Chapter 9. Molecular Geometry and Bonding Theories

Chapter 9 Molecular Geometry and Bonding Theories

MOLECULAR SHAPES 2

Molecular Shapes Lewis Structures show bonding and lone pairs do not denote shape Use Lewis Structures to determine shapes Molecular geometry describes the three dimensional shape of a molecule 3

Molecular Shapes Molecular geometry determines Physical properties Polarity Solubility Boiling and melting point Chemical properties Chemical reactivities Square planar molecule cisplatin [Pt(NH 3 ) 2 Cl 2 ]» cis isomer treats certain cancers» trans isomer can not treat cancer http://wps.prenhall.com/wps/media/objects/3085/3159106/blb2404/bl24fg17.jpg 4

Predict molecular geometry of simple molecules by using valence shell electron pair repulsion theory (VSEPR Theory) 5

Valence-Shell Electron-Pair Repulsion (VSEPR) Model Electrons try get as far apart as possible The balloon analogy demonstrates the maximum distances that minimize repulsions 2015 Pearson Education, Inc. 6

Valence-Shell Electron-Pair Repulsion (VSEPR) Model Bonding and nonbonding electron pairs repel each other Assume electron pairs are placed as far as possible 2015 Pearson Education, Inc. 7

Valence-Shell Electron-Pair Repulsion (VSEPR) Model Assume electron pairs are placed as far as possible 2015 Pearson Education, Inc. 8

Valence-Shell Electron-Pair Repulsion (VSEPR) Model Nonbonding pairs Larger than bonding pairs Repulsions are greater Compress bond angles 9

The basic principle of the VSEPR model is that bonding electron pairs in a molecule will be as far apart as possible, but nonbonding pairs can approach more closely. a. True b. False

Valence-Shell Electron-Pair Repulsion (VSEPR) Direction in which electrons point called electron domains Domains may have Two electrons Single covalent bond Nonbonding electron pair Four electrons Double bond Six electrons Triple covalent bond 2015 Pearson Education, Inc. 11

Valence-Shell Electron-Pair Repulsion (VSEPR) Count electron domains around central atom 2 electron domains 3 electron domains 4 electron domains 2015 Pearson Education, Inc. 12

Valence-Shell Electron-Pair Repulsion (VSEPR) Model Multiple bonds Greater repulsive force than single bonds 13

Valence-Shell Electron-Pair Repulsion (VSEPR) Applying VSEPR theory 1. Draw a Lewis structure 2. Determine number of electron domains 3. Predict electron domain arrangement 4. Establish molecular geometry by bonding electrons placement 14

Valence-Shell Electron-Pair Repulsion (VSEPR): Molecular Geometries Determined the electron-domain geometry Look at each electron domain to see what molecular geometries are possible Use arrangement of bonded atoms to determine molecular geometry 2015 Pearson Education, Inc. 15

What molecular geometries are possible 16

VSEPR: Electron Group Possible electron group geometries 17

VSEPR: Molecular Geometry 2 electron domains 1 molecular geometry 3 electron domains 2 molecular geometries 4 electron domains 3 molecular geometries 18

VSEPR: Molecular Geometry 5 electron domains 4 molecular geometry 2015 Pearson Education, Inc. 19

Trigonal Bipyramidal Electron Domain Two distinct positions: Axial Equatorial Lone pairs occupy equatorial positions 20

VSEPR: Molecular Geometry 6 electron domains 3 molecular geometry All positions are equivalent in the octahedral domain 21

VSEPR: Molecular Geometry Molecules with multiple central atoms Consider each atom separately, then combine results 2015 Pearson Education, Inc. 22

EXAMPLE PREDICTING GEOMETRY USING VSEPR THEORY Predict the electron and molecular geometry of PCl 3. 1. Draw a Lewis structure for the SOLUTION molecule. PCl 3 has 26 electrons. 2. Determine the total number of electron groups around the central atom. Lone pairs, single bonds, double bonds, and triple bonds each count as one group. The central atom (P) has four electron groups. 3. Determine the number of bonding groups and the number of lone pairs around the central atom. These should sum to the 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. 23

EXAMPLE PREDICTING GEOMETRY USING VSEPR THEORY Continued 4. Determine the electron geometry and molecular geometry. The electron geometry is tetrahedral (four electron groups), and the molecular geometry the shape of the molecule is trigonal pyramidal (four electron groups, three bonding groups, and one lone pair). 24

EXAMPLE PREDICTING GEOMETRY USING VSEPR THEORY Continued SKILLBUILDER Predict the molecular geometry of ClNO (N is the central atom). Answer: 25

Does molecular shape affect the polarity of a molecule? Yes How? 26

Molecular Polarity: Nonpolar Nonpolar molecules have symmetrical distribution of electronic charge 2015 Pearson Education, Inc. 28

Molecular Polarity: Polar Polar molecules have unsymmetrical electronic charge 2015 Pearson Education, Inc. 29

Molecular Polarity: Polarity To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE 30

Molecular Polarity: Comparison A NONPOLAR molecule A POLAR molecule 2015 Pearson Education, Inc. 31

COVALENT BONDING AND ORBITAL OVERLAP 32

Covalent Bond Overlap of half filled orbitals occur with bonding Produces orbital common to both atoms Electrons move throughout this new orbital This is the valence-bond theory 2015 Pearson Education, Inc. 33

Orbital Overlap and Bonding Atoms get closer together until a balance is reached between repulsions and attraction Potential energy at minimum is bond strength Internuclear distance at minimum potential energy is bond length Internuclear repulsions get too great if atoms get too close 35

VSEPR and Hybrid Orbitals Hydrogen 1s 1 Lewis Structure VSEPR Model Oxygen 1s 2 2s 2 2p 4 Draw the best Lewis structure to assign VSEPR VSEPR predicts shapes of molecules very well How does that fit with orbitals? Wouldn t that make the angle 90 o? Why is it 104.5 o? 2015 Pearson Education, Inc. 36

Hybrid Orbitals Why is it 104.5 o? Hybrid orbitals Formed by combining two or more atomic orbitals similar energy New orbitals are degenerate Energies of orbitals 'average' of the original unhybridized orbitals Orbitals created sum of original number of orbitals Have the same total electron capacity as the old orbitals Fill with electrons present in the pure atomic orbitals forming them 2015 Pearson Education, Inc. Cengage Learning. All Rights Reserved. 37

Hybrid Orbitals Hybrid orbitals Full, half filled or even empty orbitals can undergo hybridization Shapes of hybrid orbitals are identical Usually one big lobe and one small lobe Hybridization is the mixing of orbitals of same atom only Combination of orbitals belonging to different atoms is called bonding 2015 Pearson Education, Inc. Cengage Learning. All Rights Reserved. 38

Hybrid orbitals Occurs before bond formation Minimize repulsions between bonds formed Why hybridization occurs before bond formation Participate in σ bond formation Hybrid Orbitals σ bond is a covalent single bond Oxygen 2015 Pearson Education, Inc. 39

Hybrid Orbitals Why are bonds 104.5 o? Hybridization best explanation bond angles Molecule bond angles are similar to angles between the hybrid orbitals Molecule shape is influenced by the type of hybridization 2015 Pearson Education, Inc. 40

Example VSEPR Theory and the Basic Shapes Determine the molecular geometry of NO 3. Solution Determine the molecular geometry of NO 3 by counting 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 is: Use the Lewis structure, or any one of the resonance structures, to determine the number of electron groups around the central atom. The hybrid structure is intermediate between these three and has three equivalent bonds. The nitrogen atom has three electron groups.

Example VSEPR Theory and the Basic Shapes (continued) Based on the number of electron groups, determine the geometry that minimizes the repulsions between the groups. The electron geometry that minimizes the repulsions between three electron groups is trigonal planar. Because there are no lone pairs on the central atom, the molecular geometry is also trigonal planar. Since the three bonds are equivalent, they each exert the same repulsion on the other two and the molecule has three equal bond angles of 120.

Let s look at different types of hybridization 43

sp Hybridization: Be Orbital diagram for beryllium (Be) Contains only paired electrons in subshells Be makes compounds with 4 electrons in its valence Why? sp hybridization (mixing of one s orbital and one p orbital) 44

sp Orbitals sp orbitals Formed from an s and p orbitals Produces two degenerate hybrid orbitals Have two lobes like a p orbital One lobe is larger and more rounded, as in the s orbital Orbitals align themselves 180 from each other 2015 Pearson Education, Inc. 45

sp Orbitals: Position sp degenerate orbitals align at 180 Consistent with: Geometry of Be compounds (like BeF 2 ) VSEPR: linear Two electron groups gives sp hybridization 2015 Pearson Education, Inc. 46

sp 2 Hybridization: B Orbital diagram for boron (B) Contains paired and unpaired electrons Orbitals partially filled Boron makes molecules with six electrons in valence WHY? Mixing of s and p orbitals yields 3 degenerate hybrid orbitals sp 2 2p 47

sp 2 Orbitals sp 2 orbitals Formed from an s and 2 p orbitals Produces 3 degenerate hybrid orbitals Have two lobes like a p orbital One lobe is larger and more rounded, as in the s orbital larger and Orbitals align themselves 120 from each other more rounded, as in the s orbital Orbitals align themselves 120 from each other 3 electron groups gives sp 2 hybridization 2015 Pearson Education, Inc. 48

sp 3 Hybridization: C Orbital diagram for carbon (C) Contains paired and unpaired electrons Orbitals partially filled Carbon makes molecules with four bonds WHY? Mixing of s and p orbitals yields 4 degenerate hybrid orbitals sp 3 49

sp 3 orbitals Formed from an s and 3 p orbitals Produces 4 degenerate hybrid orbitals Have two lobes like a p orbital One lobe is larger and more rounded, as in the s orbital Orbitals align themselves 109.5 from each other sp 3 Orbitals 4 electron groups gives sp 3 hybridization 2015 Pearson Education, Inc. 50

Hypervalent Molecules Hypervalent Molecules Atoms have more than an octet surrounding it Valence-Bond model would use d orbitals to make bonds Correctly predicts bond orientation Theoretical studies suggest this view is incorrect Trigonal Bipyramidal Energy required to use d orbitals would be too great A more detailed bonding view is required than what we use in this course to understand 51

Hypervalent Molecules Hypervalent Molecules Atoms have more than an octet surrounding it Valence-Bond model would use d orbitals to make bonds Correctly predicts bond orientation Theoretical studies suggest this view is incorrect Trigonal Bipyramidal Energy required to use d orbitals would be too great Will show how it works because it is sometimes used to explain molecular geometry 52

dsp 3 Hybridization: P Orbital diagram for phosphorous (P) Contains paired and unpaired electrons Orbitals partially filled Phosphorous makes molecules with 5 bonds WHY? Mixing of s, p and d orbitals yields 5 degenerate hybrid orbitals 3d 3d dsp 3 3d 53

dsp 3 Orbitals dsp 3 orbitals Formed from an s, 3 p and 1 d orbitals Produces 5 degenerate hybrid orbitals Have two lobes like a p orbital One lobe is larger and more rounded, as in the s orbital Orbitals align themselves 90 o and 120 from each other Trigonal Bipyramidal 5 electron groups gives dsp 3 hybridization 2015 Pearson Education, Inc. Cengage Learning. All Rights Reserved. 54

d 2 sp 3 Hybridization: S Orbital diagram for sulfur (S) Contains paired and unpaired electrons Orbitals partially filled Sulfur makes molecules with six bonds WHY? Mixing of s, p and d orbitals yields 6 degenerate hybrid orbitals 3d 3d d 2 sp 3 3d 55

d 2 sp 3 Orbitals d 2 sp 3 orbitals Formed from an s, 3 p and 2 d orbitals Produces 6 degenerate hybrid orbitals Have two lobes like a p orbital One lobe is larger and more rounded, as in the s orbital Orbitals align themselves 90 from each other Octahedral 6 electron groups gives d 2 sp 3 hybridization 2015 Pearson Education, Inc. Cengage Learning. All Rights Reserved. 56

Which association of hybridization and molecular geometry is most unlikely? a. sp; linear b. sp 2 ; trigonal planar c. sp 3 ; tetrahedral d. sp 3 d; bent

In an sp 2 hybridized atom, we saw that there was one unhybridized 2p orbital. How many unhybridized 2p orbitals remain on an atom that has sp 3 hybrid orbitals? a. One b. Two c. Three d. None 58

Hybrid Orbital Summary 1) Draw Lewis structure 2) Use VSEPR to determine the electron group geometry 3) Specify the hybrid orbitals needed to accommodate these electron pairs d 2 sp 3 d 2 sp 3 Cengage Learning. All Rights Reserved. 59

Hybrid Orbital Summary 1) Draw Lewis structure 2) Use VSEPR to determine the electron group geometry 3) Specify the hybrid orbitals needed to accommodate these electron pairs 2015 Pearson Education, Inc. 60

Example Hybridization and Bonding Scheme Write a hybridization and bonding scheme for bromine trifluoride, BrF 3. Procedure For Hybridization and Bonding Scheme Step 1 Write the Lewis structure for the molecule. Solution 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.

Example Hybridization and Bonding Scheme (continued) Write a hybridization and bonding scheme for bromine trifluoride, BrF 3. Step 3Refer to select the correct hybridization for the central atom (or interior atoms) based on the electron geometry. A trigonal bipyramidal electron geometry corresponds to sp 3 d hybridization.

Example Hybridization and Bonding Scheme (continued) Write a hybridization and bonding scheme for bromine trifluoride, BrF 3. Step 4 Sketch the molecule, beginning with the central atom and its orbitals. Indicate overlap with the appropriate orbitals on the terminal atoms.

Example Hybridization and Bonding Scheme (continued) Write a hybridization and bonding scheme for bromine trifluoride, BrF 3. Step 5 Label all bonds using the σ notation followed by the type of overlapping orbitals.

How does a double or triple bond form? Can not form if hybridized orbitals used Only form if both hybridized and unhybridized orbitals are present 65

Types of Bonds Two types of bonds: Sigma (σ) bond Single covalent bond head-to-head overlap Pi (π) bond Multiple covalent bond side-to-side overlap electron density above and below σ bond Cengage Learning. All Rights Reserved. 66

Bonding in Molecules Single bonds are always σ-bonds 2015 Pearson Education, Inc. 67

Bonding in Molecules Single bonds are always σ-bonds Multiple bonds have one σ-bond, all other bonds are π-bonds Double bond 2015 Pearson Education, Inc. 68

Bonding in Molecules Single bonds are always σ-bonds Multiple bonds have one σ-bond, all other bonds are π-bonds Double bond Triple bond 2015 Pearson Education, Inc. 69

A molecule of crotonaldehyde (CH 3 CH=CH CH=O) contains pi bonds. a. zero b. one c. two d. four 70

Example Hybridization and Bonding Scheme Write a hybridization and bonding scheme for acetaldehyde, Procedure For Hybridization and Bonding Scheme Step 1 Write the Lewis structure for the molecule. Solution 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. 71

Example (continued) Hybridization and Bonding Scheme Step 3 Refer to Table to select the correct hybridization for the central atom (or interior atoms) based on the electron geometry. The leftmost carbon atom is sp 3 hybridized, rightmost carbon atom is sp 2 hybridized. 72

Example (continued) Hybridization and Bonding Scheme Step 4 Sketch the molecule, beginning with the central atom and its orbitals. Indicate overlap with the appropriate orbitals on the terminal atoms. Step 5 Label all bonds using the σ and π notation followed by the type of overlapping orbitals. 73

When two atoms are bonded by a triple bond, what is the hybridization of the orbitals that make up the s-bond component of the bond? a. sp hybrid atomic orbitals b. sp 2 hybrid atomic orbitals c. sp 3 hybrid atomic orbitals d. sp 3 d hybrid atomic orbitals 74

Resonance and π bonds Localized electrons Positioned on one atom or shared between two atoms Delocalized electrons Shared by multiple atoms 2015 Pearson Education, Inc. 75

Resonance and π bonds Localized electrons Positioned on one atom or shared between two atoms Delocalized electrons Shared by multiple atoms 2015 Pearson Education, Inc. 76

Hybridization Localized electrons (hybridization) Orbital overlap represents a single bond Bond (electrons) localized between atoms No information about bond energies Does not deal with unpaired electrons 2015 Pearson Education, Inc. 77

Molecular Orbital (MO) Theory Molecular Orbital Theory 1. Made from overlap of atomic orbitals that match symmetry and energy # molecular orbitals = # atomic orbitals 2015 Pearson Education, Inc. 78

Molecular Orbital (MO) Theory Molecular Orbital Theory 2. Overlap of two atomic orbitals form two s molecular orbitals Bonding orbitals Constructive combination Electrons found in region between two nuclei 2015 Pearson Education, Inc. 79

Molecular Orbital (MO) Theory Molecular Orbital Theory 2. Overlap of two atomic orbitals form two molecular orbitals Bonding orbitals Antibonding orbitals Destructive combinations Electrons found outside the region between the nuclei Contains nodal plane 2015 Pearson Education, Inc. 80

Molecular Orbital (MO) Theory Molecular Orbital Theory 2. Overlap of two atomic orbitals form two molecular orbitals Bonding orbitals Antibonding orbitals 3. Orbitals extend over the entire molecule 2015 Pearson Education, Inc. 81

Molecular Orbital (MO) Theory Molecular Orbital Theory 4. Bonding orbitals are lower in energy than antibonding orbitals 2015 Pearson Education, Inc. 82

Molecular Orbital (MO) Theory 2015 Pearson Education, Inc. 83

Molecular Orbital (MO) Theory Molecular Orbital Theory 4. Bonding orbitals are lower in energy than antibonding orbitals 5. Electrons behave in molecular orbitals much as they do in atoms Two electrons per orbital Electrons in the same orbital have opposite spin 2015 Pearson Education, Inc. 84

Molecular Orbital (MO) Theory Molecular Orbital Theory 6. Electrons are delocalized over molecule 7. Only molecular orbitals are available for occupation by valence electrons 2015 Pearson Education, Inc. 85

Molecular Orbital (MO) Diagram Energy-level diagram Also called MO diagram Visual representation of formation of MO Bond order Indicates bond strength. # bonding e Bond order # 2 antibonding e Larger the bond order stronger the bond 2015 Pearson Education, Inc. 86

Can H 2 Form? Use MO Diagram and Bond Order to Decide! What is bond order for H 2? Bond order # bonding e # 2 antibonding e Bond Order = ½(2 0) = 1 bond Which means? H 2 does exist 2015 Pearson Education, Inc. 87

Can He 2 Form? Use MO Diagram and Bond Order to Decide! What is bond order for He 2? Bond order # bonding e # 2 antibonding e Bond Order = ½(2 2) = 0 bonds Which means? He 2 does not exist 2015 Pearson Education, Inc. 88

Example Bond Order Use molecular orbital theory to predict the bond order in H 2. Is the H 2 bond stronger or weaker 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. 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. 89

Suppose one electron in H 2 is excited from the σ 1s MO to the σ* 1s MO. Would you expect the H atoms to remain bonded to each other, or would the molecule fall apart? a. Remain bonded, as the bond order remains the same. b. Fall apart, as the bond order changes from 1 to 1.5. c. Remain bonded, as the bond order changes from 1 to 1.5. d. Fall apart, as the bond order changes from 1 to zero. 90

Would you expect Be 2+ to be a stable ion? a. No, because the bond order is 0. b. Yes, because the bond order is 0.5. c. Yes, because the bond order is 1. d. Yes, because the bond order is 1.5. 91

Molecular Orbitals: s and p For atoms with both s and p orbitals, there are two types of interactions: Overlap in s fashion (direct) s orbitals 1 set p orbitals Overlap in fashion (side-ways) 2 sets of p orbitals 2015 Pearson Education, Inc. 92

Molecular Orbitals: s and p Resulting MO diagram: σ and σ * orbitals from s and p atomic orbitals π and π * orbitals from p atomic orbitals Direct overlap is stronger Effect of raising and lowering energy is greater for σ and σ * 2015 Pearson Education, Inc. 93

Molecular Orbitals: s and p s orbitals can interact with σ p orbitals more than the π p orbitals Raises energy of the σ p orbital Lowers energy of the σ s orbital 2015 Pearson Education, Inc. 94

MO Diagrams and Magnetism Diamagnetism All electrons are paired Substances repelled by a magnetic field Paramagnetism Unpaired electron is present Substances attracted to magnetic field 95

MO Diagrams and Magnetism Is oxygen (O 2 ) paramagnetic or diamagnetic? Look back at the MO diagram! http://www.wonderwhizkids.com/resources/content/imagesv4/chemistry/concept/metals_nonmetals/paramagnetism2.jpg 96

MO Diagrams: Molecules of 2 nd Period 97

Dioxygen (O 2 ) is because of its unpaired electrons. a. paramagnetic b. diamagnetic c. ferromagnetic d. gyromagnetic 98

The figure below indicates that C 2 is diamagnetic. Would that be expected if the σ 2p MO were lower in energy than the π 2p MOs? a. Yes, because the reversal of MOs does not change the electron occupation of each MO. b. No, because the reversal of MOs results in the two π 2p MOs containing one electron each. 99

Example Molecular Orbital Theory Draw an MO energy diagram and determine the bond order for the N 2 ion. Would 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. 100

Example 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. 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. 101

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