CHEMISTRY 161 Chapter 10 Chemical Bonding & Molecular Structure

1

CHEMISTRY 161

Chapter 10

Chemical Bonding & Molecular Structure

2

PREDICTING THE GEOMETRY OF MOLECULES

1. derive Lewis structure of the molecule

2. discriminate between bonding and non-bonding electron pairs

O HH

3. VALENCE SHELL ELECTRON PAIR REPULSION

3

VALENCE SHELL ELECTRON PAIR REPULSION

3. valence electron pairs stay as far apart as possible

VSEPR

1. identify in a compound the central atom

2. electrons repel each other

4. non-bonding electrons repel more than bonding electrons

4

central atom

no non-bonding pairs non-bonding pairs

5

Cl Be Cl

BeCl2

TWO ELECTRON PAIRS AROUND BERYLLIUM ATOM

AB2

6

Be

180°

180°

Be90°

270°

LINEAR ARRANGEMENT BEST

Cl Be Cl

IT PUTS ELECTRON PAIRS FURTHEST APART

7

BF3

F B F

F

THREE ELECTRON PAIRS AROUND THE BORON ATOM

AB3

8

B

120°

120°

120°

F B F

F

THREE ELECTRON PAIRS AROUND THE BORON ATOM

TRIGONAL PLANAR ARRANGEMENT BEST

9

F F

F

B

THE SHAPE OF BF3 IS TRIGONAL PLANAR.

MOLECULAR SHAPE

10

CH4

H C H

H

H

four electron pairs

90°

C

90° 90°

90°

expect square planar

AB4

11

better arrangement for four electron pairs

109.5°

C

TETRAHEDRAL

4 electron pairs tetrahedral

bigger than 90 ° in square planar

put on the H-atoms

12

109.5°

C

TETRAHEDRAL

C

H

HH

H

shape of CH4 is tetrahedral

13

PF5 FIVE ELECTRON PAIRS AROUND PHOSPHORUS

P

5 electron pairs trigonal bipyramidal

F

PFF

F F

AB5

14

PPF

F

F

F

F

Bond angle

900

1200

shape of PF5 is trigonal bipyramidal

two of the F atoms different from the others

15

PFF

F

F

FBond angle

900

1200

AXIAL

EQUATORIAL

16

F

S

F

F

F

F

F

S

six electron pairs around the sulfur atom

6 electron pairs

SF6

octahedral

AB6

17

S S

F

F

F

F

F

F

shape of SF6 is octahedral

900

18

central atom

no non-bonding pairs non-bonding pairs

19

SeO2 O Se O

AB2E AB3

20

Se

VSEPR treats double bonds like a single bondO Se O

THREE ELECTRON PAIRS AROUND SELENIUM

ELECTRON PAIR GEOMETRY

TRIGONAL PLANAR

21

Se

SeO2 IS V-SHAPED (OR BENT)

ADD OXYGENSSe

O O

THE MOLECULAR SHAPE IS THE POSITION OF

THE ATOMS

22

NH3

H N H

H

electron pairs around the nitrogen atom

AB3E AB4

23

NH3H N H

H

NPUT ON THE 3 H ATOMS

N

H

H

H

NH3 is trigonal pyramidal

24

AB2E2 AB4

H O H four electron pairs around the oxygen atom

O

PUT ON THE 2 H-ATOMS

O

HH

shape of H2O is V-shaped or bent

25

F

SFF

F

SF4

AB4E AB5

TRIGONAL BIPYRAMID

S

26

SF

F

F

F

SFF

F

F

WHERE DOES LONE PAIR GO?

OR

lone pairs occupy the trigonal plane (the “equator”) to minimize the number of 90° repulsions

27

SF4

1 lone pair

See-saw

shaped

ClF3

2 lone pairs

T-shaped

SF

F

F

F

ClF

F

F

lone pairs occupy the trigonal plane (the “equator”)

first to minimize the number of 90° repulsions

AB4E AB3E2

F

F

F

Xe

XeF2

3 lone pairs

Linear

AB2E3

28

BrF5

Br

F

F

F

F

F

Square pyramidal

Br

AB5E AB6

29

XeF4

XeF

F

F

F

XeXe

F

F

F

F

:

lone pairs MUST BE AT 1800

XeF

F

F

F

:

AB4E2AB6

30

Summary of Molecular ShapesTotal valence electron pairs

Electron Pair Geometry

Lone electron pairs

Shape of Molecule

2

3

4

Linear

Trigonal planar

Tetrahedral

0

0

1

0

1

2

Linear

Trigonal planar

V-shaped

Tetrahedral

Trigonal pyramid

V-shaped

31

Total valence electron pairs

Electron Pair Geometry

Lone electron pairs

Shape of Molecule

5

6

Trigonal bipyramidal

Octahedral

1

2

3

0

1

2

See-saw

T-shaped

Linear

Octahedral

Square pyramid

Square planar

0 Trig. bipyramid.

32

molecules with no single central atom

we apply our VSEPR rules to each atom in the chain

POLYATOMICS

Example: ETHANOL

33

ETHANOL

The atoms around the carbons form a.

The atoms around the oxygen form a

OCH H

H

H

C

H

H

tetrahedral arrangement

V-shaped structure.

C2H5OH

34

HH

H

CH

H

C

H

O

35

EXAMPLES

Cl2O BF4-

ICl4-

SO2Cl2 Cl2CO Cl2SO

NH2OH NH4+ N2F2

36

1. Lewis structures

2. VSEPR model

WHY DO MOLECULES FORM?

37

simplest molecule

H2

two H-atoms 1s1

two H-atoms approach each other and the electron waves interact

OVERLAP to form a region of increased electron density between the atoms

40

chemical bond with electron density in between the nuclei is called

bond

41

a covalent bond is formed by an

overlap of two valence atomic

orbitals that share an electron pair

VALENCE BOND THEORY

the better the overlap the stronger the bond

the orbitals need to point along the bonds

42

C

H

H

H

H

What orbitals are used?

hydrogen atoms bond using their 1s orbitals

carbon needs four orbitals to bond with.

2s, 2px , 2py, 2pz

[He] 2s22p2

CH4

43

[He] 2s22p2

1. The electronic configuration of carbon is

The orbital diagram is:

44

[He] 2s22p2

1. The electronic configuration of carbon is

the orbital diagram is: [He]

the Lewis dot structure is C ...

.

necessary to promote one 2s electron

45

[He] 2s12p3

[He]

PROMOTE AN ELECTRON

[He]

[He] 2s22p2

excited state (valence state)

Lewis dot structure

four unpaired electrons

we can use these to form chemical bonds

C

46

2. bonds formed with s orbitals will be different

to bonds formed with p orbitals

3. three p orbitals are mutually perpendicular,

suggesting 90° bond angles

combining the orbitals

1. a covalent bond is formed by an overlap of two

valence atomic orbitals that share an electron pair

Experiment shows that all four bonds are identical

experiment shows that methane has 109.5° bond

angles

47

we need four orbitals pointing to the vertices of a tetrahedron

orbitals are just mathematical functions

HYBRIDIZATION

C

H

H

H

Hwe can combine them

48

COMBINING ORBITALS TO FORM HYBRIDS

HYBRIDIZATION

number of atomic orbitals that are combined

the number of resulting hybrid orbitals

IS EQUAL TO

49

Combine one s and one p a sp- hybrid

+

+ ADD the orbitals

2s+ 2p

HYBRIDIZATION

50

Combine one s and one p a sp- hybrid

++

s + p

The positive part cancels negative part

2s+ 2p

HYBRIDIZATION

The positive part adds to positive part

CONSTRUCTIVE INTERFERENCE

DESTRUCTIVE INTERFERENCE

What do we get?

51

Combine one s and one p to give a sp- hybrid

+

s + p

REMEMBER IF WE MIX TWO WE MUST GET TWO BACK

The other combination is s - p

HYBRIDIZATION

2s+ 2p

Where is the nucleus?

52

HYBRIDIZATIONCombine one s and one p a sp- hybrid

+

+SUBTRACT the

orbitals2s- 2p

53


+

+SUBTRACT the

orbitals2s- 2p

SUBTRACTING THE p ORBITAL CHANGES ITS PHASE

54


+

+SUBTRACT the

orbitals2s- 2p

SUBTRACTING THE p ORBITAL CHANGES ITS PHASE

55


+ +

s - p2s- 2p


The positive part adds to positive part

CONSTRUCTIVE INTERFERENCE

DESTRUCTIVE INTERFERENCE

What do we get?

56


+

s - p


We get two equivalent sp orbitals

ORIENTED AT 1800

2s- 2p

Where is the nucleus?

57

sp-HYBRIDIZATION

s and p orbitals

two sp-hybrids

58

COMBINE one s-orbital and two p-orbitals

Get three sp2 - orbitals oriented at 1200

s and p orbitals

three sp2-hybrids

directed at 1200

59

COMBINE one s-orbital and three p-orbitals

three sp3- orbitals oriented at 109.50

60

four hybrid orbitals

needed to form four

bonds

s + px + py + pz

an atom with sp3 hybrid orbitals is said to be

4 sp3 hybrids

The four sp3 hybrid orbitals form a tetrahedral

arrangement.

sp3 hybridizationEPG of 4 pairs

METHANE: CH4

sp3 hybridized

What happens to the energies of the orbitals?

C

H

H

H

H

61

E2p

2s

Orbitals in free C atom

What happens to orbital energies when the are hybridized??

HYBRIDIZE

62

E2p

2s

Orbitals in free C atom

E sp3

Hybridized orbitals of C atom in methane

When orbitals are hybridized they have the same energy:

The FOUR sp3 hybrids are DEGENERATE.

HYBRIDIZE

63

x

y

z

x

y

z

x

y

z

x

y

z

Combine one s and three p orbitals…..

64

C

sp3 HYBRIDS

Now form the bonds to the H-atoms……...

sp3 orbitals

65

C

Each bond in methane results from the overlap of a

hydrogen 1s orbital and a carbon sp3 orbital.

H

HH

H

Hydrogen 1s orbital

Carbon sp3 orbitals

Each hybrid ready to overlap with H 1s orbitals

Form a chemical bond by sharing a pair of electrons.

66

VALENCE BOND MODEL

Step 1: Draw the Lewis structure(s)

Step 2: Determine the geometry of the electron

pairs around each atom using VSEPR OR

preferably use the EXPERIMENTAL

GEOMETRY

Step 3: Specify the hybrid orbitals needed to

accommodate the electron pairs on each

atom

Hybrid orbital model

67

sp3 hybrids are also employed in …...

all molecules that have a 4 pair EPG….

NH3, H2O, NH4+ , CCl4

OTHER MOLECULES USING sp3 HYBRIDS

AMMONIA…..

68

AMMONIA: NH3

2s 2p

N [He]

NVSEPR

Valence shell has four pairs

EPG is TETRAHEDRAL

Nitrogen electronic configuration

Need sp3 hybrids

HYBRIDIZE

H

H

H

69

E2p

2s

Orbitals in free N atom

E sp3

Hybridized orbitals of N atom in ammonia

When orbitals are hybridized they have the same energy:

sp3 hybridization…….

The FOUR sp3 hybrids are DEGENERATE.

HYBRIDIZE

70

N

N... .

.

2s 2p

sp3 hybrids on N in AMMONIA

Now form a bond

Overlap H 1s….. H

H

H

71

Three bonds

One lone pair in an sp3 hybrid

N

HH

H

N... .

.

2s 2p

AMMONIA

72

four bonds.N

HH

H

N... .

.

2s 2p

AMMONIUM ION NH4+

H+

ISOELECTRONIC WITH ?

CH4

73

O

2s 2p

WATER

O:.. ..

H O H

O

FOUR PAIRS

EPG?

TETRAHEDRAL!!

HYBRIDIZATION? sp3

74

Overlap of two of oxygen sp3 hybrids with …..

H atom 1s orbitals.

WATER.

Lone pairs in two of the sp3 hybrids.

To form two bonds.

Think about H3O+ !!!

O

H

H

75

Overlap of Oxygen sp3 hybrids containing a lone pair

H+ ion empty 1s orbitals.

HYDRONIUM ION.

O

H

H

H+

ISOELECTRONIC WITH?

NH3

76

QUESTIONWhich of the following molecules is uses sp3 hybrids in the valence bond description of its bonding?

A CO2

B NF3

C O3

D NO2+

E F2O

ANSWER…….

1 C and D

2 B and E

3 A and D

4 B and C

5 B and A

77

QUESTIONWhich of the following molecules is uses sp3 hybrids in the valence bond description of its bonding?

A CO2

B NF3

C O3

D NO2+

E F2O

WHAT ABOUT OTHER EPG’S …….

1 C and D

2 B and E

3 A and D

4 B and C

5 B and AOF F

C OO

O N O[ ]+

NF F

F

O O O

78

A four electron pair EPG uses sp3 hybrids

The three electron pair EPG uses sp2 hybrids

VALENCE BOND THEORY FOR OTHER ELECTRON PAIR GEOMETRIES

The two electron pair EPG uses sp hybrids

79

X

180°

180°

X

120°

120°

120°

X

109.5°

EPG’s2 3

4

HYBRIDS

sp sp2 sp3

lets look at a molecule that needs sp2

80

Ethylene: C2H4 C C

H

H

H

H

The CARBON is sp2 hybridized

3 effective electron pairs

3 sp2 hybridss + px + py

The 3 sp2 hybrid orbitals form a trigonal planar arrangement.

three hybrid orbitals on each carbon for the trigonal

planar EPG.

VSEPR trigonal planar EPG around each C-atom. a HCH angle of 1200.

sp2 hybridization

81

GROUND STATE C atom

E2p

2s

FORMATION OF sp2 hybrids

VALENCE STATE C atom

82

GROUND STATE C atom

E2p

2s


E2p

2s


HYBRIDIZE

83

GROUND STATE C atom

E2p

2s

sp2 hybridized orbitals of C


E sp2

2p

E2p

2s


HYBRIDIZE

This leaves one p orbital unhybrized…….

84

x

y

z

The unhybridized p orbital is perpendicular to sp2 plane.

sp2 - hybrid orbitalUNHYBRIDIZEDp- orbital

An sp2 hydridized C atom

Lets put it all together…….

85

x

z

y y

x

z

C C

H

H

H

H

DRAW TWO C-ATOMS

C C

Now put the orbitals on…...

86

x

y

z

x

y

z

C C

BONDING IN ETHYLENEC C

H

H

H

H

87

x

y

z

x

y

z

OVERLAP the sp2 hybrids from the two carbons to form a sigma bond between them.

C C

BONDING IN ETHYLENE

bond

PUT THE ELECTRONS IN AND…..

C C

H

H

H

H

88

x

y

z

x

y

z

overlap two sp2 hybrids on each carbon with hydrogen 1s orbitals to form sigma bonds and...

The two unhybridized p orbitals are left over to form…..

H

HH

H

C C

H

H

H

H

89

x

y

z

x

y

z

The two unhybridized p orbitals are left over to form a …..

H

HH

H

pi bond ( bond)

The second part of the carbon-carbon double bond !

C C

H

H

H

H

90

x

y

z

x

y

z

H

HH

H

pi bond ( bond)The second part of the carbon-carbon double bond !

Electrons are shared between the unhybridized p

orbitals in an area above and below the line between

nuclei.

C C

H

H

H

H

91

x

y

z

x

y

z

H

HH

H

pi bond ( bond)

sp2

sigma bonds ( bond)

THE COMPLETE PICTURE!!!!!!!

C C

H

H

H

H

SUMMARY...

92

C C

H

H

H

H

H(1s)-C(sp2)

:C(sp2)-C(sp2)C(2p)-C(2p)

H(1s)-C(sp2)

Now look at bond

bonding

93

C C

H

H

H

H

H(1s)-C(sp2)

:C(sp2)-C(sp2)C(2p)-C(2p)

H(1s)-C(sp2)

Now look at ethyne (acetylene)

bonding

94

H(1s)-C(sp)

:C(sp)-C(sp)

C(2p)-C(2p) TWO OF THESE!!

H(1s)-C(sp)

BONDING SCHEME IN ETHYNE

C CH H

What does this look like????

95

x

z

y y

x

z

DRAW TWO C-ATOMS

C C

Now put the sp-orbitals on…...

C CH H

96

x

y

z

x

y

z

OVERLAP the sp hybrids from the two carbons to form a sigma bond between them.

C C

Put in the unhybridized p orbitals

97

x

y

z

x

y

z


C C

OVERLAP the hydrogen 1s orbitals

98

x

y

z

x

y

z

OVERLAP the C sp hybrids with H 1s to form sigma bonds

C C HH

99

x

y

z

x

y

z


C C HH

sigma framework of bonds pi bonding?

100

x

y

z

x

y

z

C C HH

two pi bonds ( bonds)

LATERAL OVERLAP of p orbitals to form pi bonds.

101

x

y

z

x

y

z

LATERAL OVERLAP of p orbitals to form pi bonds.

C C HH

two pi bonds ( bonds)

SO…..

102

Single bond: One bond

SUMMARY

Double bond: One bond, one bond

Triple bond: One bond, two bonds

103

VALENCE BOND THEORY

Step 1 Lewis Dot Structure

Step 2 Get Molecular Geometry

VSEPR EXPERIMENTAL

Step 3 Choose hybrids

Step 4 Describe bonds…...

104

What about molecules with more than an octet around the central atom?

Examples: PCl5, or SF4 or SiF62-

Four pairs needs Four orbitals

Five pairs needs Five orbitals

six pairs needs six orbitals

105

PCl5

Cl

PClCl

Cl Cl

We ignore the

chlorine atoms

and just describe

central atom.

Need five hybrid orbitals on the phosphorus

d + s + px + py + pz

5 effective electron pairs dsp3 hybridization

5 dsp3 hybrids

to fit the trigonal bipyramidal EPG.

Five equivalent orbitals……..

106

dsp3 - hybrid orbitals

x

y

zTRIGONAL BIPYRAMID EPG 5 PAIRS

overlap with orbitals on chlorine to form 5 bonds.

1200

900

SIX PAIRS…..

107

SF6

F

S

F

F

F

F

F

We need six hybrid orbitals on the sulfur to

allow for the octahedral EPG and six bonds.

6 d2sp3 hybrids

d2sp3 hybridization

d + d + s + px + py + pz

6 effective electron pairs

SIX equivalent orbitals……..

We ignore the

chlorine atoms

and just describe

central atom.

108overlap with orbitals on flourine to form 6 bonds.

d2sp3 - hybrid orbitals

x

y

z 900

900

EXAMPLE

109

EXAMPLES

Xe FF

EPG 5 pairs

dsp3 hybrids

Two axial bonds at 1800

Three lone pairs in equatorial

hybrids

Describe the molecular structure and bonding in XeF2 and XeF4

Linear

110

EXAMPLES

Xe FF

EPG 5 pairs

dsp3 hybrids

Two axial bonds at 1800

Three lone pairs in equatorial

hybrids

EPG 6 pairs

d2sp3 hybrids

four bonds at 900 in a plane

Two lone pairs in axial hybrids

Describe the molecular structure and bonding in XeF2 and XeF4

Linear

XeF

F

F

F

Square planar

111

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

112

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

113

0r/a-3/2

01 e

12)(

a

rRs

114

ADDITION OF ORBITALSbuilds up electron density in overlap region

1sA + 1sB = MO1

combine them by addition

A B

115

ADDITION OF ORBITALSbuilds up electron density in overlap region.

1sA + 1sB = MO1

A Bwhat do we notice?

electron density between atoms

116

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

1sA – 1sB = MO2

A B

subtract

117

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

1sA – 1sB = MO2

A Bwhat do we notice?

no electron density between atoms

118

COMBINATION OF ORBITALS

1sA + 1sB = MO1

builds up electron density between nuclei

119

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

121

THE MO’s FORMED BY TWO 1s ORBITALS

122

1sA + 1sB = MO1

1sA – 1sB = MO2

sigma anti-bonding = 1s*

sigma bonding = 1s

1s

1s*

123

E

Energy of a 1s orbital in a free atom


A B

COMBINING TWO 1s ORBITALS

124

E



A B

1sA+1sB

MO

1s

125

E



A B

1sA-1sB

MO

1sA+1sB

MO

1s

1s*

126

E 1sAA B

1s

1s*

1sB

COMBINING TWO 1s ORBITALS

127

E1s

1s*

1s

1s

H HH2

bonding in H2

128

E1s

1s*

1s

1s

H HH2

the electrons are placed in the 1s molecular orbitals

129

E1s

1s*

1s

1s

H2: (1s)2

H HH2

130

E1s

1s*

1s

1s

He2

He HeHe2

atomic configuration of He 1s2

131

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

132

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

133

BOND ORDER

=

measure of bond strength and molecular stability

If # of bonding electrons > # of antibonding electrons

Bondorder

the molecule is predicted to be stable

134

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)

135

1s*

1s

Magnetism

Bond order

Bond energy (kJ/mol)

Bond length (pm)

H2+

E

He2+ He2H2

136

1s*

1s

Magnetism

Bond order


Bond length (pm)

H2+

E

He2+ He2H2

Dia-

1

436

74

137

1s*

1s

Magnetism

Bond order


Bond length (pm)

H2+

Para-

½

225

106

E

He2+ He2H2

Dia-

1

436

74

138

1s*

1s

Magnetism

Bond order


Bond length (pm)

H2+

Para-

½

225

106

E

He2+

Para-

½

251

108

He2H2

Dia-

1

436

74

139

1s*

1s

Magnetism

Bond order


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

140

HOMONUCLEAR DIATOMICS

Li2 Li : 1s22s1

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

second period

141

E

1s

1s*

1s

1s

2s

2s*

2s

2s

ENERGY LEVEL DIAGRAM FOR DILITHIUM

Li2

142

E

1s

1s*

1s

1s

2s

2s*

2s

2s

Li2

ELECTRONS FOR DILITHIUM

143

E

1s 1s

1s

Electron configuration for DILITHIUM

2s

2s*

2s

2s

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

Li2

Bond Order ?

144

E

1s 1s

1s


2s

2s*

2s

2s

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

Li2

nb = 4 na = 2

Bond Order = 1

single bond.

145

E

1s 1s

1s


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

146

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

147

E2s

2s*

2s

2s

Be BeBe2 Be2

148

E2s

2s*

2s

2s

Be2Be BeBe2

Electron configuration for DIBERYLLIUM

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

149

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...

150

B2

the Boron atomic configuration is

1s22s22p1

form molecular orbitals

we expect B to use 2p orbitals to

addition and subtraction

151

-molecular orbitals

152

molecular orbitals

153

ENERGY LEVEL DIAGRAM

E

2s

2s*

2s

2s

154

2p*

2p

2p

2p*

E2p 2p

155

E

expected orbital splitting

2s

2s*

2s

2s

2p

2p*

2p

2p

2p

2p*

This pushes the 2p up

156

E

MODIFIED ENERGY LEVEL DIAGRAM

2s

2s*

2s

2s

2p

2p*

2p2p

2p

2p*

Notice that the 2p and 2p

have changed places!!!!

157

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

158

E

2s

2s*

2s

2s

2p

2p*

2p2p

2p

2p*

Place electrons from 2p into 2p and 2p

Remember HUND’s RULE

159

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

160

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

161

A SUMMARY OF THE MO’s

Emphasizing nodal planes

162

ELECTRONIC CONFIGURATION OF THE HOMONUCLEAR DIATOMICS

B2 C2 N2 O2 F2Li2

163

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

164

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

165

2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)


B2

Para-

1

290

159

C2 N2 O2 F2

E

166

2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)


B2

Para-

1

290

159

C2

Dia-

2

620

131

N2 O2 F2

E

167

2p*

2p*

2p

2p

2s*

2s

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 F2

E

168

2p*

2p*

2p

2p

2s*

2s

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

E

NOTE SWITCH OF LABELS

169

2p*

2p*

2p

2p

2s*

2s

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

NOTE SWITCH OF LABELS

170

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

–

O2 :

O2+ :

O2– :

O22-:

O22-

171

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

172

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

173

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

174

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

175

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

176

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

177

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

CHEMISTRY 161 Chapter 10 Chemical Bonding & Molecular Structure

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Transcript of CHEMISTRY 161 Chapter 10 Chemical Bonding & Molecular Structure