |  1 N q = r kr rˆ πε ∑ − 2 k 4 0 k=1 | r rk |

r r r − r where rˆ = r rk k − r| r k |

9 Electric field lines

An electric is an imaginary line or curve drawn through a region of space so that its tangent at any point is in the direction of the electric field vector at that point.

Examples:

10

Electric potential φ represents the potential (stored) energy of a unit positive charge in the electric field. The difference in electric potential between two points in space is V V ≡ ∆φ = −E∆r or E = − ∆r For a given of the unit charge, the electric potential-change is the electric field strength times the distance that the charge has moved.

Thus, the potential of a point charge is φ = 1 q πε 4 0 r Units: Electric field E: V/m (equivalent to N/C ) Electric Potential φ or V: V ()

When there are N particles in the system: - the sum of overall the individual electric potential r 1  q q q  φ (r) = r 1 r + r 2r + ... + r Nr total πε  − − −  4 0 | r r1 | | r r2 | | r rN | 1 N q = r k r πε ∑ − 4 0 k=1 | r rk | 11 Electric dipole

An electric dipole consists of two +q and –q separated by a distance R.

The product of the charge q and the separation R is called the r r µ = qR

Unit: SI unit- C m Non-SI unit- Debye, D , named after , a pioneer in the study of molecular dipole moments

1D = 3.336 × 10 -30 C m

12 Examples of molecular

13 Example:

An electric dipole consists of two charges, q1 = +12 nC and q2 = -12 nC , placed 10 cm apart. Compute (i) total field at points a, b, and c (ii) the potentials at points a, b, and c by adding the potentials due to eitherr charge. (i) At point a, the field caused by the positive charge q1 r E1 and the field caused by the negative charge q2 E2 are both directed towards the right.

r r The magnitudes of E1 and E2 are

1 | q | 12 ×10 −9 C E = 1 = 0.9( ×10 9 N ⋅ m 2 / C 2 ) = 0.3 ×10 4 N / C 1 πε 2 2 4 0 r1 .0( 060 m) 1 | q | 12 ×10 −9 C E = 2 = 0.9( ×10 9 N ⋅ m 2 / C 2 ) = 8.6 ×10 4 N / C 2 πε 2 2 4 0 r2 .0( 040 m) r r E = 3.0 ×10 4 N/C; E = 0 The components of E1 and E2 are 1x 1y 4 E2x = 6.8 ×10 N/C; E 2y = 0

4 (Ea)x = E 1x + E2x = (3.0 + 6.8) ×10 N/C; (Ea)y = E 1y + E 2y = 0 At point a, the total field has magnitude 9.8 × 10 4 N/C and is directed toward the right, so r = × 4 ˆ Ea ( 9.8 10 N/C )i 14 r At point b, the field due to charge q1 is directed toward the left, and rE1 the field due to charge q2 is directed toward the right E2 r r The magnitudes of E1 and E2 are

1 | q | 12 ×10 −9 C E = 1 = 0.9( ×10 9 N ⋅ m 2 / C 2 ) = 8.6 ×10 4 N / C 1 πε 2 2 4 0 r .0( 040 m) 1 | q | 12 ×10 −9 C E = 2 = 0.9( ×10 9 N ⋅ m 2 / C 2 ) = 55.0 ×10 4 N / C 2 πε 2 2 4 0 r .0( 140 m) r r r The components of E1 and E2 , and total field Eb at point b are

4 E1x = -6.8 ×10 N/C; E 1y = 0 4 E2x = 0.55 ×10 N/C; E 2y = 0

4 (Eb)x = E 1x + E2x = (-6.8 + 0.55) ×10 N/C; (Eb)y = E 1y + E 2y = 0 Thus, the electric field at b has magnitude 6.2 × 104 N/C and is directed toward the left, so r = − × 4 ˆ Eb ( 2.6 10 N/C )i

15 At point c,

r r both the field E1 and E2 have the same magnitude, since this point is equidistant from both charges and the charge magnitudes are the same:

1 | q | 12 ×10 −9 C E = E = = 0.9( ×10 9 N ⋅ m 2 / C 2 ) = 39.6 ×10 3 N / C 1 2 πε 2 2 4 0 r .0( 130 m)

The x-components of both vectors are the same

3 3 E1x = E2x = E 1cos α = (6.39 ×10 N/C (5/13) = 2.46 ×10 N/C

From symmetry the y-components E 1y andr E 2y are equal and opposite and so add to zero.

Hence the components of the total field Ec are

3 3 (Ec)x = E 1x + E2x = 2(2.46 ×10 )N/C = 4.9 ×10 N/C; (Ec)y = E 1y + E 2y = 0 Thus, the electric field at c has magnitude 4.9× 10 3 N/C and is directed toward the right, so r = × 3 ˆ Ec ( 9.4 10 N/C )i

16 r 1  q q q  1 N q (ii). φ (r) = r 1 r + r 2r + ... + r Nr = r k r total πε  − − −  πε ∑ − 4 0 | r r1 | | r r2 | | r rN | 4 0 k =1 | r rk |

r 1  q q  12 ×10 −9 C (−12 ×10 −9 C)  φ (r ) = r1 + r 2 = ()0.9 ×10 9 N ⋅ m 2 / C 2  +  = −900 N ⋅ m / C = −900 J / C = −900 V a πε     4 0 | r1a | | r2a |  .0 060 m .0 040 m 

r 1  q q  12 ×10 −9 C (−12 ×10 −9 C)  φ (r ) = r1 + r 2 = ()0.9 ×10 9 N ⋅ m2 / C 2  +  = 1930 V b πε     4 0 | r1b | | r2b |  .0 040 m .0 140 m 

r 1  q q  12 ×10 −9 C (−12 ×10 −9 C)  φ (r) = r1 + r 2 = ()0.9 ×10 9 N ⋅ m 2 / C 2  +  = 0 c πε     4 0 | r1c | | r2c |  .0 130 m .0 130 m 

17 II. DIELECTRICS AND POLARISABILITY

Capacitor

Any two conductors separated by an (or a vacuum) form a

Q C = definition of capacitanc e Vab

Unit: , F 1F = 1 farad = 1 C/V = 1 Coulomb/

18 Parallel-plate capacitor The simplest form of capacitor consists of two parallel conducting plates, each with area A, separated by a distance d that is small in comparison with their dimensions. Surface : charges per unit area Q σ = A Since the total electric field of a charge is Q/ εo, then the field per unit is Q σ Ref: University Physics 12 th E = = edition example 21.13 or example ε ε 22.8) 0 A 0 therefore Qd σ d V = Ed = = ab ε ε 0 A 0 of a parallel-plate capacitor in vacuum = Q =ε A C0 0 Vab d

19 stored in a capacitor: Q Since V = C Let q and v be the charge and potential difference at an intermediate stage during the charging process; then v = q/C . at this stage the dW required to transfer an additional element of charge dq is qdq dW = vdq = C The total work W needed to increase the capacitor charge q from zero to a final value Q is

W 1 Q Q 2 W = ∫ dW = ∫ qdq = (work to charge a capacitor) 0 C 0 2C

If we define the potential energy of an uncharged capacitor to be zero, then W is equal to the potential energy U of the chearged capacitor. The final stored charge is Q = CV , so we can express U (which is equal to W) as 2 = Q = 1 2 = 1 U C0V QV 2C0 2 2 Energy density: the energy per unit volume in the space between the plates of a parallel-plate capacitor with plate area A and separation d. Q A 1 2 C = =ε C V 0 V 0 d 0 1 ab u = 2 = ε E 2 Ad 2 0 V E = d 20 in a capacitor When a nonconductive material, a dielectric , is placed between the parallel plates the charge in the plates is “screened” by it, and the capacitance of the capacitor increases

= C =ε Dielectric constant of the material: Κ r C0

--

ε ε ε Note: i. The ratio, relative permittivity, r = / 0 > 1 V E ii. When Q is constant, V = 0 , E = 0 ε ε r r 21 Polarization Since the electric-field magnitude is smaller when the dielectric is present, the surface charge density must be smaller as well. The surface charge on the conducting plates does not change, but an induced charge of the opposite sign appears on each surface of the dielectric. The induced charge arises as a result of redistribution of positive and negative charge within the dielectric material. This phenomenon is called polarization.

σ σ σ σ E = - i - p 0 ε E = = p - polarization 0 ε ε 0 0 Thus induced surface charge density  1  σ =σ1−  = p i  ε   r  and permittivity of the dielectric = ε ε r ε0

σ ∴ E = ε A A and C =ε C =ε ε = ε r 0 r 0 d d 22 Dielectric medium Two charges immersed in a dielectric medium: the force between is reduced in comparison to the force they feel in vacuum: r 1 q q 1 1 q q 1 1 q q F = 1 2 rˆ = 1 2 rˆ = 1 2 rˆ πε 2 πε ε ε 2 πε ε 2 4 r 4 0 / 0 r 4 0 r r ε ε ε The ratio, relative permittivity, r = / 0 > 1 therefore the force between the charges in presence of a medium is reduced.

Correspondingly, the electric field of a charge within this medium will be reduced by the same factor: r 1 q 1 1 q 1 1 q E = 1 rˆ = 1 rˆ = 1 rˆ πε 2 πε ε ε 2 πε ε 2 4 r 4 0 / 0 r 4 0 r r

the electric potential of a charge φ = 1 q = 1 q πε ε πε 4 0 r r 4 r

Energy density: 1 1 u = ε ε E 2 = εE 2 2 0 r 2

23 Notes :

1. Whenever charges are in a dielectric medium, a simple rule of thumb is to replace the ε ε ε ε 0 by = 0 r, in all the expressions found earlier.

2. Although we have an explanation to why the electric force and the electric field are reduced in the presence of dielectric media, we do not have a clear picture of why the dielectric polarises in the presence of the field. We will discuss this next.

24 III. PERMANENT AND INDUCED ELECTRIC DIPOLE MOMENTS IN MOLECULES

Polar : with permanent dipole moment

Non-polar molecule: With no permanent dipole moment

25 Induced dipole moments

In the presence of an applied electric field, like the one generated by the capacitor, the induced dipole moment is proportional to the field µ = α ⋅ ind E where α is the of the molecule. Unit: C2m2J-1 Note: 1. the greater the polarizability, the larger the induced dipole moment. 2. It is sometimes more convenient to work with the polarizability volume α α ' = , where ε is the permittivity in the vacuum 4πε 0 0 Unit: m 3 3. The value of the reflect the strengths with which the nuclear charges control the electron distribution, prevent its distortion by the field. If the molecule has few , they are tightly controlled by the nuclear charges and the polarizability is lower; On the contrary, if the molecule contains large with electrons at same distance from the nucleus, the nuclear control is weakened and the polarizability is greater. 26 Generalizations and more complicated cases: µ = α ⋅ ind E This proportionality between the induced dipole moment and the electric field is an approximation.

1. Strong field: 1 µ = α ⋅ E + β ⋅ E 2 +... ind 2 Where β is called the hyperpolarizability of the molecule

Notes: i. The stronger the filed involved, the higher order terms must be taken into account ii. We will concentrate on the simplest case ind = α * E

2. Asymmetry of the molecule:

Even for applied electric field which is not strong, the simplest equation ind = α * E is not totally accurate, and it will depend on the symmetry of the molecule.

A more general yet complicated expression is (µ ) = α ⋅ +α ⋅ +α ⋅ ind x xx Ex xy Ey xz Ez ()µ = α ⋅ +α ⋅ +α ⋅ ind y yx Ex yy Ey yz Ez ()µ = α ⋅ +α ⋅ +α ⋅ ind z zx Ex zy Ey zz Ez

27 Examples of molecular dipoles + µ 1 µ : o µ1 = µ2 = 1.52D; θ = 105

µ = µ 2 + µ 2 + µ µ θ θ 1 2 2 1 2 cos µ _ 2 + = 52.1 2 + 52.1 2 + 2× 52.1 × 52.1 × cos 105 o D = 85.1 D

i. This charge distribution is what makes the water an excellent solvent for ionic substances such as salt (NaCl). ii. Notice that an ionic substance such as NaCl is an extreme case of permanent dipole moment which leads to a molecule with positively and negatively charged ions. iii. If water molecules were not electric dipoles, water would be a poor solvent and almost all of the chemistry that occurs in aqueous solutions would be not feasible. iv. Molecules can be classified in terms of its permanent dipole moments: from a nonpolar molecule to the extreme case of an ionic substance, and a polar molecule is an intermediate case.

nonpolar molecule polar molecule ionic substance

28 Calculation of dipole moments

A better approach to the calculation of dipole moments is to take into account the locations and magnitudes of the partial charges on all the atoms. To calculate the x-component, for instance, we need to know the partial charge on each atom and the atom’s x-coordinate relative to a point in the molecule and form the sum

µ = x ∑ qJ xJ J

Here qJ is the partial charge of atom J, xJ is the x-coordinate of atom J, and the sum is over all the atoms in the molecule.

In common with all vectors, the magnitude of µ is related to the three components x , y and z by

1/2 = ( x + y + z)

29 Example Estimate the electric dipole moment of the amide group by using the partial charges and the locations of the atoms shown.

The expression for x is

x = (-0.36e) × (132 pm) + (0.45e) × (0 pm) + (0.18e) × (182 pm) + (-0.38e) × (-62.0 pm) = 8.8e pm =8.8 × (1.609 × 10 -19 C) × (10 -12 m) = 1.4 ×10 -30 C m = 0.42 D

y = (-0.36e) × (0 pm) + (0.45e) × (0 pm) + (0.18e) × (86.6 pm) + (-0.38e) × (107 pm) = -56e pm = -9.1 ×10 -30 C m = -2.7 D

Because z = 0 = {(0.42D) 2 + (-2.7D) 2}1/2 = 2.7D

Orientation: an arrow of length 2.7 units , x, y, z components of 0.42, -2.7, and 0 30 IV. MOLAR POLARISABILITY, AND OPTICAL ACTIVITY ε ε = Relative permittivity r ε 0 The relative permittivity of a substance is large if its molecules are polar or highly polarizable. The quantitative relation between the relative permittivity and the electric properties of the molecules is obtained by considering the polarization of a medium, and is expressed by the Debye equation : ε −1 ρ P r = m ε + r 2 M ρ − the mass density of the sample M − molecular mass of the molecules N  µ 2  P − the molar polarization P = A α +  m m ε   3 0  3kT   µ 2    -- the thermal averaging of the electric dipole moment in the presence of the   th  3kT  applied field, see equ 18.5 Physical Chemistry 8 Ed. N The expression without the contribution from the permanent dipole moment P = A α m ε 3 0 ε −1 ρ N α thus r = A ε + ε -- Clausius-Mossotti equation r 2 3M 0 31 Notes:

1. Conditions for using the Clausious-Mossotti equation : the Clausious-Mossotti equation is used when there is no contribution from permanent electric dipole moment to the polarization i. molecules are non-polar ii. The of the applied field is so high that the molecules cannot orientate quickly enough to follow the change in direction of the field N  µ2  P = A α +  α and µ: m ε   2. Method of measuring 3 0  3kT 

N N µ2  1  P = A α + A   m 3ε 9ε κ T  Measure relative permittivity by comparing the 0 0 capacitance of a capacitor with and without the sample µ2 presence ( C and C ) NA o Slope ε κ ε −1 ρ P 9 0 r = m ε ε +2 M ε = = C r r = Plot P vs 1/T ε C Pm Pm (T) m N 0 o A α ρ = ρ(T) Intercept ε 3 o

32 Determining dipole moment and polarizability

The relative permittivity of a substance is measured by comparing the capacitor with and without

the sample present ( C and C 0, respectively) and using εr = C/C 0. Therelative permittivity of camphor eas measured at a series of temperatures with the results given below. Determine the dipole moment and the polarizability volume of the molecule.

θ/°C 0 20 40 60 80 100 120 140 160 200

εr 12.5 11.4 10.8 10.0 9.50 8.90 8.10 7.60 7.11 6.21 ρ/(g cm -3) 0.99 0.99 0.99 0.99 0.99 0.99 0.97 0.96 0.95 0.91

M = 152.23 g mol -1

C10 H16 O

33 ε −1 ρ P -1 r = m For camphor, M = 152.23 g mol , from Debye equation ε + r 2 M

We can calculate (εr-1)/( εr+2) at each temperature, then multiply by M/ ρ to form P m θ/°C 0 20 40 60 80 100 120 140 160 200

εr 12.5 11.4 10.8 10.0 9.50 8.90 8.10 7.60 7.11 6.21 ρ/(g cm -3) 0.99 0.99 0.99 0.99 0.99 0.99 0.97 0.96 0.95 0.91 (10 3K)/T 3.66 3.41 3.19 3.00 2.83 2.68 2.54 2.42 2.31 2.11

(εr-1)/( εr+2) 0.793 0.776 0.766 0.750 0.739 0.725 0.703 0.688 0.670 0.634 3 -1 Pm/(cm mol ) 122 119 118 115 114 111 110 109 107 106

N N µ2  1  P = A α + A   m ε ε κ 3 0 9 0 T  N From the plot, the intercept lies at 82.7, Intercept = A α; 3ε so α’= 3.3 × 10 -23 cm 3. o α 3 α ' = = ⋅int ercept πε π 4 0 4 N A The slope is 10.9 (cm 3mol -1) /(10 3K) = 10.9 × 10 -6 × 10 3 (m 3mol -1/K), N µ2 slope = A 9ε k so = 4.46 ×10 -30 C m = 1.34 D 0 9ε k µ = 0 ⋅slope N (1D = 3.336 × 10 -30 C m) A 34 Polarization at high

Orientation polarization : if the applied field action on the molecules change direction periodically, the permanent dipoles moments re-orientate and follow the field. However, when the frequency of the field is high, the molecules are unable to orient themselves rapidly enough to keep up with the field and they make no contribution to the polarization of the sample. A molecule takes about 1ps to turn through about 1 radian in a fluid, then for frequencies greater than 10 11 Hz (in the region), the contribution of orientation polarization is lost.

Distortion polarization : the polarization that arises from the distortions of the nuclei by the applied field. The molecule is bent and stretched by the applied field, and the molecular dipole moment changes accordingly. The time taken for a molecule to bend is approximately the inverse of the vibrational frequency . So the distortion polarization disappears when the frequency of the radiation is increased through the infrared, 3x10 11 - 4x10 14 Hz.

Electronic polarization : in the visible region, 3.84x10 14 – 7.69x10 14 Hz, only the electrons are mobile enough to respond to the rapidly changing direction of the applied field.

35 and optical activity

Refraction : a beam of light change direction (“bends”) when it passes from one transparent medium to another, this effect, called refraction,

depends on the refractive index, nr, of the medium, the ratio of the in a vacuum, c, to its speed v in the medium. c n = r v From Maxwell equation which describe the properties of the electromagnetic radiation, it can be seen that the refractive index at a (visible or ultraviolet) specified frequency is related to the relative permittivity

ε n 2 = =ε r ε r o ε −1 ρ P n = ε r = m r r ε + Debye equation r 2 M (ε − )1 M (n 2 − )1 M P = r = r Clausius-Mossotti equation m ε + ρ 2 + ρ ( r )2 (nr )2

This gives us an alternative way of calculating the molecular polarisability and the dipole moment

Atkins P984 36 Optical activity

l l ∆t = − cL cR

l ∆t = (n −n ) R L c

2πc∆t 2πl ∆θ = 2πν∆ t = = (n − n ) λ R L λ

Atkins P985 37 V. INTERMOLECULAR FORCES AND MULTIPOLE INTERACTIONS

Van de Waals force : the attractive interaction between closed-shell molecules

Net attractive interactions between the partial charges of polar molecules

Repulsive interactions between molecules that Coulombic repulsions prevent the complete collapse of matter to nuclear densities. Pauli principle

Interactions between dipoles – the potential energy

The potential energy between two charges is given by

q q U = 1 2 πε 4 0r12

Where r12 is the distance between charges 1 and 2. 38 Point dipole and a point charge:

Point charge : is such that we are not interested in its dimensions.

Point dipole : is such that the distance separating the charges (+ q1) and (-q1) is much smaller than the distance to any other charges present in the system, l << r

The potential energy for this system: 1  q q q q  U = 1 2 − 1 2 dipole ↔ ch arg e πε  + −  4 0 r l 2/ r l 2/  q q  + − −  = 1 2 r l 2/ r l 2/ πε  + −  4 0 (r l )(2/ r l )2/  −lq q = 1 2 πε 2 4 0r − µ q since µ = q l, therefore U = 1 2 1 1 dipole ↔ch arg e πε 2 4 0 r 39 Notes:

1. This simple expression is for the case where the dipole is aligned with the point charge. 2. To take into account the dipole forming an angle, the expression should be multiplied by cos θ

− q1 θ

− µ q q = 1 2 θ 2 U ↔ cos l dipole ch arg e πε 2 4 0 r

q1 r 3. The expression refers to the potential energy of the dipole due to the charge and does not include the “self” energy of the dipole.

4. Interaction between the point dipole and the point charge vanishes when the angle θ= 90 o. This is due to fact that the separation l is considered much smaller than the distance r, therefore the point charged dipole appears a neutrally charged entity.

40 Interaction between two dipoles

1  q q q q q q q q  U =  − 1 2 + 1 2 + 1 2 − 1 2  dipole ↔dipole πε + − 4 0  r l r r r l  q q  1 1  = − 1 2  − 2 +  πε + − 4 0r 1 x 1 x  with x = l/r 2 = − 2x q1q2 πε 4 0r µ = µ = since 1 q1l, 2 q 2 l, 2x 2 q q µ µ 2 U = − 1 2 = − 1 2 dipole ↔dipole πε πε 3 4 0 r 4 0 r

41 n-pole: is an array of pint charges with an n-pole moment but no lower moment. monopole (n = 1): point charge, monopole moment is the overall charge dipole (n = 2): an array of charges that has no monopole moment (no net charge) (n = 3):consists of an array of point charges that has neither net charge nor dipole moment, as for

CO 2 molecules

42 potential energy of multipole interaction octupole: consists of an array of point charges that sum to zero and which has neither a dipole nor a quasrupole moment, as for CH 4 molecules

for the interaction of an n-pole with an m-pole, the potential energy varies with distance as

1 U ∝ r n+m−1

monopole-monopole: n =m =1, 1/r monopole-dipole: n =1, m =2, 1/r 2 dipole-dipole: n =2, m =2, 1/r 3

43 Summary of multipole interaction potential energies

44 the electric field

i. the electric field generated by a point charge:

q E = πε 2 4 0 r

ii. the electric field generated by a dipole:

1  q q  E = 1 − 1 πε  + 2 − 2  4 0 (r l )2/ (r l )2/  q  −  = 1 4rl 2/ πε  2 − 2 2  4 0 (r l )2/( )  −2µ = 1 πε 3 4 0r

Note: the electric field of multipole decreases more rapidly with distance than a monopole.

45 Dipole-dipole interaction The potential energy of interaction between two polar molecules is complicated function of their relative orientation.

When the two dipoles are parallel, the potential energy is µ µ 1( − 3cos 2 θ) U = 1 2 dipole ↔dipole πε 3 4 0 r Note: i. This expression is valid for polar molecules which have affixed orientation. ii. In a fluid, molecules are freely to rotate, or in other words, at a given time it is possible to find a molecule in every single orientation. Therefore, in average the angular function (1- 3cos 2 θ) is zero, and the interaction between freely rotating dipoles is zero.

iii. Although the interaction of two freely rotating molecules is zero, molecules do NOT rotate freely. This is due to the fact that their mutual potential energy depends on their relative orientation, and therefore the molecules do not in fact rotate completely freely even in a gas. In fact the lower energy orientations are marginally favoured so there is a non-zero average interaction between polar molecules. This means that in a fluid there will be more molecules oriented in the direction where the potential energy is minimized.

46 A measure of this preference is given by Boltzmann distribution

P ~ exp (-U/kT)

When this factor is included, the average of the interaction energy:

µ µ  µ µ  U = 1 2  1− 3cos 2 θ − 1 2 1( − 3cos 2 θ )2 + ...  πε 3  πε 3  4 0r  4 0kTr  2 µ 2µ 2 U ≈ − 1 2 πε 2 6 3 4( 0 ) kTr

Note: negative sign, means that the average interaction is attractive

47 Example:

o For pairs of molecules with µ1 = µ2 = 1 D, T = 25 C, r = 0.5nm

2 µ 2 µ 2 2 .3( 336 ×10 −30 ) 4 U ≈ − 1 2 = − ⋅ .8( 988 ×10 9 ) 2 ⋅ πε 2 6 × −23 × × × −9 6 3 4( 0 ) kTr 3 38.1 10 298 5.0( 10 ) = −1.038 ×10 −22 J 1D = 3.336 × 10 -30 C m

1 − = .8 988 ×10 9 C 2 Nm 2 For 1 mol molecules, πε 4 0 ~ -1.038 × 10 -22 × 6.023 × 10 23 = -0.0625 kJmol -1;

Which is much smaller than average molar kinetic energy

Average molar kinetic energy – energies involved in the making and breaking of chemical bonds

3 3 − kT ⋅ N = RT = 7.3 kJmol 1 2 A 2

48 Dipole – induced dipole interactions: A polar molecule with dipole moment can induce a dipole in a polarisable molecule. The induced dipole interacts with the permanent dipole of the first molecule and the two are attracted together. 2µ µ * = − 1 2 1 U * dipole ↔dipole πε 3 4 0 r The induced dipole moment depends on the field generated by the polar molecule, −2µ µ* =α E =α 1 2 2 2 πε 3 4 0r (a) A polar molecule (purple arrow) can induce a 2µ α 2µ dipole (white arrow) in a nonpolar molecule, and Then we have = − 1 2 × 1 U * (b) the latter’s orientation follows the former’s, dipole ↔dipole πε 3 πε 3 4 0 r 4 0 r so the interaction does not average to zero. 4µ 2α = − 1 2 πε 2 6 4( 0 ) r 4µ 2α ' = − 1 2 πε 6 4( 0 )r Note : the dipole – induced dipole interaction energy is independent of the temperature, because thermal motion has no effect on the averaging process.

49 Example : for a molecule with µ = 1 D, near a molecule of polarisability volume α’ = 10 -29 m3, separation is 0.3 nm, Calculate the average interaction energy

4µ 2α = − 1 2 U * dipole ↔dipole πε 2 6 4( 0 ) r µ 2α ' −30 2 −29 4 2 .3(4 336 ×10 ) ×10 = − 1 = − .8( 988 ×10 9 ) ⋅ πε 6 × −9 6 4( 0 )r 3.0( 10 ) = −5.488 ×10 −21 J 1D = 3.336 × 10 -30 C m

1 − = .8 988 ×10 9 C 2 Nm 2 πε 4 0

For 1 mol molecules,

-21 23 -1 = -5.488 × 10 ×6.02 ×10 = -3.2 kJ mol

50 Induced dipole – induced dipole interactions

Non-polar molecules attract one another even though neither has a permanent dipole moment.

Dispersion interaction: also called London interaction (in honour of Fritz London, who first described it) This interaction arises form the instantaneous transient dipoles that all molecules have as a result of fluctuations (a) In the interaction, an instantaneous in the instantaneous position of electrons. dipole on one molecule induces a dipole on another molecule, and the two dipoles then interact to lower the energy. (b) the two instantaneous dipoles are correlated and, although they occur in different orientations at different instants, the interaction does London Formula: a reasonable approximation of the not average to zero exact calculation of the dispersion interaction 2 I I 1 U = − α 'α ' 1 2 1 2 + 6 3 I1 I 2 r

Where I 1 and I 2 are the ionization energies of two molecules

51 Example:

-30 3 Two methane CH 4 molecules, α’ = 2.6 × 10 m , I ≈ 7 eV , separation r = 0.3 nm

I ≈ 7 eV = 7 × 1.6 × 10 -19 J = 1.12 × 10 -18 J

2 I I 1 U = − α 'α ' 1 2 1 2 + 6 3 I1 I 2 r −18 2 − 12.1 ×10 1 = − ⋅ 6.2( ×10 30 ) 2 ⋅ ⋅ 3 2 3.0( ×10 −9 )6 = − 46.3 ×10 −21 J

Then, for a mol molecules,

U = -3.46 × 10 -21 J × 6.02 × 10 23 mol -1= -2.082 kJmol -1

52 The total attractive interaction If we consider molecules that are unable to participate in hydrogen bond formation, then the total attractive interaction energy between rotating molecules is the sum of the three van de Waals contribution discussed:

Dipole – dipole Dipole – induced dipole Induced dipole – induced dipole

All three vary as the inverse sixth power of the separation, so we may consider the potential energy of the attractive interaction as C U = − 6 r 6

Where the coefficient C 6 depends on the identity of the molecule.

53 Repulsion and total interactions When molecules are squeezed together, the nuclear and electronic repulsions and the rising electronic kinetic energy begins to dominate the attractive forces.

The repulsion increases steeply with decreasing separation in a way that can be deduced only by very extensive molecular calculations. Hard-sphere potential : a very crude approximation, in which it is assumed that the potential energy rises abruptly to infinity as soon as the particles come within a separation d – collision diameter U = ∞, for r ≤ d U = 0, for r > d C C Mie potential : U = n − m with n > m r n r m

attractions repulsions

 r 12  r 6  with n = 12, m = 6, = 0 − 0 U 4U 0       r   r   - Lenard – Jones potential 54 Lenard – Jones potential

 r 12  r 6  = 0 − 0 U 4U 0       r   r  

When r = r 0, U = 0 1/6 When r = 2 r0, U = Umin = -U0

-U0

12 -13 6 -7 dU/dr = -12r 0 r + 6r 0 r = 0

1/6 r = 2 r0

(U 0/k)/K

55 Atomic force microscopy (AFM) With the adventure of AFM, in which the force between a molecular size probe and a surface is monitored, it has become possible to measure directly the force acting between molecules.

The force, F, is the negative slope of potential, so for a Lennard-Jones potential between individual molecules we write

13 7 In AFM, a laser beam is used to monitor the tiny dU 24 U   r   r   F = − = 0 2 0  −  0   changes in the position of a probe as it is attracted to or repelled from atoms on a surface dr r0   r   r   From dF/dr = 0 , the net attractive force is greatest at

 26  6/1 r =   r = .1 244 r  7  0 0 For typical parameters, the magnitude of this force is about 10 pN

An AFM image of bacterial DNA plasmids on a mica surface

56