# Chapter 10. (1) Lewis Theory of Bonding. (2) Lewis Symbols: . :O:.

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1 Chapter 10 (1) Lewis Theory of Bonding * Bonds between atoms form due to interactions between valence electrons (VE). (i) Ionic Bonds: Form due to a transfer of VE s. (ii) Covalent Bonds: Form due to sharing of VE s. (2) Lewis Symbols: * Show the VE s about an atom as dots. * Unpaired VE s predict ionic charges for many atoms and preferred # of covalent bonds for nonmetals.. :O:. 2 unpaired (a) 2 covalent bonds or electrons (b) predicts a 2 charge

2 (3) Lewis Structures (pp: in GCN) (a) Critical to determine Bond Order first: Bond Order = [6N + 2] [Total # VE s] (i) For every + 2 N = # non atoms (a) 1 multiple bond (b) 1 Carbon ring (5 or 6 C atoms) (c) An e deficient atom (B or Be) (ii) For every 2 Expanded octet on central atom by 2 e s (at least n = 3 atom) (b) Draw the skeleton (atomic arrangement): * Use single bonds except for specifics from Bond Order. * Remember the preferred # covalent bonds for each atom! (c) Complete Octet Rule: * Add lone pairs so that all atoms obey the Octet Rule (except )

3 Example Lewis Structures: (a) N 2 C 2 C 2 COO [6N + 2] [#VE] = [6(6) + 2] [36] = = 2 1 double bond (b) ICl 4 :O:.. : N C C C O.. (watch extra VE from the species charge!!) [6N + 2] [#VE] = [6(5) + 2] [36] = = :Cl: :Cl: : I : :Cl:.. :Cl:.. expanded octet on I atom by 4

4 (c) Resonance: * When Multiple Lewis Structures are valid. * Usually differ by the placement of multiple bonds. * Use Formal Charges to determine the best structure. FC = (#VE) (# lone pair e s) [(½)(# bonding e s] (lowest values are best; sum of FC values = overall species charge) Example: CO 2.. :O C O: :O = C = O: :O C O: best structure..

5 (4) Bond Polarity * The separation of charge between atoms in a bond. * Based on the electronegativity difference ( EN) * EN is an atom s attraction for bonding electrons (a) Nonpolar Covalent Bonds: 0 < EN < 0.4 (b) Polar Covalent Bonds: 0.4 < EN < 1.7 (c) Ionic Bonds: EN > 1.7 * Better to use % Ionic Character from a plot. * Polar bonds have a Dipole Moment with a partially positive (δ + ) side and a δ side (more EN atom)

6 (5) Molecular Shape (Geometry) * The VSEPR Model theorizes the arrangement of atoms about a central atom based on minimizing repulsion of electron groups. * Also predicts Bond Angles. (a) Draw the Lewis Structure (b) Count total # electron groups (any region of electrons) * Provides the Electron Group Geometry about the central atom (c) Find # Bonding e groups vs. nonbonding e groups * Provides the Molecular Geometry (the answer!!) * Use Table in the Textbook for Molecular Shapes and Bond Angles.

7 VSEPR Model examples 6 e groups about central Xe atom Octahedral e group Geometry 5 e groups about central S atom Trigonal Bipyramid e group Geometry 90 o Bond Angles 120 o Bond Angles 90 o Bond Angles 4 bonding e groups and 2 lone pairs Square Planar Molecular Geometry 4 bonding e groups and 1 lone pair See Saw Molecular Geometry

8 VSEPR Structures (Shapes) * These are Lewis Structures drawn with proper molecular geometries and bond angles. * Lone pairs do not need to be included. * Label the molecular shapes of each central atom. Example: Draw the VSEPR structure for C 3 CCN 2 Bond order = [6N + 2] [#VE] = = > 1 double bond Lewis Structure.. C C = C N VSEPR Structure Trigonal planar (120 o angles) C C = C N Tetrahedral Trigonal pyramid (109.5 o angles) (~ o angles)

9 (6) Molecular Polarity * The overall charge distribution about an entire molecule. * Polar Molecules have dipole moments (like bonds) * Molecular Polarity is determined by 2 factors: (i) Bond Polarity: * Polar molecules MUST AVE at least one polar bond. * If all bonds are nonpolar, then the molecule is nonpolar. (ii) Molecular Symmetry: * Use the VSEPR Model to determine molecular symmetry. * Nonpolar molecules have perfect symmetry. (all polar bonds cancel out if EN values are the same) * Polar molecules are NOT symmetric (at least 1 polar bond!)

10 (7) Bond Dissociation Energies (D) * The energy required to break a bond. * Bond energies (BE) that are averaged over several molecular environments are used and are tabulated in the text. * BE values are commonly used for rough estimates of rxn values. Example: What is rxn for the combustion of methane? C + 2 O = O -----> O = C = O + 2 O rxn = Σ BE (reactants) Σ BE (products) rxn = [(4 mol C ) + (2 mol O=O)] [(2 mol C=O) + (4 mol O )] rxn = 4 mol 414 kj + 2 mol 498 kj 2 mol 799 kj + 4 mol 464 kj mol mol mol mol rxn = (2652 kj) (3454 kj) = 802 kj

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