### 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:

© 2014 Pearson Education, Inc.

**Solution **

The molecular geometry of NO_{3}^{−} is determined by the number of electron groups around the central atom (N).

Begin by drawing a Lewis structure of NO_{3}^{−}.

NO_{3}^{−} 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 NO_{3}^{−}.

**Example 10.1 ** **VSEPR Theory and the Basic Shapes **

© 2014 Pearson Education, Inc.

*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 CCl_{4}.
Continued

**Example 10.1 ** **VSEPR Theory and the Basic Shapes **

**Procedure For… **

**Predicting Molecular Geometries **

**Solution **

**Step 1 ** Draw the Lewis structure for the molecule.

PCl_{3} 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 PCl_{3}.

**Example 10.2 ** **Predicting Molecular Geometries **

© 2014 Pearson Education, Inc.

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

**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 **

Continued

**Example 10.2 ** **Predicting Molecular Geometries **

© 2014 Pearson Education, Inc.

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

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 **

**Procedure For… **

**Predicting Molecular Geometries **

**Solution **

**Step 1 ** Draw the Lewis structure for the molecule.

ICl_{4}^{−} 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 ICl_{4}^{−}.

**Example 10.3 ** **Predicting Molecular Geometries **

© 2014 Pearson Education, Inc.

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

**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 **

Continued

**Example 10.3 ** **Predicting Molecular Geometries **

© 2014 Pearson Education, Inc.

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

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 I_{3}^{−}.
Continued

**Example 10.3 ** **Predicting Molecular Geometries **

**Solution **

Begin by drawing the Lewis structure of CH_{3}OH. CH_{3}OH 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 (CH_{3}OH) and make a sketch of the molecule.

**Example 10.4 ** **Predicting the Shape of Larger Molecules **

© 2014 Pearson Education, Inc.

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

**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 **

**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 NH_{3} is polar.

**Example 10.5 ** **Determining if a Molecule Is Polar **

© 2014 Pearson Education, Inc.

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

**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 CF_{4}is polar.

Continued

**Example 10.5 ** **Determining if a Molecule Is Polar **

**Procedure For… **

**Hybridization and Bonding Scheme **

**Solution **

**Step 1 ** **Write the Lewis structure for the molecule. **

BrF_{3} 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 sp*^{3}*d hybridization. *

Write a hybridization and bonding scheme for bromine trifluoride, BrF_{3}.

**Example 10.6 ** **Hybridization and Bonding Scheme **

© 2014 Pearson Education, Inc.

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

Continued

**Example 10.6 ** **Hybridization and Bonding Scheme **

**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 **

© 2014 Pearson Education, Inc.

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

**For Practice 10.6 **

Write a hybridization and bonding scheme for XeF_{4}.
Continued

**Example 10.6 ** **Hybridization and Bonding Scheme **

**Procedure For… **

**Hybridization and Bonding Scheme **

**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 **

© 2014 Pearson Education, Inc.

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

Continued

**Example 10.7 ** **Hybridization and Bonding Scheme **

**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 **

© 2014 Pearson Education, Inc.

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

**For Practice 10.7 **

Write a hybridization and bonding scheme for HCN.

Continued

**Example 10.7 ** **Hybridization and Bonding Scheme **

**Procedure For… **

**Hybridization and Bonding Scheme **

**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 sp*^{2} hybridization.

Use valence bond theory to write a hybridization and bonding scheme for ethene, H_{2}C＝CH_{2 }

**Example 10.8 ** **Hybridization and Bonding Scheme **

© 2014 Pearson Education, Inc.

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

Continued

**Example 10.8 ** **Hybridization and Bonding Scheme **

**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 **

© 2014 Pearson Education, Inc.

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

**For Practice 10.8 **

Use valence bond theory to write a hybridization and bonding scheme for CO_{2}.

**For More Practice 10.8 **

What is the hybridization of the central iodine atom in I_{3}^{−}?
Continued

**Example 10.8 ** **Hybridization and Bonding Scheme **

**Solution **

The H_{2}^{−} 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 H_{2}^{−}. Is the H_{2}^{−} bond a stronger or weaker bond than
the H_{2} bond?

**Example 10.9 ** **Bond Order **

© 2014 Pearson Education, Inc.

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

Since the bond order is positive, H_{2}^{−} should be stable. However, the bond order of H_{2}^{−} is lower than the bond order
of H_{2} (which is 1); therefore, the bond in H_{2}^{−} is weaker than in H_{2}.

**For Practice 10.9 **

Use molecular orbital theory to predict the bond order in H_{2}^{+}. Is the H_{2}^{+} bond a stronger or weaker bond than
the H_{2} bond?

Continued

**Example 10.9 ** **Bond Order **

**Solution **

Write an energy level diagram for the molecular orbitals in N_{2}^{−}. Use the energy ordering for N_{2}.

Draw an MO energy diagram and determine the bond order for the N_{2}^{−} ion. Do you expect the bond to be stronger or
weaker than in the N_{2} molecule? Is N_{2}^{−} diamagnetic or paramagnetic?

**Example 10.10 ** **Molecular Orbital Theory **

© 2014 Pearson Education, Inc.

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

The N_{2}^{−} 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 **

The bond order is 2.5, which is a lower bond order than in the N_{2} molecule (bond order = 3); therefore, the bond is
weaker. The MO diagram shows that the N_{2}^{−} 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 N_{2}^{+} ion. Do you expect the bond to be stronger
or weaker than in the N_{2} molecule? Is N_{2}^{+} diamagnetic or paramagnetic?

**For More Practice 10.10 **

Use molecular orbital theory to determine the bond order of Ne_{2}.
Continued

**Example 10.10 ** **Molecular Orbital Theory **

© 2014 Pearson Education, Inc.

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

**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 **

Continued

**Example 10.11 ** **Molecular Orbital Theory for Heteronuclear Diatomic **

**Molecules and Ions **

© 2014 Pearson Education, Inc.

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

**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 O_{2}.) Is the molecule
paramagnetic or diamagnetic?

Continued

**Example 10.11 ** **Molecular Orbital Theory for Heteronuclear Diatomic **

**Molecules and Ions **

**10.1 Artificial Sweeteners: Fooled by ** **Molecular Shape **

**Taste **

Chapter Ten Prentice Hall © 2005

*General Chemistry 4*^{th} edition, Hill, Petrucci, McCreary, Perry

### 38

**Molecular Geometry **

**• Molecular geometry is simply ** **the shape of a molecule. **

**• 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.

**using the Lewis structure, but**

**A water ** **molecule is **

**angular or ** **bent. **

**angular or**

**bent.**

**A carbon ** **dioxide ** **molecule is **

**linear. **

**linear.**

**10.2 VSEPR Theory: The Five Basic Shapes ** **VSEPR Theory **

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

**• Valence-Shell Electron-Pair Repulsion (VSEPR) is**

### • 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? **

**B-A-B angle?**

Chapter Ten Prentice Hall © 2005

*General Chemistry 4*^{th} edition, Hill, Petrucci, McCreary, Perry

### 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: **

**number of electron groups:**

**Electron ** **groups **

**Electron-group ** **geometry **

**2 ** **Linear **

**3 ** **Trigonal planar ** **4 ** **Tetrahedral ** **5 ** **Trigonal **

**bipyramidal **

**6 ** **Octahedral **

### 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

Chapter Ten Prentice Hall © 2005

*General Chemistry 4*^{th} edition, Hill, Petrucci, McCreary, Perry

### 42

### 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

**Valence shell electron pair repulsion (VSEPR) model: **

**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 **

**Linear Geometry **

**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

**Three Electron Groups: **

**Trigonal Planar Electron Geometry **

### Boron trifluride

### 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 **

**Tetrahedral Geometry **

### Methane CH

_{4 }

**NH**

_{4}^{+ }### 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 **

**Five Electron Groups: Trigonal Bipyramidal ** **Electron Geometry **

### Phosphorus Pentachoride

### 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 **

**Octahedral Geometry **

### Sulfur Hexafluoride

### 10.1

**10.3 VSEPR Theory: The Effect of Lone Pairs **

### Molecular Geometry

### 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 **

### 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 **

### 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 **

### 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 **

### 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

### F

**Derivatives of the Trigonal **

**Bipyramidal Electron Geometry **

### 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 **

### 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 **

### 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 **

© 2014 Pearson Education, Inc.

### 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 **

**Multiple Central Atoms **

### The shape around left C is tetrahedral.

### The shape around left N is tetrahedral–trigonal pyramidal.

© 2014 Pearson Education, Inc.

**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

### Dipole Moments and Polar Molecules

### H F

### electron rich region electron poor

### region

### d+ d-

### m = Q x r

### 10.2

© 2014 Pearson Education, Inc.

**Vector Addition **

### 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

### Does CH

_{2}

### Cl

_{2}

### have

### a dipole moment?

© 2014 Pearson Education, Inc.

**Molecular Polarity Affects Solubility **

**in Water **

**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.

© 2014 Pearson Education, Inc.

**Orbital Interaction ** **Orbital Interaction **

**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}

### 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}

**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

**sp** ^{3 } **Hybridization **

**sp**

^{3 }

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**sp** ^{3 } **Hybridization **

**sp**

^{3 }

### Predict correct

### bond angle

**sp** ^{2 } **Hybridization **

**sp**

^{2 }

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**sp** ^{2 } **Hybridization **

**sp**

^{2 }

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

### 10.5

### Hybrid orbitals overlap to form a s bond.

**Orbital Diagrams of Bonding cont. **

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**sp Hybridization **

**sp Hybridization**

**sp Hybridization **

**sp Hybridization**

### 10.5

**sp Hybridization **

**sp Hybridization**

### p s

**sp Hybridization **

**sp Hybridization**

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**sp** ^{3} **d Hybridization **

**sp**

^{3}

**d Hybridization**

**sp** ^{3} **d** ^{2 } **Hybridization**

**sp**

^{3}

**d**

^{2 }

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### # 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

### 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

**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

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**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.

**A bonding molecular orbital has lower energy and greater**

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

**An antibonding molecular orbital has higher energy and**

### formed. _{10.6 }

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**Interaction of 1s Orbitals **

**Interaction of 1s Orbitals**

### 10.6

*Interaction of p Orbitals *

### 10.6

### 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 **

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**Period Two Homonuclear Diatomic **

**Molecules **

### 10.7

**Heteronuclear Diatomic Molecules and Ions **

### s

_{2s }### bonding MO

### shows more electron

### density near O because it is

*mostly O’s 2s atomic orbital. *

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**HF **

**Polyatomic Molecules **

**Ozone, O** _{3}

_{3}

**Delocalized molecular orbitals are not confined between ** two adjacent bonding atoms, but actually extend over three **or more atoms. **

**Delocalized molecular orbitals are not confined between**

**or more atoms.**

### 10.8

### Electron density above and below the plane of the

### benzene molecule.

**Representing Three-Dimensional Shapes **

**on Paper **

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