Symmetry and Group Theory
Feature: Application for Spectroscopy
and Orbital Molecules
Dr. Indriana Kartini
Page 1
Text books
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
Page 2
1
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 3
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
Page 4
2
Simbol
Unsur
Operasi
E
Unsur identitas
Membiarkan obyek tidak
berubah
Cn
Sumbu rotasi
Rotasi seputar sumbu dengan
derajat rotasi 360/n (n adalah
bilangan bulat)
σ
Bidang simetri
Refleksi melalui bidang simetri
i
Pusat/titik inversi
Proyeksi melewati pusat
inversi ke sisi seberangnya
dengan jarak yang sama dari
pusat
Sn
Sumbu rotasi tidak
sejati (Improper
rotational axis)
Rotasi mengitari sumbu rotasi
diikuti dengan refleksi pada
bidang tegak lurus sumbu
rotasi
Page 5
Operasi Simetri
© Imperial College London
6
3
Rotations 360/n where n is an integer
BF3
Operation rotation by 360/3
around C3 axis (element)
F3
F2
Rotate 120O
B
B
F1
F2
F1
F3
Page 7
z
Reflections
Reflection is the operation
σ element is plane of symmetry
x is out of the plane
y
H2O
x
σ(xz)
H2
Page 8
H1
H1
H2
H2
H1
σ(yz)
4
Reflections for H2O
Page 9
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
Page 10
5
XeF4
Reflections σv
F
F
F
F
Xe
Xe
F
σh
F
σd
F
F
F
σd
Xe
F
F
F
Page 11
XeF4
Page 12
6
Examples: Benzene, XeF4
Ethene
Inversion
i element is a centre of symmetry
Atom at (x,y,z)
Atom at (-x,-y,-z)
Z
Z
Inversion , i
Y
Y
X
X
Centre of inversion
Page 13
H
H
C
H
S4 Improper Rotation
Rotate about C4 axis and then reflect
perpendicular to this axis
H
S4
C4
σ
Page 14
7
S4 Improper Rotation
Page 15
successive operation
Page 16
8
KULIAH MINGGU II
TEORI GRUP
Page 17
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
Page 18
9
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: Octahedral Æpoint group symbol Oh
Tetrahedral Æ point group symbol Td
Linear having no i ÆC∞υ
Linear having i Æ D∞h
Page 19
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: Octahedral Æpoint group symbol Oh
Tetrahedral Æ point group symbol Td
Linear having no i ÆC∞υ
Linear having i Æ D∞h
Page 20
10
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 Æpoint group symbol C1
If having one reflection plane Æ point group symbol Cs
If having i ÆCi
Page 21
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 elements Æpoint group symbol Cn (n is the order of
the principle axis)
If having n σh Æ point group symbol Cnh
If having n σv Æ Cnv
If having an S2n axis coaxial with principal axis Æ S2n
Page 22
11
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 Æ Dn
Having n σd bisecting the C2 axes Æ Dnd
Having one σh Æ Dnh
Page 23
Molecule
Linear?
D∞h
i?
2 or more
Cn, n>2?
C∞v
i?
Ih
C5?
Oh
Td
Y
*
Dnh
Dnd
Page 24
*
σh?
nσd?
Cn?
Select Cn with highest n,
nC2 perpendicular to Cn?
N
Dn
Cs
Cnh
σh?
Cnv
nσv?
S2n
S2n?
Ci
σ?
i?
C1
Cn
12
Benzene
Linear?
D∞h
i?
2 or more
Cn, n>2?
C∞v
i?
Ih
C5?
Oh
Td
Y
*
Benzene
is D6h
Dnh
Dnd
Page 25
n=6
σh?
nσd?
Cn?
*
Select Cn with highest n,
nC2 perpendicular to Cn?
N
Dn
Cs
Cnh
σh?
Cnv
nσv?
S2n
S2n?
Ci
σ?
i?
C1
Cn
Tugas I: Symmetry and Point Groups
Tentukan unsur simetri dan grup titik pada
molekul
a. N2F2
b. POCl3
Gambarkan geometri masing-masing
molekul tersebut
Page 26
13
KULIAH MINGGU III
Page 27
Basic Properties of Groups
•
•
Any Combination of 2 or more elements of the collection must be
equivalent to one element which is also a member of the collection
AB = C
where A, B and C are all members of the collection
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
•
•
The combination of elements in the group must be ASSOCIATIVE
A(BC) = AB(C) = ABC
Multiplication need not be commutative
(ie: AC≠CA)
Every member of the group must have an INVERSE which is also a
member of the group.
AA-1 = E
Page 28
14
Example of Group Properties
H
B(OH)3 belongs to C3 point group
O
H
It has E, C3 and C32
symmetry operations
B
O
O
H
Page 29
•Any Combination of 2 or more elements of the collection must be equivalent to
one element which is also a member of the collection
AB = C
where A, B and C are all members of the collection
H2
O2
O1
C3
B
H1
O1
H3
H1
B
B
O3
H3
H3
O3
O3
C3
H2
O2
H2
O2
O1
H1
Overall: C3 followed C3 gives C32
Page 30
15
•There must be an IDENTITY ELEMENT (E)
AE = A
for all members of the collection
E commutes with all other members of the group
H2
O2
H3
O1
O3
C32
H1
O1
C32
B
B
H1
AE= EA =A
O3
H2
O2
B
O1
H3
O3
H1
H3
O2
H2
E. C32 = C32 and C32. C3 = E and C32. C32 = C3
Page 31
•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.C32 = E ;
C3 .C3 = C32 ;
C3 .E = C3
C32 .C32 = C3
Operations are associative and E, C3 and C32 form a group
Page 32
16
Group Multiplication Table
Order of the group =3
C3
E
C3
C3
2
•Every member of the group must
E
E
C3
C32
have an INVERSE which is also
a member of the group.
C3
C3
C32
E
C32
C32
E
C3
AA-1 = E
The inverse of C32 is C3
The inverse of C3 is C32
Page 33
KULIAH MINGGU IV-V
Page 34
17
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 of
the first matrix and columns by the
column of the second matrix
© Imperial College London
35
Math based
[1 2 3]
1
0
0
0
-1
0
0
0
1
=
© Imperial College London
36
18
Representations of Groups
• Diagrams are cumbersome
• Require numerical method
– Allows mathematical analysis
– Represent by VECTORS or Mathematical Functions
– Attach Cartesian vectors to molecule
– Observe the effect of symmetry operations on these vectors
• Vectors are said to form the basis of the representation Æ each
symmetry operation is expressed as a transformation matrix
[New coordinates] = [matrix transformation] x [old coordinates]
Page 37
Constructing the Representation
Put unit vectors on each atom
z
S
O
O
y
x
C2v: [E, C2, σxz, σyz]
These are useful to describe molecular vibrations
and electronic transitions.
Page 38
19
Constructing the Representation
A unit vector on each atom represents translation in the y direction
C2
S
O
S
O
O
O
C2.(Ty) = (-1) Ty
E .(Ty) = (+1) Ty
σyz .(Ty) = (+1) Ty
σxz .(Ty) = (-1) Ty
Page 39
Constructing the Representation
A unit vector on each atom represents rotation around the z(C2) axis
S
O
O
C2.(Rz) = (+1) Rz
E .(RZ) = (+1) Rz
σyz .(Rz) = (-1) Rz
σxz .(RZ) = (-1) RZ
Page 40
20
Constructing the Representation
C2v
E
C2
σ(xz)
σ(yz)
+1
+1
+1
+1
Tz
+1
+1
-1
-1
Rz
+1
-1
+1
-1
Tx,Ry
+1
-1
-1
+1
Ty,Rx
Page 41
Constructing the Representation
Use a mathematical function
Eg: py orbital on S
S
O
C2v
Page 42
O
E
C2
σ(xz)
σ(yz)
+1
-1
-1
+1
Ty,Rx
py has the same symmetry properties as Ty and Rx vectors
21
Constructing the Representation
σh
Au
Au
C4
[AuCl4]-
σh.[d x2-y2] =
(+1) .[d x2-y2]
Au
C4.[d x2-y2] =
(-1) .[d x2-y2]
Page 43
Constructing the Representation
Effects of symmetry operations generate the
TRANSFORM MATRIX
Simple examples so far.
For all the symmetry operations of D4h on [d x2-y2]
We have:
D4h
E
2C4
C2
2C2’ 2C2” I
2S4
σh
2σv
2σd
+1
-1
+1
+1
-1
+1
+1
-1
-1
+1
Page 44
22
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.
Y
C4
X
In matrix form:
⎡ p x '⎤ ⎡0 − 1⎤ ⎡ p x ⎤
⎢ p '⎥ = ⎢
⎥⎢ ⎥
⎣ y ⎦ ⎣1 0 ⎦ ⎣ p y ⎦
px
px’ ≡ py
py
py’ ≡ px
A 2x2 transformation
matrix
Page 45
Constructing the Representation
• Vectors and mathematical functions can be used to build a
representation of point groups.
• There is no limit to the choice of these.
• Only a few have fundamental significance. These cannot be
reduced.
• The IRREDUCIBLE REPRESENTATIONS
• Any REDUCIBLE representation is the SUM of the set of
IRREDUCIBLE representations.
Page 46
23
Constructing the Representation
If a matrix belongs to a reducible representation it can be transformed
so that zero elements are distributed about the diagonal
⎡ a11
⎢a
⎢ 21
⎢⎣a 31
a12
a 22
a 32
a13 ⎤
a 23 ⎥⎥
a 33 ⎥⎦
Similarity Transformation
A goes to B
⎡b11 b21 0 ⎤
⎢b
⎥
⎢ 21 b22 0 ⎥
⎢⎣ 0
0 b33 ⎥⎦
The similarity transformation is such that
C-1 AC = B where C-1C=E
Page 47
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
⎡
⎢
⎢
⎢
⎢
⎢
⎢⎣
A
⎤
⎥
⎥
⎥
⎥
⎥
⎥⎦
⎡ B1
⎢
⎢
⎢
⎢
⎢
⎢⎣
B2
⎤
⎥
⎥
⎥
..
⎥
..
⎥
Bn ⎥⎦
This can generate very large matrices.
However, all information is held in the character of these matrices
Page 48
24
Character Tables
⎡ a11
⎢a
⎢ 21
⎢⎣a 31
a12
a 22
a 32
a13 ⎤
a 23 ⎥⎥
a33 ⎥⎦
Character , χ = a11 + a22 + a33.
n
In general
χ = ∑ a nm
i =1
And only the character χ, which is a number is required and
NOT the whole matrix.
Page 49
Character Tables an Example C3v : (NF3)
C3v
E
C 31 C 32 σ v σ v σ v
1
1
1
1
1
1
1
-1 -1 -1
Rz
2
-1
-1
0
(Tx,Ty) or (Rx,Ry)
1
0
1
0
Tz
This simplifies further. Some operations are of the same class and always have the
same character in a given irreducible representation
C31, C31
σv, σv, σv
are in the same class
are in the same class
Page 50
25
Character Tables an Example C3v : (NF3)
C3v
E
2C3 3σv
A1
1
1
1
Tz
A2
1
1
-1
Rz
E
2
-1
0
(Tx,Ty) or (Rx,Ry)
x2 + y2
(x2, y2, xy) (yz, zx)
There is a nomenclature for irreducible representations: Mulliken Symbols
A is single and E is doubly degenerate (ie x and y are indistinguishable)
Page 51
Note:
You will not be asked to generate character tables.
These can be brought/supplied in the examination
Page 52
26
KULIAH MINGGU VI-VII-VIII
Page 53
General form of Character Tables:
(a)
(b)
(f)
(c)
(d) (e)
(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 Rx, Ry, Rz, 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)Page
List
conventional symbols for irreducible representations:
54
Mulliken symbols
27
Mulliken symbols: Labelling
All one dimensional irreducible representations are labelled A or B.
All two dimensional irreducible representations are labelled E.
(Not to be confused with Identity element)
All three dimensional representations are labelled T.
For linear point groups one dimensional representations are
given the symbol Σ with two and three dimensional representations
being Π and ∆.
Page 55
Mulliken symbols: Labelling
1)
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.
2)
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)
Page 56
28
Mulliken symbols: Labelling
3)
Subscripts g (gerade) and u (ungerade) are given to irreducible representations
That are symmetric and anti-symmetric respectively, with respect to inversion
at a centre of symmetry.
4)
Superscripts ‘ and “ are given to irreducible representations that are symmetric
and anti-symmetric respectively with respect o reflection in a σh plane.
Note: Points 1) and 2) apply to one-dimensional representations only.
Points 3) and 4) apply equally to one-, two-, and three- dimensional representations.
Page 57
Generating Reducible Representations
zs
z1
O
S
xs
ys
x1→ x2
O
y1
x1
x2
σxz
For the symmetry operation σxz (a σv )
z2
x2→ x1 xs→ xs
y2
y1→ -y2 y2→ -y1 ys→ -ys
z1→ z2
z2 → z1
zs → zs
Page 58
29
Generating Reducible Representations
In matrix form
⎡ x1 ⎤ ⎡0 0
⎢ y ⎥ ⎢0 0
⎢ 1⎥ ⎢
⎢ z1 ⎥ ⎢0 0
⎢ ⎥ ⎢
⎢ x2 ⎥ ⎢1 0
σ ( xz ) ⎢ y2 ⎥ = ⎢0 − 1
⎢ ⎥ ⎢
⎢ z 2 ⎥ ⎢0 0
⎢ x ⎥ ⎢0 0
⎢ s⎥ ⎢
⎢ y y ⎥ ⎢0 0
⎢ z ⎥ ⎢0 0
⎣ s⎦ ⎣
0 1
0
0⎤ ⎡ x1 ⎤ ⎡ x2 ⎤
0 0⎥⎥ ⎢⎢ y1 ⎥⎥ ⎢⎢− y2 ⎥⎥
0 0⎥ ⎢ z1 ⎥ ⎢ z 2 ⎥
⎥
⎥⎢ ⎥ ⎢
0 0⎥ ⎢ x2 ⎥ ⎢ x1 ⎥
0 0⎥.⎢ y2 ⎥ = ⎢ − y1 ⎥
⎥
⎥⎢ ⎥ ⎢
0 0⎥ ⎢ z 2 ⎥ ⎢ z1 ⎥
0 0⎥ ⎢ xs ⎥ ⎢ xs ⎥
⎥
⎥⎢ ⎥ ⎢
− 1 0⎥ ⎢ y s ⎥ ⎢ − y s ⎥
0 1⎥⎦ ⎢⎣ z s ⎥⎦ ⎢⎣ z s ⎥⎦
0 0
0
0 0 −1 0 0
0 0
0 0
0
0
1 0
0 0
0 0
1 0
0
0
0 0
0 0
0 0
0 0
0
0
0 1
0 0
0 0
0
0 0
Page 59
⎡ x1 ⎤ ⎡0 0
⎢ y ⎥ ⎢0 0
⎢ 1⎥ ⎢
⎢ z1 ⎥ ⎢0 0
⎢ ⎥ ⎢
⎢ x 2 ⎥ ⎢1 0
σ ( xz ) ⎢ y2 ⎥ = ⎢0 − 1
⎢ ⎥ ⎢
⎢ z 2 ⎥ ⎢0 0
⎢ x ⎥ ⎢0 0
⎢ s⎥ ⎢
⎢ y y ⎥ ⎢0 0
⎢ ⎥ ⎢
⎣ z s ⎦ ⎣0 0
0 1
0
0 0
0 0 −1 0 0
0 0 0 1 0
0 0
0
0 0
0 0
1 0
0
0
0 0
0 0
0 0
0 0
0
0
0 1
0 0
0 0
0
0 0
0⎤ ⎡ x1 ⎤ ⎡ x2 ⎤
0 0⎥⎥ ⎢⎢ y1 ⎥⎥ ⎢⎢− y2 ⎥⎥
0 0⎥ ⎢ z1 ⎥ ⎢ z 2 ⎥
⎥
⎥⎢ ⎥ ⎢
0 0⎥ ⎢ x2 ⎥ ⎢ x1 ⎥
0 0⎥.⎢ y2 ⎥ = ⎢ − y1 ⎥
⎥
⎥⎢ ⎥ ⎢
0 0⎥ ⎢ z 2 ⎥ ⎢ z1 ⎥
0 0⎥ ⎢ xs ⎥ ⎢ xs ⎥
⎥
⎥⎢ ⎥ ⎢
− 1 0⎥ ⎢ y s ⎥ ⎢ − y s ⎥
0 1⎥⎦ ⎢⎣ z s ⎥⎦ ⎢⎣ z s ⎥⎦
0
Only require the characters: The sum of diagonal elements
For σ(xz) χ = + 1
Page 60
30
⎡ x1 ⎤ ⎡− 1
⎢y ⎥ ⎢ 0
⎢ 1⎥ ⎢
⎢ z1 ⎥ ⎢ 0
⎢ ⎥ ⎢
⎢ x2 ⎥ ⎢ 0
σ ( yz ) ⎢ y2 ⎥ = ⎢ 0
⎢ ⎥ ⎢
⎢ z2 ⎥ ⎢ 0
⎢x ⎥ ⎢ 0
⎢ s⎥ ⎢
⎢ yy ⎥ ⎢ 0
⎢z ⎥ ⎢ 0
⎣ s⎦ ⎣
0 0
0
0 0
1 0 0 0 0
0 1 0 0 0
0 0 −1 0 0
0 0
0 0
0 0
0
0
0
1 0
0 1
0 0
0 0
0 0
0
0
0 0
0 0
0 0⎤ ⎡ x1 ⎤ ⎡ − x1 ⎤
0 0 0⎥⎥ ⎢⎢ y1 ⎥⎥ ⎢⎢ y1 ⎥⎥
0 0 0⎥ ⎢ z1 ⎥ ⎢ z1 ⎥
⎥
⎥⎢ ⎥ ⎢
0 0 0 ⎥ ⎢ x2 ⎥ ⎢ − x2 ⎥
0 0 0⎥.⎢ y2 ⎥ = ⎢ y2 ⎥
⎥
⎥⎢ ⎥ ⎢
0 0 0⎥ ⎢ z 2 ⎥ ⎢ z 2 ⎥
− 1 0 0⎥ ⎢ xs ⎥ ⎢ − xs ⎥
⎥
⎥⎢ ⎥ ⎢
0 1 0⎥ ⎢ y s ⎥ ⎢ y s ⎥
0 0 1⎥⎦ ⎢⎣ z s ⎥⎦ ⎢⎣ z s ⎥⎦
0
For σ(yz) χ = + 3
Page 61
⎡ x1 ⎤ ⎡1
⎢ y ⎥ ⎢0
⎢ 1⎥ ⎢
⎢ z1 ⎥ ⎢0
⎢ ⎥ ⎢
⎢ x2 ⎥ ⎢0
E ⎢ y 2 ⎥ = ⎢0
⎢ ⎥ ⎢
⎢ z 2 ⎥ ⎢0
⎢ x ⎥ ⎢0
⎢ s⎥ ⎢
⎢ y y ⎥ ⎢0
⎢ z ⎥ ⎢0
⎣ s⎦ ⎣
0 0 0 0 0 0 0 0⎤ ⎡ x1 ⎤ ⎡ x1 ⎤
1 0 0 0 0 0 0 0⎥⎥ ⎢⎢ y1 ⎥⎥ ⎢⎢ y1 ⎥⎥
0 1 0 0 0 0 0 0⎥ ⎢ z1 ⎥ ⎢ z1 ⎥
⎥⎢ ⎥ ⎢ ⎥
0 0 1 0 0 0 0 0 ⎥ ⎢ x2 ⎥ ⎢ x2 ⎥
0 0 0 1 0 0 0 0⎥.⎢ y2 ⎥ = ⎢ y2 ⎥
⎥⎢ ⎥ ⎢ ⎥
0 0 0 0 1 0 0 0⎥ ⎢ z 2 ⎥ ⎢ z 2 ⎥
0 0 0 0 0 1 0 0 ⎥ ⎢ xs ⎥ ⎢ xs ⎥
⎥⎢ ⎥ ⎢ ⎥
0 0 0 0 0 0 1 0⎥ ⎢ y s ⎥ ⎢ y s ⎥
0 0 0 0 0 0 0 1⎥⎦ ⎢⎣ z s ⎥⎦ ⎢⎣ z s ⎥⎦
For E χ = + 9
Page 62
31
⎡ x1 ⎤ ⎡ 0 0
⎢y ⎥ ⎢ 0 0
⎢ 1⎥ ⎢
⎢ z1 ⎥ ⎢ 0 0
⎢ ⎥ ⎢
⎢ x2 ⎥ ⎢ − 1 0
C2 .⎢ y2 ⎥ = ⎢ 0 − 1
⎢ ⎥ ⎢
⎢ z2 ⎥ ⎢ 0 0
⎢x ⎥ ⎢ 0 0
⎢ s⎥ ⎢
⎢ yy ⎥ ⎢ 0 0
⎢z ⎥ ⎢ 0 0
⎣ s⎦ ⎣
0 −1
0
0
0
0
0 0
−1 0
0 1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0⎤ ⎡ x1 ⎤ ⎡ − x2 ⎤
0 0 0⎥⎥ ⎢⎢ y1 ⎥⎥ ⎢⎢− y2 ⎥⎥
0 0 0⎥ ⎢ z1 ⎥ ⎢ z 2 ⎥
⎥
⎥⎢ ⎥ ⎢
0 0 0⎥ ⎢ x2 ⎥ ⎢ − x1 ⎥
0 0 0⎥.⎢ y2 ⎥ = ⎢ − y1 ⎥
⎥
⎥⎢ ⎥ ⎢
0 0 0⎥ ⎢ z 2 ⎥ ⎢ z1 ⎥
− 1 0 0⎥ ⎢ xs ⎥ ⎢ − xs ⎥
⎥
⎥⎢ ⎥ ⎢
0 − 1 0⎥ ⎢ y s ⎥ ⎢ − y s ⎥
0 0 1⎥⎦ ⎢⎣ z s ⎥⎦ ⎢⎣ z s ⎥⎦
0
0
For C2 χ = -1
Page 63
Generating Reducible Representations
Summarising we get that Γ3n for this molecule is:
C2v
E
C2
σ(xz)
Γ3n
+9
-1
+1
σ(yz)
3
To reduce this we need the character table for the point groups
C2v
E
C2
σ(xz)
σ(yz)
A1
+1
+1
+1
+1
Tz
x2, y2, z2
A2
+1
+1
-1
-1
Rz
xy
B1
+1
-1
+1
-1
Tx , Rx
xz
B2
+1
-1
-1
+1
Ty , Ry
yz
Page 64
32
Reducing Reducible Representations
We need to use the reduction formula:
⎛1⎞
a p = ⎜⎜ ⎟⎟∑ n R .χ ( R).χ p (R )
⎝g⎠ R
Where ap is the number of times the irreducible representation, p,
occurs in any reducible representation.
g is the number of symmetry operations in the group
χ(R) is character of the reducible representation
χp(R) is character of the irreducible representation
nR is the number of operations in the class
Page 65
C2
1E
1C2
1σ(xz) 1σ(yz)
v
A1
+1
+1
+1
+1
Tz
x2, y2, z2
A2
+1
+1
-1
-1
Rz
xy
B1
+1
-1
+1
-1
Tx , Rx
xz
B2
+1
-1
-1
+1
Ty , Ry
yz
C2v
E
C2
σ(xz)
Γ3n
+9
-1
+1
σ(yz)
3
For C2v ; g = 4 and nR = 1 for all operations
Page 66
33
⎛1⎞
a p = ⎜⎜ ⎟⎟∑ n R .χ ( R).χ p (R )
⎝g⎠ R
C2v
E
C2
σ(xz)
Γ3n
+9
-1
+1
σ(yz)
3
aA1 = (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x1) + (1x3x1)] = (12/4) =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 67
Reducing Reducible Representations
aA1 = (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x1) + (1x3x1)] = (12/4) =3
The terms in blue represent contributions from the un-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
Page 68
34
Identity E
z
z1
E
y
y1
x
For each un-shifted atom
Page 69
x1
⎡ x1 ⎤ ⎡1 0 0⎤ ⎡ x ⎤
⎢ y ⎥ = ⎢0 1 0⎥.⎢ y ⎥
⎢ 1⎥ ⎢
⎥⎢ ⎥
⎢⎣ z1 ⎥⎦ ⎢⎣0 0 1⎥⎦ ⎢⎣ z ⎥⎦
χ(E) = +3
Inversion i
x1
z
y1
i
y
z1
x
For each un-shifted atom
⎡ x1 ⎤ ⎡− 1 0 0 ⎤ ⎡ x ⎤
⎢ y ⎥ = ⎢ 0 − 1 0 ⎥.⎢ y ⎥
⎢ 1⎥ ⎢
⎥⎢ ⎥
⎢⎣ z1 ⎥⎦ ⎢⎣ 0 0 − 1⎥⎦ ⎢⎣ z ⎥⎦
χ(i) = -3
Page 70
35
Reflection σ(xz)
(Others are same except location of –1 changes)
z1
z
σ(xz)
y1
y
x
x1
For each un-shifted atom
⎡ x1 ⎤ ⎡1 0 0⎤ ⎡ x ⎤
⎢ y ⎥ = ⎢0 − 1 0⎥.⎢ y ⎥
⎢ 1⎥ ⎢
⎥⎢ ⎥
⎢⎣ z1 ⎥⎦ ⎢⎣0 0 1⎥⎦ ⎢⎣ z ⎥⎦
χ(σ(xz)) = +1
Page 71
Rotation Cn
z1
z
Cn
y
x1
x
360/n
Page 72
⎡ ⎛ 360 ⎞
⎛ 360 ⎞ ⎤
⎢cos⎜ n ⎟ − sin ⎜ n ⎟ 0⎥
⎠ ⎥ ⎡ x⎤
⎝
⎠
⎝
x
⎡ 1⎤ ⎢
⎢ y ⎥ = ⎢ sin ⎛ 360 ⎞ cos⎛ 360 ⎞ 0⎥.⎢ y ⎥
⎜
⎟
⎢ 1 ⎥ ⎢ ⎜⎝ n ⎟⎠
⎥⎢ ⎥
⎝ n ⎠
⎢⎣ z1 ⎥⎦ ⎢
0
0
1⎥ ⎢⎣ z ⎥⎦
⎥
⎢
⎥⎦
⎢⎣
θ
θ
y1
χ(Cn) = 1 + 2.cos(360/n)
36
Improper rotation axis, Sn
z’
z
σ(xy)
Cn
y
x’
y’
x1
y1
x
⎡ ⎛ 360 ⎞
⎤
⎛ 360 ⎞
⎢cos⎜ n ⎟ − sin ⎜ n ⎟ 0 ⎥
⎠
⎝
⎠
⎡ x1 ⎤ ⎢ ⎝
⎥ ⎡ x⎤
⎢ y ⎥ = ⎢ sin ⎛ 360 ⎞ cos⎛ 360 ⎞ 0 ⎥.⎢ y ⎥
⎜
⎟
⎢ 1 ⎥ ⎢ ⎜⎝ n ⎟⎠
⎥⎢ ⎥
⎝ n ⎠
⎢⎣ z1 ⎥⎦ ⎢
0
0
− 1⎥ ⎢⎣ z ⎥⎦
⎢
⎥
Page 73
⎢⎣
⎥⎦
z1
χ(Sn) = -1 + 2.cos(360/n)
Summary of contributions from un-shifted atoms to Γ3n
R
χ(R)
E
+3
i
-3
σ
+1
1+ 2.cos(360/n) C2
-1
1+ 2.cos(360/n) C3 ,C32
1+ 2.cos(360/n) C4, C4
3
0
+1
-1 + 2.cos(360/n) S31,S32
-2
S41,S42
-1
-1 + 2.cos(360/n)
-1 + 2.cos(360/n) S61,S65
0
Page 74
37
O
Worked example: POCl3 (C3v point group)
χ(R)
R
E
σv
C3
Cl
2C3
3σv
A1
A2
E
1
1
1
1
2
1
-1
-1
0
Un-shifted
atoms
Contribution
5
2
3
3
15
0
0
1
3
Page 75
Γ3n
Cl
+3
+1
0
E
C3v
P
Cl
Number of classes,
(1 + 2 + 3 = 6)
Order of the group,
g=6
Reducing the irreducible representation for POCl3
⎛1⎞
a p = ⎜⎜ ⎟⎟∑ n R .χ ( R).χ p (R )
⎝g⎠ R
C3v
Γ3n
E
2C3
3σv
15
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
Γ3n = 4A1 + A2 + 5E
For POCl3 n= 5 therefore the number of degrees of freedom is 3n =15.
E is doubly degenerate so Γ3n has 15 degrees of freedom.
Page 76
38
KULIAH MINGGU IX-X-XI-XII
APLIKASI TEORI GRUP
Page 77
Group Theory and Vibrational Spectroscopy: SO2
C2
1E
1C2
1σ(xz) 1σ(yz)
v
A1
+1
+1
+1
+1
Tz
x2, y2, z2
A2
+1
+1
-1
-1
Rz
xy
B1
+1
-1
+1
-1
Tx , Rx
xz
B2
+1
-1
-1
+1
Ty , Ry
yz
C2v
E
C2
σ(xz)
Γ3n
+9
-1
+1
σ(yz)
3
Γ3n = 3A1 + A2 + 2B1 + 3B2
Page 78
39
Group Theory and Vibrational Spectroscopy: SO2
Γ3n = 3A1 + A2 + 2B1 + 3B2 = 3 + 1 + 2 + 3 = 9 = 3n
For non linear molecule there are 3n-6 vibrational degrees of freedom
C2
1E
1C2
1σ(xz) 1σ(yz)
v
A1
+1
+1
+1
+1
Tz
x2, y2, z2
A2
+1
+1
-1
-1
Rz
xy
B1
+1
-1
+1
-1
Tx , Rx
xz
B2
+1
-1
-1
+1
Ty , Ry
yz
Γvib = Γ3n – Γrot – Γ trans
Γ3n = 3A1 + A2 + 2B1 + 3B2
Γrot =
A2 + B1 + B2
Γtrans = A1
+ B1 + B2
Γvib = 2A1 + B2
(Degrees of freedom = 2 + 1 = 3 = 3n-6)
Page 79
O
Group Theory and Vibrational Spectroscopy: POCl3
Γ3n
P
Cl
= 4A1 + A2 + 5E
Γtrans = A1
Γrot =
Γvibe = 3A1
Cl
Cl
+ E
A2 + E
There are nine vibrational modes . (3n-6 = 9)
The E modes are doubly degenerate and
constitute TWO modes
+ 3E
There are 9 modes that transform as 3A1 + 3E.
These modes are linear combinations of the three vectors
attached to each atom.
Each mode forms a BASIS for an IRREDUCIBLE representation
of the point group of the molecule
Page 80
40
From Γ3n to Γvibe and Spectroscopy
Now that we have Γvibe what does it mean?
We have the symmetries of the normal modes of vibrations.
In terms of linear combinations of Cartesian co-ordinates.
We have the number and degeneracies of the normal modes.
Can we predict the infrared and Raman spectra?
Yes!!
Page 81
Applications in spectroscopy: Infrared Spectroscopy
• Vibrational transition is infrared active because of interaction
of radiation with the:
molecular dipole moment, µ.
• There must be a change in this dipole moment
• This is the transition dipole moment
• Probability is related to transition moment integral .
Page 82
41
Infrared Spectroscopy
TM ∝ ∫ψ i* µψ f dτ = ∫ψ i µψ f dτ
µ
Is the transition dipole moment operator and
has components: µx, µy, µz.
Note: Initial wavefunction
is always real
ψf
ψi
Wavefunction final state
Wavefunction initial state
Page 83
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 describe
the vibrational mode.
Page 84
42
The DIRECT PRODUCT representation.
TM ∝ ∫ ψ i* µψ f dτ = ∫ ψ i µψ f dτ
⎛ Γ(Tx )⎞
⎜
⎟
Γ(ψ i ) • ⎜ Γ(T y )⎟ • Γ(ψ f
⎜ Γ(T )⎟
z ⎠
⎝
For SO2 we have that:
)
Γvib = 2A1 + B2
Under C2v :
Tx, Ty and Tz transform as B1, B2 and A1 respectively.
Page 85
C2
σ(xz) σ(yz)
C2V
E
A1
+1 +1
+1
+1
µz
A2
+1 +1
−1
−1
Rz
B1
+1
−1
+1
−1
µ x, Ry
B2
+1
−1
−1
+1
µ y, R x
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 × B 2
+1 +1
−1
−1
≡A2
A1 × B2 × B 2
+1 +1
+1
+1
≡A1
A1 × A1 × B2
+1
−1
−1
+1
≡B2
Page 86
43
The DIRECT PRODUCT representation
Group theory predicts only A1 and B2 modes
⎛ B1 ⎞
⎛ B1 ⎞
⎜ ⎟
⎜ ⎟
A1 • ⎜ B2 ⎟ • A1 = ⎜ B2 ⎟
⎜A ⎟
⎜A ⎟
⎝ 1⎠
⎝ 1⎠
⎛ B1 ⎞
⎛ A2 ⎞
⎜ ⎟
⎜ ⎟
A1 • ⎜ B2 ⎟ • B2 = ⎜ A1 ⎟
⎜A ⎟
⎜B ⎟
⎝ 1⎠
⎝ 2⎠
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.
Γvib = 2A1 + B2
We predict three bands in
the infrared spectrum of SO2
Page 87
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.
Page 88
44
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 89
Raman Spectroscopy
Virtual state
Stokes Shifted
ω =ω1− ω 2
ω1
ω2
ψf
ψi
Wavefunction final state
Wavefunction initial state
Page 90
45
Raman Spectroscopy
Anti-Stokes Shifted
Virtual state
ω1
ω2
ω =ω2− ω 1
ψf
Wavefunction intial state
ψi
Wavefunction final state
Page 91
Raman Spectroscopy
Probability of a Raman transistion: ∝
∫ψ αˆψ
i
f
dτ
The operator , α , is the polarisability tensor
⎛ α xx
⎜
α = ⎜α yx
⎜α
⎝ zx
α xy α xz ⎞
⎟
α yy α yz ⎟
α zy α zz ⎟⎠
For vibrational transitions αij = αji
so there are six distinct components:
αx2, αy2, αz2, αxy, αxz and αyz
Page 92
46
Raman Spectroscopy
For C2v
αx2, αy2, αz2, αxy, αxz and αyz
Transform as:
A1, A1, A1, A2, B1 and B2
We can then evaluate the direct product representation
in a broadly analagous way
Page 93
Raman Spectroscopy The DIRECT PRODUCT representation
For SO2 group theory predicts only A1 and B2 modes
⎛ A1 ⎞
⎛ A1 ⎞
⎜ ⎟
⎜ ⎟
⎜ A2 ⎟
⎜A ⎟
A`1 • ⎜ ⎟ • A1 = ⎜ 2 ⎟
B
B
⎜ 1⎟
⎜ 1⎟
⎜B ⎟
⎜B ⎟
⎝ 2⎠
⎝ 2⎠
⎛ B2 ⎞
⎛ A1 ⎞
⎜ ⎟
⎜ ⎟
⎜B ⎟
⎜ A2 ⎟
A`1 • ⎜ ⎟ • B2 = ⎜ 1 ⎟
B
A
⎜ 2⎟
⎜ 1⎟
⎜A ⎟
⎜B ⎟
⎝ 1⎠
⎝ 2⎠
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
Page 94
Note: A1 modes are polarised
47
Raman Spectroscopy : General Rule
If a vibrational mode has the same symmetry as on or more
Of 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 95
Analysis of Vibrational Modes:
Vibrations can be classified into Stretches, Bends and Deformations
For SO2 Γvib = 2A1 + B2
We could choose more “natural” co-ordinates
z
r1
y
x
O
S
r2
O
Determine the representation for Γstretch
Page 96
48
Analysis of Vibrational Modes:
r1
How does our new basis transform
Under the operations of the group?
O
S
r2
O
Vectors shifted to new position contribute zero
Unshifted vectors contribute + 1 to χ(R)
C2v
Γstre
E
C2
σ(xz)
+2
0
0
σ(yz)
+2
This can be reduced using reduction formula or by inspection:
( 1, 1, 1, 1)(A1) + (1,-1, -1, 1) (B2) = (2, 0, 0, 2)
Γstre = A1 + B2
Page 97
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
Page 98
49
Symmetry Adapted Linear Combinations
r1
O
Pick a generating vector eg: r1
S
r2
O
How does this transform under symmetry operations?
C2v
E
C2
σ(xz)
σ(yz)
r1
r1
r2
r2
r1
Multiply this by the characters of A1 and B2
For A1 this gives:
(+1) r1+ (+1) r2 + (+1) r2 + (+1) r1 = 2r1 + 2r2
Normalise coefficients and divide by sum of squares:
=
1
2
(r1 + r2 )
Page 99
Symmetry Adapted Linear Combinations
For B2 this gives:
(+1) r1+ (-1) r2 + (-1) r2 + (+1) r1 = 2r1 - 2r2
Normalise coefficients and divide by sum of squares:
=
1
(r1 − r2 )
2
S
S
O
A1
O
O
O
B2
Sulphur must also move to maintain position of centre of mass
Page 100
50
Analysis of Vibrational Modes:
S
Remaining mode “likely” to be a bend
C2v
E
C2
σ(xz)
Γbend
1
1
1
O
O
σ(yz)
1
By inspection this bend is A1 symmetry
SO2 has three normal modes:
A1 stretch:
A1 bend:
B2 stretch:
Raman polarised and infrared active
Raman polarised and infrared active
Raman and infrared active
Page 101
Analysis of Vibrational Modes:
SO2 experimental data.
IR(Vapour)/cm-1
Raman(liquid)/cm-1
Sym
Name
518
524
A1 bend
ν1
1151
1145
A1 stretch
ν2
1362
1336
B2 stretch
ν3
Page 102
51
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 103
Analysis of Vibrational Modes: POCl3
O
O
O
P
Cl
P
Cl
Cl
Cl
P
Cl
Cl
Cl
Cl
Cl
Angle deformations
P-Cl stretch
P=O stretch
Γvibe = 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)
Page 104
Six bands, Six co-incidences
52
Analysis of Vibrational Modes: POCl3
C3v
E 2C3
3σv
Γ P=O str
1
1
1
Γ P-Cl str
3
0
1
Γ bend
6
0
2
Γvibe = 3A1 + 3E
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 105
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.
Page 106
53
Analysis of Vibrational Modes: POCl3
IR (liq)/ cm-1 Raman /cm-1 Description
Sym Label
1292
1290(pol)
P=O str( 1,4)
A1
ν1
580
582
P-Cl str(2,3)
E
ν4
487
486(pol)
P-Cl str( 1,2,3)
A1
ν2
340
337
deformation
E
ν5
267
267(pol)
A1
ν3
-
193
Sym.
Deformation(1)
deformation
E
ν6
Page 107
Analysis of Vibrational 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
Page 108
54
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 109
Symmetry, Bonding and Electronic Spectroscopy
σ bonding in AXn molecules e.g. : water
How do 2s and 2p orbitals transform?
O 2s orbital
E, C2, σxz, σyx
+
O
H
H
+
O
H
C2V
E
C2
σxz
σyz
O2s
+1
+1
+1
+1
H
a1
Page 110
55
Symmetry, Bonding and Electronic Spectroscopy
s-orbitals are spherically symmetric and when at the
most 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 111
Symmetry, Bonding and Electronic Spectroscopy
How do the 2p orbitals transform?
O 2p z orbital
+
O
H
−
+
E, C 2, σ xz, σ yx
O
H
H
C 2V
E
C2
σ xz
σ yz
O 2pz
+1
+1
+1
+1
−
H
a1
Page 112
56
Symmetry, Bonding and Electronic Spectroscopy
How do the 2p orbitals transform?
O 2p y orbital
+
H
O
+
E, σ yx
O
H
−
H
−
C 2 , σ xz
H
−
H
C 2V
E
C2
σ xz
σ yz
O 2p y
+1
−1
−1
+1
b2
O 2px
+1
−1
+1
−1
b1
O
+
H
Page 113
Symmetry, Bonding and Electronic Spectroscopy
How do the 2s and 2p orbitals transform?
Oxygen 2s and 2pz transforms as a1
2px transforms as b1 and 2py as b2
Need a set of σ-ligand orbitals of correct symmetry to interact
with Oxygen orbitals.
Construct a basis, determine the reducible representation,
reduce by inspection or using the reduction formula, estimate overlap,
draw MO diagram
Page 114
57
Symmetry, Bonding and Electronic Spectroscopy
Use the 1s orbitals on the hydrogen atoms
z
H 1s orbitals
O
H
+
y
H+
φ2
φ1
Page 115
x
C2V
E
C2
σ xz
σ yz
Γσ
2
0
0
2
a 1 + b2
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
Orbitals of like symmetry can interact
Oxygen 2px is “wrong” symmetry therefore likely to be non-bonding
Which is lower in energy a1 or b2?
Guess that it is a1 similar symmetry better interaction?
Page 116
58
Qualitative MO diagram for H2O
a1*
b2*
non-bonding (O 2px)
b1
a1 + b1 + b2
a1 + b2
b2
a1
a1
a1
non-bonding
2s
O
H2O
2H
Page 117
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)
Page 118
59
C2V
E
C2
σxz
σyz
φ1
φ1
φ2
φ2
φ1
a1
+1
+1
+1
+1
Sum and normalise
φ1
φ2
φ2
φ1
b2
+1
−1
−1
+1
Sum and normalise
φ1
−φ2
−φ2
φ1
pz
b2 = 1/√2(φ1 − φ2)
−
−
O
+
H+
φ1
Page 119
a1 = 1/√2(φ1 + φ2)
+
H
H−
φ2
φ1
Ψ(a1)
O
+
+
py
H
φ2
Ψ(b2)
Symmetry, Bonding and Electronic Spectroscopy
The overlap between the a1 orbitals (ψ1) is greater than
that for the b2 orbitals (ψ2).
Therefore a1 is lower in energy than b1.
We can use the Pauli exclusion principle and the Aufbau principle
To fill up these molecular orbitals.
This enables us to determine the symmetries of electronic states
arising from each electronic configuration.
Note: Electronic states and configurations are NOT the same thing!
Page 120
60
Qualitative MO diagram for H2O
a1*
b2*
non-bonding (O 2px)
b1
a1 + b1 + b2
a1 + b2
b2
a1
a1
a1
non-bonding
2s
O
H2O
2H
Page 121
Symmetry of Electronic States from NON-DEGENERATE MO’s.
The ground electronic configuration for water is:
(a1)2(b2)2(b1)2(b2*)0(a1*)0
The symmetry of the electronic state arising from this configuration
is given by the direct product of the symmetries of the MO’s of all
the electrons
(a1)2 = a1.a1 = A1
(b2)2 = b2.b2 = A1
(b1)2 = b1.b1 = A1
For FULL singly degenerate
MO’s, the symmetry is ALWAYS
A1
A1.A1.A1 = A1
Page 122
61
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 123
Symmetry of Electronic States from NON-DEGENERATE MO’s.
What happens if we promote an electron?
First two excitations move an electron form b1 non bonding
Into either the b2* or a1* anti-bonding orbitals .
a1 *
Anti Bonding
b2*
b1
b2
a1
Non bonding
Both of these transitions are
non bonding to anti bonding
transitions. n-π*
Bonding
Page 124
62
What electronic states do these new configurations generate?
(a1)2(b2)2(b1)1(b2*)1(a1*)0
= A1.A1.B1.B2 = A2
(a1)2(b2)2(b1)1(b2*)0(a1*)1
= 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:
3A , 1A and 3B , 1B electronic states.
2
2
1
1
Note: if S= ½ then we have a doublet state
Page 125
What electronic states do these new configurations generate?
Electronic States
Molecular Orbitals
a1 *
b2*
b1
1B
1
1A
2
3B
1
3A
2
b2
a1
1A
1
Page 126
63
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.
TM ∝ ∫ψ i* µψ f dτ = ∫ψ i µψ f dτ
Page 127
Which electronic transitions are allowed?
TMI ≈ ∫ψ *V ,iψ V , f dτ • ∫ψ * S ,iψ S , f dτ • ∫ψ *e,i µψ e, f dτ
Vibrational
Spin
Electronic
To first approximation µ can only operate on the electronic part
of 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.
Page 128
64
Which electronic transitions are allowed?
A transition is allowed if there is no change in spin and the
electronic component transforms as totally symmetric.
The intensity is modulated by Franck-Condon factors.
The electronic transition dipole moment µ transforms as the
translational species as for infrared transitions.
Page 129
Which electronic transitions are allowed?
For the example of H20 the direct products for the
electronic transition are
⎛ A2 ⎞
⎛ B1 ⎞
⎜ ⎟
⎜ ⎟
A1 • ⎜ B2 ⎟ • A2 = ⎜ B2 ⎟
⎜B ⎟
⎜A ⎟
⎝ 1⎠
⎝ 1⎠
⎛ B1 ⎞
⎛ B1 ⎞
⎜ ⎟
⎜ ⎟
A1 • ⎜ B2 ⎟ • B1 = ⎜ A1 ⎟
⎜A ⎟
⎜A ⎟
⎝ 1⎠
⎝ 2⎠
The totally symmetric species is only present for the transition
to 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”.
Page
130
65
Which electronic transitions are allowed?
1
B1
1
A2
Symmetry
forbidden
Symmetry
allowed
3
B1
3
A2
Spin forbidden
1
A1
Page 131
Which electronic transitions are allowed?
Transitions between a totally symmetric ground state and one with
an electronic state that has the same symmetry as a component
of µ, will be symmetry allowed.
Caution: The lowest energy transition may be allowed but too weak
to be observed.
Caution: Ground state is not always totally symmetric and
beware of degenerate representations.
Page 132
66
More bonding for AX6 molecules / complexes
In the case of Oh point group:
d x2-y2 and dz2 transform as eg
dxy, dyz and dzx transform as t2g
px, py and pz transform as t1u
Γσ(ligands) = a1g + eg + t1u
Γπ(ligands) = t1g + t2g + t1u + t2u
Page 133
AX6 for Oh
4p
t1u
t1u*
a1g*
4s
a1g
eg*
3d
eg + t2g
t2g
eg
a1g + eg + t1u
t1u
a1g
Page 134
67
Electronic Spectroscopy of d9 complex:
[Cu(H2O)6]2+ is a d9 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 135
Electronic Spectroscopy of d9 complex:
Need to calculate direct product representation:
2E
g
. (t1u) . 2T2g
i
6S
4
8S
6
3σ
-1
3
-1
0
-1
1
1
-1
-3
-1
0
1
1
0
0
2
2
0
-1
2
0
0
0
2
-18
0
0
-2
0
Oh
E
8C
T2g
3
0
1
-1
t1u
3
0
-1
Eg
2
-1
DP
18
0
3
6C
2
6C
4
3C
2
h
6σ
d
Page 136
68
Electronic Spectroscopy of d9 complex:
DP
18
0
0
0
2
-18
⎛1⎞
Use reduction formula: a p = ⎜⎜ ⎟⎟
⎝g⎠
0
∑n
0
R
-2
0
.χ ( R).χ p (R )
R
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 137
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.
Page 138
69
•
•
•
•
•
•
•
•
•
•
•
•
•
Are you familiar with symmetry elements operations?
Can you assign a point group?
Can you use a basis of 3 vectors to generate Γ3n ?
Do you know the reduction formula?
What is the difference between a reducible and irreducible
representation?
Can you reduce Γ3n ?
Can you generate Γvib from Γ3n ?
Can you predict IR and Raman activity for a given molecule using
direct product representation?
Can you discuss the assignment of spectra?
Can you use SALCs to describe the normal modes of SO2?
Can you discuss MO diagram in terms of SALCS?
Can you assign symmetry to electronic states and discuss whether
electronic transitions are allowed using the direct product
representation?
Given and infrared and Raman spectrum could you determine the
symmetry of the molecule?
Page 139
• http://www.chemsoc.org/exemplarchem/en
tries/2004/hull_booth/info/web_pro.htm
• http://www.hull.ac.uk/php/chsajb/symmetry
&spectroscopy/ho_1.html
• http://www.people.ouc.bc.ca/smsneil/sym
m/symmpg.htm
Page 140
70