1. What’s coming up??? Oct 25The atmosphere, part 1Ch. 8 Oct 27Midterm … No lecture Oct 29The atmosphere, part 2Ch. 8 Nov 1Light, blackbodies, BohrCh. 9 Nov 3,5Postulates of QM, p-in-a-boxCh. 9 Nov 8,10Hydrogen and multi – e atoms Ch. 9 Nov 12Multi-electron atomsCh.9,10 Nov 15Periodic propertiesCh. 10 Nov 17Periodic propertiesCh. 10 Nov 19Valence-bond; Lewis structures Ch. 11 Nov 22VSEPRCh. 11 Nov 24Hybrid orbitals; VSEPRCh. 11, 12 Nov 26Hybrid orbitals; MO theoryCh. 12 Nov 29MO theoryCh. 12 Dec 1bonding wrapupCh. 11,12 Dec 2Review for exam

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

3. 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

4. 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

6. COMBINATION OF ORBITALS 1s A + 1s B = MO 1 builds up electron density between nuclei. Remember, when we take linear combinations of orbitals we get out as many as we put in. Here, the sum of the 2 orbitals 90% probability

7. COMBINATION OF ORBITALS 1s A + 1s B = MO 1 builds up electron density between nuclei. 1s A – 1s B = MO 2 results in low electron density between nuclei

8. THE MO’s FORMED BY TWO 1s ORBITALS

9. E Energy of a 1s orbital in a free atom AB ADDITION gives an…. Energy more negative than average of original orbitals Energy more positive than average of original orbitals SUBTRACTION gives an…. 1s1s 1s*1s*

10. E 1s1s 1s*1s* 1s1s 1s1s HHH2H2 The bonding in H 2

11. E 1s1s 1s*1s* 1s1s 1s1s H2:(1s)2H2:(1s)2 HHH2H2

12. E 1s1s 1s*1s* 1s1s 1s1s He 2 :(  1s ) 2 (  1s *) 2 He He 2 The bonding effect of the (  1s ) 2 is cancelled by the antibonding effect of (  1s *) 2 The He 2 molecule is not a stable species.

13. BOND ORDER = { A high bond order indicates high bond energy and short bond length. # of bonding electrons(n b ) # of antibonding electrons (n a ) – 1/2 } A measure of bond strength and molecular stability. If # of bonding electrons > # of antibonding electrons Bond order the molecule is predicted to be stable Consider H 2 +,H 2,He 2 +,He 2 ………. = 1/2 (n b - n a )

14.  1s *  1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) First row diatomic molecules and ions H 2 + Para- ½ 225 106 E He 2 + Para- ½ 251 108 He 2 — 0 — H 2 Dia- 1 436 74

15. E 1s1s 1s*1s* 1s1s 1s1s 2s2s 2s*2s* 2s2s 2s2s Put the electrons in the MO’s Li 2 ELECTRONS FOR DILITHIUM

16. E 1s1s1s1s 1s1s Electron configuration for DILITHIUM 2s2s 2s*2s* 2s2s 2s2s (  1s ) 2 (  1s *) 2 (  2s ) 2 Li 2 Bond Order = 1/2 (n b - n a ) = 1/2(4 - 2) =1 A single bond.

17. E 2s2s 2s*2s* 2s2s 2s2s Li 2 (  2s ) 2 Li Li 2 (  1s ) 2 (  1s *) 2 assumed Only valence orbitals contribute to molecular bonding

18. E 2s2s 2s*2s* 2s2s 2s2s Be 2 Be Be 2 Electron configuration for DIBERYLLIUM Configuration: (  2s ) 2 (  2s *) 2 Bond order?

19. B2B2 The Boron atomic configuration is 1s 2 2s 2 2p 1 form molecular orbitals. So we expect B to use 2p orbitals to How do we do that??? Combine them byaddition and subtraction BUT … remember there are 3 sets of p-orbitals to combine

20.  molecular orbitals  2p * antibonding  2p bonding + - - + + - - ADD SUBTRACT

21. The  molecular orbitals.  2p * antibonding  2p bonding + - + - - + ADD SUBTRACT

22. The  molecular orbitals.

23. The M.O.’s formed by p orbitals 2p*2p* 2p2p 2p2p 2p*2p* The  do not split as much as the  because of weaker overlap. E 2p2p2p2p Combine this with the s-orbitals…..

24. E Expected orbital splitting: 2s2s 2s*2s* 2s2s 2s2s 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* The  do not split as much because of weaker overlap. But the s and p along the internuclear axis DO interact This pushes the  2p up..

25. E MODIFIED ENERGY LEVEL DIAGRAM 2s2s 2s*2s* 2s2s 2s2s 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p*  interaction Notice that the  2p and  2p have changed places!!!! Now look at B 2...

26. E 2s2s 2s*2s* 2s2s 2s2s Electron configuration for B 2 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* B is [He] 2s 2 2p 1

27. E Electron configuration for B 2 : 2s2s 2s*2s* 2s2s 2s2s 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* (  2s ) 2 (  2s *) 2 (  2p ) 2 Abbreviated configuration (  1s ) 2 (  1s *) 2 (  2s ) 2 (  2s *) 2 (  2p ) 2

28. E 2s2s 2s*2s* 2s2s 2s2s Bond order 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* Molecule is predicted to be stable and paramagnetic. 1/2(n b - n a ) = 1/2(4 - 2) =1

29. SECOND ROW DIATOMICS B2B2 C2C2 N2N2 O2O2 F2F2 E 2s2s 2s*2s* 2s2s 2s2s 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 Li 2

30. OO Back to Oxygen 2p*2p*2p2p2s*2s2p*2p*2p2p2s*2s E 12 valence electrons BO = 2 but PARAMAGNETIC BUT REMEMBER …THE LEWIS STRUCTURE WAS DIAMAGNETIC

31.  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

32. Example: Give the electron configuration and bond order for O 2, O 2 +, O 2 - & O 2 2-. Place them in order of bond strength and describe their magnetic properties. Step 1:Determine the number of valence electrons in each: O 2 + :6 + 6 - 1 = 11 O 2 – :6 + 6 + 1 = 13 O 2 2- :6 + 6 + 2 = 14 O 2 :6 + 6 = 12

33. Step 2:Determine the valence electrons configurations: O 2 :(  2s ) 2 (  2s *) 2 (  2p ) 2 (  2p ) 4 (  2p *) 2 O 2 + : O 2 – :  2p *  2p *  2p  2p  2s *  2s E O2O2 O2+O2+ O2–O2– O 2 2-

34.  2p *  2p *  2p  2p  2s *  2s E O2O2 O2+O2+ O2–O2– O 2 2- O 2 :(  2s ) 2 (  2s *) 2 (  2p ) 2 (  2p ) 4 (  2p *) 2 O 2 + :(  2s ) 2 (  2s *) 2 (  2p ) 2 (  2p ) 4 (  2p *) 1 O 2 – :(  2s ) 2 (  2s *) 2 (  2p ) 2 (  2p ) 4 (  2p *) 3 O 2 2- :(  2s ) 2 (  2s *) 2 (  2p ) 2 (  2p ) 4 (  2p *) 4

35.  2p *  2p *  2p  2p  2s *  2s E O2O2 O2+O2+ O2–O2– O 2 2- O 2 :B.O. = (8 - 4)/2 = 2 O 2 + :B.O. = (8 - 3)/2 = 2.5 O 2 – :B.O. = (8 - 5)/2 = 1.5 O 2 2- :B.O. = (8 - 6)/2 = 1 Step 3:Determine the bond orders of each species:

36. 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 HETERONUCLEAR DIATOMICS

37. 2p*2p*2p2p2s*2s2p*2p*2p2p2s*2s E NITRIC OXIDE (NO) Number of valence electrons: 5 + 6 = 11 USE THE MO DIAGRAM FOR HOMONUCLEAR DIATOMIC MOLECULES WITH s-p INTERACTION AS AN APPROXIMATION FOR < 12 ELECTRONS Put the electrons in…..

38. Molecule is stable and paramagnetic. NITRIC OXIDE (NO) 2p*2p*2p2p2s*2s2p*2p*2p2p2s*2s E Bond order Experimental data agrees. NO + and CN -

39. Ions are both stable and diamagnetic. NO + :Number of valence electrons: 5 + 6 - 1 = 10 CN – :Number of valence electrons: 4 + 5 + 1 = 10 ISOELECTRONIC 2p*2p*2p2p2s*2s2p*2p*2p2p2s*2s E Bond order Experimental data agrees. TRIPLE BOND

40. CAN NeO EXIST? How can we answer this question? Check bond order……...

41. Therefore …. It could exist. NeO:Number of valence electrons: 8 + 6 = 14 2p*2p*2p2p2s*2s2p*2p*2p2p2s*2s E Bond order SINGLE BOND