"Group Theory and its Relevance/Application to Studies of Molecular Adsorbates"

Transcription

1 "Group Theory and its Relevance/Application to Studies of Molecular Adsorbates" Professor Neville V. Richardson, EaStCHEM, School of Chemistry, University of St Andrews, UK

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3 Scope of the course This 4x3 hour graduate level course of lectures/workshops aims to provide a practical appreciation of the power of group theoretical methods in understanding the properties of well-defined, single crystal surfaces and, more specifically, the properties of molecules adsorbed on such surfaces. A particular focus will be on a symmetry-based description of electronic and vibrational transitions within the adsorbate and their links to important surface spectroscopies. The course will cover the following topics: 1. Revision of group theory of isolated objects; a) symmetry elements and operations b) properties of groups c) character tables d) reducible and irreducible representations e) symmetry adapted molecular orbitals f) group theoretical description of molecular vibrations 2. Symmetry rules governing electron excitation and emission in isolated molecules 3. Group theoretical description of molecular vibrations 4. IR and Raman activity of molecular vibrations 5. Symmetry properties of 2D periodic systems a) symmetry elements b) Bravais Lattices c) 2D space groups d) site symmetries e) factor group symmetries f) Relationship of domains to symmetry properties 6. Electronic structure of ordered arrays 7. Angle resolved photoemission from substrates and oriented molecules 8. Vibrational spectroscopy of molecular adsorbates a) adsorption induced symmetry changes b) surface selection rules c) Determination of molecular orientation 9. Other surface spectroscopies (NEXAFS, X-ray emission etc) 10. Chirality at surfaces

4 Symmetry Elements (Operations) of isolated objects (molecules) 1. Identity Element, E 2. Plane of symmetry (reflection), σ 3. Axis of symmetry (rotation), C n Rotation by 360 o /n A twofold, C 2 rotation axis involves 180 o rotation A threefold, C 3 rotation axis involves 120 o rotation A fourfold, C 4 rotation axis involves 90 o rotation A sixfold, C 5 rotation axis involves 72 o rotation etc 4. Centre or point of symmetry (inversion), i A molecule has a centre of symmetry if, on choosing any point in the molecule, then moving through the centre and for an equivalent distance on the other side, an equivalent point in the molecule is found. 5. Improper axis of symmetry (improper rotation), S n This is best viewed as a combination of two processes, which themselves may or may not be valid symmetry operations :- a)rotation by 360 o /n b) Reflect in the plane perpendicular to the rotation axis

5 Symmetry Elements (Operations) Axis of Symmetry (Rotation) Two-fold, 360/2 Three-fold, 360/3 Four-fold, 360/4 Five-fold, 360/5 C n 360/n

6 Plane of symmetry (Reflection) σ Centre or point of symmetry (Inversion) i

7 Improper Axis of Symmetry (Improper Rotation) Four-fold, 360/4 + reflection Eight-fold, 360/8 + reflection S n 360/n Ten-fold, 360/10 + reflection

8 Identify ALL the symmetry operations in the following molecules Methane Benzene Carbon Monoxide

9 Identify ALL the symmetry operations in the following molecules σ d x6 C 3 x2 x4 C 2 x3 S 4 x2 x3 Methane Benzene 2 x C σ v Carbon Monoxide 2C 6, 2C 3, C 2, 2S 6, 2S 3 C 2, C 2, i σ h, 3σ v, 3σ d

10 Symmetry Classification of Molecules C 2v H 2 O, C 5 H 5 N, CH 2 F 2 C 2h trans 1,2-dichloroethene C 3v PF 3, NH 3 D 3h BF 3, CO 3 2- Molecules with the same symmetry elements belong to the same point group. D 4h PtCl 4 2- T d SiCl 4, CH 4 O h SF 6, Cr(CO) 6

11 Combining Operations F 2 H 1 Mirror Plane, σ v (CH 2 ) Mirror Plane, σ v (CF 2 ) This is the C 2v point group F 1 H 2 Rotation, C 2 F 1 H 1 σ v F 2 F 2 H 2 H 2 σ v F 1 H 1 F 1 H 2 C 2 F 2 H 1 σ v σ v = C 2

12 Elements of a group {A, B X } Group Theory There is an element E, the identity element, such that EX=X for all elements of the group. If A and B are members of the group and AB=C then C is also a member of the group. Each element, X, has an inverse, X -1, such that XX -1 = E (AB)C = A(BC) AB may or may not be the same as BA Multiplication table Gp 4 E A B C Abstract group of 4 members E E A B C AB=C BC=A CA=B AA -1 =BB -1 =CC -1 =E A B A B E C C E B A C C B A E

13 Multiplication Table C 2v E C 2 σ v σ v E E C 2 σ v σ v C 2 C 2 E σ v σ v C 2 σ v = σ v σ v σ v σ v E C 2 σ v σ v σ v C 2 E C 2 E C 2 C s E σ E E C 2 E E σ C 2 C 2 E σ σ E

14 Mirror Plane, σ v (CH 2 ) Mirror Plane, σ v (CF 2 ) Rotation, C σ v C = C H 1 H 2 F 1 H 1 H 2 F 2 F 2 F 1 σ v = Trace =3

15 E Matrices 5 C σ v σ v Atomic Positions CH 2 F 2 C 2v E C 2 σ v σ v

16 (Ir)reducible Representations Atomic Positions CH 2 F 2 C 2v E C 2 σ v σ v σ v How many remain in position? C Position H Positions H 1 + H C H 1 +H 2 H 1 -H 2 F 1 +F 2 F 1 -F 2 = C H 1 +H 2 H 1 -H 2 F 1 +F 2 -F 1 +F 2 H 1 - H F Positions All matrices are now diagonalised F 1 + F F 1 - F Symmetry adapted combinations of atomic positions

17 C 2v E C 2 σ v σ v A 1 B 1 B 2 A Character Table Atomic positions C, H 1 +H 2, F 1 +F 2 belong to A 1 or have symmetry A 1 H 1 -H 2 belongs to B 1 or has symmetry B 1 F 1 -F 2 belongs to B 2 or has symmetry B 2 We did not find a basis set which belongs to the A 2 irreducible representation

18 C 2v E C 2 σ v σ v h=4 A 1 A Order of the group Character Tables B H 2 O, CH 2 F 2,C 4 H 4 S B T d E 8C 3 3C 2 6S 4 6σ d h=24 C 3v E 2C 3 3σ v A 1 A 2 E h=6 NH 3,PCl 3 A 1 A 2 E T T CH 4, P 4, SiCl 4, Ni(CO) 4

19 C 3v E 2C 3 3σ v h=6 How many irreducible representations are there for a given point group? A 1 A σ v σ v E Characters in the same class are equal σ v 2. There are as many irreducible representations as there are classes 3. There is always a totally symmetric irreducible representation 4. Representations are mutually orthogonal Σ R χ i (R) χ j (R) = 0 e.g. in C 3v A 1 x A 2 = (1x1) + 2(1x1) + 3(1x-1) = 0 5. The dimension/degeneracy of a representation is the character under the identity operation Singly (doubly, triply) degenerate have χ(e) = 1 (2,3). Nevertheless they cannot be reduced further! 6. The sum of the squares of the dimensions is equal to the order of the group Σ i (χ i (E)) 2 = h e.g. for C 3v (1) 2 + (1) 2 + (2) 2 = 6

20 C 2v E C 2 σ v σ v A 1 A How can we determine the combination of irreducible representations corresponding to a particular reducible representation? B 1 B n i = Σ R χ t (R) χ i (R)/h All atoms A 1 +B 1 +B 2 CH 2 F 2 n i is the number of occurrences of the irreducible representation i R labels the operations of the group χ t (R) is the character of the reducible representation χ i (R) is the character of the irreducible representation h is the order of the group n i = [χ t (E)χ i (E)+χ t (C 2 )χ i (C 2 )+χ t (σ v )χ i (σ v )+χ t (σ v )χ i (σ v )]/h n A1 = [(5x1) + (1x1) + (3x1) + (3x1)]/4 = 3 Confirm that n A2 = 0, n B1 = 1, n B2 = 1

21 C 2v E C 2 σ v σ v A 1 A 2 B 1 B Express 2A 1 + Athese 2 + 2B 1 as + a B 2 sum of irreducible 3 A 2 + 3B 2 representations For the position of the H atoms in NH 3 (point group C 3v ) determine the reducible representation and then express this as a sum of irreducible representations C 3v E 2C 3 3σ v A 1 A 2 E H atoms? 3? 0? 1??? A 1 1 +E in NH 3

22 How do we determine which function relates to which symmetry? C 2v E C 2 σ v σ v A A B 1 B H atoms A 1 +B 1 H 1 + H A 1 Ψ i = Σ R χ i (R) Rφ a H 1 - H B 1 H 1 H 1 H 2 H 1 H 2 Ψ A 1 = (1x H 1 )+ (1x H 2 )+ (1x H 1 )+ (1x H 2 ) Ψ A 1 = 2(H 1 + H 2 ) Ψ B 1 = (1x H 1 )+ (-1x H 2 )+ (1x H 1 )+ (-1x H 2 ) Ψ B 1 = 2(H 1 - H 2 )

23 Degenerate Representations! C 3v E 2C 3 3σ h=6 v A 1 A 2 E H 1s A 1 + E s 1 s 1 s 2, s 3 s 1, s 2, s 3 σ v 1 3 σ v σ v 2 ψ a = s 1 + s 2 + s 3 A 1 ψ b = 2s 1 - s 2 - s 3 E s 2 s 2 s 3, s 1 s 2, s 3, s 1 ψ c = 2s 2 - s 3 - s 1 E ψ d = 2s 3 - s 1 - s 2 E? Not independent! ψ d = -(ψ b + ψ c )

24 ψ a = s 1 + s 2 + s 3 A 1 ψ b = 2s 1 - s 2 - s 3 ψ c = 2s 2 - s 3 - s 1 E These are not orthogonal Ψ b = 2s 1 - s 2 - s 3 Ψ c = s 2 - s 3 E Ψ c = (Ψ b + 2Ψ c )/3

25 Rotations in general cos θ θ sin θ Only atoms lying on the axis are considered C n x y z = x cos θ - y sin θ y cos θ + x sin θ z θ = 360/n C n = cos θ - sin θ 0 sin θ cos θ n θ cos θ 1+ 2cos θ cos θ / / cos θ cos θ C n S n

26 What is the reducible representation based on the p z orbitals of carbon in benzene, taking the molecule to lie in the xy plane. Determine the corresponding sum of irreducible representations. Hence express, both algebraically and as sketches, the π molecular orbitals MOs) of benzene in terms of a linear combinations of atomic orbitals (LCAOs). What is the sequence of orbital energies for these MOs? Benzene D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 6 2S 3 σ h 3σ v 3σ d A 1g A 2g B 1g B 2g E 1g E 2g A 1u A 2u B 1u B 2u E 1u E 2u

27 D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 6 2S 3 σ h 3σ v 3σ d A 1g A 2g B 1g B 2g E 1g E 2g A 1u A 2u z B 1u B 2u E 1u x,y E 2u p z p 1 p 1 p 2 p 3 p 4 -p 1 -p 6 -p 4 -p 2 -p 3 -p 1 p 1 p 6 p 6 p 5 -p 3 -p 4 -p 6 -p 5 p 3 p 4 -p 5 -p 2 p 5 p 2 ψ a 2u = p 1 + p 2 + p 3 + p 4 + p 5 + p 6 σ d C 2 σ v C 6 C 3 C 2 S 6 S 3 i σ h C 2 a 2u + b 2g + e 1g + e 2u ψ b 2g = p 1 - p 2 + p 3 - p 4 + p 5 - p 6 ψ e 1g = 2p 1 + p 2 - p 3-2p 4 - p 5 + p 6, ψ e 1g = p 2 - p 3 + p 5 - p 6 ψ e 2u = 2p 1 - p 2 - p 3 + 2p 4 - p 5 - p 6, ψ e 2u = p 2 + p 3 - p 5 - p 6

28 π orbitals of benzene b 2g ψ b 2g = p 1 - p 2 + p 3 - p 4 + p 5 - p 6 e 2u ψ e 2u = 2p 1 - p 2 - p 3 + 2p 4 - p 5 - p 6, ψ e 2u = p 2 + p 3 - p 5 - p 6 e 1g ψ e 1g = 2p 1 + p 2 - p 3-2p 4 - p 5 + p 6, ψ e 2u = p 2 - p 3 + p 5 - p 6 a 2u ψ a 2u = p 1 + p 2 + p 3 + p 4 + p 5 + p 6

29 Why are we doing all this? To describe the electronic (vibrational) properties of molecules/solids and surfaces in terms of molecular orbitals (normal modes/phonons) which are symmetry adapted combinations of atomic orbitals (atomic displacements). Only molecular orbitals (normal modes/phonons) of the same symmetry interact with each other. To understand and appreciate the meaning and significance of labels such as 2 A 1g, 3 Π u, 4 T 2g, b 2g, σ g, Γ 25, Λ 3 etc To understand the splitting of atomic states in a crystal field and the influence of a structural distortion on the coupling of electron or phonon wave-functions at different points in the Brillouin zone. To understand the how symmetry based selection rules related to electronic and vibrational excitations of molecules, solids and clean or adsorbate covered surfaces determine the features observed in IR, Raman, EELS, UPS, NEXAFS etc spectroscopies.

30 Splitting of s,p,d orbitals in a tetrahedral field T d E 8C 3 3C 2 6S 4 6σ δ A z y A 2 E x T 1 T s p d A T E + T 2 C 3 x --> z y --> x z --> y C 2 x --> -x y --> -y z --> z S 4 x --> -y y --> x z --> -z σ d x --> y y --> x z --> z R d 1 = d i d 2 d j d 3 d k d 4 d l d 5 d m d r = Σ n=1-5 c n d n c r C 3 d z 2 = C 3 (3z 2 -r 2 ) = C 3 (2z 2 -x 2 -y 2 )/ 6 = (2y 2 -z 2 -x 2 )/ 6 = A((2z 2 -x 2 -y 2 )/ 6) + B(x 2 -y 2 )/ 2 A=-1/2, B=- 3/2 C 3 d z 2 = -1/2 d z 2-3/2 d x 2- y 2

31 Direct Products C 2v E C 2 σ v σ v h=4 A A B B Ψi Ψj B 1 A 2 Which states arise from a low spin (high field) d 6 configuration in a tetrahedral field? Fe (II), Co(III) ions. ford.edu/research/highli ghts_archive/electransf er.html ΨiΨj B 2 T d E 8C 3 3C 2 6S 4 6σ δ h=24 C 3v E 2C 3 3σ v A 1 A 2 E Ψi Ψj ΨiΨj h=6 E E A 1 + A 2 + E A 1 A 2 E T T T 2 xt A 1 + E + T + T 1 2

32 O h E 8C 3 6C 2 6C 4 3C 2 i 6S 4 8S 6 6σ h 6σ d Splitting and mixing in reduced symmetry O h D 3d C 3v A 1g A 1g A 1 A 2g E g T 1g T 2g A 2g E g A 2g + E g A 1g + E g A 2 E A 2 +E A 1 +E A 1u A 1u A E g + T 2g C3v E 2C3 3σv A 2u E u T 1u T 2u A 2u E u A 2u + E u A 1u + E u A 1 E A 1 +E A 2 +E D 3d E 2C 3 3C 2 ι 2S 6 3σ d A 1 + 2E A 1g + 2E g

33 Cu band structure /wcrystals.htm A 1g,L 1 A 1, Λ 1 G.A. Burdick, Phys Rev 129 (1963) 138 A 2u,L 2 E g,l 3 E g,l 3 E, Λ 3 Γ 2 E g Γ O h E, Λ 3 Γ 25, T 2g L D 3d ΓL <111> C 3v A 1, Λ 1 A 1g,L 1 X D 4h ΓX <100> C 4v A 1, Λ 1 Γ 1, A 1g K D 2d ΓK <110> C 2v L Λ Γ

34 The molecular orbitals of water, H 2 O 1s orbitals of the H atoms: 2s, 2p orbitals of the O atom H 2 ψ b = (φ 1 φ 2 ) φ 1 φ 2 ψ a = (φ 1 + φ 2 ) How do these symmetry adapted orbitals interact with the atomic orbitals of oxygen?

35 C 2v E C 2 σ xz σ yz A 1 A 2 B 1 B z y x How many remain in position? H s orbitals A 1 +B 1 φ 1 φ 1 φ 2 φ 1 φ 2 ψ a = (φ 1 + φ 2 ) ψ b = (φ 1 - φ 2 ) A 1 B 1 p z A p x B p y B Only orbitals of the same symmetry interact with each other

36 Overlap Integrals ψ 1 ψ 2 dτ > 0 Bonding interaction ψ 1 ψ 2 dτ < 0 Anti-bonding interaction A 1 ψ 1 ψ 2 dτ = 0 No interaction B 1 y Necessarily, exactly zero if ψ 1 and ψ 2 belong to different representations i.e. have different symmetries. x

37 H 2 B 1 O B 1 A 1 A 1 B 1 B 2 B 2 ψ b = (φ 1 - φ 2 ) A 1,B 1,B 2 A 1 ψ a = (φ 1 + φ 2 ) A 1 A 1 B 1 B 1 A 1 A 1 A 1

38 2b 1 4a 1 Ground electronic structure 1a 1 2, 2a 1 2, 1b 1 2, 3a 1 2, 1b 2 2 a 1 x a 1 x a 1 x a 1 x b 1 x b 1 x a 1 x a 1 x b 2 x b 2 1 A 1 Overall symmetry 1b 2 Excited state electronic structure 1a 1 2, 2a 1 2, 1b 1 2, 3a 1 2, 1b 2 1, 2b 1 1 3a 1 1b 1 O 2p, H 1s b 2 x b 1 1 A 2 Overall symmetry 2a 1 1a 1 O 2s O 1s Filled shells are totally symmetric i.e. A 1, A g etc Closed shell molecules are overall totally symmetric.

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40 Selection Rules governing electronic excitations Can electrons be excited between any pair of levels? Reminder! Atomic spectroscopy C 2v E C 2 σ xz σ yz Δl = ±1 s - p, p-d etc s-s, d-d Electric field vector A 1 A 2 B 1 B z x y I = { ψ 1 µψ 2 dψ} 2 Initial state i.e. 1b 2 Final state i.e. 2b 1 2b 1 1b 2 Water hν Dipole moment operator behaves as x,y,z For an allowed transition overall symmetry must be A 1 i.e product of initial and final state symmetries must be the same as that of x,y or z? Overall symmetry A 1 B 1 B 2 B 2 x x B > A 2 B 2 B 1 Not A 1, I=0 Transition is forbidden

41 T d E 8C 3 3C 2 6S 4 6σ δ A 1 A h=24 d 2 configuration in a tetrahedral field t 2 E T T x,y,z e E x E E x T A 1 + A 2 + E T 1 + T 2 1 T 1 A 2 x T T 1 1 T 2 3 T 1 et 2 I = <i µ f> 2 3 T 2 1 E 1 A 1 e 2 A 2 T 2 T 1 3 A 2 cf Carbon atom 1s 2 2s 2 2p 2 --> 1s 2 2s 2 2p3d

42 D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 6 2S 3 σ h 3σ v 3σ d A 1g A 2g B 1g B 2g E 1g E 2g A 1u A 2u B 1u B 2u E 1u E 2u z x,y b 2g e 2u e 1g Which transitions between the π orbitals of benzene, give rise to the UV absorption spectrum? a 2u

43 D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 6 2S 3 σ h 3σ v 3σ d A 1g A 2g B 1g B 2g E 1g E 2g A 1u A 2u B 1u B 2u E 1u E 2u z x,y Which transitions between the π orbitals of benzene, give rise to the UV absorption spectrum? b 2g Possible transitions e 1g -----> e 2u e 1g x e 2u = b 1u + b 2u + e 1u e 2u e 1g -----> b 2g a 2u -----> e 2u e 1g x b 2g = e 2g a 2u x e 2u = e 2g e 1g a 2u a 2u -----> b 2g a 2u x b 2g = b 1u This product should be the same symmetry as x,y or z i.e. a 2u or e 1u for an allowed transition. The only allowed transition is e 1g x e 2u

44 Coupling of the C 1s levels of benzene to the empty π levels (X-ray absorption spectroscopy) D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 6 2S 3 σ h 3σ v 3σ d A 1g A 2g B 1g B 2g E 1g E 2g A 1u A 2u B 1u B 2u E 1u E 2u z x,y C 1s a 1g + b 1u + e 2g + e 1u i> µ z µ x,µ y C 2p (π) a 2u + b 2g + e 1g + e 2u f> a 1g b 1u a 2u b 2g e 1u e 2g I = <i µ f> 2 e 2g e 1u e 2u e 1g b 1u + b 2u + e 1u a 1g + a 2g + e 2g µ z introduces a nodal plane parallel to the molecular (x,y) plane

45 Angle-resolved electron emission Linearly polarised I(θ) ~ cos 2 θ E z Unpolarised z y x I(θ) ~ sin 2 θ Gas-phase molecules require integration to allow for all possible orientations of the molecule in the laboratory co-ordinate frame. Contrast this with crystalline solids or adsorbed, orientated molecules.

46 Angle-resolved photoemission from oriented molecules C 2v E C 2 σ xz σ yz z y A z x A 2 B x B y I = <i µ f> 2 Symmetry of f> i> µ z µ y µ x a 1 a 1 b 2 b 1 Importance of nodal planes b 1 b 1 a 2 a 1 b 2 b 2 a 1 a 2

47 Cu band structure Angle-resolved photoemission from clean metals A 1 Γ O h X D 4h ΓX <100> C 4v E 2C 4 C 2 2σ v 2σ d z x 2 +y 2, z 2 A 1 x 2 -y 2 B 1 E x,y xy xz. yz Γ B 2 A 1 G.A. Burdick, Phys Rev 129 (1963) 138 Which bands can be observed in ARUPS, normal to a Cu{100} surface? We need to consider the symmetry properties of <f µ i>

48 HeI A 1 For a non-zero transition dipole <f µ i> The product of the symmetries of i>, f> and E must be totally symmetric i.e. A 1. For non-zero intensity in normal emission, the normal must not lie on a nodal plane. f> must therefore also belong to A 1. E f B 1 A 1 E i> and µ must therefore have the same symmetry so emission is allowed only from A 1 and E symmetry initial states, NOT from B 1 or B 2 states. B 2 A 1 Γ X

49 Ten point groups

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53 Bravais lattices in TWO dimensions

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56 Simple Clean Surfaces e.g. Cu(100) e.g. Cu(110) e.g. Cu(111) Unit cell Unit cell Unit cell Square p4mm Rectangular p2mm Hexagonal p6m C 4v + translations C 2v + translations C 6v + translations?

57 e.g. Cu(111) Bravais lattice Unit cell Hexagonal p3m C 3v + translations May need to take account of deeper layers Isolated unit cell may not have the full point group symmetry. e.g. W(110) Bravais lattice Centred rectangular Hexagonal c2mm C 2v + translations Primitive

58 e.g. Si(100) Bravais lattice Unit cell Square p2mm C 2v + translations e.g. Cu(210) Bravais lattice Unit cell Centred Rectangular c1m1 C s + translations

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60 Isolated Adsorbates - Site Symmetry

61 Isolated Adsorbates - Site Symmetry Bridge site C s /C 2v

62 Symmetry Compatibility Isolated molecule symmetry Surface symmetry Bare site symmetry Site symmetry including adsorbed species Loss of symmetry or symmetry reduction Energetically equivalent adsorption geometries are related by the missing symmetry element(s)

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64 Domains S/Cu(100) p(2x2) With thanks to Ib Chorkendorff Domain boundary or wall p(2x2) gives rise to FOUR distinct anti-phase domains These are translational domains

65 Symmetry and domain type c(2x2), p( 2x 2)R45 o Translational domains Rotational domains (2x1), p2mm p4mm p2mm Rotational domains p2 Reflectional domains 3 0 ±1 2 Rotational domains ±1

66 Equivalent (2x1), Inequivalent Inequivalent p2mm Equivalent

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68 Suppose that ethene (C 2 H 4 ) molecules were to adsorb on the (111) surface of an fcc substrate in bridge sites to form a (3x2) overlayer at θ=0.17 coverage i.e. one ethene molecule to every six substrate atoms and one per unit cell. If only a single mirror plane is retained because of tilting of the molecule, how many (energetically equivalent) domains of the structure will be present on the surface?

69 How many domains?

70 Inequivalent structures

71 Some useful definitions Point Group Symmetry properties of an isolated molecule Site Group Symmetry properties of a condensed system as seen from a particular location The site group may be relevant for low coverages of molecules on a surface (in the absence of island formation), for molecules at domain or island boundaries and in disordered systems, when the local environment can be considered to dominate the situation. Space group Symmetry properties of an ordered array including screw axes (not in 2D) and glide planes and translations Factor Group A point group which is isomorphous with (i.e. has a one-to-one relationship with) the space group without translations. Glide lines are converted to mirror planes and screw axes to rotation axes. The factor group is formally correct for the analysis of properties associated with the centre of the Brillouin zone corresponding to the periodic array. At other points in the Brillouin zone, an appropriate sub-group of the factor group is required. For a single atom or molecule per unit cell, the factor group is identical to the site group of the atom or molecule.

72 CO/Pt{110} p(2x1) p2mg space group C s site group C 2v factor group The glide line is treated as a mirror plane NB The C 2 axis does not pass through the molecular adsorption site

73 Vibrational Properties of Molecules Choose atomic displacements as the basis set for determining the symmetry of molecular vibrations z y x C 2v E C 2 σ xz σ yz A 1 A 2 B 1 B z x y An N atom molecule has 3N degrees of motional freedom of which 3 correspond to molecular translation and 3(2) correspond to rotation for non-linear (linear) molecules. There are therefore 3N-6 (3N-5) degrees of vibrational freedom (modes) Γ all A 1 + A 2 + 3B 1 + 2B 2 This includes molecular translations and rotations

74 C 2v E C 2 σ xz σ yz A 1 A z R z σ xz z y x z y x B 1 B 2 Γ all x, R y y, R x A 1 + A 2 + 3B 1 + 2B A 1 + B 1 + B 2 Γ trans A 2 + B 1 + B 2 Γ rot Γ vib A 1 + B 1 Bending and stretching vibrations?

75 Modes based on bond length and bond angle changes! Stretching modes Bending modes In water, two O-H bonds so two O-H stretching modes C 2v E C 2 σ xz σ yz 1 2 Stretch of OH 1 A 1 A 2 B 1 B OH 1 OH 1 OH 2 OH 1 OH 2 A 1 OH 1 + OH 2 Symmetric O-H stretch B 1 OH 1 - OH 2 asymmetric O-H stretch A 1 HOH bend cm cm cm -1

76 Determine the symmetry of the vibrational modes of ethene h=8 D 2h E C 2 (z) C 2 (y) C 2 (x) i σ(xy) σ(xz) σ(yz) A g y x B 1g R x B 2g R y B 3g R z A u B 1u z B 2u y B 3u x Γ all A g + 3B 1g + 2B 2g + B 3g + A u + 2B 1u + 3B 2u + 3B 3u Γ trans B 1u + B 2u + B 3u Γ rot B 1g + B 2g + B 3g Γ vibr 3Ag + 2B 1g + B 2g + A u + B 1u + 2B 2u + 2B 3u

77 D 2h E C 2 (z) C 2 (y) C 2 (x) i σ(xy) σ(xz) σ(yz) A g B 1g R x B 2g R y B 3g R z A u B 1u z B 2u y B 3u x CC CC CC CC CC CC CC CC CC 1 4 β 1 β β 2 β 3 CH 1 CH 1 CH 3 CH 2 CH 4 CH 3 CH 1 CH 4 CH 2 β 1 β 1 β 3 β 2 β 4 β 3 β 1 β 4 β 2 γ 2 γ 3 γ 1 γ 4 γ 1 γ 1 γ 3 -γ 4 -γ 2 -γ 3 -γ 1 γ 2 γ 4 CC A g CH A g + B 1g + B 2u + B 3u β A g + B 1g + B 2u + B 3u γ B 2g + B 3g + A u + B 1u

78 A g CH 1 +CH 2 +CH 3 +CH 4 A g CH 1 +CH 2 - CH 3 - CH 4 B 2u CH 1 - CH 2 +CH 3 - CH 4 B 1g CH 1 - CH 2 - CH 3 +CH 4 B 3u β 1 + β 2 + β 3 + β 4 A g β 1 + β 2 - β 3 - β 4 B 2u γ 1 + γ 2 + γ 3 + γ 4 B 1u γ 1 + γ 2 - γ 3 - γ 4 B 2g β 1 - β 2 + β 3 - β 4 B 1g β 1 - β 2 - β 3 + β 4 B 3u γ 1 - γ 2 + γ 3 - γ 4 A u γ 1 - γ 2 - γ 3 + γ 4 B 3g ROTATION!

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80 A g CH 1 +CH 2 +CH 3 +CH 4 A g CH 1 +CH 2 - CH 3 - CH 4 B 2u CH 1 - CH 2 +CH 3 - CH 4 B 1g CH 1 - CH 2 - CH 3 +CH 4 B 3u β 1 + β 2 + β 3 + β 4 A g β 1 + β 2 - β 3 - β 4 B 2u γ 1 + γ 2 + γ 3 + γ 4 B 1u γ 1 + γ 2 - γ 3 - γ 4 B 2g β 1 - β 2 + β 3 - β 4 B 1g β 1 - β 2 - β 3 + β 4 B 3u γ 1 - γ 2 + γ 3 - γ 4 A u γ 1 - γ 2 - γ 3 + γ 4 B 3g ROTATION!

81 Infrared spectroscopy Selection rule? Vibration should give rise to a change in dipole moment!! I = <i µ f> 2 initial final Vibrational wavefunctions ψ n vib = f(q) ~ qn Dipole moment operator - a vector which behaves as x,y,z Totally symmetric - A 1, A g etc Symmetry of the mode A mode is infrared active if it has the same symmetry as x,y or z

82 D 2h E C 2 (z) C 2 (y) C 2 (x) i σ(xy) σ(xz) σ(yz) A g B 1g R x B 2g R y B 3g R z A u B 1u z B 2u y B 3u x B 2u A mode is infrared active if it has the same symmetry as x,y or z B 3u B 1u

83 Infrared cm bands B 1u CH 2 wag B 2u B 2u CH 2 rock B 3u CH 2 scissors B 3u Symmetric C-H stretch Asymmetric C-H stretch

84 Raman spectroscopy Selection rule? Vibration should give rise to a change in polarisability!! I = <i α f> 2 initial final Vibrational wavefunctions ψ n vib = f(q) ~ qn Polarisability operator Totally symmetric - A 1, A g etc Symmetry of the mode µ ind = α E E is the electric vector of the radiation with components E x, E y and E z µ is the induced dipole with components µ x, µ y and µ z µ x µ y µ z = α xx α yx α zx α yx α yy α zy α zx α yz α zz E x E y E z α is the polarisability tensor with components α xx, α yy, α zz, α xy, α xz, α yz A mode is Raman active if it has the same symmetry as x 2,y 2, z 2, xy, xz, yz

85 D 2h E C 2 (z) C 2 (y) C 2 (x) i σ(xy) σ(xz) σ(yz) A g x 2,y 2,z 2 B 1g R x, yz B 2g R y, xz B 3g R z, xy A u B 1u z B 2u y B 3u x A g A g A g A mode is Raman active if it has the same symmetry as x 2,y 2, z 2, xy, xz, yz B 2g B 1g B 1g

86 Raman cm bands B 2g A g CH 2 wag A g B 1g B 1g A g C=C stretch CH 2 rock CH 2 scissors Symmetric C-H stretch Asymmetric C-H stretch A u This mode is inactive in both Infrared and Raman spectroscopy

87 Determine the symmetry of the vibrational modes of cyclopropane. Determine their activity in IR and Raman spectroscopy. Sketch the atomic motion for the modes based on C-H stretching. D 3h E 2C 3 3C 2 σ h 2S 3 3σ v A 1 A 2 E A 1 A 2 E x 2 +y R z (x,y)(x 2 -y 2 ) z (R x,r y )(xz,yz)

88 D 3h E 2C 3 3C 2 σ h 2S 3 3σ v h=12 A 1 A 2 E A 1 A 2 E x 2 +y R z (x,y)(x 2 -y 2 ) z (R x,r y )(xz,yz) Γ all A 1 + 2A 2 + 5E + A 1 + 3A 2 + 4E Γ trans Γ rot A 2 + E A 2 + E Γ vib 3A 1 + A 2 + 4E + A 1 + 2A 2 + 3E 14 bands (21 modes) IR active Raman active Inactive active 2A 2 + 4E 6 bands (10 modes) 3A 1 + 4E + 3E 10 bands (17 modes) A 2 + A 1 2 modes

89 D 3h E 2C 3 3C 2 A 1 A 2 E A 1 A 2 E σ h 2S 3 3σ v CH 1 CH 1 CH 2 CH 4 CH 4 CH 5 CH CH 1 + CH 2 + CH 3 + CH 4 + CH 5 + CH 6 A 1 Raman 3038cm -1 CH 3 CH 5 CH 6 CH 2 CH 6 CH 3 CH 1 + CH 2 + CH 3 - CH 4 - CH 5 - CH 6 A 2 IR 3103cm -1 2CH 1 - CH 2 - CH 3-2CH 4 + CH 5 + CH 6 E CH 2 - CH 3 - CH 5 + CH 6 2CH 1 - CH 2 - CH 3 + 2CH 4 - CH 5 - CH 6 E Raman 3082cm -1 CH 2 - CH 3 + CH 5 - CH 6 IR and Raman 3025cm -1

90 RAIRS, IRAS FT (Fourier transform) PM (polarisation modulation) ATR (attenuated total reflection) Reflectivity, ΔR/R x 100% Resolution ~1cm -1 < 1/τc on metal surfaces Does not require vacuum condiitons

91 R. Greenler J. Chem Phys. (1966) ε 2 /cos θ Only totally symmetric modes are observable by IR spectroscopy at a metal (semiconductor) surface)

92

93 Steps in considering the IR activity of molecules at metallic surfaces 1. Identify the IR active modes of the gas-phase species. 2. Given the orientation of the species on the surface, determine which of the gas-phase modes have their dipole moment change perpendicular to the surface. 3. Given the interaction of the species, in this orientation with an unstructured (jellium!) surface, determine a new point group (NB this will be one of the ten point groups compatible with the presence of an interface close to the molecule). Determine which modes become newly dipole allowed under this new point group and, in particular, which of these modes have their dynamic dipole perpendicular to the surface. NB since translation along the surface (usually taken as the z direction) necessarily belongs to the totally symmetric representation, this is equivalent to determining which modes belong to the totally symmetric representation of the new point group. 4. Given the full surface structure, identify the relevant Factor Group (or, at low coverage, site group), then determine which further modes now become dipole allowed and which belong to the totally symmetric representation, permitting them to be active at the surface. 5. Worry about species at defects and domain boundaries, where the symmetry is lower still!!! The intensity of an IR band is directly related to the strength of the perturbation which reduces the symmetry and thereby permits the band to become dipole allowed.

94 C 2 factor group or C 1 with second layer interactions or if tilted D 2h gas-phase molecule C s (2) C 2v site group or C s (2) with second layer interactions or if tilted C s (1) C 2v site group or C s (1) with second layer interactions or if tilted C 2v site group for molecule lying flat on an unstructured surface

95 Slater et al Surf. Sci. (1994) Intense feature at 950cm -1 assigned to b 1u mode. At higher coverages, additional band at 3075cm -1 assigned to b 2u (C-H) stretch implying C s (2) symmetry because of tilted configuration - 950cm -1 feature decreases in intensity as the dynamic dipole tilts.

96 Ethene/Cu(111) a g a g a g b 1u E.M. McCash, Vacuum (1990) See also discussion in Priebe et al, J. Phys. Chem. (2006)

97 Dipole active modes with displacements parallel to the surface Ethene/Ethyne π-bond Electron response on timescale compared with for nuclear motion determined by the relevant vibrational frequency H C C H d-π back bonding H C C H For strong adsorbate substrate interactions there is a change of configuration: π bonding to di-σ bonding Coupling and intensity borrowing between modes now of the same symmetry (A 1 ). ν(c-c), ν(c-h) etc initially of Ag symmetry and γ(c-h), ν(m-c) i.e frustrated z translation initially of B 1u symmetry

98 Planar aromatic hydrocarbons (e.g. benzene, perylene, tetracene) show strong, diagnostic γ(c-h) out-of-plane bending modes, signifying an orientation with the plane of the aromatic ring parallel to the substrate. ~700cm -1 There is one such mode, with the same symmetry as translation perpendicular to the ring, for each group of equivalent carbon atoms. These modes are surface IR active if the ring is parallel to the substrate. In the case of tetracene, there are three such modes. Tetracene/Cu{110} A B C ν/cm -1 Int % Wavenumber /cm -1

99 Low coverage of phthalic anhydride on Cu{110} Wavenumber cm Two out-of-plane bending modes of B 2 symmetry in the gas-phase species (C 2v ) become A in the highest possible symmetry of a flat-lying molecule

100 0.1 o C O 1420 cm ν (OCO) s - 1 H H H H H cm γ (CH) O O H Frequency (cm -1 )

101 Space group p3m Factor group C 3v

102

103 Symmetric and antisymmetric M-CO stretching vibrations Symmetric and antisymmetric C-O stretching vibrations

104 Hydrogen on Tungsten - H/W(100) θ=2 All bridge sites occupied. Two H atoms per unit cell. Site group Space group Factor group C 2v p4m C 4v Vibrational modes of hydrogen? Three - frustrated (hindered) translation in x, y and z directions! Wrong!!

105 Atoms on opposite sides of the unit cell are related by translational symmetry M.R. Barnes and R.F. Willis, Phys. Rev. Letters (1978) Specular A E E Off-specular + B

106

107 First and second layer coupling

108 c(2x2), p( 2x 2)R45 o Factor group C 4v Factor group C 2v

109 On top Four-fold hollow A B E E B 1 A 1 A 2 Bridge - + B 2 A 2 A S. Andersson and M. Persson, Phys. Rev. B (1981)

110 Ibach and Bruchmann, Phys. Rev. Letters (1980) Displacements of substrate atoms related by translational periodicity must be identical (modes at Γ). Displacements of substrate atoms related by point group symmetry elements (including glide lines) must be such that the mode belongs to the totally symmetric representation of the relevant factor group.

111 Importance of surface order Three scattering mechanisms

112 Electron Scattering Mechanisms Dipole Scattering Long Range z > 20Å Forward Scattering (Specular/Bragg scattering) Low energy electrons E < 5eV Surface dipole selection rule (as IR but a requirement for good surface order) Impact Scattering Short range z 5Å High energy electrons I ~ E 1/2 Forbidden mode Symmetry plane Scattering plane Modes polarised perpendicular to the scattering plane are forbidden if this is a symmetry plane of the system Resonance Scattering Electron capture, resonant process Angular distribution determined by capturing orbital Totally symmetric modes only are excited

113 Benzoate/Cu(110)-c(8x2)

114 Resonant Scattering in EELS q

115

116 HCOO - Exposure of late transition metals to formic acid results in an adsorbed formate species C 2v E C 2 σ xz σ yz Describe the vibrational modes of the isolated species (symmetry, polarisation and dipole activity) A 1 A 2 B 1 B z R z x, R y y, R x For a species adsorbed with its molecular plane perpendicular to the surface, which of these (internal, or intrinsic) modes are observable by IR spectroscopy or specular EELS? Which external or extrinsic modes are similarly observable?

117 Which additional modes are observable in off-specular EELS when the scattering plane is a) Aligned with the molecular plane b) Perpendicular to the molecular plane On Ni{110} formate forms a c(2x2) structure with the oxygen atoms on top of Ni atoms and the C-H bond projecting onto a short bridge site. Which substrate phonon mode becomes dipole active?

118 DCOO/Ni(110) A 1 ν NiNi 8eV specular ν NiO ν COs δ COO Tz A 1 ν CD Energy Loss/cm -1 Active phonon mode B 1 B2 X S Ry Γ Y B 1 B 1 B 2 A 2 B 2 c(2x2)

119 A.<i µ f>

120 Coupling of the C 1s levels of benzene to the empty π levels (X-ray absorption spectroscopy) D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 6 2S 3 σ h 3σ v 3σ d A 1g A 2g B 1g B 2g E 1g E 2g A 1u A 2u B 1u B 2u E 1u E 2u z x,y b 2g e 2u e 1g a 2u C 1s a 1g + b 1u + e 2g + e 1u i> µ z µ x,µ y C 2p (π) a 2u + b 2g + e 1g + e 2u a 1g a 2u e 1u f> b 1u b 2g e 2g I = A.<i µ f> 2 e 2g e 2u b 1u + b 2u + e 1u e 1u e 1g a 1g + a 2g + e 2g

121 NEXAFS of Benzene/Ag{110} e 2u θ E =90 o θ E =20 o b 2g Tilted? Distorted, C 6v? J. Stöhr, "NEXAFS Spectroscopy" Springer Series in Surface Sciences 25, Springer, Berlin 1992.

122 What are the main features of the Carbon and Oxygen NEXAFS spectra of the adsorbed Formate (HCOO - /Cu{110}) species under the following conditions? θ E =90 o θ E =20 o z y, <100> x, <110> x, <110> I = A.<i µ f> 2 I α [A.µ] 2 = const. cos 2 θ C 2v E C 2 σ xz σ yz A 1 A 2 B 1 B z R z x, R y y, R x

123 NEXAFS at the Carbon and Oxygen K (1s) edges of adsorbed formate species Formate electronic structure? π π 1a 12,1b 12, 2a 12, 3a 12,2b 12, 4a 12, 5a 12,6a 12,1b 22,1a 22,3b 12,4b 1 2 2b 20,7a 10,8a 10,5b 1 0 O 1s C 1s O 2s C 2s C 2p O 2p H 1s Symmetry adapted orbitals of the formate ion H 1s C 1s,2s O 1s,2s O 1s,2s O 2pz C 2pz O 2pz O 2px C 2px O 2px C 2py O 2py O 2py

124 θ E =90 o θ E =20 o z y, <100> I = A.<i µ f> 2 x, <110> x, <110> C 2v E C 2 σ xz σ yz A 1 A 2 B 1 B z x y

125 Θ E =20 o Θ E =20 o y, <100> Θ E =90 o x, <110> Θ E =90 o J. Somers, A.W. Robinson, Th. Lindner, D. Ricken and A.M. Bradshaw, Phys. Rev. B40 (1989) y z z x

126 y z z x * x x z

127 Angle-resolved photoemission from adsorbed species Dipole coupling links initial and final states, i> and f> I = A.<i µ f> 2 The relative importance of the components µ x, µ y and µ z is determined by the photon beam polarisation and angle of incidence through the A.µ The detector position determines if and to what extent the final state is observable a) In normal emission the final state must belong to the totally symmetric representation of the relevant surface point group to be observable. In spherical harmonic terms l,0>. b) In mirror plane emission, the final state must be symmetric with respect to the plane to be observable. NB for photons incident in the mirror emission plane, s- polarised radiation excites antisymmetric (odd) initial states into the detector <- - +> while p-polarised radiation excites symmetric (even) initial states into the detector <+ + +>

128 Angle-resolved photoemission from S/Ni{100} p(2x2) p x E p y p z A c(2x2) - ( 2x 2)R45 o P x +P y P x -P y p z E 2C 4 C 2 2σ v 2σ d z I = A.<i µ f> 2 x,y f>

129 E.W. Plummer, B. Tonner, N. Holzwarth and A. Liebsch Phys Rev B21 (1980) 4306

130 P x P y p z Γ Γ X <100> X Γ X

131 Angle resolved photoemission from Chemisorbed CO I = A.<i µ f> 2 f>

132 Predict the angle resolved photoemission from chemisorbed CO under the following experimental conditions a) Normal emission and s-polarised radiation b) Normal emission and p-polarised radiation at high incidence angle Angle-resolved photoemission from CO/Ni

133 Tetracene A B C D E y x a u D 2h E C 2 (z) C 2 (y) C 2 (x) i σ(xy) σ(xz) σ(yz) A g B 1g B 2g b 3g B 3g A u B 1u z b 2g b 1u B 2u y B 3u x p z B 2g +5B 3g +4A u +5B 1u

134 For a flat-lying molecule, which bands are observable in normal emission? f>? I = A.<i µ f> 2 In spherical harmonic terms l,0>? D 2h E C 2 (z) C 2 (y) C 2 (x) i σ(xy) σ(xz) σ(yz) A g B 1g B 2g B 3g A u B 1u z B 2u y B 3u x µ x <? µ b y 1u µ a > z g b 3u b 2u b 1u b 1u a g <? >

135 < b 2g b 3u b 1u > b 3u b 2u < b 3g b 2u b 1u > b 1u b 1u a g <? > < a g b 1u b 1u > < b 3u b 3u a g > < b 2u b 2u a g > < b 1u b 1u a g > D 2h E C 2 (z) C 2 (y) C 2 (x) i σ(xy) σ(xz) σ(yz) A g B 1g B 2g B 3g A u B 1u z B 2u y B 3u x

136 Chemisorbed benzene D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 6 2S 3 σ h 3σ v 3σ d A 1g A 2g B 1g B 2g E 1g E 2g A 1u A 2u B 1u B 2u E 1u E 2u z x,y Valence electronic structure a 1g2, b 1u2, b 2u2, e 1u4, a 2u2, e 2g4, e 1g 4 Which bands can be observed at normal emission? What is the dependence on photon polarisation? µ x µ a y 2u µ a z 1g < > e 1u a 2u a 2u a 1g <? >

137 < a 2u a 2u a 1g > < e 1u e 1u a 1g > < a 1g a 2u a 2u > < e 1g e 1u a 2u > z, high photon incidence angle xy, low photon incidence angle D 6h E 2C 6 2C 3 C 2 3C 2 3C 2 i 2S 6 2S 3 σ h 3σ v 3σ d A 1g A 2g B 1g B 2g E 1g E 2g A 1u A 2u B 1u B 2u E 1u E 2u z x,y

138 C 6 H 6 /Pd(100) < > e 1g e 1u a 2u < e 1u a 1g > e 1u a 2u a 2u a 1g < > < a 1g a 2u a 2u > Nyberg et al, Surface Science 85 (1979) 335

139 Huber et al, Surface Science 253 (1991) 72-98

140 C 6 D 6 /Ni(110) Huber et al, Surface Science 253 (1991) The ARUPS measurements for a dilute layer suggest an orientation of the molecular plane parallel to the surface with C 2v symmetry of the adsorption complex benzene/ni(llo) and the molecules azimuthally oriented with their corners along [OOl]. For the saturated layer the molecular plane is still parallel to the surface but the symmetry of the adsorption complex is lowered to C 1 due to an azimuthal rotation of the benzene molecules induced by strong lateral interactions in the densely packed saturated layer.

141 Photoemission from oriented benzene i> l,0> z x,y a 2 1g 0,0>, 2,0> 1,0>, 3,0> 1,±1>, 3,±1> b 2 1u 3,±3> 4,±3> 2,±2> b 2 2u 3,±3> 4,±3> 2,±2> e 4 1u 1,±1>, 3,±1> 2,±1> 0,0>, 2,0>, 2,±2> a 2 2u 1,0>, 3,0> 2,0>, 4,0> 2,±1> e 4 2g 2,±2> 3,±2> 1,±1>, 3,±1>, 3,±3> e 4 1g 2,±1> 1,±1>, 3,±1> 1,0>, 3,0>, 2,±2>

142 Point symmetry elements/ operations Reducible/irreducible representations Mathematical Groups? Symmetry adapted linear combinations Electronic structure of molecules (H 2 O) ψdτ =0 ψ i ψ j dτ =0 ψ f µψ i dτ =0 Direct products Orbital symmetries/degeneracies (C 6 H 6, d orbitals) Symmetry of molecular vibrations Stretching and bending modes (H 2 O, C 2 H 4 ) IR and Raman activity of modes (C 3 H 6, C 2 H 4 ) Electronic excitations (H 2 O, Cu, C 6 H 6 ) Photoemission Polarisation A.<i µ f>

143 Symmetry elements at surfaces (inc. glide lines) Point groups/ space groups Bravais lattices Symmetry of isolated molecules at surfaces (C 2 H 4 ) Correlation tables Domains Factor groups Reciprocal space (phonons and orbitals) Umklapp processes Reflection infrared spectroscopy Metal screening effects Surface dipole activity of vibrations (C 2 H 4 ) Adsorbate induced phonons EELS (cf IR, mechanisms, HCO 2- ) NEXAFS (C 6 H 6, HCO 2- ) Photoemission (S/Ni, CO/Ni, ads. Tetracene, C 6 H 6 )

144 Merci - Au revoir