II-1 1

PART TWO: Electrostatic Interactions In the first section of this course we were more concerned with structural aspects of molecules. In this section the emphasis is on bonding. Bonds in molecules are formed by the interactions between electrons. One way of probing the bonds is to “look” at the electrons in the bonds. We can do this by the process of ionization. Ionization is the basis of Photoelectron Spectroscopy. You will meet this again in 19-207 (Structure and Spectroscopy). Ionization Energies (p. 73, DeKock & Gray and p. 15-16, DeKock & Gray) When light (photons) of energy E = hν strikes a gas (or solid, or liquid) it can cause electrons to move from one orbital to another.

this is called excitation

II-2 2

If the energy is high enough the electron can escape from the proton.

this is called ionization For the hydrogen atom, the ionization process is:

H(g) + E(=hν) → H+(g) + e-(21 mev2)

∆EH = EnII - EnI The ejected electron can have a Kinetic energy of 2

1 mev2.

The Ionization Energy, IE, of an atom or molecule is the minimum energy required to remove an electron from the gaseous atom or molecule in its ground state. The ejected electron will then have zero kinetic energy.

II-3 3

Einstein: Correctly explained the Photoelectric

effect. hν = IE + 2

1 mev2

NOBEL PRIZE IN PHYSICS

Einstein Photoelectric Law

When Ionization occurs as a result of the interaction of photons with the molecule it is called photoionization. The resulting electrons are called photoelectrons. See DeKand G, Ch. 2 & Ch. 1, p. 16.

II-4 4

The Photoelectric Effect is a very important process in Physics and Chemistry. Basis of Photoelectron Spectroscopy (19-207, Analytical 2/3) To calculate the ionization energy for the H atom we must first recognize that when the atom is ionized its change in energy is the ionization energy. nII ≡ H+

∆EH = IE = EnII - EnI nI ≡ ground State. If we know the photon energy and can measure the KE of the electron (this is photoelectron spectroscopy). We can measure the ionization energy. IE = hν - 2

1 mν2 = Eorb.

This is known as Koopmans’ Theorem.

II-5 5

The experiment: simplified

The energies of the photoelectrons can be measured by placing a negative voltage on the mesh grid. This will repel electrons unless they have enough kinetic energy to get through. This is over-simplified but it will do for now.

II-6 6

Ionization Energies are almost always given in electron volts: This is a useful unit of energy.

The electron volt is the energy acquired byan electron when it is accelerated by apotential difference of 1 VOLT (See APPENDIX1 DEK and G for More.

The Ionization Energy for H is 13.59 eV. (This is sometimes called a Rydberg) 1eV = 96.49 kJ mol-1

For Sodium Na(g) → Na+(g) + e- IE1 = 5.139 eV In all atoms except hydrogen, further ionizations are possible (1 for each electron).

II-7 7

For example Li has 3 ionization energies from the configuration 1s22s1: Li(g) → Li+(g) + e- 2s IE1 = 5.392 eV Li+(g) → Li2+(g) + e- 1s IE2 = 75.638 eV Li2+(g) → Li3+(g) + e- 1s IE3 = 122.45 eV

http://www.webelements.com/webelements/elements/text/Li/econ.html

II-8 8

At the moment we can interpret this by saying that the two electrons in the 1s orbital are closer to the nucleus than the 2s electron and are harder to remove. Why is it harder to remove the last 1s electron than the first? We will need part 4 of this course to really understand this properly. We will return to this again…….

II-9 9

Periodic Trends In Ionization Energies

First ionization energies vary systematically through the periodic table.

DeKock and Gray, p. 78

II-10 10General Points

â In any row IE’s ↓ as Z ↑ (Z = atomic

number). Largest change He→ Xe: Smallest Li → Rb.

ã Across a period IE ↑ (e.g. Na → Ar). (but some exceptions)

II-11 11

Exceptions IE Be > B and IE N > O

Atoms with half full e.g., N: 2s22p3

and full shells e.g., Be: 2s2

have larger IE’s than one would expect. why?

B has a higher nuclear charge than Be.

II-12 12

However in B, the outermost electron is in a 2p orbital and is less strongly bound. i.e., 2s1 2s2 2s22p1

Li Be B IE/eV 5.39 9.32 8.30 Recall the difference between s and p orbitals. s p p electrons “further” from nucleus See part 4 of course.

II-13 13

What About N And O?

IE N > O Valence configurations are: N 2s22p1 2p1 2p1 x y z

O 2s22p22p1 2px y 1z this is the cause: The 2 electrons are close and repel each other and so, the electron is “helped” out by repulsion of the other pX electron. electon-electron repulsions are extremely important in many aspects of bonding as we will see during this course.

II-14 14

Let’s Look At The General Trends Lithium → Neon (e.g. across) Increase in IE. due to a steady increase in the

+ve nuclear charge. effective

Li → Rb (down)

Gradual (but slight decrease) in IE. Electrons in Rb are further away from nucleus and are

screened by the electrons closer to the nucleus.

Screened: “Don’t feel full force of extra nuclear charge, because other electrons are in the way”.

II-15 15

Ionization Energies For Core Electrons (p. 79, DeKock & Gray) • Core electrons are those situated close to the

nucleus and not in the valence shell. • So far we have measured how hard it is to

remove a core electron after the valence electrons have been removed.

• Ionization energies can be measured for

removing an electron from a neutral atom. Ionization of Lithium 1s.

Li(1s22s1) → Li+(1s12s1) IE = 64.84 eV Now Look at Figure 2-3 DeKock & Gray

II-16 16

Further Insights

Li(1s22s1) → Li+(1s2) + e IE1 = 5.392 eV Li+(1s2) → Li2+(1s1) + e IE2 = 75.638 eV Li(1s22s1) → Li+(1s12s1) + e IE1s = 64.84 eV • In the case of IE1s. The energy is less than IE2. Why is it harder to remove the 1s electron when the 2s is not there?? It is much more complicated but we must wait for Part Four (Quantum Chemistry)

II-17 17

So far, we have discussed the basics of ionization energy. We will meet some more advanced material on this later. . Two key characteristics of an atom or molecule that are the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO). Together, these two orbitals are called the frontier orbitals. The HOMO can be found by locating the outer most orbital containing an electron. The LUMO then is the first orbital that does not contain an electron. See the diagram below:

We will see later how these orbitals dictate the chemical reactivity of organic molecules

II-18 18

Electron Affinity Chemical reactivity is largely controlled by electrons. We are trying to understand how strongly these are held by atoms. We have looked at IE’s …now EA’s (Electron Affinities).

Electron Affinity

(p. 81, DeKock and Gray)

The electron affinity (EA) of an atom is defined as the energy required to remove an

electron from the negative ion.

i.e. ions(g) → atom (g) + e- ∆E = EA

II-19 19

Be Very Careful With Signs

if energy is required for

A-(g) → A(g) + e- EA is +ve if energy is released A-(g) → A(g) + e EA is –ve then A(g) is more stable than A-(g).

II-20 20 Element Name/SymbolZ Electron Element Name/Symbol Z Electron

Affinity Affinity

eV eV Hydrogen H 1 0.754209 Ruthenium Ru 44 1.05 Helium He 2 -0.5 Rhodium Rh 45 1.137 Lithium Li 3 0.6180 Palladium Pd 46 0.557 Beryllium Be 4 -0.5 Silver Ag 47 1.302 Boron B 5 0.277 Cadmium Cd 48 NA Carbon C 6 1.2629 Indium In 49 0.30 Nitrogen N 7 -0.07 Tin Sn 50 1.112 Oxygen O 8 1.461125 Antimony Sb 51 1.07 Fluorine F 9 3.399 Tellurium Te 52 1.9708 Neon Ne 10 -1.2 Iodine I 53 3.0591 Sodium Na 11 0.547930 Xenon Xe 54 -0.8 Magnesium Mg 12 -0.4 Cesium Cs 55 0.471630 Aluminum Al 13 0.441 Barium Ba 56 NA Silicon Si 14 1.385 Lanthanum La 57 0.5 Phosphorus P 15 0.7465 Cerium Ce 58 0.5 Sulfur S 16 2.077120 Praseodymium Pr 59 0.5 Chlorine Cl 17 3.617 Neodymium Nd 60 0.5 Argon Ar 18 -1.0 Promethium Pm 61 0.5 Potassium K 19 0.50147 Samarium Sm 62 0.5 Calcium Ca 20 -0.3 Europium Eu 63 0.5 Scandium Sc 21 0.188 Titanium Ti 22 0.079 Vanadium V 23 Chromium Cr 24 0.666 Manganese Mn 25 -0.4 Iron Fe 26 0.151 Cobalt Co 27 0.662 Nickel Ni 28 1.156 Copper Cu 29 1.228 Zinc Zn 30 ~0 Gallium Ga 31 0.3 Germanium Ge 32 1.233 Arsenic As 33 0.81 Selenium Se 34 2.0206 Bromine Br 35 3.365 Krypton Kr 36 NA Rubidium Rb 37 0.48592 Strontium Sr 38 -0.3 Yttrium Y 39 0.307 Zirconium Zr 40 0.426 Niobium Nb 41 0.893 Molybdenum Mo 42 0.746 Technetium Tc 43 0.55

II-21 21

General Trends halogens −large electron affinities

(~3.5 eV) −completes closed shell s2p6config.

closed shells – small EA’s

− He, Ne, Be, Mg, Zn

IE’s and EA’s: closed shells or subshells give very stable electronic

First Row

II-22 22

Overall

And now….. Electronegativity (pp. 90-92, Dekock and Gray) • Electron Affinity is a very useful

thermodynamic quantity – you will use it again.

II-23 23

• However electron distribution in molecules involves the question of the relative tendencies of each atom to acquire control of shared electrons.

• Molecular Property is usually discussed using

Atomic property of Electronegativity.

(symbol is χ )

“power of an atom to attract electrons when part of a molecule” My grandmother had a Boston Terrier that was obsessed with playing tug of war. You couldn't sit down in her house without having him push a nasty damp sock into your hand. If you were bored enough to accept it, he'd growl (ferociously, he thought) and pull. He was a little dog, though, and you could easily pick up the sock with dog still attached. He'd dangle like a little Christmas tree ornament until he got tired and let go. The contest was futile because my mass was about ten times his. If I'd been a Boston Terrier, the match would have been different. A tug-of-war also goes on between atoms involved in a chemical bond. The bonding electrons are the sock. The atom that can pull on the bonding electrons more strongly will get them. The winner is expected to be the atom with the higher effective nuclear charge.

II-24 24

FOR EXAMPLE: almost complete transfer of electron (ionic bond) Sm (

all X Large Xve)+ (_ ve)

Na Cl

Mulliken Electronegativity Mulliken proposed that electronegativity was proportional to the sum of the IE and EA.

EN = c(IE + EA) (c = proportionality constant.)

IE – ability of atom to hold electron

EA – ability of atom to attract an electron.

− Seems to make sense Cl : large IE and large EA − large χ

II-25 25

However this doesn’t always work that well â EA’s are not known that accurately. ã This can artificially give noble gases a large

χ because of their large IE. Compare χMULL for F & Ne EN = c(IE + EA) F Ne IE + EA = 17.4 + 3.34 IE + EA = 21.56 + 0 = 20.74 = 21.56 Since (IENe + EANe) > (IEF + EAF) χNe > χF

II-26 26

χPAULING

↑

Pauling Electronegativity

- Uses comparison of bond energies. - A2 molecule and B2 molecule versus AB

molecule.

II-27 27

- Bond Energy would be average unless there

is ionic bonding. Example HF Bond Energy HF 135 kcalmol-1

1-

2

2 kcalmol 62 Av37 F

103 H =

⎪⎪⎭

⎪⎪⎬

⎫

Extra bond energy assumed to be a consequence of ionic bonding because of electronegativity differences in H and F. Pauling then set the electronegativity difference as ENA – ENB = 0.208∆1/2

converts to eV where ∆ = DEAB = 1/2

BA )])(DE[(DE22

D = Bond Dissociation Energy

II-28 28

PAULING set χF = 3.98 • Pauling AND Mulliken, χ’s agree fairly well. • one way to convert them is

37.1X1.35 1/2Mp −=χ

Q: Fluorine is the most electronegativeelement and yet chlorine has a larger electronaffinity: Why? Hint: think about the relativesizes.

II-29 29

Electronic Configurations • We can now see that bonding is affected

strongly by ideas of how electrons are held by atoms in molecules.

• We know that electrons are held in orbitals. • Now we review this material and see how

electrons affect each other. Configurations • We will meet the mathematical basis for

orbitals and what they look like in part four of this course. Now we merely state the results in a rather pictorial way.

Atomic Quantum Numbers

II-30 30

Each Atomic Orbital (AO) is defined by 3 Quantum Numbers n, R, mR n → Energies

R → shape

mR → (orientation) subsets of R

• We will define these somewhat more accurately later.

n: 1, 2, 3, 4, … R : 0, 1 ,…, n-1 mR : R, R-1, R-2, … , -R

R = 0, 1, 2, 3, 4, s p d f g

II-31 31

s Subshell R = 0, mR = 0 p : R = 1 mR = +1, 0, -1 (p orbitals) d : R = 2 mR = +2, +1, 0, -1, -2 (d orbitals) So the number of orbitals in a particular shell is n2 where n is the principal quantum number.

Configurations For Atoms And Ions So we can write down the configuration for any atom or ion (or can we?). - In first year we based this on the building up

or AUFBAU principle. - Method for determining the lowest energy

configuration for an atom (ground state).

II-32 32

However it is not quite as simple.

Pauli exclusion principle: “no more than 2 electrons may occupy a single orbital and if two occupy a single orbital their spins must be paired.”

Penetration And Shielding - orbital energies differ from those in

hydrogen. - Electron in orbital experiences.

â coulombic attraction of Nucleus. ã coulombic repulsion from other

electrons.

- we approximate: Electons experience a CENTRAL FIELD.

- sum of field from nucleus and average of

all electrons.

II-33 33

• reduces Nuclear Charge from true value (Ze)

to effective nuclear charge Zeff e (for a particular electron)

Shielding These electrons are screened by Shielding of inner electrons.

+

NUCLEUS Shielding Parameter Zeff = Z - σ

II-34 34

If you like think of it this way: electrons in outer orbitals experience attraction from nucleus but repulsion from other electrons. Effective nuclear charges Zeff H He Z 1 2 1s 1.00 1.69 Li Be B C N O F Ne Z 3 4 5 6 7 8 9 10 1s 2.69 3.68 4.68 5.67 6.66 7.66 8.65 9.64 2s 1.28 1.91 2.58 3.22 3.85 4.49 5.13 5.76 2p 2.42 3.14 3.83 4.45 5.10 5.76 Na Mg Al Si P S Cl Ar Z 11 12 13 14 15 16 17 18 1s 10.63 11.61 12.59 13.57 14.56 15.54 16.52 17.512s 6.57 7.39 8.21 9.02 9.82 10.63 11.43 12.232p 6.80 7.83 8.96 9.94 10.96 11.98 12.99 14.013s 2.51 3.31 4.12 4.90 5.64 6.37 7.07 7.763p 4.07 4.29 4.89 5.48 6.12 6.76 NOTE: ns electrons are less shielded than np

electrons.

II-35 35

- Due to penetration s electrons get closer to nucleus than p electrons.

- s electrons less shielded. Similar differences for the orbitals. s orbitals most penetrating followed by p, d, f. - However the ordering depends on the

number of electrons present. - Penetration effects v. marked for 4s e’s of K

and Ca.

II-36 36

Penetration effects are very marked for 4s electrons of K and Ca. Q: Why is 2s and 2p same energy on the y-axis (no electrons)

II-37 37

Ground State Electron Configurations (DeKock and Gray, p. 46-56)

In your reading of DeKock and Gray you willcome across the Schrödinger equation. Thisis the mathematical basis of quantummechanics. For now do not need theequation. We will cover it in PART FOUR.

GROUND STATE for 1st 5 elements. H He Li Be B 1s1 1s2 1s22s1 1s22s2 1s22s22p1

What About Carbon?

CARBON (Z = 6): Removed from chem* 2060

Fall 2005

1s22s22p2

II-38 38

WAIT! - We have a problem. - where do we put the electron? Let’s look at the 2p shell THESE ARE OF DIFFERENT ENERGY - choice can be made by using Hund’s rule. - maximum no of parallel spins results in

lowest e- - e- repulsion.

II-39 39

Ground State can be 1s22s2 11 yx2p2p . We will return to p2 a bit later. electron-electron repulsions and the Aufbau principle

(DeKock and Gray, p. 51) • We have already seen that e-e repulsions

affect. â Ionization Energies (See Li)

ã Electron Affinities (See Cl & F) ä Ground State Configurations (e.g. C)

We now look briefly at the ordering of 3d and 4s orbitals of transition metal ions.

II-40 40

- Transition Elements - Importance of d electrons, also see 19-365,

19-463 (Bioinoganic Chemistry.)

II-41 41

The electronic Configurations of the Elements.

Electronic Configuration Z neutral +ve ion 1 H 1s^1 - 2 He 1s^2 1s^1 3 Li [He] 2s^1 1s^2 4 Be [He] 2s^2 [He] 2s^1 5 B [He] 2s^2 2p^1 [He] 2s^2 6 C [He] 2s^2 2p^2 [He] 2s^2 2p^1 7 N [He] 2s^2 2p^3 [He] 2s^2 2p^2 8 O [He] 2s^2 2p^4 [He] 2s^2 2p^3 9 F [He] 2s^2 2p^5 [He] 2s^2 2p^4 10 Ne [He] 2s^2 2p^6 [He] 2s^2 2p^5 11 Na [Ne] 3s^1 [He] 2s^2 2p^6 12 Mg [Ne] 3s^2 [Ne] 3s^1 13 Al [Ne] 3s^2 3p^1 [Ne] 3s^2 14 Si [Ne] 3s^2 3p^2 [Ne] 3s^2 3p^1 15 P [Ne] 3s^2 3p^3 [Ne] 3s^2 3p^2 16 S [Ne] 3s^2 3p^4 [Ne] 3s^2 3p^3 17 Cl [Ne] 3s^2 3p^5 [Ne] 3s^2 3p^4

II-42 42

18 Ar [Ne] 3s^2 3p^6 Ne] 3s^2 3p^5 19 K [Ar] 4s^1 [Ne] 3s^2 3p^6 20 Ca [Ar] 4s^2 [Ar] 4s^1 21 Sc [Ar] 3d^1 4s^2 [Ar] 3d^1 4s^1 22 Ti [Ar] 3d^2 4s^2 [Ar] 3d^2 4s^1 23 V [Ar] 3d^3 4s^2 [Ar] 3d^4 24 Cr [Ar] 3d^5 4s^1 [Ar] 3d^5 25 Mn [Ar] 3d^5 4s^2 [Ar] 3d^5 4s^1 26 Fe [Ar] 3d^6 4s^2 [Ar] 3d^6 4s^1 27 Co [Ar] 3d^7 4s^2 [Ar] 3d^8 28 Ni [Ar] 3d^8 4s^2 [Ar] 3d^9 29 Cu [Ar] 3d^10 4s^1 [Ar] 3d^10 30 Zn [Ar] 3d^10 4s^2 [Ar] 3d^10 4s^1 31 Ga [Ar] 3d^10 4s^2 4p^1 [Ar] 3d^10 4s^2 32 Ge [Ar] 3d^10 4s^2 4p^2 [Ar] 3d^10 4s^2 4p^1 33 As [Ar] 3d^10 4s^2 4p^3 [Ar] 3d^10 4s^2 4p^2 34 Se [Ar] 3d^10 4s^2 4p^4 [Ar] 3d^10 4s^2 4p^3 35 Br [Ar] 3d^10 4s^2 4p^5 [Ar] 3d^10 4s^2 4p^4 36 Kr [Ar] 3d^10 4s^2 4p^6 [Ar] 3d^10 4s^2 4p^5 37 Rb [Kr] 5s^1 [Ar] 3d^10 4s^2 4p^6 38 Sr [Kr] 5s^2 [Kr] 5s^1 39 Y [Kr] 4d^1 5s^2 [Kr] 5s^2 40 Zr [Kr] 4d^2 5s^2 [Kr] 4d^2 5s^1 41 Nb [Kr] 4d^4 5s^1 [Kr] 4d^4 42 Mo [Kr] 4d^5 5s^1 [Kr] 4d^5 43 Tc [Kr] 4d^5 5s^2 [Kr] 4d^5 5s^1 44 Ru [Kr] 4d^7 5s^1 [Kr] 4d^7 45 Rh [Kr] 4d^8 5s^1 [Kr] 4d^8 46 Pd [Kr] 4d^10 [Kr] 4d^9 47 Ag [Kr] 4d^10 5s^1 [Kr] 4d^10 48 Cd [Kr] 4d^10 5s^2 [Kr] 4d^10 5s^1 49 In [Kr] 4d^10 5s^2 5p^1 [Kr] 4d^10 5s^2 50 Sn [Kr] 4d^10 5s^2 5p^2 [Kr] 4d^10 5s^2 5p^1 Here we see several exceptions to the Aufbau principle.

II-43 43

• We have seen (p. 89) that the filling rules that

we learnt in first year chemistry are not quite accurate.

• The simple picture for filling orbitals depends on

â atomic number (see p. 86) ã charge

II-44 44

Let’s Look At The 3d & 4s More Closely (P. 52, DeKock and Gray)

• 4s is lower in energy than 3d

(so fills first) but: Chromium [Ar] 3d54s1

Copper [Ar] 3d104s1

Also Scandium [Ar] 3d14s2

Scandium2+ [Ar] 3d1

s fills first d fills first

II-45 45

The situation is even stranger Ionization Energies

Scandium has 1 d electron and 2s electrons in the valence orbitals.

IE’s for these are:

Sc(3d14s2) → Sc+(3d14s1) + e- 6.62 eV Sc(3d14s2) → Sc+(3d04s2) + e- 7.98 eV - d orbital is more stable. - Then why does s shell fill first? - Larger e-e repulsion in d orbitals.

(Less diffuse than 4s orbital) s electrons penetrate better.

(Wait until part 4)

II-46 46

• So for most of the d block the ground state is 3dn4s2 (s fills first) because of large e-e repulsions in d orbitals.

• Note exceptions usually occur with ½ and full

shells (see Chromium d5s1 ) (see Copper d10s1)

• Complications are not important when orbital

energy of 3d is much less than 4s.

II-47 47

Electrostatics Of Atoms & Molecules • In the previous section we looked at

Coulombic attractions and repulsions and their importance in electrons in atoms.

• We now look at coulombic attractions

between atoms and molecules. Interacting Charges

• Interactions between atoms and molecules are

electrical in nature. • Classical Electrostatics can be used to predict

energy of interaction. • All based on simple coulombic law.

II-48 48

Types Of Interaction a) Monopole – Monopole b) Monopole – Dipole c) Dipole – Dipole a) Ion-Ion Bonds (Monopole – Monopole) For correct use and units see Prob. Set (Q’s 1, 4)

r coulombic interaction =

r4qq -

πε+

PLEASE NOTE 4πε - This gets units right. We leave it out in the notes.

II-49 49

Note Signs:

attractive has –ve sign.

rqq- E 21= note:

επ41 term left out

for convenience Monopole – Monopole

- Non – Directional

- Strong over atomic distances.

- Good model for ionic bonding

e.g. LiF estimate 686 kJ mol-1 (measured 755 kJ mol-1)

(see later) b) monopole – dipole

II-50 50

r

RMonopole Dipole

Z+

4Rr

ZqR 2R r

Zq 2R -r

Zq- E 22 −=

++= • 2r

Z- µ

↑ ↑ attraction repulsion

i.e. , )2

Rr)(2R-r(

)2R-r(Zq)2

Rr(Zq-

2Rr

Zq

2R-r

Zq+

++=

++−

qR) (but

4Rr

ZqR- 22=

−= µ

and if R is small compared to r : E = 2r

Zµ−

II-51 51

net attraction: even though there is no charge on the dipole. • Monopole orients the dipole in favourable

direction. Example: Solvated Ions: When NaCl dissolves in water Na+ is surrounded by H2O because of dipole-monopole interaction.

In aqueous solution Na+ ishydrated by an octahedron ofwater molecules. Note thenegatively charged Cl- (fromNaCl) does not participate in thehydration shell.

H

H

O

O

O

H H

Na

δδ

δδ

δ

δ

δδ

δ

δ

δ

δ

O

O

H δ

δ

δ

HH

O

δ

δ

δH

H

H

H

H

II-52 52

• Polar nature of water makes it an excellent solvent for ionic solids like NaCl.

• Energy needed to separate ions is provided by

formation of hydrated ions (monopole-dipole interactions).

• Non polar solvents (e.g., gasoline) cannot

form such strong bonds. So NaCl and other salts are insoluble in gasoline.

Strengths Of Bonds

Monopole – Monopole ~ 400 kJ mol-1

Monopole – Dipole ~ 40 kJ mol-1

Smaller Energy because of distance dependence.

II-53 53

Soaps (and detergents) work because their structures combine in one molecule a hydrocarbon chain which is

hydrophobic (rejects water) and lipophilic (attracts oily materials) with an end which is hydrophilic (attracts water)

and lipophobic (rejects oily materials). The long "fatty" chains provide solubility in hydrocarbons (grease) and the

polar, usually ionic, heads provide solubility in water. If both oily and watery materials are present, a soap provides

a "bridge" by dissolving its hydrocarbon chain in a droplet of oil in such a way that the ionic, hydrophilic, end sticks

out into the surrounding water. This arrangement is called a micelle and permit soapy water to "wash away" greasy

materials.

II-54 54

The hydrophilic (polar) end of the detergent binds strongly to water – the non-polar part bonds to the grease.

c) dipole-dipole µ = qR r

R

II-55 55

rq

REPRr

q Rr

q rq- E

2

22

2

22

22−

++

++=

4444444 34444444 21

ATT. ATT E •

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+ 2

rRrRq

23

22 for R << r

Magnitude ~ 5 kJ mol-1E – 3

2

VERY SHORT RANGE INTERACTION (AND WEAK)

• E = 32

rµ− for dipole-dipole is a special case. If

there is an angle θ between the dipoles then:

)3cos-(1 r

- E 23

21 θµµ=

rµ−

II-56 56

IMPORTANCE: Polar Liquids, Dissolving of Molecular Solids (see Later)

+__+ _+

+__+ _+

_+

_++_

+_

+

_

+

_ +_

+_

EnergyRequired

Energy Released

SOLVATION

II-57 57

Induced Dipoles & Van Der Waals Forces • We have seen now that ionic type bonds can

occur when a dipole moment exists in a molecule.

• A dipole can be set up in a non-polar molecule

by an electric field. For example:

_+_ _

__

__

+

++

+

++

++++

++

+

__

____

_

+ _

induced dipole moment

voltage off

voltage on

Electrons attracted to r Nuclei attracted to s

II-58 58

electric field can be from neighbouring atom or ion

_+_

__

__

++

+

+q+

Monopole

µ • The magnitude of the induced dipole moment

(µ) is related to the polarizability of the molecule.

µIND = α E

Polarizability Strength of

Electric field. • α is the ease of deformation of the electron

“cloud” around the molecule. It is a measure of how “floppy” the electrons are.

II-59 59

note: interaction of induced dipole with another charge can only be an attraction since it is automatically created with the correct geometry.

• If species is atom – no orienting effect. • If species is molecule, α may be different in

one direction. Molecule will tend to orient itself to create largest induced dipole moment.

Polarizability And Electronegativity • Polarizability increases with volume. • Also depends on electronegativity. Electronegative atoms – not very polarizable.

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Monopole-Induced Dipole E = 42

21

2rq- α

(~ 5 kJ mol-1)

Dipole Induced Dipole 62

21r

- E αµ=

~ 0.05 kJ mol-1

Induced Dipole-Induced Dipole

621

21

21r

IIII

23- E αα

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

+=

I = ionization energy

E ~ 0.5 kJ mol-1

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• In solids where there are no permanent dipoles (e.g. Xe, Kr, He, H2, O2, N2) the solid is held together by weak forces.

• These are called Van der Waals Forces.

Van Der Waals Forces (DeKock and Gray, p. 431) attractive and repulsive Induced Dipole Repulsion between

6r1 electrons on

neighboring atoms (London or dispersion forces)

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Attractive Part

Interaction (Potential) Energies that have a 6r1

dependence (very short range) are usually lumped together as “Van der Waals Forces”. Repulsive Part

Must be a repulsion (or everything would collapse into itself).

Repulsion Energy = be-ar

Importance Of Fluctuating Dipoles (Also Fig.7-13 DeK &G)

For the instant that thissituation occurs thereis an attractionbetween theinstantaneous andinduced dipoles. Theeffect is felt by both –each induces apolarization in theother.

• - + - +

F

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• The electron “clouds” repel each other at very small distances.

- end up with a balance.

• One way of writing this is the Lennard-Jones

6-12 potential.

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡= 612 r

1 - r1 C E

Repulsive Attractive • Writing the repulsive part as an exponential is

actually more realistic because of the exponential nature of radial wave functions (see part four)

i.e. E = be-ar - 6r

d

(SEE DeKock and Gray, p. 432)

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IT HELPS TO LOOK AT THIS PICTORIALLY POTENTIAL

ENERGY

ATTRACTIVE

BA

TOTAL

Equilibrium Internuclear Distance

REPULSIVE

req

rAB

r

DAB

DAB = BOND ENERGY

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• Actually well Depth for a crystal refers to the enthalpy of sublimation.

solid → gas

Short & Long Range “Forces”

Potential Energy 12

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

12r1 Shortest Range

6 ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

6r1

4 ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

4r1

2 ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

2r1 Longest Range

12642

Distance

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Van Der Waals Radii • Size of Molecule (or Atom) revealed by LJ

potential is much larger than that revealed by other measures of atomic size (see PART 3) such as covalent radius or ionic radius.

• Because electron clouds stop interpenetration

for a non-bonding interaction (or no electron sharing).

• Covalent and Ionic Bonding will be discussed

later PARTS 3, 4 and 5.

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Van Der Waals Solids

Good reading here: http://www.chemguide.co.uk/atoms/bonding/vdw.html

A comparison between a covalent molecular bond and a much weaker VDW bond in He

II-68 68

• molecular solids in which only Van der Waals intermolecular bonding exists generally melt at low temperatures.

• Because thermal energy is able to overcome VDW attraction very easily.

• Liquid and solid He exist only below 4.6 K . • Van der Waals bonds get stronger as atoms

get bigger (more polarizable).

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(see DeK&G 435 et seq)

• DW attractions responsible for liquid state.

II-70 70

Hydrogen Bonding (DeKock & Gray, p. 436) • Polar Molecules held together in molecular

solids by dipoles. • i.e. Opposite Ends of Dipole attract each

other. • A Kind of Polar

Interaction is the . - Relatively Weak ~ 20 kJ mol-1

(Covalent & Ionic Bonds ~ 400 kJ mol-1.

VERY IMPORTANT

Hydrogen Bond

EXAMPLES:

HF H F

0.92 D1.87D

HYDROGEN BOND

θ δ + δ_δ_δ +

II-71 71

OH

H

H O

H

D0.96104.5 o

104.5 o

HYDROGEN BOND

2.02D

F

H

NH H

H

HYDROGEN BOND

ALL GASEOUS DIMERS

II-72 72

Features Common To H-Bonded Systems â Molecular Units Retain Their Integrity.

e.g. H-F bond Lengths are same as in monomer. ã Fa - - Hb – Fb bond is linear. ä H atom is asymmetric (Only in very strong

H bonds e.g. FHF- is this untrue). å Angle θ (see HF) is between 100 and 120°. In solid HF the bonding is zig-zag.

H

F

H

F

H

F

H

Fa

a

b

b

solid is held together by H-bonds

II-73 73

ICE • Each H2O molecule is bonded to 4 others

(Tetrahedral). • Although bonds are weak they are important. • Hydrogen bonding in water is responsible for

many of its important properties.

II-74 74

• Melting & Boiling Temperatures of water are

unexpectedly high due to H bonding.

II-75 75

• Since H-bonding creates an open network ice is less dense than water.

• Only about 1/3 of H-bonds are broken when

ice melts. • In liquid phase water there are still H-bonds. • As T↑ clusters of H bonded water break up.

∴ volume continues to shrink.

• As T↑ further thermal expansion occurs. - molecules need more room.

(This then dominates over shrinkage caused by collapse of H-bonds).

So liquid water has minimum volume (Max

Density) at 4°C. (WHY Lakes Don’t Freeze Solid) We will look at H-bonds again when we know more about bonding.

II-76 76

unbelievably HYDROGEN Bonds are

important in biochemistry. e.g. DNA Sequence and Replication

A CCC

N CNN

C N

N

H NC N

CCC

O H N H

H

HO

H

H

HH

H

A CCC

N CNN

C N

H NC N

CCC

N HH

O CH3

H

HO

H

H

Thymine

Cytostine

One of the biggest sources of difficulty for a chemistry student is the distinction between chemical bonds and intermolecular forces. If you are having trouble try: http://www.bcpl.net/~kdrews/interactions/interactions.html

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