anu chemistry i hpo lect 2012
TRANSCRIPT
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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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15Absorption Spectra
Absorption Spectra of Cyanine Dyes
Wavelength, #/ nm
Absorba
nce
k=2
k=3
k=4 k=5
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16Absorption Spectra
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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17Absorption Spectra
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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18Absorption Spectra
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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19Absorption Spectra
Energy Levels for (CH3)2N-(-CH=CH-)3-CH=N(CH3)2
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20Absorption Spectra
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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21Absorption Spectra
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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22Absorption Spectra
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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23Absorption Spectra
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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24Absorption Spectra
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
$.
!2p*
2p2p
!2p
!2s*
!2s
2s2s
E
"
2p*
"2p
AO AOMOs
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34Diatomic Molecules
MO diagrams for O2and F2
E
!2s*
!2s
2s2s
O atom O atom
!2p*
2p2p
!
2p
"2p*
"2p
!2s*
!2s
2s2s
F atom F atom
!2p*
2p2p
!
2p
"2p*
"2p
O2
MOs
F2
MOs
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35Diatomic Molecules
Except for O2and F2, all other 2ndrow diatomics have
the'
2pand$
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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36Diatomic Molecules
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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37Diatomic Molecules
!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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39Heteronuclear Diatomics
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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40Heteronuclear Diatomics
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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42Polyatomic Molecules
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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43Polyatomic Molecules
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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44Polyatomic Molecules
2p
2s
!"
!+
#g
#u
#u*
$unb
#g*
Be AOsBeH2
MOs
H 1s
LCAOs
2px,2py
2pz