1. 1 This molecular ion exits and has been experimentally measured ; its dissociation equals 2.791 eV and its H-H distance is 2.0 a 0 (1.06Å). There is no other 1 e - 2 nuclei stable system than H 2 + Hydrogenoids exist even if they might be exotic HeH 2+ is unstable relative to dissociation into He + + H +. H 2 +, the one electron system

2. 2 Write the Schrodinger Equation for H 2 + Tell whether terms are simple or difficult

3. 3 Simple, 1/R does not depend on the electron position !

4. 4 There are exact solutions of the equation. We will consider an approximate one, open to generalization  is a one-electron wave function: a molecular orbital We will consider that  is Linear Combination of Atomic Orbitals. For H 2 +, it is possible to find them using symmetry. Mirror or Inversion center: A single atomic function is not a solution I  a =  b and I  b =  a The Molecular orbitals must have the molecular symmetry.  g =  a +  b and  u =  a -  b are solutions: I  g =  a +  b =  g I  u =  a -  b =  u   g =  a +  b =  g   u =  a -  b =  u Molecular orbitals -LCAO gerade ungerade

5. 5  Normalization  g I  g  1s a +1s b I1s a +1s b   1s a I1s a  1s a I1s b  1s b I1s a  1s b I1s b   2 + 2S  u I  u  1s a - 1s b I1s a - 1s b   1s a I1s a  -  1s a I1s b  -  1s b I1s a  1s b I1s b   2 - 2S S AB

6. 6  Normalization  g I  g  1s a +1s b I1s a +1s b   1s a I1s a  1s a I1s b  1s b I1s a  1s b I1s b   2 + 2S  u I  u  1s a  -  1s b I1s a - 1s b   1s a I1s a  -  1s a I1s b  -  1s b I1s a  1s b I1s b   2 - 2S Neglecting S:  g  √  1s a +1s b ) and  u  √  1s a -1s b ) With S:  g  √  S  1s a +1s b ) and  u  √  S  1s a -1s b )

7. 7 Density partition  g I  g  1s a +1s b I1s a +1s b   1s a I1s a  1s a I1s b  1s b I1s a  1s b I1s b   2 + 2S ¼ On atom A¼ On atom B ½ On the AB bond ½ On atoms

8. 8 Neglecting S:1  g  √  1s a +1s b ) With S: 1  g  √  S  1s a +1s b ) No node, the whole space is in-phase Symmetric with respect to  h  v C ∞ C 2 and I

9. 9 Neglecting S:1  g  √  1s a +1s b ) With S: 1  g  √  S  1s a +1s b )

10. 10 Neglecting S:1  u  √  1s a -1s b ) With S: 1  u  √  S  1s a -1s b ) Nodal plane

11. 11 Charge, Bond index: Without S  g I  g    1s a +1s b I1s a +1s b     1s a I1s a    1s b I1s b    1s a I1s b    1s b I1s a   √  1/2 On atom A D = C 2 = 1/2 Q = 1 – D = +1/2 1/2 On atom B By symmetry Half on each atoms = 0 Square of the coefficient, square of amplitude L = 1/√2 1/√2 = 1/2 L AB = C A C B

12. 12 Charge, Bond index: With S  g I  g    1s a +1s b I1s a +1s b     1s a I1s a    1s b I1s b    1s a I1s b    1s b I1s a   √  S  1/2 On atom A D A = C A 2 + C A C B S AB = 1/2 Q = 1 – D = +1/2 1/2 On atom B 1/(2+2S) 1/(2+2S) S/(2+2S) S/(2+2S) Half of the contribution For bonds L = 1/√2 1/√2 S = S/2 L AB = C A C B S AB

13. 13 Energies E g and E u From the number of nodal planes, it follows that  g is below  u  E g =(  )/(1+S) E u =(  )/(1-S)

14. 14 E g and E u, bonding and antibonding states      

15. 15 E g and E u, bonding and antibonding states  2s  1s Rydberg States Valence states The bonding and antibonding levels are referred to “dissociation” Not to the “free electron” ; an antibonding level could be higher than a bonding one if referred to a higher reference level. ’’

16. 16  From the number of nodal planes, it follows that  g is below  u The atomic energy level Remember !

17. 17 This term represents the difference between H mol and H at. Either the electron is close to A: R and r b are nearly the same and [1/R - 1/r b ] is small Or the electron is far from A and 1s a 2 is small ~  or H aa, the atomic level   -13.6 eV for H

18. 18 This is a natural reference for a bond formation. For a system involving similar AOs,  = 0 This is not the usual reference (free electron) For conjugated systems of unsaturated hydrocarbon It is the Atomic energy of a 2p orbital  or H aa, the atomic level   -11.4 eV for C (2p level)

19. 19  or H ab,the bond interaction From the number of nodal planes, it follows that  g is below  u  represents the interaction energy between A and B 2  represents half of the energy gap ( E g - E u )  is the resonance integral (~ -3 eV) negative It should be roughly proportional to the overlap E g = (  )/(1+S) E u = (  )/(1-S) E g - E u = (  1-S)/(1-S 2 ) - (  )(1+S)/(1-S 2 ) E g - E u = (  - 2S  ) /(1-S 2 ) ~ 

20. 20 This is a natural unit for a bond formation. For a system involving similar bonds,  is the unit We define the unit including the negative sign. For conjugated systems of unsaturated hydrocarbon It represents half of a C=C bond (2 electrons gain the energy of the splitting)  or H ab, the value of the splitting A C=C bond is 2 

21. 21 E g and E u, with S    S  S The gap is ~2  ; the average E M value is close to  above it.

22. 22 The average E M value is above  E mean = (E g +E u )/2= [(  )/(1+S) +(  )/(1-S)]/2 E mean = [ (  )(1-S)/(1-S 2 ) +(  )(1+S)/(1-S 2 )]/2 E mean =(  S)/(1-S 2 )  S > 0 The mean value corresponds to 0 a destabilization (energy loss) The antibonding level is more antibonding than the bonding level is bonding! Small E mean

23. 23 Energy A-B distance The bonding level is stable for the equilibrium distance Electron in the antibonding level should lead to dissociation At small d, e 2 /R dominates

24. 24 The molecule with several electrons The orbitalar approximation: Molecular configurations. H 2 1g21g2 Ground state 1u21u2 1g1u1g1u 2g22g2 Excited states Rydberg states Three rules: Pauli, Stability and Hund diagram of states

25. 25 The molecule with several electrons The orbitalar approximation: Molecular configurations. H 2 1g21g2 Ground state 1u21u2 1g1u1g1u 2g22g2 Excited states Rydberg states Three rules: Pauli, Stability and Hund diagram of states     2s  1s  2s2s

26. 26 Diagram of orbitals    S  S AO right AO left MO center

27. 27 Diagram of orbitals  2e : best situation 1e  2e  3e  4e 0 (-4  S) # e energy gain Positive (4e - repulsion)

28. 28  u  diexcited state S=0 E=  First excited states:  g  u  ↓ ± ↓  ;   and ↓↓ E=  One is alone =singlet state S=0 E=  S 3 are degenerate = triplet spin S=1  g  Ground state S=0 E=  orbitals: Diagram of States:

29. 29 Mulliken charge, Bond index: ground state  g I  g    1s a +1s b I1s a +1s b   2  1s a I1s a  2  1s b I1s b  2  1s a I1s b  2  1s b I1s a  1 On atom A D A =  i i C A 2 +  i i C A C B S AB = 1 Q = 1 – D = 0 1 On atom B Half of the contribution For bonds L = 2 1/√2 1/√2 S = 1 S L = 2 1/√2 1/√2 = 1 L AB =  i i C A C B S AB i : occupancy of orbital i

30. 30 Mulliken charge, Overlap population: excited states D A =  i i C A 2 +  i i C A C B S AB = 1 Q = 1 – D = 0 First Excited States OP = 2 1/√2 (-1)/√2 S = -1 S OP AB =  i i C A C B S AB i : orbital i occupancy diexcited State D A =  i i C A 2 +  i i C A C B S AB = 1 Q = 1 – D = 0 OP = 1 1/√2 1/√2 S + 1 1/√2 (-1)/√2 S = 0 OP AB =  i i C A C B S AB i : orbital i occupancy

31. 31 Rydberg states, from 2s and 2p To find M.O.s First construct Symmetry orbitals Each atom A or B does not have the molecular symmetry, It is necessary to pair atomic orbitals between symmetry related atoms.

32. 32 Rydberg states

33. 33 The drawings or the symmetry labels are unambiguous Mathematic expression is Ambiguous; it requires defining S + for positive S, good for pedagogy - for similar direction on the z axis, better for generalization better for computerization.

34. 34 Sigma overlap S-S S- d p-dp-d s-p

35. 35 ug

36. 36 ug  overlap is lateral; it concerns p or d orbitals that have a nodal plane p-d in-phase d-d in-phase p-d out-of-phase d-d out-of-phase

37. 37 Bonding and antibonding d-d orbitals

38. 38  overlap for d orbitals

39. 39 Rydberg states

40. 40 Rydberg states

41. 41 There is no interaction no overlap no mixing between orbital of different symmetry s is symmetric relative to z p is antisymmetric relative to z

42. 42  and  separation Linear molecule: symmetry relative to C ∞ :  SYM and  Planar molecule: symmetry relative to  :  SYM and   orbitals in linear molecule: 2 sets of degenerate E g and E u orbitals. Degenerate for H, not for C ∞ not for  V ; appropriate combination shows symmetry. WARNING! Do not confuse  and  orbitals and   and  overlaps.

43. 43  orbitals in linear molecule: 2 sets of degenerate E g and E u orbitals. * Bonding Antibonding Real Complex

44. 44 Euler transformation Complex real

45. 45 Bonding Antibonding  orbitals. The  overlap (the  resonance integral) is weaker than the  one.

46. 46 Bonding Antibonding  orbitals: Lateral overlap.

47. 47  orbitals. 2S and 2P Z mix non bonding Bonding Antibonding M. O. Symmetry Orbitals g are bonding s-p hybrization u are antibonding Symmetry g Symmetry u

48. 48 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.

49. 49 Hybrid orbitals on A non bonding Bonding Antibonding M. O. Hybrid orbitals on B The non bonding hybrids Can be symmetryzed

50. 50 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).

51. 51 LCAO  a  This is a unitary transformation; n AO → n MO MO AO

52. 52 Combination of 2 AOs of same symmetry They mix to generate a bonding combination and an antibonding one. The bonding orbital is the in-phase combination looks more like the orbital of lowest energy (larger coefficient of mixing) has an energy lower than this orbital The antibonding orbital is the out-of-phase combination looks more to the orbital of highest energy (larger coefficient of mixing) has an energy higher than this orbital Antibonding Bonding

53. 53 Combination of 3 AOs of same symmetry One bonding combination, one non-bonding and one antibonding. Either 2 bonding and 2 antibonding, or 1 bonding, 2 non-bonding and an antibonding In general, Combination of 4 AOs of same symmetry

54. 54 Populating MOs 1.Fill the in increasing order, respecting the Pauli principle. 2.Do not consider where the electron originate ! This is a different problem « correlating » the « initial distribution » to the final one. To determine the ground state just respect rule 1! 2 CH 2 → H 2 C=CH 2 may be a fragment analysis to build ethene in the ground state, not an easy reaction leading directly to the ground state!

55. 55 A-A Homonuclear diatomic molecules Generalization of the LCAO approach: Build the symmetry orbitals and classify them by symmetry If E 2s -E 2p < 10 eV combine orbitals of same symmetry If E 2s -E 2p > 10 eV do not

56. 56 For Homonuclear diatomics E 2s (A) = E 2s (B)  E 2p (A) = E 2p (B) Making symmetry orbitals, we combine symmetry related orbital first!

57. 57 Symmetry orbitals  type Hybridization: 2s g and 3s g may mix; 2s u and 3s u may mix

58. 58 Symmetry orbitals 2  Due to hybridization, 3  g goes up The relative order of E 3  g and E 1  u may change

59. 59 Cas du lithium à L'azote.3  g en dessous de 1  u. Li-N: 3  g above 1  u.

60. 60 Lithium: Li 2 2 valence electrons : one occupied MO Configuration: (core)2  g 2. Li-Li single bond . Beryllium: Be 2 4 valence electrons configuration : Configuration: ((core)2  g 2 2  u 2. 2 occupied MOs no bond (excepting weak polarization). Boron: B 2 6 valence electrons configuration: (core)2  g 2 2  u 2 2  u 2  ’ u. 2 occupied MOs + 2 unpaired electrons (Hund’s rule). B 2 is paramagnetic. Bonding equivalent to a single  bond Carbon: C 2 8 valence electrons configuration: (core)2  g 2 2  u 2 2  u 2 2  ’ u 2. 4 occupied MOs 3 bonding, one antibonding Strong bonding :C=C: 2  bonds.

61. 61 Nitrogen N 2 10 valence electrons : 5 occupied MO Configuration: (core)2  g 2 2  u 2 2  u 2 2  ’ u 2 3  g 2. 1 bond and 2 bonds: This is the most stable case with the maximum of bonding electrons. It corresponds to the shortest distance and to the largest dissociation energy. It is very poorly reactive, inert most of the time (representing 80% of atmosphere). 3  g close but above 1  u. A N-N elongation weakens the bonding and the hybridization; 3 3  g passes below 1  u.

62. 62 Cas de l'oxygène et du fluor. 3  g en dessous de 1  u. O-F: 3  g below 1  u

63. 63 Oxygen O 2 10 valence electrons : 5 occupied MO plus 2 unpaired electrons Configuration: 1 s bond and 1 p bond (2 halves). paramagnetic. Fluor F 2 12 valence electrons : 7 occupied MO (core)2  g 2 2  u 2 2  u 2 2  ’ u 2 3  g 2 3  g 2 3  ’ g 2 A single bond Neon Ne 2. No bond.

64. 64 Li-H Antibonding Bonding  = 0.9506 (2s Li ) - 0.3105(1s H ).  = 0.3105 (2s Li ) +0.9506 (1s H ). d H = 1.807 Q H = -.807 d Li = 0.193 Q Li = +.807 Li-H is 80.7% ionic, 19.7% covalent. There is a dipole moment Li  + -H  -. Large coefficient on Li in  antibonding Large coefficient on H in  bonding

65. 65 HF Only one s bond Large coefficient on 2p Z (F) Dipole H  + –F  -

66. 66 CO C O N2N2

67. 67  orbitals of CO Antibonding: 2/3 2p Z (C) -1/3 2p Z (O) Non bonding: 1/3 2p Z (C)-1/3 2p Z (C)+1/3 2p Z (O) It accounts for the electron pair on C Bonding: 2/3 2s(C) +1/3 2p Z (O)