1. Lecture 7

2. Reducible representation: How the basis objects transform. By adding and subtracting the basis objects the reducible representation can be reduced to a combination irreducible representations. # of irreducible representations = # of classes of symmetry operations Basis of a Representation: A set of objects capable of demonstrating the effects of the all the symmetry operations in a Point Group. A set of functions, atomic orbitals on a central atom or ligands, or common objects. Representation: How the operations affect the basis objects by transforming them into themselves or each other by the operations. The representation is an array showing how the basis objects transform. Ususally we do not need the full matrices but only the trace of the matrices. These values are called the characters. The collected set of character represntations is called the character table. Irreducible representations: A representation using a minimal set of basis objects. A single atomic orbital (2s, 2p z ), a linear combination of atomic orbitals or hybrids (h 1 + h 2 ), etc. An irreducible representation may be minimally based on A single object, one dimensional, (A or B) Two objects, two dimensional, (E) Three objects, three dimensional, (T)

3. Effect of the 4 operations in the point group C 2v on a translation in the x direction. The translation is simply multiplied by 1 or -1. It forms a 1 dimensional basis to show what the operators do to an object. OperationEC2C2 vv v’v’ Transformation11

4. Character Table Verify this character. It represents how a function that behaves as x, R y, or xz behaves for C 2.

5. x, y, z Symmetry of translations (p orbitals) R x, R y, R z : rotations Classes of operations d xy, d xz, d yz, as xy, xz, yz d x 2 - y 2 behaves as x 2 – y 2 d z 2 behaves as 2z 2 - (x 2 + y 2 ) p x, p y, p z behave as x, y, z s behaves as x 2 + y 2 + z 2 Another point group, C 3v.

6. Symmetry of Atomic Orbitals

7. Naming of Irreducible representations One dimensional (non degenerate) representations are designated A or B. (A basis object is only changed into itself or the negative of itself by the symmetry operations) Two-dimensional (doubly degenerate) are designated E. (Two basis object are required to repesent the effect of the operations for an E representation. In planar PtCl 4 2- the p x and p y orbitals of the Pt, an E representation, are transformed into each other by the C 4 rotation, for instance.) Three-dimensional (triply degenerate) are designated T. (Three objects are interconverted by the symmetry operations for the T representations. In tetrahedral methane, T d, all three p orbitals are symmetry equivalent and interchanged by symmetry operations) Any 1-D representation symmetric with respect to C n is designated A; antisymmétric ones are designated B Subscripts 1 or 2 (applied to A or B refer) to symmetric and antisymmetric representations with respect to C 2  C n or (if no C 2 ) to   v respectively Superscripts ‘ and ‘’ indicate symmetric and antisymmetric behavior respectively with respect to  h. In groups having a center of inversion, subscripts g (gerade) and u (ungerade) indicate symmetric and antisymmetric representations with respect to i But note that while this rationalizes the naming, the behavior with respect to each operation is provided in the character table.

8. Character Tables You have been exposed to symmetry considerations for diatomic molecules:  or  bonding. Characters indicate the behavior of an orbital or group of orbitals under the corresponding operations (+1 = orbital does not change; -1 = orbital changes sign; anything else = more complex change) Characters in the E column indicate the dimension of the irreducible representation (of degenerate orbitals having same energy) Irrecible representations are represented by CAPITAL LETTERS (A, B, E, T,...) whereas orbitals of that symmetry behavior are represented in lowercase (a, b, e, t,...) The identity of orbitals which a row represents is found at the extreme right of the row Pairs in brackets refer to groups of degenerate orbitals and, in those cases, the characters refer to the properties of the set

9. Definition of a Group A group is a set, G, together with a binary operation, *, such that the “product” of any two members of the group is a member of the group, usually denoted by a*b, such that the following properties are satisfied : –(Associativity) (a*b)*c = a*(b*c) for all a, b, c belonging to G. –(Identity) There exists e belonging to G, such that e*g = g = g*e for all g belonging to G. –(Inverse) For each g belonging to G, there exists the inverse of g, g -1, such that g -1 *g = g*g -1 = e. If commutativity ( a*b = b*a) for all a, b belonging to G, then G is called an Abelian group. The symmetry operations of a Point Group comprise a “group”.

10. Example Is there an identity element, e? a*e = a Yes, 0 Consider the set of all integers and the operation of addition (“*” = +) Is this set of objects (all integers) associative under the operation? (a*b)*c = a*(b*c) Yes, (3 + 4) + 5 = 3 + (4+5) For each element is there an inverse element, a -1 ? a -1 * a = e Yes, 4 + (-4) = 0 Abelian? Is commutativity satisfied for each element? a * b = b * a We have a group. Yes. 3 + (-5) = (-5) + 3

11. As applied to our symmetry operators. For the C 3v point group What is the inverse of each operator? A * A -1 = E E C 3 (120) C 3 (240)  v (1)  v (2)  v (3) E C 3 (240) C 3 (120)  v (1)  v (2)  v (3)

12. Examine the matrix representation of the elements of the C 2v point group C2C2  v (xz) E -  ’ v (yz )

13. Multiplying two matrices (a reminder) Most of the transformation matrices we use have the form

14. C2C2 E  v (xz)  ’ v (yz ) What is the inverse of C 2 ?C2C2 What is the inverse of  v ? vv = =

15. C2C2 E  v (xz)  ’ v (yz ) What of the products of operations? E * C 2 = ? C2C2 =  v * C 2 = ? ’v’v =

16. Classes Two members, c 1 and c 2, of a group belong to the same class if there is a member, g, of the group such that g*c 1 *g -1 = c 2 Consider PtCl 4 C2C2 C2C2 C 2 (x)C 2 (y) So these operations belong to the same class?

17. C2C2 C2C2 C4C4 C 2 (x) = C 2 (y) = Since C 4 moves C 2 (x) on top of C 2 (y) it is an obvious choice for g C 4 = C 4 3 =

18. C 2 (x) C 2 (y)C4C4 C 4 3 = Belong to same class! How about the other two C 2 elements?

19. Properties of Characters of Irreducible Representations in Point Groups Total number of symmetry operations in the group is called the order of the group (h). For C 3v, for example, it is 6. 1 + 2 + 3 = 6 Symmetry operations are arranged in classes. Operations in a class are grouped together as they have identical characters. Elements in a class are related. This column represents three symmetry operations having identical characters.

20. Properties of Characters of Irreducible Representations in Point Groups - 2 The number of irreducible reps equals the number of classes. The character table is square. 3 by 3 The sum of the squares of the dimensions of the each irreducible rep equals the order of the group, h. 1 + 2 + 3 = 6 = h 1 2 6 = h

21. Properties of Characters of Irreducible Representations in Point Groups - 3 For any irreducible rep the squares of the characters summed over the symmetry operations equals the order of the group, h. A 1 : 1 2 + 2 (1 2 ) + 3 (1 2 ) = 6 = # of sym operations = 1+2+3 A 2 : 1 2 + 2 (1 2 ) + 3((-1) 2 ) = 6 E: 2 2 + 2 (-1) 2 + 3 (0) 2 = 6

22. Properties of Characters of Irreducible Representations in Point Groups - 4 Irreducible reps are orthogonal. The sum over the symmetry operations of the products of the characters for two different irreducible reps is zero. For A 1 and E: 1 * 2 + 2 (1 *(-1)) + 3 (1 * 0) = 0 Note that for any single irreducible rep the sum is h, the order of the group.

23. Properties of Characters of Irreducible Representations in Point Groups - 5 Each group has a totally symmetric irreducible rep having all characters equal to 1

24. Reduction of a Reducible Representation. Given a Reducible Rep how do we find what Irreducible reps it contains? Irreducible reps may be regarded as orthogonal vectors. The magnitude of the vector is h -1/2 Any representation may be regarded as a vector which is a linear combination of the irreducible representations. Reducible Rep =  (a i * IrreducibleRep i ) The Irreducible reps are orthogonal. Hence for the reducible rep and a particular irreducible rep i  (character of Reducible Rep)(character of Irreducible Rep i ) = a i * h Or a i =  (character of Reducible Rep)(character of Irreducible Rep i ) / h Sym ops

25. Reducible Representations in C s = E and  h Use the two sp hybrids as the basis of a representation h1h1 h2h2 E operation.  h operation. = = The hybrids are unaffected by the E operation. The reflection operation interchanges the two hybrids. Proceed using the trace of the matrix representation. 1 + 1 = 2 0 + 0 = 0 h 1 becomes h 1 ; h 2 becomes h 2.h 1 becomes h 2 ; h 2 becomes h 1.

26. = = The hybrids are unaffected by the E operation. The reflection operation interchanges the two hybrids. Proceed using the trace of the matrix representation. 1 + 1 = 2 0 + 0 = 0 h 1, h 1 ; become themselves h 1 h 2 ; do not become themselves, interchange Let’s observe one helpful thing here. Only the objects (hybrids) that remain themselves, appear on the diagonal of the transformation of the symmetry operation, contribute to the trace. They commonly contribute +1 or -1 to the trace depending whether or not they are multiplied by -1.

27. The Irreducible Representations for C s. CsCs E  h A’ A” 1 1 -1 x, y,R z z, R x,R y x 2,y 2,z 2,xy yz, xz The reducible representation derived from the two hybrids can be attached to the table.  2 0(h 1, h 2 ) Note that  = A’ + A”

28. The Irreducible Representations for C s. CsCs E  h A’ A” 1 1 -1 x, y,R z z, R x,R y x 2,y 2,z 2,xy yz, xz The reducible representation derived from the two hybrids can be attached to the table.  2 0(h 1, h 2 ) Let’s verify some things. Order of the group = # sym operations = 2 A’ and A’’ are orthogonal: 1*1 + 1+(-1) = 0 Sum of the squares over sym operations = order of group = h The magnitude of the A’ and A’’ vectors are each (2) 1/2 : magnitude 2 = ( 1 2 + (+/- 1) 2 )

29. Now let’s do the reduction. We assume that the reducible rep G can be expressed as a linear combination of A’ and A’’  = a A’ A’ + a A’’ A’’; our task is to find out the coefficients a A’ and a A’’ CsCs E  h A’ A” 1 1 -1 x, y,R z z, R x,R y x 2,y 2,z 2,xy yz, xz  2 0(h 1, h 2 ) a A’ = (1 * 2 + 1 *0)/2 = 1 a A’’ = (1 * 2 + 1 *0)/2 = 1 Or again G = 1*A’ + 1*A’’. Note that this holds for any reducible rep G as above and not limited to the hybrids in any way.

30. These are block-diagonalized matrices (x, y, z coordinates are independent of each other) Irreducible representations Reducible Rep

31. Point group Symmetry operations Characters +1 symmetric behavior -1 antisymmetric Mülliken symbols Each row is an irreducible representation Water is C 2v. Let’s use the Character Table

32. # of atomsdegrees of freedom Translational modes Rotational modes Vibrational modes N (linear)3 x N323N-5 Example 3 (HCN) 9324 N (non- linear) 3N333N-6 Example 3 (H 2 O) 9333 Let’s determine how many independent vibrations a molecule can have. It depends on how many atoms, N, and whether the molecule is linear or non-linear.

33. Symmetry and molecular vibrations A molecular vibration is IR active only if it results in a change in the dipole moment of the molecule A molecular vibration is Raman active only if it results in a change in the polarizability of the molecule In group theory terms: A vibrational mode is IR active if it corresponds to an irreducible representation with the same symmetry of a x, y, z coordinate (or function) and it is Raman active if the symmetry is the same as A quadratic function x 2, y 2, z 2, xy, xz, yz, x 2 -y 2 If the molecule has a center of inversion, no vibration can be both IR & Raman active

34. How many vibrational modes belong to each irreducible representation? You need the molecular geometry (point group) and the character table Use the translation vectors of the atoms as the basis of a reducible representation. Since you only need the trace recognize that only the vectors that are either unchanged or have become the negatives of themselves by a symmetry operation contribute to the character.

35. Apply each symmetry operation in that point group to the molecule and determine how many atoms are not moved by the symmetry operation. Multiply that number by the character contribution of that operation: E = 3  = 1 C 2 = -1 i = -3 C 3 = 0 That will give you the reducible representation A shorter method can be devised. Recognize that a vector is unchanged or becomes the negative of itself if the atom does not move. A reflection will leave two vectors unchanged and multiply the other by -1 contributing +1. For a rotation leaving the position of an atom unchanged will invert the direction of two vectors, leaving the third unchanged. Etc.

36. 3x3 9 1x-1 3x1 3 1x1 1  Finding the reducible representation (# atoms not moving x char. contrib.) E = 3  = 1 C 2 = -1 i = -3 C 3 = 0

37. Now separate the reducible representation into irreducible ones to see how many there are of each type A 1 = 1/4 (1x9x1 + 1x(-1)x1 + 1x3x1 + 1x1x1) = 3 A 2 =1/4 (1x9x1 + 1x(-1)x1 + 1x3x(-1) + 1x1x(-1)) = 1 931  

38. Symmetry of molecular movements of water Vibrational modes

39. IR activeRaman active Which of these vibrations having A 1 and B 1 symmetry are IR or Raman active?

40. Often you analyze selected vibrational modes (CO) Find: # vectors remaining unchanged after operation. 2 x 1 2 0 x 1 0 2 x 1 2 0 x 1 0  Example: C-O stretch in C 2v complex.

41. A 1 = 1/4 (1x2x1 + 1x0x1 + 1x2x1 + 1x0x1) = 1 A 2 = 1/4 (1x2x1 + 1x0x1 + 1x2x-1 + 1x0x-1) = 0 2020  B 1 = 1/4 (1x2x1 + 1x0x1 + 1x2x1 + 1x0x1) = 1 B 2 = 1/4 (1x2x1 + 1x0x1 + 1x2x-1 + 1x0x1) = 0 Both A 1 and B 1 are IR and Raman active = A 1 + B 1

42. What about the trans isomer? Only one IR active band and no Raman active bands Remember cis isomer had two IR active bands and one Raman active

43. Symmetry and NMR spectroscopy The # of signals in the spectrum corresponds to the # of types of nuclei not related by symmetry The symmetry of a molecule may be determined From the # of signals, or vice-versa

44. Molecular Orbitals

45. Atomic orbitals interact to form molecular orbitals Electrons are placed in molecular orbitals following the same rules as for atomic orbitals In terms of approximate solutions to the Scrödinger equation Molecular Orbitals are linear combinations of atomic orbitals (LCAO)  c a  a  c b  b (for diatomic molecules) Interactions depend on the symmetry properties and the relative energies of the atomic orbitals

46. As the distance between atoms decreases Atomic orbitals overlap Bonding takes place if: the orbital symmetry must be such that regions of the same sign overlap the energy of the orbitals must be similar the interatomic distance must be short enough but not too short If the total energy of the electrons in the molecular orbitals is less than in the atomic orbitals, the molecule is stable compared with the atoms

47. Combinations of two s orbitals (e.g. H 2 ) Antibonding Bonding More generally:  c a  (1s a )  c b  (1s b )] n A.O.’sn M.O.’s

48. Electrons in bonding orbitals concentrate between the nuclei and hold the nuclei together (total energy is lowered) Electrons in antibonding orbitals cause mutual repulsion between the atoms (total energy is raised)

49. Both  and  notation means symmetric/antisymmetric with respect to rotation  Not 

50. Combinations of two p orbitals (e.g. H 2 )  and  notation means change of sign upon C 2 rotation  and  notation means no change of sign upon rotation

51. Combinations of two p orbitals

52. Combinations of two sets of p orbitals

53. Combinations of s and p orbitals

54. Combinations of d orbitals No interaction – different symmetry  means change of sign upon C 4

55. NO YES Is there a net interaction?

56. Relative energies of interacting orbitals must be similar Strong interaction Weak interaction

57. Molecular orbitals for diatomic molecules From H 2 to Ne 2 Electrons are placed in molecular orbitals following the same rules as for atomic orbitals: Fill from lowest to highest Maximum spin multiplicity Electrons have different quantum numbers including spin (+ ½, - ½)

59. O 2 (2 x 8e) 1/2 (10 - 6) = 2 A double bond Or counting only valence electrons: 1/2 (8 - 4) = 2 Note subscripts g and u symmetric/antisymmetric upon i

60. Place labels g or u in this diagram gg  g  u uu

61. gg  u  g uu  u gg g or u?

62. Orbital mixing Same symmetry and similar energies ! shouldn’t they interact?

63.  orbital mixing When two MO’s of the same symmetry mix the one with higher energy moves higher and the one with lower energy moves lower

64. H 2  g 2 (single bond) He 2  g 2  u 2 (no bond) Molecular orbitals for diatomic molecules From H 2 to Ne 2

65. E (Z*)  E  >  E  Paramagnetic due to mixing C 2  u 2  u 2 (double bond) C 2 2-  u 2  u 2  g 2 (triple bond) O 2  u 2  u 2  g 1  g 1 (double bond) paramagnetic O 2 2-  u 2  u 2  g 2  g 2 (single bond) diamagnetic

66. Bond lengths in diatomic molecules Filling bonding orbitals Filling antibonding orbitals

67. Photoelectron Spectroscopy

68. N2N2 O2O2  * u (2s)  u (2p)  g (2p)  * u (2s)  g (2p)  u (2p)  u (2p) (Energy required to remove electron, lower energy for higher orbitals) Very involved in bonding (vibrational fine structure)