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) Theory and Molecular Shape 10.4 Molecular Shape and Molecular Polarity

Rules for Lewis Structures 1. Attach atoms in reasonable fashion with single bonds Consider ABn formulas, atom with lower group number goes in center with others attached to it. If same group number, place atom with highest period in center. The atom with the lowest E.N. goes in middle Consider # of free valence e-, this is often the number of bonds this atom will make. Atoms making more bonds will be central in molecules. 2. Sum valence electrons (adding for neg charge and sub for pos.) 3. Complete octets of peripheral atoms (not He like atoms) 4. Place leftover e- on central atom 5. If necessary use multiple bonds to fill center atom's octet Last Slide

For NF3 : : : : : : : : : : N 5e- F F N F 7e- X 3 = 21e- Total 26e- F 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

SAMPLE PROBLEM 10.1 Writing Lewis Structures for Molecules with One Central Atom PROBLEM: Write a Lewis structure for CCl2F2, one of the compounds responsible for the depletion of stratospheric ozone. PLAN: Follow the steps outlined in Figure 10.1 . 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-. : C Cl F : : Make bonds and fill in remaining valence electrons placing 8e- around each atom. :

H : H C O H : H SAMPLE PROBLEM 10.2 Writing Lewis Structure for Molecules with More than One Central Atom 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. 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

: C C H H : : : 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 a Step 5 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. C H : C H (b) N2 has 2(5) = 10 valence e-. Therefore a triple bond is required to make the octet around each N. N : . N : . N :

Resonance: Delocalized Electron-Pair Bonding O3 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. Resonance structures have the same relative atom placement but a difference in the locations of bonding and nonbonding electron pairs.

SAMPLE PROBLEM 10.4 Writing Resonance Structures PROBLEM: Write resonance structures for the nitrate ion, NO3-. PLAN: After Steps 1-4, go to 5 and then see if other structures can be drawn in which the electrons can be delocalized over more than two atoms. SOLUTION: Nitrate has 1(5) + 3(6) + 1 = 24 valence e- N does not have an octet; a pair of e- will move in to form a double bond.

Resonance (continued) Formal charge of atom = # valence e- - (# unshared electrons + 1/2 # shared electrons) Three criteria for choosing the more important resonance structure: 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.

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 OC # valence e- = 6 # nonbonding e- = 6 # bonding e- = 2 X 1/2 = 1 Formal charge = -1 For OA # valence e- = 6 # nonbonding e- = 4 # bonding e- = 4 X 1/2 = 2 Formal charge = 0 For OB # valence e- = 6 # nonbonding e- = 2 # bonding e- = 6 X 1/2 = 3 Formal charge = +1

Resonance (continued) EXAMPLE: NCO- has 3 possible resonance forms - A B C formal charges -2 +1 -1 -1 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.

Exceptions to the Octet Rule Electron deficient – have fewer than eight ex BeCl2, BF3 may attain an octet by coordinate covalent bond Odd number of electrons – aka free radicals ex NO2 May attain an octet by pairing with another free radical Expanded Octets – only on period 3 and higher Expanded octets form when an atom can decrease (or maintain at 0) it’s formal charge ex SF6, PCl5, H2SO4, SO2, SO3, SO4

SAMPLE PROBLEM 10.5 Writing Lewis Structures for Exceptions to the Octet Rule. PROBLEM: Write Lewis structures for (a) H3PO4 and (b) BFCl2. In (a) decide on the most likely structure. PLAN: Draw the Lewis structures for the molecule and determine if there is an element which can be an exception to the octet rule. Note that (a) contains P which is a Period-3 element and can have an expanded valence shell. SOLUTION: (a) H3PO4 has two resonance forms and formal charges indicate the more important form. -1 +1 (b) BFCl2 will have only 1 Lewis structure. more stable lower formal charges

Bond Energies and Hrxn In chemical reactions, reactant bonds are broken and new product bonds are formed. The overall energy change of a reaction is the energy change of this process, plus the energy associated with changes of state. In gaseous reactions (no state changes) Hºrxn = Hºbonds broken + Hºbonds formed

DH0rxn = DH0reactant bonds broken + DH0product bonds formed Figure 10.2 Using bond energies to calculate DH0rxn DH0rxn = DH0reactant bonds broken + DH0product bonds formed Enthalpy, H DH01 = + sum of BE DH02 = - sum of BE DH0rxn

Using bond energies to calculate DH0rxn of methane Figure 10.3 Using bond energies to calculate DH0rxn of methane BOND BREAKAGE 4BE(C-H)= +1652kJ 2BE(O2)= + 996kJ DH0(bond breaking) = +2648kJ BOND FORMATION 2[-BE(C O)]= -1598kJ Enthalpy,H 4[-BE(O-H)]= -1868kJ DH0(bond forming) = -3466kJ DH0rxn= -818kJ

SAMPLE PROBLEM 10.6 Calculating Enthalpy Changes from Bond Energies PROBLEM: Use Table 9.2 (button at right) 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. SOLUTION: bonds broken bonds formed

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

Molecular Shapes Lewis structures show which atoms are connected where, and by how many bonds, but they don't properly show 3-D shapes of molecules. To find the actual shape of a molecule, first draw the Lewis structure, and then use VSEPR Theory.

Valence Shell Electron-Pair Repulsion Theory or VSEPR molecular shape is determined by the repulsions of electron pairs Electron pairs around the central atom stay as far apart as possible. electron pair geometry - based on number of regions of electron density Consider non-bonding (lone pairs) as well as bonding electrons. Electron pairs in single, double and triple bonds are treated as single electron clouds. molecular geometry - based on the electron pair geometry, this is the shape of the molecule

A balloon analogy for the mutual repulsion of electron groups. Figure 10.4 A balloon analogy for the mutual repulsion of electron groups.

Electron-group repulsions and the five basic molecular shapes. Figure 10.5 Electron-group repulsions and the five basic molecular shapes.

The single molecular shape of the linear electron-group arrangement. Figure 10.6 The single molecular shape of the linear electron-group arrangement. Examples: CS2, HCN, BeF2

Figure 10.7 The two molecular shapes of the trigonal planar electron-group arrangement. Class Shape Examples: SO2, O3, PbCl2, SnBr2 Examples: SO3, BF3, NO3-, CO32-

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. 1200 1220 1160 real Effect of Double Bonds larger EN 1200 ideal greater electron density Effect of Nonbonding Pairs lone pairs repel bonding pairs more strongly than bonding pairs repel each other 950

Figure 10.8 The three molecular shapes of the tetrahedral electron-group arrangement. Examples: CH4, SiCl4, SO42-, ClO4- NH3 PF3 ClO3 H3O+ H2O OF2 SCl2

Lewis structures and molecular shapes Figure 10.9 Lewis structures and molecular shapes

Figure 10.10 The four molecular shapes of the trigonal bipyramidal electron-group arrangement. PF5 AsF5 SOF4 SF4 XeO2F2 IF4+ IO2F2- XeF2 I3- IF2- ClF3 BrF3

Figure 10.11 The three molecular shapes of the octahedral electron-group arrangement. SF6 IOF5 BrF5 TeF5- XeOF4 XeF4 ICl4-

The steps in determining a molecular shape. Figure 10.12 The steps in determining a molecular shape. See Figure 10.1 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)

SAMPLE PROBLEM 10.7 Predicting Molecular Shapes with Two, Three, or Four Electron Groups PROBLEM: Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF3 and (b) COCl2. SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding pair The shape is based upon the tetrahedral arrangement. The F-P-F bond angles should be <109.50 due to the repulsion of the nonbonding electron pair. The final shape is trigonal pyramidal. <109.50 The type of shape is AX3E

SAMPLE PROBLEM 10.7 Predicting Molecular Shapes with Two, Three, or Four Electron Groups continued (b) For COCl2, C has the lowest EN and will be the center atom. There are 24 valence e-, 3 atoms attached to the center atom. C does not have an octet; a pair of nonbonding electrons will move in from the O to make a double bond. Type AX3 The shape for an atom with three atom attachments and no nonbonding pairs on the central atom is trigonal planar. 124.50 1110 The Cl-C-Cl bond angle will be less than 1200 due to the electron density of the C=O.

SAMPLE PROBLEM 10.8 Predicting Molecular Shapes with Five or Six Electron Groups PROBLEM: Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b) BrF5. SOLUTION: (a) SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs; shape is AX5 - trigonal bipyramidal. (b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX5E, square pyramidal.

The tetrahedral centers of ethane. Figure 10.13 The tetrahedral centers of ethane.

The tetrahedral centers of ethanol. Figure 10.13 The tetrahedral centers of ethanol.

SAMPLE PROBLEM 10.9 Predicting Molecular Shapes with More Than One Central Atom 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. SOLUTION: tetrahedral tetrahedral trigonal planar >1200 <1200

The orientation of polar molecules in an electric field. Figure 10.14 The orientation of polar molecules in an electric field. Electric field ON Electric field OFF

SAMPLE PROBLEM 10.10 Predicting the Polarity of Molecules PROBLEM: From electronegativity (EN) values (button) 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: (a) Ammonia, NH3 (b) Boron trifluoride, BF3 (c) Carbonyl sulfide, COS (atom sequence SCO) PLAN: Draw the shape, find the EN values and combine the concepts to determine the polarity. SOLUTION: (a) NH3 The dipoles reinforce each other, so the overall molecule is definitely polar. ENN = 3.0 ENH = 2.1 molecular dipole bond dipoles

SAMPLE PROBLEM 10.10 Predicting the Polarity of Molecules continued (b) BF3 has 24 valence e- and all electrons around the B will be involved in bonds. The shape is AX3, trigonal planar. F (EN 4.0) is more electronegative than B (EN 2.0) and all of the dipoles will be directed from B to F. Because all are at the same angle and of the same magnitude, the molecule is nonpolar. 1200 (c) COS is linear. C and S have the same EN (2.0) but the C=O bond is quite polar(DEN) so the molecule is polar overall.