1. 1 He and hydrogenoid ions The one nucleus-electron system

2. 2 topic Mathematic required. Schrödinger for a hydrogenoid Orbital s Orbital p

3. 3 Two prerequisites Our world is 3D! We need to calculate integrals and derivatives in full space (3D). A system of one atom has spherical symmetry. Spherical units are appropriate.  r  rather than  x,y,z  except that  was defined in cartesian units.

4. 4 Spherical units x = r sin . cos  y = r sin . sin  z = r cos 

5. 5 derivation d  /dr for fixed  and 

6. 6 derivation If we know making one derivation, we know how to make others, to make second derivatives, and the we know calculating the Laplacian, 

7. 7 Integration in space dV = r 2 sin  drd  d 

8. 8 Integration limits     r = 0 r = ∞

9. 9 Integration in space

10. 10 Integration in space Integration over ,  and r gives V = 4/3  r 4 The integration over  and  gives the volume between two spheres of radii r and r+dr: dV = 4  r 2 dr Volume of a sphere Volume between two concentric spheres

11. 11 Dirac notation triple integrals

12. 12 Spherical symmetry dV = 4  r 2 dR

13. 13 Radial density dP/dR = 4  r 2  *  It is the density of probability of finding a particule (an electron) at a given distance from a center (nucleus) It is not the density of probability per volume dP/dV=    It is defined relative to a volume that increases with r. The unit of  is L -3/2

14. 14 Schrödinger for a hydrogenoid (1 nucleus – 1 electron) The definition of an orbital atomic orbital: any function  e (x,y,z) representing a stationary state of an atomic electron. Born-Oppenheimer approximation: decoupling the motion of N and e  (x N,y N,z N,x e,y e,z e )=  (x N,y N,z N )  (x e,y e,z e ) m H =1846 m e : When e - covers 1m H covers 2.4 cm, C 6.7 mm and Au 1.7 mm

15. 15 Schrödinger for a hydrogenoid (1 nucleus – 1 electron)

16. 16 Schrödinger for a hydrogenoid (1 nucleus – 1 electron)

17. 17 Schrödinger for a hydrogenoid (1 nucleus – 1 electron) We first look for solutions valid for large r

18. 18 Solution for large r Which of the 2 would you chose ?

19. 19 Solution  e -  r still valid close to the nucleus Already set to zero by taking Ne -  r New To be set to zero leading to a condition on  : a quantification due to the potential

20. 20 The quantification of  is a quantification on E This energy is negative. The electron is stable referred to the free electron

21. 21 Energy units 1 Rydberg = 21.8 10 -19 Joules =14.14 10 5 J mole -1 1 Rydberg = 13.606 eV = 0.5 Hartree (atomic units) 1eV (charge for an electron under potential of 1 Volt) 1eV = 1.602 10 -19 Joules = 96.5 KJoules mole -1 (→ = 8065.5 cm -1 ) 1eV = = 24.06 Kcal mole -1.

22. 22 Atomic units The energy unit is that of a dipole +/- e of length a 0 It is the potential energy for H which is not the total energy for H (-1/2 a.u.) (E=T+V) Atomic units : h/2  =1 and 1/4  0 =1 –The Schrödinger equation becomes simpler lengthchargemassenergy a 0, Bohr radius e, electron charge 1.602 10 -19 C m, electron mass 9.11 10 -41 Kg Hartree

23. 23 Normalization of Ne -  r From math textbooks The density of probability is maxima at the nucleus and decreases with the distance to the nucleus.

24. 24 Radial density of probability a 0 /Z is the most probable distance to the nucleus; it was found by Niels Bohr using a planetary model. The radial density close to zero refers to a dense volume but very small; far to zero, it corresponds to a large volume but an empty one

25. 25 Orbital 1s

26. 26 Average distance to the nucleus In an average value, the weight of heavy values dominates: (half+double)/2 = 1.25 > 1) larger than a 0 /Z Distance: Operator r From math textbooks

27. 27 Distances the nucleus

28. 28 Excited states We have obtained a solution using  e -  r ; it corresponds to the ground state. There are other quantified levels still lower than E=0 (classical domain where the e is no more attached to the nucleus)o We can search for other spherical function  N n P n (r)e -  r where P n (r) is a polynom of r of degree n-1

29. 29 Orbital ns E ns = Z 2 /n 2 E 1s (H) E 2s = Z 2 /4 E 1s (H) Nodes: spheres for solution of equation P n =O n-1 solutions Average distance Principal quantum number

30. 30 Orbital 2s = E 2s = Z 2 /4 E 1s (H) Average distance 5 a 0 /Z A more diffuse orbital: One nodal surface separating two regions with opposite phases: the sphere for r=2a 0 /Z. Within this sphere the probability of finding the electron is only 5.4%. The radial density of probability is maximum for r=0.764a 0 /Z and r=5.246a 0 /Z. Between 4.426 et 7.246 the probability of finding the electron is 64%.

31. 31 Orbital 2s

32. 32 Radial distribution

33. 33 Resolution of Schrödinger equation in , Solving the equation in  and  leads to define two other quantum numbers. They also can be defined using momentum instead of energy. For  : Angular Momentum (Secondary, Azimuthal) Quantum Number (l): l = 0,..., n-1. For  : Magnetic Quantum Number (ml): m l = -l,..., 0,..., +l.

34. 34 Resolution of Schrödinger equation in ,the magnetic quantum number

35. 35 Quantum numbers Principal Quantum Number (n): n = 1, 2,3,4, …, ∞ Specifies the energy of an electron and the size of the orbital (the distance from the nucleus of the peak in a radial probability distribution plot). All orbitals that have the same value of n are said to be in the same shell (level). Angular Momentum (Secondary, Azimunthal) Quantum Number (l): l = 0,..., n-1. Magnetic Quantum Number (m): m = -l,..., 0,..., +l. Spin Quantum Number (m s ): m s = +½ or -½. Specifies the orientation of the spin axis of an electron. An electron can spin in only one of two directions (sometimes called up and down).

36. 36 Name of orbitals 1s 2s 2p (2p +1, 2p 0, 2p -1 ) 3s 3p (3p +1, 3p 0, 3p -1 ) 3d (3p +2, 3p +1, 3p 0, 3p -1 3p -2 ) 4s 4p (4p +1, 4p 0, 4p -1 ) 4d (4d +2, 4d +1, 4d 0, 4d -1 4d -2 ) 4f (4f +3, 4f +2, 4f +1, 4f 0, 4f -1 4f -2, 4f -3 The letter indicates the secondary Quantum number, l The index indicates the magnetic Quantum number The number indicates the Principal Quantum number

37. 37 Degenerate orbitals E= Z 2 /n 2 (E 1sH ) 1s 2s 2p (2p +1, 2p 0, 2p -1 ) 3s 3p (3p +1, 3p 0, 3p -1 ) 3d (3p +2, 3p +1, 3p 0, 3p -1 3p -2 ) 4s 4p (4p +1, 4p 0, 4p -1 ) 4d (4d +2, 4d +1, 4d 0, 4d -1 4d -2 ) 4f (4f +3, 4f +2, 4f +1, 4f 0, 4f -1 4f -2, 4f -3 Depends only on the principal Quantum number 1 function 4 functions 9 functions 16 functions Combination of degenerate functions: still OK for hydrogenoids. New expressions (same number); real expressions; hybridization.

38. 38 Functions 2p

39. 39 symmetry of 2p Z Nodes: No node for the radial part (except 0 and ∞ ) cos  = 0 corresponds to  =  /2 : the xy plane or z/r=0 : the xy plane The 2p z orbital is antisymmetric relative to this plane cos(-  )=-cos  The z axis is a C ∞ axis

40. 40 Directionality of 2p Z This is the product of a radial function (with no node) by an angular function cos . It does not depend on  and has the z axis for symmetry axis. The angular contribution to the density of probability varies like cos  2 Within cones: Full space 2 cones: a diabolo

41. 41 angle Density inside The half cone (1-cos(  l ) 3 )/2 Part of the volume (1-cos  l )/2 15° 0.049 0.017 30°0.1750.067 45°0.3230.146 60°0.4370.25 75°0.4910.37 90°0.5

42. 42 Directionality of 2p Z This is the product of a radial function (with no node) by an angular function cos . It does not depend on  and has the z axis for symmetry axis. The angular contribution to the density of probability varies like cos  2 Within cones: angle Density inside The half cone (cos  l ) 3 -1)/2 Part of the volume (cos  l -1)/2 15° 0.049 0.017 30°0.1750.067 45°0.3230.146 60°0.4370.25 75°0.4910.37 90°0.5 Probability is 87.5% in half of the space 22.5% in the other half

43. 43 Spatial representation of the angular part. Let us draw all the points M with the same contribution of the angular part to the density The angular part of the probability is OM = cos  2 All the M points belong to two spheres that touch at O

44. 44 Isodensities, isolevels

45. 45 2p orbital

46. 46 3p orbital

47. 47 The 2p x and 2p y orbitals are equivalent

48. 48 One electron equally distributed on the three 2p levels  2 is proportional to x 2 /r 2 + y 2 /r 2 + z 2 /r 2 =1 and thus does not depend on r: spherical symmetry An orbital p has a direction, like a vector. A linear combination of 3 p orbitals, is another p orbital with a different axis: The choice of the x,y, z orbital is arbitrary

49. 49 orbitals  = N radial function angular function r l (polynom of degree n-l-r) n-l-r nodes l nodes

50. 50 d orbitals Clover, the forth lobe is the lucky one; clubs have three

51. 51

52. 52 3d orbitals

53. 53 Compare the average radius of 1s for the hydrogenoides whose nuclei are H and Pb. Pb 207 82 Make comments on the 1s orbital the atom Pb?

54. 54 Name these orbitals Spherical coordinates x = r sin  cos  ; y = r sin  sin  ;  z = r cos 

55. 55 Paramagnetism Is an atom with an odd number of electron necessarily diamagnetic? Is an atom with an even number of electron necessarily paramagnetic? What is the (l, m l ) values for Lithium? Is Li dia or para? Why?

56. 56 Summary

57. 57