Applications of Molecular Symmetry and Group Theory

Paper No and Title 13 Applications of molecular symmetry and group theory

Module No and Title 26 and group theory and vibrational spectroscopy part-III

Module Tag CHE_P13_M26

CHEMISTRY PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 26: Group theory and vibrational spectroscopy part-III

TABLE OF CONTENTS

1. Learning Outcomes 2. Introduction 3. Reducible and irreducible representations recollection 3.1 More examples and find Γ3N 4. How to get vibrational modes? 4.1 Example of C2v point group 4.2 Example of C3v point group 4.4 Example of Td point group 4.5 Example of D4h point group 5. Summary

CHEMISTRY PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 26: Group theory and vibrational spectroscopy part-III

1. Learning Outcomes

After studying this module, you shall be able to

• Recollect about reducible and irreducible representation • Learn how to find Γ3N using Cartesian coordinates as basis • Get ΓT and ΓR • Find Γvib after subtracting ΓT and ΓR from Γ3N • How to find Γvib for C2v,C3v,Td,D4h point groups

2. Introduction

In this we will continue with vibrational spectroscopy and symmetry aspects of it. We will recollect reducible representations and irreducible representations as these will be needed for normal mode of analysis and classification of vibrational modes according to symmetry species.

3 Reducible and irreducible representations recollection

We will take more examples to find irreducible representations as these are very important in dealing with the normal vibration mode analysis and finding the symmetry species associated with these normal modes of vibrations.

3.1 Let us take few more examples and find Γ3N

2- (I) We take the example of XeF4, or [PtCl4] . These molecules belong to D4h point group.The symmetry elements and 3N vectors are shown in case of XeF4 in fig.1.

, C4 C2 F

F σh Xe

F σv F

Fig.1 3 N vectors and some symmetry elements in XeF4 molecule CHEMISTRY PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 26: Group theory and vibrational spectroscopy part-III

The character table in part for D4h point group together with T3N is given in Table .1 Table.1 Character table and Γ3N for D4h point group

D4h E 2C4 C2 2C2’ 2C2” i 2S4 σh 2σv 2σd A1g 1 1 1 1 1 1 1 1 1 1 A2g 1 1 1 -1 -1 1 1 1 -1 -1 B1g 1 -1 1 1 -1 1 -1 1 1 -1 B2g 1 -1 1 -1 1 1 -1 1 -1 1 Eg 2 0 -2 0 0 2 0 -2 0 0 A1u 1 1 1 1 1 -1 -1 -1 -1 -1

A2u 1 1 1 -1 -1 -1 -1 -1 1 1

B1u 1 -1 1 1 -1 -1 1 -1 -1 1 B2u 1 -1 1 -1 1 -1 1 -1 1 -1 Eu 2 0 -2 0 0 -2 0 2 0 0 Number of 5 1 1 3 1 1 1 5 3 1 unshifted atoms Multiplication x3 x1 x(-1) x(-1) x(-1) x(-3) x(-1) x1 x1 x1 factor

Γ3N 15 1 -1 -3 -1 -3 -1 5 3 1

Γ3N can be reduced to irreducible representations by following standard reduction formula.

(A1g)=1/16[1x15x1+2x1x1+1x-1x1+2x-3x1+2x-1x1+1x-3x1+2x-1x1+1x5x1 +2x3x1 +2x1x1] 1/16[16=1 i.e. A1g Similarly other irreducible representations can be obtained and the net result is

Γ3N =A1g+A2g+B1g+B2g+Eg+2A2u+B2u+3Eu Total irreducible representations are; [1+1+1+1+2+2+1+2x3]=15 i.e. equals to sum of characters of E in all irreducible representations of Γ3N

CHEMISTRY PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 26: Group theory and vibrational spectroscopy part-III

2- (II) Let us take examples of molecules CCl4 CH4, RuO4, SO4 .These belong to Td. point group. Total number of symmetry operations in Td point group are 24. Therefore, h=24.Some of the symmetry elements and 3N vectors are shown in case of CCl4 in fig.1.

C3 Cl

Cl Cl Cl

Fig.2 3N vectors and some symmetry elements in CCl4 molecule

The character table for Td point group together with 3N vectors on each atom is shown in case is give in table .2 Table .2 Γ3N and character table of Td point group

Td E 8C3 3C2 6S4 6σd A1 1 1 1 1 1 A2 1 1 1 -1 -1 E 2 -1 2 0 0 T1 3 0 -1 1 -1 T2 3 0 -1 -1 1 Number of 5 2 1 1 3 unshifted atom Multiplication x3 x0 x(-1) x(-1) x1 factor Γ3N 15 0 -1 -1 3

Γ3N can be reduced to irreducible representations by making use of standard reduction formula.

A1=1/24[1x15x1+8x0x1+3x-1x1+6x-1x1+6x3x1]=1/24[24] =1 i.e. A1

T1=1/24[1x15x3+8x0x1+3x-1x-1+6x-1x1+6x3x-1] =1/24[24] =1 i.e. T1 Similarly other irreducible representations can be obtained and over all result is CHEMISTRY PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 26: Group theory and vibrational spectroscopy part-III

Γ3N =A1+E+T1+3T2. Total irreducible representations are: 1+2x1+3x1+3x3=15 which is equal to sum of character of E in irreducible representations in Γ3N The results of Γ3N and its reduction are summarized as follows:

(A) SO2,H2O (C2v) ; Γ3N = 3A1+A2+2B1+3B2 (B ) POCl3,NH3,CHCl3 (C3v) ; Γ3N = 4A1+A2+5E 2- (I )XeF4, [PtCl4] ( D4h) ; Γ3N = A1g+A2g+B1g+B2g+Eg+2A2u+B2u+3Eu 2- (II) CCl4 CH4, RuO4,SO4 (Td) ; Γ3N = A1+E+T1+3T2,

4. How to get vibrational modes?

In section 3.1 results that we have obtained are based on 3N vectors as basis. This 3N base vectors include translational and rotational degrees of vectors also. But we are interested in irreducible representations that are given only by the vibration/stretching/ deformation vectors. Thus results for Γ3N include the results for rotational and translational vectors also i.e. Γ3N =Γvib+ ΓR+ ΓT.. Where Γvib represents results due to stretching/ deformation vibrations only, ΓR represents results for rotational vector and ΓT represents results for translational vectors. In order to get pure Γvib we have to subtract (ΓR+ ΓT.) from Γ3N

Therefore, Γvib = Γ3N -(ΓR+ ΓT.) = Γ3N -ΓR- ΓT. How to find ΓR+ ΓT? For this we will consider the four examples we have discussed. In modules on representations we have learnt that how x, y, z vectors along Cartesian coordinate form the basis for irreducible representation. We have also seen the results for Rx, Ry, Rz rotational vectors. All these information that for which irreducible representations the x, y, z, Rx, Ry, Rz vectors form the basis is given in the character table of the point group of the molecule. So we have not to bother for finding these irreducible representations every time by performing symmetry operations of the point group in question. 4.1 (i) Let us take H2O molecule example and write down Γ3N . Γ3N = 3A1+A2+2B1+3B2 In table.3 of C2v point group against irreducible representations A1, A2, B1, B2 in column I are listed basis set for that irreducible representation. We have to now find irreducible representations for which x, y, z, Rx, Ry, Rz are the basis

Table.3 Character table of C2v point group C2v E C2 σxz σxz I II 2 2 2 A1 1 1 1 1 Tz, z x ,y ,z A2 1 1 -1 -1 Rz xy B1 1 -1 1 -1 Tx, Ry zx B2 1 -1 -1 1 Ty, Rx, yz First let us look for Rx, Ry, Rz rotational vectors as basis and find irreducible representation

CHEMISTRY PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 26: Group theory and vibrational spectroscopy part-III

Let us take Rx rotational vector first see for which irreducible representation it is the basis. In character table it represents C2v E C2 σxz σxz I II 1 -1 -1 1 B2 Rx , Ty, So Rx represents B2 irreducible representation Let us take Ry rotational vector first see for which irreducible representation it is the basis. In character table it represents C2v E C2 σxz σxz I II B1 1 -1 1 -1 zx Ry, Tx,

So Ry represents B1 irreducible representation

Let us take Rz rotational vector first see for which irreducible representation it is the basis. In character table it represents

C2v E C2 σxz σxz I II A2 1 1 -1 -1 Xy Rz

Thus Rx,Ry, Rz vectors correspond to B2+B1+A2 irreducible representations Thus ΓR = B2+B1+A2 ------1

Similarly let us take translational vectors Tx (or x) ,Ty (or y) ,Tz (or z ) as basis and find irreducible representation for which these are the basis.

Let us first take Tx or x translational vector and find corresponding irreducible representation.

C2v E C2 σxz σxz I II B1 1 -1 1 -1 x ,Tx, Ry Zx

Thus Tx or x represents B1

Let us now take Ty or y translational vector and find corresponding irreducible representation.

C2v E C2 σxz σxz I II B2 1 -1 -1 1 yz Ty, Rx, CHEMISTRY PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 26: Group theory and vibrational spectroscopy part-III

Thus Ty or y represents B2

Let us now take Tz or z translational vector and find corresponding irreducible representation

C2v E C2 σxz σxz I II 1 1 1 1 x2,y2,z2

A1 Tz, z

Thus Tx (x) ,Ty( y), Tz (y) vectors correspond to B1+B2+A1 irreducible representations

Thus ΓT = B1+B2+A1 ------2

From equations 1 and 2

ΓR + ΓT = (B2+B1+A1 ) +(B1+B2+A1 )

ΓR + ΓT =2B1+2B2+A1+A2

Γvib = Γ3N -(ΓR+ ΓT.) = Γ3N -ΓR- ΓT

But Γ3N = 3A1+A2+2B1+3B2

Therefore, Γvib = Γ3N -(ΓR+ ΓT.) = Γ3N -ΓR- ΓT

Γvib = (3A1+A2+2B1+3B2 )-( 2B1+2B2+A1+A2)

Γvib =2A1+B2

There are three atoms in H2O so there should be 3N-6 vibrational modes of i.e. there should be three vibrational modes. The irreducible representations, 2A1+B2, equals to three modes as these are one dimensional irreducible representations. 4.2(ii) Let us take another example of C3v point group and follow the above procedure for getting Γvib Let us take POCl3 molecule example and write down Γ3N . Γ3N = 4A1+A2+5E In Table 4 of C3v point group against irreducible representations A1, A2, E in column under linear functions or rotations are listed basis set for these irreducible representations. We have to now find irreducible representations for which CHEMISTRY PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 26: Group theory and vibrational spectroscopy part-III

x, y, z, Rx, Ry, Rz are the basis

Table.4 Character table of C3v point group

C3v E 2C3 (z) 3 v linear functions, rotations

A1 +1 +1 +1 z

A2 +1 +1 -1 Rz

E +2 -1 0 (x, y) (Rx, Ry)

First let us look for Rx, Ry, Rz rotational vectors as basis and find irreducible representation Let us take Rx and R y rotational vectors and see for which irreducible representation these are the basis. In character table it represents

C3v E 2C3 (z) 3 v linear functions,

rotations

E +2 -1 0 (x, y) (Rx, Ry)

Thus Rx and Ry together form the basis for E representation

Similarly x and y (Tx,Ty) form the basis for E representation

Rz vector is the basis for A2

C3v E 2C3 (z) 3 v linear functions,

rotations

A2 +1 +1 -1 Rz

Z orTz vector is the basis for A1

CHEMISTRY PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 26: Group theory and vibrational spectroscopy part-III

C3v E 2C3 (z) 3 v linear functions,

rotations

A1 +1 +1 +1 Rz

ΓR+ ΓT equals to ΓT+R =A1+A2+2E

and Γ3N = 4A1+A2+5E

Γvib = Γ3N -(ΓR+ ΓT.) = Γ3N -ΓR- ΓT

Γvib =(4A1+A2+5E )-( A1+A2+2E )

Γvib =3A1+3E There are five atoms in POCl3 so there should be 3N-6 vibrational modes i.e. there should be nine vibrational modes. The irreducible representations, 3A1+3E, equals to nine modes (3+3x2) as there are three one dimensional irreducible representations and three two dimensional representations. 4.3(iii) Let us take the example of CCl4 of Td point group and work out total number of vibrational modes for this molecules . For CCl4, Γ3N = A1+E+T1+3T2, In table.5 of Td point group against irreducible representations A1, A2, E, T1, T2 in column under linear functions or rotations are listed basis set for these irreducible representation. We have to now find irreducible representations for which x, y, z, Rx, Ry, Rz vectors are the basis

Table .5 Character table for Td point group

Td E 8C3 3C2 6S4 6 d linear functions, rotations A +1 +1 +1 +1 +1 - 1 A2 +1 +1 +1 -1 -1 -

E +2 -1 +2 0 0 -

T1 +3 0 -1 +1 -1 (Rx, Ry, Rz)

T2 +3 0 -1 -1 +1 (x, y, z)

CHEMISTRY PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 26: Group theory and vibrational spectroscopy part-III

Let us take Rx, Ry, Rz rotational vectors and see for which irreducible representation these are the basis. In character table these Rx, Ry, Rz rotational vectors represent T1 representation

Td E 8C3 3C2 6S4 6 d linear functions, rotations +3 0 -1 +1 -1

T1 (Rx, Ry, R ) z

Let us take Tx, Ty, Tz translational vectors and see for which irreducible representation these are the basis. In character table these Tx, Ty, Tz translational vectors represent T2 representation

Γ + Γ equals to Γ + Γ Td E 8C3 3C2 6S4 6 d linear functions, R T T rotations R =T1+T2

and Γ3N = +3 0 -1 1 -1 A1+E+T1+3T2 T2 (Tx,Ty, Tz)

Γvib = Γ3N -(ΓR+ ΓT.) = Γ3N -ΓR- ΓT

Γvib =( A1+E+T1+3T2)-( T1+T2 )

Γvib =A1+E+2T2

There are five atoms in CCl4, molecule so there should be 3N-6 vibrational modes i.e. there should be nine vibrational modes. The irreducible representations, A1+E +2T2, equals to nine modes (1+2+2x3) as there is one dimensional irreducible representations ,one two dimensional and two three dimensional representations .

2- 4.4(iv) Similarly one can work out Γvib for molecules XeF4,PtCl4 (D4h) These are given as: Γvib =[ T3N=A1g+A2g+B1g+B2g+Eg+2A2u+B2u+3Eu]-[ ΓT+R =A2g+Eg+A2u+E]

CHEMISTRY PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 26: Group theory and vibrational spectroscopy part-III

Γvib = [A1g+ +B1g+B2g +A2u+B2u+2Eu

5. Summary

• More facts about reducible and irreducible representation summarized • How to find Γ3N using Cartesian coordinates as basis explained • How to get ΓT and ΓR using character table of that point group • How to find Γvib after subtracting ΓT and ΓR from Γ3N explained in detail • Getting of Γvib for C2v, C3v, Td, D4h point groups explained in detailed manner.

CHEMISTRY PAPER No.13 :Applications of molecular symmetry and group theory MODULE No. 26: Group theory and vibrational spectroscopy part-III