Page 1 Symmetry and Group Theory Feature: Application for Spectroscopy and Orbital Molecules Dr....

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Symmetry and Group TheoryFeature: Application for Spectroscopy and Orbital Molecules

Dr. Indriana Kartini

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P. H. Walton “Beginning Group Theory for Chemistry”Oxford University Press Inc., New York, 1998ISBN 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

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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

Simbol Unsur Operasi

E Unsur identitas Membiarkan obyek tidak berubah

CnSumbu rotasi Rotasi seputar sumbu dengan

derajat rotasi 360/n (n adalah bilangan bulat)

Bidang simetri Refleksi melalui bidang simetri

i Pusat/titik inversi Proyeksi melewati pusat inversi ke sisi seberangnya dengan jarak yang sama dari pusat

SnSumbu rotasi tidak sejati (Improper rotational axis)

Rotasi mengitari sumbu rotasi diikuti dengan refleksi pada bidang tegak lurus sumbu rotasi

© 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

H1H2

H1H2

H1 H2

(xz)

(yz)

z

y

x

x is out of the plane

Reflection is the operation 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

(xz) in plane perpendicular to molecular plane

(yz) in plane parallel to molecular plane

both examples of v

v : reflection in plane containing highest order axis

h : reflection in plane perpendicular to highest order axis

d : dihedral plane generally bisecting v

Page 11

XeF F

F F

Xe

F F

F F

Xe

F F

F F

Reflections v

h

d

d

XeF4

Page 12

XeF4

Page 13

Z

Y

X

Z

Y

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 H

HH

C4

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 IITEORI GRUP

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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 canbe 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: Octahedral point group symbol Oh

Tetrahedral point group symbol Td

Linear having no i CLinear having i Dh

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Question 1:• Is the molecule one of the following recognisable

groups ?

NO: Go to the Question 2

YES: Octahedral point group symbol Oh

Tetrahedral point group symbol Td

Linear having no i CLinear having i Dh

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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 i Ci

Page 22


Question 3:• Has the molecule more than one rotation axis ?

YES: Go to the Question 4NO:

If no other symmetry elements point group symbol Cn (n is the order of the principle axis)

If having n h point group symbol Cnh

If having n v 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 d bisecting the C2 axes Dnd

Having one h Dnh

Page 24

Molecule

Linear?

i?Dh Cv2 or moreCn, n>2?

i?

TdC5?Ih Oh

Cn?

Select Cn with highest n,nC2 perpendicular to Cn?

**h?Dnh

nd?Dnd Dn ?Cs

i?Ci C1h?Cnh

nv?Cnv

S2n?S2n Cn

Y

N

Page 25

Benzene

Linear?

i?Dh Cv2 or moreCn, n>2?

i?

TdC5?Ih Oh

Cn?

Select Cn with highest n,nC2 perpendicular to Cn?

**h?Dnh

nd?Dnd Dn ?Cs

i?Ci C1h?Cnh

nv?Cnv

S2n?S2n Cn

Y

Nn = 6Benzene

is D6h

Page 26

Tugas I: Symmetry and Point Groups

Tentukan unsur simetri dan grup titik pada molekul

a. N2F2

b. POCl3

Gambarkan geometri masing-masing molekul tersebut

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KULIAH MINGGU III

Page 28

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 collectionE 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: ACCA)• Every member of the group must have an INVERSE which is also a

member of the group.

AA-1 = E

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 collection

AB = C where A, B and C are all members of the collection

B

O 2

O 3O 1

H2

H1

H3

B

O 1

O 2O 3

H1

H3

H2

B

O 3

O 1O 2

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

O 2

O 3O 1

H2

H1

H3

B

O 3

O 1O 2

H3

H2

H1

B

O 1

O 2O 3

H1

H3

H2

C32 C3

2

E. C32

= C3

2 and C32. C3 = E and C3

2. C3

2 = C3

Page 32

•The combination of elements in the group must be ASSOCIATIVE

A(BC) = AB(C) = ABC

Multiplication need not be commutative (ie: ACCA)

C3 .(C3 .C32 )=

(C3 .C3) C3

2

(Do RHS First)

C3.C3

2 = E ; C3 .E = C3

C3 .C3 = C32 ; C3

2 .C32 = C3

Operations are associative and E, C3 and C32

form a group

Page 33

Group Multiplication Table

C3 E C3 C32

E E C3 C32

C3 C3 C32 E

C32 C3

2 E C3

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 C32

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 of the 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 37

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 each symmetry operation is expressed as a transformation matrix

[New coordinates] = [matrix transformation] x [old coordinates]

Page 38

S

O O

z

y

x

Constructing the Representation

Put unit vectors on each atom

C2v: [E, C2, xz, yz]

These are useful to describe molecular vibrations and electronic transitions.

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

yz .(Ty) = (+1) Ty xz .(Ty) = (-1) Ty


Page 40

S

O O


A unit vector on each atom represents rotation around the z(C2) axis

C2.(Rz) = (+1) Rz E .(RZ) = (+1) Rz

yz .(Rz) = (-1) Rz xz .(RZ) = (-1) RZ

Page 41


C2vE C2 (xz) (yz)

+1 +1 +1 +1 Tz

+1 +1 -1 -1 Rz

+1 -1 +1 -1 Tx,Ry

+1 -1 -1 +1 Ty,Rx

Page 42

S

O O


Use a mathematical function Eg: py orbital on S

C2vE C2 (xz) (yz)

+1 -1 -1 +1 Ty,Rx

py has the same symmetry properties as Ty and Rx vectors

Page 43


Au Au

Au

h

h.[d x2-y2] = (+1) .[d x2-y2]

C.[d x2-y2] = (-1) .[d x2-y2]

C4[AuCl4]-

Page 44


D4h E 2C4 C2 2C2’ 2C2” I 2S4 h 2v 2d

+1 -1 +1 +1 -1 +1 -1 +1 +1 -1

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 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 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

nB

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 , = a11 + a22 + a33.

n

inma

1

In general

And only the character , which is a number is required and NOT the whole matrix.

Page 50

Character Tables an Example C3v : (NF3)

C3v E C31 C3

2 v v v

1 1 1 1 1 1 Tz

1 1 1 -1 -1 -1 Rz

2 -1 -1 0 0 0 (Tx,Ty) or (Rx,Ry)

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

v, v, v are in the same class

Page 51

Character Tables an Example C3v : (NF3)

C3vE 2C3 v

A1 1 1 1 Tz x2 + y2

A2 1 1 -1 Rz

E 2 -1 0 (Tx,Ty) or (Rx,Ry) (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 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 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) List conventional symbols for irreducible representations: Mulliken symbols

Page 55

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 with two and three dimensional representations beingand

Page 56


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 = + 1 for the operation.)

If it is anti-symmetric with respect to the operation = - 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 vplane. 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 57


Subscripts g (gerade) and u (ungerade) are given to irreducible representationsThat are symmetric and anti-symmetric respectively, with respect to inversion at a centre of symmetry.

Superscripts ‘ and “ are given to irreducible representations that are symmetric and anti-symmetric respectively with respect o reflection in a h plane.

3)

4)

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 58

S

O O

z2

y2

x2

Generating Reducible Representations

x1

xsy1

ys

zs

z1

xz

For the symmetry operation xz (a v )

x1 x2 x2 x1 xs xs

y1 -y2 y2 -y1 ys -ys

z1 z2 z2 z1 zs zs

Page 59


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

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

Only require the characters: The sum of diagonal elements

For (xz) = + 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

For (yz) = + 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 = + 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 = -1

Page 64


C2v

3n

E C2(xz) (yz)

+9 -1 +1 3

Summarising we get that 3n for this molecule is:

C2v E C2 (xz) (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

To reduce this we need the character table for the point groups

Page 65

Reducing Reducible Representations

We need to use the reduction formula: RRng

a pR

Rp ).(.1

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

(R) is character of the reducible representation

p(R) is character of the irreducible representation

nR is the number of operations in the class

Page 66

C2v 1E 1C2 (xz) (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

C2v

3n

E C2(xz) (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

RRng

a pR

Rp ).(.1

C2v

3n

E C2(xz) (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

3n = 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 the un-shifted atomsOnly 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

(E) = +3

z

y

x

z1

y1

x1

Page 70

Inversion i

z

y

xz1

y1

x1

i


(i) = -3

z

y

x

z

y

x

.

100

010

001

1

1

1

Page 71


z1

y1

x1x

z

y

(xz)

Reflection (xz) (Others are same except location of –1 changes)

((xz)) = +1

z

y

x

z

y

x

.

100

010

001

1

1

1

Page 72

360/n

x1 y1

z1z

y

x

Cn

Rotation Cn

z

y

x

nn

nn

z

y

x

.

100

0360

cos360

sin

0360

sin360

cos

1

1

1

(Cn) = 1 + 2.cos(360/n)

Page 73

z

y

x

nn

nn

z

y

x

.

100

0360

cos360

sin

0360

sin360

cos

1

1

1

(Sn) = -1 + 2.cos(360/n)

Improper rotation axis, Sn

Cn

(xy)z

y

x

z’

y’x’

y1x1

z1

Page 74

Summary of contributions from un-shifted atoms to 3n

R (R)

E +3

i -3

+1

1+ 2.cos(360/n) C2 -1

1+ 2.cos(360/n) C3 ,C32 0

1+ 2.cos(360/n) C4, C43 +1

-1 + 2.cos(360/n) S31,S3

2 -2

-1 + 2.cos(360/n) S41,S4

2 -1

-1 + 2.cos(360/n) S61,S6

5 0

Page 75

P

O

C lC lC l

Worked example: POCl3 (C3v point group)

R (R)

E

v

2C3

+3

+1

0

C3vE v

A1

A2

E

1 1 1

1 1 -1

2 -1 0

C3

Un-shiftedatoms

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

RRng

a pR

Rp ).(.1

2C3C3vE v

3n15 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-XIIAPLIKASI TEORI GRUP

Page 78


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

3n

E C2(xz) (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


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

For non linear molecule there are 3n-6 vibrational degrees of freedom

rot = A2 + B1 + B2

trans = A1 + B1 + B2

vib = 3n – rot – trans

vib = 2A1 + B2 (Degrees of freedom = 2 + 1 = 3 = 3n-6)

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

Page 80

P

O

C lC lC l

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 vectors attached to each atom.

Each mode forms a BASIS for an IRREDUCIBLE representation of the point group of the molecule

Page 81

From 3n to vibe and Spectroscopy

Now that we have vibe 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 82

Applications in spectroscopy: Infrared Spectroscopy

• Vibrational transition is infrared active because of interaction of radiation with the:

molecular dipole moment, .

• There must be a change in this dipole moment

• This is the transition dipole moment

• Probability is related to transition moment integral .

Page 83

f

i

ddTM fifi *

Infrared Spectroscopy

Is the transition dipole moment operator and has components: x, y, z.

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: f i contains the totally symmetric representation.

• IE all numbers for in representation are +1

• The ground state i 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 85

The DIRECT PRODUCT representation.

f

z

y

x

i

T

T

T

ddTM fifi *

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(xz) (yz)

A1 +1 +1 +1 +1 z

A2 +1 +1 1 1 Rz

B1 +1 1 +1 1 x, Ry

B2 +1 1 1 +1 y, 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 containthe totally symmetric species so they are symmetry allowed.

This does not tell us the intensity only whether they are allowed or not.

vib = 2A1 + B2We 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 .• 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 1.

• Scattered light at 2.

• Can be Stokes(lower energy) or Anti-Stokes shifted• Much weaker effect than direct absorption.

Page 90

f

i


Wavefunction initial state

Virtual state

Raman Spectroscopy

Stokes Shifted

Page 91

f

i

Wavefunction intial state


Virtual stateRaman Spectroscopy

Anti-Stokes Shifted

Page 92

Raman Spectroscopy

dfi ˆProbability of a Raman transistion:

zzzyzx

yzyyyx

xzxyxx

The operator , , is the polarisability tensor

For vibrational transitions ij = ji

so there are six distinct components:

x2, y2, z2, xy, xz and yz

Page 93

x2, y2, z2, xy, xz and yz

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 phenomenaand 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 SO2vib = 2A1 + B2

We could choose more “natural” co-ordinates

S

O O

z

y

x

r2r1

Determine the representation for stretch

Page 97

Analysis of Vibrational Modes: S

O Or2r1

How does our new basis transformUnder the operations of the group?

Vectors shifted to new position contribute zeroUnshifted vectors contribute + 1 to (R)

C2v

stre

E C2(xz) (yz)

+2 0 0 +2

This can be reduced using reduction formula or by inspection:

stre = 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 adaptedlinear combinations (SALCs) of the two stretching vectors.

Our intuition tells us that we might have a symmetric and ananti-symmetric stretching vibration

A1 and B2

Page 99

Symmetry Adapted Linear Combinations S

O Or2r1

C2v

r1

E C2(xz) (yz)

r1 r2 r2 r1

Pick a generating vector eg: r1

How does this transform under symmetry operations?

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:

)(2

121 rr

Page 100

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:

)(2

121 rr

S

O O

S

O OA1

B2

Sulphur must also move to maintain position of centre of mass

Page 101


S

O ORemaining mode “likely” to be a bend

C2v

bend

E C2(xz) (yz)

1 1 1 1

By inspection this bend is A1 symmetry

SO2 has three normal modes:

A1 stretch: Raman polarised and infrared activeA1 bend: Raman polarised and infrared activeB2 stretch: Raman and infrared active

Page 102

Analysis of Vibrational Modes: SO2 experimental data.

IR(Vapour)/cm-1 Raman(liquid)/cm-1 Sym Name

518 524 A1 bend 1

1151 1145 A1 stretch 2

1362 1336 B2 stretch 3

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 todiffering 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

C lC lC l

P

O

C lC lC l

P

O

C lC lC l

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 + 3EC3v

E 2C3 3v

P=O str

P-Cl str

bend

1 1 1

3 0 1

6 0 2

Using reduction formulae or by inspection:

P=O str = A1 and P-Cl str = A1 + E

bend = vibe - P=O str - P-Cl str = 3A1 + 3E – 2A1 – E = A1 + 2E

Reduction of the representation for bends gives: bend = 2A1 + 2E

Page 106


bend = vibe - P=O str - P-Cl str = 3A1 + 3E – 2A1 – E = A1 + 2E

Reduction of the representation for bends gives: bend = 2A1 + 2E

One of the A1 terms is REDUNDANT as not all the angles can symmetrically increase

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

IR (liq)/ cm-1 Raman /cm-1 Description Sym Label

1292 1290(pol) P=O str( 1,4) A1 1

580 582 P-Cl str(2,3) E 4

487 486(pol) P-Cl str( 1,2,3) A1 2

340 337 deformation E 5

267 267(pol) Sym. Deformation(1)

A1 3

- 193 deformation E 6


Page 108


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

OH H

+O

H H

+E , C 2 , x z , y x

O 2 s o r b i t a l

C 2 V E C 2

x z

y z

O 2 s a 1


bonding in AXn molecules e.g. : water

How do 2s and 2p orbitals transform?

Page 111


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 112

OH H

E , C 2 , x z , y x

O 2 p z o r b i t a l

OH H

C 2 V E C 2

x z

y z

O 2 p z a 1


How do the 2p orbitals transform?

Page 113

E , y x

OH H

O 2 p y o rb i ta l

C 2 , x z

OH H

OH H

C 2 V E C 2

x z

y z

O 2 p y b 2

O 2 p x b 1


How do the 2p orbitals transform?

Page 114


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 -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 115

OH H

H 1 s o rb ita ls

+ +

z

y

x

C 2 V E C 2

x z

yz

a 1 + b 2


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 or 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 119

C 2 V E C 2 x z

y z

1

1

2

2

1

a 1

S u m a n d n o r m a l i s e 1

2

2

1 a 1 = 1 / 2 (

1 +

2 )

b 2

S u m a n d n o r m a l i s e 1

2

2

1 b 2 = 1 / 2 (

1

2 )

O

H H+ +

+

( a 1 )

O

H H +

( b 2 )

pz

py

Page 120


The overlap between the a1 orbitals ( is greater than that for the b2 orbitals (

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 statesarising from each electronic configuration.

Note: Electronic states and configurations are NOT the same thing!

Page 121

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 122

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 configurationis 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

A1.A1.A1 = A1

For FULL singly degenerateMO’s, the symmetry is ALWAYSA1

Page 123


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


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 bondingInto either the b2* or a1* anti-bonding orbitals .

Both of these transitions arenon bonding to anti bondingtransitions. n-*

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 forinfrared transitions.

ddTM fifi *

Page 128

Electronic

dddTMI feiefSiSfViV ,,*

,,*

,,*

Which electronic transitions are allowed?

Vibrational

Spin

To first approximation can only operate on the electronic partof the wavefunction.

Vibrational part is overlap between ground and excited state nuclearwavefunctions. 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 the electronic component transforms as totally symmetric. The intensity is modulated by Franck-Condon factors.

The electronic transition dipole moment 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 131

1A1

1A2

1B1

3B1

3A2

Symmetry

forbidden

Spin forbiddenSymmetry

allowed


Page 132


Transitions between a totally symmetric ground state and one with an electronic state that has the same symmetry as a componentof , will be symmetry allowed.

Caution: Ground state is not always totally symmetric and beware of degenerate representations.

Caution: The lowest energy transition may be allowed but too weakto be observed.

Page 133

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

(ligands) = a1g + eg + t1u

(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 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 2T2gUnder 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

Oh E 8C36C2

6C43C2 i 6S4

8S63h

6d

T2g 3 0 1 -1 -1 3 -1 0 -1 1

t1u 3 0 -1 1 -1 -3 -1 0 1 1

Eg 2 -1 0 0 2 2 0 -1 2 0

DP 18 0 0 0 2 -18 0 0 -2 0

Page 137


DP 18 0 0 0 2 -18 0 0 -2 0

Use reduction formula: RRng

a pR

Rp ).(.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 partiallyallowed.

Page 139

• Are you familiar with symmetry elements operations?• Can you assign a point group?• Can you use a basis of 3 vectors to generate 3n ?• Do you know the reduction formula?• What is the difference between a reducible and irreducible

representation?• Can you reduce 3n ?• Can you generate vib from 3n ?• 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 140

• http://www.chemsoc.org/exemplarchem/entries/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/symm/symmpg.htm

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