2. 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

4. Describing the bonding in more complicated molecules…... PROPYNE: CH 3 CCH 16 valence electrons Lewis Dot Structure….. C CC H H H H CENTRAL CARBON NEEDS OCTET

5. PROPYNE: CH 3 CCH has the structure CCCH H H H Tetrahedral Carbon: Hybridization? Linear Carbons Hybridization? sp 3 sp

6. PROPYNE: CH 3 CCH has the bonding CCCH H H H C sp 3 overlaps with H 1s  bond C sp 3 overlaps with C sp  bond C sp overlaps with C sp C p overlaps with C p  bond  bonds

7. H H H C C C H These four atoms are in a straight line FREE ROTATION of Methyl Group.

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

9. QUESTION C CH H H H C O H O O H THE BOND ANGLE SHOWN IS 1120° 290° 4109° 3180°

10. QUESTION C CH H H H C O H O O H THE BOND ANGLE SHOWN IS 1120° 290° 4109° 3180°

11. QUESTION C CH H H H C O H O O H THE BOND ANGLE SHOWN IS 1109° 290° 4180° 3120°

12. H H H C C O H H C O O LACTIC ACID H

14. OK, so we have it all figured out! … or do we?? Let’s think about oxygen, O 2 12 electrons … Lewis picture gives :O::O:.... 3 electron groups around each … sp 2 and 120 0

15. O O  bond  bond All electrons paired … diamagnetic BUT O 2 is paramagnetic  unpaired electrons!!

16. Atoms, molecules or ions with at least one unpaired electron are paramagnetic. Paramagnetic materials are strongly attracted to magnetic fields. MAGNETIC PROPERTIES 1s2s2p1s2s2p N1s22s22p3N1s22s22p3 Nitrogen atom is PARAMAGNETIC

18. Let’s try another approach …

19. Combining the two 1S orbitals from two hydrogen atoms will give another way to think about bonding

20. 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

21. 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

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

23. 1s A + 1s B = MO 1 1s A – 1s B = MO 2 BONDING ANTI-BONDING 1s*1s* 1s1s Each orbital can hold up to two electrons.

24. E Energy of a 1s orbital in a free atom AB COMBINING TWO 1s ORBITALS

25. E Energy of a 1s orbital in a free atom AB Energy of 1s A +1s B MO COMBINING TWO 1s ORBITALS ADDITION gives an Energy more negative than average of original orbitals 1s1s

26. 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*

27. E 1s A AB 1s1s 1s*1s* 1s B COMBINING TWO 1s ORBITALS

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

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

30. E 1s1s 1s*1s* 1s1s 1s1s He 2 He He 2 One pair of electrons goes in  1s Atomic configuration of He 1s 2 and the next pair in  1s *

31. 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.

32. BOND ORDER The net number of bonds existing after the cancellation of bonds by antibonds. the two bonding electrons were cancelled out by the two antibonding electrons. In He 2 (  1s ) 2 (  1s *) 2 Useful concept: the electronic configuration is…. There is no BOND! BOND ORDER = 0

33. 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 )

34.  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