anu chemistry i hpo lect 2012

8/13/2019 ANU Chemistry I HPO lect 2012

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1Atomic Structure Particle in a Box

The vibrations of a guitar string produce standing

waves because the string is bound at both ends sothat the waveform cannot propagate along the string.

The electron in an atom is also bound (i.e. its

movement is restricted to a small region of space due

to its attraction to the nucleus) and therefore itsmotion also leads to standing waves and quantization

of energy.

Travelling WaveStanding Wave


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2Particle in a Box

The so-called Particle-in-a-Box model serves to

illustrate how the wave nature of a bound electron inan atom naturally leads to quantized energy levels.

Consider an electron confined to move along a wire

fixed between two rigid walls at a distance Lapart.

0 L


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3Particle in a Box

A classical description of the electrons motion

allows the electron to move back and forwards along

the wire. The kinetic energy, E, of the electron is given

by

2

2

1muE =

Since the electron can have any velocity, u, its

energy, E,can take any value, even zero.

No position along the wire is more favorable than anyother and the exact position and velocity of the

electron can be known simultaneously.


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4Particle in a Box

A wavedescription of the electrons motion inside the

one-dimensional box is analogous to the vibrations ofa fixed guitar string.

The wavelength ofthe vibrations are

restricted to whole

numbers of half-

wavelengths.

L


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5Particle in a Box

Like the vibrations of a guitar string, the wavefunction

of the electron must fit exactly inside the box and thusthe wavelength, !, of the electron is restricted.

The length, L, of the box must be a whole number of

half-wavelengths, i.e.

!#

$&

=

2

'nL

where the integer ncan be considered as a quantum

number. Rearranging to solve for the wavelength of

the electron gives

n

L2=!


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6Particle in a Box

To calculate the kinetic energy of the electron we

need to know its velocity, u. This can be obtained

from rearranging the de Broglie equation to give

Substituting for u in the equation for the kinetic

energy of the electron gives

!m

hu =

2

2

2

22

1

!m

hmuE ==


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7Particle in a Box

Substituting for !using the earlier expression relating

!to the box length, L, gives

Since n is an integer, the kinetic energy of the

electron is restricted to multiples of h2/8mL2 and istherefore quantized.

2

22

2

2

82 mL

hn

m

hE ==

!


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8Particle in a Box

Energy Levels & Wave Functions for a Particle in a Box

E

nergy/(h2/8m

L2

)

n

n

L2=!


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9Particle in a Box

The smallest value of n is 1, therefore the lowest

energy level (ground state) is at h2/8mL2. Unlike theclassical model, the kinetic energy of the electron

cannot be zero, i.e. the electron is never at rest.

This lowest, irremovable energy is known as the

zero-point energy and corresponds to a continuous

fluctuating motion of the electron between the two

walls of the box.


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10Particle in a Box

Since the energy levels are proportional to 1/mL2,

they become more closely spaced as the box getsbigger or as the mass of the particle gets larger.

For macroscopic particles with large relative mass,

the levels are so closely spaced that the energy is

effectively no longer quantized.

E1

E2

E3

E4E5

Energy

Small mor L Large mor L


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11Particle in a Box

The wavefunctions " for the particle-in-a-box model

do not correspond to an oscillation of the wire butinstead describe the shape of the electron wave

inside the box.

The amplitude of the electron wave at any point along

the wire is related to the probability of finding theelectron and is proportional to "2.

Unlike the classical model, the probability is not

identical everywhere and in fact can be zero at certain

points (nodes) along the wire. The number of nodes

is given by n 1.


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12Particle in a Box

Wavefunctions for the Particle-in-a-Box Model

Wavefunction Square of Wavefunction


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13Particle in a Box

The particle in a box model has a number of parallels

with the quantum model of the atom:

1. The motion of the electron is described bystanding waves.

2. The energy of the electron is quantized.3. The wavefunctions and energy levels are

dependent on the integer n, analogous to the

principal quantum number in an atom.

4. The probability of finding the electron is related tothe square of the wavefunction and is not the

same at all points.


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14Absorption Spectra

The particle in a box model can be used to rationalise

the absorption spectra of cyanine dyes which havethe general formula

These dyes are positively charged linear molecules

containing conjugated double bonds. Their very

intense colours make them useful in many

applications including tunable lasers.

The absorption maxima of the dyes show a strong

correlation with the number of conjugated bonds, k.

(CH3)2N!(!CH=CH!)k!C=N(CH3)2


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Absorption Spectra of Cyanine Dyes

Wavelength, #/ nm

Absorba

nce

k=2

k=3

k=4 k=5


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If the chain of conjugated double bonds are

approximated as a wire of length, L, along which the

$electrons are free to travel,

2

22

8mL

hnE = n= 1, 2, 3, .

then the energy levels for the system can be

calculated using

(CH3)2N!CH=CH!CH=CH!CH=CH!C=N(CH3)2

L


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The energy of the dye absorption is due to an

electron being excited from the highest level occupied

by the $electrons to the lowest unoccupied level.

The energy difference between two adjacent levels

can be calculated using

2

2

18

)12(mL

hnEEE

nn +=!="

+

For the dye absorption,n

andn+

1are the quantumnumbers for the highest occupied and lowestunoccupied levels, respectively.


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The quantum number nis connected with the number

of $ electrons, N. Since each level accommodates

two electrons, the highest occupied level has n= N/2.

For the cyanine dyes the lone pair of electrons on the

outer N atom are also included as $electrons due totwo possible resonance structures. Thus, N= 2k + 4

(CH3)2N!(!CH=CH!)k!C=N(CH3)2

(CH3)2N=C!(!CH=CH!)k!N(CH3)2


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Energy Levels for (CH3)2N-(-CH=CH-)3-CH=N(CH3)2


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The length, L, of the conjugated $ system can be

expressed in terms of the number of $ electrons, N,and the bond length, d

o, as L= Nd

o

Substituting for nand Lin the expression for %Egives

22

2

8)1(

odmN

hNE +=!

Using the above expression with h= 6.63 x 10-34J s,

m= 9.11 x 10-31kg and do= 140 x 10-12m allows theexcitation energy associated with the absorption

spectra of cyanine dyes to be calculated.


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Using the simple relations

E

hc

v

coo

!==

max"

where vis the frequency and co= 3.0 x 108m/s is the

speed of light, the wavelength of the absorptionmaxima, #max, can be calculated from

Ehv != and!

oc

v =


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Calculated and Experimental Absorption Maxima #max

of Cyanine Dyes (CH3)2N-(-CH=CH-)k-CH=N(CH3)2

k N max/nm Colour

of SolnCalc Expn

0 4 206 224 Colourless1 6 332 313 Colourless

2 8 459 416 Yellow

3 10 587 519 Red

4 12 716 625 Blue5 14 844 735 Green

6 16 973 848 Colourless


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Substituting for %Ein the expression for !maxgives

h

cmd

N

N

E

hcooo

22

max

8

1!!"

#$$%

&

+

=

'=(

As the number of $electrons increases, N2/N+1 &N.

This implies that #maxshould increase linearly with Nin agreement with the experimental data.

nm7.641

2

!!"#$$

%&

+

=

N

N


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Experimental Absorption Maxima (#max) vs

Number of $Electrons (N) for Cyanine Dyes

N

2 4 6 8 10 12 14 16 18

!max

100

200

300

400

500

600

700

800

900

k=0

k=1

k=2

k=3

k=4

k=5

k=6(CH3)2N!(!CH=CH!)k!C=N(CH3)2

(N = 2k + 4)


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25Molecular Orbital Model

The molecular-orbital (MO) model is the only

bonding model which has its foundations based onthe solution of the Schrdinger Wave Equation

(SWE).

Whereas atomic orbitals are solutions to the SWE in

the atomic case, molecular-orbitals (MOs) are thesolutions in the molecular case.

MOs result from the overlap of atomic orbitals on

adjacent atoms. The valence electrons involved in

bonding are no longer localised on individual atoms

but delocalised over the entire molecule.


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26

The general rules which apply in using the

molecular-orbital (MO) model are:

Only valence orbitals need be considered whenforming MOs.

The number of MOs which form is equal to thenumber of atomic orbitals that interact. If two orbitals interact, an in-phase (bonding)MO

and an out-of-phase (antibonding)MO result.

The energy of the bonding MO is lower than theenergy of the interacting orbitals while the energy

of the antibonding MO is higher.

General Rules


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27

Wavefunctions for orbitalshave the same phase in

the region of overlap and

therefore reinforce one

another

Wavefunctions for orbitals

have opposite phase in

the region of overlap andtherefore cancel one

another

General Rules


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28

Overlap of H 1s atomic orbitals to form bonding(in-

phase) and antibonding(out-of-phase) MOs in H2:

General Rules

+

+

Out-of-Phase

In-Phase


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29

Each MO accommodates

at most only two electronswith opposite spin.

General Rules

H H

!1s"

!1s

H2

AO AOMOs

From the Aufbau Principle, the lowest energyconfiguration (ground state) for H2 corresponds to

both electrons in the 'bonding MO, i.e. '1s2.

Since both electrons are

lowered in energy relative

to the isolated atoms, the

H2molecule is stable.


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30

The MO configuration for H2- is '1s

2 '*1s1. Since two

electrons are lowered in energy while one is raised,

it only has half the stability of H2. Similarly for H2+,

with a MO configuration of '1s1, it also has only half

the stability of H2.

General Rules

H H

!1s"

!1s

H H

!1s"

!1s

H2#

H2+


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31

An indication of the bond strength in a molecule is

given by the bond order (BO)where

General Rules

2

)egantibondinof(N)ebondingof(NBO

oo !!!

=

H2+ H2 He2

+ He2

Stability stable stable stable unstableBond Energy (kJ/mol) 255 456 251 N/A

Bond Length () 1.06 0.74 1.08 N/A

Bond Order ! 1 ! 0

!"

!

1s

1s

For a molecule

to be stable, thebond order mustbe greater thanzero (BO > 0)


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32General Rules

Overlap of s orbitals

results in only 'and

'* MOs.

Overlap of p orbitalsresults in ', '*, $

and $* MOs.

Overlap of d orbitals

results in'

,'

*,$,

$*, (and (* MOs.


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33Diatomic Molecules

Qualitative MO diagram

for 2nd row homonuclear

diatomics which assumes' overlap is greater than

$.

!2p*

2p2p

!2p

!2s*

!2s

2s2s

E

"

2p*

"2p

AO AOMOs


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MO diagrams for O2and F2

E

!2s*

!2s

2s2s

O atom O atom

!2p*

2p2p

!

2p

"2p*

"2p

!2s*

!2s

2s2s

F atom F atom

!2p*

2p2p

!

2p

"2p*

"2p

O2

MOs

F2

MOs


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Except for O2and F2, all other 2ndrow diatomics have

the'

2pand$

2pMOs inverted due to mixing of 2s and2pzorbitals.!2p*

2p2p

!2p

!2s*

!2s

2s2s

E

"2p*

"2p

!2p*

2p2p

!2p

"2p*

"2p

!2s*

!2s

2s2s

MOsMOs


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Inversion of '2p and $2p MOs explains paramagnetic

behavior of B2due to presence of unpaired electrons.!2p*

2p2p

!2p

!2s*

!2s

2s2s

E

"2p*

"2p

!2p*

2p2p!2p

"2p*

"2p

!2s*

!2s

2s2s

B2MOsB2MOs

Unpairedelectrons


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!2p*

!2s*

!2s

"2p*

"2p

!2p

!2p*

!2s*

!2s

"2p*

"2p

!2p

B2 C2 N2 O2 F2Bond Order 1 2 3 2 1

Bond Energy 290 620 942 495 154 kJ/mol

Bond Length 159 131 110 121 143

Magnetism paramag diamag diamag paramag diamag

MO model rationalizes bond energy, bond lengths andmagnetic properties of 2ndrow diatomics.


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38Heteronuclear Diatomics

In heteronuclear diatomicsthe overlapping orbitalsare at different energies. The valence orbitals on the

more electronegative atom lie at lower energy.

!"

!

More Electro-negative Atom

Less Electro-negative Atom

AO AOMOs


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A modified MO scheme is used to describe

heteronuclear diatomics involving atoms with similarelectronegativities such as CO and NO.

C atom O atom

!"

#"

!

#

!"

!

2p

2s

2s

2p

CO MOsN atom O atom

!"

!

2s

2s

NO MOs

!"

#"

!

#

2p

2p


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If the atoms have very different electronegativities,

the MOs are largely localized on one or the otheratoms and the bonding becomes more ionic in

character, e.g. HF.

2s

2p

1s

!"

#nb

!

H atom F atomHF MOs

The 2s orbital on F is so

low in energy that only

the 2pz orbital overlapswith the H 1s orbital to

form 'and '* MOs.


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41Polyatomic Molecules

In polyatomic molecules, the MOs are built up by

forming combinations of equivalent atomic orbitalson symmetry-related atoms known as linear

combinations of atomic orbitals (LCAOs).

In BeH2 the two hydrogen atoms are related by

symmetry and therefore their 1s orbitals can becombined to form two linear combinations given by:

"+= HA(1s) + HB(1s)

"= HA(1s) HB(1s)

HA HBBe


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LCAOs formed fromH 1s orbitals overlap

with atomic orbitals

on Be which have the

same symmetry.

Since H 1s and Be 2s

and 2pz orbitals are

orientated along the

internuclear axis, only

' bonding and anti-

bonding MOs result.

Be (2s)+!+

Be (2s)"!+

Be (2pz)+!"

Be (2pz)"!"

H HBe

#g

#g*

#u

#u*


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The Be (2s) + " and

Be (2pz) + "+ comb-inat ions are non-

bonding as in-phase

overlap is cancelled by

out-of-phase overlap.

The Be 2px and 2pyorbitals are non bond-

ing as they do not

have the co r rec tsymmetry to overlap

with either "+or ".

Be (2s)+ !

"

Be (2pz)+ !+

Be (2px)+!"

H HBe

Be (2px)+ !+

Be (2py)+ !+

Be (2py)+!"


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2p

2s

!"

!+

#g

#u

#u*

$unb

#g*

Be AOsBeH2

MOs

H 1s

LCAOs

2px,2py

2pz

anu chemistry i hpo lect 2012

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