10-1 Chapter 10 The Shapes of Molecules

10-2 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

10-3 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.

10-4 The steps in converting a molecular formula into a Lewis structure. Molecular formula Atom placement Sum of valence e - Remaining valence e - Lewis structure Place atom with lowest EN in center Add A-group numbers Draw single bonds. Subtract 2e - for each bond. Give each atom 8e - (2e - for H) Step 1 Step 2 Step 3 Step 4

10-5 Molecular formula Atom placement Sum of valence e - Remaining valence e - Lewis structure For NF 3 FF F N 5e - F 7e - X 3 = 21e - Total 26e - : :: :::: : : : N

10-6 SAMPLE PROBLEM 10.1Writing Lewis Structures for Molecules with One Central Atom SOLUTION: PROBLEM:Write a Lewis structure for CCl 2 F 2, one of the compounds responsible for the depletion of stratospheric ozone. Step 1: Carbon has the lowest EN and is the central atom. The other atoms are placed around it. C Steps 2-4: C has 4 valence e -, Cl and F each have 7. The sum is 4 + 4(7) = 32 valence e -. Cl F F C F F Make bonds and fill in remaining valence electrons placing 8e - around each atom. :: : :: : : : : :: :

10-7 SAMPLE PROBLEM 10.2Writing Lewis Structure for Molecules with More than One Central Atom PROBLEM:Write the Lewis structure for methanol (molecular formula CH 4 O), an important industrial alcohol that is being used as a gasoline alternative in car engines. 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) = 14 valence e -. C has 4 bonds and O has 2. O has 2 pair of nonbonding e -. COH H H H : :

10-8 SAMPLE PROBLEM 10.3Writing Lewis Structures for Molecules with Multiple Bonds. PLAN: SOLUTION: PROBLEM: Write Lewis structures for the following: (a) Ethylene (C 2 H 4 ), the most important reactant in the manufacture of polymers (b) Nitrogen (N 2 ), the most abundant atmospheric gas 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. (a) There are 2(4) + 4(1) = 12 valence e -. H can have only one bond per atom. CC H HH H CC H HH H (b) N 2 has 2(5) = 10 valence e -. Therefore a triple bond is required to make the octet around each N. N : N :.... N : N :.. N : N :.

10-9 Resonance: Delocalized Electron-Pair Bonding 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. Ozone, O 3 can be drawn in 2 ways 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 O : O : O O O O ::....

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10-11 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) For O A # valence e - = 6 # nonbonding e - = 4 # bonding e - = 4 X 1/2 = 2 Formal charge = 0 For O B # valence e - = 6 # nonbonding e - = 2 # bonding e - = 6 X 1/2 = 3 Formal charge = +1 For O C # valence e - = 6 # nonbonding e - = 6 # bonding e - = 2 X 1/2 = 1 Formal charge = -1 OAOA : OBOB :.. OCOC

10-12 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. Three criteria for choosing the more important resonance structure:

10-13 EXAMPLE: NCO - has 3 possible resonance forms - Resonance (continued) A BC formal charges 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. N C O ::::::. N C O ::::::.

10-14 Lewis Structures - Exceptions to the Octet Rule

10-15 Using bond energies to calculate  H 0 rxn  H 0 rxn =  H 0 reactant bonds broken +  H 0 product bonds formed  H 0 1 = + sum of BE  H 0 2 = - sum of BE  H 0 rxn Enthalpy, H

10-16 Using bond energies to calculate  H 0 rxn for combustion of methane Enthalpy,H BOND BREAKAGE 4BE(C-H)= +1652kJ 2BE(O 2 )= + 996kJ  H 0 (bond breaking) = +2648kJ BOND FORMATION 4[-BE(O-H)]= -1868kJ  H 0 (bond forming) = -3466kJ  H 0 rxn = -818kJ 2[-BE(C O)]= -1598kJ

10-17 SAMPLE PROBLEM 10.6Calculating Enthalpy Changes from Bond Energies SOLUTION: PROBLEM: Use average bond energies to calculate  H 0 rxn for the following reaction: CH 4 ( g ) + 3Cl 2 ( g ) CHCl 3 ( g ) + 3HCl( g ) PLAN:Write the Lewis structures for all reactants and products and calculate the number of bonds broken and formed. bonds brokenbonds formed

10-18 SAMPLE PROBLEM 10.6Calculating Enthalpy Changes from Bond Energies continued bonds brokenbonds formed 4 C-H = 4 mol(413 kJ/mol) = 1652 kJ 3 Cl-Cl = 3 mol(243 kJ/mol) = 729 kJ  H 0 bonds broken = 2381 kJ 3 C-Cl = 3 mol(-339 kJ/mol) = kJ 1 C-H = 1 mol(-413 kJ/mol) = -413 kJ  H 0 bonds formed = kJ 3 H-Cl = 3 mol(-427 kJ/mol) = kJ  H 0 reaction =  H 0 bonds broken +  H 0 bonds formed = 2381 kJ + (-2711 kJ) = kJ

10-19 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.

10-20 Electron-group repulsions and the five basic molecular shapes.

10-21 Looking at the Five Shapes in Detail Examples: CS 2, HCN, BeF 2

10-22 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. Likewise lone pairs repel bonding pairs more strongly than bonding pairs repel each other also compressing the other angles. Effect of Double Bonds Effect of Nonbonding Pairs Multiple bonds count just as one group but are larger than a single bond. The result is some compression of the other bond angles.

10-23 The two molecular shapes of the trigonal planar electron- group arrangement. Class Shape Examples: SO 3, BF 3, NO 3 -, CO 3 2- Examples: SO 2, O 3, PbCl 2, SnBr 2

10-24 The three molecular shapes of the tetrahedral electron- group arrangement. Examples: CH 4, SiCl 4, SO 4 2-, ClO 4 - NH 3 PF 3 ClO 3 H 3 O + H 2 O OF 2 SCl 2

10-25 The four molecular shapes of the trigonal bipyramidal electron-group arrangement. SF 4 XeO 2 F 2 I F 4 + I O 2 F 2 - ClF 3 BrF 3 XeF 2 I 3 - I F 2 - PF 5 AsF 5 SOF 4

10-26 The three molecular shapes of the octahedral electron- group arrangement. SF 6 I OF 5 BrF 5 TeF 5 - XeOF 4 XeF 4 I Cl 4 -

10-27 The steps in determining a molecular shape. Molecular formula Lewis structure Electron-group arrangement Bond angles Molecular shape (AX m E n ) Count all e - groups around central atom (A) Note lone pairs and double bonds Count bonding and nonbonding e - groups separately. Step 1 Step 2 Step 3 Step 4

10-28 Lewis structures and molecular shapes

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10-31 Molecular Shapes with More Than One Central Atom The tetrahedral centers of ethane.

10-32 The tetrahedral centers of ethanol. Molecular Shapes with More Than One Central Atom

10-33 SAMPLE PROBLEM 10.9Predicting Molecular Shapes with More Than One Central Atom SOLUTION: PROBLEM:Determine the shape around each of the central atoms in acetone, (CH 3 ) 2 C=O. PLAN:Find the shape of one atom at a time after writing the Lewis structure. tetrahedral trigonal planar

10-34 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.

10-35 SAMPLE PROBLEM 10.10Predicting the Polarity of Molecules (a) Ammonia, NH 3 (b) Boron trifluoride, BF 3 (c) Carbonyl sulfide, COS (atom sequence SCO) 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.

10-36 SAMPLE PROBLEM 10.10Predicting the Polarity of Molecules

10-37 End of Chapter 10