simetri s1 lengkap 2 sks
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Symmetry and Group Theory
Feature: Appl ication for Spectroscopy
and Orbi tal Molecules
Dr. Indriana Kartini
Page 2
P. H. Walton “Beginning Group Theory for Chemistry”
Oxford University Press Inc., New York, 1998
ISBN 019855964
A.F.Cotton “ Chemical Applications of Group Theory”
ISBN 0471510947
Text books
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Marks
• 80% exam: – 40% mid
– 50% final
• 10% group assignments of 4 students
• Syllabus pre-mid: Prinsip dasar – Operasi dan unsur simetri
– Sifat grup titik dan klasifikasi molekul dalam suatu grup titik
– Matriks dan representasi simetri
– Tabel karakter
• Syllabus Pasca-mid: Aplikasi – prediksi spektra vibrasi molekul: IR dan Raman
– prediksi sifat optik molekul – prediksi orbital molekul ikatan molekul
Page 4
Unsur simetri dan operasi simetri molekul
• Operasi simetri
– Suatu operasi yang dikenakan pada suatu molekul
sedemikian rupa sehingga mempunyai orientasi
baru yang seolah-olah tak terbedakan dengan
orientasi awalnya
• Unsur simetri – Suatu titik, garis atau bidang sebagai basis
operasi simetri
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Rotasi mengitari sumbu rotasi
diikuti dengan refleksi pada
bidang tegak lurus sumbu
rotasi
Sumbu rotasi tidak
sejati (Improper
rotational axis)Sn
Proyeksi melewati pusat
inversi ke sisi seberangnya
dengan jarak yang sama dari
pusat
Pusat/titik inversii
Refleksi melalui bidang simetriBidang simetriσ
Rotasi seputar sumbu dengan
derajat rotasi 360/n (n adalah
bilangan bulat)
Sumbu rotasiCn
Membiarkan obyek tidak
berubah
Unsur identitas
E
OperasiUnsur Simbol
© Imperial College London 6
Operasi Simetri
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B B
Rotate 120O
F1 F1
F2F3
F3F2
Operation rotation by 360/3
around C3 axis (element)
BF3
Rotations 360/n where n is an integer
Page 8
H1
H2
H1
H2
H1 H
2
σ(xz)
σ(yz)
z
y
x
x is out of the plane
Reflection is the operationσ element is plane of symmetry
H2O
Reflections
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Reflections for H2O
Page 10
Reflections
• Principle (highest order) axis is defined as Z axis
– After Mulliken
σ(xz) in plane perpendicular to molecular plane
σ(yz) in plane parallel to molecular plane
both examples of σv
σv : reflection in plane containing highest order axis
σh : reflection in plane perpendicular to highest
order axis
σd : dihedral plane generally bisecting σv
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Xe
F
F F
Xe
F F
F F
Xe
F F
F F
Reflections σv
σ
h
σd
σd
XeF4
Page 12
XeF4
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Z
X
Z
X
Atom at (-x,-y,-z) Atom at (x,y,z)
Inversion , i
Centre of inversion
i element is a centre of symmetry
InversionExamples: Benzene, XeF4
Ethene
Page 14
C
H
HH
C4σ
S4 Improper Rotation
Rotate about C4 axis and then reflect
perpendicular to this axis
S4
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S4 Improper Rotation
Page 16
successive operation
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Page 17
KULIAH MINGGU II
TEORI GRUP
Page 18
Mathematical Definition: Group Theory
A group is a collection of elements having certain properties
that enables a wide variety of algebraic manipulations to be
carried out on the collection
Because of the symmetry of molecules they can
be assigned to a point group
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Steps to classify a molecule into a point group
Question 1:• Is the molecule one of the following recognisable
groups ?
NO: Go to the Question 2
YES:Octahedralpoint group symbol Oh
Tetrahedral point group symbol Td
Linear having no iC∞υ
Linear having i D∞h
Page 20
Steps to classify a molecule into a point group
Question 1:
• Is the molecule one of the following recognisablegroups ?
NO: Go to the Question 2
YES:Octahedralpoint group symbol Oh
Tetrahedral point group symbol Td
Linear having no iC∞υLinear having i D∞h
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Steps to classify a molecule into a point group
Question 2:• Does the molecule possess a rotation axis of order ≥ 2 ?
YES: Go to the Question 3
NO:
If no other symmetry elements point group symbol C1
If having one reflection plane point group symbol Cs
If having iCi
Page 22
Steps to classify a molecule into a point group
Question 3:
• Has the molecule more than one rotation axis ?
YES: Go to the Question 4
NO:
If no other symmetry elementspoint group symbol Cn (n is the order ofthe principle axis)
If having n σh point group symbol Cnh
If having n σv Cnv
If having an S2n axis coaxial with principal axis S2n
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Steps to classify a molecule into a point group
Question 4:• The molecule can be assigned a point group as
follows:
No other symmetry elements present Dn
Having n σd bisecting the C2 axes Dnd
Having one σh Dnh
Page 24
Molecule
Linear ?
i?D∞hC∞v
2 or moreCn, n>2?
i?
Td C5?Ih Oh
Cn?
Select Cn with highest n,
nC2 perpendicular to Cn?
**σh?Dnh
nσd ?Dnd Dn σ?Cs
i?Ci C1σh?Cnh
nσv?Cnv
S2n?S2n Cn
Y
N
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Benzene
Linear ?
i?D∞hC∞v
2 or more
Cn, n>2?
i?
Td C5?Ih Oh
Cn?
Select Cn with highest n,
nC2 perpendicular to Cn?
**σh?Dnh
nσd ?Dnd Dn σ?Cs
i?Ci C1σh?Cnh
nσv?Cnv
S2n?S2n Cn
Y
N
n = 6Benzene
is D6h
Page 26
Tugas I: Symmetry and Point Groups
Tentukan unsur simetri dan grup titik pada
molekul
a. N2F2
b. POCl3Gambarkan geometri masing-masing
molekul tersebut
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B
O
OO
H
H
H
Example of Group Properties
B(OH)3 belongs to C3 point group
It has E, C3 and C32
symmetry operations
Page 30
•Any Combination of 2 or more elements of the collection must be equivalent to
one element which is also a member of the collectionAB = C where A, B and C are all members of the collection
B
O2
O3
O1
H2
H1
H3
B
O1
O2
O3
H1
H3
H2
B
O3
O1
O2
H3
H2
H1
C3C3
Overall: C3 followed C3 gives C32
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•There must be an IDENTITY ELEMENT (E)
AE = A for all members of the collection
E commutes with all other members of the group AE= EA =A
B
O2
O3
O1
H2
H1
H3
B
O3
O1
O2
H3
H2
H1
B
O1
O2
O3
H1
H3
H2
C32 C3
2
E. C32
= C32 andC3
2. C3 = E and C32
. C32 = C3
Page 32
•The combination of elements in the group must be ASSOCIATIVE
A(BC) = AB(C) = ABC
Multiplication need not be commutative (ie: AC≠CA)
C3 .(C3 .C32 )= (C3
.C3) C32
(Do RHS First)
C3.C3
2 = E ; C3 .E = C3
C3 .C3 = C32 ; C3
2 .C32 = C3
Operations are associative and E, C3 andC32 form a group
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Group Multiplication Table
C3EC32C3
2
EC32C3C3
C32C3EE
C32C3EC3
Order of the group =3
•Every member of the group must
have an INVERSE which is also
a member of the group.
AA-1 = E
The inverse of C32 is C3
The inverse of C3
is C3
2
Page 34
KULIAH MINGGU IV-V
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© Imperial College London 35
Math Based
Matrix math is an integral
part of Group Theory;
however, we will focus
on application of the
results.
For multiplication:
Number of vertical columns in the
first matrix = number of horisontal
rows of the second matrix
Product:
Row is determined by the row ofthe first matrix and columns by the
column of the second matrix
© Imperial College London 36
Math based
[1 2 3]
1 0 0
0 -1 0
0 0 1
=
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S
O O
S
O O
C2
A unit vector on each atom represents translation in the y direction
C2.(Ty) = (-1) Ty E .(Ty) = (+1) Ty
σyz .(Ty) = (+1) Ty σxz .(Ty) = (-1) Ty
Constructing the Representation
Page 40
S
O O
Constructing the Representation
A unit vector on each atom represents rotation around the z(C2) axis
C2.(R z) = (+1) R z E .(R Z) = (+1) R z
σyz .(R z) = (-1) R z σxz .(R Z) = (-1) R Z
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Page 41
Constructing the Representation
Ty,Rx+1-1-1+1
Tx,Ry-1+1-1+1
Rz-1-1+1+1
Tz+1+1+1+1
σ(yz)σ(xz)C2EC2v
Page 42
S
O O
Constructing the Representation
Use a mathematical function
Eg: py orbital on S
Ty,Rx+1-1-1+1
σ(yz)σ(xz)C2EC2v
py has the same symmetry properties as Ty and R x vectors
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Constructing the Representation
AuAu
Au
σh
σh.[d x2-y2] =
(+1) .[d x2-y2]
C4.[d x2-y2] =
(-1) .[d x2-y2]
C4[AuCl4]-
Page 44
Constructing the Representation
-1+1+1-1+1-1+1+1-1+1
2σd2σvσh2S4I2C2”2C2’C22C4ED4h
Effects of symmetry operations generate the
TRANSFORM MATRIX
For all the symmetry operations of D4h on [d x2-y2]
We have:
Simple examples so far.
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Constructing the Representation:
The TRANSFORMATION MATRIX
Examples can be more complex:e.g. the px and py orbitals in a system with a C4 axes.
X
Y
C4 px px’ ≡ py
py py’ ≡ px
⎥⎦⎤⎢
⎣⎡⎥
⎦⎤⎢
⎣⎡ −=⎥
⎦⎤⎢
⎣⎡
y
x
y
x
p
p
p
p
01
10''In matrix form: A 2x2 transformation
matrix
Page 46
Constructing the Representation
• Vectors and mathematical functions can be used to build arepresentation of point groups.
• There is no limit to the choice of these.
• Only a few have fundamental significance. These cannot bereduced.
• The IRREDUCIBLE REPRESENTATIONS
• Any REDUCIBLE representation is the SUM of the set ofIRREDUCIBLE representations.
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⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
333231
232221
131211
aaa
aaa
aaa
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
33
2221
2111
00
0
0
b
bb
bb
Constructing the Representation
If a matrix belongs to a reducible representation it can be transformed so that zero elements are distributed about the diagonal
Similarity Transformation
A goes to B
The similarity transformation is such that
C-1 AC = B where C-1C=E
Page 48
⎥⎥⎥
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎢⎢⎢
⎣
⎡
A
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
n B
B
B
..
..
2
1
Constructing the Representation
Generally a reducible representation A can be reduced such
That each element Bi is a matrix belonging to an irreducible representation.All elements outside the Bi blocks are zero
This can generate very large matrices.
However, all information is held in the character of these matrices
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Page 49
Character Tables
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
333231
232221
131211
aaa
aaa
aaa
Character , χ = a11 + a22 + a33.
∑=
=n
i
nma1
χ In general
And only the characterχ, which is a number is required and NOT the whole matrix.
Page 50
Character Tables an Example C3v : (NF3)
(Tx,Ty) or (Rx,Ry)000-1-12
Rz-1-1-1111
Tz111111
σvvvC32C3
1EC3v
This simplifies further. Some operations are of the same class and always have the
same character in a given irreducible representation
C31, C3
1 are in the same class
σv,σv, σv are in the same class
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Page 51
Character Tables an Example C3v : (NF3)
(x2, y2, xy) (yz, zx)(Tx,Ty) or (Rx,Ry)0-12E
Rz-111 A2
x2 + y2Tz111 A1
σv2C3EC3v
There is a nomenclature for irreducible representations: Mulliken Symbols
A is single and E is doubly degenerate (ie x and y are indistinguishable)
Page 52
Note:
You will not be asked to generate character tables.
These can be brought/supplied in the examination
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Page 53
KULIAH MINGGU VI-VII-VIII
Page 54
General form of Character Tables:
(a) (b)
(c) (d) (e)(f)
(a) Gives the Schonflies symbol for the point group.
(b) Lists the symmetry operations (by class) for that group.
(c) Lists the characters, for all irreducible representations for each class
of operation.
(d) Shows the irreducible representation for which the six vectors
Tx, Ty, Tz, and R x, R y, R z, provide the basis.
(e) Shows how functions that are binary combinations of x,y,z (xy or z2)
provide bases for certain irreducible representation.(Raman d orbitals)
(f) List conventional symbols for irreducible representations:
Mulliken symbols
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Page 55
Mulliken symbols: Labelling
All one dimensional irreducible representations are labelledA or B.
All two dimensional irreducible representations are labelledE.
(Not to be confused with Identity element)
All three dimensional representations are labelled T.
For linear point groups one dimensional representations aregiven the symbol Σ with two and three dimensional representations
being Π and ∆.
Page 56
Mulliken symbols: Labelling
A one dimensional irreducible representation is labelled A if it is symmetric
with respect to rotation about the highest order axis Cn.
(Symmetric means that χ = + 1 for the operation.)
If it is anti-symmetric with respect to the operation χ = - 1 and it is labelled B.
A subscript 1 is given if the irreducible representation is symmetric with respect
to rotation about a C2 axis perpendicular to Cn or (in the absence of such an axis)
to reflection in a σv plane. An anti-symmetric representation is given the subscript 2.
For linear point groups symmetry with respect to s is indicated by a superscript
+ (symmetric) or – (anti-symmetric)
1)
2)
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Generating Reducible Representations
⎥⎥
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−
−
=
⎥⎥
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−
−
=
⎥⎥
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
s
s
s
s
s
s
s
y
s
xz
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
1
1
1
2
2
2
2
2
2
1
1
1
2
2
2
1
1
1
)( .
100000000010000000
001000000
000000100
000000010
000000001
000100000
000010000
000001000
σ
In matrix form
Page 60
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−
−
=
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−
−
=
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
s
s
s
s
s
s
s
y
s
xz
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
1
1
1
2
2
2
2
2
2
1
1
1
2
2
2
1
1
1
)( .
100000000
010000000
001000000
000000100
000000010
000000001
000100000
000010000
000001000
σ
Only require the characters: The sum of diagonal elements
For σ(xz) χ = + 1
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Page 61
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−
−
=
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−
−
=
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
s
s
s
s
s
s
s
y
s
yz
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
2
2
2
1
1
1
2
2
2
1
1
1
2
2
2
1
1
1
)( .
100000000
010000000
001000000
000100000
000010000
000001000
000000100
000000010
000000001
σ
For σ(yz) χ = + 3
Page 62
⎥
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎥
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎥
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
s
s
s
s
s
s
s
y
s
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
E
2
2
2
1
1
1
2
2
2
1
1
1
2
2
2
1
1
1
.
100000000
010000000
001000000
000100000
000010000
000001000
000000100
000000010
000000001
For E χ = + 9
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Page 63
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−
−
−
−
−
=
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−
−
−
−
−
=
⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
s
s
s
s
s
s
s
y
s
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
C
1
1
1
2
2
2
2
2
2
1
1
1
2
2
2
1
1
1
2 .
100000000
010000000
001000000
000000100
000000010
000000001
000100000
000010000
000001000
.
For C2 χ = -1
Page 64
Generating Reducible Representations
C2v
Γ3n
E C2σ(xz) σ(yz)
+9 -1 +1 3
Summarising we get that Γ3n for this molecule is:
yzTy , Ry+1-1-1+1B2
xzTx , Rx-1+1-1+1B1
xyRz-1-1+1+1 A2
x2, y2, z2Tz+1+1+1+1 A1
σ(yz)σ(xz)C2EC2v
To reduce this we need the character table for the point groups
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Page 65
Reducing Reducible Representations
We need to use the reduction formula: ( ) R Rng
a p
R
R p χ χ ).(.1 ∑⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ =
Where a p is the number of times the irreducible representation, p,
occurs in any reducible representation.
g is the number of symmetry operations in the group
(R) is character of the reducible representation
p(R) is character of the irreducible representation
n R is the number of operations in the class
Page 66
yzTy , Ry+1-1-1+1B2
xzTx , Rx-1+1-1+1B1
xyRz-1-1+1+1 A2
x2, y2, z2Tz+1+1+1+1 A1
1σ(yz)1σ(xz)1C21EC2
v
C2v
Γ3n
E C2σ(xz) σ(yz)
+9 -1 +1 3
For C2v ; g = 4 and nR = 1 for all operations
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Page 67
aA1= (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x1) + (1x3x1)] = (12/4) =3
( ) R Rng
a p
R
R p χ χ ).(.1
∑⎟⎟ ⎠
⎞⎜⎜⎝
⎛ =
C2v
Γ3n
E C2σ(xz) σ(yz)
+9 -1 +1 3
aA2= (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x-1) + (1x3x-1)] = (4/4) =1
aB1= (1/4)[ ( 1x9x1) + (1x-1x-1) + (1x1x1) + (1x3x-1)] = (8/4) =2
aB2= (1/4)[ ( 1x9x1) + (1x-1x-1) + (1x1x-1) + (1x3x1)] = (12/4) =3
Γ3n = 3A1 + A2 + 2B1 + 3B2
Page 68
aA1= (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x1) + (1x3x1)] = (12/4) =3
Reducing Reducible Representations
The terms in blue represent contributions from theun-shifted atoms
Only these actually contribute to the trace.
If we concentrate only on these un-shifted atoms we can
simplify the problem greatly.
For SO2 (9 = 3 x 3) ( -1 = 1 x –1) (1 = 1 x 1) and ( 3 = 3 x 1)
Number of un-shifted atoms Contribution from these atoms
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Page 69
Identity E
E
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
z
y
x
z
y
x
.
100
010
001
1
1
1
For each un-shifted atom
χ(E) = +3
z
y
x
z1
y1
x1
Page 70
Inversion i
z
y
xz1
y1
x1
i
For each un-shifted atom
χ(i) = -3
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
−
=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
z
y
x
z
y
x
.
100
010
001
1
1
1
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Page 71
For each un-shifted atom
z1
y1
x1x
z
y
σ(xz)
Reflection σ(xz) (Others are same except location of –1 changes)
χ(σ(xz)) = +1
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
z
y
x
z
y
x
.
100
010
001
1
1
1
Page 72
θθ
360/n
x1y1
z1z
y
x
Cn
Rotation Cn
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎟ ⎠
⎞⎜⎝
⎛ ⎟ ⎠
⎞⎜⎝
⎛
⎟ ⎠
⎞⎜⎝
⎛ −⎟ ⎠
⎞⎜⎝
⎛
=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
z
y
x
nn
nn
z
y
x
.
100
0360
cos360
sin
0360
sin360
cos
1
1
1
χ(Cn) = 1 + 2.cos(360/n)
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Page 73
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
−
⎟ ⎠ ⎞⎜
⎝ ⎛ ⎟
⎠ ⎞⎜
⎝ ⎛
⎟ ⎠
⎞⎜⎝
⎛ −⎟ ⎠
⎞⎜⎝
⎛
=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
z
y
x
nn
nn
z
y
x
.
100
0360cos360sin
0360
sin360
cos
1
1
1
χ(Sn) = -1 + 2.cos(360/n)
Improper rotation axis, Sn
Cn
σ(xy)z
y
x
z’
y’x’
y1x1
z1
Page 74
Summary of contributions from un-shifted atoms toΓ3n
0-1 + 2.cos(360/n) S61,S6
5
-1-1 + 2.cos(360/n) S41,S4
2
-2-1 + 2.cos(360/n) S31,S32
+11+ 2.cos(360/n) C4, C43
01+ 2.cos(360/n) C3 ,C32
-11+ 2.cos(360/n) C2
+1σ
-3i
+3E
χ(R)R
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Page 75
P
O
Cl
Cl
Cl
Worked example: POCl3 (C3v point group)
R χ(R)E
σv
2C3
+3
+1
0
C3vE 3σv
A1
A2
E
1 1 1
1 1 -1
2 -1 0
C3
Un-shifted
atoms
Contribution
3n
5 2 3
3 0 1
15 0 3
Number of classes,
(1 + 2 + 3 = 6)
Order of the group,
g = 6
Page 76
Reducing the irreducible representation for POCl3
( ) R Rng
a p
R
R p χ χ ).(.1
∑⎟⎟ ⎠
⎞⎜⎜⎝
⎛ =
2C3C3vE 3σv
Γ3n15 0 3
a(A1) = 1/6[(1x 15x1) + (2 x 0 x 1) + (3 x 3x 1)] = 1/6 [15 + 0+ 9] = 4
a(A2) = 1/6[(1 x 15 x 1) + ( 2 x 0 x 1) + (3 x 3x –1)] = 1/6 [15 + 0 -9] = 1
a(E) = 1/6[ (1 x 15 x 2) + (2 x 0 x –1) + (3 x 3 x 0)] = 1/6[30 + 0 + 0 ] =5
3n = 4A1 + A2 + 5E
For POCl3 n= 5 therefore the number of degrees of freedom is 3n =15.
E is doubly degenerate so 3n has 15 degrees of freedom.
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Page 77
KULIAH MINGGU IX-X-XI-XII
APLIKASI TEORI GRUP
Page 78
yzTy , Ry+1-1-1+1B2
xzTx , Rx-1+1-1+1B1
xyRz-1-1+1+1 A2
x2, y2, z2Tz+1+1+1+1 A1
1σ(yz)1σ(xz)1C21EC2
v
C2v
Γ3n
E C2σ(xz) σ(yz)
+9 -1 +1 3
3n = 3A1 + A2 + 2B1 + 3B2
Group Theory and Vibrational Spectroscopy: SO2
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Page 79
Group Theory and Vibrational Spectroscopy: SO2
3n
= 3A1
+ A2
+ 2B1
+ 3B2
= 3 + 1 + 2 + 3 = 9 = 3n
yzTy , Ry+1-1-1+1B2
xzTx , Rx-1+1-1+1B1
xyRz-1-1+1+1 A2
x2, y2, z2Tz+1+1+1+1 A1
1σ(yz)1σ(xz)1C21EC2
v
For non linear molecule there are 3n-6 vibrational degrees of freedom
Γrot = A2 + B1 + B2
Γtrans = A1 + B1 + B2
Γvib = Γ3n – Γrot – Γ trans
vib = 2A1 + B2 (Degrees of freedom = 2 + 1 = 3 = 3n-6)
Γ3n = 3A1 + A2 + 2B1 + 3B2
Page 80
P
O
ClCl
Cl
Group Theory and Vibrational Spectroscopy: POCl3
3n = 4A1 + A2 + 5E
trans = A1 + E
rot = A2 + E
vibe = 3A1 + 3E
There are nine vibrational modes . (3n-6 = 9)
The E modes are doubly degenerate and
constitute TWO modes
There are 9 modes that transform as 3A1 + 3E.
These modes are linear combinations of the three vectorsattached to each atom.
Each mode forms a BASIS for an IRREDUCIBLE representation
of the point group of the molecule
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Page 83
ψf
ψi
τ ψ µ ψ τ ψ µ ψ d d TM f i f i ∫∫ =∝ *
Infrared Spectroscopy
µ Is the transition dipole moment operator and
has components: µx, µy, µz.
Wavefunction final state
Wavefunction initial state
Note: Initial wavefunction
is always real
Page 84
Infrared Spectroscopy
• Transition is forbidden if TM = 0
• Only non zero if direct product: ψf µ ψi
contains the totally symmetric representation.
• IE all numbers for χ in representation are +1
• The ground state ψi is always totally symmetric
• Dipole moment transforms as Tx, Ty and Tz.
• The excited state transforms the same as the vectors that describethe vibrational mode.
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Page 85
The DIRECT PRODUCT representation.
( )( )( )( )
( ) f
z
y
x
i
T
T
T
ψ ψ Γ•⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
Γ
Γ
Γ
•Γ
τ ψ µ ψ τ ψ µ ψ d d TM f i f i ∫∫ =∝ *
vib = 2A1 + B2For SO2 we have that:
Under C2v :
Tx, Ty and Tz transform as B1, B2 and A1 respectively.
Page 86
C2V E C2 σ(xz) σ(yz)
A1 +1 +1 +1 +1 µz
A2 +1 +1 −1 −1 Rz
B1 +1 −1 +1 −1 µx, Ry
B2 +1 −1 −1 +1 µy, Rx
A1 × B1 × A1 +1 −1 +1 −1 ≡B1
A1 × B2 × A1 +1 −1 −1 +1 ≡B2
A1 × A1 × A1 +1 +1 +1 +1 ≡A1
A1 × B1 × B2 +1 +1 −1 −1 ≡A2
A1 × B2 × B2 +1 +1 +1 +1 ≡A1
A1 × A1 × B2 +1 −1 −1 +1 ≡B2
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Page 87
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ =•
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ •
1
2
1
1
1
2
1
1
A
B
B
A
A
B
B
A⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ =•
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ •
2
1
2
2
1
2
1
1
B
A
A
B
A
B
B
A
The DIRECT PRODUCT representation
Group theory predicts only A1
and B2
modes
Both of these direct product representations contain
the totally symmetric species so they are symmetry allowed.
This does not tell us the intensity only whether they are allowed
or not.
vib = 2A1 + B2
We predict three bands in
the infrared spectrum of SO2
Page 88
Infrared Spectroscopy : General Rule
If a vibrational mode has the same symmetry properties
as one or more translational vectors(Tx,Ty, or Tz) for that
point group, then the totally symmetric representation is
present and that transitions will be symmetry allowed.
Note:
Selection rule tells us that the dipole changes during a vibration
and can therefore interact with electromagnetic radiation.
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Page 89
Raman Spectroscopy
• Raman effect depends on change in polarisability α.• Measures how easily electron cloud can be distorted
• How easy it is to induce a dipole
• Intermediate is a virtual state
• THIS IS NOT AN ABSORPTION
• Usually driven by a laser at ω1.
• Scattered light at ω2.
• Can be Stokes(lower energy) or Anti-Stokes shifted
• Much weaker effect than direct absorption.
Page 90
ψf
ψi
Wavefunction final state
Wavefunction initial state
ω1 ω2
ω =ω1− ω 2
Virtual state
Raman Spectroscopy
Stokes Shifted
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Page 91
ψf
ψi
Wavefunction intial state
Wavefunction final state
ω1 ω2
ω =ω2− ω 1
Virtual stateRaman Spectroscopy
Anti-Stokes Shifted
Page 92
Raman Spectroscopy
τ ψ α ψ d f i
ˆ∫∝Probability of a Raman transistion:
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
=
zz zy zx
yz yy yx
xz xy xx
α α α
α α α
α α α
α
The operator , α , is the polarisability tensor
For vibrational transitions αij = α ji
so there are six distinct components:
αx2, αy2, αz2, αxy, αxz and αyz
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Page 93
αx2, αy2, αz2, αxy, αxz and αyz
Raman Spectroscopy
For C2v
Transform as:
A1, A1, A1, A2, B1 and B2
We can then evaluate the direct product representation
in a broadly analagous way
Page 94
Raman Spectroscopy The DIRECT PRODUCT representation
For SO2 group theory predicts only A1 and B2 modes
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
=•
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
•
2
1
2
1
1
2
1
2
1
1̀
B
B
A
A
A
B
B
A
A
A
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
=•
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
•
1
2
1
2
2
2
1
2
1
1̀
A
A
B
B
B
B
B
A
A
A
Both of these direct product representations contain
the totally symmetric species so they are symmetry allowed.
We predict three bands in the Raman spectrum of SO2
Note: A1 modes are polarised
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Page 95
Raman Spectroscopy : General Rule
If a vibrational mode has the same symmetry as on or moreOf the binary combinations of x,y and z the a transition from
this mode will be Raman active.
Any Raman active A1 modes are polarised.
Infrared and Raman are based on two DIFFERENT phenomena
and therefore there is no necessary relationship
between the two activities.
The higher the molecular symmetry the fewer “co-incidences”
between Raman and infrared active modes.
Page 96
Analysis of Vibrational Modes:
Vibrations can be classified into Stretches, Bends and Deformations
For SO2 vib = 2A1 + B2
We could choose more “natural” co-ordinates
S
O O
z
y
x
r 2r 1
Determine the representation for Γstretch
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Page 97
Analysis of Vibrational Modes: S
O O
r 2r 1How does our new basis transform
Under the operations of the group?
Vectors shifted to new position contribute zero
Unshifted vectors contribute + 1 to χ(R)
C2v
Γstre
E C2σ(xz) σ(yz)
+2 0 0 +2
This can be reduced using reduction formula or by inspection:
Γstre = A1 + B2
( 1, 1, 1, 1)(A1) + (1,-1, -1, 1) (B2) = (2, 0, 0, 2)
Page 98
Analysis of Vibrational Modes:
Two stretching vibrations exist that transform as A1 and B2.
These are linear combinations of the two vectors along the bonds.
We can determine what these look like by using symmetry adapted
linear combinations (SALCs) of the two stretching vectors.
Our intuition tells us that we might have a symmetric and an
anti-symmetric stretching vibration
A1 and B2
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Page 99
Symmetry Adapted Linear Combinations S
O O
r 2r 1
C2v
r 1
E C2σ(xz) σ(yz)
r 1 r 2 r 2 r 1
Pick a generating vector eg: r 1
How does this transform under symmetry operations?
Multiply this by the characters of A1 and B2
For A1 this gives: (+1) r 1+ (+1) r 2 + (+1) r 2 + (+1) r 1 = 2r 1 + 2r 2
Normalise coefficients and divide by sum of squares:
)(2
121 r r +=
Page 100
Symmetry Adapted Linear Combinations
For B2 this gives: (+1) r 1+ (-1) r 2 + (-1) r 2 + (+1) r 1 = 2r 1 - 2r 2
Normalise coefficients and divide by sum of squares:
)(2
121 r r −=
S
O O
S
O O
A1 B2
Sulphur must also move to maintain position of centre of mass
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Page 101
Analysis of Vibrational Modes:
S
O ORemaining mode “likely” to be a bend
C2v
Γ bend
E C2σ(xz) σ(yz)
1 1 1 1
By inspection this bend is A1 symmetry
SO2 has three normal modes:
A1 stretch: Raman polarised and infrared active
A1 bend: Raman polarised and infrared active
B2 stretch: Raman and infrared active
Page 102
Analysis of Vibrational Modes: SO2 experimental data.
ν3B2 stretch13361362
ν2 A1 stretch11451151
ν1 A1 bend524518
NameSymRaman(liquid)/cm-1IR(Vapour)/cm-1
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Page 103
Analysis of Vibrational Modes: SO2 experimental data.
Notes:
Stretching modes usually higher in frequency than bending modes
Differences in frequency between IR and Raman are due to
differing phases of measurements
“Normal” to number the modes According to how the Mulliken term
symbols appear in the character table, ie. A1 first and then B2
Page 104
Analysis of Vibrational Modes: POCl3
P
O
ClCl
Cl
P
O
ClCl
ClP
O
ClCl
Cl
P=O stretch P-Cl stretch
Angle deformations
vibe = 3A1 + 3E
3 A1 vibrations IR active(Tz) + Raman active polarised( x2 + y2 and z2)
3 E vibrations IR active(Tx,Ty) + Raman active ( x2 - y2 , xy) (yz,zx)
Six bands, Six co-incidences
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Page 105
Analysis of Vibrational Modes: POCl3
vibe = 3A1 + 3E
C3v E 2C3 3σv
Γ P=O str
Γ P-Cl str
Γ bend
1 1 1
3 0 1
6 0 2
Using reduction formulae or by inspection:
Γ P=O str = A1 and Γ P-Cl str = A1 + E
Γ bend = Γvibe - Γ P=O str - Γ P-Cl str = 3A1 + 3E – 2A1 – E = A1 + 2E
Reduction of the representation for bends gives:Γ bend = 2A1 + 2E
Page 106
Analysis of Vibrational Modes: POCl3
Γ bend = Γvibe - Γ P=O str - Γ P-Cl str = 3A1 + 3E – 2A1 – E = A1 + 2E
Reduction of the representation for bends gives:Γ bend = 2A1 + 2E
One of the A1 terms is REDUNDANT as not
all the angles can symmetrically increase
Γ bend = A1 + 2E
Note:
It is advisable to look out for redundant co-ordinates and think
about the physical significance of what you are representing.
Redundant co-ordinates can be quite common and can lead to a
double “counting” for vibrations.
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Page 107
ν6Edeformation193-
ν3 A1Sym.
Deformation(1)
267(pol)267
ν5Edeformation337340
ν2 A1P-Cl str( 1,2,3)486(pol)487
ν4EP-Cl str(2,3)582580
ν1 A1P=O str( 1,4)1290(pol)1292
LabelSymDescriptionRaman /cm-1IR (liq)/ cm-1
Analysis of Vibrational Modes: POCl3
Page 108
Analysis of Vib rat ional Modes: POCl3
1) All polarised bands are Raman A1 modes.
2) Highest frequencies probably stretches.
3) P-Cl stretches probably of similar frequency.
4)Double bonds have higher frequency than similar single bonds.
A1 modes first. P=O – highest frequency
Then P-Cl stretch, then deformation.
581 similar to P-Cl stretch so assym. stretch.
Remaining modes must therefore be deformations
Could now use SALCs to look more closely at the normal modes
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Page 109
Symmetry, Bonding and Electronic Spectroscopy
• Use atomic orbitals as basis set.
• Determine irreducible representations.
• Construct QULATITATIVE molecular orbital diagram.
• Calculate symmetry of electronic states.
• Determine “allowedness” of electronic transitions.
Page 110
O
H H
+
O
H H
+E, C2, σxz, σyx
O 2s orbital
C2V E C2 σxz σyz
O2s +1 +1 +1 +1 a1
Symmetry, Bonding and Electronic Spectroscopy
σ bonding in AXn molecules e.g. : water
How do 2s and 2p orbitals transform?
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Page 111
Symmetry, Bonding and Electronic Spectroscopy
s-orbitals are spherically symmetric and when at themost symmetric point always transform as the totally
symmetric species
For electronic orbitals, either atomic or molecular,
use lower case characters for Mulliken symbols
Oxygen 2s orbital has a1 symmetry in the C2v point group
Page 112
OH H
E, C2, σxz, σyx
O 2pz orbital
−
+
OH H−
+
C2V E C2 σxz σyz
O2pz +1 +1 +1 +1 a1
Symmetry, Bonding and Electronic Spectroscopy
How do the 2p orbitals transform?
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Page 115
OH H
H 1s orbitals
+ +
φ1 φ2
z
y
x
C2V E C2 σxz σyz
σ
2 0 0 2 a1 + b2
Symmetry, Bonding and Electronic Spectroscopy
Use the 1s orbitals on the hydrogen atoms
Page 116
Symmetry, Bonding and Electronic Spectroscopy
Assume oxygen 2s orbitals are non bonding
Oxygen 2pz is a1, px is b1 and py is b2
Ligand orbitals are a1 and b2
Which is lower in energy a1 or b2?
Guess that it is a1 similar symmetry better interaction?
Orbitals of like symmetry can interact
Oxygen 2px is “wrong” symmetry therefore likely to be non-bonding
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Page 117
2s
O 2H
a1
a1 + b1 + b2
a1 + b2
non-bonding
non-bonding (O 2px)
Qualitative MO diagram for H2O
a1
a1
a1*
b2
b2*
b1
H2O
Page 118
Symmetry, Bonding and Electronic Spectroscopy
Is symmetry sufficient to determine ordering of a1 and b2 orbitals?
Construct SALC and asses degree of overlap.
Take one basis that maps onto each other
Use φ1 or φ2 as a generating function.
(These functions must be orthogonal to each other)
Observe the effect of each symmetry operation on the function
Multiply this row by each irreducible representation of the point
Group and then normalise.
(Here the irreducible representation is already known)
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Page 123
Symmetry of Electronic States from NON-DEGENERATE MO’s.
For FULL singly degenerate MO’s, the symmetry is ALWAYS A1
(The totally symmetric species of the point group)
For orbitals with only one electron:
(a1)1 = A1, (b2)
1 = B2, (b1)1 =B1
General rule:
For full MO’s the ground state is always totally symmetric
Page 124
Symmetry of Electronic States from NON-DEGENERATE MO’s.
What happens if we promote an electron?
a1
b1
b2
b2*
a1*
Bonding
Non bonding
Anti Bonding
First two excitations move an electron form b1 non bonding
Into either the b2* or a1* anti-bonding orbitals .
Both of these transitions are
non bonding to anti bonding
transitions. n-π*
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Page 125
What electronic states do these new configurations generate?
(a1
)2(b2
)2(b1
)1(b2
*)1(a1
*)0
(a1)2(b2)
2(b1)1(b2*)0(a1*)1
= A1
.A1
.B1
.B2
= A2
= A1.A1.B1.A1 = B1
In these states the spins can be paired or not.
IE: S the TOTAL electron spin can equal to 0 or 1.
The multiplicity of these states is given by 2S+1
These configurations generate:3A2 , 1A2 and 3B1 , 1B1 electronic states.
Note: if S= ½ then we have a doublet state
Page 126
a1
b1
b2
b2*
a1*
What electronic states do these new configurations generate?
Molecular Orbitals
1A1
1B1
3B1
1A2
3A2
Electronic States
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Page 127
What electronic states do these new configurations generate?
Triplet states are always lower than the related singlet states
Due to a minimisation of electron-electron interactions and
thus less repulsion
Between which of these states are electronic transitions
symmetry allowed?
Need to evaluate the transition moment integral like we did for
infrared transitions.
τ ψ µ ψ τ ψ µ ψ d d TM f i f i ∫∫ =∝ *
Page 128
Electronic
τ ψ µ ψ τ ψ ψ τ ψ ψ d d d TMI f eie f S iS f V iV ,,*
,,*
,,* ∫∫∫ ••≈
Which electronic transitions are allowed?
Vibrational Spin
To first approximation µ can only operate on the electronic partof the wavefunction.
Vibrational part is overlap between ground and excited state nuclear
wavefunctions. Franck-Condon factors.
Spin selection rules are strict. There must be NO change in spin
Direct product for electronic integral must contain the totally
symmetric species.
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Page 129
Which electronic transitions are allowed?
A transition is allowed if there is no change in spin and theelectronic component transforms as totally symmetric.
The intensity is modulated by Franck-Condon factors.
The electronic transition dipole momentµ transforms as the
translational species as for infrared transitions.
Page 130
Which electronic transitions are allowed?
For the example of H20 the direct products for the
electronic transition are
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ =•
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ •
1
2
2
2
1
2
1
1
B
B
A
A
A
B
B
A
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ =•
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ •
2
1
1
1
1
2
1
1
A
A
B
B
A
B
B
A
The totally symmetric species is only present for the transitionto the B1 state. Therefore the transition to the A2 state is
“symmetry forbidden”
Transitions between singlet states are “spin allowed”.
transitions between singlet and triplet state are “spin forbidden”.
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Page 133
More bonding for AX6 molecules / complexes
In the case of Oh point group:
d x2-y2 and d z2 transform as eg
d xy, d yz and d zx transform as t2g
px, py and pz transform as t1u
Γσ(ligands) = a1g + eg + t1u
Γπ(ligands) = t1g + t2g + t1u + t2u
Page 134
t1u
a1g
eg + t2g
t1u
t1u
*
a1g
a1g*
eg*
eg
t2g
a1g + eg + t1u
AX6 for Oh
4p
4s
3d
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Page 135
Electronic Spectroscopy of d9 complex:
[Cu(H2O)6]2+
is a d 9
complex. That is approximately Oh.
Ground electronic configuration is: (t2g)6(eg
*)3
Excited electronic configuration is : (t2g)5(eg
*)4
The ground electronic state is 2Eg
Excited electronic state is 2T2g
Under Oh the transition dipole moment transforms as t1u
Are electronic transitions allowed between these states?
Page 136
Electronic Spectroscopy of d9 complex:
Need to calculate direct product representation:
2Eg . (t1u) .2T2g
0-200-18200018DP
02-102200-12Eg
110-1-3-11-103t1u
1-10-13-1-1103T2g
6σd3σh
8S66S4i3C2
6C46C2
8C3EOh
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Page 137
Electronic Spectroscopy of d 9 complex:
0-200-18200018DP
Use reduction formula: ( ) R Rng
a p
R
R p χ χ ).(.
1∑⎟⎟
⎠
⎞⎜⎜⎝
⎛ =
aa1g= 1/48 .[( 1x18x1)+(3x2x1) +(1x-18x1) +(3x-2x1)] = 0
The totally symmetric species is not present in this direct product.
The transition is symmetry forbidden.
We knew this anyway as g-g transitions are forbidden.
Transition is however spin allowed.
Page 138
Electronic Spectroscopy of d9 complex:
Groups theory predicts no allowed electronic transition.
However, a weak absorption at 790nm is observed.
There is a phenomena known as vibronic coupling where the
vibrational and electronic wavefunctons are coupled.
This effectively changes the symmetry of the states involved.
This weak transition is vibronically induced and therefore is partially
allowed.
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