1 CHEMISTRY 161 Chapter 10 Chemical Bonding II
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 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
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5 ADDITION OF ORBITALS builds up electron density in overlap region 1s A + 1s B = MO 1 combine them by addition AB
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 SUBTRACTION OF ORBITALS results in low electron density in overlap region.. 1s A – 1s B = MO 2 A B subtract
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 COMBINATION OF ORBITALS 1s A + 1s B = MO 1 builds up electron density between nuclei
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
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12 THE MO’s FORMED BY TWO 1s ORBITALS
13 1s A + 1s B = MO 1 1s A – 1s B = MO 2 sigma anti-bonding = 1s * sigma bonding = 1s 1s1s 1s*1s*
14 E Energy of a 1s orbital in a free atom AB COMBINING TWO 1s ORBITALS
15 E Energy of a 1s orbital in a free atom AB 1s A +1s B MO 1s1s
16 E Energy of a 1s orbital in a free atom AB 1s A -1s B MO 1s A +1s B MO 1s1s 1s*1s*
17 E 1s A AB 1s1s 1s*1s* 1s B COMBINING TWO 1s ORBITALS
18 E 1s1s 1s*1s* 1s1s 1s1s HHH2H2 bonding in H 2
19 E 1s1s 1s*1s* 1s1s 1s1s HHH2H2 the electrons are placed in the 1s molecular orbitals
20 E 1s1s 1s*1s* 1s1s 1s1s H2:(1s)2H2:(1s)2 HHH2H2
21 E 1s1s 1s*1s* 1s1s 1s1s He 2 He He 2 atomic configuration of He 1s 2
22 E 1s1s 1s*1s* 1s1s 1s1s 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 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 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 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 1s * 1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2+H2+ E He 2 + He 2 H2H2
27 1s * 1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H2+H2+ E He 2 + He 2 H 2 Dia
28 1s * 1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H 2 + Para- ½ E He 2 + He 2 H 2 Dia
29 1s * 1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) H 2 + Para- ½ E He 2 + Para- ½ He 2 H 2 Dia
30 1s * 1s Magnetism Bond order Bond energy (kJ/mol) Bond length (pm) First row diatomic molecules and ions H 2 + Para- ½ E He 2 + Para- ½ He 2 — 0 — H 2 Dia
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 E 1s1s 1s*1s* 1s1s 1s1s 2s2s 2s*2s* 2s2s 2s2s ENERGY LEVEL DIAGRAM FOR DILITHIUM Li 2
33 E 1s1s 1s*1s* 1s1s 1s1s 2s2s 2s*2s* 2s2s 2s2s Li 2 ELECTRONS FOR DILITHIUM
34 E 1s1s1s1s 1s1s Electron configuration for DILITHIUM 2s2s 2s*2s* 2s2s 2s2s ( 1s ) 2 ( 1s *) 2 ( 2s ) 2 Li 2 Bond Order ?
35 E 1s1s1s1s 1s1s Electron configuration for DILITHIUM 2s2s 2s*2s* 2s2s 2s2s ( 1s ) 2 ( 1s *) 2 ( 2s ) 2 Li 2 n b = 4 n a = 2 Bond Order = 1 single bond.
36 E 1s1s1s1s 1s1s Electron configuration for DILITHIUM 2s2s 2s*2s* 2s2s 2s2s ( 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 E 2s2s 2s*2s* 2s2s 2s2s 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 E 2s2s 2s*2s* 2s2s 2s2s Be Be 2
39 E 2s2s 2s*2s* 2s2s 2s2s Be 2 Be Be 2 Electron configuration for DIBERYLLIUM Configuration: ( 2s ) 2 ( 2s *) 2 Bond order = 0
40 E 2s2s 2s*2s* 2s2s 2s2s ( 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 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 -molecular orbitals
43 molecular orbitals
44 ENERGY LEVEL DIAGRAM E 2s2s 2s*2s* 2s2s 2s2s
45 2p*2p* 2p2p 2p2p 2p*2p* E 2p2p2p2p
46 E expected orbital splitting 2s2s 2s*2s* 2s2s 2s2s 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* This pushes the 2p up
47 E MODIFIED ENERGY LEVEL DIAGRAM 2s2s 2s*2s* 2s2s 2s2s 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* Notice that the 2p and 2p have changed places!!!!
48 E 2s2s 2s*2s* 2s2s 2s2s Electron configuration for B 2 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* Place electrons from 2s into 2s and 2s * B is [He] 2s 2 2p 1
49 E 2s2s 2s*2s* 2s2s 2s2s 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* Place electrons from 2p into 2p and 2p Remember HUND’s RULE
50 E 2s2s 2s*2s* 2s2s 2s2s 2p2p 2p*2p* 2p2p 2p2p 2p2p 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 E 2s2s 2s*2s* 2s2s 2s2s Electron configuration for B 2 : Bond order 2p2p 2p*2p* 2p2p 2p2p 2p2p 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 A SUMMARY OF THE MO’s Emphasizing nodal planes
53 ELECTRONIC CONFIGURATION OF THE HOMONUCLEAR DIATOMICS B2B2 C2C2 N2N2 O2O2 F2F2 Li 2
54 B2B2 C2C2 N2N2 O2O2 F2F2 E 2s2s 2s*2s* 2s2s 2s2s 2p2p 2p*2p* 2p2p 2p2p 2p2p 2p*2p* 2s2s 2s*2s* 2s2s 2s2s 2p*2p* 2p2p 2p2p 2p2p 2p*2p* 2p2p Li 2
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 2p * 2p * 2p 2p 2s * 2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) Second row diatomic molecules B 2 Para C2C2 N2N2 O2O2 F2F2 E
57 2p * 2p * 2p 2p 2s * 2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) Second row diatomic molecules B 2 Para C 2 Dia N2N2 O2O2 F2F2 E
58 2p * 2p * 2p 2p 2s * 2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) Second row diatomic molecules B 2 Para C 2 Dia N 2 Dia O2O2 F2F2 E
59 2p * 2p * 2p 2p 2s * 2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) Second row diatomic molecules B 2 Para C 2 Dia N 2 Dia O 2 Para F2F2 E NOTE SWITCH OF LABELS
60 2p * 2p * 2p 2p 2s * 2s Magnetism Bond order Bond E. (kJ/mol) Bond length(pm) Second row diatomic molecules B 2 Para C 2 Dia N 2 Dia O 2 Para F 2 Dia E NOTE SWITCH OF LABELS
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 2p * 2p * 2p 2p 2s * 2s E O2O2 O2+O2+ O2–O2– O 2 2-
63 2p * 2p * 2p 2p 2s * 2s E O2O2 O2+O2+ O2–O2– O 2 2-
64 2p * 2p * 2p 2p 2s * 2s E O2O2 O2+O2+ O2–O2– O 2 2-
65 2p * 2p * 2p 2p 2s * 2s E O2O2 O2+O2+ O2–O2– O 2 2-
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 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 OO OXYGEN How does the Lewis dot picture correspond to MOT? 2p*2p*2p2p2s*2s2p*2p*2p2p2s*2s E 12 valence electrons BO = 2 but PARAMAGNETIC
69 Homework Chapter 10 p ages , problem sets