Chapter 9

Chapter 9Molecular Geometry

Bonding Theories

Molecular Shape• A bond angle is the

angle defined by lines joining the centers of two atoms to a third atom to which they are covalently bonded

• The molecular geometry or shape is defined by the lowest energy arrangement of its atoms in three-dimensional space.

VSEPR

The geometric arrangement of atoms bonded to a given atom is determined principally by minimizing electron pair repulsions of bonding and non-bonding electrons.

Valence-Shell Electron-Pair Repulsion Theory

Central Atoms without Lone Pairs

Steric number (SN) is the number of volumes of space occupied by electrons surrounding a central atom

Geometric Forms

Predicting a VSEPR Structure1. Draw Lewis structure.

2. Determine the steric number of the central atom.

3. Use the SN to determine the geometry around the central atom.

4. The name for molecular structure is determined by the number of volumes of space occupied by bonding electrons.

Examples

• What is the molecular geometry of BF3?

• What is the molecular geometry of CH4

Examples



Lewis Structure (exception to Law of Octaves)

B

F

F

F

Examples




B

F

F

F

Bond Angles = 120°

Trigonal Planar

Examples




B

F

F

F

Bond Angles = 120°

Trigonal Planar

C

H

HH

H

Examples




B

F

F

F

Bond Angles = 120°

Trigonal Planar

C

H

HH

HBond Angles = 120°

Tetrahedral

Central Atoms with Lone Pairs

• Electron-pair geometry describes the arrangement of atoms and lone pairs of electrons about a central atom.

The electron-pair geometry will always be one of the five geometries presented previously.

• The molecular geometry in these molecules describes the shape of the atoms present (it excludes the lone pairs).

Lone Pairs• Lone pairs of electrons occupy more space

around a central atom than do bonding electrons.• Lone pair-lone pair repulsion is the largest.• Lone pair-bonding pair repulsion is the next

largest.• Bonding pair-bonding pair repulsion is the smallest.• In structures with lone pairs on the central atom,

the bond angles are a little smaller than predicted based on the electron-pair geometry.

SN = 3, Electron-pair Geometry = Trigonal Planar

No. of Bonded Atoms

No. of Lone Pairs

Molecular Geometry

Bond Angles

3 0 Trigonal Planar 120o

2 1 Bent<120o

Like 119.6o

Non-bonding Electrons & Shape

B

H

H H

HB

H H Angles = 120

Trigonal Planar

Angles < 120

Bent

SN = 4, Electron-pair Geometry = Tetrahedral

No. of Bonded Atoms

No. of Lone Pairs

Molecular Geometry

Bond Angles

4 0 Tetrahedral 109.5o

3 1Trigonal

Pyramidal<109.5o

2 2 Bent <109.5o



Tetrahedral Trigonal Pyramid v-shape

SN = 5, Electron-pair Geometry = Trigonal Bipyramidal

No. of Bonded Atoms

No. of Lone Pairs

Molecular Geometry

Bond Angles

5 0Trigonal

Bipyramidal120o & 90o

4 1 Seesaw <120o & 90o

3 2 T-shaped <120o & 90o

2 3 Linear 180o

The lone pairs of electrons are always found in the trigonalplanar part of the structure to minimize repulsion.

SN = 6, Electron-pair Geometry = Octahedral

No. of Bonded Atoms

No. of Lone Pairs

Molecular Geometry

Bond Angles

6 0 Octahedral 90o

5 1Square

Pyramidal<90o

4 2 Square Planar 90o

3 3 Although these arrangements are possible, we will not encounter any molecules with these arrangements.2 4

Polar Bonds and Polar Molecules• Two covalently bonded atoms with different

electronegativities have partial electric charges of opposite sign creating a bond dipole.

• A molecule is called a polar molecule when it has polar bonds and a shape where the bond dipoles don’t offset each other.

Examples

Measuring PolarityThe permanent dipole moment () is a measured value that defines the extent of separation of positive and negative charge centers in a covalently bonded molecule.

Valence Bond Theory• Hybridization is the mixing of atomic orbitals to

generate new sets of orbitals that are then available to overlap and form covalent bonds with other atoms.

• A hybrid atomic orbital is one of a set of equivalent orbitals about an atom created when specific atomic orbitals are mixed.

Valence-Bond Theory• Valence-bond theory assumes that covalent

bonds form when orbitals on different atoms overlap or occupy the same region of space.

• A sigma () bond is a covalent bond in which the highest electron density lies between the two atoms along the bond axis connecting them.

Atomic Orbitals and Bonds• A tetrahedral molecule requires that four

orbitals of the central atom must overlap with an orbital of an outer atom to form a bond.

• The central atom would use its s orbital and its three p orbitals, but these orbitals would not yield the 109° bond angles observed in the tetrahedral molecule.

Hybrid OrbitalsYou may have noticed that the electron pairs in molecules have different orientations in space compared to atomic orbitals. Wave equations mathematically generated volumes of space where electrons spend most of their time, but what about molecules?

Hybrid OrbitalsYou may have noticed that the electron pairs in molecules have different orientations in space compared to atomic orbitals. Wave equations mathematically generated volumes of space where electrons spend most of their time, but what about molecules? This brings us to the concept of hybrid orbitals, combinations of atomic orbitals, or molecular orbitals (from wave equations of electrons in molecules)

Hybrid OrbitalsYou may have noticed that the electron pairs in molecules have different orientations in space compared to atomic orbitals. Wave equations mathematically generated volumes of space where electrons spend most of their time, but what about molecules?

Hybrid OrbitalsYou may have noticed that the electron pairs in molecules have different orientations in space compared to atomic orbitals. Wave equations mathematically generated volumes of space where electrons spend most of their time, but what about molecules? This brings us to the concept of hybrid orbitals, combinations of atomic orbitals, or molecular orbitals (from wave equations of electrons in molecules)

Hybrid OrbitalsHybridization is a concept you might be familiar with. For example a grapefruit is a hybrid of what two fruits?

Hybrid OrbitalsHybridization is a concept you might be familiar with. For example a grapefruit is a hybrid of what two fruits?

Hybrid OrbitalsHybridization is a concept you might be familiar with. For example a grapefruit is a hybrid of what two fruits? Lemon and orange

Hybrid OrbitalsHow about a nectarine?

Hybrid OrbitalsHow about a nectarine? Plumb and a peach.


Broccoaflower? Broccoli and cauliflower



And a Cocapoo?



And a Cocapoo? Cocker spaniel and poodle

Hybrid OrbitalsOn to Chemistry! How about an s-orbital and a p-orbital? Yes, sp orbital.


How about one s-orbital and two p-orbitals?


How about one s-orbital and two p-orbitals? Yes an sp2 orbital.

Examples

Tetrahedral Geometry: sp3 Hybrid Orbitals

A tetrahedral orientation of valence electrons is achieved by forming four sp3 hybrid orbitals form one s and three p atomic orbitals.

Other sp3 Hybrid Examples

sp2 Hybridization

• In a covalent pi () bond, electron density is greatest above and below the bonding axis.

sp Hybridization

• Pi bonds will not exist between two atoms unless a sigma bond forms first.

The Bonding in Carbon Dioxide

The carbon atom is sp hybridized and these orbitals form the two sigma bonds. The bonds are rotated 90° from one another.

d2sp3 Hybridization

dsp3 Hybridization

Delocalization of Electrons

• The electrons in the system with alternating single and double bonds can be delocalized over several atoms or even an entire molecule.

Localized

Delocalized

Hybrid Orbital NotationIn order to construct hybrid orbital notation, we need to separate the central atom from the surrounding electrons, usually the central atom is the largest, the most electronegative, or the one that there is one of.

Hybrid Orbital NotationIn order to construct hybrid orbital notation, we need to separate the central atom from the surrounding electrons, usually the central atom is the largest, the most electronegative, or the one that there is one of. When constructing a hybrid orbital diagram, all of the valence electrons of the central atom are used and only the single electrons of the atoms attached to the central atom are use.

Hybrid Orbital ExampleSuppose we want to make a diagram of SF6

First we separate the central atom from the other atoms. The central atom is A and the other atoms are called X’s

SF6

A X’s



SF6

A X’s

Then we generate a set of degenerate hybrid orbitals to house the valence electrons



SF6

A X’s

Then we generate a set of degenerate hybrid orbitals to house the valence electrons F

Insert single electrons into the degenerate hybrid orbitalsF F F F FF



SF6

A X’s

Then we generate a set of degenerate hybrid orbitals to house the valence electrons F

Insert single electrons into the degenerate hybrid orbitalsF F F F FF

Structure of Sulfurhexafluoride

sF

FF

F

F

F

Shape- Octahedral

Hybrid Orbitals- sp3d2

Bond angles- 90°

Polarity- Nonpolar

Noble Gas CompoundsCritics of the hybrid orbital theory argued that the hybrid orbital theory suggests that compounds of Noble gases should exist or be made. In 1962 Neil Bartlett created a compound of xenon, platinum and fluorine. Today there are now several hundred Noble gas compounds known.

University of British Columbia

Noble Gas Compounds

Practice

XeO2 KrF4 XeO2F22+

Problems with Bonding Theories

• Lewis structure and valence bond theory help us understand the bonding capacities of elements.

• VSEPR and valence bond theories account for the observed molecular geometries.

• None of these models enables us to explain why O2 is attracted to a magnetic field while N2 is repelled slightly.

Molecular Orbital (MO) Theory• The wave functions of atomic orbitals of

atoms are combined to create molecular orbitals (MOs) in molecules. Each MO is associated with an entire

molecule, not just a single atom. MOs are spread out, or delocalized over all the atoms in a molecule.

Types of MOs• Electrons in bonding orbitals serve to hold

atoms together in molecules by increasing the electron density between nuclear centers.

• Electrons in antibonding orbitals in a molecule destabilize the molecule because they do not increase the electron density between nuclear centers.

MO Guidelines1. The total number of MO formed equals the number of

atomic orbitals used in the mixing process.

2. Orbitals with similar energy and shape mix more effectively than do those that are different.

3. Orbitals of different principal quantum numbers have different sizes and energies resulting in less effective mixing.

4. A MO can accommodate two electrons with opposite spin.

5. Electrons are placed in MO diagrams according to Hund’s rule.

MOs for H2

• The two 1s orbitals may be added or subtracted to yield two MOs.

Bond OrderBond Order = 1/2 (# bonding electrons - # antibonding

electrons)

The bond order is zero in He2 and the molecule is not stable.

67

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.

So……..

In He2

(1s)2(1s*)2

Useful concept:

the electronic configuration is….

There is no BOND! BOND ORDER = 0!!!!!

Bond Types

• A sigma, , bond is a covalent bond in which the highest electron density lies along the bond axis.

• A pi, , bond is formed by the mixing of atomic orbitals that are not oriented along the bonding axis in a molecule.

69

BOND ORDER

=

A measure of bond strength and molecular stability.

If # of bonding electrons > # of antibonding electrons

Bondorder

the molecule is predicted to be stable

70

BOND ORDER

= {

A high bond order indicates high bond energy and short bond length.

# of bonding electrons(nb)

# of antibonding electrons (na) –

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 H2+,H2,He2

+,He2……….

= 1/2 (nb - na)

71

1s*

1s

Magnetism

Bond order

Bond energy (kJ/mol)

Bond length (pm)

H2+

E

He2+ He2H2

First row diatomic molecules and ions

72

1s*

1s

Magnetism

Bond order


Bond length (pm)

H2+

E

He2+ He2H2

Dia-

1

436

74


73

1s*

1s

Magnetism

Bond order


Bond length (pm)

H2+

Para-

½

225

106

E

He2+ He2H2

Dia-

1

436

74


74

1s*

1s

Magnetism

Bond order


Bond length (pm)

H2+

Para-

½

225

106

E

He2+

Para-

½

251

108

He2H2

Dia-

1

436

74


75

1s*

1s

Magnetism

Bond order


Bond length (pm)


H2+

Para-

½

225

106

E

He2+

Para-

½

251

108

He2

—

0

—

—

H2

Dia-

1

436

74

76

HOMONUCLEAR DIATOMICS

First is Li2 Li : 1s22s1

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

Now look at second period…..

This is the energy level diagram…..

77

E

1s

1s*

1s

1s

2s

2s*

2s

2s Put the electrons in…….

DI-LITHIUMLi2

78

E

1s

1s*

1s

1s

2s

2s*

2s

2s Put the electrons in the MO’s

Li2

ELECTRONS FOR DILITHIUM

79

E

1s 1s

1s

Electron configuration for DILITHIUM

2s

2s*

2s

2s

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

Li2

Bond Order ??????

What do we need?

80

E

1s 1s

1s


2s

2s*

2s

2s

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

Li2

nb = 4 na = 2

Bond Order = 1/2 (nb - na)

= 1/2(4 - 2) =1

A single bond.

81

E

1s 1s

1s


2s

2s*

2s

2s

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

The 1s and 1s* orbitals cancel!

Li2

nb = 4 na = 2

Note:

So…….

82

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

We often omit the inner shell!

83

E2s

2s*

2s

2s

Li2 (2s)2

Li LiLi2

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

Only valence orbitals contribute to molecular bonding

But can be included. The Li2 configuration….

84

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

What is the bond order????

85

E2s

2s*

2s

2s

Li LiLi2


nb = 2

na = 0

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

A single bond.

Ignoring the filled (1s)2(1s*)2

Is the molecule stable or unstable???

86

E2s

2s*

2s

2s

Li LiLi2


nb = 2

na = 0

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

A single bond.

Ignoring the filled (1s)2(1s*)2

STABLE! Now Be2…….

87

E2s

2s*

2s

2s

Be BeBe2 Be2

VALENCE ELECTRONS FOR DIBERYLLIUM

Put the electrons in the MO’s...

88

E2s

2s*

2s

2s

Be2Be BeBe2

Electron configuration for DIBERYLLIUM

Configuration: (2s)2(2s*)2 Bond order?

89

E2s

2s*

2s

2s

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

Be2


nb = 2

na = 2

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

No bond!!! We conclude???????

90

E2s

2s*

2s

2s

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

Be2


nb = 2

na = 2

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

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

91

B2

The Boron atomic configuration is

1s22s22p1

form molecular orbitals.

So we expect B to use 2p orbitals to

How do we do that???

Combine them by ???

Addition and subtraction….

This is what they look like…….

92

molecular orbitals

2p* antibonding

2p bonding PHH picture

+-

-+ +

-

-

ADD

SUBTRACT

93

-molecular orbitals

-+-

-+ +-

PHH picture - MO’s

94

The molecular orbitals.

2p* antibonding

2p bonding

+ -

+-

-

+PHH...

ADD

SUBTRACT

95

The molecular orbitals.+ -

+-

-+

ENERGY LEVELS?

96

ENERGY LEVEL DIAGRAM

When we form MO’s we get orbitals of different energies

Example the 2s 2s* From 2sE

2s

2s*

2s

2s

The p- MO’s……..

97

The M.O.’s formed by p orbitals

2p*

2p

2p

2p*

The do not split as much because of weaker overlap.

E2p 2p

Combine this with the s-orbitals…..

Modified Molecular Orbital DiagramIt should be noted that both sigma 1s and sigma 2p orbitals have similar shape and energy, thus mixing occurs lowering the sigma 1s orbital and elevating the sigma 2p orbital above the bonding pi molecular orbitals.

This is illustrated on the next slide.

99

E

Expected orbital splitting:

2s

2s*

2s

2s

2p

2p*

2p

2p

2p

2p*The do not split as much because of weaker overlap.

But the s and p along the internuclear axis interact…….

This pushes the 2p up..

100

E

MODIFIED ENERGY LEVEL DIAGRAM

2s

2s*

2s

2s

2p

2p*

2p2p

2p

2p* Shows additional interaction.

Notice that the 2p and 2p

have changed places!!!!

Now look at B2...

101

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

102

E

2s

2s*

2s

2s

Electron configuration for B2:

2p

2p*

2p2p

2p

2p*

Place electrons from 2p into 2p and 2p

Remember HUND’s RULE

103

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

Bond order????

ELECTRONS ARE UNPAIRED

104

E

2s

2s*

2s

2s


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

105

HOMONUCLEAR DIATOMICS

B2 C2 N2 O2 F2

Which energy level diagram???

The one with s and p interaction or the one without?

We find……….

Li2

106

SECOND ROW DIATOMICSB2 C2 N2 O2 F2

USEE

2s

2s*

2s

2s

2p

2p*

2p2p

2p

2p*

2s

2s*

2s

2s

2p*

2p

2p

2p

2p*

2p

USELi2

107

2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)

Second row diatomic moleculesB2 C2 N2 O2 F2

E

108

2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)

Second row diatomic moleculesB2

Para-

1

290

159

C2 N2 O2 F2

E

109

2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)


Para-

1

290

159

C2

Dia-

2

620

131

N2 O2 F2

E

110

2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)


Para-

1

290

159

C2

Dia-

2

620

131

N2

Dia-

3

942

110

O2 F2

E

111

2p*

2p*

2p

2p

2s*

2s

Magnetism

Bond order

Bond E. (kJ/mol)

Bond length(pm)


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

112

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

113

QUESTIONThe molecule X+

2 that has the molecular orbital configuration (1s)2(1s*)2(2s)2(2s*)2(2p)4 (2p)2 (2p*)1 is

1 B2+

2 C2+

3 N2+

4 O2+

5 F2+

ANSWER…...

114

QUESTIONThe molecule X+

2 that has the molecular orbital configuration (1s)2(1s*)2(2s)2(2s*)2(2p)4 (2p)2 (2p*)1 is

1 B2+

2 C2+

3 N2+

4 O2+

5 F2+

ANSWER…...

How?The molecule X2 has 16 electrons

Therefore Z = 8

Therefore O2+

PARAMAGNETIC? YES!

BOND ORDER 2.5

EXAMPLE…..

115

Example: Give the electron configuration and bond order for O2, O2+

, O2- & O2

2-. Place them in order of bond strength and describe their

magnetic properties.

Step 1:Determine the number of valence electrons in each:

O2+ : 6 + 6 - 1 = 11

O2– : 6 + 6 + 1 = 13

O22- : 6 + 6 + 2 = 14

O2 : 6 + 6 = 12

116

Step 2: Determine the valence electrons configurations:

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

–

O2 :

O2+ :

O2– :

O22-:

O22-

117


O2 : (2s)2(2s*)2 (2p)2(2p)4 (2p*)2

O2+ :

O2– :

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

118


O2 : (2s)2(2s*)2 (2p)2 (2p)4(2p*)2

O2+ : (2s)2(2s*)2 (2p)2 (2p)4(2p*)1

O2– :

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

119


O2 : (2s)2(2s*)2 (2p)2 (2p)4(2p*)2

O2+ : (2s)2(2s*)2 (2p)2 (2p)4(2p*)1

O2– : (2s)2(2s*)2 (2p)2 (2p)4(2p*)3

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

120

Step 2: What about O22- ???:

2p*

2p*

2p

2p

2s*

2s

E

O2 O2+ O2

– O22-

O2 : (2s)2(2s*)2 (2p)2 (2p)4(2p*)2

O2+ : (2s)2(2s*)2 (2p)2 (2p)4(2p*)1

O2– : (2s)2(2s*)2 (2p)2 (2p)4(2p*)3

O22- : (2s)2(2s*)2 (2p)2 (2p)4(2p*)4

121

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

Step 3: Determine the bond orders of each species:

122

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

Step 4: Use the bond orders to place the species in order of

bond strength:

O2+ >O2 >O2

– > O22-

HETERONUCLEAR DIATOMICS…...

BOND ENERGY ORDER

123

2s

2s*

2s

2s

SECOND ROW DIATOMICS

E

2p

2p*

2p2p

2p

2p*

2s

2s*

2s

2s

2p*

2p

2p

2p

2p*

2p

Can be used for

heteronuclear diatomics

Which????

124

2p*

2p*

2p

2p

2s*

2s

E

NITRIC OXIDE (NO)Number of valence electrons: 5 + 6 = 11use the mo diagram for homonuclear diatomic molecules with s-p interaction as an approximation

Put the electrons in…..

125

NITRIC OXIDE (NO)Number of valence electrons: 5 + 6 = 11use the mo diagram for homonuclear diatomic molecules with s-p interaction as an approximation2p*

2p*

2p

2p

2s*

2s

E

126

Bond order

NITRIC OXIDE (NO)Number of valence electrons: 5 + 6 = 11

use the mo diagram for homonuclear diatomic molecules with s-p interaction as an approximation2p*

2p*2p

2p

2s*2s

E 522

38.

127

Number of valence electrons: 5 + 6 = 11

Molecule is stable and paramagnetic.

NITRIC OXIDE (NO)

use the mo diagram for homonuclear diatomic molecules with s-p interaction as an approximation

2p

2p

2s*

2s

E

Bond order 522

38.

Experimental data agrees.

NO+ and CN-

2p*

2p*

Localized/MO CombinationRecalling the molecular orbital diagram for carbon we notice that the sigma and sigma antibonding fill before the pi bonds. Since this does not contribute to the bond order it makes sense that we can use localized bonding (line) to describe a sigma bond and use molecular orbitals on just the pi bonding electrons.

Localized/MO CombinationConsider ozone, O3, we can use lines to connect the three oxygen atoms together and do a molecular orbital diagram on just the unhybridized p-orbitals. Then we arrange the three combinations of p-orbitals in increasing potential energy order by counting the nodes. Potential energy is inversely proportional to the number of nodes.

Localized/MO CombinationConstructive Interference

+

+

+

+

-

-

Destructive Interference

+ +

+- -

-

Makes bonging orbitals Makes bonging orbitals

A review of adding orbitals

Localized/MO Combination

O O O

one node

zero Node

one node

P=2 , which means two pi bonding electrons

Ozone Molecular Orbital Diagram

Pi bonding

non bonding

Pi antibonding

Note: Odd electron systems will always have a nonbonding molecular orbital, which is equal in potential energy of the p-orbital they were derived from.


O O O

one node

zero Node

one node



Pi bonding

non bonding

Pi antibonding

Now fill Pi MO with electrons


O O O

one node

zero Node

one node



Pi bonding

non bonding

Pi antibonding



O O O

one node

zero Node

one node



Pi bonding ( π)

non bonding

Pi antibonding ( π*)


Comparison of Theories

• MO theory may provide the most complete picture of covalent bonding, but it is also the most difficult to apply to large molecules and it does not account for molecular shape.

ChemTour: Partial Charges and Bond Dipoles

Click to launch animation

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Students learn that covalent bonds often include unequal distribution of electrons leading to partial charges on atoms, bond dipole moments, and molecule polarity. Interactive Practice Exercises ask students to calculate dipole moments of polar molecule.

ChemTour: Greenhouse Effect


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This unit explores how excess carbon dioxide and CFCs in the atmosphere contribute to global warming.

ChemTour: Vibrational Modes


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This tutorial illustrates the three vibrational modes: bending, symmetric stretching, and asymmetric stretching. Students learn that molecules can absorb specific wavelengths of infrared radiation by converting this energy into molecular vibrations.

ChemTour: Hybridization


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This tutorial animates the formation of hybrid orbitals from individual s and p orbitals, shows examples of their geometry, and describes how they can produce single, double, and triple bonds. Includes Practice Exercises.

ChemTour: Chemistry of the Upper Atmosphere


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This ChemTour examines how particles of the upper atmosphere absorb and emit electromagnetic radiation.

ChemTour: Molecular Orbitals


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This animated tutorial offers a patient explanation of molecular orbital theory, an alternative to the bonding theory depicted by Lewis dot structures. Includes Practice Exercises.

Ethylene, which has the molecular formula C2H4, is a rigid molecule in which all 6 atoms lie in a plane. Which of the following molecules also has a rigid planar structure?

A) H2C=C=CH2 B) H2C=C=C=CH2 C) Neither

Planar Hydrocarbons

Planar Hydrocarbons

Consider the following arguments for each answer and

vote again:

A. A combination of 3 carbons and 4 hydrogens can form the rigid planar molecule H2C=C=CH2.

B. The orientations of the π bonds in H2C=C=C=CH2 alternate in such a way as to create a planar structure.

C. The hybridization of the atomic orbitals on the carbons prevents the retention of a planar structure in molecules longer than C2H4.

What is the bond order of the N-O bond in nitrate, NO3-?

A) 1 B) 11/3 C) 2

Bond Order of Nitrate


Consider the following arguments for each answer and

vote again:

A. The majority of the bonds in NO3- are single bonds, so the bond order is 1.

B. The N-O bond is twice as likely to be a single bond as it is to be a double bond, so the bond order should be 11/3.

C. The bond order is dictated by the strongest bond, which in NO3- is a double bond.

Which of the following species is not paramagnetic in its ground state?

A) NO+ B) NO C) NO-



Consider the following arguments for each answer

and vote again:

A. NO+ is isoelectronic with N2, which has no unpaired electrons and hence is not paramagnetic.

B. NO has no electrical charge and thus cannot be paramagnetic.

C. By pairing an additional electron with the one unpaired electron in NO, a diamagnetic anion, NO-, is formed.

Molecular Geometry of SF2, SF3-, and SF4

According to Valence Shell Electron Pair Repulsion (VSEPR) theory, 4 objects around a central atom will have the tetrahedral arrangement shown to the left with bond angles of ~109.5º. Which of the following compounds has a bond angle of ~109.5º?

A) SF2 B) SF3- C) SF4

Molecular Geometry of SF2, SF3-, and SF4

Please consider the following arguments for each answer and

vote again:

A. SF2 consists of a sulfur atom surrounded by 2 lone electron pairs and bonded to 2 fluorine atoms, therefore, it has an approximately tetrahedral bond angle.

B. The tetrahedral VSEPR arrangement of SF3- is formed by a sulfur atom surrounded by 3 fluorine atoms and by the additional electron (from the negative charge).

C. Sulfur tetrafluoride is the only molecule with a central atom (sulfur) surrounded by 4 additional atoms (4 fluorines) and so is the only molecule with a bond angle of ~109.5º.

Bond Angles of BrF2 and ICl2

Which of the following is true of the bond angle (θ1) in BrF2

+ compared to the bond angle (θ2) in ICl2

-?

A) θ1 = θ2 B) θ1 > θ2 C) θ1 < θ2

Bond Angles of BrF2 and ICl2

Please consider the following arguments for each

answer and vote again:

A. Both BrF2+ and ICl2

- consist of a central halogen atom bonded to two halogen atoms, and therefore should have the same arrangement of atoms.

B. ICl2- has 1 more lone pair of electrons than BrF2

+, which forces the chlorine atoms closer together.

C. ICl2-, with 3 lone pairs, is linear whereas BrF2

+, with 2 lone pairs, is bent.

Reaction of Boron Trifluoride

Boron trifluoride (BF3), which has the structure shown to the left, is capable of reacting with an unknown compound to form a new compound without breaking any bonds. Which of the following could be the unknown compound?

A) BF3 B) CH4 C) NH3

Reaction of Boron Trifluoride

Please consider the following arguments for each answer and vote again:

A. BF3 can dimerize to BF3-BF3 by forming a boron-boron single bond.

B. By forming a boron-carbon bond, the carbon atom in CH4 will increase its steric number to 5, thus expanding its octet to compensate for boron's incomplete octet.

C. The nitrogen lone electron pair can form a nitrogen-boron bond yielding BF3-NH3, isoelectronic with CH3-CH3.

Dipole Moments of Dichloroethylene

Pictured to the left is the planar molecule ethylene, C2H4, which does not have a permanent electric dipole moment.

A) 0 B) 1 C) 2

If chlorine atoms were substituted for two hydrogen atoms, how many of the possible structures would also not possess a dipole moment?

Dipole Moments of Dichloroethylene

Consider the following arguments for each answer

and vote again:

A. Chlorine atoms always draw electron density away from carbon atoms, so all possible structures will possess a dipole moment.

B. Only if the chlorine atoms are diagonally opposite will the two carbon-chlorine dipole moments cancel each other.

C. So long as the two chlorine atoms are on different carbon atoms, no permanent dipole moment will form.

For which central atom "X" does the anion pictured to the left have a square planar geometry?

A) C B) S C) Xe

Molecular Geometry of XF42-

Molecular Geometry of XF42-

Please consider the following arguments for each

answer and vote again:

A. CF42- forms a structure in which the 4 fluorine

atoms form a square plane with one negative charge on either side of the plane.

B. With 2 lone electron pairs on the sulfur in SF42-, its

steric number is 6.

C. To maximize fluorine-fluorine distances, the 4 fluorine atoms in XeF4

2- will lie in a plane.

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Chapter 9