11/7/2018 CHAPTER 5: Bonding Theories - Explaining Molecular Geometry 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 1 11/7/2018 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 2 11/7/2018 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 3 11/7/2018 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 4 11/7/2018 SN 3 = Trigonal Planar Bond Angle = 120o SN 4 = Tetrahedral Bond Angle = 109.5o 5 11/7/2018 SN 5 = Trigonal Bipyramidal Bond Angles = 120o and 90o Axial position Top View Equatorial Plane Side View SN 6 = Octahedral Bond Angles = 90o 6 11/7/2018 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 7 11/7/2018 Central Atoms with No Lone Pairs Example II: Boron Trifluoride, BF3 Central Atoms with No Lone Pairs Example III: Carbon Tetrachloride, CCl4 8 11/7/2018 Central Atoms with No Lone Pairs Example IV: Phosphorus Pentafluoride, PF5 Central Atoms with No Lone Pairs Example V: Sulfur hexafluoride, SF6 9 11/7/2018 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 10 11/7/2018 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 11 11/7/2018 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. 12 11/7/2018 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 13 11/7/2018 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 14 11/7/2018 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 15 11/7/2018 Central Atoms with Lone Pairs: SN = 5 Trigonal Bipyramidal Electron Group Geometry Central Atoms with Lone Pairs: SN = 5 Trigonal Bipyramidal Electron Group Geometry 16 11/7/2018 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 17 11/7/2018 Central Atoms with Lone Pairs: SN = 6 Octahedral Electron Group Geometry Central Atoms with Lone Pairs: SN = 6 Octahedral Electron Group Geometry 18 11/7/2018 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 19 11/7/2018 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 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! 20 11/7/2018 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 ) © 2012 by W. W. Norton & Company 21 11/7/2018 Examples Determine whether or not the following molecules are polar - (a) CH2O (b) CHCl3 (c) CCl3F Permanent Dipole Moments © 2012 by W. W. Norton & Company 22 11/7/2018 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 23 11/7/2018 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! 24 11/7/2018 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 25 11/7/2018 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 26 11/7/2018 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 27 11/7/2018 Sigma () bonds = end-to-end overlap -Bonds = side by side overlap 28 11/7/2018 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 29 11/7/2018 sp2 Hybridization = Trigonal Planar H2CO 30 11/7/2018 31 11/7/2018 sp Hybridization = Linear C2H2 sp hybridization on each C atom - 32 11/7/2018 33 11/7/2018 Sigma () Bonding in Acetylene Unhybridized p-orbitals Pi () Bonding in Acetylene 34 11/7/2018 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 35 11/7/2018 Pi Bonding in CO2 sp3d = Trigonal Bipyramidal Hybidization e.g. PBr5 Needed to form 5 sigma bonds 36 11/7/2018 sp3d2 = Octahedral Hybidization e.g. SF6 Needed to form 6 sigma bonds 37 11/7/2018 Summary of Hybridization 38 11/7/2018 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 39 11/7/2018 Molecules With Multiple “Central Atoms” Ethylene, C2H4 Molecules With Multiple “Central Atoms” 40 11/7/2018 Molecules With Multiple “Central Atoms” Molecules With Multiple “Central Atoms” Acrolein, CH2CHCHO 41 11/7/2018 Molecules With Multiple “Central Atoms” Benzene, C6H6 42 11/7/2018 Chapter Outline .