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Chapter 10

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

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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. Figure 10.1 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

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

  5. 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 . : : : 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.

  6. 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

  7. 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 a Step 5which 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.

  8. 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.

  9. 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.

  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)

  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 B C Resonance (continued) EXAMPLE: NCO- has 3 possible resonance forms - formal charges -2 0 +1 -1 0 0 0 0 -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.

  13. PROBLEM: Write Lewis structures for (a) H3PO4 and (b) BFCl2. In (a) decide on the most likely structure. SAMPLE PROBLEM 10.5 Writing Lewis Structures for Exceptions to the Octet Rule. 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 0 +1 0 (b) BFCl2 will have only 1 Lewis structure. 0 0 0 0 0 0 0 0 0 0 more stable 0 0 lower formal charges

  14. 10.2 Using Lewis Structures and Bond Energies to Calculate Heats of Reaction

  15. DH0 = 436.4 kJ H2 (g) H (g) + H (g) DH0 = 242.7 kJ Cl2 (g) Cl (g) + Cl (g) DH0 = 431.9 kJ HCl (g) H (g) + Cl (g) DH0 = 498.7 kJ O2 (g) O (g) + O (g) O DH0 = 941.4 kJ N2 (g) N (g) + N (g) O Bond Energies Single bond < Double bond < Triple bond N N The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy. Bond Energy

  16. DH0 = 502 kJ H2O (g) H (g) + OH (g) DH0 = 427 kJ OH (g) H (g) + O (g) = 464.5 kJ Average OH bond energy = 502 + 427 2 Average bond energy in polyatomic molecules

  17. Bond Energies (BE) and Enthalpy changes in reactions Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. DH0 = total energy input – total energy released = SBE(reactants) – SBE(products)

  18. Use bond energies to calculate the enthalpy change for: H2 (g) + F2 (g) 2HF (g) Type of bonds broken Number of bonds broken Bond energy (kJ/mol) Energy change (kJ) 1 436.4 436.4 1 156.9 156.9 Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) H H F H F F 2 568.2 1136.4 DH0 = SBE(reactants) – SBE(products) DH0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ

  19. 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

  20. 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. SAMPLE PROBLEM 10.6 Calculating Enthalpy Changes from Bond Energies SOLUTION: bonds broken bonds formed

  21. 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

  22. Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f DH0 (O2) = 0 DH0 (O3) = 142 kJ/mol DH0 (C, graphite) = 0 DH0 (C, diamond) = 1.90 kJ/mol f f f f Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest? The standard enthalpy of formation of any element in its most stable form is zero.

  23. The standard enthalpy of reaction (DH0 ) as the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD - [ + ] [ + ] = - S S = DH0 DH0 rxn rxn mDH0 (reactants) dDH0 (D) nDH0 (products) cDH0 (C) aDH0 (A) bDH0 (B) f f f f f f Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. (Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

  24. C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ rxn S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ rxn CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ rxn 2S(rhombic) + 2O2(g) 2SO2(g)DH0 = -296.1x2 kJ C(graphite) + 2S(rhombic) CS2 (l) C(graphite) + 2S(rhombic) CS2 (l) rxn rxn C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ + CO2(g) + 2SO2(g) CS2(l) + 3O2(g)DH0 = +1072 kJ rxn DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn Calculate the standard enthalpy of formation of CS2 (l) given that: 1. Write the enthalpy of formation reaction for CS2 2. Add the given rxns so that the result is the desired rxn.

  25. 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) - S S = DH0 DH0 DH0 - [ ] [ + ] = rxn rxn rxn [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ = 12DH0 (CO2) 2DH0 (C6H6) f f = 2973 kJ/mol C6H6 6DH0 (H2O) -5946 kJ f 2 mol mDH0 (reactants) nDH0 (products) f f Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. What is the heat released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

  26. The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents Which could be used for melting ice? Which could be used for a cold pack?

  27. Valence shell electron pair repulsion(VSEPR) model:

  28. # of atoms bonded tocentral atom # lone pairs on central atom Arrangement ofelectron pairs Molecular Geometry Class linear linear B B Valence shell electron pair repulsion (VSEPR) model: Predict the geometry of the molecule from the electrostatic repulsion between the electron (bonding and nonbonding) pairs. AB2 2 0 Examples: CS2, HCN, BeF2

  29. 0 lone pairs on central atom Cl Be Cl 2 atoms bonded to central atom

  30. # of atoms bonded tocentral atom # lone pairs on central atom trigonal planar trigonal planar Arrangement ofelectron pairs Molecular Geometry Class VSEPR AB2 2 0 linear linear AB3 3 0 Examples: SO3, BF3, NO3-, CO32-

  31. 10.1

  32. # of atoms bonded tocentral atom # lone pairs on central atom trigonal planar trigonal planar AB3 3 0 Arrangement ofelectron pairs Molecular Geometry Class tetrahedral tetrahedral VSEPR AB2 2 0 linear linear AB4 4 0 Examples: CH4, SiCl4, SO42-, ClO4-

  33. # of atoms bonded tocentral atom # lone pairs on central atom trigonal planar trigonal planar AB3 3 0 Arrangement ofelectron pairs Molecular Geometry Class trigonal bipyramidal trigonal bipyramidal VSEPR AB2 2 0 linear linear tetrahedral tetrahedral AB4 4 0 AB5 5 0 Examples: PF5 AsF5 SOF4

  34. # of atoms bonded tocentral atom # lone pairs on central atom trigonal planar trigonal planar AB3 3 0 Arrangement ofelectron pairs Molecular Geometry Class trigonal bipyramidal trigonal bipyramidal AB5 5 0 octahedral octahedral VSEPR AB2 2 0 linear linear tetrahedral tetrahedral AB4 4 0 AB6 6 0 SF6 IOF5

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

  36. 1220 Effect of Double Bonds 1160 real 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. 1200 larger EN 1200 ideal greater electron density lone pairs repel bonding pairs more strongly than bonding pairs repel each other 950

  37. lone-pair vs. lone pair repulsion lone-pair vs. bonding pair repulsion bonding-pair vs. bonding pair repulsion > >

  38. # of atoms bonded tocentral atom # lone pairs on central atom Arrangement ofelectron pairs Molecular Geometry Class trigonal planar bent VSEPR trigonal planar trigonal planar AB3 3 0 AB2E 2 1

  39. # of atoms bonded tocentral atom # lone pairs on central atom Arrangement ofelectron pairs Molecular Geometry Class trigonal pyramidal tetrahedral VSEPR tetrahedral tetrahedral AB4 4 0 AB3E 3 1 NH3 PF3 ClO3 H3O+

  40. # of atoms bonded tocentral atom # lone pairs on central atom trigonal pyramidal Arrangement ofelectron pairs Molecular Geometry AB3E 3 1 tetrahedral Class bent tetrahedral O H H VSEPR tetrahedral tetrahedral AB4 4 0 AB2E2 2 2 H2O OF2 SCl2

  41. # of atoms bonded tocentral atom # lone pairs on central atom trigonal bipyramidal distorted tetrahedron Arrangement ofelectron pairs Molecular Geometry Class VSEPR trigonal bipyramidal trigonal bipyramidal AB5 5 0 AB4E 4 1 SF4 XeO2F2 IF4+ IO2F2-

  42. # of atoms bonded tocentral atom # lone pairs on central atom trigonal bipyramidal distorted tetrahedron Arrangement ofelectron pairs Molecular Geometry AB4E 4 1 Class trigonal bipyramidal T-shaped F F Cl F VSEPR trigonal bipyramidal trigonal bipyramidal AB5 5 0 AB3E2 3 2 ClF3 BrF3

  43. # of atoms bonded tocentral atom # lone pairs on central atom trigonal bipyramidal distorted tetrahedron Arrangement ofelectron pairs Molecular Geometry AB4E 4 1 Class trigonal bipyramidal T-shaped AB3E2 3 2 trigonal bipyramidal linear I I I VSEPR trigonal bipyramidal trigonal bipyramidal AB5 5 0 AB2E3 2 3 H2O OF2 SCl2

  44. octahedral octahedral AB6 6 0 # of atoms bonded tocentral atom # lone pairs on central atom square pyramidal octahedral Arrangement ofelectron pairs Molecular Geometry Class F F F Br F F VSEPR AB5E 5 1 BrF5 TeF5- XeOF4

  45. octahedral octahedral AB6 6 0 # of atoms bonded tocentral atom # lone pairs on central atom square pyramidal octahedral AB5E 5 1 Arrangement ofelectron pairs Molecular Geometry Class square planar octahedral F F Xe F F VSEPR AB4E2 4 2 XeF4 ICl4-

  46. What is the molecular geometry of SO2 and SF4? F S F F O O S F Predicting Molecular Geometry • Draw Lewis structure for molecule. • Count number of lone pairs on the central atom and number of atoms bonded to the central atom. • Use VSEPR to predict the geometry of the molecule. AB4E AB2E distorted tetrahedron bent

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