1. 1 Molecular Orbitals several atoms Any group of atoms: Molecules, Molecular ions, Fragments, Supermolecules

2. 2 Describing molecular properties as a whole Molecular orbital theory is a method for determining molecular structure in which electrons are not assigned to individual bonds between atoms, but are treated as moving under the influence of the nuclei in the whole molecule.

3. 3 The orbitalar approximation Molecular orbital is a function that describes the wave-like behavior of a single electron in a molecule. A polyelectronic wave-function is expressed in terms (as a product or a determinant) of MOs. The use of the term "orbital" was first used in English by Robert S. Mulliken in 1925 as the English translation of Schrödinger's use of the German word, 'Eigenfunktion'. Robert Sanderson Mulliken 1996-1986 Nobel 1966

4. 4 Generalizing the LCAO approach: A linear combination of atomic orbitals or LCAO It was introduced in 1929 by Lennard-Jones with the description of bonding in the diatomic molecules of the first main row of the periodic table, but had been used earlier by Pauling for H 2 +. Sir John Lennard-Jones 1894-1954 Linus Carl Pauling 1901-1994 Nobel 1962

5. 5 Input Definition of the system Total number of electrons Choice of state (ground state or excited state) Nature of atoms involved (Z) Atomic Orbitals Geometry to define Potential Output Energy, Molecular Orbitals Electronic density Spin, Magnetism Search for minimization of energy The calculated geometry is the minimum found by optimization Any calculation starts by inputs and provides outputs

6. 6 Structure and Reactivity How the energy of a system respond to a structural variation? Changing distance or angle.

7. 7 Symmetry To describe a molecule is describing Its geometry and its symmetry Its energy MOs are eigenfunctions of the symmetry operator and of H For H 2, we have use symmetries without solving the Schrodinger equation, For a larger system, symmetry helps simplifying or analyzing. One common symmetry is the plane for planar molecules; this causes   and  separation. To be considered when some parameter varies (structure modification, structure optimization, reactivity) symmetry has to be preserved during the variation

8. 8 Valence orbitals size of the orbitals Usually, the atomic orbitals that participate to the MOs are valence orbitals. These are the valence orbitals of the neutral atom. For a cation in its highest oxidation state, this might be the unoccupied shell whereas the size of the cation is given by the outermost occupied shell See the case of Li +

9. 9 Valence orbitals size of the orbitals, the Li + example Size of the cation Li + according to Slater? The last occupied shell is the 1s Z=(3-0.31) = 1.5/2.69 a 0 = 0.558 a 0 = 0.295 Ả This is small (The Pauling ionic radius is also small, 0.60 Ả) The valence orbital 2s is the one involved in forming bonds Z=(3-2*0.85) = 5/1.3 a 0 = 3.846 a 0 = 2.034 Ả Atomic radius (empirical)Atomic radius (empirical): 1.45 Ả Atomic radius (calculated)Atomic radius (calculated): 1.67 Ả Covalent radius (2008 values)Covalent radius (2008 values): 1.28 Ả Covalent radius (empirical)Covalent radius (empirical): 1.34 Ả van der Waals radius: 1.82 Ả The covalent radius for the 2s orbital is such that the orbitals overlap; it is less than the value

10. 10 Method to build M.O.s Determine the symmetry elements of the molecule Make the list of the functions involved (valence atomic orbitals) Classify them according to symmetry (build symmetry orbitals if necessary by mixing in a combination the set of orbitals related by symmetry) Combine orbitals of the same symmetry (whose overlap is significant and whose energy levels differ by less than 10 eV).

11. 11 Valence orbitals of H 2 O yOz Molecular plane xOz symmetry plane C2 z axis 2sOSSSsigma 2pxOASApi 2pyOSAAsigma 2pzOSSSsigma 1s 1 S--sigma 1s 2 S--sigma Symmetry relative to the molecular plane, sigma-pi separation

12. 12 Valence orbitals of H 2 O yOz Molecular plane xOz symmetry plane C2 z axis 2sOSSS 2pxOASA 2pyOSAA 2pzOSSS 1s 1 S-- 1s 2 S-- Symmetry relative to xOz, necessitates building symmetry orbitals Build them ! What are their energy levels?

13. 13 Valence orbitals of H 2 O yOz Molecular plane xOz symmetry plane C2 z axis 2sOSSSA1 2pxOASAB1 2pyOSAAB2 2pzOSSSA1 (1s 1 +1s 2 )SSSA1 (1s 1 -1s 2 )SAAB2 Redundancy: Three groups; A2 is not present

14. 14 C2vEyOz molecular plane xOz symmetry plane C2 axis A11111 A21 1 B111 B211

15. 15 Symmetry orbitals (1s 1 +1s 2 ) ( 1s 1 -1s 2 ) The H-H distance is 2*sin(105/2)*1.09 Ả = 1.73 Ả This is long relative to 0.74 Ả ! The energy level of the symmetry orbital is the atomic level for H, -13.6 eV

16. 16 Values from Extended Hückel 1962 (-eV) HHe 1s13.624.25 LiBeBCNOF 2s5.41015.221.42632.340 2p3.568.511.413.414.818.1 NaMgAlSiPSCl 3s5.1912.317.318.62030 3p3.4.56.59.21413.315 Roald Hoffmann Nobel 1981 William Nunn Lipscomb, Jr. American 1919- Nobel 1976

17. 17 1 Valence orbital of H 2 O Symmetry B1 2sOA1 2pxOB1 2pyOB2 2pzOA1 (1s 1 +1s 2 )A1 (1s 1 -1s 2 )B2 The atomic orbital 2pxO is a lone in its group, It is a molecular orbital It has  symmetry. E 2p (O)= -14.8 eV

18. 18 2 Valence orbitals of H 2 O Symmetry B2 2sOA1 2pxOB1 2pyOB2 2pzOA1 (1s 1 +1s 2 )A1 (1s 1 -1s 2 )B2 E 2p (O)= -14.8 eV and E 1S (H) = -13.6 eV

19. 19 3 Valence orbitals of H 2 O Symmetry A1 2sOA1 2pxOB1 2pyOB2 2pzOA1 (1s 1 +1s 2 )A1 (1s 1 -1s 2 )B2 E 2s (O)= -32.3 eV E 2p (O)= -14.8 eV And E 1S (H) = -13.6 eV E 2s is low in energy an d does not mix

20. 20 6 AO → 6 MOs 3 A1 → 3 A1 2B2 → 2B2 1 B1 → 1 B1 What is the ordering of the levels?

21. 21 Overlap involved in B2 Overlap involved in A1 At 45° same overlap angle HOH/2 = 105/2 = 52°5 > 45° Splitting in B2 > Splitting in A1

22. 22 6 AO → 6 MOs 3 A1 → 3 A1 2B2 → 2B2 1 B1 → 1 B1 In the Lewis formula, two electron pairs, two bonds The electron pairs are 2s and 2p Spectroscopy shows different electron levels The bonds are the result of two contributing MO A1 and B2 Orbital numbering: 1A1 for the core? 2A1 3A1 4A1 1B2 2B2

23. 23 Hybridization sp 3 agreement with VSEPR and Total density sp 2 closer to MOs and spectroscopy: two different pairs MOs

24. 24 The Walsh diagram

25. 25 Reduction of symmetry

26. 26 Typical Bond Angles in AH 2 Molecules molecule electronic configuration H-A-H bond angle 1a 1 = 1  g ; 1b 2 = 1  u BeH 2 (1a 1 ) 2 (1b 2 ) 2 180° BH 2 (1a 1 ) 2 (1b 2 ) 2 (2a 1 ) 1 127° CH 2 (1a 1 ) 2 (1b 2 ) 2 (2a 1 ) 1 (b 1 ) 1 134° CH 2 (1a 1 ) 2 (1b 2 ) 2 (2a 1 ) 2 102° NH 2 (1a 1 ) 2 (1b 2 ) 2 (2a 1 ) 2 (b 1 ) 1 103° OH 2 (1a 1 ) 2 (1b 2 ) 2 (2a 1 ) 2 (b 1 ) 2 104° MgH 2 (1a 1 ) 2 (1b 2 ) 2 180° AlH 2 (1a 1 ) 2 (1b 2 ) 2 (2a 1 ) 1 119° SiH 2 (1a 1 ) 2 (1b 2 ) 2 (2a 1 ) 1 (b 1 ) 1 118° SiH 2 (1a 1 ) 2 (1b 2 ) 2 (2a 1 ) 2 93° PH 2 (1a 1 ) 2 (1b 2 ) 2 (2a 1 ) 2 (b 1 ) 1 92° SH 2 (1a 1 ) 2 (1b 2 ) 2 (2a 1 ) 2 (b 1 ) 2 92°

27. 27 F. Walsh diagram for H 2 S

28. 28 This is called an orbital correlation diagram:

29. 29 3 Valence orbitals of CH 2 Symmetry A1 2sCA1 2pxCB1 2pyCB2 2pzCA1 (1s 1 +1s 2 )A1 (1s 1 -1s 2 )B2 E 2s (C)= -21.4 eV E 2p (C)= -11.4 eV And E 1S (H) = -13.6 eV E 2sC – E 1sH < 10 eV C A Hybrid orbitals Not atomic eigenfunctions But part of the MOs eigenfunctions for the molecule 2s-2p Z 2s+2p Z 2A1 is bonding 3A1 is non-bonding 4A1 is antibonding

30. 30 3 Valence orbitals of CH 2 Symmetry A1 2sCA1 2pxCB1 2pyCB2 2pzCA1 (1s 1 +1s 2 )A1 (1s 1 -1s 2 )B2 E 2s (C)= -21.4 eV E 2p (C)= -11.4 eV And E 1S (H) = -13.6 eV E 2sC – E 1sH < 10 eV C A Among the 3 A1 valence MO, the lowest one is bonding and the middle one non-bonding; this differs from H 2 O!

31. 31 hydridization Justification: s and p Z both are of same symmetry and appear in the same linear combinations (MOs); hybridization allows anticipating. hybridization aims combining AOs, each hybrid maximizing the interaction with a partner (here the symmetry orbital on Hs) and minimizing those with others. The idea is that we can thus separate the interactions with different partners. If correct, this is useful and allows transferability (replacing a substituent by another one involving the same hybrid). Mathematically a set of hybrids is equivalent to a set of canonic orbitals Failure: It is not possible to rigorously separate the interactions with different partners. Hybrids interact with each other. Existing symmetries reappear if interactions between hybrids is included.

32. 32 Methane The MO orbital description conflicts with sp 3 hybridization

33. 33

34. 34 sp 3 hybridization 2s2p x 2p y 2p z t1t1 1/2 t2t2 - 1/2 t3t3 1/2 - 1/2 1/2 - 1/2 t4t4 1/2 - 1/2 1/2 = δ ij = (E 2s -E 2p )/4 E ti = = (E 2s +3E 2p )/4

35. 35 sp 3 hybridization 3 (E 2s -E 2p )/4 (E 2s +3E 2p )/4 E 2s E 2p 1(E 2s -E 2p )/4 x   x   x   x = 0 With  =(E 2s -E 2p )/4

36. 36 sp 3 hybrids along C 3 2s2p x 2p y 2p z t1t1 1/200 √  /2 t2t2 1/2√(2/3) 0 -1/√12 t3t3 1/2-1/√61/√2-1/√12 t4t4 1/2 - 1/√6 - 1/√2 -1/√12 ¼ of 2s 3/4 of 2p

37. 37 Interest of hybridization Pictorial Anticipating AO combinations within MOs Global density Transferable: Analysis through substituents Writing VB structures What is bad with hybridization? Not describing symmetry Not good for spectroscopy Lack of orthogonality between hybrid orbitals belonging to the same center (tails of localized orbitals)

38. 38 sp 3 hybrids along C 2 2s2p x 2p y 2p z t1t1 1/21/√201/2 t2t2 -1/√201/2 t3t3 01/√2-1/2 t4t4 1/20 - 1/√2 -1/2

39. 39

40. 40

41. 41 hybridization Mixing 2s and 2p: requires degeneracy to maintain eigenfunctions of AOs. Otherwise, the hybrid orbital is an average value for the atom, not an exact solution. This makes sense when ligands impose directionality: guess of the mixing occurring in OMs.

42. 42 Angular dependence Expression along C 2 (z is the main axis) t = a s + √(1-a 2 ) [p z cos  ± p x sin  ] t = a’s + √(1-a’ 2 ) [p z cos  ’ ± p y sin  ’] 2a 2 +2a’ 2 =1 √(1-a 2 ) sin  √2 √(1-a’ 2 ) sin  ’  √2  → a = 0.459 a’ = √(1/2-a 2 ) = 0.548 sin  ’  √2)/(1/ √(1-a’ 2 )=0.8453  ’  : :   ’

43. 43 Angular dependence a 2 +a’ 2 = ½ √(1-a 2 ) sin  √2 √(1-a’ 2 ) sin  ’  √2 (1-a 2 ) sin  2  2  (1-a’ 2 ) sin  ’ 2 1-  2sin  2  a 2 1-  2sin  ’ 2  a’ 2 1-  2sin  2 +1-  2sin  ’ 2  a 2  a’ 2 2 =  2sin  2 +  2sin  ’ 2  ½ 3 =  sin  2 +  sin  ’ 2 When  decreases  ’  increases! : :   ’

44. 44 Weight of s orbital  = 105°5 water a is smaller than ½ a’ is larger than ½ The weight of s is larger in the lone pairs. The s level is lower in energy than the p level. It has to participate to the stabilization of lone pairs. L -, CH 3 + 90° pure p sp 3 ligand 109°28’ ¼ s ¾ p CH 3 -, L + >109° 28’ >1/4 s

45. 45 For the reconstructed Silicon(100) surface, the s character is higher for the dangling bonds of the outmost atoms. + -

46. 46 H3H3 The orbitals of H 3 (equilateral triangle in the xy plane) Are symmetry orbitals of NH 3 and CH 3 matching p orbitals 2p z 2p x 2p y  1/√3 -- 1/√2-1/√2

47. 47 H3H3 The orbitals of H 3 (equilateral triangle in the xy plane) are symmetry orbitals of NH 3 and CH 3 matching p orbitals 2p z 2p x 2p y  1/√3 1/√2-1/√2 --

48. 48 H3H3 Pseudo symmetry : for H 3 the OM are degenerate. They remain degenerate interacting with the 2p orbitals (E) 2p x 2p y 1/√2-1/√2 -- √(2/3) 2/3 = 1/√2 + c 2 Normalization 1 = (2/√3) 2 +2 c 2

49. 49 H3H3 --  The orbitals of H 3 (equilateral triangle in the xy plane) are symmetry orbitals of NH 3 and CH 3 matching p orbitals 1/√3 1/√2 √(2/3) -1/√6-1/√2-1/√6 2p z 2p x 2p y

50. 50 NH 3  system Bonding Lone pair

51. 51 NH 3  system 1/√2 √(2/3) -1/√6 -1/√2 -1/√6 2p x 2p y

52. 52 Ammonia

53. 53 Free rotation of a methyl group adjacent to a  system Hyperconjugation CH 3 always has one orbital conjugating with the  system Mirror symmetry pseudosymmetry

54. 54 Free rotation of a methyl group x y ZX x y ZX    

55. 55 Orientation of CH 2 D conjugated with a CH 2 + + + D D or

56. 56 x + + D A D CH 2 With a substituent, the 2 orbitals are not equivalent. That of highest energy level interacts more strongly with the CH 2 group

57. 57 Ethane

58. 58 Ethene (ethylene)

59. 59 Ethyne (acetylene)

60. 60 BF 3

61. 61 Octet rule, eighteen electron rule N in NH 3 4 atomic orbitals 4 bonding orbitals or non bonding accommodating 8 electrons 3 antibonding orbitals C in CH 4 4 atomic orbitals 4 bonding orbitals accommodating 8 electrons 4 antibonding orbitals W in W(CO) 6 9 atomic orbitals 9 bonding orbitals or non bonding accommodating 8 electrons 6 antibonding orbitals

62. 62 Ligand field for octahedral environment W in W(CO) 6 9 atomic orbitals 9 bonding orbitals or non bonding accommodating 8 electrons 6 antibonding orbitals The  interaction raises the d X 2 and d X 2 -d Y 2 levels The  with  * CO stabilizes the “non bonding levels”

63. 63 Ligand field for octahedral environment W in W(CO) 6 9 atomic orbitals 9 bonding orbitals or non bonding accommodating 8 electrons 6 antibonding orbitals The  interaction raises the d X 2 and d X 2 -d Y 2 levels The  with  * CO stabilizes the “non bonding levels”

64. 64 Ligand field for octahedral environment Ti in Ti(H 2 O) 2 (HO) 4 Or Ti in TiO 2 Ti in 9 atomic orbitals 9 bonding orbitals or non bonding accommodating 8 electrons 6 antibonding orbitals The  interaction raises the d X 2 and d X 2 -d Y 2 levels The  interaction with  O (oxygen pairs) is a stabilizing interaction. The same occurs with PH 3 ligands These 6 electrons count for the eighteen electron rule Rutile TiO 2

65. 65 Ligand field for various environments

66. 66 Ni(PH 3 ) 2...CH 2 =CH 2 NiD 2 where D is a donor substituent (group PH 3 ). Ni will be only represented by the set of d (x 2 -y 2, z 2, xy, xz et yz) orbitals taking E(3d) = - 12 eV. D is modeled by an s orbital with 2 electrons: E(D)= - 15 eV. What is the best orientation for the C=C bond? 1) Draw an energy diagram for NiD 2 2) Why the ethylene molecule is represented above by its  * orbital? 3) Draw an energy diagram for the interaction and tell which orientation is the best. nickel (Z= 28)

67. 67 Ni(PH 3 ) 2...CH 2 =CH 2 Interaction with dx 2 -y 2 Largest interaction Interaction with d(x+y),z weaker

68. 68 z y CH 2 CH 2 O O Formaldehyde *

69. 69   orbital With  overlap