CHAPTER 5: Bonding Theories - Explaining Molecular Geometry

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Chapter Outline

. 5.1 Molecular Shape . 5.2 Valence-Shell Electron-Pair Repulsion Theory (VSEPR) . 5.3 Polar Bonds and Polar Molecules » What Makes a Molecule Polar? » Dipole Moments . 5.4 Valence Bond Theory . 5.5 Shape and Interactions with Large Molecules . 5.6 Chirality and Molecular Recognition . 5.7 Molecular Orbital Theory

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Bar Code Reader App (IOS) QR Code Reader and Scanner 4+ ShopSavvy, Inc.

https://tinyurl.com/clj7f7u

VSEPR Molecular Shapes Website https://phet.Colorado.edu https://tinyurl.com/pcwl97v

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Predict the geometry of the molecule from the negative repulsions between the electron pairs (both bonding and nonbonding lone pairs) .

Linear Trigonal Tetrahedral Trigonal Octahedral 180o planar 109.5o Bipyramidal 90o 120o 120 and 90o

Drawing VSEPR Figures in 3D

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Number of atoms Number of lone Steric Number (SN) = bonded to the + pairs bonded to central atom central atom

Used to select the correct VSEPR geometry

https://tinyurl.com/pcwl97v

Central Atoms with No Lone Pairs Electron Group Geometry = Molecular Geometry

SN 2 = Linear Bond Angle = 180o

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SN 3 = Trigonal Planar Bond Angle = 120o

SN 4 = Tetrahedral Bond Angle = 109.5o

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SN 5 = Trigonal Bipyramidal Bond Angles = 120o and 90o

Axial position

Top View Equatorial Plane

Side View

SN 6 = Octahedral Bond Angles = 90o

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Applying VSEPR Theory

1. Determine the Lewis structure 2. Determine the Steric Number around the CENTRAL ATOM 3. Multiple bonds count as one electron group 4. Find out the appropriate VSEPR geometry for the specified number of electron groups 5. Use the positions of atoms to establish the resulting molecular geometry.

Central Atoms with No Lone Pairs

Example I: Carbon Dioxide, CO2

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Central Atoms with No Lone Pairs

Example II: Boron Trifluoride, BF3

Central Atoms with No Lone Pairs

Example III: Carbon Tetrachloride, CCl4

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Central Atoms with No Lone Pairs

Example IV: Phosphorus Pentafluoride, PF5

Central Atoms with No Lone Pairs

Example V: Sulfur hexafluoride, SF6

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Sample Exercise 9.1 –a double bond counts as one pair

Formaldehyde, CH2O, is a gas at room temperature and is an ingredient in solutions used to preserve biological samples. Use VSEPR to predict the molecular geometry of formaldehyde.

Central Atoms with Lone Pairs: SN = 2 Linear Electron Group Geometry

Electron Group Geometry Molecular Geometry

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Central Atoms with Lone Pairs: SN = 2 Linear Electron Group Geometry e.g. HCN

Central Atoms with Lone Pairs: SN = 3 Trigonal Planar Electron Group Geometry

Electron Group Geometry Molecular Geometry

Trigonal Planar

Ben

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Central Atoms with Lone Pairs: SN = 3 Trigonal Planar Electron Group Geometry

e.g. O3

Electron Pair Repulsions

. Electron Pair Repulsion Order: • Lone pair—Lone pair = greatest repulsion. • Lone pair—Bonding pair is next. • Bonding pair—Bonding pair = least repulsion. » Double bonds exert more repulsion than single bonds. . Bond angles around central atom decrease as repulsive forces increase.

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Sample Exercise 9.2

Rank NH3, CH4, and H2O in order of decreasing bond angles in their molecular structures.

Central Atoms with Lone Pairs: SN = 4 Tetrahedral Electron Group Geometry

Electron Group Geometry Molecular Geometry

Tetrahedral Pyramidal

Bent

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Central Atoms with Lone Pairs: SN = 4 Tetrahedral Electron Group Geometry

e.g. NH3 and H2O

Central Atoms with Lone Pairs: SN = 5 Trigonal Bipyramidal Electron Group Geometry

Electron Group Geometry Molecular Geometry

Trigonal Bipyramidal See-saw

T-shaped Linear

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Central Atoms with Lone Pairs: SN = 5 Trigonal Bipyramidal Electron Group Geometry

Lone pairs are always equatorial

Central Atoms with Lone Pairs: SN = 5 Trigonal Bipyramidal Electron Group Geometry

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Central Atoms with Lone Pairs: SN = 5 Trigonal Bipyramidal Electron Group Geometry

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Central Atoms with Lone Pairs: SN = 5 Trigonal Bipyramidal Electron Group Geometry

- e.g. SF4, ClF3, and I3

Central Atoms with Lone Pairs: SN = 6 Octahedral Electron Group Geometry

Electron Group Geometry Molecular Geometry

Square Octahedral pyramidal

Square planar

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Central Atoms with Lone Pairs: SN = 6 Octahedral Electron Group Geometry

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Central Atoms with Lone Pairs: SN = 6 Octahedral Electron Group Geometry

e.g. SF6, BrF5, and XeF4

Summary of all Geometries (with and without lone pairs)

Electron Cloud Bond Angles Molecular Geometry Geometry Linear 180o Linear Trigonal planar 120o Trigonal Planar Bent Tetrahedral 109.5o Tetrahedral Trigonal Pyramidal Bent Trigonal Bipyramidal equatorial = Trigonal Bipyramidal 120o, axial = 90o “seesaw” NOTE: lone pairs are to the equatorial T-shaped equatorial plane Linear Octahedral 90o Octahedral Square Pyramidal Square Planar

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Chapter Outline

Polar Bonds and Polar Molecules

. Requirements for Polar Molecule: • 1. Molecule must contain polar bonds (i.e., covalent bond between atoms with ΔEN). • 2. Orientation of polar bonds results in charge separation from one part of the molecule to another.

Polar bonds… but linear shape results in partial charges canceling out; nonpolar!

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Polar Bonds and Polar Molecules (cont.)

Bond Dipole: •Separation of charge within a covalent bond.

Polar Molecule: •Vectors of bond dipoles sum > zero. Polar!

Measuring Polarity

. Dipole moment (μ): • Measured value defining extent of separation of + and − charge centers in a molecule. (Units = debyes (D); 1 D = 3.34 × 10−30 coul∙m )

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Examples

Determine whether or not the following molecules are polar -

(a) CH2O

(b) CHCl3

Permanent Dipole Moments

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Chapter Outline

. 5.1 Molecular Shape . 5.2 Valence-Shell Electron-Pair Repulsion Theory (VSEPR) . 5.3 Polar Bonds and Polar Molecules . 5.4 Valence Bond Theory » Orbital Overlap and Hybridization. » Hybridization and Molecular Geometries . 5.5 Shape and Interactions with Large Molecules . 5.6 Chirality and Molecular Recognition . 5.7 Molecular Orbital Theory

The Orbital Overlap Model of Bonding

H-H H-F

End to end overlap = sigma () bond

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Predicted Bonding and VSEPR Geometry for CH4

109.5 o

Lewis Electron Group Geometry Structure around the central C atom is tetrahedral.

Problem: the available s and p-orbitals are at 90o angles, not at the predicted 109.5o!

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Valence Bond Theory (hybrid orbitals)

• A quantum mechanics-based theory of bonding that assumes covalent bonds form when half-filled orbitals on different atoms overlap or occupy the same region of space.

• In order for the bonding to match the VSEPR geometry, atomic orbitals must be combined into new “hybrid” orbitals that do result in the correct geometry.

Hybridization Rules – will be upgraded as we go

New orbitals are constructed from pre-existing s, p, and d-orbitals = hybrid orbitals

1. Hybridize the CENTRAL ATOM ONLY (others as needed) 2. Only use valence shell electrons 3. The number of hybrid orbitals formed = number of atomic orbitals used

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sp3 Hybridization = Tetrahedral EG

For CH4, we need 4 hybrid orbitals, so 4 atomic orbitals are required as follows: (s + p + p + p) = sp3

Needed to form 4 sigma bonds

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Hybridization Rules – cont’d

1. Hybridize the CENTRAL ATOM ONLY (others as needed) 2. Only use valence shell electrons 3. The number of hybrid orbitals formed = number of atomic orbitals used 4. Hybrid orbitals get 1 electron for a -bond, 2 electrons for a lone pair.

Other Examples – NH3 and H2O

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Sigma () bonds = end-to-end overlap

-Bonds = side by side overlap

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C - C 1  bond

C = C 1  bond 1  bond

C C 1  bond 2  bonds

H - C C - H

(upgraded – more will be added)

1. Hybrid orbitals get 1 electron for a -bond, 2 electrons for a lone pair. 2. Remaining electrons go into unhybridized orbitals =  bonds

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sp2 Hybridization = Trigonal Planar

H2CO

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sp Hybridization = Linear

C2H2

sp hybridization on each C atom -

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Sigma () Bonding in Acetylene

Unhybridized p-orbitals

Pi () Bonding in Acetylene

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Sample Exercise 5.5

Use Valence Bond Theory to explain the linear molecular geometry of CO2. Determine the hybridization of the carbon and oxygen in this molecule, and describe the orbitals that overlap to form the bonds.

solution:

Sigma Bonding in CO2

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Pi Bonding in CO2

sp3d = Trigonal Bipyramidal Hybidization

e.g. PBr5

Needed to form 5 sigma bonds

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sp3d2 = Octahedral Hybidization

e.g. SF6

Needed to form 6 sigma bonds

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Summary of Hybridization

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Chapter Outline

. 5.1 Molecular Shape . 5.2 Valence-Shell Electron-Pair Repulsion Theory (VSEPR) . 5.3 Polar Bonds and Polar Molecules . 5.4 Valence Bond Theory . 5.5 Molecules With Multiple Central Atoms . 5.6 Chirality and Molecular Recognition . 5.7 Molecular Orbital Theory

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Molecules With Multiple “Central Atoms”

Ethylene, C2H4

Molecules With Multiple “Central Atoms”

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Molecules With Multiple “Central Atoms”

Acrolein, CH2CHCHO

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Molecules With Multiple “Central Atoms”

Benzene, C6H6

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Chapter Outline

Problems With Valence Bond Theory

VSEPR Theory - electron pair repulsions influence molecular shape

Valence Bond Theory - atoms form bonds by overlapping hybrid orbitals

Applied to O2 O = O

There’s a problem with this Lewis structure for oxygen, however, and it has to do with magnetism in atoms -

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Cause of Magnetism = ELECTRON SPIN

Diamagnetic materials = all electron N S spins are “paired”, so the magnetic fields cancel out =   S N Zn = [Ar]3d104s2

Paramagnetic materials = unpaired N electrons so there is a net magnetism =  Mn = [Ar]3d54s2 S

In the Lewis structure for oxygen, all of the electrons are paired up, so the molecule should be diamagnetic, but experiments prove that it is PARAMAGNETIC.

An additional refinement in bonding theory is necessary =

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Molecular Orbitals - Preliminary Ideas

Don’t forget that electrons behave like WAVES, and there are WAVE FUNCTIONS () that describe the electron position in space = ATOMIC ORBITALS (2)

Waves (electrons) can interfere with each other, either CONSTRUCTIVELY or DESTRUCTIVELY

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Types of Molecular Orbitals Bonding and Antibonding

Molecular Orbital Diagrams - 1s and *1s

AO’s MO’s AO’s

*1s E

1s 1s atom 1 atom 2

1s

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Molecular Orbital Diagrams - 2s and *2s

AO’s MO’s AO’s

*2s E

2s 2s atom 1 atom 2

2s

-bond formation involving p-orbitals

*  2p

2p

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-bond formation involving p-orbitals

*  2p

2p

*  2p

2p

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Applying Molecular Orbital Theory

1. The total number of molecular orbitals = total number of atomic orbitals contributed by the bonding atoms 2. Bonding MO’s are lower in energy (more stable) than anti-bonding MO’s 3. Electrons occupy molecular orbitals following the Pauli Exclusion Principle (spins pair up) and Hund’s Rule (remain unpaired as long as an empty orbital is available of the same energy)

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Molecular Orbital Diagram: H2

The two 1s orbitals may be added or subtracted to yield two sigma MOs (1 bonding/1 antibonding).

Bond Order and Stability

Bond Order = 1/2 (number or bonding e− − number or antibonding e−).

Bond order in H2 = 1 Bond order in He2 = 0 (Stable). (Not Stable).

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MO Diagrams for Z  7 and Z  8 (the filled 1s MO’s are not shown)

Z  7 Z  8

MO Scheme for N2

. Electron configuration

for N2 =

. Bond order =

• N2 has three bonds.

• N2 has no unpaired electrons.

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MO Scheme for O2

. MO Electron configuration for

O2 =

. Bond order=

• O2 has two bonds

• O2 has two unpaired * electrons in π2p

Other Diatomic Molecules MO Theory correctly predicts both the magnetic properties and bond orders of diatomic molecules.

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Sample Exercise 5.9 – Using MO theory for Charged Homonuclear Diatomics

In which of the molecules in the proceeding diagram would there be an increase in bond order if they lost one electron each, forming + singly charged diatomic cations (X2 )?

Using MO Theory to Explain Fractional Bond Orders and Resonance

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Using MO Theory to Explain Fractional Bond Orders and Resonance

MO Theory: Summary

. Advantages: • Provides the most complete picture of covalent bonding, including bond types and bond orders. • Accounts for magnetic properties. . Disadvantages: • The most difficult to apply to large molecules; does not account for molecular shape.