Bond Theory, and

(1)

Sherril Soman

Grand Valley State University

Lecture Presentation

Chapter 10 Chemical

Bonding II: Molecular Shapes, Valence

Bond Theory, and Molecular Orbital

Theory

10.1 Artificial Sweeteners: Fooled by Molecular Shape 425

10.2 VSEPR Theory: The Five Basic Shapes 426

10.3 VSEPR Theory: The Effect of Lone Pairs 430

10.4 VSEPR Theory: Predicting Molecular Geometries 435

10.5 Molecular Shape and Polarity 438 Polarity 438

10.6 Valence Bond Theory: Orbital

Orbital Overlap as a Chemical Bond 443 Bond 443

10.7 Valence Bond Theory:

(2)

(3)

Solution

The molecular geometry of NO₃⁻ is determined by the number of electron groups around the central atom (N).

Begin by drawing a Lewis structure of NO₃⁻.

NO₃⁻ has 5 + 3(6) + 1 = 24 valence electrons. The Lewis structure has three resonance structures:

The hybrid structure is intermediate between these three and has three equivalent bonds.

Use any one of the resonance structures to determine the number of electron groups around the central atom.

Determine the molecular geometry of NO₃⁻.

Example 10.1 VSEPR Theory and the Basic Shapes

(4)

Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

The electron geometry that minimizes the repulsions between three electron groups is trigonal planar.

Since the three bonds are equivalent (because of the resonance structures), they each exert the same repulsion on the other two and the molecule has three equal bond angles of 120°.

For Practice 10.1

Determine the molecular geometry of CCl₄. Continued

Example 10.1 VSEPR Theory and the Basic Shapes

(5)

Procedure For…

Predicting Molecular Geometries

Solution

Step 1 Draw the Lewis structure for the molecule.

PCl₃ has 26 valence electrons.

Step 2 Determine the total number of electron groups around the central atom. Lone pairs, single bonds, double bonds, triple bonds, and single electrons each count as one group.

The central atom (P) has four electron groups.

Predict the geometry and bond angles of PCl₃.

Example 10.2 Predicting Molecular Geometries

(6)

Step 3 Determine the number of bonding groups and the number of lone pairs around the central atom.

These should sum to your result from step 2. Bonding groups include single bonds, double bonds, and triple bonds.

Three of the four electron groups around P are bonding groups and one is a lone pair.

Step 4 Refer to Table 10.1 to determine the electron geometry and molecular geometry. If no lone pairs are present around the central atom, the bond angles will be that of the ideal geometry. If lone pairs are present, the bond angles may be smaller than the ideal geometry.

Continued

Example 10.2 Predicting Molecular Geometries

(7)

Continued

Example 10.2 Predicting Molecular Geometries

(8)

The electron geometry is tetrahedral (four electron groups) and the molecular geometry—the shape of the molecule—is trigonal pyramidal (three bonding groups and one lone pair). Because of the presence of a lone pair, the bond angles are less than 109.5°.

For Practice 10.2

Predict the molecular geometry and bond angle of ClNO.

Continued

Example 10.2 Predicting Molecular Geometries

(9)

Procedure For…

Predicting Molecular Geometries

Solution

Step 1 Draw the Lewis structure for the molecule.

ICl₄⁻ has 36 valence electrons.

Step 2 Determine the total number of electron groups around the central atom. Lone pairs, single bonds, double bonds, triple bonds, and single electrons each count as one group.

The central atom (I) has six electron groups.

Predict the geometry and bond angles of ICl₄⁻.

Example 10.3 Predicting Molecular Geometries

(10)

Step 3 Determine the number of bonding groups and the number of lone pairs around the central atom.

These should sum to your result from step 2. Bonding groups include single bonds, double bonds, and triple bonds.

Four of the six electron groups around I are bonding groups and two are lone pairs.

Step 4 Refer to Table 10.1 to determine the electron geometry and molecular geometry. If no lone pairs are present around the central atom, the bond angles will be that of the ideal geometry. If lone pairs are present, the bond angles may be smaller than the ideal geometry.

Continued

Example 10.3 Predicting Molecular Geometries

(11)

Continued

Example 10.3 Predicting Molecular Geometries

(12)

The electron geometry is octahedral (six electron groups) and the molecular geometry—the shape of the molecule—is square planar (four bonding groups and two lone pairs). Even though lonepairs are present, the bond angles are 90° because the lone pairs are symmetrically arranged and do not compress the I⁻ Cl bond angles.

For Practice 10.3

Predict the molecular geometry of I₃⁻. Continued

Example 10.3 Predicting Molecular Geometries

(13)

Solution

Begin by drawing the Lewis structure of CH₃OH. CH₃OH contains two interior atoms: one carbon atom and one oxygen atom. To determine the shape of methanol, determine the geometry about each interior atom as follows:

Using the geometries of each of these, draw a three-dimensional sketch of the molecule as shown here:

Predict the geometry about each interior atom in methanol (CH₃OH) and make a sketch of the molecule.

Example 10.4 Predicting the Shape of Larger Molecules

(14)

For Practice 10.4

Predict the geometry about each interior atom in acetic acid and make a sketch of the molecule.

Continued

Example 10.4 Predicting the Shape of Larger Molecules

(15)

Solution

Draw the Lewis structure for the molecule and determine its molecular geometry.

The Lewis structure has three bonding groups and one lone pair about the central atom. Therefore the molecular geometry is trigonal pyramidal.

Determine if the molecule contains polar bonds. Sketch the molecule and superimpose a vector for each polar bond. The relative length of each vector should be proportional to the electronegativity difference between the atoms forming each bond. The vector should point in the direction of the more electronegative atom.

The electronegativities of nitrogen and hydrogen are 3.0 and 2.1, respectively. Therefore, the bonds are polar.

Determine if NH₃ is polar.

Example 10.5 Determining if a Molecule Is Polar

(16)

Determine if the polar bonds add together to form a net dipole moment. Examine the symmetry of the vectors (representing dipole moments) and determine if they cancel each other or sum to a net dipole moment.

The three dipole moments sum to a net dipole moment. The molecule is polar.

For Practice 10.5

Determine if CF₄ is polar.

Continued

Example 10.5 Determining if a Molecule Is Polar

(17)

Procedure For…

Hybridization and Bonding Scheme

Solution

Step 1 Write the Lewis structure for the molecule.

BrF₃ has 28 valence electrons and the following Lewis structure:

Step 2 Use VSEPR theory to predict the electron geometry about the central atom (or interior atoms).

The bromine atom has five electron groups and therefore has a trigonal bipyramidal electron geometry.

Step 3 Select the correct hybridization for the central atom (or interior atoms) based on the electron geometry (see Table 10.3).

A trigonal bipyramidal electron geometry corresponds to sp³d hybridization.

Write a hybridization and bonding scheme for bromine trifluoride, BrF₃.

Example 10.6 Hybridization and Bonding Scheme

(18)

Continued

Example 10.6 Hybridization and Bonding Scheme

(19)

Step 4 Sketch the molecule, beginning with the central atom and its orbitals. Show overlap with the appropriate orbitals on the terminal atoms.

Step 5 Label all bonds using the σ or π notation followed by the type of overlapping orbitals.

Continued

Example 10.6 Hybridization and Bonding Scheme

(20)

For Practice 10.6

Write a hybridization and bonding scheme for XeF₄. Continued

Example 10.6 Hybridization and Bonding Scheme

(21)

Procedure For…

Solution

Step 1 Write the Lewis structure for the molecule.

Acetaldehyde has 18 valence electrons and the following Lewis structure:

Step 2 Use VSEPR theory to predict the electron geometry about the central atom (or interior atoms).

The leftmost carbon atom has four electron groups and a tetrahedral electron geometry. The rightmost carbon atom has three electron groups and a trigonal planar geometry.

Write a hybridization and bonding scheme for acetaldehyde,

Example 10.7 Hybridization and Bonding Scheme

(22)

Continued

Example 10.7 Hybridization and Bonding Scheme

(23)

Step 4 Sketch the molecule, beginning with the central atom and its orbitals. Show overlap with the appropriate orbitals on the terminal atoms.

Step 5 Label all bonds using the σ or π notation followed by the type of overlapping orbitals.

Continued

Example 10.7 Hybridization and Bonding Scheme

(24)

For Practice 10.7

Write a hybridization and bonding scheme for HCN.

Continued

Example 10.7 Hybridization and Bonding Scheme

(25)

Procedure For…

Solution

Step 1 Write the Lewis structure for the molecule.

Step 2 Use VSEPR theory to predict the electron geometry about the central atom (or interior atoms).

The molecule has two interior atoms. Since each atom has three electron groups (one double bond and two single bonds), the electron geometry about each atom is trigonal planar.

Step 3 Select the correct hybridization for the central atom (or interior atoms) based on the electron geometry (see Table 10.3).

A trigonal planar geometry corresponds to sp² hybridization.

Use valence bond theory to write a hybridization and bonding scheme for ethene, H₂C＝CH₂

Example 10.8 Hybridization and Bonding Scheme

(26)

Continued

Example 10.8 Hybridization and Bonding Scheme

(27)

Step 4 Sketch the molecule, beginning with the central atom and its orbitals. Show overlap with the appropriate orbitals on the terminal atoms.

Step 5 Label all bonds using the σ or π notation followed by the type of overlapping orbitals.

Continued

Example 10.8 Hybridization and Bonding Scheme

(28)

For Practice 10.8

Use valence bond theory to write a hybridization and bonding scheme for CO₂.

For More Practice 10.8

What is the hybridization of the central iodine atom in I₃⁻? Continued

Example 10.8 Hybridization and Bonding Scheme

(29)

Solution

The H₂⁻ ion has three electrons. Assign the three electrons to the molecular orbitals, filling lower energy orbitals first and proceeding to higher energy orbitals.

Calculate the bond order by subtracting the number of electrons in antibonding orbitals from the number in bonding orbitals and dividing the result by two.

Use molecular orbital theory to predict the bond order in H₂⁻. Is the H₂⁻ bond a stronger or weaker bond than the H₂ bond?

Example 10.9 Bond Order

(30)

Since the bond order is positive, H₂⁻ should be stable. However, the bond order of H₂⁻ is lower than the bond order of H₂ (which is 1); therefore, the bond in H₂⁻ is weaker than in H₂.

For Practice 10.9

Use molecular orbital theory to predict the bond order in H₂⁺. Is the H₂⁺ bond a stronger or weaker bond than the H₂ bond?

Continued

Example 10.9 Bond Order

(31)

Solution

Write an energy level diagram for the molecular orbitals in N₂⁻. Use the energy ordering for N₂.

Draw an MO energy diagram and determine the bond order for the N₂⁻ ion. Do you expect the bond to be stronger or weaker than in the N₂ molecule? Is N₂⁻ diamagnetic or paramagnetic?

Example 10.10 Molecular Orbital Theory

(32)

The N₂⁻ ion has 11 valence electrons (5 for each nitrogen atom plus 1 for the negative charge). Assign the electrons to the molecular orbitals beginning with the lowest energy orbitals and following Hund’s rule.

Calculate the bond order by subtracting the number of electrons in antibonding orbitals from the number in bonding orbitals and dividing the result by two.

Continued

Example 10.10 Molecular Orbital Theory

(33)

The bond order is 2.5, which is a lower bond order than in the N₂ molecule (bond order = 3); therefore, the bond is weaker. The MO diagram shows that the N₂⁻ ion has one unpaired electron and is therefore paramagnetic.

For Practice 10.10

Draw an MO energy diagram and determine the bond order for the N₂⁺ ion. Do you expect the bond to be stronger or weaker than in the N₂ molecule? Is N₂⁺ diamagnetic or paramagnetic?

For More Practice 10.10

Use molecular orbital theory to determine the bond order of Ne₂. Continued

Example 10.10 Molecular Orbital Theory

(34)

Solution

Determine the number of valence electrons in the molecule or ion.

Number of valence electrons

= 4 (from C) + 5 (from N) +

1 (from negative charge) = 10

Write an energy level diagram using Figure 10.15 as a guide. Fill the orbitals beginning with the lowest energy orbital and progressing upward until all electrons have been assigned to an orbital. Remember to allow no more an two electrons (with paired spins) per orbital and to fill degenerate orbitals with single electrons (with parallel spins) before pairing.

Use molecular orbital theory to determine the bond order of the CN⁻ ion. Is the ion paramagnetic or diamagnetic?

Example 10.11 Molecular Orbital Theory for Heteronuclear Diatomic

Molecules and Ions

(35)

Continued

Example 10.11 Molecular Orbital Theory for Heteronuclear Diatomic

Molecules and Ions

(36)

Solution

Calculate the bond order using the appropriate formula:

If the MO diagram has unpaired electrons, the molecule or ion is paramagnetic. If the electrons are all paired, the molecule or ion is diamagnetic.

Since the MO diagram has no unpaired electrons, the ion is diamagnetic.

For Practice 10.11

Use molecular orbital theory to determine the bond order of NO. (Use the energy ordering of O₂.) Is the molecule paramagnetic or diamagnetic?

Continued

Example 10.11 Molecular Orbital Theory for Heteronuclear Diatomic

Molecules and Ions

(37)

10.1 Artificial Sweeteners: Fooled by Molecular Shape

Taste

(38)

General Chemistry 4^th edition, Hill, Petrucci, McCreary, Perry

38 Molecular Geometry

• Molecular geometry is simply the shape of a molecule.

• Molecular geometry is

described by the geometric

figure formed when the atomic nuclei are joined by (imaginary) straight lines.

• Molecular geometry is found using the Lewis structure, but the Lewis structure itself does NOT necessarily represent the molecule’s shape.

A water molecule is

angular or bent.

A carbon dioxide molecule is

linear.

(39)

10.2 VSEPR Theory: The Five Basic Shapes VSEPR Theory

• Valence-Shell Electron-Pair Repulsion (VSEPR) is a simple method for determining geometry.

• Basis: pairs of valence electrons in bonded atoms repel one another.

• These mutual repulsions push electron pairs as far from one another as possible.

B A B

B B A

When the electron pairs (bonds) are as far apart as they can

get, what will be the

B-A-B angle?

(40)

40 Electron-Group Geometries

• An electron group is a collection of valence electrons, localized in a

region around a central atom.

• One electron group:

– an unshared pair of valence electrons or – a bond (single, double, or triple)

• The repulsions among electron

groups lead to an orientation of the groups that is called the electron- group geometry.

• These geometries are based on the number of electron groups:

Electron groups

Electron-group geometry

2 Linear

3 Trigonal planar 4 Tetrahedral 5 Trigonal

bipyramidal

6 Octahedral

(41)

41 A Balloon Analogy

• Electron groups repel one another in the same way that balloons push one another apart.

• When four balloons, tied at the middle, push themselves apart as much as

possible, they make a

(42)

42

(43)

O N O • •

• •

• • • • There are three electron groups on N:

• Three lone pair

• One single bond

• One double bond

Electron Groups

• In the VSEPR notation

• A, central atom

• X, terminal atoms, bond pairs.

• E, electrons ,lone pairs

• The H ₂ O molecule would

(44)

Valence shell electron pair repulsion (VSEPR) model:

Predict the geometry of the molecule from the electrostatic

repulsions between the electron (bonding and nonbonding) pairs.

AB

₂

2 0

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

10.1 linear linear

B B

Two Electron Groups: Linear Electron

Geometry

(45)

Linear Geometry

(46)

Three Electron Groups:

Trigonal Planar Electron Geometry

AB

₂

2 0 linear linear

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

VSEPR

AB

₃

3 0 trigonal

planar

trigonal planar

10.1

(47)

Three Electron Groups:

Trigonal Planar Electron Geometry

Boron trifluride

(48)

AB

₂

2 0 linear linear

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

VSEPR

AB

₃

3 0 trigonal

planar

trigonal planar

10.1 AB

₄

4 0 tetrahedral tetrahedral

Four Electron Groups:

Tetrahedral Electron Geometry

(49)

Tetrahedral Geometry

Methane CH

₄

NH

₄⁺

(50)

AB

₂

2 0 linear linear

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

VSEPR

AB

₃

3 0 trigonal

planar

trigonal planar

10.1 AB

₄

4 0 tetrahedral tetrahedral

AB

₅

5 0 trigonal

bipyramidal

trigonal bipyramidal

Five Electron Groups: Trigonal

Bipyramidal Electron Geometry

(51)

Five Electron Groups: Trigonal Bipyramidal Electron Geometry

Phosphorus Pentachoride

(52)

AB

₂

2 0 linear linear

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

VSEPR

AB

₃

3 0 trigonal

planar

trigonal planar

10.1 AB

₄

4 0 tetrahedral tetrahedral

AB

₅

5 0 trigonal

bipyramidal

trigonal bipyramidal

AB

₆

6 0 octahedral octahedral

Octahedral Electron

Geometry

(53)

Octahedral Geometry

Sulfur Hexafluoride

(54)

10.1

(55)

10.3 VSEPR Theory: The Effect of Lone Pairs

Molecular Geometry

(56)

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

VSEPR

AB

₃

3 0 trigonal

planar

trigonal planar

AB

₂

E 2 1 trigonal

planar bent

10.1 Bent Molecular Geometry: Derivative of

Trigonal Planar Electron Geometry

(57)

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

VSEPR

AB

₃

E 3 1

AB

₄

4 0 tetrahedral tetrahedral

tetrahedral trigonal pyramidal

Pyramidal and Bent Molecular Geometries:

Derivatives of Tetrahedral Electron

Geometry

(58)

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

VSEPR

AB

₄

4 0 tetrahedral tetrahedral

10.1 AB

₃

E 3 1 tetrahedral trigonal

pyramidal

AB

₂

E

₂

2 2 tetrahedral bent

H O

H

Pyramidal and Bent Molecular Geometries:

Derivatives of Tetrahedral Electron

Geometry

(59)

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

VSEPR

AB

₅

5 0 trigonal

bipyramidal

trigonal bipyramidal

AB

₄

E 4 1 trigonal

bipyramidal

distorted tetrahedron

Derivatives of the Trigonal

Bipyramidal Electron Geometry

(60)

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

VSEPR

10.1 AB

₅

5 0 trigonal

bipyramidal

trigonal bipyramidal

AB

₄

E 4 1 trigonal

bipyramidal

distorted tetrahedron

AB

₃

E

₂

3 2 trigonal

bipyramidal T-shaped

F Cl

F

Derivatives of the Trigonal

Bipyramidal Electron Geometry

(61)

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

VSEPR

AB

₅

5 0 trigonal

bipyramidal

trigonal bipyramidal

AB

₄

E 4 1 trigonal

bipyramidal

distorted tetrahedron

AB

₃

E

₂

3 2 trigonal

bipyramidal T-shaped

AB

₂

E

₃

2 3 trigonal

bipyramidal linear

Derivatives of the Trigonal

Bipyramidal Electron Geometry

(62)

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

VSEPR

10.1 AB

₆

6 0 octahedral octahedral

AB

₅

E 5 1 octahedral square

pyramidal

Br F F

F F

F

Derivatives of the Octahedral

Geometry

(63)

Class

# of atoms bonded to central atom

# lone pairs on central atom

Arrangement of electron pairs

Molecular Geometry

VSEPR

AB

₆

6 0 octahedral octahedral

AB

₅

E 5 1 octahedral square

pyramidal

AB

₄

E

₂

4 2 octahedral square

planar

Xe F F

Derivatives of the Octahedral

Geometry

(64)

(65)

(66)

1. Draw Lewis structure for molecule.

2. Determine the number of electron groups around the central atom.

3. Classify each electron group as a bonding or lone pair, and count each type.

– Remember, multiple bonds count as one group.

4. Use VSEPR to predict the geometry of the molecule.

What are the molecular geometries of SO

₂

and SF

₄

? S

O O

AB

₂

E bent

S F

F

F F

AB

₄

E distorted tetrahedron

10.4 10.4 Predicting the Shapes around Central Atoms

(67)

Multiple Central Atoms

The shape around left C is tetrahedral.

The shape around left N is tetrahedral–trigonal pyramidal.

(68)

10.5 Molecular Shape and Polarity Polarity of Molecules

m = Q x r

Q is the charge

r is the distance between charges

1 D = 3.36 x 10

^-30

C m

(69)

Dipole Moments and Polar Molecules

H F

electron rich region electron poor

region

d+ d-

m = Q x r

(70)

10.2

(71)

(72)

Vector Addition

(73)

(74)

10.2 Which of the following molecules have a dipole moment?

H

₂

O, CO

₂

, SO

₂

, and CH

₄

O

dipole moment polar molecule

S

C

O O

no dipole moment nonpolar molecule

dipole moment polar molecule

C H

H

H H

no dipole moment

nonpolar molecule

(75)

Does CH

₂

Cl

₂

have

a dipole moment?

(76)

Molecular Polarity Affects Solubility

in Water

(77)

10.6 Valence Bond Theory: Orbital Overlap as a Chemical Bond

Bond Dissociation Energy Bond Length

H ₂ F ₂

436.4 kJ/mole 150.6 kJ/mole

74 pm 142 pm

Overlap Of 2 1s

2 2p How does Lewis theory explain the bonds in H

₂

and F

₂

?

Sharing of two electrons between the two atoms.

(78)

Orbital Interaction Orbital Interaction

(79)

10.7 Valence Bond Theory: Hybridization of Atomic Orbitals─混成只有在中心原子

Valence Bond Theory and NH ₃ N – 1s ² 2s ² 2p ³

3 H – 1s ¹

If the bonds form from overlap of 3 2p orbitals on nitrogen with the 1s orbital on each hydrogen atom, what would the molecular geometry of NH

₃

be? If use the

3 2p orbitals

predict 90

⁰

(80)

Hybridization – mixing of two or more atomic orbitals to form a new set of hybrid orbitals.

1. Mix at least 2 nonequivalent atomic orbitals (e.g. s and p). Hybrid orbitals have very different shape from original atomic orbitals.

2. Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process.

3. Covalent bonds are formed by:

a. Overlap of hybrid orbitals with atomic orbitals b. Overlap of hybrid orbitals with other hybrid

orbitals

10.4

(81)

sp ³ Hybridization

(82)

sp ³ Hybridization

Predict correct

bond angle

(83)

sp ² Hybridization

(84)

sp ² Hybridization

(85)

Sigma bond (s) – electron density between the 2 atoms

(86)

10.5

(87)

Hybrid orbitals overlap to form a s bond.

Orbital Diagrams of Bonding cont.

(88)

sp Hybridization

(89)

sp Hybridization

(90)

10.5 sp Hybridization

(91)

p s

sp Hybridization

(92)

sp ³ d Hybridization

(93)

sp ³ d ² Hybridization

(94)

(95)

# of Lone Pairs +

# of Bonded Atoms Hybridization Examples 2

3 4

sp sp

²

sp

³

BeCl

₂

BF

₃

CH

₄

, NH

₃

, H

₂

O How do I predict the hybridization of the central atom?

Count the number of lone pairs AND the number

of atoms bonded to the central atom

(96)

Sigma ( s ) and Pi Bonds ( p )

Single bond 1 sigma bond

Double bond 1 sigma bond and 1 pi bond Triple bond 1 sigma bond and 2 pi bonds

How many s and p bonds are in the acetic acid (vinegar) molecule CH

₃

COOH?

C H

H

C H

O

O H s bonds = 6 + 1 = 7 p bonds = 1

10.5

(97)

Problems with Valence Bond (VB) Theory

Molecular orbital theory – bonds are formed from

O O

No unpaired e

^-

Should be diamagnetic

Experiments show O

₂

is paramagnetic

(98)

10.8 Molecular Orbital (MO) Theory

Energy levels of bonding and antibonding molecular orbitals in hydrogen (H ₂ ).

A bonding molecular orbital has lower energy and greater stability than the atomic orbitals from which it was formed.

An antibonding molecular orbital has higher energy and lower stability than the atomic orbitals from which it was

formed. _10.6

(99)

(100)

Interaction of 1s Orbitals

(101)

(102)

10.6

(103)

Interaction of p Orbitals

(104)

10.6

(105)

1. The number of molecular orbitals (MOs) formed is always equal to the number of atomic orbitals combined.

2. The more stable the bonding MO, the less stable the corresponding antibonding MO.

3. The filling of MOs proceeds from low to high energies.

4. Each MO can accommodate up to two electrons.

5. Use Hund’s rule when adding electrons to MOs of the same energy.

Molecular Orbital (MO) Configurations

(106)

Period Two Homonuclear Diatomic

Molecules

(107)

(108)

10.7

(109)

Heteronuclear Diatomic Molecules and Ions

s

_2s

bonding MO

shows more electron

density near O because it is

mostly O’s 2s atomic orbital.

(110)

HF

(111)

Polyatomic Molecules

Ozone, O ₃

(112)