HW7 NME220 Solution

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Introduction toMolecular and Nanoscale Principles (NME 220)

Homework 7 Solution due:May 18, 2012

1

Reading Assignment in Course Textbook:

Course Textbook: Chapter 4 (Section 4.1-4.2.3); Chapter 10 (Section 10.2.1)

Lectures: Lecture notes on Interactions in particular: Dipolar Interaction

A. Narrative Overview Questions

1.1. Provide approximate magnitudes of the interaction energies of ionic,covalent, hydrogen and Van der Waals (VdW) bonds.

Answer:Approximate interaction energies are as follows:

- Ionic: ~150-250 kcal/mol

- Covalent: ~150-300 kcal/mol

- Hydrogen: ~2-40 kcal/mol

NH:O ~2 kcal/mol,NH:N ~3 kcal/mol,

OH:N ~7 kcal/mol,

OH:O 5 kcal/mol,

FH:F ~39 kcal/mol

- VdW: ~1-5 kcal/mol

CH4 ~2 kcal/mol

N-butane ~5 kcal/mol

1.2. Briefly explain what is involved in hydrogen bonding. Whichelectronegative atoms are involved in hydrogen bonds? What are theirelectronegativities based on the Pauling scale.

Answer: Hydrogen bonding leads to a strong attractive interaction between

- a very electronegative atom (from another molecule or chemical group)

- a tiny partially positive proton in form a hydrogen atom (covalently bonded

to another electronegative atom)

The following electronegative atoms are involved in hydrogen bonding:

Electronegativity Values

- Nitrogen 3.04- Oxygen 3.44

- Fluorine 3.98


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1.3. List at least three molecular groups, in which their compounds can interactvia hydrogen bonds.

Answer:

Acid Amide

Carboxylic Acid

Alcohol

Amine

1.4. What kind of interaction holds the -sheet in proteins together?

Answer: Hydrogen bonds. The sheet is a

regular secondary structure in proteins that

connects beta strands laterally by five or more

hydrogen bonds, forming a generally twisted,

pleated sheet.

1.5. What is the relative permittivity of a medium?

Answer: The relative permittivity of a medium is a measure of how an

external electric field is modified, and in turn affects the medium.

1.6. List the three groups of VdW interactions.Answer:

- Dynamic dipole-dipole forces (Keesom forces)

- Induced dipole-dipole forces (Debye)

- Dispersion forces (London)

1.7. What do they have in common concerning their interaction lengthdependence?

Answer: all terms are proportional to 1/r6

.

1.8. Sketch the interaction length dependence V(r) between attractive ionicinteractions and attractive VdW interactions.

Answer: Key is that ionic is proportional to 1/r while VdW are proportional to

1/r6. Attractive interactions are those for which the potential V(r) is negative.

Repulsive interactions are those for which the potential V(r) is positive.


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1.9. Can VdW interactions be repulsive?

Answer: Yes. VdW interactions can be repulsive depending on the sign and

strength of the mediums permittivity. In air, Van der Waals forces are

typically attractive, which explains the universal shape of the Lennard-Jones(LJ) potential.

1.10. What is the definition for the dipole moment?

Answer: The dipole moment can be defined as a charge, q, displaced over a

distance, l:

qlu =

1.11. What is a Debye?

Answer: A Debye (D) is the CGS unit of the electric dipole moment.1D = 3.3310

-30Cm.


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1.12. How does the dipole moment compare between HCl, H2O and CO2?Explain the deviating dipole moment of CO2 in comparison to the twoother molecules

Answer:from Yaws Chemical Properties Handbook:

Chemical Dipole Moment (D)

CO2 0.00

H2O 1.85

HCl 1.08

CO2 has a zero dipole moment (non-polar molecule) due to its perfect

symmetry.

1.13. Provide formally more insight into the Van der Waals interactionparameter C considering the variety of dipolar interactions. Theexpressions have to indicate the dependence on dipole moments,dielectric permittivity, polarizability, thermal dependence and quantumfluctuations.

Answer:

Dispersion/LondonDebyeKeesom CCCC ++=

In the following, identical molecules are assumed.

( ) Tk

uC

Bo

Keesom 2

4

43 =

( )2

2

4

2

o

o

Debye

uC =

( )22

44

3

o

oDispersion/London

hC

=


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B. Worked Problems

2.1. For a monovalent ion such as Na+, the electric field at a distance of 0.4 nmfrom its center is E=e/(4or

2) = 9 109 V/m. Estimate the distance bywhich the electron cloud of a methane (CH4) molecule is shifted relative tothe center of the molecule due to the imposed electric field of the ion.Methane polarizability from the Electronic Polarizability Table: o = 2.6 in units of 4o

3.

Solution

nmC

mC

e

ul

mCCV

J

m

Vm

mJ

Cu

EEu

ind

ind

oind

016.0106.62.1

106.2

106.2109

1

106.21085.84

4

19

30

309330212

0

=

==

=

=

==

This is about 8% of the molecular radius of methane (0.2 nm).

2.2. Determine the orientationalpolarizability of water at 300 K, and compare itto the value of the electronic polarizability. The dipole moment of water is1.85 D, and its electronic polarizability is

( ) ( ) 330o m104481. or ( ) 12240 JmC10111481 .. .

Note the units:11210

o JmC101114

= .

Solution

uwater= 1.85 D(ebye) = (1.85 D) (3.336 10-30 C m D-1) = 6.17 10-30 Cm.

The orientational polarizability is

( )( )( )

12239 JmC103.07

=

==

K300KJ103813

Cm10176

Tk3

u123

230

B

2

orient.

.


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or ( ) 329o m1042.76

( )

( ) 18.64=

=

330o

329

o

o

orient

m104481

m104762

.

.

and exceeds the electronic polarizability by a factor of ~ 20 at room temperature

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HW7 NME220 Solution