1 CHEMISTRY 161 Chapter 10 Chemical Bonding II .

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CHEMISTRY 161

Chapter 10

Chemical Bonding II

www.chem.hawaii.edu/Bil301/welcome.html

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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

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H2 molecule:

1s orbital on Atom A 1s orbital on Atom B

the H2 molecule’s molecular orbitals can be

constructed from the two 1s atomic orbitals

1sA + 1sB = MO1

1sA – 1sB = MO2

constructive interference

destructive interference

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0r/a-3/2

01 e

12)(

a

rRs

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ADDITION OF ORBITALSbuilds up electron density in overlap region

1sA + 1sB = MO1

combine them by addition

A B

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ADDITION OF ORBITALSbuilds up electron density in overlap region.

1sA + 1sB = MO1

A Bwhat do we notice?

electron density between atoms

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SUBTRACTION OF ORBITALSresults in low electron density in overlap region..

1sA – 1sB = MO2

A B

subtract

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SUBTRACTION OF ORBITALSresults in low electron density in overlap region..

1sA – 1sB = MO2

A Bwhat do we notice?

no electron density between atoms

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COMBINATION OF ORBITALS

1sA + 1sB = MO1

builds up electron density between nuclei

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COMBINATION OF ORBITALS

1sA + 1sB = MO1

builds up electron density between nuclei

1sA – 1sB = MO2

results in low electron density between nuclei

BONDING

ANTI-BONDING

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THE MO’s FORMED BY TWO 1s ORBITALS

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1sA + 1sB = MO1

1sA – 1sB = MO2

sigma anti-bonding = 1s*

sigma bonding = 1s

1s

1s*

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E

Energy of a 1s orbital in a free atom

Energy of a 1s orbital in a free atom

A B

COMBINING TWO 1s ORBITALS

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E

Energy of a 1s orbital in a free atom

Energy of a 1s orbital in a free atom

A B

1sA+1sB

MO

1s

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E

Energy of a 1s orbital in a free atom

Energy of a 1s orbital in a free atom

A B

1sA-1sB

MO

1sA+1sB

MO

1s

1s*

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E 1sAA B

1s

1s*

1sB

COMBINING TWO 1s ORBITALS

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E1s

1s*

1s

1s

H HH2

bonding in H2

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E1s

1s*

1s

1s

H HH2

the electrons are placed in the 1s molecular orbitals

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E1s

1s*

1s

1s

H2: (1s)2

H HH2

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E1s

1s*

1s

1s

He2

He HeHe2

atomic configuration of He 1s2

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E1s

1s*

1s

1s

He2: (1s)2(1s*)2

He HeHe2

bonding effect of the (1s)2 is cancelled by the

antibonding effect of (1s*)2

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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

He2

(1s)2(1s*)2

the electronic configuration is….

BOND ORDER = 0

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BOND ORDER

=

measure of bond strength and molecular stability

If # of bonding electrons > # of antibonding electrons

Bondorder

the molecule is predicted to be stable

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BOND ORDER

= {

high bond order indicates high bond energy and short bond length

# of bonding electrons(nb)

# of antibonding electrons (na)

– 1/2 }

measure of bond strength and molecular stability

If # of bonding electrons > # of antibonding electrons

Bondorder

the molecule is predicted to be stable

H2+,H2,He2

+

= 1/2 (nb - na)

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1s*

1s

Magnetism

Bond order

Bond energy (kJ/mol)

Bond length (pm)

H2+

E

He2+ He2H2

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1s*

1s

Magnetism

Bond order

Bond energy (kJ/mol)

Bond length (pm)

H2+

E

He2+ He2H2

Dia-

1

436

74

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1s*

1s

Magnetism

Bond order

Bond energy (kJ/mol)

Bond length (pm)

H2+

Para-

½

225

106

E

He2+ He2H2

Dia-

1

436

74

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1s*

1s

Magnetism

Bond order

Bond energy (kJ/mol)

Bond length (pm)

H2+

Para-

½

225

106

E

He2+

Para-

½

251

108

He2H2

Dia-

1

436

74

30

1s*

1s

Magnetism

Bond order

Bond energy (kJ/mol)

Bond length (pm)

First row diatomic molecules and ions

H2+

Para-

½

225

106

E

He2+

Para-

½

251

108

He2

—

0

—

—

H2

Dia-

1

436

74

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HOMONUCLEAR DIATOMICS

Li2 Li : 1s22s1

both the 1s and 2s overlap to produce bonding and anti-bonding orbitals

second period

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E

1s

1s*

1s

1s

2s

2s*

2s

2s

ENERGY LEVEL DIAGRAM FOR DILITHIUM

Li2

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E

1s

1s*

1s

1s

2s

2s*

2s

2s

Li2

ELECTRONS FOR DILITHIUM

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E

1s 1s

1s

Electron configuration for DILITHIUM

2s

2s*

2s

2s

(1s)2(1s*)2(2s)2

Li2

Bond Order ?

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E

1s 1s

1s

Electron configuration for DILITHIUM

2s

2s*

2s

2s

(1s)2(1s*)2(2s)2

Li2

nb = 4 na = 2

Bond Order = 1

single bond.

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E

1s 1s

1s

Electron configuration for DILITHIUM

2s

2s*

2s

2s

(1s)2(1s*)2(2s)2

the 1s and 1s* orbitals can be ignored when

both are FILLED!

Li2

omit the inner shell

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E2s

2s*

2s

2s

Li LiLi2

The complete configuration is: (1s)2(1s*)2 (2s)2

Li2 (2s)2 only valence orbitals contribute to molecular bonding

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E2s

2s*

2s

2s

Be BeBe2 Be2

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E2s

2s*

2s

2s

Be2Be BeBe2

Electron configuration for DIBERYLLIUM

Configuration: (2s)2(2s*)2 Bond order = 0

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E2s

2s*

2s

2s

(2s)2(2s*)2Be BeBe2

Be2

Electron configuration for DIBERYLLIUM

nb = 2

na = 2

Bond Order = 1/2(nb - na) = 1/2(2 - 2) =0

No bond!!! The molecule is not stable! Now B2...

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B2

the Boron atomic configuration is

1s22s22p1

form molecular orbitals

we expect B to use 2p orbitals to

addition and subtraction

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-molecular orbitals

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molecular orbitals

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ENERGY LEVEL DIAGRAM

E

2s

2s*

2s

2s

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2p*

2p

2p

2p*

E2p 2p

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E

expected orbital splitting

2s

2s*

2s

2s

2p

2p*

2p

2p

2p

2p*

This pushes the 2p up

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E

MODIFIED ENERGY LEVEL DIAGRAM

2s

2s*

2s

2s

2p

2p*

2p2p

2p

2p*

Notice that the 2p and 2p

have changed places!!!!

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E

2s

2s*

2s

2s

Electron configuration for B2

2p

2p*

2p2p

2p

2p*

Place electrons from 2s into 2s and 2s*

B is [He] 2s22p1

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E

2s

2s*

2s

2s

2p

2p*

2p2p

2p

2p*

Place electrons from 2p into 2p and 2p

Remember HUND’s RULE

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E

2s

2s*

2s

2s

2p

2p*

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

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E

2s

2s*

2s

2s

Electron configuration for B2:

Bond order

2p

2p*

2p2p

2p

2p*(2s)2(2s*)2(2p)2

Molecule is predicted to be stable and paramagnetic.

na = 2

nb = 4

1/2(nb - na)

= 1/2(4 - 2) =1

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A SUMMARY OF THE MO’s

Emphasizing nodal planes

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ELECTRONIC CONFIGURATION OF THE HOMONUCLEAR DIATOMICS

B2 C2 N2 O2 F2Li2

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B2 C2 N2

O2 F2

E

2s

2s*

2s

2s

2p

2p*

2p2p

2p

2p*

2s

2s*

2s

2s

2p*

2p

2p

2p

2p*

2p

Li2

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2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)

Second row diatomic molecules

B2 C2 N2 O2 F2

E

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2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)

Second row diatomic molecules

B2

Para-

1

290

159

C2 N2 O2 F2

E

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2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)

Second row diatomic molecules

B2

Para-

1

290

159

C2

Dia-

2

620

131

N2 O2 F2

E

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2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)

Second row diatomic molecules

B2

Para-

1

290

159

C2

Dia-

2

620

131

N2

Dia-

3

942

110

O2 F2

E

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2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)

Second row diatomic molecules

B2

Para-

1

290

159

C2

Dia-

2

620

131

N2

Dia-

3

942

110

O2

Para-

2

495

121

F2

E

NOTE SWITCH OF LABELS

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2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)

Second row diatomic molecules

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

NOTE SWITCH OF LABELS

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2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

–

O2 :

O2+ :

O2– :

O22-:

O22-

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2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

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2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

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2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

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2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

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2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

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

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2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

O2 : B.O. = 2

O2+ : B.O. = 2.5

O2– : B.O. = 1.5

O22- : B.O. = 1

O2+ >O2 >O2

– > O22-

BOND ENERGY ORDER

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O O

OXYGEN

How does the Lewis dot picture correspond to MOT?

2p*

2p*

2p

2p

2s*

2s

E

12 valence electrons

BO = 2 but PARAMAGNETIC

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Homework

Chapter 10

pages 397-409, problem sets