Presentation on theme: "Page 1 Symmetry and Group Theory Feature: Application for Spectroscopy and Orbital Molecules Dr. Indriana Kartini."— Presentation transcript:
Page 1 Symmetry and Group Theory Feature: Application for Spectroscopy and Orbital Molecules Dr. Indriana Kartini
Page 2 P. H. Walton “Beginning Group Theory for Chemistry” Oxford University Press Inc., New York, 1998 ISBN A.F.Cotton “ Chemical Applications of Group Theory” ISBN 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
Page 4 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 SimbolUnsurOperasi E Unsur identitasMembiarkan obyek tidak berubah CnCn Sumbu rotasiRotasi seputar sumbu dengan derajat rotasi 360/n (n adalah bilangan bulat) Bidang simetriRefleksi melalui bidang simetri i Pusat/titik inversiProyeksi melewati pusat inversi ke sisi seberangnya dengan jarak yang sama dari pusat SnSn Sumbu rotasi tidak sejati (Improper rotational axis) Rotasi mengitari sumbu rotasi diikuti dengan refleksi pada bidang tegak lurus sumbu rotasi
Page 7 Rotate 120 O F1F1 F1F1 F2F2 F3F3 F3F3 F2F2 Operation rotation by 360/3 around C 3 axis (element) BF 3 Rotations 360/n where n is an integer
Page 8 (xz) (yz) z y x x is out of the plane Reflection is the operation element is plane of symmetry H2OH2O Reflections
Page 9 Reflections for H 2 O
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 Reflections vv hh dd dd XeF4
Page 12 XeF4
Page 13 Atom at (-x,-y,-z) Atom at (x,y,z) Inversion, i Centre of inversion i element is a centre of symmetry Inversion Examples: Benzene, XeF 4 Ethene
Page 14 C4C4 S 4 Improper Rotation Rotate about C 4 axis and then reflect perpendicular to this axis S4S4
Page 15 S 4 Improper Rotation
Page 16 successive operation
Page 17 KULIAH MINGGU II TEORI GRUP
Page 18 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
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 O h Tetrahedral point group symbol T d Linear having no i C Linear having i D h
Page 20 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 O h Tetrahedral point group symbol T d Linear having no i C Linear having i D h
Page 21 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 C 1 If having one reflection plane point group symbol C s If having i C i
Page 22 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 C n (n is the order of the principle axis) If having n h point group symbol C nh If having n v C nv If having an S 2n axis coaxial with principal axis S 2n
Page 23 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 D n Having n d bisecting the C 2 axes D nd Having one h D nh
Page 24 Molecule Linear ? i? DhDh CvCv 2 or more C n, n>2? i? TdTd C5?C5?IhIh OhOh Cn?Cn? Select C n with highest n, nC 2 perpendicular to C n ? * * h?h? D nh nd?nd? D nd DnDn ?? CsCs i? CiCi C1C1 h?h? C nh nv?nv? C nv S 2n ?S 2n CnCn Y N
Page 25 Benzene Linear ? i? DhDh CvCv 2 or more C n, n>2? i? TdTd C5?C5?IhIh OhOh Cn?Cn? Select C n with highest n, nC 2 perpendicular to C n ? * * h?h? D nh nd?nd? D nd DnDn ?? CsCs i? CiCi C1C1 h?h? C nh nv?nv? C nv S 2n ?S 2n CnCn Y N n = 6Benzene is D 6h
Page 26 Tugas I: Symmetry and Point Groups Tentukan unsur simetri dan grup titik pada molekul a. N 2 F 2 b. POCl 3 Gambarkan geometri masing-masing molekul tersebut
Page 27 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 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
Page 29 Example of Group Properties B(OH) 3 belongs to C 3 point group It has E, C 3 and C 3 2 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 C3C3 C3C3 Overall: C 3 followed C 3 gives C 3 2
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 groupAE= EA =A C32C32 C32C32 E. C 3 2 = C 3 2 and C 3 2. C 3 = E and C 3 2. C 3 2 = C 3
Page 32 The combination of elements in the group must be ASSOCIATIVE A(BC) = AB(C) = ABC Multiplication need not be commutative (ie: AC CA) C 3.(C 3.C 3 2 ) = (C 3. C 3 ) C 3 2 (Do RHS First) C 3. C 3 2 = E ; C 3.E = C 3 C 3.C 3 = C 3 2 ; C 3 2. C 3 2 = C 3 Operations are associative and E, C 3 and C 3 2 form a group
Page 33 Group Multiplication Table C3C3 E C3C3 C32C32 EE C3C3 C32C32 C3C3 C3C3 C32C32 E C32C32 C32C32 E C3C3 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 C 3 2 is C 3 The inverse of C 3 is C 3 2
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 C 2v : [E, C 2, xz, yz ] These are useful to describe molecular vibrations and electronic transitions.
Page 39 S O O S O O C2C2 A unit vector on each atom represents translation in the y direction C 2.(T y ) = (-1) T y E.(T y ) = (+1) T y yz.(T y ) = (+1) T y xz.(T y ) = (-1) T y Constructing the Representation
Page 40 S O O Constructing the Representation A unit vector on each atom represents rotation around the z(C 2 ) axis C 2.(R z ) = (+1) R z E.(R Z ) = (+1) R z yz.(R z ) = (-1) R z xz.(R Z ) = (-1) R Z
Page 41 Constructing the Representation C 2v EC2C2 (xz) (yz) +1 TzTz RzRz +1+1T x,R y +1 +1T y,R x
Page 42 S O O Constructing the Representation Use a mathematical function Eg: p y orbital on S C 2v EC2C2 (xz) (yz) +1 +1T y,R x p y has the same symmetry properties as T y and R x vectors
Page 43 Constructing the Representation Au hh h.[d x 2 -y 2 ] = (+1).[d x 2 -y 2 ] C .[d x 2 -y 2 ] = (-1).[d x 2 -y 2 ] C4C4 [AuCl 4 ] -
Page 44 Constructing the Representation D 4h E2C 4 C2C2 2C 2 ’2C 2 ”I2S 4 hh 2v2v 2d2d Effects of symmetry operations generate the TRANSFORM MATRIX For all the symmetry operations of D 4h on [d x 2 -y 2 ] We have: Simple examples so far.
Page 45 Constructing the Representation: The TRANSFORMATION MATRIX Examples can be more complex: e.g. the p x and p y orbitals in a system with a C 4 axes. X Y C4C4 p x p x ’ p y p y p y ’ p x In matrix form: A 2x2 transformation matrix
Page 46 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.
Page 47 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 -1 C=E
Page 48 Constructing the Representation Generally a reducible representation A can be reduced such That each element B i is a matrix belonging to an irreducible representation. All elements outside the B i blocks are zero This can generate very large matrices. However, all information is held in the character of these matrices
Page 49 Character Tables Character, = a 11 + a 22 + a 33. In general And only the character , which is a number is required and NOT the whole matrix.
Page 50 Character Tables an Example C 3v : (NF 3 ) C 3v EC31C31 C32C32 vv vv vv TzTz 111 RzRz 2 000(T x,T y ) or (R x,R y ) This simplifies further. Some operations are of the same class and always have the same character in a given irreducible representation C 3 1, C 3 1 are in the same class v, v, v are in the same class
Page 51 Character Tables an Example C 3v : (NF 3 ) C 3v E2C 3 v A1A1 111TzTz x 2 + y 2 A2A2 11RzRz E2 0(T x,T y ) or (R x,R y )(x 2, y 2, 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 T x, T y, T z, and R x, R y, R z, provide the basis. (e) Shows how functions that are binary combinations of x,y,z (xy or z 2 ) 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 being and
Page 56 Mulliken symbols: Labelling A one dimensional irreducible representation is labelled A if it is symmetric with respect to rotation about the highest order axis C n. (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 C 2 axis perpendicular to C n or (in the absence of such an axis) to reflection in a v 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) 1) 2)
Page 57 Mulliken symbols: Labelling 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. 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 z2z2 y2y2 x2x2 Generating Reducible Representations x1x1 xsxs y1y1 ysys zszs z1z1 xz For the symmetry operation xz (a v ) x 1 x 2 x 2 x 1 x s x s y 1 -y 2 y 2 -y 1 y s -y s z 1 z 2 z 2 z 1 z s z s
Page 59 Generating Reducible Representations In matrix form
Page 60 Only require the characters: The sum of diagonal elements For (xz) = + 1
Page 61 For (yz) = + 3
Page 62 For E = + 9
Page 63 For C 2 = -1
Page 64 Generating Reducible Representations C 2v 3n EC2C2 (xz) (yz) Summarising we get that 3n for this molecule is: C 2v EC2C2 (xz) (yz) A1A1 +1 TzTz x 2, y 2, z 2 A2A2 +1 RzRz xy B1B1 +1+1T x, R x xz B2B T y, R y 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: Where a p 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 n R is the number of operations in the class
Page 66 C2vC2v 1E1E1C21C2 (xz) (yz) A1A1 +1 TzTz x 2, y 2, z 2 A2A2 +1 RzRz xy B1B1 +1+1T x, R x xz B2B T y, R y yz C 2v 3n EC2C2 (xz) (yz) For C 2v ; g = 4 and n R = 1 for all operations
Page 67 a A 1 = (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x1) + (1x3x1)] = (12/4) =3 C 2v 3n EC2C2 (xz) (yz) a A 2 = (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x-1) + (1x3x-1)] = (4/4) =1 a B 1 = (1/4)[ ( 1x9x1) + (1x-1x-1) + (1x1x1) + (1x3x-1)] = (8/4) =2 a B 2 = (1/4)[ ( 1x9x1) + (1x-1x-1) + (1x1x-1) + (1x3x1)] = (12/4) =3 3n = 3A 1 + A 2 + 2B 1 + 3B 2
Page 68 a A 1 = (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x1) + (1x3x1)] = (12/4) =3 Reducing Reducible Representations 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 SO 2 (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 For each un-shifted atom (E) = +3 z y x z1z1 y1y1 x1x1
Page 70 Inversion i z y x z1z1 y1y1 x1x1 i For each un-shifted atom (i) = -3
Page 71 For each un-shifted atom z1z1 y1y1 x1x1 x z y (xz) Reflection (xz) (Others are same except location of –1 changes) ( (xz) ) = +1
Page 72 360/n x1x1 y1y1 z1z1 z y x CnCn Rotation C n (C n ) = cos(360/n)
Page 73 (S n ) = cos(360/n) Improper rotation axis, S n CnCn (xy) z y x z’ y’x’ y1y1 x1x1 z1z1
Page 74 Summary of contributions from un-shifted atoms to 3n R (R) E+3 i-3 cos(360/n) C cos(360/n) C 3,C cos(360/n) C 4, C cos(360/n) S 3 1,S cos(360/n) S 4 1,S cos(360/n) S 6 1,S 6 5 0
Page 75 Worked example: POCl 3 (C 3v point group) R (R) E vv 2C C 3v E v A1A1 A2A2 E C3C3 Un-shifted atoms Contribution 3n Number of classes, ( = 6) Order of the group, g = 6
Page 76 Reducing the irreducible representation for POCl 3 2C 3 C 3v E v 3n 1503 a(A 1 ) = 1/6[(1x 15x1) + (2 x 0 x 1) + (3 x 3x 1)] = 1/6 [ ] = 4 a(A 2 ) = 1/6[(1 x 15 x 1) + ( 2 x 0 x 1) + (3 x 3x –1)] = 1/6 [ ] = 1 a(E) = 1/6[ (1 x 15 x 2) + (2 x 0 x –1) + (3 x 3 x 0)] = 1/6[ ] =5 3n = 4A 1 + A 2 + 5E For POCl 3 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-XII APLIKASI TEORI GRUP
Page 78 C2vC2v 1E1E1C21C2 (xz) (yz) A1A1 +1 TzTz x 2, y 2, z 2 A2A2 +1 RzRz xy B1B1 +1+1T x, R x xz B2B T y, R y yz C 2v 3n EC2C2 (xz) (yz) 3n = 3A 1 + A 2 + 2B 1 + 3B 2 Group Theory and Vibrational Spectroscopy: SO 2
Page 79 Group Theory and Vibrational Spectroscopy: SO 2 3n = 3A 1 + A 2 + 2B 1 + 3B 2 = = 9 = 3n C2vC2v 1E1E1C21C2 (xz) (yz) A1A1 +1 TzTz x 2, y 2, z 2 A2A2 +1 RzRz xy B1B1 +1+1T x, R x xz B2B T y, R y yz For non linear molecule there are 3n-6 vibrational degrees of freedom rot = A 2 + B 1 + B 2 trans = A 1 + B 1 + B 2 vib = 3n – rot – trans vib = 2A 1 + B 2 (Degrees of freedom = = 3 = 3n-6) 3n = 3A 1 + A 2 + 2B 1 + 3B 2
Page 80 Group Theory and Vibrational Spectroscopy: POCl 3 3n = 4A 1 + A 2 + 5E trans = A 1 + E rot = A 2 + E vibe = 3A 1 + 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 3A 1 + 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 ff ii 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 T x, T y and T z. The excited state transforms the same as the vectors that describe the vibrational mode.
Page 85 The DIRECT PRODUCT representation. vib = 2A 1 + B 2 For SO 2 we have that: Under C 2v : T x, T y and T z transform as B 1, B 2 and A 1 respectively.
Page 87 The DIRECT PRODUCT representation Group theory predicts only A 1 and B 2 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. vib = 2A 1 + B 2 We predict three bands in the infrared spectrum of SO 2
Page 88 Infrared Spectroscopy : General Rule If a vibrational mode has the same symmetry properties as one or more translational vectors(T x,T y, or T z ) 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 ff ii Wavefunction final state Wavefunction initial state Virtual state Raman Spectroscopy Stokes Shifted
Page 91 ff ii Wavefunction intial state Wavefunction final state Virtual state Raman Spectroscopy Anti-Stokes Shifted
Page 92 Raman Spectroscopy Probability of a Raman transistion: The operator, , is the polarisability tensor For vibrational transitions ij = ji so there are six distinct components: x 2, y 2, z 2, xy, xz and yz
Page 93 x 2, y 2, z 2, xy, xz and yz Raman Spectroscopy For C 2v Transform as: A 1, A 1, A 1, A 2, B 1 and B 2 We can then evaluate the direct product representation in a broadly analagous way
Page 94 Raman Spectroscopy The DIRECT PRODUCT representation For SO 2 group theory predicts only A 1 and B 2 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 SO 2 Note: A 1 modes are polarised
Page 95 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 A 1 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.
Page 96 Analysis of Vibrational Modes: Vibrations can be classified into Stretches, Bends and Deformations For SO 2 vib = 2A 1 + B 2 We could choose more “natural” co-ordinates S OO z y x r2r2 r1r1 Determine the representation for stretch
Page 97 Analysis of Vibrational Modes: S OO r2r2 r1r1 How does our new basis transform Under the operations of the group? Vectors shifted to new position contribute zero Unshifted vectors contribute + 1 to (R) C 2v stre EC2C2 (xz) (yz) +200 This can be reduced using reduction formula or by inspection: stre = A 1 + B 2 ( 1, 1, 1, 1)(A 1 ) + (1,-1, -1, 1) (B 2 ) = (2, 0, 0, 2)
Page 98 Analysis of Vibrational Modes: Two stretching vibrations exist that transform as A 1 and B 2. 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 A 1 and B 2
Page 99 Symmetry Adapted Linear Combinations S OO r2r2 r1r1 C 2v r1r1 EC2C2 (xz) (yz) r1r1 r2r2 r2r2 r1r1 Pick a generating vector eg: r 1 How does this transform under symmetry operations? Multiply this by the characters of A 1 and B 2 For A 1 this gives:(+1) r 1 + (+1) r 2 + (+1) r 2 + (+1) r 1 = 2r 1 + 2r 2 Normalise coefficients and divide by sum of squares:
Page 100 Symmetry Adapted Linear Combinations For B 2 this gives:(+1) r 1 + (-1) r 2 + (-1) r 2 + (+1) r 1 = 2r 1 - 2r 2 Normalise coefficients and divide by sum of squares: S OO S OO A1A1 B2B2 Sulphur must also move to maintain position of centre of mass
Page 101 Analysis of Vibrational Modes: S OO Remaining mode “likely” to be a bend C 2v bend EC2C2 (xz) (yz) 1111 By inspection this bend is A 1 symmetry SO 2 has three normal modes: A 1 stretch: Raman polarised and infrared active A 1 bend: Raman polarised and infrared active B 2 stretch: Raman and infrared active
Page 102 Analysis of Vibrational Modes: SO 2 experimental data. IR(Vapour)/cm -1 Raman(liquid)/cm -1 SymName A 1 bend A 1 stretch B 2 stretch 3
Page 103 Analysis of Vibrational Modes: SO 2 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. A 1 first and then B 2
Page 104 Analysis of Vibrational Modes: POCl 3 P=O stretch P-Cl stretch Angle deformations vibe = 3A 1 + 3E 3 A 1 vibrations IR active(T z ) + Raman active polarised( x 2 + y 2 and z 2 ) 3 E vibrations IR active(T x,T y ) + Raman active ( x 2 - y 2, xy) (yz,zx) Six bands, Six co-incidences
Page 105 Analysis of Vibrational Modes: POCl 3 vibe = 3A 1 + 3E C 3v E2C 3 3v3v P=O str P-Cl str bend Using reduction formulae or by inspection: P=O str = A 1 and P-Cl str = A 1 + E bend = vibe - P=O str - P-Cl str = 3A 1 + 3E – 2A 1 – E = A 1 + 2E Reduction of the representation for bends gives: bend = 2A 1 + 2E
Page 106 Analysis of Vibrational Modes: POCl 3 bend = vibe - P=O str - P-Cl str = 3A 1 + 3E – 2A 1 – E = A 1 + 2E Reduction of the representation for bends gives: bend = 2A 1 + 2E One of the A 1 terms is REDUNDANT as not all the angles can symmetrically increase bend = A 1 + 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 108 Analysis of Vibrational Modes: POCl 3 1) All polarised bands are Raman A 1 modes. 2) Highest frequencies probably stretches. 3) P-Cl stretches probably of similar frequency. 4)Double bonds have higher frequency than similar single bonds. A 1 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 Symmetry, Bonding and Electronic Spectroscopy bonding in AX n molecules e.g. : water How do 2s and 2p orbitals transform?
Page 111 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 a 1 symmetry in the C 2v point group
Page 112 Symmetry, Bonding and Electronic Spectroscopy How do the 2p orbitals transform?
Page 113 Symmetry, Bonding and Electronic Spectroscopy How do the 2p orbitals transform?
Page 114 Symmetry, Bonding and Electronic Spectroscopy How do the 2s and 2p orbitals transform? Oxygen 2s and 2p z transforms as a 1 2p x transforms as b 1 and 2p y as b 2 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 Symmetry, Bonding and Electronic Spectroscopy Use the 1s orbitals on the hydrogen atoms
Page 116 Symmetry, Bonding and Electronic Spectroscopy Assume oxygen 2s orbitals are non bonding Oxygen 2p z is a 1, p x is b 1 and p y is b 2 Ligand orbitals are a 1 and b 2 Which is lower in energy a 1 or b 2 ? Guess that it is a 1 similar symmetry better interaction? Orbitals of like symmetry can interact Oxygen 2px is “wrong” symmetry therefore likely to be non-bonding
Page 118 Symmetry, Bonding and Electronic Spectroscopy Is symmetry sufficient to determine ordering of a 1 and b 2 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 pzpz pypy
Page 120 Symmetry, Bonding and Electronic Spectroscopy The overlap between the a 1 orbitals ( is greater than that for the b 2 orbitals ( Therefore a 1 is lower in energy than b 1. 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!
Page 122 Symmetry of Electronic States from NON-DEGENERATE MO’s. The ground electronic configuration for water is: (a 1 ) 2 (b 2 ) 2 (b 1 ) 2 (b 2 *) 0 (a 1 *) 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 (a 1 ) 2 = a 1.a 1 = A 1 (b 2 ) 2 = b 2.b 2 = A 1 (b 1 ) 2 = b 1.b 1 = A 1 A 1.A 1.A 1 = A 1 For FULL singly degenerate MO’s, the symmetry is ALWAYS A 1
Page 123 Symmetry of Electronic States from NON-DEGENERATE MO’s. For FULL singly degenerate MO’s, the symmetry is ALWAYS A 1 (The totally symmetric species of the point group) For orbitals with only one electron: (a 1 ) 1 = A 1, (b 2 ) 1 = B 2, (b 1 ) 1 =B 1 General rule: For full MO’s the ground state is always totally symmetric
Page 124 Symmetry of Electronic States from NON-DEGENERATE MO’s. What happens if we promote an electron? a1a1 b1b1 b2b2 b2*b2* a1*a1* Bonding Non bonding Anti Bonding First two excitations move an electron form b 1 non bonding Into either the b 2 * or a 1 * anti-bonding orbitals. Both of these transitions are non bonding to anti bonding transitions. n- *
Page 125 What electronic states do these new configurations generate? (a 1 ) 2 (b 2 ) 2 (b 1 ) 1 (b 2 *) 1 (a 1 *) 0 (a 1 ) 2 (b 2 ) 2 (b 1 ) 1 (b 2 *) 0 (a 1 *) 1 = A 1.A 1.B 1.B 2 = A 2 = A 1.A 1.B 1.A 1 = B 1 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: 3 A 2, 1 A 2 and 3 B 1, 1 B 1 electronic states. Note: if S= ½ then we have a doublet state
Page 126 a1a1 b1b1 b2b2 b2*b2* a1*a1* What electronic states do these new configurations generate? Molecular Orbitals 1A11A1 1B11B1 3B13B1 1A21A2 3A23A2 Electronic States
Page 127 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.
Page 128 Electronic Which electronic transitions are allowed? VibrationalSpin To first approximation 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.
Page 129 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 transforms as the translational species as for infrared transitions.
Page 130 Which electronic transitions are allowed? For the example of H 2 0 the direct products for the electronic transition are The totally symmetric species is only present for the transition to the B 1 state. Therefore the transition to the A 2 state is “symmetry forbidden” Transitions between singlet states are “spin allowed”. transitions between singlet and triplet state are “spin forbidden”.
Page 131 Which electronic transitions are allowed?
Page 132 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 , 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 weak to be observed.
Page 133 More bonding for AX 6 molecules / complexes In the case of O h point group: d x 2 -y 2 and d z 2 transform as e g d xy, d yz and d zx transform as t 2g p x, p y and p z transform as t 1u (ligands) = a 1g + e g + t 1u (ligands) = t 1g + t 2g + t 1u + t 2u
Page 134 t 1u a 1g e g + t 2g t 1u t 1u * a 1g a 1g * eg*eg* egeg t 2g a 1g + e g + t 1u AX 6 for O h 4p 4s 3d
Page 135 Electronic Spectroscopy of d 9 complex: [Cu(H 2 O) 6 ] 2+ is a d 9 complex. That is approximately O h. Ground electronic configuration is: (t 2g ) 6 (e g * ) 3 Excited electronic configuration is : (t 2g ) 5 (e g * ) 4 The ground electronic state is 2 E g Excited electronic state is 2 T 2g Under O h the transition dipole moment transforms as t 1u Are electronic transitions allowed between these states?
Page 136 Electronic Spectroscopy of d 9 complex: Need to calculate direct product representation: 2 E g. (t 1u ). 2 T 2g OhOh E 8C38C3 6C26C2 6C46C4 3C23C2 i 6S46S4 8S68S6 3h3h 6d6d T 2g t 1u EgEg DP
Page 137 Electronic Spectroscopy of d 9 complex: DP Use reduction formula: a a 1g = 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 Electronic Spectroscopy of d 9 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.
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 SO 2 ? 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?