Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring.

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Presentation transcript:

Philip Dutton University of Windsor, Canada N9B 3P4 Prentice-Hall © 2002 General Chemistry Principles and Modern Applications Petrucci Harwood Herring 8 th Edition Chapter 10: Chemical Bonding I: Basic Concepts

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 2 of 43 Contents 10-1Lewis Theory: An Overview 10-2Covalent Bonding: An Introduction 10-3Polar Covalent Bonds 10-4Writing Lewis Structures 10-7The Shapes of Molecules 10-8Bond Order and Bond Lengths 10-9Bond Energies Focus on Polymers— Macromolecular Substances

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 3 of Lewis Theory: An Overview Valence e - play a fundamental role in chemical bonding. e - transfer leads to ionic bonds. Sharing of e - leads to covalent bonds. e - are transferred of shared to give each atom a noble gas configuration –the octet.

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 4 of 43 Lewis Symbols A chemical symbol represents the nucleus and the core e -. Dots around the symbol represent valence e -. Si N P As Sb Bi Al Se Ar I

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 5 of 43 Lewis Structures for Ionic Compounds Ba O O Ba Mg Cl Cl Cl Mg BaO MgCl 2

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 6 of Covalent Bonding

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 7 of 43 Coordinate Covalent Bonds H N H H H N H H H H + Cl Cl -

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 8 of 43 Multiple Covalent Bonds C O O C O O C O O C O O

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 9 of 43 Multiple Covalent Bonds N N N N N N N N

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 10 of Polar Covalent Bonds H Cl δ+δ+δ-δ-

General Chemistry: Chapter 11Slide 11 of 43 Electronegativity Mulliken: EN = (I 1 + A 1 )/2 (known for 57 elements) Pauling (1932): empiric, Allred-Rochow (1958): Z* eff

Bond ionic character Pauling: the ionic character of a bond depends on the electronegativity difference of the two bonded atoms: x = EN 1  EN 2 General Chemistry: Chapter 10Slide 12 of 43

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 13 of 43 Percent Ionic Character x = EN 1  EN 2

General Chemistry: Chapter 11Slide 14 of 43 Electronegativity of the Elements

General Chemistry: Chapter 11Slide 15 of 43 Classification of Compounds

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 16 of Writing Lewis Structures All the valence e - of atoms must appear. Usually, the e - are paired. Usually, each atom requires an octet. –H only requires 2 e -. Multiple bonds may be needed. –Readily formed by C, N, O, S, and P.

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 17 of 43 Skeletal Structure Identify central and terminal atoms. C H H H H C H H O

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 18 of 43 Skeletal Structure Hydrogen atoms are always terminal atoms. Central atoms are generally those with the lowest electronegativity. Carbon atoms are always central atoms. Generally structures are compact and symmetrical.

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 19 of 43 Strategy for Writing Lewis Structures

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 20 of 43 Formal Charge FC = # valence e- - # lone pair e- - # bond pair e- 2 1

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 21 of 43 Example 10-6 Writing a lewis Structure for a Polyatomic Ion. Write the Lewis structure for the nitronium ion, NO 2 +. Step 1:Total valence e - = – 1 = 16 e - Step 2:Plausible structure:O—N—O Step 3:Add e - to terminal atoms:O—N—O Step 4:Determine e - left over:16 – 4 – 12 = 0

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 22 of 43 Example 10-6 Step 5:Use multiple bonds to satisfy octets. O—N—O O=N=O Step 6:Determine formal charges: FC(O) = – (4) = FC(N) = – (8) =

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 23 of The Shapes of Molecules H O H

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 24 of 43 Terminology Bond length – distance between nuclei. Bond angle – angle between adjacent bonds. VSEPR (Valence Shell Electron Pair Repulsion) Theory –Electron pairs repel each other whether they are in chemical bonds (bond pairs) or unshared (lone pairs). Electron pairs assume orientations about an atom to minimize repulsions. Electron group geometry – distribution of e - pairs. Molecular geometry – distribution of nuclei.

Torsion angle:  ijkl Torsion angle – angle between intramolecular planes: 4 atoms (i, j, k, l): Prentice-Hall © 2002General Chemistry: Chapter 11Slide 25 of 43

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 26 of 43 Balloon Analogy

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 27 of 43 Methane, Ammonia and Water

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 28 of 43 Table 11.1 Molecular Geometry as a Function of Electron Group Geometry

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 29 of 43 Applying VSEPR Theory Draw a plausible Lewis structure. Determine the number of e - groups and identify them as bond or lone pairs. Establish the e - group geometry. Determine the molecular geometry. Multiple bonds count as one group of electrons. More than one central atom can be handled individually.

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 30 of 43 Dipole Moments

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 31 of 43 Dipole Moments

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 32 of 43 Bond Order and Bond Length Bond Order –Single bond, order = 1 –Double bond, order = 2 Bond Length –Distance between two nuclei Higher bond order –Shorter bond –Stronger bond

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 33 of 43 Bond Length

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 34 of 43 Bond Energies

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 35 of 43 Bond Energies

Prentice-Hall © 2002General Chemistry: Chapter 11Slide 36 of 43 Bond Energies and Enthalpy of Reaction ΔH rxn = -  ΔH(product bonds) +  ΔH(reactant bonds) = -  ΔH bonds formed +  ΔH bonds broken = -770 kJ/mol + (657 kJ/mol) = -113 kJ/mol