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Chapter 10 The Shapes of Molecules

Chapter 10 The Shapes of Molecules. The Shapes of Molecules. 10.1 Depicting Molecules and Ions with Lewis Structures. 10.2 Using Lewis Structures and Bond Energies to Calculate Heats of Reaction. 10.3 Valence-Shell Electron-Pair Repulsion (VSEPR)

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Chapter 10 The Shapes of Molecules

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  1. Chapter 10 The Shapes of Molecules

  2. The Shapes of Molecules 10.1 Depicting Molecules and Ions with Lewis Structures 10.2 Using Lewis Structures and Bond Energies to Calculate Heats of Reaction 10.3 Valence-Shell Electron-Pair Repulsion (VSEPR) Theory and Molecular Shape 10.4 Molecular Shape and Molecular Polarity

  3. Lewis Dot Structures A bookkeeping way to keep track of valence electrons and bonding in a molecule. Lines are used to represent a pair of bonding electrons. Dots are used to represent lone pair electrons. In many cases (not H) the octet rule is followed, having eight electrons (including bonding and lone pairs) around each atom. We’ll see later that the octet rule is a natural consequence of having four valence orbitals (s, 3p’s) and the limitation of two electron’s per orbital.

  4. The steps in converting a molecular formula into a Lewis structure. Place atom with lowest EN in center Molecular formula Step 1 Atom placement Add A-group numbers Step 2 Sum of valence e- Draw single bonds. Subtract 2e- for each bond. Step 3 Give each atom 8e- (2e- for H) Remaining valence e- Step 4 Lewis structure

  5. Molecular formula For NF3 Atom placement : : N 5e- : F : : F : : Sum of valence e- N F 7e- X 3 = 21e- Total 26e- : F : : Remaining valence e- Lewis structure

  6. PROBLEM: Write a Lewis structure for CCl2F2, one of the compounds responsible for the depletion of stratospheric ozone. : : : Cl : : : Cl : C F : : : : F : SAMPLE PROBLEM 10.1 Writing Lewis Structures for Molecules with One Central Atom SOLUTION: Cl Step 1: Carbon has the lowest EN and is the central atom. The other atoms are placed around it. Cl C F F Steps 2-4: C has 4 valence e-, Cl and F each have 7. The sum is 4 + 4(7) = 32 valence e-. Make bonds and fill in remaining valence electrons placing 8e- around each atom.

  7. PROBLEM: Write the Lewis structure for methanol (molecular formula CH4O), an important industrial alcohol that is being used as a gasoline alternative in car engines. SAMPLE PROBLEM 10.2 Writing Lewis Structure for Molecules with More than One Central Atom SOLUTION: Hydrogen can have only one bond so C and O must be next to each other with H filling in the bonds. There are 4(1) + 4 + 6 = 14 valence e-. C has 4 bonds and O has 2. O has 2 pair of nonbonding e-. H : H C O H : H

  8. H H H H C C C C H H H H N : N : . . N : N : N : N : . . . . SAMPLE PROBLEM 10.3 Writing Lewis Structures for Molecules with Multiple Bonds. PROBLEM: Write Lewis structures for the following: (a) Ethylene (C2H4), the most important reactant in the manufacture of polymers (b) Nitrogen (N2), the most abundant atmospheric gas PLAN: For molecules with multiple bonds, there is an additional step which follows the other steps in Lewis structure construction. If a central atom does not have 8e-, an octet, then e- can be moved in to form a multiple bond. SOLUTION: (a) There are 2(4) + 4(1) = 12 valence e-. H can have only one bond per atom. . . (b) N2 has 2(5) = 10 valence e-. Therefore a triple bond is required to make the octet around each N.

  9. Resonance: Delocalized Electron-Pair Bonding Ozone, O3 can be drawn in 2 ways . . . . . . . . O : O O : : O O O : . . . . . . . . Neither structure is actually correct but can be drawn to represent a structure which is a hybrid of the two - a resonance structure.Not a single bond - double bond, but a bond and a half for both bonds . . . . : O O O : . . . . Resonance structures have the same relative atom placement but a difference in the locations of bonding and nonbonding electron pairs.A double headed arrow is used to indicate resonance structures.

  10. For OC For OA # valence e- = 6 # valence e- = 6 # nonbonding e- = 6 # nonbonding e- = 4 # bonding e- = 2 X 1/2 = 1 # bonding e- = 4 X 1/2 = 2 For OB Formal charge = -1 Formal charge = 0 # valence e- = 6 # nonbonding e- = 2 # bonding e- = 6 X 1/2 = 3 Formal charge = +1 Formal Charge: Selecting the Best Resonance Structure An atom “owns” all of its nonbonding electrons and half of its bonding electrons. Formal charge of atom = # valence e- - (# unshared electrons + 1/2 # shared electrons) . . . . OA : OB OC : . . . .

  11. Three criteria for choosing the more important resonance structure: Resonance (continued) Smaller formal charges (either positive or negative) are preferable to larger charges; Avoid like charges (+ + or - - ) on adjacent atoms; A more negative formal charge should exist on an atom with a larger EN value.

  12. A Resonance (continued) EXAMPLE: NCO- has 3 possible resonance forms - -1 -1 -1 . . . . . . . . : N C O : : N C O : : N C O : . . . . B C formal charges -2 0 +1 -1 0 0 0 0 -1 -1 -1 -1 . . . . . . . . : N C O : : N C O : : N C O : . . . . Forms B and C have negative formal charges on N and O; this makes them more important than form A. Form C has a negative charge on O which is the more electronegative element, therefore C contributes the most to the resonance hybrid.

  13. Lewis Structures - Exceptions to the Octet Rule

  14. Using bond energies to calculate DH0rxn DH0rxn = DH0reactant bonds broken + DH0product bonds formed Enthalpy, H DH01 = + sum of BE DH02 = - sum of BE DH0rxn

  15. 2[-BE(C O)]= -1598kJ DH0rxn= -818kJ Using bond energies to calculate DH0rxn for combustion of methane BOND BREAKAGE 4BE(C-H)= +1652kJ 2BE(O2)= + 996kJ DH0(bond breaking) = +2648kJ BOND FORMATION 4[-BE(O-H)]= -1868kJ Enthalpy,H DH0(bond forming) = -3466kJ

  16. PROBLEM: Use average bond energies to calculate DH0rxn for the following reaction: CH4(g) + 3Cl2(g) CHCl3(g) + 3HCl(g) PLAN: Write the Lewis structures for all reactants and products and calculate the number of bonds broken and formed. SAMPLE PROBLEM 10.6 Calculating Enthalpy Changes from Bond Energies SOLUTION: bonds broken bonds formed

  17. SAMPLE PROBLEM 10.6 Calculating Enthalpy Changes from Bond Energies continued bonds broken bonds formed 4 C-H = 4 mol(413 kJ/mol) = 1652 kJ 3 C-Cl = 3 mol(-339 kJ/mol) = -1017 kJ 3 Cl-Cl = 3 mol(243 kJ/mol) = 729 kJ 1 C-H = 1 mol(-413 kJ/mol) = -413 kJ 3 H-Cl = 3 mol(-427 kJ/mol) = -1281 kJ DH0bonds broken = 2381 kJ DH0bonds formed = -2711 kJ DH0reaction = DH0bonds broken + DH0bonds formed = 2381 kJ + (-2711 kJ) = - 330 kJ

  18. Valence-Shell Electron-Pair Repulsion Theory (VSEPR) VSEPR is a very good theory for predicting the shape of molecules. It involves any group of valence electrons around an atom. These groups can be lone pairs, single bonds, or multiple bonds. In essence, these groups of negatively charge particles will be arranged as far apart as possible around the atom.

  19. Electron-group repulsions and the five basic molecular shapes.

  20. Looking at the Five Shapes in Detail Examples: CS2, HCN, BeF2

  21. Effect of Double Bonds Effect of Nonbonding Pairs Factors Affecting Actual Bond Angles Bond angles are consistent with theoretical angles when the atoms attached to the central atom are the same and when all electrons are bonding electrons of the same order. Multiple bonds count just as one group but are larger than a single bond. The result is some compression of the other bond angles. Likewise lone pairs repel bonding pairs more strongly than bonding pairs repel each other also compressing the other angles.

  22. Class Shape The two molecular shapes of the trigonal planar electron-group arrangement. Examples: SO2, O3, PbCl2, SnBr2 Examples: SO3, BF3, NO3-, CO32-

  23. The three molecular shapes of the tetrahedral electron-group arrangement. Examples: CH4, SiCl4, SO42-, ClO4- NH3 PF3 ClO3 H3O+ H2O OF2 SCl2

  24. The four molecular shapes of the trigonal bipyramidal electron-group arrangement. PF5 AsF5 SOF4 SF4 XeO2F2 IF4+ IO2F2- XeF2 I3- IF2- ClF3 BrF3

  25. The three molecular shapes of the octahedral electron-group arrangement. SF6 IOF5 BrF5 TeF5- XeOF4 XeF4 ICl4-

  26. The steps in determining a molecular shape. Molecular formula Step 1 Lewis structure Count all e- groups around central atom (A) Step 2 Electron-group arrangement Note lone pairs and double bonds Step 3 Count bonding and nonbonding e- groups separately. Bond angles Step 4 Molecular shape (AXmEn)

  27. Lewis structures and molecular shapes

  28. Molecular Shapes with More Than One Central Atom The tetrahedral centers of ethane.

  29. Molecular Shapes with More Than One Central Atom The tetrahedral centers of ethanol.

  30. PROBLEM: Determine the shape around each of the central atoms in acetone, (CH3)2C=O. PLAN: Find the shape of one atom at a time after writing the Lewis structure. tetrahedral tetrahedral trigonal planar SAMPLE PROBLEM 10.9 Predicting Molecular Shapes with More Than One Central Atom SOLUTION:

  31. Molecular Polarity Knowing the shape of the molecule, plus knowing the polarity (dipole) of the individual bonds allows the determination of the overall polarity of the molecule.

  32. PROBLEM: From electronegativity (EN) values and their periodic trends, predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable: PLAN: Draw the shape, find the EN values and combine the concepts to determine the polarity. SAMPLE PROBLEM 10.10 Predicting the Polarity of Molecules (a) Ammonia, NH3 (b) Boron trifluoride, BF3 (c) Carbonyl sulfide, COS (atom sequence SCO)

  33. SAMPLE PROBLEM 10.10 Predicting the Polarity of Molecules

  34. End of Chapter 10

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