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What’s coming up???

What’s coming up???. Oct 25 The atmosphere, part 1 Ch. 8 Oct 27 Midterm … No lecture Oct 29 The atmosphere, part 2 Ch. 8 Nov 1 Light, blackbodies, Bohr Ch. 9 Nov 3,5 Postulates of QM, p-in-a-box Ch. 9 Nov 8,10 Hydrogen and multi – e atoms Ch. 9 Nov 12 Multi-electron atoms Ch.9,10

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What’s coming up???

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  1. What’s coming up??? • Oct 25 The atmosphere, part 1 Ch. 8 • Oct 27 Midterm … No lecture • Oct 29 The atmosphere, part 2 Ch. 8 • Nov 1 Light, blackbodies, Bohr Ch. 9 • Nov 3,5 Postulates of QM, p-in-a-box Ch. 9 • Nov 8,10 Hydrogen and multi – e atoms Ch. 9 • Nov 12 Multi-electron atoms Ch.9,10 • Nov 15 Periodic properties Ch. 10 • Nov 17 Periodic properties Ch. 10 • Nov 19 Valence-bond; Lewis structures Ch. 11 • Nov 22 VSEPR Ch. 11 • Nov 24 Hybrid orbitals; VSEPR Ch. 11, 12 • Nov 26 Hybrid orbitals; MO theory Ch. 12 • Nov 29 MO theory Ch. 12 • Dec 1 bonding wrapup Ch. 11,12 • Dec 2 Review 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 … everythingexcept 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

  5. COMBINATION OF ORBITALS Remember, when we take linear combinations of orbitals we get out as many as we put in. Here, the sum of the 2 orbitals 1sA + 1sB = MO1 builds up electron density between nuclei. 90% probability

  6. COMBINATION OF ORBITALS 1sA – 1sB = MO2 results in low electron density between nuclei 1sA + 1sB = MO1 builds up electron density between nuclei.

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

  8. SUBTRACTION gives an…. Energy more positive than average of original orbitals s1s* Energy of a 1s orbital in a free atom Energy of a 1s orbital in a free atom A B E s1s ADDITION gives an…. Energy more negative than average of original orbitals

  9. The bonding in H2 H H2 H s1s* E 1s 1s s1s

  10. H H2 H s1s* E 1s 1s s1s H2: (s1s)2

  11. He2: (s1s)2(s1s*)2 The He2 molecule is not a stable species. He He2 He s1s* E 1s 1s s1s The bonding effect of the (s1s)2 is cancelled by the antibonding effect of (s1s*)2

  12. BOND ORDER A measure of bond strength and molecular stability. If # of bonding electrons > # of antibonding electrons the molecule is predicted to be stable { Bond order } # of bonding electrons(nb) = # of antibonding electrons (na) 1/2 – = (nb - na) 1/2 A high bond order indicates high bond energy and short bond length. Consider H2+,H2,He2+,He2……….

  13. First row diatomic molecules and ions s1s* s1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2 Dia- 1 436 74 H2+ Para- ½ 225 106 He2+ Para- ½ 251 108 He2 — 0 — — E

  14. ELECTRONS FOR DILITHIUM s2s* Li2 2s 2s E s2s Put the electrons in the MO’s s1s* 1s 1s s1s

  15. Electron configuration for DILITHIUM s2s* Li2 (s1s)2(s1s*)2(s2s)2 2s 2s E s2s Bond Order = (nb - na) 1/2 = 1/2(4 - 2) =1 A single bond. 1s 1s s1s

  16. Only valence orbitals contribute to molecular bonding Li2(s2s)2 Li Li2 Li s2s* E 2s 2s s2s (s1s)2(s1s*)2 assumed

  17. Electron configuration for DIBERYLLIUM Be2 Be Be2 Be s2s* E 2s 2s s2s Bond order? (s2s)2(s2s*)2 Configuration:

  18. B2 The Boron atomic configuration is 1s22s22p1 So we expect B to use 2p orbitals to form molecular orbitals. How do we do that??? Combine them by addition and subtraction BUT … remember there are 3 sets of p-orbitals to combine

  19. s molecular orbitals SUBTRACT s2p*antibonding - - + + ADD - - + s2p bonding

  20. The p molecular orbitals. p2p* antibonding SUBTRACT - + + - + p2p bonding ADD -

  21. The p molecular orbitals.

  22. 2p 2p E The M.O.’s formed by p orbitals The p do not split as much as the s because of weaker overlap. s2p* p2p* p2p s2p Combine this with the s-orbitals…..

  23. Expected orbital splitting: s2p* p2p* p2p 2p s2p s2s* 2s 2s s2s The p do not split as much because of weaker overlap. 2p But the s and p along the internuclear axis DO interact E This pushes the s2p up..

  24. MODIFIED ENERGY LEVEL DIAGRAM s2p* s interaction p2p* s2p 2p 2p p2p E Notice that the s2p and p2p have changed places!!!! s2s* 2s 2s Now look at B2... s2s

  25. Electron configuration for B2 s2p* B is [He] 2s22p1 p2p* s2p 2p 2p p2p E s2s* 2s 2s s2s

  26. Electron configuration for B2: s2p* p2p* s2p 2p 2p p2p s2s* 2s 2s s2s (s1s)2(s1s*)2(s2s)2(s2s*)2(p2p)2 E Abbreviated configuration (s2s)2(s2s*)2(p2p)2

  27. Bond order 1/2(nb - na) s2p* = 1/2(4 - 2) =1 p2p* Molecule is predicted to be stable and paramagnetic. s2p 2p 2p p2p E s2s* 2s 2s s2s

  28. SECOND ROW DIATOMICS s2p* s2p* p2p* p2p* s2p p2p 2p 2p 2p 2p p2p s2p s2s* s2s* 2s 2s 2s 2s s2s s2s Li2 B2 C2 N2 O2 F2 E

  29. Back to Oxygen O O s2p* p2p* s2p p2p s2s* s2s E 12 valence electrons BO = 2 but PARAMAGNETIC BUT REMEMBER …THE LEWIS STRUCTURE WAS DIAMAGNETIC

  30. Second row diatomic molecules NOTE SWITCH OF LABELS s2p* p2p* s2porp2p p2pors2p s2s* s2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) B2 Para- 1 290 159 C2 Dia- 2 620 131 N2 Dia- 3 942 110 O2 Para- 2 495 121 F2 Dia- 1 154 143 E

  31. Example: Give the electron configuration and bond order for O2, O2+ , O2- & O22-. Place them in order of bond strength and describe their magnetic properties. Step 1:Determine the number of valence electrons in each: O2 : 6 + 6 = 12 O2+ : 6 + 6 - 1 = 11 O2– : 6 + 6 + 1 = 13 O22- : 6 + 6 + 2 = 14

  32. Step 2: Determine the valence electrons configurations: O2 :(s2s)2(s2s*)2 (s2p)2(p2p)4 (p2p*)2 O2+ : O2– : s2p* p2p* p2p s2p s2s* s2s O2 O2+ O2– O22- E

  33. s2p* p2p* p2p s2p s2s* s2s O22- O2 O2+ O2– E O2 :(s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)2 O2+ :(s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)1 O2– :(s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)3 O22- :(s2s)2(s2s*)2 (s2p)2 (p2p)4(p2p*)4

  34. Step 3: Determine the bond orders of each species: s2p* p2p* p2p s2p s2s* s2s O22- O2 O2+ O2– E O2 : B.O. = (8 - 4)/2 = 2 O2+ : B.O. = (8 - 3)/2 = 2.5 O2– : B.O. = (8 - 5)/2 = 1.5 O22- : B.O. = (8 - 6)/2 = 1

  35. s2p* s2p* p2p* p2p* s2p p2p 2p 2p 2p 2p p2p s2p s2s* s2s* 2s 2s 2s 2s s2s s2s HETERONUCLEAR DIATOMICS E

  36. NITRIC OXIDE (NO) s2p* p2p* s2p p2p s2s* s2s E 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…..

  37. NITRIC OXIDE (NO) s2p* p2p* s2p p2p s2s* s2s E Bond order Molecule is stable and paramagnetic. Experimental data agrees. NO+ and CN-

  38. NO+: Number of valence electrons: 5 + 6 - 1 = 10 CN–: Number of valence electrons: 4 + 5 + 1 = 10 ISOELECTRONIC s2p* p2p* s2p p2p s2s* s2s E Bond order TRIPLE BOND Ions are both stable and diamagnetic. Experimental data agrees.

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

  40. s2p* p2p* p2p s2p s2s* s2s E NeO: Number of valence electrons: 8 + 6 = 14 Bond order SINGLE BOND Therefore …. It could exist.

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