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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



Page 3

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



Page 5

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



Page 7

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



Page 9

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



Page 11

Xe

F

F F

Xe

F F

F F

Xe

F F

F F

Reflections σv

σ

h

σd

σd

XeF4

Page 12

XeF4



Page 13

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



Page 15

S4 Improper Rotation

Page 16

successive operation



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



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:Octahedralpoint group symbol Oh

Tetrahedral point group symbol Td

Linear having no iC∞υ

Linear having i D∞h

Page 20


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



Page 21


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


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



Page 23


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



Page 25

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



Page 29

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



Page 31

•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



Page 33

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




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


Math based

[1 2 3]

1 0 0

0 -1 0

0 0 1

=



Page 39

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


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



Page 41


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


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



Page 43


AuAu

Au

σh

σh.[d x2-y2] =

(+1) .[d x2-y2]

C4.[d x2-y2] =

(-1) .[d x2-y2]

C4[AuCl4]-

Page 44


-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.



Page 45

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


• 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.



Page 47

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

333231

232221

131211

aaa

aaa

aaa

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

33

2221

2111

00

0

0

b

bb

bb


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


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



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



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



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



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)



Page 59

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



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



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



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



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



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


χ(i) = -3

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

−

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

z

y

x

z

y

x

.

100

010

001

1

1

1



Page 71


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)



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



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.



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


v

C2v

Γ3n

E C2σ(xz) σ(yz)

+9 -1 +1 3

3n = 3A1 + A2 + 2B1 + 3B2

Group Theory and Vibrational Spectroscopy: SO2



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


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



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.



Page 85

The DIRECT PRODUCT representation.

( )( )( )( )

( ) f

z

y

x

i

T

T

T

ψ ψ Γ•⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

Γ

Γ

Γ

•Γ


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



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.



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 initial state

ω1 ω2

ω =ω1− ω 2

Virtual state

Raman Spectroscopy

Stokes Shifted



Page 91

ψf

ψi

Wavefunction intial 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



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



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



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


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 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



Page 101


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



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



Page 105


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


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 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


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



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


σ bonding in AXn molecules e.g. : water

How do 2s and 2p orbitals transform?



Page 111


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


How do the 2p orbitals transform?



Page 115

OH H

H 1s orbitals

+ +

φ1 φ2

z

y

x

C2V E C2 σxz σyz

σ

2 0 0 2 a1 + b2


Use the 1s orbitals on the hydrogen atoms

Page 116


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



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


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 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*



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*


Molecular Orbitals

1A1

1B1

3B1

1A2

3A2

Electronic States



Page 127


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.


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.



Page 129


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


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”.



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



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


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



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


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.

Simetri s1 Lengkap 2 Sks

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