Presentation on theme: "Symmetry and Group Theory Feature: Application for Spectroscopy and Orbital Molecules Dr. Indriana Kartini."— Presentation transcript:
1 Symmetry and Group Theory Feature: Application for Spectroscopy and Orbital Molecules Dr. Indriana Kartini
2 Text books P. H. Walton “Beginning Group Theory for Chemistry” Oxford University Press Inc., New York, 1998ISBNA.F.Cotton “ Chemical Applications of Group Theory”ISBN
3 Marks 80% exam: 10% group assignments of 4 students 40% mid50% final10% group assignments of 4 studentsSyllabus pre-mid: Prinsip dasarOperasi dan unsur simetriSifat grup titik dan klasifikasi molekul dalam suatu grup titikMatriks dan representasi simetriTabel karakterSyllabus Pasca-mid: Aplikasiprediksi spektra vibrasi molekul: IR dan Ramanprediksi sifat optik molekulprediksi orbital molekul ikatan molekul
4 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 awalnyaUnsur simetriSuatu titik, garis atau bidang sebagai basis operasi simetri
5 E Cn s i Sn Simbol Unsur Operasi Unsur identitas Membiarkan obyek tidak berubahCnSumbu rotasiRotasi seputar sumbu dengan derajat rotasi 360/n (n adalah bilangan bulat)sBidang simetriRefleksi melalui bidang simetriiPusat/titik inversiProyeksi melewati pusat inversi ke sisi seberangnya dengan jarak yang sama dari pusatSnSumbu rotasi tidak sejati (Improper rotational axis)Rotasi mengitari sumbu rotasi diikuti dengan refleksi pada bidang tegak lurus sumbu rotasi
10 Reflections Principle (highest order) axis is defined as Z axis After Mullikens(xz) in plane perpendicular to molecular planes(yz) in plane parallel to molecular planeboth examples of svsv : reflection in plane containing highest order axissh : reflection in plane perpendicular to highest order axissd : dihedral plane generally bisecting sv
18 Mathematical Definition: Group Theory A group is a collection of elements having certain propertiesthat enables a wide variety of algebraic manipulations to becarried out on the collectionBecause of the symmetry of molecules they canbe assigned to a point group
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 2YES: Octahedral point group symbol OhTetrahedral point group symbol TdLinear having no i CLinear having i Dh
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 2YES: Octahedral point group symbol OhTetrahedral point group symbol TdLinear having no i CLinear having i Dh
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 3NO:If no other symmetry elements point group symbol C1If having one reflection plane point group symbol CsIf having i Ci
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 4NO:If no other symmetry elements point group symbol Cn (n is the order of the principle axis)If having n h point group symbol CnhIf having n v CnvIf having an S2n axis coaxial with principal axis S2n
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 DnHaving n d bisecting the C2 axes DndHaving one h Dnh
24 Molecule * * Y N Dh Cv i? i? Select Cn with highest n, Linear?DhCv2 or moreCn, n>2?i?i?IhC5?OhTdY*Cn?*Dnhsh?Select Cn with highest n,nC2 perpendicular to Cn?NDndnsd?Dns?CsCnhsh?Cii?C1Cnvnsv?S2nS2n?Cn
25 Benzene * * Y Benzene is D6h n = 6 N Dh Cv i? i? Linear?DhCv2 or moreCn, n>2?i?i?IhC5?OhTdY*Cn?Benzeneis D6hn = 6*Dnhsh?Select Cn with highest n,nC2 perpendicular to Cn?NDndnsd?Dns?CsCnhsh?Cii?C1Cnvnsv?S2nS2n?Cn
26 Tugas I: Symmetry and Point Groups Tentukan unsur simetri dan grup titik pada molekulN2F2POCl3Gambarkan geometri masing-masing molekul tersebut
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 collectionAB = C where A, B and C are all members of the collectionThere must be an IDENTITY ELEMENT (E)AE = A for all members of the collectionE commutes with all other members of the groupAE= EA =AThe combination of elements in the group must be ASSOCIATIVEA(BC) = AB(C) = ABCMultiplication need not be commutative (ie: ACCA)Every member of the group must have an INVERSE which is also a member of the group.AA-1 = E
29 Example of Group Properties B(OH)3 belongs to C3 point groupIt has E, C3 and C32symmetry operations
30 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 collectionAB = C where A, B and C are all members of the collectionC3C3Overall: C3 followed C3 gives C32
31 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 collectionE commutes with all other members of the group AE= EA =AC32C32E. C32 = C32 and C32. C3 = E and C32. C32 = C3
32 Operations are associative and E, C3 and C32 form a group The combination of elements in the group must be ASSOCIATIVEA(BC) = AB(C) = ABCMultiplication need not be commutative (ie: ACCA)C3 .(C3 .C32 )= (C3 .C3) C32(Do RHS First)C3.C32 = E ; C3 .E = C3C3 .C3 = C32 ; C32 .C32 = C3Operations are associative and E, C3 and C32 form a group
33 Group Multiplication Table Order of the group =3C3EC32Every member of the group musthave an INVERSE which is alsoa member of the group.AA-1 = EThe inverse of C32 is C3The inverse of C3 is C32
37 Representations of Groups Diagrams are cumbersomeRequire numerical methodAllows mathematical analysisRepresent by VECTORS or Mathematical FunctionsAttach Cartesian vectors to moleculeObserve the effect of symmetry operations on these vectorsVectors 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]
38 Constructing the Representation Put unit vectors on each atomzSOOyxC2v: [E, C2, sxz, syz]These are useful to describe molecular vibrationsand electronic transitions.
39 Constructing the Representation A unit vector on each atom represents translation in the y directionC2SSOOOOC2.(Ty) = (-1) Ty E .(Ty) = (+1) Tysyz .(Ty) = (+1) Ty sxz .(Ty) = (-1) Ty
40 Constructing the Representation A unit vector on each atom represents rotation around the z(C2) axisSOOC2.(Rz) = (+1) Rz E .(RZ) = (+1) Rzsyz .(Rz) = (-1) Rz sxz .(RZ) = (-1) RZ
41 Constructing the Representation C2vEC2s(xz)s(yz)+1Tz-1RzTx,RyTy,Rx
42 C2v +1 -1 Ty,Rx Constructing the Representation Use a mathematical functionEg: py orbital on SSOOC2vEC2s(xz)s(yz)+1-1Ty,Rxpy has the same symmetry properties as Ty and Rx vectors
44 Constructing the Representation Effects of symmetry operations generate theTRANSFORM MATRIXSimple examples so far.For all the symmetry operations of D4h on [d x2-y2]We have:D4hE2C4C22C2’2C2”I2S4sh2sv2sd+1-1
45 Constructing the Representation: The TRANSFORMATION MATRIXExamples can be more complex:e.g. the px and py orbitals in a system with a C4 axes.YC4px px’ pypy py’ pxXA 2x2 transformationmatrixIn matrix form:
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 REPRESENTATIONSAny REDUCIBLE representation is the SUM of the set of IRREDUCIBLE representations.
47 Constructing the Representation If a matrix belongs to a reducible representation it can be transformedso that zero elements are distributed about the diagonalSimilarity TransformationA goes to BThe similarity transformation is such thatC-1 AC = B where C-1C=E
48 Constructing the Representation Generally a reducible representation A can be reduced suchThat each element Bi is a matrix belonging to an irreducible representation.All elements outside the Bi blocks are zeroThis can generate very large matrices.However, all information is held in the character of these matrices
49 Character TablesCharacter , = a11 + a22 + a33.In generalAnd only the character , which is a number is required andNOT the whole matrix.
50 Character Tables an Example C3v : (NF3) sv1Tz-1Rz2(Tx,Ty) or (Rx,Ry)This simplifies further. Some operations are of the same class and always have thesame character in a given irreducible representationC31, C31 are in the same classsv, sv, sv are in the same class
51 Character Tables an Example C3v : (NF3) 3svA11Tzx2 + y2A2-1Rz2(Tx,Ty) or (Rx,Ry)(x2, y2, xy) (yz, zx)There is a nomenclature for irreducible representations: Mulliken SymbolsA is single and E is doubly degenerate (ie x and y are indistinguishable)
52 Note:You will not be asked to generate character tables.These can be brought/supplied in the examination
54 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 classof operation.(d) Shows the irreducible representation for which the six vectorsTx, 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
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 aregiven the symbol S with two and three dimensional representationsbeing P and D.
56 Mulliken symbols: Labelling 1)A one dimensional irreducible representation is labelled A if it is symmetricwith 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 respectto 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)
57 Mulliken symbols: Labelling 3)Subscripts g (gerade) and u (ungerade) are given to irreducible representationsThat are symmetric and anti-symmetric respectively, with respect to inversionat a centre of symmetry.4)Superscripts ‘ and “ are given to irreducible representations that are symmetricand 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.
58 Generating Reducible Representations zsz1z2For the symmetry operation sxz (a sv )ysSx1 x x2 x1 xs xsy1 -y y2 -y1 ys -ysz1 z z2 z zs zsxsOy1Oy2x1x2sxz
59 Generating Reducible Representations In matrix form
60 Only require the characters: The sum of diagonal elements Only require the characters: The sum of diagonal elementsFor s(xz) c = + 1
64 Generating Reducible Representations Summarising we get that G3n for this molecule is:s(xz)s(yz)C2vEC2G3n+9-1+13To reduce this we need the character table for the point groupsC2vEC2s(xz)s(yz)A1+1Tzx2, y2, z2A2-1RzxyB1Tx , RxxzB2Ty , Ryyz
65 g is the number of symmetry operations in the group Reducing Reducible RepresentationsWe 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 groupc(R) is character of the reducible representationcp(R) is character of the irreducible representationnR is the number of operations in the class
66 For C2v ; g = 4 and nR = 1 for all operations 1s(xz)1s(yz)A1+1Tzx2, y2, z2A2-1RzxyB1Tx , RxxzB2Ty , Ryyzs(xz)s(yz)C2vEC2G3n+9-1+13For C2v ; g = 4 and nR = 1 for all operations
68 Reducing Reducible Representations aA1 = (1/4)[ ( 1x9x1) + (1x-1x1) + (1x1x1) + (1x3x1)] = (12/4) =3The terms in blue represent contributions from the un-shifted atomsOnly these actually contribute to the trace.If we concentrate only on these un-shifted atoms we cansimplify 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 atomsNumber of un-shifted atoms
69 Identity Ezz1Eyy1xx1For each un-shifted atomc(E) = +3
70 Inversion ix1zy1iyz1xFor each un-shifted atomc(i) = -3
71 Reflection s(xz) (Others are same except location of –1 changes) y1yxx1For each un-shifted atomc(s(xz)) = +1
73 Improper rotation axis, Sn z’zs(xy)Cnx’y’yx1y1xz1c(Sn) = cos(360/n)
74 Summary of contributions from un-shifted atoms to G3n c(R)E+3i-3s+11+ 2.cos(360/n) C2-11+ 2.cos(360/n) C3 ,C321+ 2.cos(360/n) C4, C43cos(360/n) S31,S32-2cos(360/n) S41,S42cos(360/n) S61,S65
75 Worked example: POCl3 (C3v point group) c(R)RE+3sv+1C3C3vE2C33svNumber of classes,( = 6)Order of the group,g = 6A1111A211-1E2-1Un-shiftedatoms523Contribution31G3n153
76 Reducing the irreducible representation for POCl3 3svC3v153G3na(A1) = 1/6[(1x 15x1) + (2 x 0 x 1) + (3 x 3x 1)] = 1/6 [ ] = 4a(A2) = 1/6[(1 x 15 x 1) + ( 2 x 0 x 1) + (3 x 3x –1)] = 1/6 [ ] = 1a(E) = 1/6[ (1 x 15 x 2) + (2 x 0 x –1) + (3 x 3 x 0)] = 1/6[ ] =5G3n = 4A1 + A2 + 5EFor POCl3 n= 5 therefore the number of degrees of freedom is 3n =15.E is doubly degenerate so G3n has 15 degrees of freedom.
78 Group Theory and Vibrational Spectroscopy: SO2 C2v1E1C21s(xz)1s(yz)A1+1Tzx2, y2, z2A2-1RzxyB1Tx , RxxzB2Ty , Ryyzs(xz)s(yz)C2vEC2G3n+9-1+13G3n = 3A1 + A2 + 2B1 + 3B2
79 Group Theory and Vibrational Spectroscopy: SO2 G3n = 3A1 + A2 + 2B1 + 3B2 = = 9 = 3nFor non linear molecule there are 3n-6 vibrational degrees of freedomC2v1E1C21s(xz)1s(yz)A1+1Tzx2, y2, z2A2-1RzxyB1Tx , RxxzB2Ty , RyyzGvib = G3n – Grot – G transG3n = 3A1 + A2 + 2B1 + 3B2Grot = A2 + B B2Gtrans = A B B2Gvib = 2A1 + B2(Degrees of freedom = = 3 = 3n-6)
80 Group Theory and Vibrational Spectroscopy: POCl3 G3n = 4A1 + A2 + 5EGtrans = A EThere are nine vibrational modes . (3n-6 = 9)The E modes are doubly degenerate andconstitute TWO modesGrot = A2 + EGvibe = 3A EThere are 9 modes that transform as 3A1 + 3E.These modes are linear combinations of the three vectorsattached to each atom.Each mode forms a BASIS for an IRREDUCIBLE representationof the point group of the molecule
81 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!!
82 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 momentThis is the transition dipole momentProbability is related to transition moment integral .
83 Infrared Spectroscopy Is the transition dipole moment operator andhas components: mx, my, mz.Note: Initial wavefunctionis always realyfWavefunction final stateyiWavefunction initial state
84 Infrared Spectroscopy Transition is forbidden if TM = 0Only non zero if direct product: yf m yicontains the totally symmetric representation.IE all numbers for c in representation are +1The ground state yi is always totally symmetricDipole moment transforms as Tx, Ty and Tz.The excited state transforms the same as the vectors that describe the vibrational mode.
85 The DIRECT PRODUCT representation. For SO2 we have that:Gvib = 2A1 + B2Under C2v :Tx, Ty and Tz transform as B1, B2 and A1 respectively.
87 The DIRECT PRODUCT representation Group theory predicts only A1 and B2 modesBoth of these direct product representations containthe totally symmetric species so they are symmetry allowed.This does not tell us the intensity only whether they are allowedor not.We predict three bands inthe infrared spectrum of SO2Gvib = 2A1 + B2
88 Infrared Spectroscopy : General Rule If a vibrational mode has the same symmetry propertiesas one or more translational vectors(Tx,Ty, or Tz) for thatpoint group, then the totally symmetric representation ispresent and that transitions will be symmetry allowed.Note:Selection rule tells us that the dipole changes during a vibrationand can therefore interact with electromagnetic radiation.
89 Raman effect depends on change in polarisability a. Raman SpectroscopyRaman effect depends on change in polarisability a.Measures how easily electron cloud can be distortedHow easy it is to induce a dipoleIntermediate is a virtual stateTHIS IS NOT AN ABSORPTIONUsually driven by a laser at w1.Scattered light at w2.Can be Stokes(lower energy) or Anti-Stokes shiftedMuch weaker effect than direct absorption.
90 Raman Spectroscopy Virtual state Stokes Shifted w1 w2 yf w =w1- w 2 yi Wavefunction final statew =w1- w 2yiWavefunction initial state
91 Raman Spectroscopy Virtual state w1 w2 Anti-Stokes Shifted yf Wavefunction intial statew =w2- w 1yiWavefunction final state
92 Raman Spectroscopy Probability of a Raman transistion: The operator , a , is the polarisability tensorFor vibrational transitions aij = ajiso there are six distinct components:ax2, ay2, az2, axy, axz and ayz
93 Raman Spectroscopy For C2v ax2, ay2, az2, axy, axz and ayz Transform as:A1, A1, A1, A2, B1 and B2We can then evaluate the direct product representationin a broadly analagous way
94 Raman Spectroscopy The DIRECT PRODUCT representation For SO2 group theory predicts only A1 and B2 modesBoth of these direct product representations containthe totally symmetric species so they are symmetry allowed.We predict three bands in the Raman spectrum of SO2Note: A1 modes are polarised
95 Raman Spectroscopy : General Rule If a vibrational mode has the same symmetry as on or moreOf the binary combinations of x,y and z the a transition fromthis mode will be Raman active.Any Raman active A1 modes are polarised.Infrared and Raman are based on two DIFFERENT phenomenaand therefore there is no necessary relationshipbetween the two activities.The higher the molecular symmetry the fewer “co-incidences”between Raman and infrared active modes.
96 S O O Analysis of Vibrational Modes: Vibrations can be classified into Stretches, Bends and DeformationsFor SO2 Gvib = 2A1 + B2We could choose more “natural” co-ordinateszSr1r2OOyxDetermine the representation for Gstretch
97 S O O Analysis of Vibrational Modes: r1 r2 How does our new basis transformUnder the operations of the group?OOVectors shifted to new position contribute zeroUnshifted vectors contribute + 1 to c(R)s(xz)s(yz)C2vEC2Gstre+2+2This 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
98 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 adaptedlinear combinations (SALCs) of the two stretching vectors.Our intuition tells us that we might have a symmetric and ananti-symmetric stretching vibrationA1 and B2
99 S O O Symmetry Adapted Linear Combinations r1 r2 Pick a generating vector eg: r1How does this transform under symmetry operations?s(xz)s(yz)C2vEC2r1r1r2r2r1Multiply this by the characters of A1 and B2For A1 this gives: (+1) r1+ (+1) r2 + (+1) r2 + (+1) r1 = 2r1 + 2r2Normalise coefficients and divide by sum of squares:
100 S S O O O O Symmetry Adapted Linear Combinations For B2 this gives: (+1) r1+ (-1) r2 + (-1) r2 + (+1) r1 = 2r1 - 2r2Normalise coefficients and divide by sum of squares:SSOOOOB2A1Sulphur must also move to maintain position of centre of mass
101 S O O Analysis of Vibrational Modes: Remaining mode “likely” to be a bendOOs(xz)s(yz)C2vEC2Gbend1111By inspection this bend is A1 symmetrySO2 has three normal modes:A1 stretch: Raman polarised and infrared activeA1 bend: Raman polarised and infrared activeB2 stretch: Raman and infrared active
103 Analysis of Vibrational Modes: SO2 experimental data. Notes:Stretching modes usually higher in frequency than bending modesDifferences in frequency between IR and Raman are due todiffering phases of measurements“Normal” to number the modes According to how the Mulliken termsymbols appear in the character table, ie. A1 first and then B2
104 Analysis of Vibrational Modes: POCl3 Angle deformationsP-Cl stretchP=O stretchGvibe = 3A1 + 3E3 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
105 Analysis of Vibrational Modes: POCl3 Gvibe = 3A1 + 3EC3vE2C33svG P=O str111G P-Cl str31G bend62Using reduction formulae or by inspection:G P=O str = A1 and G P-Cl str = A1 + EG bend = Gvibe - G P=O str - G P-Cl str = 3A1 + 3E – 2A1 – E = A1 + 2EReduction of the representation for bends gives: G bend = 2A1 + 2E
106 Analysis of Vibrational Modes: POCl3 G bend = Gvibe - G P=O str - G P-Cl str = 3A1 + 3E – 2A1 – E = A1 + 2EReduction of the representation for bends gives: G bend = 2A1 + 2EOne of the A1 terms is REDUNDANT as notall the angles can symmetrically increaseG bend = A1 + 2ENote:It is advisable to look out for redundant co-ordinates and thinkabout the physical significance of what you are representing.Redundant co-ordinates can be quite common and can lead to adouble “counting” for vibrations.
107 Analysis of Vibrational Modes: POCl3 IR (liq)/ cm-1Raman /cm-1DescriptionSymLabel12921290(pol)P=O str( 1,4)A1n1580582P-Cl str(2,3)En4487486(pol)P-Cl str( 1,2,3)n2340337deformationn5267267(pol)Sym. Deformation(1)n3-193n6
108 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 frequencyThen P-Cl stretch, then deformation.581 similar to P-Cl stretch so assym. stretch.Remaining modes must therefore be deformationsCould now use SALCs to look more closely at the normal modes
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.
110 Symmetry, Bonding and Electronic Spectroscopy s bonding in AXn molecules e.g. : waterHow do 2s and 2p orbitals transform?
111 Symmetry, Bonding and Electronic Spectroscopy s-orbitals are spherically symmetric and when at themost symmetric point always transform as the totallysymmetric speciesFor electronic orbitals, either atomic or molecular,use lower case characters for Mulliken symbolsOxygen 2s orbital has a1 symmetry in the C2v point group
112 Symmetry, Bonding and Electronic Spectroscopy How do the 2p orbitals transform?
113 Symmetry, Bonding and Electronic Spectroscopy How do the 2p orbitals transform?
114 Symmetry, Bonding and Electronic Spectroscopy How do the 2s and 2p orbitals transform?Oxygen 2s and 2pz transforms as a12px transforms as b1 and 2py as b2Need a set of s-ligand orbitals of correct symmetry to interactwith Oxygen orbitals.Construct a basis, determine the reducible representation,reduce by inspection or using the reduction formula, estimate overlap,draw MO diagram
115 Symmetry, Bonding and Electronic Spectroscopy Use the 1s orbitals on the hydrogen atoms
116 Symmetry, Bonding and Electronic Spectroscopy Assume oxygen 2s orbitals are non bondingOxygen 2pz is a1, px is b1 and py is b2Ligand orbitals are a1 and b2Orbitals of like symmetry can interactOxygen 2px is “wrong” symmetry therefore likely to be non-bondingWhich is lower in energy a1 or b2?Guess that it is a1 similar symmetry better interaction?
118 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 otherUse f1 or f2 as a generating function.(These functions must be orthogonal to each other)Observe the effect of each symmetry operation on the functionMultiply this row by each irreducible representation of the pointGroup and then normalise.(Here the irreducible representation is already known)
120 Symmetry, Bonding and Electronic Spectroscopy The overlap between the a1 orbitals (y1) is greater thanthat for the b2 orbitals (y2).Therefore a1 is lower in energy than b1.We can use the Pauli exclusion principle and the Aufbau principleTo fill up these molecular orbitals.This enables us to determine the symmetries of electronic statesarising from each electronic configuration.Note: Electronic states and configurations are NOT the same thing!
122 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*)0The symmetry of the electronic state arising from this configurationis given by the direct product of the symmetries of the MO’s of allthe electrons(a1)2 = a1.a1 = A1(b2)2 = b2.b2 = A1(b1)2 = b1.b1 = A1A1.A1.A1 = A1For FULL singly degenerateMO’s, the symmetry is ALWAYSA1
123 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 =B1General rule:For full MO’s the ground state is always totally symmetric
124 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 bondingInto either the b2* or a1* anti-bonding orbitals .a1*Anti BondingBoth of these transitions arenon bonding to anti bondingtransitions. n-p*b2*b1Non bondingb2Bondinga1
125 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 = B1In 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+1These configurations generate:3A2 , 1A2 and 3B1 , 1B1 electronic states.Note: if S= ½ then we have a doublet state
126 What electronic states do these new configurations generate? Molecular OrbitalsElectronic Statesa1*1B1b2*1A23B1b13A2b2a11A1
127 What electronic states do these new configurations generate? Triplet states are always lower than the related singlet statesDue to a minimisation of electron-electron interactions andthus less repulsionBetween which of these states are electronic transitionssymmetry allowed?Need to evaluate the transition moment integral like we did forinfrared transitions.
128 Which electronic transitions are allowed? VibrationalSpinElectronicTo first approximation m can only operate on the electronic partof the wavefunction.Vibrational part is overlap between ground and excited state nuclearwavefunctions. Franck-Condon factors.Spin selection rules are strict. There must be NO change in spinDirect product for electronic integral must contain the totallysymmetric species.
129 Which electronic transitions are allowed? A transition is allowed if there is no change in spin and theelectronic component transforms as totally symmetric.The intensity is modulated by Franck-Condon factors.The electronic transition dipole moment m transforms as thetranslational species as for infrared transitions.
130 Which electronic transitions are allowed? For the example of H20 the direct products for theelectronic transition areThe totally symmetric species is only present for the transitionto 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”.
132 Which electronic transitions are allowed? Transitions between a totally symmetric ground state and one withan electronic state that has the same symmetry as a componentof m, will be symmetry allowed.Caution: The lowest energy transition may be allowed but too weakto be observed.Caution: Ground state is not always totally symmetric andbeware of degenerate representations.
133 More bonding for AX6 molecules / complexes In the case of Oh point group:d x2-y2 and dz2 transform as egdxy, dyz and dzx transform as t2gpx, py and pz transform as t1uGs(ligands) = a1g + eg + t1uGp(ligands) = t1g + t2g + t1u + t2u
135 Electronic Spectroscopy of d9 complex: [Cu(H2O)6]2+ is a d9 complex. That is approximately Oh.Ground electronic configuration is: (t2g)6(eg*)3Excited electronic configuration is : (t2g)5(eg*)4The ground electronic state is 2EgExcited electronic state is 2T2gUnder Oh the transition dipole moment transforms as t1uAre electronic transitions allowed between these states?
136 Electronic Spectroscopy of d9 complex: Need to calculate direct product representation:2Eg . (t1u) . 2T2gOhE8C36C26C43C2i6S48S63sh6sdT2g31-1t1u-3Eg2DP18-18-2
137 Electronic Spectroscopy of d9 complex: DP182-18-2Use reduction formula:aa1g= 1/48 .[( 1x18x1)+(3x2x1) +(1x-18x1) +(3x-2x1)] = 0The 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.
138 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 thevibrational and electronic wavefunctons are coupled.This effectively changes the symmetry of the states involved.This weak transition is vibronically induced and therefore is partiallyallowed.
139 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?