1. The Final Exam December 13 (Monday) 9:00 – 12:00 Cumulative (covers everything!!) Worth 50% of total mark Multiple choice

2. The Final Exam From my portion, you are responsible for: –Chapter 8 … material from my lecture notes –Chapter 9 … everything –Chapter 10 … everything –Chapter 11 … everything –Chapter 12 … everything except 12.7

3. The Final Exam You will need to remember –Relationship between photon energy and frequency / wavelength –De Broglie AND Heisenberg relationships –Equations for energies of a particle-in-a-box AND of the hydrogen atom –VSEPR shapes AND hybribizations which give them

4. My office hours next week Wednesday Dec 8: 10-12 AND 2-4 Friday Dec 10: 10-12 AND 2-4

5. Planck Postulated “forced to have only certain discrete values” “Energy can only be transferred in discrete quantities.” is the frequency of the energy h is Planck’s constant, 6.626 x 10 -34 J s. Energy is not continuous Planck……. Energy is quantized

6. THE PHOTOELECTRIC EFFECT light electron metal KE of electron Frequency of light ( ) 0 When < 0, no electrons are ejected at any light intensity. KE of the ejected electrons depends only on the light’s frequency When > 0, the number of electrons is proportional to the light intensity. This lead Einstein to use Planck’s idea of quanta

7. EINSTEIN POSTULATED “Electromagnetic radiation can be viewed as a stream of particle-like units called photons.” Energy of a Photon: The energy of the photon depends upon the frequency

8. Diffraction of an electron beam…. We can relate these spacings to the electron wavelength De Broglie:

9. SPECTROSCOPY EMISSION Sample heated. Many excited states populated

10. SPECTROSCOPY EMISSION Sample heated. Many excited states populated n = 1 Ground state n = 2 n = 3 n = 4 n = Ion 8 Excited states... Energy The spectrum…..

11. We will describe atoms and molecules using wavefunctions, which we will give symbols … like this:  These wavefunctions contain all the information about the item we are trying to understand Since they are waves, they will have wave properties: amplitude, frequency, wavelength, phase, etc.

12. We obtain the energy by performing the “energy operation” on the wavefunction – the result is a constant (the energy) times the wavefunction H  This equation is called the Schrodinger wave equation (SWE) Let’s see how this might work

13. So H = KE operator + PE operator H = H  2  (h 2 / 8   m) + V 2  (h 2 / 8   m) + V {} 

14. PARTICLE IN A BOX ENERGY 0 L  (0) = 0  (L) = 0 BOUNDARY CONDITION x  (x)A kx  sin

15. PARTICLE IN A BOX The SCHRODINGER WAVE EQUATION for a for a 1-D particle with no potential acting on it!

19. THE HEISENBERG UNCERTAINTY PRINCIPLE  p is the uncertainty in the particle’s momentum  x is the uncertainty in the particle’s position For a particle like an electron, we cannot know both the position and velocity to any meaningful precision simultaneously.

20. Electrons have both wavelike and particle like properties. Because of the wavelike character of electron we CANNOT say that an electron WILL be found at certain point in an atom! WAVE-PARTICLE DUALITY What does  Mean????? The answer lies in

21. PARTICLE IN A BOX The probability of finding the particle on a segment of the x-axis of length dx surrounding the point x is…. If we sum all of these infinitesimal probabilities, the total must be equal to one, since there must be some finite probability of finding the particle somewhere..  n (x) A n L x  sin 

22. 0 L ENERGY OF A PARTICLE IN A BOX n =1 n = 2 n = 3 ENERGY HALF-WAVELENGTHS INTEGRAL NUMBER OF

23. n =1 n = 2 n = 3 ENERGY PROBABILITIES 22 ZERO NO CHANCE OF FINDING ELECTRON AND….

24. x Proton at (0,0,0) y z Electron at (r  2  (h 2 / 8   m) + V { }  Where now has terms in {d 2 /dr 2 ; d 2 /d  2 ; d 2 /d  2 } and V =  Ze 2 / r 2 After the transformation we still have the Schrodinger equation

25. The result of solving the Schrodinger equation this way is that we can split the hydrogen wavefunction into two:  (x,y,z)   (r,  ) = R(r) x Y(  ) Depends on r only Depends on angular variables

26. The solutions have the same features we have seen already: –Energy is quantized E n =  R Z 2 / n 2 =  2.178 x 10 -18 Z 2 / n 2 J [ n = 1,2,3 …] –Wavefunctions have shapes which depend on the quantum numbers –There are (n-1) nodes in the wavefunctions

27. Because we have 3 spatial dimensions, we end up with 3 quantum numbers: n, l, m l n = 1,2,3, …; l = 0,1,2 … (n  1); m l =  l,  l+1, …0…l  1, l n is the principal quantum number – gives energy and level l is the orbital angular momentum quantum number – it gives the shape of the wavefunction m l is the magnetic quantum number – it distinguishes the various degenerate wavefunctions with the same n and l

28. nlm l 10 (s)0 20 (s)0 1 (p)-1, 0, 1 30 (s)0 1 (p)-1, 0, 1 2 (d)-2, -1, 0, 1, 2

29. The result (after a lot of math!)

30. A more interesting way to look at things is by using the radial probability distribution, which gives probabilities of finding the electron within an annulus at distance r (think of onion skins) max. away from nucleus

32. The Radial Probability Distribution for the 3s, 3p, and 3d Orbitals

34. Another quantum number! Electrons are influenced by a magnetic field as though they were spinning charges. They are not really, but we think of them as having “spin up” or “spin down” levels. These are labeled by the 4 th quantum number: m s, which can take 2 values.

35. 1s1s E 2s2s 2p2p 3s3s 3p3p 3d3d 4s4s 4p4p 5s5s 4d4d Remember the energies are < 0 THE MULTI-ELECTRON ATOM ENERGY LEVEL DIAGRAM

36. THE PAULI PRINCIPLE An orbital is described by three quantum numbers, each orbital may contain a maximum of two electrons, and they must have opposite spins. No two electrons in the same atom can have the same set of four quantum numbers ( n, l, m l, m s ). Then each electron in a given orbital must have a different m s HOW MANY ELECTRONS IN AN ORBITAL?

37. 1s2s2p Ne:1s 2 2s 2 2p 6 1s2s2p F:1s 2 2s 2 2p 5 1s2s2p O:1s 2 2s 2 2p 4 1s2s2p N:1s 2 2s 2 2p 3 Nitrogen Oxygen Fluorine Neon THE ELECTRON CONFIGURATIONS FOR NITROGEN TO NEON

38. SCREENING AND PENETRATION THE 1s close to the nucleus PENETRATES WELLSEES A CHARGE OF Z=2 EFFECTIVE NUCLEAR CHARGE…... Is 3p THE 3p DOES NOT SCREEN THE NUCLEUS

39. THE 3s orbital penetrates better than 3p orbital 3s 3p 3d The 3p orbital penetrates better than 3d orbital Z eff (s) > Z eff (p) > Z eff (d)

40. the metals that fill the d orbitals in their valence shell. TRANSITION METALS HUND’S RULE OBEYED FOR ALL EXCEPT Cr and Cu Cr:[Ar] 4s 2 3d 4 EXPECTED OBSERVED Cr:[Ar] 4s 1 3d 5 …. The d-shell is ½ filled this way; all spin up WHEN n=3 FOR COPPER

41. Z eff INCREASES RADIUS DECREASES DOWN GROUP...

42. Z eff DECREASES RADIUS INCREASES

43. TRENDS IN EA ELECTRON AFFINITY MORE NEGATIVE ELECTRON AFFINITY MORE NEGATIVE ELECTRON AFFINITY MORE NEGATIVE

44. TRENDS IN FIRST IE Electrons closer to nucleus more tightly held Z eff DECREASES Z eff INCREASES UP THE GROUP IONIZATION ENERGY First Ionization energies decrease down the group

45. TRENDS IN FIRST IE Greater effective nuclear charge across period Poor shielding by electrons added Z eff INCREASES IONIZATION ENERGY

48. Based on the idea that all electron pairs repel each other. we can find the molecular shape! Lets see how it works…... VALENCE SHELL ELECTRON PAIR REPULSION: VSEPR The bonding and lone pairs push apart as far as possible…….. This means that atoms bound to a central atom are as far apart as possible…….

51. COMBINING ORBITALS TO FORM HYBRIDS HYBRIDIZATION : the combination of two or more “native” atomic orbitals on an atom to produce “hybrid” orbitals RULE: the number of atomic orbitals that are combined must equal the number which are formed All resulting hybrid orbitals are identical.

53. The bonding  frame work of lactic acid LACTIC ACID C CH H H H C O H O O H

54. H H H C C O H H C O O H

56. THE MO’s FORMED BY TWO 1s ORBITALS

57. The  molecular orbitals.

58.  2p *  2p *  2p or  2p  2p  or  2p  2s *  2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) Second row diatomic molecules B 2 Para- 1 290 159 C 2 Dia- 2 620 131 N 2 Dia- 3 942 110 O 2 Para- 2 495 121 F 2 Dia- 1 154 143 E NOTE SWITCH OF LABELS

59. 2s2s 2s*2s* 2s2s 2s2s E 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* 2s2s 2s*2s* 2s2s 2s2s 2p*2p* 2p2p 2p2p 2p2p 2p*2p* 2p2p

60. Double bonds involve interacting p orbitals, outside of the bonding line p-  bonding spread over whole molecule p-  antibonding p-  non-bonding

62. Benzene - aromatic molecules

64. Figure 9.48 The Pi System for Benzene