13, Applications of molecular symmetry and group theory

Transcription

1 Subject Paper No and Title Module No and Title Module Tag Chemistry 13, Applications of molecular symmetry and group theory 27, Group theory and vibrational spectroscopy: Part-IV(Selection rules for IR and Raman spectroscopy) CHE_P13_M27

2 TABLE OF CONTENTS 1. Learning Outcomes 2. Introduction 3.Symmetry Selection rules for Infrared Spectroscopy 3.1 Examples of H 2 O/SO 2 (C 2v ) point group molecules 3.2 Examples of POCl 3 / CHCl 3 (C 3v ) point group molecules 4. Symmetry Selection Rules for Raman Spectroscopy 4.1 Example of C 2v point group molecule H 2 O/SO Example of C 3v point group molecule POCl 3 /CHCl 3 5. Summary

3 1. Learning Outcomes After studying this module, you shall be able to: Know the interaction of internal electric field due to vibration of molecule and external field. Learn the selection rules for IR spectroscopy. Learn about IR vibrational active modes for H 2 O molecule. Learn about vibrational active modes for POCl 3 molecule. Learn the selection rules for Raman spectroscopy. Learn about Raman vibrational active modes for H 2 O molecule. Learn about Raman vibrational active modes for POCl 3 molecule.. 2. Introduction You already know about reducible and irreducible representations and how to get these representations by taking 3N vectors as basis. You have also learnt how to get irreducible representations corresponding to x, y, z, R x, R y, R z vectors i.e. how to find Γ T and Γ R.The Γ vib were then obtained by subtracting Γ T and Γ R from Γ 3N. In this module we will learn about selection rules for infrared and Raman spectroscopy. In spectroscopy selection rule, or transition rule, formally constrains the possible transitions of a system from ground to excited state. Here selection rules have been derived for vibrational spectroscopy. The selection rules may differ according to the technique used to observe the transition. 3.Symmetry Selection rules for Infrared Spectroscopy A vibrating bond in a molecule behaves like oscillating electric field and there is a change electric dipole moment. Electromagnet waves in the region of infrared also have electric field component with it. When internal electric field component due to vibrating molecule interacts with the external electric field component of electromagnet waves, the transitions from ground vibrational state to higher vibrational excited state may occur or not i.e. this transition is allowed or not. In this section we learn selection rules about these types of transitions. According to quantum mechanical rule υ =±1 i.e. transition υ 0 υ +1 is allowed. In quantum mechanics the basis for infrared spectroscopic selection rule is the value of the transition moment integral of the type ψ g *µ ψ e dt equation 1 Transition occurs from vibrational ground state ( initial ) ψ g to vibrationally excited state ψ e by absorbing one quanta of energy. This is called fundamental transition. The µ is molecular dipole moment vector. The probability of such transition from ψ g ground vibrational state to ψ e first

4 excited vibrational state is directly proportional to transition moment given by the integral of the type given in equation.1.since ψ g ground state wave function is always real we have ψ g *= ψ g. Thus equation 1 can be written as ψ g µψ e dt dt means integration is carried over all variables of the wave function. The spectroscopic activity of a vibrational mode depends upon the value of transition moment integral ψ g µψ e dt. If this integral has zero value for a particular transition, then the probability of occurring this transition will be zero and no absorption of IR radiations will occur. Thus the transition is forbidden and vibration will vibrationally inactive. If the integral is non zero, then there will be absorption of IR radiations by the molecule and the transition will be allowed.hence it will be IR active. In practice, the integral itself does not need to be calculated to determine a selection rule. It is sufficient to determine the symmetry of transition moment function, ψ g µψ e dt. The idea of symmetry is important when considering the integral of odd functions. Since ψ g ground state wave function is always real and symmetric. It follows that, for a vibrational transition to be allowed, the symmetry of the excited state wave function must be the same as the symmetry of the transition moment operator. Solving of integrals of the type ψ g µψ e dt is a cumbersome process. Group theory helps and simplifies the solutions of such integrals. According to group theory and symmetry considerations, if the direct product in this integral contains totally symmetrical representation ( an irreducible representation for which χ =+1 for all symmetry operations of the point group) then the integral will be non-zero and there will be interaction between electric field component of electromagnetic radiation(in IR region)and electric field component of oscillating electric dipole moment of the vibrating molecule and hence absorption of IR radiation will occur and vibration will be IR active. Thus only difficulty left, in assigning the spectroscopic activity to vibrational modes, is to solvethis integral. The transition moment integral ψ g µψ e dt Can be resolved into three components along X, Y, Z axes and these resolved integrals can be written as: ψ g µ x ψ e dt equation 2 ψ g µ y ψ e dt equation3 ψ g µ z ψ e dt equation 4 If any of these integrals given by equations 2, 3, 4 is nonzero then according to symmetry point of view this transition will be allowed. So problem now is how to check the symmetries of direct

5 product in these three component integrals? We know that vibrational ground state υ 0 =0, i.e. ψ g has mathematical similarity to that of s-orbital. Since a sphere is totally symmetrical in a point group, the ground vibrational state ψ g will be totally symmetrical for all molecules. Symmetries of components of transition moment vectors µ x,, µ y, µ z is the same as that of X, Y, Z axes or that of translational vectors T x,t y,t z. Their symmetries can be directly obtained from the character table of the point group to which the molecule belongs. The symmetry properties of vibrationally excited states ψ e are the same as that of the vector which describes/forms the basis, for the particular vibrational mode (Γ vib ). So we have to evaluate the direct product of ψ g ( totally symmetric), the irreducible representation for which X,Y,Z axes or T x,t y,t z form the basis for representations and ψ e, irreducible representation of the vibrational mode to be considered for IR activity, spectroscopic activity of which we want to determine. In the integral ψ g µψ e dtthe direct product of ψ g.ψ e will give the product as ψ e.the direct product of a totally symmetric irreducible representation with any non symmetrical irreducible representation equals to non symmetric irreducible representation. For example in C 2v point group A 1 xb 1 =B 1.So direct productµ ψ e must contain totally symmetric irreducible representation in it. So we have to check the symmetries of µ or µ x, µ y, µ z. The symmetry of ψ e is the same as that of irreducible representation the activity of which we want to know and it is given. A normal mode belonging to any symmetry species as that of X,Y,Z axes or T x,t y,t z translational vectors will be IR active. Symmetry characteristics of transition moment operator are summarized as: Symmetry characteristics of transition moment operator Type of Transition µ transforms as Electric dipole x, y, z Infrared spectra Electric polarizability x 2, y 2, z 2, xy, xz, yz Raman spectra To explain the points made so far about vibrational transition let us take few examples 3.1 Examples of H 2 O/SO 2 (C 2v ) point group molecules In module -30 we have obtained Γ 3N and then Γ vib as 2A 1 +B 2 for water molecule. Now we will find the activity of A 1 and B 2 modes i.e. that of ψ e which belongs to either A 1 or B 2 symmetries. In C 2v point group the totally symmetric representation is A 1.We have to find the irreducible representations for which X,Y, Z axes or T x, T y, T z form the basis for representations i.e. symmetry species of µ x,, µ y, µ z components. Using character table of C 2v point group the symmetries of these vectors are: C 2v E C 2 (z) linear functions,rotations

6 v(xz) v(yz) A z A R z B x, R y B y, R x µ x = T x or x belongs to B 1 symmetry µ y = T y or y belongs to B 2 symmetry µ z = T z or z belongs to A 1 symmetry In order to check the activity of 2A 1 and B 2 modes we have to evaluate the following direct products. Check whether the direct products equal to or contain totally symmetric irreducible representation in it. ψ g µ x ψ e dt A 1 B 1 A 1 B equ.5 ψ g µ y ψ e dt A 1 B 2 A 1 B equ.6 ψ g µ z ψ e dt A 1 A 1 A 1 A equ.7 Here ψ g hasa 1 symmetry; µ x, µ y, µ z have B 1, B 2, A 1 symmetries respectively and ψ e has the symmetries of the type A 1, B 2. Out of the three integrals given by equations 5, 6, 7 if any of these contain totally symmetric representation then the vibrational mode will be IR active. Out of these three equations 5, 6, 7 the direct product integral in equation-7 contains a totally symmetric representation A 1. Thus the overall integral ψ g µψ e dt has A 1 representation in it and thus integral is nonzero. Hence the vibrational mode is IR active i.e. A 1 mode is IR active. Let us check the activity of B 2 mode for this we have to consider the following direct products and evaluate these. ψ g µ x ψ e dt A 1 B 1 B 2 A equ.8 ψ g µ y ψ e dt A 1 B 2 B 2 A equ.9 ψ g µ z ψ e dt A 1 A 1 B 2 B equ.10 Out of the equations 8, 9, 10; equation-9 contains totally symmetric representations. Hence over all intergral contains A 1 mode and thus it is non zero and the B 2 mode is also IR active. Thus in Γ vib = 2A 1 +B 2 both A 1 and B 2 modes are IR active for H 2 O/SO 2 (C 2v ) point group molecules.

7 3.2 Examples of POCl 3 / CHCl 3 (C 3v ) point group molecules We have obtained Γ 3N and then Γ vib as Γ vib =3A 1 +3E for POCl 3 molecule. Now we will find the activity of A 1 and E modes i.e. ψ e belongs to either A 1 or E symmetries. In C 3v point group the totally symmetric representation is A 1.We have to find the irreducible representations for which X,Y,Z axes or T x,t y,t z form the basis for representations i.e. symmetry species of µ x,, µ y, µ z components. Using character table of C 3v point group symmetries of these vectors i.e., of these X,Y,Z axes or T x,t y,t z are: C 3v E 2C 3 (z) 3 v linear functions, A Z rotations A R z E (x, y) (R x, R y ) µ x =T x or x belongs E µ y =T y or y belongs E µ z =T z or z belongs A 1 In order to find the activity of A 1 and E modes we have to evaluate the following direct product integrals i.e. have to check the activity when ψ e has the symmetry as 3A 1 and 3E.Whether these are IR active or not. For this we have to consider the integrals of the type given below for A1 mode and solve them ψ g µ x ψ e dt A 1 E A 1 E equation 11 ψ g µ y ψ e dt A 1 E A 1 E equation 12 ψ g µ z ψ e dt A 1 A 1 A 1 A equation 13 Out of the equations 11, 12, 13, the equation 13 contains totally symmetric representation. Overall integral is non zero. Hence A 1 mode is IR active. Now let us take E mode and check its activity. For this we have to consider the following direct products and evaluate these. ψ g µ x ψ e dt A 1 E E A equation 14 ψ g µ y ψ e dt A 1 E E A equation 15

8 ψ g µ z ψ e dt 16 A 1 A 1 E E equation Out of the integral in equations 14, 15 and16, the equations 14 and 15 contain the totally symmetric representation. So over all integral is non zero. So the vibrational mode E is also IR active Thus in Γ vib = 3A 1 +3E, both A 1 and E modes are IR active for POCl 3 /CHCl 3 /(C 3v ) point group molecules 4. Symmetry Selection Rule For Raman spectroscopy In Raman spectroscopy transition between vibrational states can be described in similar manner as described in case of IR spectroscopy, by the integral of the type ψ g α ψ e dt equ.17 Where ψ g and ψ e represent the ground and first excited vibrational states respectively. The α describes the polarizability of molecule under investigation. Raman effect depends upon the extent of induced dipole by laser beam. Induced dipole is directly proportional to polarizability of the molecule. P=α E Polarizability means the extent to which electron distribution in the molecule can be distorted by the applied laser beam. As mentioned earlier α is a tensor with a 3 x 3 array of component terms such as ; α xx α xy, α xz, α yx, α yy, α yz α zx, α zy α zz.. P x P y P z = α xx α xy α xz α yx α yy α yz α zx α zy α zz E x E y E z For vibrational transition in Raman spectrum α jk = α kj ( where j,k = x,y,or z). Therefore, out of nine terms only six terms are to be considered. α xy = α yx, α zx, = α xz, α yz =α zy, α xx (α x 2 ), α yy (α y 2 ), α zz. (α z 2 ). Thus there are six distinct components of the integral ψ g α ψ e dt. These integrals are: (1) ψ g (α 2 xx=x ) ψ e dt (4) ψ g (α xy=yx )ψ e dt (2) ψ g (α 2 yy=y ) ψ e dt (5) ψ g (α yz=zy )ψ e dt 2 (3) ψ g (α zz=z ) ψ e dt (6) ψ g ( α zx=xz )ψ je dt For a vibration to be Raman active any one of the above six integral should be non-zero. We need to find the symmetry of direct product of these integrals.ψ g is totally symmetrical and symmetry of α xy = α yx ; α yz = α zy ; α zx = α xz ; α xx (α 2 x ); α yy (α 2 y ); α zz (α 2 z ) is the same as the symmetry of binary functions ; xy =yx ; yz=zy ; zx=xz ; xx(x 2 ); yy(y 2 ) ; zz(z 2 ). Symmetries of

9 these binary functions can be noted down from column ( which has binary functions as basis) of character table of that point group. 4.1 Example of C 2v point group molecule H 2 O/SO 2. In earlier modules we have obtained Γ 3N and then Γ vib as 2A 1 +B 2 for water molecule. Now we will find the Raman activity of A 1 and B 2 modes i.e. that of ψ e which belongs to either A 1 or B 2 symmetries. In C 2v point group the totally symmetric representation is A 1. For Raman activity we have to find the irreducible representations for which xy =yx ; yz=zy ; zx=xz ; xx(x 2 ); yy( y 2) ; zz(z 2 ) binary functions form the basis for irreducible representations i.e. symmetry species of α xy = α yx, α zx, = α xz, α yz =α zy, α xx (α x 2 ), α yy (α y 2 ), α zz. (α z 2 ) components. Using character table of C 2v point group the symmetries of these vectors can be obtained easily. C 2v E C 2 (z) v(xz) v(yz) quadratic functions A x 2, y 2, z 2 A xy B xz B yz Symmetries of α xy = α yx =A 2 ; α yz = α zy =B 2 ; α zx,= α xz, = B 1 ; α xx (α x 2 )=A 1 ;α yy (α y 2 )=A 1; α zz. (α z 2 )=A 1. For checking the Raman activity of A 1 mode above six direct product integrals can be rewritten as: (1a) A 1 A 1 A 1 A 1 (4d) A 1 A 2 A 1 A 2 (2b) A 1 A 1 A 1 A 1 (5e) A 1 B 2 A 1 B 2 (3c) A 1 A 1 A 1 A 1 (6f) A 1 B 1 A 1 B 1, Integrals 1a, 2b, 3c all contain A 1 representation and thus the whole integral ψ g α ψ e dt is non zero or A 1 mode is Raman active. Symmetric modes are Raman active and always polarized.

10 For checking the Raman activity of B 2 mode above six direct product integrals can be rewritten as: (7) A 1 A 1 B 2 B 2 (10) A 1 A 2 B 2 B 1 (8) A 1 A 1 B 2 B 2 (11) A 1 B 2 B 2 A 1 (9) A 1 A 1 B 2 B 2 (12) A 1 B 1 B 2 A 2 Of these six integrals (11) contains totally symmetric representation. So the whole integral is non zero. Hence B 2 mode is also Raman active. So for H 2 O molecule both vibrational modes 2A 1 and B 2 are Raman Active 4.2 Example of C 3v point group molecule POCl 3 /CHCl 3. For C 3v point group Γ vib =3A 1 +3E. Using character table of C 3v point group symmetries of components of α are noted as: α xy = α yx =E ;α yz = α zy =E; α zx,= α xz, = E; α xx (α x 2 )=A 1 ; α yy (α y 2 )=A 1; α zz. (α z 2 )=A 1 and ψ e is either of A 1 or E symmetry i.e. we have to check the Raman activity of A 1 and E modes. C 3v E 2C 3 (z) 3 v quadratic functions A x 2 +y 2, z 2 A E (x 2 -y 2, xy) (xz, yz) For this we have to consider the following six direct products integrals of ψ g α ψ e dt. For checking the activity of A 1 mode these can be written in direct product form as: (13) A 1 A 1 A 1 A 1 (16) A 1 E A 1 E (14) A 1 A 1 A 1 A 1 (17) A 1 E A 1 E (15) A 1 A 1 A 1 A 1 (18) A 1 E A 1 E

11 The integrals in equations 13, 14, 15 contain totally symmetric representation A 1. So the whole integral is non- zero.hence A 1 mode is Raman active. For checking the activity of E mode 6 similar integrals can be written in direct product form as: (19) A 1 A 1 E E(22) A 1 E E A 1 (20) A 1 A 1 E E(23) A 1 E E A 1 (21) A 1 A 1 E E(24) A 1 E E A 1 Integrals in 22, 23, 24 contain totally symmetric representation. So whole integral is non zero.hence E mode is also Raman active So for POCl 3 /CHCl 3 molecule all vibrational modes 3A 1 and 3E are Raman Active. For C 2v and C 3v point group molecules we have noticed that all modes are active in both IR and Raman. For centrosymmetric molecules i.e. which have centre of inversion i the, mutual Exclusion Principle applies. According to this rule, there exists no common vibrational mode which is active in both IR and Raman. A vibration which is active in IR will be Raman inactive and vice-versa. The reason for this is that, T x,t y,t z, which form the basis for vibrational modes, are antisymmetric with repect to i i.e. these are ungerade, where as the product x 2,y 2,z 2 etc. are always symmetric with respect to i and cannot show IR active bands. 5. Summary How interaction of internal electric field due to vibration of molecule and external field is responsible infrared radiation absorption? The selection rules for IR spectroscopy have been explained How to find IR vibrational active modes for H 2 O molecule has been explained in detail? How to find IR vibrational active modes for POCl 3 molecule has been explained in detail? The selection rules for Raman spectroscopy have been explained How to find Raman vibrational active modes for H 2 O molecule has been explained in detail? How to find Raman vibrational active modes for POCl 3 molecule has been explained in detail?