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: Hybridization of Atomic Orbitals 445 10.8 Molecular Orbital Theory: Electron Electron Delocalization 458 Lecture Presentation Chapter 10 Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory Sherril Soman Grand Valley State University

Example 10.1 VSEPR Theory and the Basic Shapes Determine the molecular geometry of NO 3. 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. The nitrogen atom has three electron groups. Based on the number of electron groups, determine the geometry that minimizes the repulsions between the groups. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.1 VSEPR Theory and the Basic Shapes Continued 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.2 Predicting Molecular Geometries Predict the geometry and bond angles of PCl 3. 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.2 Predicting Molecular Geometries Continued 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.2 Predicting Molecular Geometries Continued Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.2 Predicting Molecular Geometries Continued 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.3 Predicting Molecular Geometries Predict the geometry and bond angles of ICl 4. 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.3 Predicting Molecular Geometries Continued 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.3 Predicting Molecular Geometries Continued Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.3 Predicting Molecular Geometries Continued 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.4 Predicting the Shape of Larger Molecules Predict the geometry about each interior atom in methanol (CH 3 OH) and make a sketch of the molecule. 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: Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.4 Predicting the Shape of Larger Molecules Continued For Practice 10.4 Predict the geometry about each interior atom in acetic acid and make a sketch of the molecule. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.5 Determining if a Molecule Is Polar Determine if NH 3 is polar. 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.5 Determining if a Molecule Is Polar Continued 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.6 Hybridization and Bonding Scheme Write a hybridization and bonding scheme for bromine trifluoride, BrF 3. 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 Step 3 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. 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.6 Hybridization and Bonding Scheme Continued Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.6 Hybridization and Bonding Scheme Continued 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.6 Hybridization and Bonding Scheme Continued For Practice 10.6 Write a hybridization and bonding scheme for XeF 4. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.7 Hybridization and Bonding Scheme Write a hybridization and bonding scheme for acetaldehyde, 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 Step 3 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. Select the correct hybridization for the central atom (or interior atoms) based on the electron geometry (see Table 10.3). The leftmost carbon atom is sp 3 hybridized, and the rightmost carbon atom is sp 2 hybridized. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.7 Hybridization and Bonding Scheme Continued Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.7 Hybridization and Bonding Scheme Continued 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.7 Hybridization and Bonding Scheme Continued For Practice 10.7 Write a hybridization and bonding scheme for HCN. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.8 Hybridization and Bonding Scheme Use valence bond theory to write a hybridization and bonding scheme for ethene, H 2 C=CH 2 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.8 Hybridization and Bonding Scheme Continued Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.8 Hybridization and Bonding Scheme Continued 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.8 Hybridization and Bonding Scheme Continued 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? Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.9 Bond Order 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? 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.9 Bond Order Continued 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? Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.10 Molecular Orbital Theory 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? Solution Write an energy level diagram for the molecular orbitals in N 2. Use the energy ordering for N 2. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.10 Molecular Orbital Theory Continued 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.10 Molecular Orbital Theory Continued 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.11 Molecular Orbital Theory for Heteronuclear Diatomic Molecules and Ions Use molecular orbital theory to determine the bond order of the CN ion. Is the ion paramagnetic or diamagnetic? 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. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.11 Molecular Orbital Theory for Heteronuclear Diatomic Molecules and Ions Continued FIGURE 10.15 Molecular Orbital Energy Diagrams for Second-Row p-block Homonuclear Diatomic Molecules Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

Example 10.11 Molecular Orbital Theory for Heteronuclear Diatomic Molecules and Ions Continued 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? Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro

10.1 Artificial Sweeteners: Fooled by Molecular Shape Taste Using Lewis Theory to Predict Molecular Shapes

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. Prentice Hall 2005 A carbon dioxide molecule is linear. A water molecule is angular or bent. General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Ten

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 A B When the electron pairs (bonds) are as far apart as they can get, what will be the B-A-B angle? B B

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 electrongroup geometry. These geometries are based on the number of electron groups: Electron groups 2 Linear Electron-group geometry 3 Trigonal planar 4 Tetrahedral 5 Trigonal bipyramidal 6 Octahedral Prentice Hall 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Ten

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 tetrahedral shape. Prentice Hall 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Ten

42 Prentice Hall 2005 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Chapter Ten

Electron Groups O N O There are three electron groups on N: In the VSEPR notation A, central atom X, terminal atoms, bond pairs. E, electrons,lone pairs Three lone pair One single bond One double bond The H 2 O molecule would therefore carry the designation AX 2 E 2.

Two Electron Groups: Linear Electron Geometry Valence shell electron pair repulsion (VSEPR) model: Predict the geometry of the molecule from the electrostatic repulsions between the electron (bonding and nonbonding) pairs. Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB 2 2 0 linear linear B B 10.1

Linear Geometry Cl Be Cl 2 0 atoms lone bonded pairs to central central atom atom 10.1

Three Electron Groups: Trigonal Planar Electron Geometry VSEPR Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB 2 2 0 linear linear AB 3 3 0 trigonal planar trigonal planar 10.1

Three Electron Groups: Trigonal Planar Electron Geometry Boron trifluride

Four Electron Groups: Tetrahedral Electron Geometry Class # of atoms bonded to central atom VSEPR # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB 2 2 0 linear linear AB 3 3 0 trigonal planar trigonal planar AB 4 4 0 tetrahedral tetrahedral 10.1

Tetrahedral Geometry CH 4 NH 4 + Methane

Five Electron Groups: Trigonal Bipyramidal Electron Geometry VSEPR Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB 2 2 0 linear linear AB 3 3 0 trigonal planar trigonal planar AB 4 4 0 tetrahedral tetrahedral AB 5 5 0 trigonal bipyramidal trigonal bipyramidal 10.1

Five Electron Groups: Trigonal Bipyramidal Electron Geometry Phosphorus Pentachoride

Octahedral Electron Geometry Class # of atoms bonded to central atom VSEPR # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB 2 2 0 linear linear AB 3 3 0 trigonal planar trigonal planar AB 4 4 0 tetrahedral tetrahedral AB 5 5 0 trigonal bipyramidal trigonal bipyramidal AB 6 6 0 octahedral octahedral 10.1

Octahedral Geometry Sulfur Hexafluoride

10.1

10.3 VSEPR Theory: The Effect of Lone Pairs Molecular Geometry lone-pair vs. lone pair repulsion lone-pair vs. bonding > > pair repulsion bonding-pair vs. bonding pair repulsion

Bent Molecular Geometry: Derivative of Trigonal Planar Electron Geometry VSEPR Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB 3 3 0 AB 2 E 2 1 trigonal planar trigonal planar trigonal planar bent 10.1

Pyramidal and Bent Molecular Geometries: Derivatives of Tetrahedral Electron Geometry Class # of atoms bonded to central atom # lone pairs on central atom VSEPR Arrangement of electron pairs Molecular Geometry AB 4 4 0 tetrahedral tetrahedral AB 3 E 3 1 tetrahedral trigonal pyramidal 10.1

Pyramidal and Bent Molecular Geometries: Derivatives of Tetrahedral Electron Geometry Class # of atoms bonded to central atom VSEPR # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB 4 4 0 tetrahedral tetrahedral AB 3 E 3 1 tetrahedral trigonal pyramidal AB 2 E 2 2 2 tetrahedral bent H O H 10.1

Derivatives of the Trigonal Bipyramidal Electron Geometry Class # of atoms bonded to central atom VSEPR # lone pairs on central atom AB 5 5 0 AB 4 E 4 1 Arrangement of electron pairs trigonal bipyramidal trigonal bipyramidal Molecular Geometry trigonal bipyramidal distorted tetrahedron seesaw shape 10.1

Derivatives of the Trigonal Bipyramidal Electron Geometry Class # of atoms bonded to central atom VSEPR # lone pairs on central atom AB 5 5 0 AB 4 E 4 1 AB 3 E 2 3 2 Arrangement of electron pairs trigonal bipyramidal trigonal bipyramidal trigonal bipyramidal Molecular Geometry trigonal bipyramidal distorted tetrahedron T-shaped F F Cl F 10.1

Derivatives of the Trigonal Bipyramidal Electron Geometry Class # of atoms bonded to central atom VSEPR # lone pairs on central atom AB 5 5 0 AB 4 E 4 1 AB 3 E 2 3 2 AB 2 E 3 2 3 Arrangement of electron pairs trigonal bipyramidal trigonal bipyramidal trigonal bipyramidal trigonal bipyramidal Molecular Geometry trigonal bipyramidal distorted tetrahedron T-shaped linear I I I 10.1

Derivatives of the Octahedral Geometry Class # of atoms bonded to central atom VSEPR # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB 6 6 0 octahedral octahedral AB 5 E 5 1 octahedral square pyramidal F F F F Br F 10.1

Derivatives of the Octahedral Geometry Class # of atoms bonded to central atom VSEPR # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB 6 6 0 octahedral octahedral AB 5 E 5 1 octahedral AB 4 E 2 4 2 octahedral square pyramidal square planar F F Xe F F 10.1

10.4 Predicting the Shapes around Central Atoms 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? O S O F AB 4 E AB 2 E bent F S F F distorted tetrahedron 10.4

Multiple Central Atoms The shape around left N is tetrahedral trigonal pyramidal. The shape around left C is tetrahedral. The shape around center C is trigonal planar. The shape around right O is tetrahedral bent.

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 electron poor region H electron rich region F d+ d- m = Q x r Q is the charge r is the distance between charges 1 D = 3.36 x 10-30 C m 10.2

10.2

10.2

Vector Addition

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 dipole moment polar molecule H O C O H C H no dipole moment nonpolar molecule H no dipole moment nonpolar molecule 10.2

Does CH 2 Cl 2 have a dipole moment? 10.2

Molecular Polarity Affects Solubility in Water

10.6 Valence Bond Theory: Orbital Overlap as a Chemical Bond How does Lewis theory explain the bonds in H 2 and F 2? Sharing of two electrons between the two atoms. Bond Dissociation Energy Bond Length Overlap Of H 2 436.4 kj/mole 74 pm 2 1s F 2 150.6 kj/mole 142 pm 2 2p Valence bond theory bonds are formed by sharing of e - from overlapping atomic orbitals.

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 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 Actual H-N-H bond angle is 107.3 0 7

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 Hybridization Predict correc bond angle

sp 2 Hybridization

sp 2 Hybridization

Sigma bond (s) electron density between the 2 atoms Pi bond (p) electron density above and below plane of nuclei of the bonding atoms 10.5

10.5

Orbital Diagrams of Bonding cont. Hybrid orbitals overlap to form a s bond. Unhybridized p orbitals overlap to form a p bond.

sp Hybridization

sp Hybridization

sp Hybridization 10.5

sp Hybridization p s p

sp 3 d Hybridization

sp 3 d 2 Hybridization

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 # of Lone Pairs + # of Bonded Atoms Hybridization Examples 2 sp BeCl 2 3 sp 2 BF 3 4 sp 3 CH 4, NH 3, H 2 O 5 sp 3 d PCl 5 6 sp 3 d 2 SF 6 10.4

Sigma (s) and Pi Bonds (p) Single bond Double bond Triple bond 1 sigma bond 1 sigma bond and 1 pi bond 1 sigma bond and 2 pi bonds How many s and p bonds are in the acetic acid (vinegar) molecule CH 3 COOH? H O H C C O H s bonds = 6 + 1 = 7 p bonds = 1 H 10.5

Problems with Valence Bond (VB) Theory O O No unpaired e - Should be diamagnetic Experiments show O 2 is paramagnetic Molecular orbital theory bonds are formed from interaction of atomic orbitals to form molecular orbitals. 10.8

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

10.6

Interaction of 1s Orbitals

bond order ½ 1 ½ 0 10.7

10.6

Interaction of p Orbitals

10.6

Molecular Orbital (MO) Configurations 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. 6. The number of electrons in the MOs is equal to the sum of all the electrons on the bonding atoms. 10.7

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.

HF

Polyatomic Molecules Ozone, O 3

Delocalized molecular orbitals are not confined between two adjacent bonding atoms, but actually extend over three or more atoms. Electron density above and below the plane of the benzene molecule. 10.8

Representing Three-Dimensional Shapes on Paper 10.4

Bond Rotation 10.7

10.7