13 Applications of molecular symmetry and group theory

Transcription

1 Subject Chemistry Paper No and Title Module No and Title Module Tag 13 Applications of molecular symmetry and 26 and and vibrational spectroscopy part-iii CHE_P13_M26

2 TABLE OF CONTENTS 1. Learning Outcomes 2. Introduction 3. Reducible and irreducible representations recollection 3.1 More examples and find Γ 3N 4. How to get vibrational modes? 4.1 Example of C 2v point group 4.2 Example of C 3v point group 4.4 Example of Td point group 4.5 Example of D 4h point group 5. Summary

3 1. Learning Outcomes After studying this module, you shall be able to Recollect about reducible and irreducible representation Learn how to find Γ 3N using Cartesian coordinates as basis Get Γ T and Γ R Find Γ vib after subtracting Γ T and Γ R from Γ 3N How to find Γ vib for C 2v,C 3v,T d,d 4h 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 Γ 3N (I) We take the example of XeF 4, or [PtCl 4 ] 2-. These molecules belong to D 4h point group.the symmetry elements and 3N vectors are shown in case of XeF 4 in fig.1. C 2, F F Xe C 4 F F σ h σ v Fig.1 3 N vectors and some symmetry elements in XeF 4 molecule

4 The character table in part for D 4h point group together with T 3N is given in Table.1 Table.1 Character table and Γ 3N for D 4h point group D 4h E 2C 4 C 2 2C 2 2C 2 i 2S 4 σ h 2σ v 2σ d A 1g A 2g B 1g B 2g E g A 1u A 2u B 1u B 2u E u Number of unshifted atoms Multiplication factor x3 x1 x(-1) x(-1) x(-1) x(-3) x(-1) x1 x1 x1 Γ 3N Γ 3N can be reduced to irreducible representations by following standard reduction formula. (A 1g )=1/16[1x15x1+2x1x1+1x-1x1+2x-3x1+2x-1x1+1x-3x1+2x-1x1+1x5x1 +2x3x1 +2x1x1] 1/16[16=1 i.e. A 1g Similarly other irreducible representations can be obtained and the net result is Γ 3N = A 1g +A 2g +B 1g +B 2g +E g +2A 2u +B 2u +3E u Total irreducible representations are; [ x3]=15 i.e. equals to sum of characters of E in all irreducible representations of Γ 3N

5 (II) Let us take examples of molecules CCl 4 CH 4, RuO 4, SO 4 2-.These belong to T d. point group. Total number of symmetry operations in T d point group are 24. Therefore, h=24.some of the symmetry elements and 3N vectors are shown in case of CCl 4 in fig.1. Cl C 3 C Cl Cl Cl Fig.2 3N vectors and some symmetry elements in CCl 4 molecule The character table for T d point group together with 3N vectors on each atom is shown in case is give in table.2 Table.2 Γ 3N and character table of T d point group T d E 8C 3 3C 2 6S 4 6σ d A A E T T Number of unshifted atom Multiplication x3 x0 x(-1) x(-1) x1 factor Γ 3N Γ 3N can be reduced to irreducible representations by making use of standard reduction formula. A 1 =1/24[1x15x1+8x0x1+3x-1x1+6x-1x1+6x3x1]=1/24[24] =1 i.e. A 1 T 1 =1/24[1x15x3+8x0x1+3x-1x-1+6x-1x1+6x3x-1] =1/24[24] =1 i.e. T 1 Similarly other irreducible representations can be obtained and over all result is

6 Γ 3N = A 1 +E+T 1 +3T 2. Total irreducible representations are: 1+2x1+3x1+3x3=15 which is equal to sum of character of E in irreducible representations in Γ 3N The results of Γ 3N and its reduction are summarized as follows: (A) SO 2,H 2 O (C 2v ) ; Γ 3N = 3A 1 +A 2 +2B 1 +3B 2 (B ) POCl 3,NH 3,CHCl 3 (C 3v ) ; Γ 3N = 4A 1 +A 2 +5E (I )XeF 4, [PtCl 4 ] 2- ( D 4h ) ; Γ 3N = A 1g +A 2g +B 1g +B 2g +E g +2A 2u +B 2u +3E u (II) CCl 4 CH 4, RuO 4,SO 2-4 (T d ) ; Γ 3N = A 1 +E+T 1 +3T 2, 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 Γ 3N include the results for rotational and translational vectors also i.e. Γ 3N =Γ vib + Γ R + Γ T.. Where Γ vib represents results due to stretching/ deformation vibrations only, Γ R represents results for rotational vector and Γ T represents results for translational vectors. In order to get pure Γ vib we have to subtract (Γ R + Γ T. ) from Γ 3N Therefore, Γ vib = Γ 3N -(Γ R + Γ T. ) = Γ 3N -Γ R - Γ T. How to find Γ R + Γ T? 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 R x, R y, R z rotational vectors. All these information that for which irreducible representations the x, y, z, R x, R y, R z 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 H 2 O molecule example and write down Γ 3N. Γ 3N = 3A 1 +A 2 +2B 1 +3B 2 In table.3 of C 2v point group against irreducible representations A 1, A 2, B 1, B 2 in column I are listed basis set for that irreducible representation. We have to now find irreducible representations for which x, y, z, R x, R y, R z are the basis Table.3 Character table of C 2v point group A T z, z x 2,y 2,z 2 A R z xy B T x, R y zx B T y, R x, yz First let us look for R x, R y, R z rotational vectors as basis and find irreducible representation

7 Let us take R x rotational vector first see for which irreducible representation it is the basis. In character table it represents B 2 R x, T y, So R x represents B 2 irreducible representation Let us take R y rotational vector first see for which irreducible representation it is the basis. In character table it represents B zx R y, T x, So R y represents B 1 irreducible representation Let us take R z rotational vector first see for which irreducible representation it is the basis. In character table it represents A Xy Thus R x,r y, R z vectors correspond to B 2 +B 1 +A 2 irreducible representations Similarly let us take translational vectors T x (or x),t y (or y),t z (or z ) as basis and find irreducible representation for which these are the basis. R z Thus Γ R = B 2 +B 1 +A Let us first take T x or x translational vector and find corresponding irreducible representation. B x,t x, R y Zx Thus T x or x represents B 1 Let us now take T y or y translational vector and find corresponding irreducible representation. B yz T y, R x,

8 Thus T y or y represents B 2 Let us now take T z or z translational vector and find corresponding irreducible representation x 2,y 2,z 2 T z, z A 1 Thus T x (x),t y ( y), T z (y) vectors correspond to B 1 +B 2 +A 1 irreducible representations Thus Γ T = B 1 +B 2 +A From equations 1 and 2 Γ R + Γ T = (B 2 +B 1 +A 1 ) +(B 1 +B 2 +A 1 ) Γ R + Γ T =2B 1 +2B 2 +A 1 +A 2 Γ vib = Γ 3N -(Γ R + Γ T. ) = Γ 3N -Γ R - Γ T But Γ 3N = 3A 1 +A 2 +2B 1 +3B 2 Therefore, Γ vib = Γ 3N -(Γ R + Γ T. ) = Γ 3N -Γ R - Γ T Γ vib = (3A 1 +A 2 +2B 1 +3B 2 )-( 2B 1 +2B 2 +A 1 +A 2 ) Γ vib =2A 1 +B 2 There are three atoms in H 2 O so there should be 3N-6 vibrational modes of i.e. there should be three vibrational modes. The irreducible representations, 2A 1 +B 2, equals to three modes as these are one dimensional irreducible representations. 4.2(ii) Let us take another example of C 3v point group and follow the above procedure for getting Γ vib Let us take POCl 3 molecule example and write down Γ 3N. Γ 3N = 4A 1 +A 2 +5E In Table 4 of C 3v point group against irreducible representations A 1, A 2, 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

9 x, y, z, R x, R y, R z are the basis Table.4 Character table of C 3v point group C 3v E 2C 3 (z) 3 v linear functions, rotations A z A R z E (x, y) (R x, R y ) First let us look for R x, R y, R z rotational vectors as basis and find irreducible representation Let us take R x and R y rotational vectors and see for which irreducible representation these are the basis. In character table it represents C 3v E 2C 3 (z) 3 v linear functions, rotations E (x, y) (R x, R y ) Thus R x and R y together form the basis for E representation Similarly x and y (T x,t y ) form the basis for E representation R z vector is the basis for A 2 C 3v E 2C 3 (z) 3 v linear functions, rotations A R z Z ort z vector is the basis for A 1

10 C 3v E 2C 3 (z) 3 v linear functions, rotations A R z Γ R + Γ T equals to Γ T + R =A 1 +A 2 +2E and Γ 3N = 4A 1 +A 2 +5E Γ vib = Γ 3N -(Γ R + Γ T. ) = Γ 3N -Γ R - Γ T Γ vib =(4A 1 +A 2 +5E )-( A 1 +A 2 +2E ) Γ vib =3A 1 +3E There are five atoms in POCl 3 so there should be 3N-6 vibrational modes i.e. there should be nine vibrational modes. The irreducible representations, 3A 1 +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 CCl 4 of T d point group and work out total number of vibrational modes for this molecules. For CCl 4, Γ 3N = A 1 +E+T 1 +3T 2, In table.5 of T d point group against irreducible representations A 1, A 2, E, T 1, T 2 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, R x, R y, R z vectors are the basis Table.5 Character table for T d point group T d E 8C 3 3C 2 6S 4 6 d linear functions, rotations A A E T (R x, R y, R z ) T (x, y, z)

11 Let us take R x, R y, R z rotational vectors and see for which irreducible representation these are the basis. In character table these R x, R y, R z rotational vectors represent T 1 representation T 1 T d E 8C 3 3C 2 6S 4 6 d linear functions, rotations (R x, R y, R z ) Let us take T x, T y, T z translational vectors and see for which irreducible representation these are the basis. In character table these T x, T y, T z translational vectors represent T 2 representation T d E 8C 3 3C 2 6S 4 6 d linear functions, rotations Γ R + Γ T equals to Γ T + Γ R =T 1 +T 2 T (T x,t y, T z ) and Γ 3N = A 1 +E+T 1 +3T 2 Γ vib = Γ 3N -(Γ R + Γ T. ) = Γ 3N -Γ R - Γ T Γ vib =( A 1 +E+T 1 +3T 2 )-( T 1 +T 2 ) Γ vib =A 1 +E+2T 2 There are five atoms in CCl 4, molecule so there should be 3N-6 vibrational modes i.e. there should be nine vibrational modes. The irreducible representations, A 1 +E +2T 2, equals to nine modes (1+2+2x3) as there is one dimensional irreducible representations,one two dimensional and two three dimensional representations. 4.4(iv) Similarly one can work out Γ vib for molecules XeF 4,PtCl 4 2- (D 4h ) These are given as: Γ vib =[ T 3N= A 1g +A 2g +B 1g +B 2g +E g +2A 2u +B 2u +3E u ]-[ Γ T + R =A 2g +E g +A 2u +E]

12 Γ vib = [A 1g + +B 1g +B 2g +A 2u +B 2u +2E u 5. Summary More facts about reducible and irreducible representation summarized How to find Γ 3N using Cartesian coordinates as basis explained How to get Γ T and Γ R using character table of that point group How to find Γ vib after subtracting Γ T and Γ R from Γ 3N explained in detail Getting of Γ vib for C 2v, C 3v, T d, D 4h point groups explained in detailed manner.