1. Multi-Electron Atoms

2. Complete Description of a Ground State Wavefunction ψ A total of three quantum numbers appear from the solution of n = principal quantum number l = angular momentum quantum number m l = magnetic quantum number  nlm (r,θ,Φ)

3. What will a ground state wave- function be called??

4. Correlation of Wavefunctions to Orbitals Using the terminology of chemists:   100 (r, ,  ) is instead called the 1s orbital.  n designates the shell (1, 2, 3, 4,…..)  l designates the sub-shell (s, p, d, f….)  m l completes the description of the orbital l = 0 (s orbital) l = 1 (p orbital) l = 2 (d orbital)l = 3 (f orbital) When l =1 and m l = 0 then orbital is p z When l=1 and m l = ±1 then orbital is p x or p y

5. Hydrogen Atom Wavefunctions n = 2 l = 1 m = 0 n = 2 l = 0 m = 0 n = 2 l = 1 m = +1 n = 1 l = 0 m = 0 n = 2 l = 1 m = -1 100100 200200 211211 210210 21-1  100 1s1s -R H / 1 2 -2.18 x 10 -18 J  200  211  210  21-1 2s2s 2p x (or 2p y ) 2pz2pz 2p y (or 2p x ) -R H / 2 2 -5.45 x 10 -19 J

6. What is the corresponding orbital for a 4,1,0 state? 1.1s 2.2s 3.4s 4.5s 5.4p x 6.4p y 7.4p z 8.4d z

7. Arrangement of Shells/Subshells/Orbitals and Corresponding Quantum Numbers For a H-atom, orbitals with same value of n have equal energy. E n = -R H n 2 For any shell n there are n 2 degenerate orbitals.

8. Energy Levels For a Hydrogen Atom: 1s 2s 3s 2p x 2p z 2p y 3p x 3p y 3p z 3d xy 3d yz 3d z 2 3d xz 3d x 2 -y 2

9. Concept Check! How many orbitals in a single atom can have the following two quantum numbers: n = 4, m l = -2 1.one 2.two 3.three 4.four 5.five 6.six 7.seven 8.eight 9.zero

10. Degeneracy of states States having the same energy are called degenerate. For every value of n there are n 2 degenerate states.

11. Physical Interpretation of Ψ Quantum world is very different from the macroscopic world that we are used to seeing. Therefore unfortunately a physical interpretation of Ψ does not exist. However a physical interpretation for Ψ 2 does exist! |  nlm (r, ,  )| 2 = Probability Density Max Born Probability / Volume Figure from MIT Open CourseWare

12. Electron clouds Although we cannot know how the electron travels around the nucleus we can know where it spends the majority of its time (thus, we can know position but not trajectory). The “probability” of finding an electron around a nucleus can be calculated. Relative probability is indicated by a series of dots, indicating the “electron cloud”. 90% electron probability or cloud for 1s orbital (notice higher probability toward the centre)

13. Solution to the Wave function for a H- atom Any wave function Ψ can be divided into two components radial Ψangular Ψ R nl (r) Y lm ( ,  ) R nl (r) Y lm ( ,  ) Where a 0 = Bohr radius (constant) = 52.9 pm For all s orbitals (1s, 2s, 3s, etc,) Y is a constant.

14. Shape of an s-orbital The shape of an s-orbital is spherically symmetrical, independent of  and . Figure 1.23 Chem Principles

15. Probability Density Plots of s-orbitals Figures from MIT OCW NODE a value of r,  and  for which both  and  2 = 0 Radial Nodes = n - 1 - l For 1s radial nodes = 1 – 1 – 0 = 0 For 2s radial nodes = 2 – 1 – 0 = 1 For 3s radial nodes = 3 – 1 – 0 = 2 2a 0 RADIAL NODE: a value of r for which both  and  2 = 0

16. How many radial nodes does a hydrogen atom 3d orbital have? 1.One 2.Two 3.Three 4.Four 5.Five 6.Six 7.Seven 8.Eight 9.Zero

17. Radial Probability Distribution The probability of finding an electron in a shell of thickness dr at a distance r from the nucleus.  For s-orbitals RPD = 4πr 2  2 dr r mp = Bohr radius = 0.529Å http://www.emu.edu.tr/mugp101/PHYSLETS/physletprob/ch10_modern/radial.html

18. RPD for a 2s and a 3s orbital for H- atom: volume r mp = 6a 0 r mp = 11.5a 0 node as n increases r mp also increases.

19. Concept Check Identify the correct RPD plot (and radial node number) for a 4s orbital 1. 2. 3. 4.

20. Radial Probability Distributions for other orbitals: < 3d 3p 3s

21. Quantum Tunneling Scanning Tunneling Microscope

22. p-orbitals (l = 1): For any sub-shell l = 1 there are three p orbitals;  m = +1 or -1 (p x or p y ) and m = 0 (p z ) Difference from the s-orbitals lies in the fact that p-orbitals wave-functions depend on  and . p-orbitals are not spherically symmetrical!

23. p and d-orbitals p-orbitals look like a dumbell with 3 orientations: p x, p y, p z (“p sub z”). p-orbitals consist of two lobes seperated by a nodal plane. There is zero probability of finding a p-electron at the nucleus. Only electrons in the s orbitals have a substantial probability of being very close to the nucleus Electrons in the s orbitals are LEAST shielded.

24. d-orbitals Four of the d orbitals resemble two dumbells in a clover shape. The last d orbital resembles a p-orbital with a donut wrapped around the middle!

25. Multi-Electron Atoms and the spin quantum number m s When the Schrodinger wave equation is solved for multielectron atoms, a fourth quantum number m s the spin quantum number also appears. m s = +1/2 (spin up) or m s = -1/2 (spin down) m s completes the description of an electron and is NOT dependant on the orbital.

26. Atoms with Many Electrons and the Periodic Table The underlying physical laws necessary for …the whole of chemistry are thus completely known, and the difficulty is only that the exact application of these laws leads to equations much too complicated to be soluble! Paul Dirac (1929) Paul Dirac at a Super-Collider workshop in the early 1930s.

27. Discovery of Electron Spin: 1s 2 2s 2 2p x 2 2p y 2 2p z 2 Ne http://www.ilorentz.org/history/spin/goudsmit.html Uhlenbeck and Goudsmit Wolfgang Pauli

28. Pauli Exclusion Principle No two electrons in the same atom can have the same four quantum numbers. The Pauli exclusion principle limits us to two electrons per orbital. 1s 2 2s 2 2p x 2 2p y 2 2p z 2 Ne

29. Concept Check! How many electrons in a single atom can be in a 2p state? 1.one 2.two 3.three 4.four 5.five 6.six 7.seven 8.eight 9.zero

30. How many electrons in a single atom can have the following two quantum numbers: n = 4, m l = -2 1.One 2.Two 3.Three 4.Four 5.Five 6.Six 7.Seven 8.Eight 9.Zero

31. Shrodinger Equation for Multielectron Atoms (r 1  1  1 r 2  2  2 ) (r 1  1  1 r 2  2  2 r 3  3  3 ) NEED AN APPROXIMATION!

32. Hartree Orbitals One electron orbital approximation: e - # 1e - # 2 1s (1)1s (2) 2s(1)  100+1/2  100-1/2  200+1/2

33. Electronic Configurations Electronic configurations are basically short hand notations for different wavefunctions, using the “1 electron orbital approximation”. 1s 2 2s 2 1s 2 2s 2 2p 1

34. Multi-electron vs. Hydrogen Atom Wave Functions e.g. Ar 1s 2 2s 2 2p 6 3s 2 3p 6 Similarities to H-atom Wave functions: Differences to H-atom Wave functions: Similar in shape Identical nodal structure Each multi-electron orbital is smaller than the corresponding hydrogen atom orbital. In multi-electron atoms, orbital energies depend not only on n (shell) they also depend on l (sub-shell).

35. Multi-electron vs. Hydrogen Atom Energy Levels more negative

36. Z eff ≠ Z Z eff differs from Z because of shielding.

37. Shielding and Z eff Case A Electron #2 cancels part of the charge experienced by electron #1. Electron #1 experiences a force on average of Zeff = ___, not Zeff = +2e. The energy of electron #1 is that of an electron in a H (1-electron) atom. (2.18 x 10 -18 J) +1 total Shielding # 2 2 # 1

38. Shielding and Z eff Case B Electron #2 does not cancel the charge experienced by electron #1. Electron #1 experiences a force on average of Zeff = ___ The energy of electron #1 is that of an electron in a He +1 (1-electron) ion. (8.72 x 10 -18 J) +2 No Shielding # 1 2 # 2

39. Extreme case A: Z eff = 1, IE He = 2.18 x 10 –18 J total shielding Extreme case B: Z eff = 2, IE He = 8.72 x 10 –18 J no shielding Experimental IE He = 3.94 x 10 –18 J So the reality is somewhere between total shielding and no shielding.

40. We can calculate the Z eff from the experimentally determined IE: Our calculated Z eff should be a reasonable value, it should fall between total shielding and no shielding.

41. Which value(s) below is a possible Z eff for the 2s electron in a Li (Z = 3) atom? 1.Z eff = 0.39 2.Z eff = 0.87 3.Z eff = 1.42 4.Z eff = 3.19 5.Option 1 and 2 6.Option 1, 2, and 3 7.Option 2 and 4

42. Why is E 2s <E 2p and E 3s <E 3p <E 3d ?

43. Energy differences of s and p-orbitals: can get Also for a given n state, electrons in the s-orbitals are less shielded from the nucleus as compared to the p-electrons and hence experience a greater Z eff. single multi

44. Radial Probability Distributions for other orbitals: < 3d 3p 3s

45. Consider why the electronic configuration for Li is 1s 2 2s 1 and not 1s 2 2p 1. The s-orbital is less shielded. Averaging over the RPD yields Z eff 2p < Z eff 2s E 2s < E 2p

46. Aufbau (building up) principle Fill energy states that depend on (n & l) one electron at a time, starting at the lowest energy state. parallel O (Z = 8) 1s 2 2s 2 2p 4

47. Identify the correct electron configuration for the carbon (Z = 6) atom. 1s 2s 3s 1.1s 2 2s 2 3s 2 2.1s 2 2s 2 2p x 2 3.1s 2 2s 2 2p y 2 4.1s 2 2s 2 2p z 2 5.1s 2 2s 2 2p x 1 2p z 1 6.1s 2 2s 2 2p x 1 2p z 1 2p y 1 2p x 2p z 2p y E

48. Periods in the Periodic Table Period in a periodic table refers to the value of the principal quantum number “n”. core valence 3d 10 3d 5

49. Writing Down the Electronic Configurations:

50. Electron Configurations for Ions 3d 2 4s 2 3d 2

51. Concept Check! Refer to the periodic table and determine which element has the following electron configuration: [Ar]4s 1 3d 10 1.Cu 2.Zn 3.Ga 4.Ag 5.Cd 6.In

52. Concept Check! Select the correct electronic configuration for V 1+ (Z = 23) 1.[Ar] 4s 2 3d 3 2.[Ar] 3d 2 3.[Ar] 4s 1 3d 3 4.[Ar] 4s 2 3d 2 5.[Ar]4s 1 4d 3 6.[Ar]3d 3

53. Photo-Electron Spectroscopy (PES) K.E. 1s 2 2s 2 2p 6 1s 2 2s 2 2p 5 + K.E. 1s 2 2s 2 2p 6 1s 2 2s 1 2p 6 + 1s 2 2s 2 2p 6 1s 1 2s 2 2p 6 + K.E.

54. (Recall for the photoelectric effect: Ei = φ + KE) IE = _____ -_____EiEi KE 22 48 870

55. Which electron ejection requires the least amount of energy? 1.Si ([Ne]3s 2 3p 2 )  Si + ([Ne]3s 1 3p 2 ) + e - 2.Si ([Ne]3s 2 3p 2 )  Si + ([Ne]3s 2 3p 1 ) + e - 3.Si + ([Ne]3s 2 3p 1 )  Si +2 ([Ne]3s 2 ) + e -

56. If a certain element being studies by x-ray PES displays an emission spectrum with five distinct kinetic energies. What are all of the possible elements that could produce this spectrum?