Symmetry and Group Theory Feature: Application for Spectroscopy and Orbital Molecules Dr. Indriana Kartini

Text books 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

Marks 80% exam: 10% group assignments of 4 students 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

Unsur simetri dan operasi simetri molekul 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

E Cn s i Sn Simbol Unsur Operasi Unsur identitas Membiarkan obyek tidak berubah Cn Sumbu rotasi Rotasi seputar sumbu dengan derajat rotasi 360/n (n adalah bilangan bulat) s Bidang simetri Refleksi melalui bidang simetri i Pusat/titik inversi Proyeksi melewati pusat inversi ke sisi seberangnya dengan jarak yang sama dari pusat Sn Sumbu rotasi tidak sejati (Improper rotational axis) Rotasi mengitari sumbu rotasi diikuti dengan refleksi pada bidang tegak lurus sumbu rotasi

© Imperial College London Operasi Simetri Chemistry Chemistry Chemistry yrtsimehC Chemistry Inversi Refleksi Rotasi © Imperial College London

BF3 Rotations 360/n where n is an integer Operation rotation by 360/3 around C3 axis (element) F3 F2 Rotate 120O F1 F1 F3 F2

H2O Reflections Reflection is the operation s element is plane of symmetry z x is out of the plane y H2O x s(xz) s(yz)

Reflections for H2O

Reflections Principle (highest order) axis is defined as Z axis After Mulliken s(xz) in plane perpendicular to molecular plane s(yz) in plane parallel to molecular plane both examples of sv sv : reflection in plane containing highest order axis sh : reflection in plane perpendicular to highest order axis sd : dihedral plane generally bisecting sv

XeF4 Reflections sv sh sd sd

XeF4

Inversion i element is a centre of symmetry Examples: Benzene, XeF4 Ethene Inversion i element is a centre of symmetry Atom at (x,y,z) Atom at (-x,-y,-z) Inversion , i Centre of inversion

S4 Improper Rotation s C4 Rotate about C4 axis and then reflect perpendicular to this axis S4 s C4

S4 Improper Rotation

successive operation

KULIAH MINGGU II TEORI GRUP

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

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 C Linear having i  Dh

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 C Linear having i  Dh

Steps to classify a molecule into a point group 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

Steps to classify a molecule into a point group Question 3: Has the molecule more than one rotation axis ? YES: Go to the Question 4 NO: 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

Steps to classify a molecule into a point group 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

Molecule * * Y N Dh Cv i? i? Select Cn with highest n, Linear? Dh Cv 2 or more Cn, n>2? i? i? Ih C5? Oh Td Y * Cn? * Dnh sh? Select Cn with highest n, nC2 perpendicular to Cn? N Dnd nsd? Dn s? Cs Cnh sh? Ci i? C1 Cnv nsv? S2n S2n? Cn

Benzene * * Y Benzene is D6h n = 6 N Dh Cv i? i? Linear? Dh Cv 2 or more Cn, n>2? i? i? Ih C5? Oh Td Y * Cn? Benzene is D6h n = 6 * Dnh sh? Select Cn with highest n, nC2 perpendicular to Cn? N Dnd nsd? Dn s? Cs Cnh sh? Ci i? C1 Cnv nsv? S2n S2n? Cn

Tugas I: Symmetry and Point Groups Tentukan unsur simetri dan grup titik pada molekul N2F2 POCl3 Gambarkan geometri masing-masing molekul tersebut

KULIAH MINGGU III

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 collection E 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: ACCA) Every member of the group must have an INVERSE which is also a member of the group. AA-1 = E

Example of Group Properties B(OH)3 belongs to C3 point group It has E, C3 and C32 symmetry operations

Overall: C3 followed C3 gives C32 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 C3 C3 Overall: C3 followed C3 gives C32

C32 C32 E. C32 = C32 and C32. C3 = E and C32. C32 = C3 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 C32 C32 E. C32 = C32 and C32. C3 = E and C32. C32 = C3

Operations are associative and E, C3 and C32 form a group 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.C32 = E ; C3 .E = C3 C3 .C3 = C32 ; C32 .C32 = C3 Operations are associative and E, C3 and C32 form a group

Group Multiplication Table Order of the group =3 C3 E C32 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

KULIAH MINGGU IV-V

© Imperial College London 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 © Imperial College London

© Imperial College London Math based 0 0 0 -1 0 0 0 1 [1 2 3] = © Imperial College London

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]

Constructing the Representation Put unit vectors on each atom z S O O y x C2v: [E, C2, sxz, syz] These are useful to describe molecular vibrations and electronic transitions.

Constructing the Representation A unit vector on each atom represents translation in the y direction C2 S S O O O O C2.(Ty) = (-1) Ty E .(Ty) = (+1) Ty syz .(Ty) = (+1) Ty sxz .(Ty) = (-1) Ty

Constructing the Representation A unit vector on each atom represents rotation around the z(C2) axis S O O C2.(Rz) = (+1) Rz E .(RZ) = (+1) Rz syz .(Rz) = (-1) Rz sxz .(RZ) = (-1) RZ

Constructing the Representation C2v E C2 s(xz) s(yz) +1 Tz -1 Rz Tx,Ry Ty,Rx

C2v +1 -1 Ty,Rx Constructing the Representation Use a mathematical function Eg: py orbital on S S O O C2v E C2 s(xz) s(yz) +1 -1 Ty,Rx py has the same symmetry properties as Ty and Rx vectors

Constructing the Representation sh Au Au C4 [AuCl4]- sh.[d x2-y2] = (+1) .[d x2-y2] Au C4.[d x2-y2] = (-1) .[d x2-y2]

Constructing the Representation Effects of symmetry operations generate the TRANSFORM MATRIX Simple examples so far. For all the symmetry operations of D4h on [d x2-y2] We have: D4h E 2C4 C2 2C2’ 2C2” I 2S4 sh 2sv 2sd +1 -1

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. Y C4 px px’  py py py’  px X A 2x2 transformation matrix In matrix form:

Constructing the Representation 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.

Constructing the Representation 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

Constructing the Representation 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

Character Tables Character ,  = a11 + a22 + a33. In general And only the character , which is a number is required and NOT the whole matrix.

Character Tables an Example C3v : (NF3) sv 1 Tz -1 Rz 2 (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, C31 are in the same class sv, sv, sv are in the same class

Character Tables an Example C3v : (NF3) 3sv A1 1 Tz x2 + y2 A2 -1 Rz 2 (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)

Note: You will not be asked to generate character tables. These can be brought/supplied in the examination

KULIAH MINGGU VI-VII-VIII

General form of Character Tables: (d) (e) (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

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 S with two and three dimensional representations being P and D.

Mulliken symbols: Labelling 1) 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 c = + 1 for the operation.) If it is anti-symmetric with respect to the operation c = - 1 and it is labelled B. 2) 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 sv 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)

Mulliken symbols: Labelling 3) Subscripts g (gerade) and u (ungerade) are given to irreducible representations That are symmetric and anti-symmetric respectively, with respect to inversion at a centre of symmetry. 4) Superscripts ‘ and “ are given to irreducible representations that are symmetric and anti-symmetric respectively with respect o reflection in a sh plane. Note: Points 1) and 2) apply to one-dimensional representations only. Points 3) and 4) apply equally to one-, two-, and three- dimensional representations.

Generating Reducible Representations zs z1 z2 For the symmetry operation sxz (a sv ) ys S x1 x2 x2 x1 xs xs y1 -y2 y2 -y1 ys -ys z1 z2 z2 z1 zs zs xs O y1 O y2 x1 x2 sxz

Generating Reducible Representations In matrix form  

Only require the characters: The sum of diagonal elements   Only require the characters: The sum of diagonal elements For s(xz) c = + 1

For s(yz) c = + 3

For E c = + 9

For C2 c = -1

Generating Reducible Representations Summarising we get that G3n for this molecule is: s(xz) s(yz) C2v E C2 G3n +9 -1 +1 3 To reduce this we need the character table for the point groups C2v E C2 s(xz) s(yz) A1 +1 Tz x2, y2, z2 A2 -1 Rz xy B1 Tx , Rx xz B2 Ty , Ry yz

g is the number of symmetry operations in the group Reducing Reducible Representations We need to use the reduction formula: 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 c(R) is character of the reducible representation cp(R) is character of the irreducible representation nR is the number of operations in the class

For C2v ; g = 4 and nR = 1 for all operations 1s(xz) 1s(yz) A1 +1 Tz x2, y2, z2 A2 -1 Rz xy B1 Tx , Rx xz B2 Ty , Ry yz s(xz) s(yz) C2v E C2 G3n +9 -1 +1 3 For C2v ; g = 4 and nR = 1 for all operations

s(xz) s(yz) C2v E C2 G3n +9 -1 +1 3 aA1 = (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x1) + (1x3x1)] = (12/4) =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 G3n = 3A1 + A2 + 2B1 + 3B2

Reducing Reducible Representations aA1 = (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x1) + (1x3x1)] = (12/4) =3 The terms in blue represent contributions from the un-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) Contribution from these atoms Number of un-shifted atoms

Identity E z z1 E y y1 x x1 For each un-shifted atom c(E) = +3

Inversion i x1 z y1 i y z1 x For each un-shifted atom c(i) = -3

Reflection s(xz) (Others are same except location of –1 changes) y1 y x x1 For each un-shifted atom c(s(xz)) = +1

Rotation Cn z1 z Cn y q x1 q y1 x 360/n c(Cn) = 1 + 2.cos(360/n)

Improper rotation axis, Sn z’ z s(xy) Cn x’ y’ y x1 y1 x z1 c(Sn) = -1 + 2.cos(360/n)

Summary of contributions from un-shifted atoms to G3n c(R) E +3 i -3 s +1 1+ 2.cos(360/n) C2 -1 1+ 2.cos(360/n) C3 ,C32 1+ 2.cos(360/n) C4, C43 -1 + 2.cos(360/n) S31,S32 -2 -1 + 2.cos(360/n) S41,S42 -1 + 2.cos(360/n) S61,S65

Worked example: POCl3 (C3v point group) c(R) R E +3 sv +1 C3 C3v E 2C3 3sv Number of classes, (1 + 2 + 3 = 6) Order of the group, g = 6 A1 1 1 1 A2 1 1 -1 E 2 -1 Un-shifted atoms 5 2 3 Contribution 3 1 G3n 15 3

Reducing the irreducible representation for POCl3 3sv C3v 15 3 G3n 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 G3n = 4A1 + A2 + 5E For POCl3 n= 5 therefore the number of degrees of freedom is 3n =15. E is doubly degenerate so G3n has 15 degrees of freedom.

KULIAH MINGGU IX-X-XI-XII APLIKASI TEORI GRUP

Group Theory and Vibrational Spectroscopy: SO2 C2v 1E 1C2 1s(xz) 1s(yz) A1 +1 Tz x2, y2, z2 A2 -1 Rz xy B1 Tx , Rx xz B2 Ty , Ry yz s(xz) s(yz) C2v E C2 G3n +9 -1 +1 3 G3n = 3A1 + A2 + 2B1 + 3B2

Group Theory and Vibrational Spectroscopy: SO2 G3n = 3A1 + A2 + 2B1 + 3B2 = 3 + 1 + 2 + 3 = 9 = 3n For non linear molecule there are 3n-6 vibrational degrees of freedom C2v 1E 1C2 1s(xz) 1s(yz) A1 +1 Tz x2, y2, z2 A2 -1 Rz xy B1 Tx , Rx xz B2 Ty , Ry yz Gvib = G3n – Grot – G trans G3n = 3A1 + A2 + 2B1 + 3B2 Grot = A2 + B1 + B2 Gtrans = A1 + B1 + B2 Gvib = 2A1 + B2 (Degrees of freedom = 2 + 1 = 3 = 3n-6)

Group Theory and Vibrational Spectroscopy: POCl3 G3n = 4A1 + A2 + 5E Gtrans = A1 + E There are nine vibrational modes . (3n-6 = 9) The E modes are doubly degenerate and constitute TWO modes Grot = A2 + E Gvibe = 3A1 + 3E 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

From G3n to Gvibe and Spectroscopy Now that we have Gvibe 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!!

Applications in spectroscopy: Infrared Spectroscopy Vibrational transition is infrared active because of interaction of radiation with the: molecular dipole moment, m. There must be a change in this dipole moment This is the transition dipole moment Probability is related to transition moment integral .

Infrared Spectroscopy Is the transition dipole moment operator and has components: mx, my, mz. Note: Initial wavefunction is always real yf Wavefunction final state yi Wavefunction initial state

Infrared Spectroscopy Transition is forbidden if TM = 0 Only non zero if direct product: yf m yi contains the totally symmetric representation. IE all numbers for c in representation are +1 The ground state yi 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.

The DIRECT PRODUCT representation. For SO2 we have that: Gvib = 2A1 + B2 Under C2v : Tx, Ty and Tz transform as B1, B2 and A1 respectively.

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. We predict three bands in the infrared spectrum of SO2 Gvib = 2A1 + B2

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.

Raman effect depends on change in polarisability a. Raman Spectroscopy Raman effect depends on change in polarisability a. 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 w1. Scattered light at w2. Can be Stokes(lower energy) or Anti-Stokes shifted Much weaker effect than direct absorption.

Raman Spectroscopy Virtual state Stokes Shifted w1 w2 yf w =w1- w 2 yi Wavefunction final state w =w1- w 2 yi Wavefunction initial state

Raman Spectroscopy Virtual state w1 w2 Anti-Stokes Shifted yf Wavefunction intial state w =w2- w 1 yi Wavefunction final state

Raman Spectroscopy Probability of a Raman transistion: The operator , a , is the polarisability tensor For vibrational transitions aij = aji so there are six distinct components: ax2, ay2, az2, axy, axz and ayz

Raman Spectroscopy For C2v ax2, ay2, az2, axy, axz and ayz Transform as: A1, A1, A1, A2, B1 and B2 We can then evaluate the direct product representation in a broadly analagous way

Raman Spectroscopy The DIRECT PRODUCT representation For SO2 group theory predicts only A1 and B2 modes 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

Raman Spectroscopy : General Rule If a vibrational mode has the same symmetry as on or more Of 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.

S O O Analysis of Vibrational Modes: Vibrations can be classified into Stretches, Bends and Deformations For SO2 Gvib = 2A1 + B2 We could choose more “natural” co-ordinates z S r1 r2 O O y x Determine the representation for Gstretch

S O O Analysis of Vibrational Modes: r1 r2 How does our new basis transform Under the operations of the group? O O Vectors shifted to new position contribute zero Unshifted vectors contribute + 1 to c(R) s(xz) s(yz) C2v E C2 Gstre +2 +2 This can be reduced using reduction formula or by inspection: ( 1, 1, 1, 1)(A1) + (1,-1, -1, 1) (B2) = (2, 0, 0, 2) Gstre = A1 + B2

Analysis of Vibrational Modes: 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

S O O Symmetry Adapted Linear Combinations r1 r2 Pick a generating vector eg: r1 How does this transform under symmetry operations? s(xz) s(yz) C2v E C2 r1 r1 r2 r2 r1 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:

S S O O O O 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: S S O O O O B2 A1 Sulphur must also move to maintain position of centre of mass

S O O Analysis of Vibrational Modes: Remaining mode “likely” to be a bend O O s(xz) s(yz) C2v E C2 Gbend 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

Analysis of Vibrational Modes: SO2 experimental data. IR(Vapour)/cm-1 Raman(liquid)/cm-1 Sym Name 518 524 A1 bend n1 1151 1145 A1 stretch n2 1362 1336 B2 stretch n3

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

Analysis of Vibrational Modes: POCl3 Angle deformations P-Cl stretch P=O stretch Gvibe = 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

Analysis of Vibrational Modes: POCl3 Gvibe = 3A1 + 3E C3v E 2C3 3sv G P=O str 1 1 1 G P-Cl str 3 1 G bend 6 2 Using reduction formulae or by inspection: G P=O str = A1 and G P-Cl str = A1 + E G bend = Gvibe - G P=O str - G P-Cl str = 3A1 + 3E – 2A1 – E = A1 + 2E Reduction of the representation for bends gives: G bend = 2A1 + 2E

Analysis of Vibrational Modes: POCl3 G bend = Gvibe - G P=O str - G P-Cl str = 3A1 + 3E – 2A1 – E = A1 + 2E Reduction of the representation for bends gives: G bend = 2A1 + 2E One of the A1 terms is REDUNDANT as not all the angles can symmetrically increase G 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.

Analysis of Vibrational Modes: POCl3 IR (liq)/ cm-1 Raman /cm-1 Description Sym Label 1292 1290(pol) P=O str( 1,4) A1 n1 580 582 P-Cl str(2,3) E n4 487 486(pol) P-Cl str( 1,2,3) n2 340 337 deformation n5 267 267(pol) Sym. Deformation(1) n3 - 193 n6

Analysis of Vibrational 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

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.

Symmetry, Bonding and Electronic Spectroscopy s bonding in AXn molecules e.g. : water How do 2s and 2p orbitals transform?

Symmetry, Bonding and Electronic Spectroscopy 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

Symmetry, Bonding and Electronic Spectroscopy How do the 2p orbitals transform?

Symmetry, Bonding and Electronic Spectroscopy How do the 2p orbitals transform?

Symmetry, Bonding and Electronic Spectroscopy 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 s-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

Symmetry, Bonding and Electronic Spectroscopy Use the 1s orbitals on the hydrogen atoms

Symmetry, Bonding and Electronic Spectroscopy Assume oxygen 2s orbitals are non bonding Oxygen 2pz is a1, px is b1 and py is b2 Ligand orbitals are a1 and b2 Orbitals of like symmetry can interact Oxygen 2px is “wrong” symmetry therefore likely to be non-bonding Which is lower in energy a1 or b2? Guess that it is a1 similar symmetry better interaction?

Symmetry, Bonding and Electronic Spectroscopy 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 f1 or f2 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)

pz py

Symmetry, Bonding and Electronic Spectroscopy The overlap between the a1 orbitals (y1) is greater than that for the b2 orbitals (y2). 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 states arising from each electronic configuration. Note: Electronic states and configurations are NOT the same thing!

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 configuration is 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 degenerate MO’s, the symmetry is ALWAYS A1

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

Symmetry of Electronic States from NON-DEGENERATE MO’s. What happens if we promote an electron? First two excitations move an electron form b1 non bonding Into either the b2* or a1* anti-bonding orbitals . a1* Anti Bonding Both of these transitions are non bonding to anti bonding transitions. n-p* b2* b1 Non bonding b2 Bonding a1

What electronic states do these new configurations generate? (a1)2(b2)2(b1)1(b2*)1(a1*)0 = A1.A1.B1.B2 = A2 (a1)2(b2)2(b1)1(b2*)0(a1*)1 = 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

What electronic states do these new configurations generate? Molecular Orbitals Electronic States a1* 1B1 b2* 1A2 3B1 b1 3A2 b2 a1 1A1

What electronic states do these new configurations generate? 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.

Which electronic transitions are allowed? Vibrational Spin Electronic To first approximation m can only operate on the electronic part of 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.

Which electronic transitions are allowed? 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 m transforms as the translational species as for infrared transitions.

Which electronic transitions are allowed? For the example of H20 the direct products for the electronic transition are The totally symmetric species is only present for the transition to 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”.

Which electronic transitions are allowed?

Which electronic transitions are allowed? Transitions between a totally symmetric ground state and one with an electronic state that has the same symmetry as a component of m, will be symmetry allowed. Caution: The lowest energy transition may be allowed but too weak to be observed. Caution: Ground state is not always totally symmetric and beware of degenerate representations.

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 Gs(ligands) = a1g + eg + t1u Gp(ligands) = t1g + t2g + t1u + t2u

AX6 for Oh 4p 4s 3d t1u* a1g* t1u a1g eg* t2g eg + t2g eg a1g + eg + t1u t1u a1g

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 2T2g Under Oh the transition dipole moment transforms as t1u Are electronic transitions allowed between these states?

Electronic Spectroscopy of d9 complex: Need to calculate direct product representation: 2Eg . (t1u) . 2T2g Oh E 8C3 6C2 6C4 3C2 i 6S4 8S6 3sh 6sd T2g 3 1 -1 t1u -3 Eg 2 DP 18 -18 -2

Electronic Spectroscopy of d9 complex: DP 18 2 -18 -2 Use reduction formula: 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.

Electronic Spectroscopy of d9 complex: 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.

Are you familiar with symmetry elements operations? Can you assign a point group? Can you use a basis of 3 vectors to generate G3n ? Do you know the reduction formula? What is the difference between a reducible and irreducible representation? Can you reduce G3n ? Can you generate Gvib from G3n ? 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?

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