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1 CHEMISTRY 161 Chapter 10 Chemical Bonding II www.chem.hawaii.edu/Bil301/welcome.html.

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Presentation on theme: "1 CHEMISTRY 161 Chapter 10 Chemical Bonding II www.chem.hawaii.edu/Bil301/welcome.html."— Presentation transcript:

1 1 CHEMISTRY 161 Chapter 10 Chemical Bonding II www.chem.hawaii.edu/Bil301/welcome.html

2 2 MOLECULAR ORBITAL THEORY electrons occupy orbitals each of which spans the entire molecule molecular orbitals each hold up to two electrons and obey Hund’s rule, just like atomic orbitals

3 3 H 2 molecule: 1s orbital on Atom A 1s orbital on Atom B the H 2 molecule’s molecular orbitals can be constructed from the two 1s atomic orbitals 1s A + 1s B = MO 1 1s A – 1s B = MO 2 constructive interference destructive interference

4 4

5 5 ADDITION OF ORBITALS builds up electron density in overlap region 1s A + 1s B = MO 1 combine them by addition AB

6 6 ADDITION OF ORBITALS builds up electron density in overlap region. 1s A + 1s B = MO 1 AB what do we notice? electron density between atoms

7 7 SUBTRACTION OF ORBITALS results in low electron density in overlap region.. 1s A – 1s B = MO 2 A B subtract

8 8 SUBTRACTION OF ORBITALS results in low electron density in overlap region.. 1s A – 1s B = MO 2 A B what do we notice? no electron density between atoms

9 9 COMBINATION OF ORBITALS 1s A + 1s B = MO 1 builds up electron density between nuclei

10 10 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 BONDING ANTI-BONDING

11 11

12 12 THE MO’s FORMED BY TWO 1s ORBITALS

13 13 1s A + 1s B = MO 1 1s A – 1s B = MO 2 sigma anti-bonding =  1s * sigma bonding =  1s 1s1s 1s*1s*

14 14 E Energy of a 1s orbital in a free atom AB COMBINING TWO 1s ORBITALS

15 15 E Energy of a 1s orbital in a free atom AB 1s A +1s B MO 1s1s

16 16 E Energy of a 1s orbital in a free atom AB 1s A -1s B MO 1s A +1s B MO 1s1s 1s*1s*

17 17 E 1s A AB 1s1s 1s*1s* 1s B COMBINING TWO 1s ORBITALS

18 18 E 1s1s 1s*1s* 1s1s 1s1s HHH2H2 bonding in H 2

19 19 E 1s1s 1s*1s* 1s1s 1s1s HHH2H2 the electrons are placed in the  1s molecular orbitals

20 20 E 1s1s 1s*1s* 1s1s 1s1s H2:(1s)2H2:(1s)2 HHH2H2

21 21 E 1s1s 1s*1s* 1s1s 1s1s He 2 He He 2 atomic configuration of He 1s 2

22 22 E 1s1s 1s*1s* 1s1s 1s1s He 2 :(  1s ) 2 (  1s *) 2 He He 2 bonding effect of the (  1s ) 2 is cancelled by the antibonding effect of (  1s *) 2

23 23 BOND ORDER net number of bonds existing after the cancellation of bonds by antibonds the two bonding electrons were cancelled out by the two antibonding electrons He 2 (  1s ) 2 (  1s *) 2 the electronic configuration is…. BOND ORDER = 0

24 24 BOND ORDER = measure of bond strength and molecular stability If # of bonding electrons > # of antibonding electrons Bond order the molecule is predicted to be stable

25 25 BOND ORDER = { high bond order indicates high bond energy and short bond length # of bonding electrons(n b ) # of antibonding electrons (n a ) – 1/2 } measure of bond strength and molecular stability If # of bonding electrons > # of antibonding electrons Bond order the molecule is predicted to be stable H 2 +,H 2,He 2 + = 1/2 (n b - n a )

26 26  1s *  1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2+H2+ E He 2 + He 2 H2H2

27 27  1s *  1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2+H2+ E He 2 + He 2 H 2 Dia- 1 436 74

28 28  1s *  1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H 2 + Para- ½ 225 106 E He 2 + He 2 H 2 Dia- 1 436 74

29 29  1s *  1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H 2 + Para- ½ 225 106 E He 2 + Para- ½ 251 108 He 2 H 2 Dia- 1 436 74

30 30  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

31 31 HOMONUCLEAR DIATOMICS Li 2 Li : 1s 2 2s 1 both the 1s and 2s overlap to produce  bonding and anti-bonding orbitals second period

32 32 E 1s1s 1s*1s* 1s1s 1s1s 2s2s 2s*2s* 2s2s 2s2s ENERGY LEVEL DIAGRAM FOR DILITHIUM Li 2

33 33 E 1s1s 1s*1s* 1s1s 1s1s 2s2s 2s*2s* 2s2s 2s2s Li 2 ELECTRONS FOR DILITHIUM

34 34 E 1s1s1s1s 1s1s Electron configuration for DILITHIUM 2s2s 2s*2s* 2s2s 2s2s (  1s ) 2 (  1s *) 2 (  2s ) 2 Li 2 Bond Order ?

35 35 E 1s1s1s1s 1s1s Electron configuration for DILITHIUM 2s2s 2s*2s* 2s2s 2s2s (  1s ) 2 (  1s *) 2 (  2s ) 2 Li 2 n b = 4 n a = 2 Bond Order = 1 single bond.

36 36 E 1s1s1s1s 1s1s Electron configuration for DILITHIUM 2s2s 2s*2s* 2s2s 2s2s (  1s ) 2 (  1s *) 2 (  2s ) 2 the  1s and  1s * orbitals can be ignored when both are FILLED! Li 2 omit the inner shell

37 37 E 2s2s 2s*2s* 2s2s 2s2s Li Li 2 The complete configuration is: (  1s ) 2 (  1s *) 2 (  2s ) 2 Li 2 (  2s ) 2 only valence orbitals contribute to molecular bonding

38 38 E 2s2s 2s*2s* 2s2s 2s2s Be Be 2

39 39 E 2s2s 2s*2s* 2s2s 2s2s Be 2 Be Be 2 Electron configuration for DIBERYLLIUM Configuration: (  2s ) 2 (  2s *) 2 Bond order = 0

40 40 E 2s2s 2s*2s* 2s2s 2s2s (  2s ) 2 (  2s *) 2 Be Be 2 Electron configuration for DIBERYLLIUM n b = 2 n a = 2 Bond Order = 1/2(n b - n a ) = 1/2(2 - 2) =0 No bond!!!The molecule is not stable! Now B 2...

41 41 B2B2 the Boron atomic configuration is 1s 2 2s 2 2p 1 form molecular orbitals we expect B to use 2p orbitals to addition and subtraction

42 42  -molecular orbitals

43 43  molecular orbitals

44 44 ENERGY LEVEL DIAGRAM E 2s2s 2s*2s* 2s2s 2s2s

45 45 2p*2p* 2p2p 2p2p 2p*2p* E 2p2p2p2p

46 46 E expected orbital splitting 2s2s 2s*2s* 2s2s 2s2s 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* This pushes the  2p up

47 47 E MODIFIED ENERGY LEVEL DIAGRAM 2s2s 2s*2s* 2s2s 2s2s 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* Notice that the  2p and  2p have changed places!!!!

48 48 E 2s2s 2s*2s* 2s2s 2s2s Electron configuration for B 2 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* Place electrons from 2s into  2s and  2s * B is [He] 2s 2 2p 1

49 49 E 2s2s 2s*2s* 2s2s 2s2s 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* Place electrons from 2p into  2p and  2p Remember HUND’s RULE

50 50 E 2s2s 2s*2s* 2s2s 2s2s 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* (  2s ) 2 (  2s *) 2 (  2p ) 2 Abbreviated configuration Complete configuration (  1s ) 2 (  1s *) 2 (  2s ) 2 (  2s *) 2 (  2p ) 2 ELECTRONS ARE UNPAIRED

51 51 E 2s2s 2s*2s* 2s2s 2s2s Electron configuration for B 2 : Bond order 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* (  2s ) 2 (  2s *) 2 (  2p ) 2 Molecule is predicted to be stable and paramagnetic. n a = 2 n b = 4 1/2(n b - n a ) = 1/2(4 - 2) =1

52 52 A SUMMARY OF THE MO’s Emphasizing nodal planes

53 53 ELECTRONIC CONFIGURATION OF THE HOMONUCLEAR DIATOMICS B2B2 C2C2 N2N2 O2O2 F2F2 Li 2

54 54 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

55 55  2p *  2p *  2p  2p  2s *  2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) Second row diatomic molecules B2B2 C2C2 N2N2 O2O2 F2F2 E

56 56  2p *  2p *  2p  2p  2s *  2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) Second row diatomic molecules B 2 Para- 1 290 159 C2C2 N2N2 O2O2 F2F2 E

57 57  2p *  2p *  2p  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 N2N2 O2O2 F2F2 E

58 58  2p *  2p *  2p  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 O2O2 F2F2 E

59 59  2p *  2p *  2p  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 F2F2 E NOTE SWITCH OF LABELS

60 60  2p *  2p *  2p  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

61 61  2p *  2p *  2p  2p  2s *  2s E O2O2 O2+O2+ O2–O2– O 2 : O 2 + : O 2 – : O 2 2- : O 2 2-

62 62  2p *  2p *  2p  2p  2s *  2s E O2O2 O2+O2+ O2–O2– O 2 2-

63 63  2p *  2p *  2p  2p  2s *  2s E O2O2 O2+O2+ O2–O2– O 2 2-

64 64  2p *  2p *  2p  2p  2s *  2s E O2O2 O2+O2+ O2–O2– O 2 2-

65 65  2p *  2p *  2p  2p  2s *  2s E O2O2 O2+O2+ O2–O2– O 2 2-

66 66  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

67 67  2p *  2p *  2p  2p  2s *  2s E O2O2 O2+O2+ O2–O2– O 2 2- O 2 :B.O. = 2 O 2 + :B.O. = 2.5 O 2 – :B.O. = 1.5 O 2 2- :B.O. = 1 O 2 + >O 2 >O 2 – > O 2 2- BOND ENERGY ORDER

68 68 OO OXYGEN How does the Lewis dot picture correspond to MOT? 2p*2p*2p2p2s*2s2p*2p*2p2p2s*2s E 12 valence electrons BO = 2 but PARAMAGNETIC

69 69 Homework Chapter 10 p ages 397-409, problem sets

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