Chapter 6. Chemical Reactions
Faculty Resource and Organizational Guide (FROG)
Table of Contents
Materials for Chapter 6 Activities ................................................................................................6
Reagents for Chapter 6 Activities: ...............................................................................................7
Section 6.1. Classifying Chemical Reactions ..............................................................................9
Learning objectives for Section 6.1: ............................................................................................9
Consider This 6.1. What chemical reactions do you know? .......................................................9
Investigate This 6.2. What changes do you observe? .................................................................9
Investigate This 6.3. How can you analyze the gas from a reaction? .......................................11
Consider This 6.4. What is the gas from the NaHCO3 + KHC4H5O6 reaction? .......................13
Consider This 6.5. How can you classify chemical reactions? .................................................13
Section 6.2. Ionic Precipitation Reactions.................................................................................14
Learning Objectives for Section 6.2: .........................................................................................14
Investigate This 6.6. How can you find the stoichiometry of a precipitate? ............................14
Consider This 6.7. What is the stoichiometry of the calcium oxalate precipitate? ...................16
Consider This 6.8. Does ionic precipitation fit the definition of a chemical reaction? ............17
Sample Worksheet for Section 6.2: ...........................................................................................17
Section 6.3. Lewis Acids and Bases: Definition .......................................................................19
Learning Objectives for Section 6.3: .........................................................................................19
Consider This 6.12. How do ionic precipitation and Lewis acid-base reactions differ? ..........19
Section 6.4. Lewis Acids and Bases: Brønsted-Lowry Acid-Base Reactions .........................19
Learning Objectives for Section 6.4: .........................................................................................19
Investigate This 6.13. Do the acidities of acids differ?.............................................................20
Consider This 6.14. How do the acidities of acids differ? ........................................................21
Consider This 6.17. What is the molecular level interpretation of expression (6.11)? .............21
Section 6.5. Predicting Strengths of Lewis/Lowry-Lowry Bases and Acids ..........................22
Learning Objectives for Section 6.5 ..........................................................................................22
Consider This 6.27. Is electronegativity a reliable predictor of Lewis acid-base strength? .....22
Consider This 6.28. Does electronegativity or size dominate Lewis acid-base strength? ........23
Investigate This 6.30. Do the acidities of alcohols and carboxylic acids differ? ......................24
Consider This 6.31. How do the acidities of alcohols and carboxylic acids differ? .................26
Consider This 6.34. How can you explain the relative acidities of the chlorine oxyacids? .....26
Consider This 6.37. How do oxyacids/oxyanions of different central atoms compare? ..........28
Section 6.6. Lewis Acids and Bases: Metal Ion Complexes .....................................................29
Learning Objectives for Section 6.6 ..........................................................................................29
Investigate This 6.41. Do Lewis bases react with metal ion? ...................................................29
Consider This 6.42. How do you know a Lewis acid-base complex ion has reacted? .............30
Investigate This 6.43. Do calcium ions react with complexing ligands?..................................32
Consider This 6.44. Do calcium ions react with complexing ligands? .....................................33
Investigate This 6.45. What is the three-dimensional structure of EDTA? ..............................34
February 2005
ACS Chemistry FROG
1
Chapter 6
Chemical Reactions
Consider This 6.48. What is the structure of the nickel-dimethylglyoxime complex?.............34
Section 6.7. Lewis Acids and Bases: Electrophiles and Nucleophiles....................................35
Learning Objectives for Section 6.7 ..........................................................................................35
Investigate This 6.50. Does an acid react with an alcohol? ......................................................35
Consider This 6.51. What is the product of reaction of an acid with an alcohol? ....................37
Section 6.8. Formal Charge ........................................................................................................38
Learning Objectives for Section 6.8 ..........................................................................................38
Consider This 6.59. How do you assign formal charge? ..........................................................38
Consider This 6.61. How are the correlations in Table 6.3 applied? ........................................40
Section 6.9. Reduction-Oxidation Reactions: Electron Transfer ..........................................41
Learning Objectives for Section 6.9 ..........................................................................................41
Investigate This 6.62. Do ion-metal reactions between copper and silver occur? ....................42
Consider This 6.63. What copper and silver ion-metal reactions occur? .................................43
Investigate This 6.66. Does Cu(s) react with nitric and/or hydrochloric acids? .......................44
Consider This 6.67. What are the products of Cu(s) reactions with nitric and/or hydrochloric
acid? ...........................................................................................................................................46
Consider This 6.68. What is the reaction of Cu(s) with H3O+(aq)? ..........................................47
Consider This 6.74. Are there patterns in oxidation numbers?.................................................47
Section 6.10. Balancing Reduction Oxidation Reaction Equations ........................................48
Learning Objectives for Section 6.10 ........................................................................................48
Investigate This 6.80. What happens when bleach and iodide are mixed?...............................48
Consider This 6.81. What is the reaction between bleach and iodide? .....................................49
Section 6.11. Reduction-Oxidation Reactions of Carbon-containing Molecules...................50
Learning Objectives for Section 6.11. .......................................................................................50
Investigate This 6.86. Does methanal react with silver ion? ....................................................51
Consider This 6.87. What is the reaction of methanal with silver ion? ....................................53
Consider This 6.88. What oxidation numbers are available for carbon? ..................................53
Section 6.13. Extension: Titration .............................................................................................54
Learning Objectives for Section 6.13. .......................................................................................54
Investigate This 6.93. How can you analyze an acid-base reaction? ........................................54
Consider This 6.94. How can you follow an acid-base reaction? .............................................56
Investigate This 6.98. What other titrations are possible? ........................................................57
Consider This 6.99. What is the stoichiometry of the iodine-ascorbic acid reaction?..............58
Solutions for Chapter 6 Check This Activities ...........................................................................60
Check This 6.10. Stoichiometric calculation for calcium oxalate formation ...........................60
Check This 6.11. Nickel–dimethylglyoxime reaction stoichiometry .......................................60
Check This 6.15. [H3O+(aq)] in 0.1 M ethanoic acid solution ..................................................61
Check This 6.16. Fraction of proton transfer between ethanoic acid and water .......................61
Check This 6.18. [OH–(aq)] in an aqueous sodium amide solution..........................................62
Check This 6.19. Writing Lewis acid–base reactions ...............................................................62
Check This 6.21. Relative basicities of chloride and hydroxide ions .......................................62
Check This 6.23. Relative basicities of ethanoate and hydroxide ions .....................................62
Check This 6.25. pH of an aminium (ammonia-like) chloride solution ...................................62
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ACS Chemistry FROG
Chemical Reactions
Chapter 6
Check This 6.26. Relative Lewis base strengths.......................................................................63
Check This 6.29. Relative Lewis base strengths.......................................................................63
Check This 6.32. Relative basicities of ethanoate, ethoxide, and hydroxide anions ................63
Check This 6.33. Relative basicities of oxyanions and acidities of oxyacids...........................64
Check This 6.35. Acid–base properties of bleach solutions .....................................................64
Check This 6.36. Another way to predict relative strengths of oxyacids and oxyanions .........64
Check This 6.38. Relative acid and base strengths ...................................................................65
Check This 6.40. Predicting the direction of reaction for acid–base reactions.........................65
Check This 6.46. Keeping the shine in your hair ......................................................................65
Check This 6.47. Names and structures of Pt(NH3)2Cl2 ...........................................................66
Check This 6.49. The porphine ring system in chlorophyll and heme .....................................66
Check This 6.52. Centers of positive and negative charge in molecules ..................................66
Check This 6.53. More nucleophile-electrophile reactions ......................................................67
Check This 6.54. A condensation reaction ...............................................................................67
Check This 6.55. Polyamide condensation polymers ...............................................................67
Check This 6.56. Water as a nucleophile ..................................................................................68
Check This 6.57. Polymeric structure of DNA .........................................................................68
Check This 6.60. Using formal charge .....................................................................................69
Check This 6.65. Balancing simple reduction–oxidation reaction expressions........................70
Check This 6.69. Oxidation numbers for elements and monatomic ions .................................71
Check This 6.70. The procedure and the definition of oxidation number ................................71
Check This 6.72. Assigning oxidation numbers .......................................................................71
Check This 6.73. Reduction–oxidation reactions .....................................................................72
Check This 6.75. Oxidation number for oxygen in peroxides ..................................................72
Check This 6.77. Assigning oxidation numbers .......................................................................73
Check This 6.79. Balancing redox equations by the oxidation-number method ......................73
Check This 6.83. Balancing redox equations by the half-reactions method .............................74
Check This 6.85. Balancing redox equations in basic solution ................................................75
Check This 6.89. Balancing the methanal-silver ion reaction equation ...................................76
Check This 6.90. Oxidation numbers for carbons in glucose fermentation..............................77
Check This 6.91. Net glycolysis reaction .................................................................................77
Check This 6.92. Reduction–oxidation of glucose ...................................................................78
Check This 6.95. Acid–base equivalence in Investigate This 6.93 ..........................................78
Check This 6.97. An acid–base titration ...................................................................................78
Check This 6.100. Ascorbic acid oxidation ..............................................................................79
Solutions for Chapter 6 End-of-Chapter Problems....................................................................80
Problem 6.1. ...............................................................................................................................80
Problem 6.2. ...............................................................................................................................80
Problem 6.3. ...............................................................................................................................80
Problem 6.4. ...............................................................................................................................81
Problem 6.5. ...............................................................................................................................81
Problem 6.6. ...............................................................................................................................82
Problem 6.7. ...............................................................................................................................82
Problem 6.8. ...............................................................................................................................83
Problem 6.9. ...............................................................................................................................84
Problem 6.10. .............................................................................................................................84
ACS Chemistry FROG
3
Chapter 6
Chemical Reactions
Problem 6.11. .............................................................................................................................85
Problem 6.12. .............................................................................................................................86
Problem 6.13. .............................................................................................................................86
Problem 6.14. .............................................................................................................................86
Problem 6.15. .............................................................................................................................87
Problem 6.16. .............................................................................................................................87
Problem 6.17. .............................................................................................................................88
Problem 6.18. .............................................................................................................................88
Problem 6.19. .............................................................................................................................89
Problem 6.20. .............................................................................................................................89
Problem 6.21. .............................................................................................................................90
Problem 6.22. .............................................................................................................................90
Problem 6.23. .............................................................................................................................91
Problem 6.24. .............................................................................................................................91
Problem 6.25. .............................................................................................................................91
Problem 6.26. .............................................................................................................................92
Problem 6.27. .............................................................................................................................93
Problem 6.28. .............................................................................................................................93
Problem 6.29. .............................................................................................................................94
Problem 6.30. .............................................................................................................................95
Problem 6.31. .............................................................................................................................95
Problem 6.32. .............................................................................................................................96
Problem 6.33. .............................................................................................................................96
Problem 6.34. .............................................................................................................................96
Problem 6.35. .............................................................................................................................97
Problem 6.36. .............................................................................................................................97
Problem 6.37. .............................................................................................................................98
Problem 6.38. .............................................................................................................................98
Problem 6.39. .............................................................................................................................98
Problem 6.40. .............................................................................................................................99
Problem 6.41. ...........................................................................................................................100
Problem 6.42. ...........................................................................................................................100
Problem 6.43. ...........................................................................................................................101
Problem 6.44. ...........................................................................................................................102
Problem 6.45. ...........................................................................................................................103
Problem 6.46. ...........................................................................................................................104
Problem 6.47. ...........................................................................................................................106
Problem 6.48. ...........................................................................................................................106
Problem 6.49. ...........................................................................................................................106
Problem 6.50. ...........................................................................................................................107
Problem 6.51. ...........................................................................................................................107
Problem 6.52. ...........................................................................................................................107
Problem 6.53. ...........................................................................................................................108
Problem 6.54. ...........................................................................................................................108
Problem 6.55. ...........................................................................................................................109
Problem 6.56. ...........................................................................................................................110
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ACS Chemistry FROG
Chemical Reactions
Chapter 6
Problem 6.57. ...........................................................................................................................111
Problem 6.58. ...........................................................................................................................111
Problem 6.59. ...........................................................................................................................112
Problem 6.60. ...........................................................................................................................112
Problem 6.61. ...........................................................................................................................113
Problem 6.62. ...........................................................................................................................113
Problem 6.63. ...........................................................................................................................113
Problem 6.64. ...........................................................................................................................114
Problem 6.65. ...........................................................................................................................115
Problem 6.66. ...........................................................................................................................115
Problem 6.67. ...........................................................................................................................116
Problem 6.68. ...........................................................................................................................117
Problem 6.69. ...........................................................................................................................117
ACS Chemistry FROG
5
Chapter 6
Chemical Reactions
Materials for Chapter 6 Activities
Activity
Material
Total Quantity
6.2
250-mL beakers or clear,
colorless plastic tumblers
4
6.2, 6.3
Hot plate
1
6.3
Stoppered, 250-mL filter
flask
1
6.3
Rubber tubing attached to
filter flask.
1
6.3, 6.41, 6.43, 6.86, 6.98
Test tubes
1-3
6.6
Centrifuge test tubes
5
6.6
Centrifuge
1
6.6
Graduated pipets
A few.
6.13, 6.30
pH meter and combination
pH electrode
1
6.13, 6.30, 6.62, 6.93
Small beakers or well plate
2-5
6.43, 6.98
Thin-stem plastic pipets
3
6.45
Model kit
6.50
25200-mm test tubes
2
6.50
Water bath
1
6.66
400-mL beaker
2
6.66
100-mL beaker
2
6.66
10-mL graduated cylinder
2
6.66
1-gallon plastic zip-closure
bag
2
6.80
Small flask
1
6
ACS Chemistry FROG
Chemical Reactions
Chapter 6
Reagents for Chapter 6 Activities:
Activity
Reagents
Total Quantity
6.2
Dried yeast
2 tsp
6.2
Glucose
1 tsp
6.2, 6.3
Potassium hydrogen
tartrate, KHC4H5O6
1 tsp
6.2
Calcium chloride, CaCl2
1 tsp
6.2, 6.3
Sodium hydrogen
carbonate, NaHCO3
2 tsp
6.2, 6.3
Water
6.3
Limewater [saturated
Ca(OH)2 solution]
50 mL
6.6, 6.43
0.1M calcium nitrate,
Ca(NO3)2.4H2O
2.4 g in 100 mL water
6.6, 6.43
0.1 M sodium oxalate,
Na2C2O4
1.3 g in 100mL water
6.13
0.1 M hydrochloric acid,
HCl
Dilute from stock
6.13
0.1 M sodium chloride,
NaCl
0.6 g in 100 mL water
6.13, 6.30, 6.92
0.1 M ethanoic (acetic) acid
CH3C(O)OH
Dilute from stock
6.13
0.1 M sodium ethanoate
(acetate), CH3C(O)ONa
0.8 g in 100 mL water
6.13, 6.30
Buffers to standardize the
pH meter and electrode
About 50 mL of each buffer
6.30
0.1 M ethanol
0.4 mL of ethanol in 100
mL of water
6.41
0.1 M nickel (II) chloride,
NiCl2
1.3 g in 100 mL water
6.41
1 M ammonia, NH3
Dilute from stock
6.43
0.1 M sodium oleate
3.0 g in 100 mL water
6.43
0.1 M EDTA
3.8 g in 100 mL water
6.50
Salicylic acid
4g
6.50
Methanol
10 mL
ACS Chemistry FROG
7
Chapter 6
Chemical Reactions
6.50
Concentrated sulfuric acid,
H2SO4
4 drops
6.50
Saturated sodium hydrogen
carbonate, NaHCO3
2 mL
6.62
0.1 M silver nitrate, AgNO3
1.7 g in 100 mL water
6.62
0.1 M copper sulfate,
CuSO4.5H2O
2.5 g in 100 mL water
6.62,6.66
Copper wire, 20-24 gauge
2-3 cm length
6.62
Silver wire, 20-24 gauge
2-3 cm length
6.66, 6.86
Concentrated Nitric acid,
HNO3
5 mL
6.66
Concentrated Hydrochloric
acid, HCl
5 mL
6.80
6 M sulfuric acid, H2SO4
1 mL; dilute from stock
6.80
20% potassium iodide, KI
10 g in 50 mL water
6.80
Household bleach
Few drops
6.86
Silver nitrate, AgNO3
1 g in 10 mL water
6.86
Sodium hydroxide, NaOH
1 g in 10 mL water
6.86
Ammonia, NH3
Dilute concentrated
ammonia 1:10 with water
6.86
0.1 M hydrochloric acid
Dilute from stock.
6.86
5-10% formaldehyde in
water
Few drops
6.92
Bromocresol green acidbase indicator
6.92
0.1 M sodium hydroxide,
NaOH
0.4 g in 100 mL of water
6.98
0.0050 M ascorbic acid
0.22 g in 250 mL of water
6.98
0.0050 M iodine
See prep on p.xx.
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ACS Chemistry FROG
Chemical Reactions
Chapter 6
Section 6.1. Classifying Chemical Reactions
Learning objectives for Section 6.1:
 Develop criteria for deciding whether a chemical reaction has occurred in a chemical system.
 Describe the classification of chemical reactions in terms of the interactions of positive and
negative centers and/or the transfer of electrons from one reactant to another.
Consider This 6.1. What chemical reactions do you know?
Goal:
Suggest a criterion or criteria for deciding whether a chemical reaction has occurred in a
chemical system.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to complete this activity. Then, the
instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so, perhaps by having a spokesperson from each group
present the group’s list.
 This activity can be conducted as an open class discussion.
Time for activity:
 5-10 minutes,
Instructor notes:
 Use this activity to introduce students to this chapter on chemical reactions. Reinforce the idea
that they already know a lot about chemical reactions, and that it is useful to categorize them
in order to have only a few categories to remember.
 Students should reason and conclude:
(a) A few examples might include rusting, burning, systems produce color changes,
precipitates, or bubbles.
(b) Criteria might include evidence that the products are not the same as the reactants, such
as temperature change, color change, precipitation, and production of a gas (bubbles or
foam). Some might suggest molecular level criteria such as bonds broken are not the same as
bonds formed. These are not right or wrong responses, but a way for you and the class to
think about the variety of criteria that might be used to answer the question, “How do you
know a chemical reaction has occurred?”
Follow-up discussion:
 Define “chemical reaction” based upon follow-up discussion.
Follow-up activities:
 Investigate This 6.2. What changes do you observe?
 Investigate This 6.3. How can you analyze the gas from the NaHCO3 + KHC4H5O6 reaction?
 Consider This 6.4. What is the gas from the NaHCO3 + KHC4H5O6 reaction?
 End of chapter problems 6.1 through 6.4.
Investigate This 6.2. What changes do you observe?
Goal:
Observe if any changes occur in four possible reaction mixtures.
Set-up time:
Approximately 15-20 minutes.
ACS Chemistry FROG
9
Chapter 6
Chemical Reactions
Time for activity:
 From about 10 minutes to more than 30 minutes, depending on the amount of discussion
about the nature of chemical reactions and the changes that occur here you wish to
incorporate.
Materials:
 4 250-mL beakers or clear, colorless plastic tumblers.
 1 tsp measuring spoon or equivalent for dispensing the solid reagents.
 Hot plate for warming water.
Reagents:
 Dried yeast packets.
 Glucose (sucrose can be substituted for glucose).
 Calcium chloride, CaCl2.
 Sodium hydrogen carbonate, NaHCO3.
 Potassium hydrogen tartrate (cream of tartar), KHC4H5O6.
 Water, heated to 45-55 ºC.
Procedure:
SAFETY NOTE
Wear safety goggles

Conduct this as a class investigation with student volunteers, but have students work in small
groups to discuss and analyze the results.
 Prepare the beakers with the contents shown in this table:
Beaker
Contents
1
tsp dried yeast
2
3
tsp dried yeast + tsp CaCl2 + tsp
tsp glucose
NaHCO3




4
tsp NaHCO3 +
tsp KHC4H5O6
Place the beakers on paper or plastic plates, in case they spill over.
To each beaker, add about 40 mL of warm (45-55 ºC) water and swirl to mix the ingredients.
Observe the beakers for about two minutes and record any evidence of change(s) occurring.
If there is no apparent change in one or more of the beakers, wait for a few minutes and check
them again.
 Continue this checking until you are convinced no change is going to occur. The reaction of
yeast and sugar produces a great deal of bubbling foam, but does not occur immediately.
Warm water is used in the investigation to decrease the amount of the induction period before
this reaction becomes apparent.
Anticipated results:
 Set up before addition of water:
10
ACS Chemistry FROG
Chemical Reactions

Chapter 6
Results after sufficient time for the yeast-sugar mixture (2) to begin reaction. The evolution of
bubbles is essentially over in reactions (3) and (4).
Disposal:
 Wash down the drain with copious amounts of water.
Follow-up discussion:
 Reactions (2), (3), and (4) all produce the same gas, carbon dioxide. To analyze this gas,
conduct Investigate This 6.3 first before further discussing the reactions.
 The reactions are:
(2) C6H12O6(aq) + (yeast)  2CH3CH2OH(l) + 2CO2(g)
(3) Ca2+(aq) + 2HCO3–(aq)  CaCO3(s) + CO2(g) + H2O(l)
(4) HC4H3O6–(aq) + HCO3–(aq)  C4H3O62–(aq) + CO2(g) + H2O(l)
Follow-up activities:
 Investigate This 6.3. How can you analyze the gas from the NaHCO3 + KHC4H5O6 reaction?
 End of chapter problems 6.1 through 6.4.
Investigate This 6.3. How can you analyze the gas from a reaction?
Goal:
Determine whether the gas from reaction (4) in Investigate This 6.2 produces a precipitate when
bubbled into limewater.
Set-up time:
 Approximately 15 minutes.
Time for activity:
 10-15 minutes (including discussion).
Materials:
 250-mL filter flask with stopper.
 Rubber tubing.
 Test tube.
 Appropriate ring stands and clamps.
 1 tsp measuring spoon or equivalent for dispensing the solid reagents.
 Hot plate for warming water. (Warm water is unnecessary for this activity, but can be used to
mimic the conditions in the fourth reaction in Investigate This 6.2.)
Reagents:
 Sodium hydrogen carbonate, NaHCO3.
 Potassium hydrogen tartrate, KHC4H5O6.
 Limewater (saturated solutions of Ca(OH)2).
 Water heated to 45-55 ºC. (Warm water is unnecessary for this activity, but can be used to
mimic the conditions in the fourth reaction in Investigate This 6.2.)
ACS Chemistry FROG
11
Chapter 6
Chemical Reactions
Procedure:
SAFETY NOTE
Wear safety goggles

Conduct as a class investigation with student volunteers, but have students work in small
groups to discuss and analyze the results.
 Repeat the reaction in the fourth beaker from Investigate This 6.2, but in a stoppered, 250-mL
reaction flask with a length of rubber tubing attached to the filter arm. Clamp the open end of
the tubing below the surface of the limewater in the test tube. A picture of the set-up is shown
below.
 Add the solid reactants to the flask and then stopper it as soon as the water is been added.
 Swirl the flask to mix the reactants.
 Observe and record what happens.
Anticipated results:
 Set up before addition of water to reaction flask:

Results after water has been added to the reaction flask and a close-up of the test tube of
limewater after gas from the reaction has bubbled through:
NOTE: The CaCO3(s) formed in the test tube may begin to dissolve if a small amount of
limewater is used and a large volume of CO2 is produced. It is unlikely that enough will
dissolve to be noticeable in a few minutes. (See Investigate This 2.89, Chapter 2, Section 2.16,
pp. 138-139.
Disposal:
 Wash down the drain with copious amounts of water.
Follow-up discussion:
 Use Consider This 6.4 to initiate discussion the results of this activity.
Follow-up activities:
 Consider This 6.5. How can you classify chemical reactions?
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ACS Chemistry FROG
Chemical Reactions

Chapter 6
End of chapter problems 6.1 through 6.4.
Consider This 6.4. What is the gas from the NaHCO3 + KHC4H5O6 reaction?
Goal:
Based upon their observation of the precipitate formation in Investigate This 6.3, determine that
the gas formed is CO2.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to complete this activity. Then, the
instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so.
 This activity can be conducted as an open class discussion.
Time for activity:
 Approximately 10 minutes.
Instructor notes:
 If necessary, review limewater reactions from Chapter 2, Section 2.16.
Students should reason and conclude:
(a) When the gas bubbled through the limewater solution, a white precipitate formed. In
Investigate This 2.89, we found that a white precipitate of CaCO3(s) was formed when
CO2(g) was bubbled into limewater. We can probably conclude that the gas from the reaction
in Investigate This 6.3 is CO2(g), because a white precipitate is formed when the gas bubbles
into limewater.
(b) In Chapter 2, Section 2.14, we found that a compound that can transfer a proton to the
hydrogen carbonate anion, HCO3–(aq), produces carbonic acid, H2CO3(aq), reaction (2.29).
The H2CO3(aq) can dissociate to water and carbon dioxide CO2(g), which bubbles out of the
solution, reaction (2.30). The hydrogen tartrate anion, HC4H3O6–(aq), can transfer a proton to
the hydrogen carbonate anion. The overall reaction forming CO2(g) is:
HC4H3O6–(aq) + HCO3–(aq)  C4H3O62–(aq) + CO2(g) + H2O(l)
Follow-up discussion:
 Review the chemical reactions that occurred in Investigate This 6.2 and 6.3.
Follow-up activities:
 Consider This 6.5. How can you classify chemical reactions?
 End of chapter problems 6.1 through 6.4.
Consider This 6.5. How can you classify chemical reactions?
Goal:
Students discuss how to classify chemical reactions based upon previous completed activities in
this section.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to complete this activity. Then, the
instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so.
 This activity can be conducted as an open class discussion.
Instructor notes:
 Review outcomes from Consider This 6.1, Investigate This 6.2, Investigate This 6.3, and
Consider This 6.4. Consider having students lead the discussion.
Students should reason and conclude:
ACS Chemistry FROG
13
Chapter 6
Chemical Reactions
(a) It will probably be possible to classify the reactions into groups that have similar
properties and/or behaviors.
(b) Maybe, but we will almost certainly need more information on the molecular structures
of the reactants as well as an understanding of the polarity of the molecules.
Follow-up discussion:
 Discuss the criteria for classifying chemical reactions.
(1) Center of a positive charge in one reactant molecule or ion is attracted to a center of
negative charge in another.
(2) Rearrangement of electrons and/or atoms to form products that are different from the
reactants.
 Review the three broad classes of chemical reactions we will use: ionic precipitation, Lewis
acid-base, and oxidation-reduction.
Alternate Activity:
 One approach is to illustrate each of the types of reactions with a demonstration at this
point, instead of waiting until later in the chapter. The reactions used by one faculty member
are:
(1) precipitation of calcium carbonate by reaction of sodium carbonate and calcium nitrate.
(2) synthesis of nylon from adipoyl chloride and 1,6-diaminohexane using interfacial
polymerization (the “nylon rope trick”) to illustrate Lewis acid-base reactions. The synthesis
of nylon polymers is a nice example of condensation polymerization, but should be done
with some care, since the reagents are relatively corrosive and the best methods use
chlorinated solvents that are generally suspect carcinogens. T. L. Bieber, J. Chem. Educ.
1979, 56, 409-410, gives a good discussion of the demonstration. G. C. East and S. Hassell,
J. Chem. Educ. 1983, 60, 69, suggest an alternative way to prepare the necessary acid
chloride(s) as needed.
(3) reduction of silver ion with copper metal.
Follow-up activities:
 End of chapter problems 6.1 through 6.4.
Section 6.2. Ionic Precipitation Reactions
Learning Objectives for Section 6.2:
 Classify a chemical change as a precipitation based on known reactants and products or from
experimental observations on the change.
 Predict probable precipitation reactions based on the cation and anion charges.
 Use the stoichiometry of continuous variations studies to determine the formula of a reaction
product or the ratio in which reactants react.
Investigate This 6.6. How can you find the stoichiometry of a precipitate?
Goal:
Students determine the stoichiometry of CaC2O4(s).
Set-up time:
 15 minutes (assumes that solutions are prepared previously)
Time for activity:
 10-15 minutes (including discussion).
14
ACS Chemistry FROG
Chemical Reactions
Chapter 6
Materials:
 5 centrifuge tubes.
 1 centrifuge to accommodate the tubes.
 Graduated pipets (plastic pipets with a 1-mL graduated stem are adequate).
Reagents:
 0.1 M Ca(NO3)2(aq) (or CaCl2)
 0.1 M Na2C2O4(aq)
Procedure:
SAFETY NOTE
Wear safety goggles

Conduct this as a class investigation with student volunteers, but have students work in small
groups to discuss and analyze the results. It also can be completed in a laboratory or
discussion section.
 Prepare the centrifuge tubes with the contents shown in this table:
tube number
1
2
3
4
5
calcium nitrate, mL
0.5
1.5
2.5
3.5
4.5
sodium oxalate, mL
4.5
3.5
2.5
1.5
0.5

Centrifuge the tubes for 1-2 minutes. Centrifuge time may vary depending on type of
centrifuge. If a centrifuge is not available, the tubes can be allowed to stand for 10-20 minutes
and then observed.
Anticipated results:
 After centrifugation, the tubes look like this:
Disposal:
 Place precipitates in a properly labeled waste container.
Follow-up discussion:
 Use Consider This 6.7 to initiate discussion of the results of this activity.
Follow-up activities:
 Consider This 6.8. Does ionic precipitation fit the definition of a chemical reaction?
 Worked Example 6.9. Stoichiometric calculation for calcium oxalate formation.
 Check This 6.10. Stoichiometric calculation for calcium oxalate formation.
 End of chapter problems 6.5 through 6.9.
ACS Chemistry FROG
15
Chapter 6
Chemical Reactions
Consider This 6.7. What is the stoichiometry of the calcium oxalate precipitate?
Goal:
Based on their results from Investigate This 6.6, determine the stoichiometry of the calcium
oxalate precipitate.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to complete this activity. Then, the
instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so.
 This activity can be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class session.
Time for activity:
 Approximately 10 minutes.
Instructor notes:
 Make sure the class agrees on the observations from Investigate This 6.6.
Students should reason and conclude:
(a) The concentration of the reagents in each stock solution is the same, 0.1 M. The ratio of
the volume of the calcium cation solution to oxalate anion solution varies while the total
volume of the solution in each tube is the same. Thus the ratio of the number of moles of the
calcium cation to number of moles of oxalate anion varies from one tube to the next while
the total number of moles of the two reactants remains constant.
(b) The amount of precipitate is not the same in each tube. Tube 3 has the most precipitate
while tubes 1 and 5 have the least precipitate. Graphs should reflect the data. A bar graph is
probably easiest to draw, because the data themselves look like a bar graph.
(c) As you observe the differences in precipitate from tubes 1 through tube 5, the amount of
precipitate increases until tube 3 and then decreases. Tubes 1 and 5 contain about the same
amount of precipitate and tubes 2 and 4 contain about the same amount of precipitate. It
seems likely that amount of precipitate formed in each tube is related to the ratio of cation to
anion in each one. When the ratio is one-to-one, in tube 5, we get the largest amount of
precipitate, which makes sense if the net ionic reaction is:
Ca2+(aq) + C2O42-(aq)  CaC2O4(s)
Follow-up discussion:
 Point out that multiply-charged cations and anions often form relatively insoluble salts.
 The attractions among multiply-charged cations and anions are stronger than the hydration
energy. The hydration energy simply cannot overcome the lattice energy.
 Review the solubility rules listed in Table 6.1
Follow-up activities:
 Consider This 6.8. Does ionic precipitation fit the definition of a chemical reaction?
 Worked Example 6.9. Stoichiometric calculation for calcium oxalate formation.
 Check This 6.10. Stoichiometric calculation for calcium oxalate formation.
 Check This 6.11. Nickel-dimethylglyoxime reaction stoichiometry.
 End of chapter problems 6.5 through 6.9.
16
ACS Chemistry FROG
Chemical Reactions
Chapter 6
Consider This 6.8. Does ionic precipitation fit the definition of a chemical reaction?
Goal:
Decide whether rearrangements of electrons and/or atoms (our definition of a chemical reaction)
occur in precipitation reactions
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to complete this activity. Then, the
instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so.
 This activity can be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class session.
Time for activity:
 Approximately 5-10 minutes.
Instructor notes:
 Review results from Investigate This 6.6, including the net ionic equation [reaction (6.4)]:
Ca2+(aq) + C2O42–(aq)  CaC2O4(s)
Students should reason and conclude:
Reaction (6.4) does fit the description that chemical reactions involves rearrangements of
atoms and/or electrons to form products different than reactants, if we take into account the
attractions of the ions to water molecules that have to be broken in order for a precipitate to
form in which the attractions are among the ions themselves. Ca2+(aq) are attracted to C2O42–
(aq) and visa versa. The lattice energy for forming CaC2O4(s) is stronger than the hydration
energy to keep these ions apart.
Follow-up discussion:
 Discuss that the stoichiometry of the precipitate formed and, hence, the amount of precipitate
formed, is based on charge balance.
 Use the discussion to lead to Worked Example 6.9 and the quantification of these
stoichiometric arguments.
Follow-up activities:
 Worked Example 6.9. Stoichiometric calculation for calcium oxalate formation.
 Check This 6.10. Stoichiometric calculation for calcium oxalate formation.
 Check This 6.11. Nickel-dimethylglyoxime reaction stoichiometry.
 End of chapter problems 6.5 through 6.10.
Sample Worksheet for Section 6.2:
This worksheet was contributed by Dr. Laura Eisen, The Mount Vernon College of George
Washington University, Washington, DC. It has been edited to conserve space, but no content
has been changed.
Chemistry 12, Section MV
Spring 2003
Background: A bicycle shop is making bicycles according to the following "equation":
1 frame + 2 wheels  1 bike [or F + 2 W  B]
On 11 different days, the shop makes bikes, starting with different numbers of frames and
wheels, always keeping the sum of the # frames + # wheels constant. Complete the following
table, assuming that the shop makes the maximum number of bikes possible with the frames and
wheels that are available.
ACS Chemistry FROG
17
Chapter 6
Chemical Reactions
Day
1
2
3
4
5
6
7
8
9
10
11
# frames (F)
available
6
12
18
24
30
36
42
48
54
60
66
# wheels (W)
available
66
60
54
48
42
36
30
24
18
12
6
Total F + W
72
# bikes made
6
Unused
frames
0
Unused
wheels
54
"limiting"
component
F
Under what conditions can the maximum number of bicycles be made?
Using graph paper, graph number of bikes made vs. wheels available. What do the two line
segments of the graph represent? What does the intersection of the two line segments represent?
How could you use the graph to determine the maximum number of bikes that could be made?
APPLY what you have learned:
This table represents data from another bike shop.
Day
1
2
3
4
5
6
7
8
9
10
11
# frames
6
available
12
18
24
30
36
42
48
54
60
66
# wheels
66
available
60
54
48
42
36
30
24
18
12
6
# bikes
made
12
18
16
14
12
10
8
6
4
2
6
What can you say about the bikes made in this shop? What is the recipe for the bikes made in
this shop? How do you know?
Do Check This 6.11, and End-of-Chapter problems 6.9 and 6.10.
In what way does the chemical problem differ from the bicycle problem? (Hint: Assume that
you have bikes made according to F + 2W  B. If you have 10 frames and 19 wheels, how
many bikes can you make? Now assume that you have a chemical reaction that has the
following stoichiometry: A + 2B  C. Assuming that the reaction goes to completion, how
many moles of C can you make if you start with 10 moles of A and 19 moles of B? Is there a
difference between the bicycle result and the chemical result? Why?)
18
ACS Chemistry FROG
Chemical Reactions
Chapter 6
Section 6.3. Lewis Acids and Bases: Definition
Learning Objectives for Section 6.3:
 Define Lewis acid-base reactions and begin to recognize three types of Lewis acid-base
reactions: Brønsted-Lowry proton transfers, metal ion complexation, and nucleophileelectrophile reactions.
Consider This 6.12. How do ionic precipitation and Lewis acid-base reactions differ?
Goal:
Distinguish the difference between ionic precipitation and Lewis acid-base reactions.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to answer these questions. Then,
the instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so.
 This activity can be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class session. The discussion is very important, because this distinction may be subtle enough
to be confusing.
Time for activity:
 Approximately 10 minutes.
Instructor notes:
 Briefly define the Lewis acid-base model with an emphasis on the formation of electron pair
bonds between a donor molecule and an acceptor molecule.
Students should reason and conclude:
 In an ionic precipitation reaction, cations and anions are attracted to each other to form an
ordered crystalline solid in which several cations interact with each anion and vice versa.
None of the nonbonding pairs of electrons on the anions are associated with one particular
cation. In Lewis acid-base reactions, a new molecule is formed in which the Lewis base
electron pair forms a covalent, electron-pair bond with the Lewis acid.
Follow-up discussion:
 Use the discussion to lead into and introduce the three types of Lewis acid-base reactions we
consider: Brønsted-Lowry acid-base reactions, formation of metal ion complexes, and
nucleophile (Lewis base)–electrophile (Lewis acid) reactions.
Follow-up activities:
 End of chapter problems 6.11 and 6.12.
Section 6.4. Lewis Acids and Bases: Brønsted-Lowry Acid-Base Reactions
Learning Objectives for Section 6.4:
 Define and give examples of strong and weak Lewis/Brønsted-Lowry acids and bases.
 Use the observed pH and stoichiometry of a solution to determine the relative basicity
(acidity) of the species in the solution.
 Use relative basicities (acidities) to predict the predominant species in solutions of
Lewis/Brønsted-Lowry bases (acids).
ACS Chemistry FROG
19
Chapter 6
Chemical Reactions
Investigate This 6.13. Do the acidities of acids differ?
Goal:
Determine the pH of several solutions (strong and weak acid and their sodium salts).
Set-up time:
 10-15 minutes (assuming reagents and buffers are prepared for standardizing the pH meter
and electrode).
Time for activity:
 10-15 minutes (including discussion)
Materials:
 pH meter and combination electrode.
 Small vials (place reagents in vials before class)
Reagents:
 Distilled water.
 0.1 M hydrochloric acid, HCl.
 0.1 M sodium chloride, NaCl.
 0.1 M ethanoic (acetic) acid, CH3C(O)OH
 0.1 M sodium ethanoate (acetate), CH3C(O)ONa
Procedure:
SAFETY NOTE
Wear safety goggles

Use a pH meter to determine the pH of the reagents listed above.
Alternative procedures:
 pH paper can be used to determine the approximate pH of the solutions. These pHs will be
fine for the purpose of these analyses.
 Students can do the activity in small groups. Each group gets a beaker (or other container)
containing the reagents in small test tubes, pH paper, and a small stirring rod. Students place a
drop of solution from the stirring rod onto the pH paper. A scanned copy of the pH paper
color indicator chart is projected so the students can determine the pH of the solution.
Anticipated results:
 Typical results look something like these:

20
Solution
H2O
HCl
NaCl
CH3C(O)OH
CH3C(O)ONa
pH
6.9
1.0
6.9
2.6
8.6
For the alternative procedure, the results might look like these. The pH paper is shown in the
solutions only for photographic purposes. pH paper should never be dipped in the solution to
be tested, but rather a drop of the solution transferred to the paper with a stirring rod (or
toothpick would work for these samples).
ACS Chemistry FROG
Chemical Reactions
Chapter 6
Disposal:
 Rinse reagents down the drain with copious amounts of water.
Follow-up discussion:
 Use Consider This 6.14 to initiate discussion of the results of this activity and lead to the
calculation of [H3O+] from pH.
Follow-up activities:
 Check This 6.15. [H3O+(aq)] in 0.1M ethanoic (acid) solution.
 Check This 6.16. Fraction of ethanoic acid that transfers a proton to water.
 Check This 6.17. [OH–(aq)] in an aqueous sodium amide solution.
 End of chapter problems 6.13 through 6.19.
Consider This 6.14. How do the acidities of acids differ?
Goal:
Categorize the solutions in Investigate This 6.13 into those in which H3O+(aq) predominates
(compared to water as a standard) and those in which OH–(aq) predominates.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to answer these questions. Then,
the instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so.
 This activity can be conducted as an open class discussion coupled with Investigate This 6.13.
 This activity could also be assigned as a homework problem and then discussed at the next
class session.
Instructor notes:
 Be sure the class agrees on the results from Investigate This 6.13.
Students should reason and conclude:
 The order of increasing pH is: HCl < CH3C(O)OH < NaCl ≈ H2O < CH3C(O)ONa.
 The NaCl(aq) solution has nearly the same pH as water. The hydronium ion predominates in
the HCl(aq) and CH3C(O)OH(aq) solutions. The hydroxide ion predominates in the
CH3C(O)ONa(aq) solution. Both HCl(aq) and CH3C(O)OH(aq) must transfer protons to
water to produce solutions in which H3O+(aq) predominates while water must transfer protons
to CH3C(O)ONa(aq) to produce a solution in which OH–(aq) predominates.
Follow-up activities:
 Check This 6.15. [H3O+(aq)] in 0.1 M ethanoic (acetic) acid solution.
 Check This 6.16. Fraction of proton transfer between ethanoic acid and water.
 Consider This 6.17. What is the molecular level interpretation of expression (6.11)?
 Check This 6.18. [OH–(aq)] in an aqueous sodium amide solution.
 Check This 6.19. Writing Lewis acid-base reactions.
 Check This 6.21. Relative basicities of chloride and hydroxide ions.
 Check This 6.23. Relative basicities of ethanoate (acetate) and hydroxide ions.
 Check This 6.25. pH of an aminium (ammonia-like) chloride solution.
 End of chapter problems 6.13 through 6.19.
Consider This 6.17. What is the molecular level interpretation of expression (6.11)?
Goal:
Give a molecular level description of the symbolic equation for an equilibrium reaction based
on the animation in the Web Companion, Chapter 6, Section 6.4, page 1.
ACS Chemistry FROG
21
Chapter 6
Chemical Reactions
Classroom options:
 This activity can be conducted as an open class discussion, if facilities are available for
viewing the Web Companion in the classroom.
 This activity could also be assigned as a homework problem and then discussed at the next
class session. The discussion is essential so you can try to be sure that the class correctly
interprets the symbolic equilibrium expression at the molecular level.
Instructor notes:
 Students need to have access to the Web Companion, Chapter 6, Section 6.4, page 1, either in
the class or on their own (preferably both).
Students should reason and conclude:
 The animation shows that proton transfer between water molecules can occur after ethanoic
acid donates a proton to a water molecule, the forward reaction in expression (6.11), forming
H3O+ and ethanoate ion. Since water molecules make up the vast majority of molecules in the
solution, transfer of a proton from this H3O+ ion to another water molecule is a likely event.
 Transfer of a proton from a H3O+ to an ethanoate ion, the reverse reaction in expression
(6.11), can occur to form ethanoic acid and water.
 The proton originally transferred from the ethanoic acid to water is unlikely to be the one that
is transferred from hydronium ion to an ethanoate ion to form ethanoic acid.
 Although expression (6.11) is a representation of the molecular level processes of proton
transfer between molecules of ethanoic acid and water and hydronium ion and ethanoate ion,
it has to be interpreted in the context of all the other molecules in the system, which are not
shown explicitly in the expression, but among which proton transfers can occur. Any one of
the ethanoic acid molecules can transfer a proton to any one of the surrounding water
molecules and, in turn, any of the ethanoate anions formed in the solution can gain a proton
from any of the hydronium ions it comes near.
Follow-up activities:
 Check This 6.18. [OH–] in an aqueous sodium amide solution.
 Check This 6.19. Writing Lewis acid-base reactions.
 Check This 6.21. Relative basicities of chloride and hydroxide ions.
 Check This 6.23. Relative basicities of ethanoate (acetate) and hydroxide ions.
 Check This 6.25. pH of an aminium (ammonia-like) chloride solution.
 End of chapter problems 6.13 through 6.19.
Section 6.5. Predicting Strengths of Lewis/Lowry-Lowry Bases and Acids
Learning Objectives for Section 6.5
 Predict the relative basicities (acidities) of a series of Lewis/Brønsted-Lowry bases (acids) of
known structure.
 Use relative basicities (acidities) to predict the predominant species in solutions of
Lewis/Brønsted-Lowry bases (acids).
Consider This 6.27. Is electronegativity a reliable predictor of Lewis acid-base strength?
Goal:
Determine that electronegativity is not a predictor of Lewis acid-base strength in a family of the
periodic table.
22
ACS Chemistry FROG
Chemical Reactions
Chapter 6
Time for activity:
 From 5 to 15 minutes, depending on how much of the size argument you wish to bring into
the discussion.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to answer the questions. Then, the
instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so.
 This activity can be conducted as an open class discussion.
Instructor notes:
 Try to be sure students understand the basis of electronegativity (attraction of the nucleus for
electrons in its chemical bonds).
 Show or direct students to the charge density models in Consider This 6.27.
Students should reason and conclude:
(a) Fluorine is the most electronegative element and the electronegativity values decrease
down the halogen group. Thus, the order of electronegativities is F > Cl > Br > I. The
argument we make in Check This 6.26 and the preceding text is that the more electronegative
elements hold their electrons more tightly, so they are not as available to bond with a proton
and therefore form stronger acids (or are weaker bases). Fluorine is the most electronegative
of the halogens, so we predict (on the basis of electronegativity) that it would form the
strongest Lewis acid. The relative order of the hydrohalic acid strengths (from most to least
acidic) would be: HF > HCl > HBr > HI.
(b) The same reasoning as in part (a) predicts the relative base strengths (from least to most
basic) would be: :F– < :Cl–- < :Br– < :I–.
(c) The answers in parts (a) and (b) are not consistent with the data in Table 6.2. The table
shows that HI is the strongest acid while HF is the weakest. The :I– anion is the weakest base
and the :F– anion is the strongest base. The electronegativity argument predicts the wrong
order of acidity and basicity.
Follow-up discussion:
 In determining the factors that influence the strength of a Lewis base, the size of the atom
with the nonbonding electrons must be considered.
 Ions are more stable when their valence electrons occupy larger volumes. Also, the charge
density is not as large when the same charge occupies a larger volume, so the attraction for a
proton is reduced as we go down a family in the periodic table.
Follow-up activities:
 Consider This 6.28. Does electronegativity or size dominate Lewis acid-base strength?
 Check This 6.29. Relative Lewis base strengths.
 End of chapter problems 6.20 through 6.25.
Consider This 6.28. Does electronegativity or size dominate Lewis acid-base strength?
Goal:
Determine that size dominates Lewis acid-base strength in a family of the periodic table.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to answer these questions. Then,
the instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so.
 This activity can be conducted as an open class discussion.
ACS Chemistry FROG
23
Chapter 6
Chemical Reactions

This activity could also be assigned as a homework problem and then discussed at the next
class session, either in small groups initially or in open class discussion. The discussion is
important to be sure the class understands why size (and thus more spread out electrons) are
important in determining the acid-base strength.
Time for activity:
 Approximately 10 minutes.
Instructor notes:
 Show or direct students to the charge density models in Consider This 6.28.
 Recall the order of basicity for the halide ions.
Students should reason and conclude:
(a) Assume that acid strength is directly proportional to electronegativity of the central atom.
Based on the electronegativities of the oxygen family elements, the relative order of acid
strength of the oxygen-family hydrides (from most to least acidic) is: H2O > H2S ≈ H2Se >
Electronegativity values generally decrease as one goes down a family, but Figure 4.33, page
257, shows that the electronegativities of S and Se are essentially the same, so we have made
their acidities about the same.
(b) Assume that acid strength is proportional to the size of the central atom. Based on the
sizes of the oxygen family elements, the relative order of acid strength of the oxygen-family
hydrides (from most to least acidic) is: H2Te > H2Se > H2S > H2O. The size of the atoms
generally increase down a family and Figure 4.32, page 256, seems to show this trend for the
oxygen family. Thus, H2Te and H2Se are the largest of the group and :TeH- and :SeH- are the
weakest Lewis bases and least attractive to a proton due to their relatively large size.
(c) H-OH and H-SH are found in Table 6.2. H-SH is more acidic than H-OH. Assuming that
this relative order persists down the family, the relative acid strengths (from most to least
acidic) is: H2Te > H2Se > H2S > H2O. The relative order of base strengths for the anions
(from least to most basic) is: :TeH– < :HSe– < :HS– < :HO–.
(d) We see that the probable experimental order of acidities (or basicities) in part (c) is the
same as the order predicted in part (b) on the basis of size. Size, not electronegativity,
dominates Lewis acid-base strength in hydrides of a family of the periodic table.
Follow-up activities:
 Check This 6.29. Relative Lewis base strengths
 End of chapter problems 6.20 through 6.25.
Investigate This 6.30. Do the acidities of alcohols and carboxylic acids differ?
Goal:
Measure the pH of ethanol and ethanoic acid solutions and reason from the results whether
alcohols and carboxylic acids differ in acidity.
Set-up time:
 5-10 minutes (assuming reagents and buffers are prepared for standardizing the pH meter and
electrode).
Time for activity:
 Less than 10 minutes.
Materials:
 pH meter and combination electrode (or pH paper).
 Small beakers/vials or well plate.
24
ACS Chemistry FROG
Chemical Reactions
Chapter 6
Reagents:
 Distilled water
 0.1 M ethanoic (acetic) acid.
 0.1 M ethanol.
Procedure:
SAFETY NOTE
Wear safety goggles

Use a pH meter to determine the pH of the reagents listed above.
Alternative procedures:
 pH paper can be used to determine the approximate pH of the solutions. These pHs will be
fine for the purpose of these analyses.
 Students can do the activity in small groups. Each group gets a beaker (or other container)
containing the reagents in small test tubes, pH paper, and a small stirring rod. Students place a
drop of solution from the stirring rod onto the pH paper. A scanned copy of the pH paper
color indicator chart is projected so the students can determine the pH of the solution.
Anticipated results:
 Typical results look something like these:
Solution
H2O
CH3C(O)OH
CH3CH2OH
pH
6.9
2.6
6.9

The pH of the ethanoic acid solution is lower than that for water. The pH of the ethanol
solution is essentially the same as for water.
 For the alternative procedure, the results might look like these. The pH paper is shown in the
solutions only for photographic purposes. pH paper should never be dipped in the solution to
be tested, but rather a drop of the solution transferred to the paper with a stirring rod (or
toothpick would work for these samples).
Disposal:
 Rinse reagents down the drain with copious amounts of water.
Follow-up discussion:
 Students should reason and conclude that the acidity of the alcohol is less than the acidity of
the acid. The pH of the acid solution is lower than the pH of water, because the acid is able to
transfer protons to water. The pH of the alcohol solution is the same as water, because the
alcohol apparently does not transfer protons to water.
 Use Consider This 6.30 to initiate further discussion of the results of this activity.
Follow-up activities:
 Check This 6.32. Relative basicities of ethanoate, ethoxide, and hydroxide anions.
 Check This 6.33. Relative basicities of oxyanions and acidities of oxyacids.
ACS Chemistry FROG
25
Chapter 6

Chemical Reactions
End of chapter problems 6.24 through 6.27.
Consider This 6.31. How do the acidities of alcohols and carboxylic acids differ?
Goal:
Based on the results of Investigate This 6.30, show that the transfer of a proton from ethanol to
water does not occur.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to answer the questions. Then, the
instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so.
 This activity can be conducted as an open class discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class session, either in small groups initially or in open class discussion.
Time for activity:
 From less than 10 to 15 minutes, depending on how the discussion goes in terms of ranking of
acid and base strengths.
Instructor notes:
 Be sure the class agrees on the results from Investigate This 6.30.
 Display reactions (6.17), (6.19), and (6.27) so students can consider them in their discussions.
Students should reason and conclude:
(a) No detectable extra hydronium ion is formed in an ethanol solution compared to water
alone, so the results provide no evidence for reaction (6.27).
(b) Since the results show that ethanoic acid can transfer a proton to water, but ethanol does
not, we can conclude that the acid strengths (from strongest to weakest) are: ethanoic acid >
water ≈ ethanol. We set the water and ethanol about equal because there is also no evidence
that water can transfer a proton to ethanol, so they must be about matched in acid strength.
The base strengths of the anions derived from these acids are in the reverse order, since the
strongest acid will produce the weakest base. The base strengths of the anions (from weakest
to strongest) are: ethanoate anion < hydroxide anion ≈ ethoxide anion
Follow-up discussion:
 Use the discussion to have students relate the reasoning here to that done in Section 6.4 in
terms of arguments based on the predominant species in the solution. Focus here on the
relative amounts of hydronium ion in the three solutions.
Follow-up activities:
 Check This 6.32. Relative basicities of ethanoate, ethoxide, and hydroxide anions.
 Check This 6.33. Relative basicities of oxyanions and acidities of oxyacids.
 End of chapter problems 6.24 through 6.27.
Consider This 6.34. How can you explain the relative acidities of the chlorine oxyacids?
Goal:
Show that the relative strengths of the chlorine oxyacids parallel the number of equivalent
Lewis structures that can be written for the anions, and, hence, that electron delocalization in the
anions is responsible for the increasing acid strength of the oxyacids.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to answer these questions. Then, the
instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so.
26
ACS Chemistry FROG
Chemical Reactions
Chapter 6

This activity can be conducted as an open class discussion.
NOTE: Some may wish to introduce resonance (or reinforce it, if you have introduced it
previously in Chapter 5) stabilization of the conjugate bases and perhaps encourage students to
explore Consider This 6.34 through Check This 6.36 on their own. Some of the generality of the
electron delocalization model (which includes the size effects discussed earlier in this section) is
lost in this way, but the nomenclature may be more familiar to you.
Time for activity:
 Approximately 10-15 minutes.
Instructor notes:
 Show or direct students to examine Figure 6.4 as this activity is carried out.
Students should reason and conclude:
(a) The number of equivalent Lewis structures for the oxyanions of the oxyacids of chlorine
are:
hypochlorite anion (1 structure)
Cl
O
chlorite anion (2 structures)
O
Cl O
O
Cl O
chlorate anion (3 structures)
O
O
O
Cl O
O
O
Cl O
Cl O
O
perchlorate anion (4 structures)
O
O
O Cl O
O Cl O
O
O
O
O Cl
O
O
O
O Cl O
O

(b) There is a correlation between the answers in part (a) and the relative strengths of the
oxyacids. The larger the number of equivalent structures for the oxyanions, the stronger the
Lewis oxyacid and the weaker the Lewis base for the oxyanion. The more equivalent Lewis
structures, the greater the amount of electron delocalization (more volume for the electron
waves to occupy), so the greater electron delocalization, the more stable the oxyanion and the
less basic it is.
Follow-up discussion:
 As part of the discussion, help students see that the delocalization effect here is related to the
size effect for the anions discussed earlier in the section. The objective is to reinforce the idea
that electron delocalization lowers energies and makes species more stable.
Follow-up activities:
 Check This 6.35. Acid-base properties of bleach solutions.
 Check This 6.36. Another way to predict the strengths of oxyacids and oxyanions.
 Consider This 6.37. How do oxyacids/oxyanions of different central atoms compare?
ACS Chemistry FROG
27
Chapter 6


Chemical Reactions
Check This 6.38. Relative acid and base strengths.
End of chapter problems 6.25 through 6.27.
Consider This 6.37. How do oxyacids/oxyanions of different central atoms compare?
Goal:
Conclude, by comparing the oxyacids/oxyanions of different central atoms, that the central atom
acidities/basicities are about the same for comparable compounds in a period, but that a higher
period central atom leads to higher acidity and weaker basicity.
Classroom options:
 Allow about 5 minutes for students, working in small groups, to answer the questions. Then,
have the groups share their structures and conclusions with the class.
 Students can draw the Lewis structures as a homework assignment to discuss in small groups
or open class discussion at the next class session.
Time for activity:
 From approximately 10-15 minutes to 15-20 minutes, depending on how much of the
succeeding paragraph you wish to incorporate in the discussion.
Instructor notes:.
Students should reason and conclude:
(a) The reaction of proton transfer from phosphoric acid to water shown as Lewis structures
is:
O
HO
P
O
OH
O
HO
P
OH
H
O
H
H
O
H
H
OH
O
HO
P
O
OH
(b) The rule from Check This 6.36(a) predicts that oxyacids with the same number of oxygen
atoms double bonded to the central atom will have about the same acid strength. Chlorous
and phosphoric acid both have one oxygen double bonded to the central atom, so we expect
their acid strengths to be comparable, as they are. This is also what the delocalization of
electrons in the oxyanions predicts, because both phosphoric and chlorous acid form anions
that have two equivalent Lewis structures, that is, about the same amount of delocalization.
These examples suggest that the central atom of elements from the same period does not
influence the strength of their oxyacids.
(c) The Lewis structures for the oxyanions of nitrous acid, HONO, and chlorous acid,
HOClO are:
O
N
O
O
N
O
nitrite anion
28
O
Cl O
O
chlorite anion
ACS Chemistry FROG
Cl O
Chemical Reactions
Chapter 6
The rule from Check This 6.36(a) does not predict that the nitrite oxyanion is a somewhat
stronger base than the chlorite oxyanion. The delocalization of electrons in the oxyanions
also does not predict this observation. Since third period atoms are somewhat larger than
second period atoms, the delocalized electrons in oxyanions of third period atoms will
occupy a larger volume. The third period (larger) oxyanion will, therefore, be more stable
and thus a weaker base than similar oxyanions from second period elements even though
they have the same electron delocalization.
Follow-up discussion:
 Use the discussion to further reinforce the idea that greater the delocalization of electrons the
more stable the species and, in the case of oxyanions, the weaker the bases.
Follow-up activities:
 Check This 6.38. Relative acid and base strengths.
 Worked Example 6.39. Predicting the direction of reaction for acid-base reactions.
 Check This 6.40. Predicting the direction of reaction for acid-base reactions.
 End of chapter problems 6.25 through 6.27.
Section 6.6. Lewis Acids and Bases: Metal Ion Complexes
Learning Objectives for Section 6.6
 Classify a chemical change as a Lewis acid-base reaction based on known reactants and
products, or from experimental observations on the change.
 Define and give examples of Lewis acid-base reactions that are metal ion complexation
reactions.
 Recognize the formation of a metal ion complex between a Lewis base and a metal ion in
solution by observations on mixtures of the reactants (or in competitions between the Lewis
base and another reactant for the metal ion) and suggest a structure for the complex.
Investigate This 6.41. Do Lewis bases react with metal ion?
Goal:
Decide if a reaction occurs when ammonia, a Lewis base, is added to a Ni2+(aq) solution.
Set-up time:
 From 5-10 minutes to longer than 20 minutes if solutions have to be prepared.
Time for activity:
 Less than 10 minutes.
Materials:
 2 test tubes.
Reagents:
 0.1 M NiCl2(aq)
 1 M NH3(aq). Note that a higher concentration of ammonia is recommended (compared to the
direction in the text) in order to have an excess of the ligand in the solution.
Procedure:
SAFETY NOTE
Wear your safety goggles



Record the colors of the solutions.
Mix equal volumes of the two solutions.
Record the color of the mixture.
ACS Chemistry FROG
29
Chapter 6
Chemical Reactions
Anticipated results:
 Photographs of the solutions before (left) and after (right) mixing are:

A reaction does occur between the ingredients of the mixture. Before the reaction, the nickel
solution was green. After the ammonia was added, the solution turned blue.
Disposal:
 Dispose of the nickel solution according to local ordinances.
Supplemental activity:
 An extension/supplement to this activity is to add 1 M ammonia solution to aqueous solutions
of sodium chloride and calcium chloride. The lack of color change in these mixtures,
compared to the distinct color change for nickel chloride in ammonia, provides a good point
of departure for a discussion of the differences between transition metals and the groups 1 and
2 metals, if you wish to include these ideas in your course.
Follow-up discussion:
 Use Consider This 6.42 to initiate discussion of the results of this activity.
Follow-up activities:
 Investigate This 6.43. Do calcium ions react with complexing ligands?
 Consider This 6.44. How do calcium ions react with complexing ligands?
 Investigate This 6.45. What is the three-dimensional structure of EDTA?
 Check This 6.46. Keeping the shine in your hair.
 Check This 6.47. - Names and structures of Pt(NH3)2Cl2.
 Consider This 6.48. What is the structure of the nickel-dimethylglyoxime complex?
 Check This 6.49. The porphine ring system in chlorophyll and heme.
 End of chapter problems 6.28 through 6.31.
Consider This 6.42. How do you know a Lewis acid-base complex ion has reacted?
Goal:
Based on their results in Investigate This 6.41, determine if a Lewis acid-base complex ion
reaction has occurred.
Classroom options:
 Students can build the models as a homework assignment. During the next class session,
allow 3-5 minutes for students, working in small groups, to compare their models and answer
the questions. Then, the groups can share their models and answers with the class.
30
ACS Chemistry FROG
Chemical Reactions
Chapter 6

You can build models of the complexes and have students answer the questions in small
groups or during an open class discussion. This is not as valuable an experience for your
students.
Time for activity:
 5-10 minutes (if models are made by students before class).
Instructor notes:
 Be sure the class agrees on the results from Investigate This 6.41. Show or refer to students to
Figure 6.5 and/or a model to demonstrate what an octahedral complex looks like.
Students should reason and conclude:
(a) The model for Ni2+(aq) should mimic the drawing of Cr3+(aq) = Cr(H2O)63+ in Figure 6.5.
One of the properties of this complex is its green color. When a new Lewis base is added to
the solution, the color changes from green to blue. The interaction with the added Lewis base
is probably responsible for the color change, so complexes of the nickel cation with different
Lewis bases have different colors. The evidence is that the complex with water is green.
(b) In this competition, ammonia wins because it is a stronger Lewis base than water (see
Table 6.2). We observed a color change when ammonia was added to the Ni2+(aq) solution.
The color change suggests that a new species has been formed in the solution. Since the only
change was to add a colorless solution containing NH3, we conclude that the stronger base,
ammonia, has won the competition for the Lewis acid, Ni2+ cation, and formed a new
complex ion.
(c) The non-bonding electron pair from the nitrogen atom in ammonia (Lewis base) will bond
to the Ni2+(Lewis acid). Six ammonia molecules bond to Ni2+ forming another octahedral
complex like the one with water. One of the properties of this new complex is its blue color.
Follow-up activities:
 End of chapter problems 6.28 through 6.30.
ACS Chemistry FROG
31
Chapter 6
Chemical Reactions
Investigate This 6.43. Do calcium ions react with complexing ligands?
Goal:
Students observe if calcium ions react with complexing ligands.
Set-up time:
 10-15 minutes (assuming solutions are made previously)
Time for activity:
 Approximately 10 minutes.
Materials:
 Two small plastic test tubes.
 Four thin-stem plastic pipets.
Reagents:
 Water.
 0.1 M calcium nitrate, Ca(NO3)2.
 0.1 M sodium oxalate, NaC2O4.
 0.1 M sodium oleate, NaO(O)C(CH2)7CH=CH(CH2)7CH3 (an ingredient in some soaps). The
0.1 M sodium oleate is 3.0 g of the sodium salt of oleic acid in 100 mL of solution.
Alternative (1): slowly add 2.7 g of oleic acid to a gently stirred 100 mL of a warm 0.1 M
NaOH solution. Alternative (2): Use 0.1 M sodium dodecylsulfate, SDS [2.9 g of SDS
(technical grade is fine) in 100 mL of solution] instead of the oleate. Either treat the solution
as if it is oleate (since the precipitate is all you are interested in observing) or tell students it’s
SDS (a detergent used extensively in biological research and readily available in practically
any biochemistry or cell biology lab) and give them its structure.
O
O
S O
O

0.1 M tetrasodium ethylenediaminetetraacetic acid (EDTA), Na4C10H12N2O8. The 0.1 M
EDTA is 3.7 g of disodium EDTA tetrahydrate + 0.80 g NaOH in 100 mL of solution.
(Equivalently, 4.52 g of the slightly less common tetrasodium EDTA tetrahydrate can be
dissolved in 100 mL of water.)
NOTE: Place each solution in a labeled, thin-stem plastic pipet for easy use in the classroom.
Procedure:
 Do this activity as a class investigation with student volunteers while students work in small
groups to discuss and analyze the results. The leading questions included here integrate
Consider This 6.44 with this activity.
SAFETY NOTE
Wear your safety goggles

Add about 2 mL of water and 4 drops of 0.1 M calcium nitrate solution to each of the test
tubes.
 To one of the test tubes, add 5 drops of the 0.1 M sodium oxalate solution. Swirl gently to mix
the solution.
 Ask your students: "What do you observe? Is this what you expected? Why or why not?"
 To this same test tube add 5 drops of the 0.1 M EDTA solution and gently swirl to mix the
solution.
32
ACS Chemistry FROG
Chemical Reactions
Chapter 6


Ask your students: "What do you observe? How would you explain all your observations?"
To the other test tube, add 5 drops of the 0.1 M sodium oleate solution. Swirl gently to mix the
solution without creating suds.
 Ask your students: "What do you observe? Is this what you expected? Why or why not?"
 Add 5 drops of the 0.1 M EDTA solution and gently swirl to mix the solution.
 Ask your students: "What do you observe? How would you explain your observations?"
Anticipated results:
 These photographs show the calcium nitrate plus oleate solution before (left) and after (right)
EDTA has been added:
Disposal:
 Dispose reagents down the drain with copious amounts of water.
Follow-up discussion:
 Use Consider This 6.44 to initiate discussion as you do this activity.
Follow-up activities:
 Investigate This 6.45. What is the three-dimensional structure of EDTA?
 Check This 6.46. Keeping the shine in your hair.
 End of chapter problems 6.28 through 6.30.
Consider This 6.44. Do calcium ions react with complexing ligands?
Goal:
Based on their results from Investigate This 6.43, students discuss if calcium ions react with
complexing ligands.
Classroom options:
 This activity is probably best conducted as a part of Investigate This 6.43 with designated
group members responding to the questions.
Instructor notes:
 Review results from Investigate This 6.43 before conducting this activity.
Students should reason and conclude:
(a) When oxalate or oleate is added to the calcium ion solutions, white precipitates form. We
have observed the calcium oxalate precipitate previously. However, calcium oleate is also
insoluble in water because the long nonpolar, hydrocarbon tails are insoluble in water.
(b) When EDTA solution is added to the mixtures, the precipitates disappear. The Ca2+(aq)
ions now react with the EDTA ion, leaving behind the soluble oxalate and oleate ions in
solution.
Follow-up activities:
ACS Chemistry FROG
33
Chapter 6



Chemical Reactions
Investigate This 6.45. What is the three-dimensional structure of EDTA?
Check This 6.46. Keeping the shine in your hair.
End of chapter problems 6.28 through 6.30.
Investigate This 6.45. What is the three-dimensional structure of EDTA?
Goal:
Show that a model of EDTA can be twisted around to a hexadentate structure.
Time for activity:
 Approximately 15-20 minutes. To save time, have students construct the model or parts of the
model as homework before the class session.
Materials:
 Molecular modeling kit.
Procedure:
 Make a model of the ethylenediaminetetraacetate tetraanion.
 Leave off all the hydrogen atoms.
 Use the long yellow connectors to represent the nonbonding electron pairs on nitrogen and
one of the nonbonding electron pairs on each charged oxygen.
 Twist the model to show that all six of the yellow connectors (nonbonding electron pairs)
direct toward a central point.
Anticipated results:
Follow-up discussion:
 Show or direct students to Figure 6.6 to facilitate discussion of the structure of the complex.
Follow-up activities:
 Check This 6.46. Keeping the shine in your hair.
 Check This 6.47. Names and structures of Pt(NH3)2Cl2.
 Consider This 6.48. What is the structure of the nickel-dimethylglyoxime complex?
 Check This 6.49. The porphine ring system in chlorophyll and heme.
 End of chapter problem 6.31.
Consider This 6.48. What is the structure of the nickel-dimethylglyoxime complex?
Goal:
Build a molecular model of the nickel-dimethylglyoxime complex that explains the
stoichiometric and other data for the complex.
Time for activity
 5-10 minutes (if students make models or parts of them before class).
34
ACS Chemistry FROG
Chemical Reactions
Chapter 6
Classroom options:
 Students can build the model (or parts of the model) as a homework assignment. During the
next class session, allow 3-5 minutes for students, working in small groups, to complete the
activity and then share their model(s) and rationales with the class.
 You can bring in a model of the Ni(dmg)2 complex and have students discuss how the model
fits the data. This is not as valuable an experience for your students and does not force them to
use the data to build a satisfactory model.
Instructor notes:
 Review the results from Check This 6.11 before conducting this activity.
Students should reason and conclude:
The structure of the complex should be planar with a square planar array of the nitrogen
atoms around the central metal cation, and the deprotonated ends of the dmg anions opposite
one another with hydrogen bonds between the two oxygen atoms at opposite sides of the
complex.
Follow-up discussion:
 The data from Check This 6.11 give a 2:1 ratio for dmg to Ni. The complex is square planar.
The dmg molecules each lose a proton (Check This 6.11 says the reaction is carried out in
basic solution.), so the complex is neutral. The dmg ions are aligned head (still with proton) to
tail (proton lost) in the complex, so that H bonding from the proton still remaining on each
dmg to the oxygen that has lost its proton works nicely to help spread out the electron charge
density in the two negative centers and further reduce the solubility.
 Make students aware of the large number of metal-ion complexes in biological systems. Show
or refer students to Figure 6.9 and Figure 6.10 for examples.
Follow-up activities:
 End of chapter problem 6.31.
Section 6.7. Lewis Acids and Bases: Electrophiles and Nucleophiles
Learning Objectives for Section 6.7
 Classify a chemical change as a Lewis acid-base reaction based on known reactants and
products, or from experimental observations on the change.
 Define and give examples of Lewis acid-base nucleophile-electrophile reactions.
 Identify the nucleophilic and electrophilic sites in a pair of reactants that react to form a
condensation product, including condensation polymers.
 Predict the product of a nucleophile-electrophile reaction, including condensation
polymerization, between reactants with the functional groups we have introduced.
Investigate This 6.50. Does an acid react with an alcohol?
Goal:
Determine from the change in odor that an acid (salicylic) reacts with an alcohol (methanol) to
yield a new product (methyl salicylate—oil of wintergreen).
Set-up time:
From 5-10 minutes to 15-20 minutes, depending on ready availability of materials and reagents.
Time for activity:
 From less than 10 minutes to 20-30 minutes, depending on the amount of discussion
incorporated with the activity.
ACS Chemistry FROG
35
Chapter 6
Chemical Reactions
NOTE: Many faculty members report that students enjoy this activity. It is the first one where a
reaction product is detected by its pleasant odor.
Materials:
 Two 25200-mm test tubes.
 Water bath at 70 ºC.
Reagents:
 4 g of salicylic acid, C7H6O3.
 10 mL of methanol
 4 drops of concentrated sulfuric acid
 2 ml of saturated aqueous sodium hydrogen carbonate solution
Procedure:
 This activity should be done as a class investigation with students working in small groups to
observe and analyze the results.
SAFETY NOTE
Wear your safety goggles
Methanol is flammable—no flames
Concentrated sulfuric acid is corrosive—wear protective
clothing

Mix about 2 g of solid salicylic acid, 5 mL of methanol, and 2 small drops of concentrated
sulfuric acid in each of the two large test tubes.
 Place one of the test tubes in a water bath at about 70 ºC.
 In about 5 minutes, you can smell the oil of wintergreen. You can continue the reaction for
another 10 minutes.
 When it cools to room temperature, add 1 mL of saturated aqueous sodium hydrogen
carbonate solution to each test tube. Do this with care. The solutions will bubble as carbon
dioxide is evolved by the reaction of the sulfuric acid with the hydrogen carbonate anion. The
gas evolution also carries some the ester out of the solution and helps to make the odor more
easily detectable.
 Carefully smell the contents of both test tubes by wafting air across the mouths of the test
tubes to your nose or wet cotton balls with the mixture from each test tube, place them in
labeled vials and pass them around the class. As a control, a cotton ball wetted with the
original mixture might be included.
Extensions:
 To test whether acid is necessary for the reaction, a third sample containing no acid can be
placed in the hot water bath and treated like the other two.
 This activity can be extended to a mini-lab about esters by exploring other combinations or
possibly done together with a synthesis of aspirin. If you do the latter, you have to be prepared
to explain why the acid carbon acts as the electrophile in one reaction (methyl salicylate
synthesis) while the phenolic –OH on the salicylic acid acts as the nucleophile in the other
(aspirin synthesis).
 Have students look for methyl salicylate in consumer products.
Disposal:
 Dispose in a properly labeled organic waste container according to local ordinances.
Follow-up discussion:
 Use Consider This 6.51 to initiate discussion of this activity.
36
ACS Chemistry FROG
Chemical Reactions
Chapter 6
Follow-up activities:
 Check This 6.52. Centers of positive and negative charge in molecules.
 Check This 6.53. More nucleophile-electrophile reactions.
 Check This 6.54. A condensation reaction.
 Check This 6.55. Polyamide condensation polymers.
 Check This 6.56. Water as a nucleophile.
 Check This 6.57. Polymeric structure of DNA.
 End of chapter problems 6.32 through 6.43.
Consider This 6.51. What is the product of reaction of an acid with an alcohol?
Goal:
Identify the product of the reaction of salicylic acid with methanol by its odor and suggest the
conditions necessary for the reaction.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to answer the questions and then
share their suggested product identity and required conditions with the class.
 This activity can be conducted as an open class discussion.
Time for activity
 Approximately 10 minutes.
Instructor notes:
 Be sure all the students have an opportunity to observe (smell) the results from Investigate
This 6.50.
Students should reason and conclude:
 The heated test tube containing salicylic acid, methanol, and a drop of sulfuric acid has a very
different odor than the starting reaction mixture or the sample that has not been heated. (If you
do a further control experiment heating the salicylic acid and methanol mixture, that sample
will also not have the new odor.)
 The new odor is quite distinctive and many (if not all) students will recognize the wintergreen
odor.
 The conditions required for the reaction appear to be heating and perhaps a large amount of
alcohol compared to salicylic acid as well as the presence of a strong acid (probably to
provide hydronium ion to make the reaction proceed more readily). (If you do the further
control experiment, the conclusion is reinforced that the hydronium ion, as well as heat, is
required.)
Follow-up discussion:
 Introduce and define nucleophile and electrophile.
 Review the identification of reaction centers found for alcohol and aldehyde functional groups
in Chapter 5, Section 5.10, Consider This 5.68(b). Point out that this is a nucleophileelectrophile interaction.
 Discuss the reaction pathway represented in equations (6.33) and (6.34). Try to assure that
students recognize that these are two different ways (molecular level and symbolic) to
represent the same reaction. The structural drawings in equation (6.34) are designed to look
similar in geometry to the space filling structures in equation (6.33). Also, show that the
interaction shown in the first step of equation (6.34) is analogous to the interaction in
Consider This 5.68(b).
ACS Chemistry FROG
37
Chapter 6

Chemical Reactions
The mechanism is more complex than is presented in equation (6.34) but, for general
chemistry students, we felt it was adequate. For your information and perhaps for some
interested students, a more complete mechanism is:
H
H
O
H
C
R
OH
OH2
O
O
(– H2O)
R
C
OH R
R'
O
H
OH2
O
C
R
OR'
R
(– H 2O)
R'
C
C
OH
OH
H
H
O
R
OH2
C
OH
O
O
R'
H
O
H
C
R

OR'
OH2
The product is methyl salicylate and students are asked to give its structure in Check This
6.53(a). You might incorporate Check This 6.52 and 6.53 into the discussion initiated by
Consider This 6.51.
OH
O
C
CH3
O
Follow-up activities:
 Check This 6.52. Centers of positive and negative charge in molecules.
 Check This 6.53. More nucleophile-electrophile reactions.
 Check This 6.54. A condensation reaction.
 Check This 6.55. Polyamide condensation polymers.
 Check This 6.56. Water as a nucleophile.
 Check This 6.57. Polymeric structure of DNA.
 End of chapter problems 6.32 through 6.43.
Section 6.8. Formal Charge
Learning Objectives for Section 6.8
 Use formal charge to explain the rearrangements that some reaction intermediates undergo or
to explain the relative stability of different isomeric Lewis structures.
Consider This 6.59. How do you assign formal charge?
Goal:
Assign formal charges to the cyanate ion, the isocyanate ion, and the intermediate structure
shown in equation (6.35).
38
ACS Chemistry FROG
Chemical Reactions
Chapter 6
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to complete this activity. Then, the
groups can share their Lewis structures.
 This activity could also be assigned as a homework problem and then discussed at the next
class session, either in small groups initially or open class discussion.
 This activity can be conducted as an open class discussion.
Time for activity:
 Ranged from 5-10 minutes to 15-20 minutes.
Instructor notes:
 Answer questions about Worked Example 6.58 before beginning this activity.
Students should reason and conclude:
(a) Three possible Lewis structures for the cyanate ion that obey the octet rule (shown with
the formal charge on each atom and overall charge on the ion) are:
0
0
–1
–1 0
0
–2 0
+1
N
C
O
N
O
N
O
C
C
There has to be a formal charge on at least one atom in the Lewis structures, since the overall
ion has a 1– charge. The first two structures have a –1 formal charge on either O or N, both
of which are electronegative atoms. Since O is more electronegative, we would expect the
first structure to have lower energy (be more stable) than the second. The third structure puts
a –2 formal charge on N and +1 formal charge on O, which is a sum of 1– for the overall
charge. This distribution of the electrons is likely to result in a much higher energy structure
than either of the other two. (We will find later that structures with low formal charges are
more stable than those(a)
with higher
formal
-1 charges.)
0
0
cyanate ion is more stable
N
C
O
Two possible Lewis structures that obey the octet rule for the isocyanate ion (shown with the
formal charge on each atom and overall charge on the ion) are:
–1
-1 +1+1–1 -1
CC N N O O
-
–2
0
There +1
is too much charge separation
in C
this Lewis
N Ostructure for the
isocyanate ion.
In both these structures, there are multiple
atoms with formal charge and/or atoms with
0
-1 - In this Lewis structure for the
0
higher than 1 formal charge.C(Both
of Othese conditions lead to instability, so isocyanate is
N
isocyanate ion, the octet rule is
predicted to be less stable than cyanate. See Check
This 6.60.)
not satisfied.
(b) If the structure of the intermediate in reaction (6.34) is “exploded,” we get:
(b)
H
-1
O
H
H
O
C
C
H
H
H
0
H
C
C
+1
H
0
H
O
H
The formal charges on the oxygens are –1, 0, and +1, so their sum is zero, which is the
overall charge on this electrically neutral intermediate.
ACS Chemistry FROG
39
Chapter 6
Chemical Reactions
Follow-up discussion:
 Point out that an interpretation of formal charge is a change in polarization of an atom when
compared to its polarization in a different bonding situation.
 Reaction intermediates often contain formal charges on several atoms, but when they undergo
electron and atomic core redistribution, the number of formal charges is usually reduced, as
observed in the resulting products.
 You might present examples such as the nitrate ion and the carbonate ion to tie the material
back to the previous chapter that involved delocalized pi orbitals (and/or resonance structures,
if you have chosen to introduce this piece of nomenclature).
Follow-up activities:
 Check This 6.60. Using formal charge.
 Consider This 6.61. How are the correlations in Table 6.3 applied?
 End of chapter problems 6.44 through 6.47.
Consider This 6.61. How are the correlations in Table 6.3 applied?
Goal:
Apply the information presented in Table 6.3 to the intermediate in reaction (6.34), water
reacting with carbon dioxide to produce carbonic acid, and to reaction (6.8).
Classroom options:
 This activity can be conducted as an open class discussion.
Time for activity:
 From 10-15 minutes to 15-20 minutes, depending on the depth of the discussion you wish to
have.
Instructor notes:
 As a part of this activity, discuss the information listed in Table 6.3.
 The reaction and rearrangements asked for here in part (b) are shown in Problem 6.26 and
those for part (c) in Problem 6.39.
Students should reason and conclude:
(a) The intermediate from reaction (6.34) is reproduced below with electron rearrangements
shown and the oxygen atoms labeled for reference. Oxygen (1) has a –1 formal charge and
one of its nonbonding electron pairs changes to become a bonding pair, thus creating a
double bond in the product. Oxygen (iii) has a +1 formal charge and changes a bonding
electron pair (to the hydrogen atom) to a nonbonding pair with the transfer of a proton to a
base, one of the nonbonding electron pairs on oxygen (ii). Although oxygen (ii) has zero
formal charge in the structure shown, it would have a +1 formal charge if this proton transfer
occurs and it would also change a bonding electron pair (to the carbon atom) to a nonbonding
pair and leave the intermediate as a water molecule. Although the reaction is a bit more
complex than this, we can think of the reaction occurring as a concerted series of electron
and atomic core rearrangements that lead to the final product with each oxygen acting as
suggested in Table 6.3.
40
ACS Chemistry FROG
Chemical Reactions
Chapter 6
H O (i)
(ii)
C C
O H
H O
(iii) H
C H
H C
H
H H
H
(b) The intermediate for the nucleophile-electrophile reaction of water with carbon dioxide
is:
(ii) O
(iii)
(i) O C O
H
H
Oxygen atom (i) has a –1 formal charge and is likely to change a nonbonding electron pair to
a bonding pair by gaining a proton from an acid to form an –OH group. Oxygen (ii) has no
formal charge and is likely to remain as it is. Oxygen (iii) has a +1 formal charge and is
likely to change a bonding electron pair to a nonbonding pair by losing a proton to a base.
The acid and base required for these changes could be other water molecules, or the changes
to form carbonic acid could be internal, as shown here:
(ii) O
(iii)
(i) O C O
H
H
(c) Reaction (6.8) written as an electrophile-nucleophile reaction is:
–
O
C
H3CH2C + H
(i) O
O CH3
H
H
H3CH2C C O (ii)
CH3
H
O
H
H3CH2C C O
H
CH3
In the intermediate, oxygen atom (i) has a –1 formal charge and is likely to change a
nonbonding electron pair to a bonding pair by gaining a proton from an acid to form an –OH
group. Oxygen (ii) has a +1 formal charge and is likely to change a bonding electron pair to a
nonbonding pair by losing a proton to a base. As in part (b) the acid and base required for
these changes could be other water molecules, or the changes to form carbonic acid could be
internal, as shown at the end of part (b).
Follow-up discussion:
 Review the main points in the "Reflection and Projection."
Follow-up activities:
 End of chapter problems 6.44 through 6.47.
Section 6.9. Reduction-Oxidation Reactions: Electron Transfer
Learning Objectives for Section 6.9
 Classify a chemical change as an oxidation-reduction reaction based on known reactants and
products, or from experimental observations on the change.
ACS Chemistry FROG
41
Chapter 6
Chemical Reactions

Identify the molecules or ions that are reduced and oxidized and their respective products in
an oxidation-reduction reaction.
 Assign oxidation numbers to all the atoms in a given molecule or ion.
 Balance a given oxidation-reduction reaction in acidic or basic solution by inspection, the
oxidation-number method, or the half-reactions method.
Investigate This 6.62. Do ion-metal reactions between copper and silver occur?
Goal:
Observe that change occurs if copper metal is immersed in a solution of silver ion but not if
silver metal is placed in a solution of copper ion.
Set-up time:
 Ranges from 5-10 minutes to 15-20 minutes, depending on whether the solutions are prepared.
Time for activity:
 Less than 10 minutes.
Materials:
 6-well microtiter plate (2 Petri dishes or 2 small beakers can be used instead).
Reagents:
 0.1 M aqueous AgNO3 solution. (Silver solutions stain the skin; wear gloves.)
 0.1 M aqueous CuSO4·5H2O solution.
 Copper wire.
 Silver wire.
Procedure:
SAFETY NOTE
Wear your safety goggles

To one well of the well plate, add enough 0.1 M aqueous silver nitrate, AgNO3, solution to
give a layer of liquid 2-3 mm deep. To an adjacent well, add about the same volume of 0.1 M
aqueous copper sulfate, CuSO4, solution.
 Use an overhead projector to show the plate on the screen.
 Bend a clean copper wire into a zigzag shape, so it will lie flat in the bottom of a well.
 Place it in the well containing the silver nitrate solution.
 Do the same with a piece of clean silver wire and place it in the copper sulfate well.
NOTE: Clean the copper wire by dipping it in a 1M HCl solution for several seconds and then
rinsing well with water. The silver can be cleaned with a fine emery cloth. Since no reaction
occurs with the silver metal, it is fully recoverable to be used another day.
Anticipated results:
 The blue solution of copper(II) ion containing the silver metal will stay the same; no reaction
occurs. Soon after copper metal is placed in the clear colorless silver(I) ion solution, the
projected image of the wire begins to become fatter and soon looks fuzzy. Direct observation
of the wire in the solution shows that a gray layer builds up on the wire with little silvery
needles covering the surface. After several minutes, the original colorless solution begins to
look light blue. The photo shows the unreacted silver wire-copper(II) ion solution on the left.
The copper wire-silver(I) ion solution reaction on the right is shown after silver metal has
begun to build up on a coil of copper wire, but before the solution is appreciably blue.
42
ACS Chemistry FROG
Chemical Reactions
Chapter 6

Another illustration of this reaction is found in the movie in the Web Companion, Chapter 6,
Section 6.9, page 1.
Disposal:
 Dispose of solutions in a properly labeled heavy metal waste container according to local
ordinances. The silver wire can be used again. Dispose of the solid copper-silver metal
according to local ordinances for metal waste.
Follow-up discussion:
 Use Consider This 6.63 to initiate discussion of the results of this activity.
Follow-up activities:
 Worked Example 6.64. A balanced chemical equation for the Cu/Ag+ reaction.
 Check This 6.65. Balancing simple oxidation-reduction reaction expressions
 Investigate This 6.66. Does Cu(s) react with nitric and/or hydrochloric acid?
 Consider This 6.67. What are the products of Cu(s) reactions with hydrochloric and nitric
acids?
 Consider This 6.68. What is the reaction of Cu(s) with H3O+(aq)?
 End of chapter problems 6.48 through 6.55.
Consider This 6.63. What copper and silver ion-metal reactions occur?
Goal:
Based upon their observations from Investigate This 6.62, students decide that copper metal
reacts with silver ion to produce silver metal and copper ion.
Classroom options:
 This activity can be conducted as an open class discussion while Investigate This 6.62 is
taking place.
Time for activity:
 5-10 minutes.
Instructor notes:
 Make sure students agree on the results of Investigate This 6.62 before discussing them.
Students should reason and conclude:
(a) In the well containing silver metal and copper ion, no changes occur, so we conclude that
there is no apparent reaction. In the well containing copper metal and silver ion, silvery/black
metal strands appear on the copper surface and the solution becomes pale blue in color.
(b) The movie in the Web Companion, Chapter 6, Section 6.9, page 1, shows a copper coin
immersed in silver nitrate solution and how the spongy silver-gray coating that forms on the
coin can be scraped off to reveal that the surface of the coin has been eroded as the copper
reacted. Still pictures show the contents of the solutions, a representation of the metals, as
well as the net ionic equation for the reaction.
(c) The net ionic reaction for the process that proceeds is:
ACS Chemistry FROG
43
Chapter 6
Chemical Reactions
Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
Follow-up discussion:
 Define oxidation-reduction reactions (or lead students to propose definitions) and how these
definitions apply to Investigate This 6.61.
Follow-up activities:
 Worked Example 6.64. A balanced chemical equation for the Cu/Ag+ reaction.
 Check This 6.65. Balancing simple oxidation-reduction reaction expressions
 Investigate This 6.66. Does Cu(s) react with nitric and/or hydrochloric acid?
 Consider This 6.67. What are the products of Cu(s) reactions with hydrochloric and nitric
acids?
 Consider This 6.68. What is the reaction of Cu(s) with H3O+(aq)?
 End of chapter problems 6.48 through 6.55.
Investigate This 6.66. Does Cu(s) react with nitric and/or hydrochloric acids?
Goal:
Observe that Cu(s) reacts vigorously with nitric acid to produce a colored solution and the
colored gas, but that no apparent reaction occurs with hydrochloric acid.
Set-up time:
 Approximately 15 minutes.
Time for activity:
 From less than 10 minutes to 15-20 minutes, depending on the amount of discussion and
interpretation that is done during the investigation.
Materials:
 Two 400-mL beakers.
 Two 100-mL beakers.
 Two 10-mL graduated cylinders (or plastic transfer pipets that will hold about 5 mL).
 Four 1-gallon zip-closure plastic bags.
Reagents:
 5 mL concentrated nitric acid, HNO3.
 5 mL concentrated hydrochloric acid, HCl.
 Two 0.25 g lengths of copper wire. (Do NOT use a larger amount.)
NOTE: Clean the copper wire before class by briefly immersing it in 1 M HCl and then rinsing
it with water. This process removes the coating of oxide and other salts that form on copper.
Thus, this change in appearance will not occur during the investigation to confuse students.
Procedure:
SAFETY NOTES
Wear your safety goggles and gloves to handle the
reagents.
Concentrated nitric acid and hydrochloric acid are
powerfully corrosive.
Some products of the reaction are toxic and must be kept
in the plastic bags and disposed properly in a fume hood.
NOTE: If you have the facilities to do so, you can conduct this activity in a hood.
 Carry out this reaction in a "double-bagged" plastic zip-closure bag by placing one zip closure
bag inside another. This set-up will contain any fumes and odors.
 Place a 400-mL beaker about three-quarters full of water in the bag.
44
ACS Chemistry FROG
Chemical Reactions
Chapter 6

Being careful not to spill it, place a 100-mL beaker containing 5 mL of concentrated nitric
acid into the bag.
 Wind 0.25 gram of copper wire into a flat spiral that will fit on the bottom of the 100-mL
beaker and add the copper to the bag.
 Seal the bags, checking that each is well sealed.
 Using your hands on the outside of the bag, carefully pick up the spiral of copper wire and
drop it into the 100-mL beaker of nitric acid.
 Students in small groups should observe and record any changes that occur.
 Repeat with concentrated hydrochloric acid in place of the nitric acid.
Notes/Warnings:
 DO NOT USE A PENNY for this experiment, even if you have a real copper, pre-1982
penny. There is enough copper in the penny to create more NO2(g) than you would want and
you would probably find yourself in a similar quandary to that of the young Ira Remsen when
he determined to find out what it meant when he read that “nitric acid acts upon copper.” In
the gallon-size zip-seal bag, there is plenty of slack in the plastic to allow the manipulations
described in the activity and to contain the NO2 that is formed from 0.25 g of copper.
Anticipated results:
 This series of photos shows the set up (in a single plastic bag) and the reaction with nitric acid
shortly after it begins and when it is finished a minute or two later:

In the nitric acid reaction with copper, the reddish-brown color of NO2(g) is obvious during
the reaction. The blue-green color of the Cu2+(aq) in the small beaker [combination of the
color of the still-dissolved NO2 and the copper(II) ion] and blue color of the solution after it is
diluted is the clue to its formation.
 No apparent reaction occurs with hydrochloric acid (much to the disappointment of the class).
Disposal:
 To dispose of the contents of the bag, you can open it in a hood to vent NO2(g) and pour the
diluted nitric acid (less than about 0.25 M at this point) down the drain with lots of running
water. If your institution has a Cu2+(aq) disposal requirement, dispose in a properly labeled
container. NO2(g) will dissolve in the water in the baggie, if given enough time.
 Dispose of the diluted hydrochloric acid in the same way. No Cu2+(aq) contaminates this
solution.
Follow-up discussion:
 Use Consider This 6.67 to initiate discussion of the results of this activity and Consider This
6.68 to continue the discussion, once the redox reaction between copper and nitric acid has
been established.
Follow-up activities:
 Consider This 6.68. What is the reaction of Cu(s) with H3O+(aq)?
ACS Chemistry FROG
45
Chapter 6

Chemical Reactions
End of chapter problems 6.48 through 6.55.
Consider This 6.67. What are the products of Cu(s) reactions with nitric and/or
hydrochloric acid?
Goal:
Based on their observations from Investigate This 6.66, students determine the products of
reaction of Cu(s) with nitric and hydrochloric acids.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to complete this activity. Then, the
instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so.
 This activity could also be assigned as a homework problem and then discussed at the next
class session, either in small groups initially or open class discussion.
 This activity can be conducted as an open class discussion.
Instructor notes:
 Be sure the class agrees on the results from Investigate This 6.66 before conducting this
activity and discussion.
Students should reason and conclude:
(a) Copper reacts with nitric acid. The evidence is that a reddish-brown gas and bubbles are
observed when copper metal is placed in concentrated nitric acid. The solution turns bluegreen and when diluted is blue. In Investigate This 6.62, we saw that an aqueous solution of
copper(II) ions is blue, so we can identify one of the reaction products as Cu2+(aq). [Few
students will know that the brown gas is NO2, but the text following this activity says that
NO2 is a reddish-brown gas, so the identification is made quickly.] There was no visible
change when copper was added to hydrochloric acid, so we conclude that copper does not
react with hydrochloric acid under these conditions.
(b) The vigorous reaction between copper and nitric acid must be a reduction-oxidation
reaction, because copper metal is oxidized to copper(II) ion in the reaction. Whatever is
reduced could be the source of the reddish-brown gas. There is no apparent reaction of any
kind between copper and hydrochloric acid.
Follow-up discussion:
 Discuss which reactant is oxidized and which is reduced and the evidence for the answers.
 Use the discussion to lead up to the overall balanced redox reaction, after ascertaining in
Consider This 6.68 that copper does not react with hydronium ion.
NOTE: The balanced reduction-oxidation reaction is given in Worked Example 6.78.
Follow-up activities:
 Consider This 6.68. What is the reaction of Cu(s) with H3O+(aq)?
 Check This 6.69. Oxidation numbers for elements and monatomic ions
 Check This 6.70. The procedure and definition of oxidation number.
 Worked Example 6.71. Assigning oxidation numbers.
 Check This 6.72. Assigning oxidation numbers.
 Check This 6.73. Oxidation-reduction reactions.
 Consider This 6.74. Are there patterns in oxidation numbers?
 Check This 6.75. Oxidation number for oxygen in peroxides.
 Worked Example 6.76. Assigning oxidation numbers.
 Check This 6.77. Assigning oxidation numbers.
46
ACS Chemistry FROG
Chemical Reactions

Chapter 6
End of chapter problems 6.48 through 6.55.
Consider This 6.68. What is the reaction of Cu(s) with H3O+(aq)?
Goal:
Conclude that there is no reaction of Cu(s) with H3O+(aq).
Classroom options:
 This activity can be conducted as an open class discussion following the finding that copper
and nitric acid undergo a redox reaction.
 This activity can also be assigned as a homework problem and then discussed at the next class
session, either in small groups initially or open class discussion.
Time for activity:
 5-10 minutes
Instructor notes:
 Review results from Investigate This 6.65 before conducting this activity.
Students should reason and conclude:
 There is no apparent reaction when Cu(s) is added to hydrochloric acid. In particular, no
bubbles of gas are formed. Therefore, we can conclude that Cu(s) does not react with
H3O+(aq) to produce hydrogen gas, H2(g), by a reaction analogous to that for Zn(s).
Follow-up discussion:
 Introduce and define oxidation numbers, in order to move toward understanding the reaction
between copper metal and nitric acid.
Follow-up activities:
 Check This 6.69. Oxidation numbers for elements and monatomic ions
 Check This 6.70. The procedure and definition of oxidation number.
 Worked Example 6.71. Assigning oxidation numbers.
 Check This 6.72. Assigning oxidation numbers.
 Check This 6.73. Oxidation-reduction reactions.
 Consider This 6.74. Are there patterns in oxidation numbers?
 Check This 6.75. Oxidation number for oxygen in peroxides.
 Worked Example 6.76. Assigning oxidation numbers.
 Check This 6.77. Assigning oxidation numbers.
 End of chapter problems 6.48 through 6.55.
Consider This 6.74. Are there patterns in oxidation numbers?
Goal:
Conclude that oxygen atoms in molecules usually have a –2 oxidation number and hydrogen
atoms a +1 oxidation number.
Classroom options:
 This activity can be conducted as an open class discussion.
 This activity can also be assigned as a homework problem and then discussed at the next class
session, either in small groups initially or open class discussion.
Time for activity:
 10-15 minutes
Instructor notes:
 Be sure the class agrees on the correct results from Worked Example 6.71 and Check This
6.72 and 6.73 before conducting this activity.
Students should reason and conclude:
ACS Chemistry FROG
47
Chapter 6
Chemical Reactions
(a) The oxygen atom has an oxidation number of –2 except in H2O2 where the oxygen atoms
have an oxidation number of –1. The oxygen atom is more electronegative than any other
atom (except fluorine), so, in determining oxidation numbers, oxygen is assigned all the
bonding electrons, which gives it a –2 oxidation number. The exception in H2O2 is because
the oxygen atoms are bonded to each other and each is assigned only one of the bonding
electron pair, thus making the oxidation number of each oxygen atom increase by 1 to +1.
(b) In the H-containing compounds that we have considered, oxidation number for H is +1,
except for when it is bonded to carbon when its oxidation number is 0. In most of the
compounds we have met so far, hydrogen atoms are bonded to carbon or elements more
electronegative than carbon. For the more electronegative elements, the bonding electron pair
between H and the other element is always assigned to the more electronegative element,
leaving H+, with an oxidation number of +1. Our rule for carbon assigns one of the bonding
pair of electrons to H and other to C, so the hydrogen atom formed has a zero oxidation
number.
Follow-up discussion:
 Discuss Table 6.5, which provides an alternative set of rules for assigning oxidation numbers
based on initial assignments of the oxidation numbers for oxygen and hydrogen atoms in
compounds.
Follow-up activities:
 Check This 6.75. Oxidation number for oxygen in peroxides.
 Worked Example 6.76. Assigning oxidation numbers.
 Check This 6.77. Assigning oxidation numbers.
 End of chapter problems 6.48 through 6.55.
Section 6.10. Balancing Reduction Oxidation Reaction Equations
Learning Objectives for Section 6.10
 Identify the molecules or ions that are reduced and oxidized and their respective products in
an oxidation-reduction reaction.
 Balance a given oxidation-reduction reaction in acidic or basic solution by inspection, the
oxidation-number method, or the half-reactions method.
 Use your knowledge of the oxidation numbers of atoms in various molecules or ions to predict
possible reduced or oxidized products from a reaction.
Investigate This 6.80. What happens when bleach and iodide are mixed?
Goal:
Determine that an acidified mixture of clear, colorless bleach and iodide produce a reddish
orange solution.
Set-up time:
 5 minutes (assuming solutions have been prepared previously).
Time for activity:
 5-10 minutes including discussion.
Materials:
 One 50- or 100-mL Erlenmeyer flask.
 Two long-stem plastic pipets containing the acid and bleach solutions.
48
ACS Chemistry FROG
Chemical Reactions
Chapter 6
Reagents:
 1 mL of 6 M sulfuric acid solution, (HO)2SO2(aq).
 20 mL of 20% aqueous potassium iodide, KI. (If this is not made up fresh, it may appear
yellowish, due to air oxidation of the iodide to iodine. Add a tiny crystal of sodium
thiosulfate, Na2S2O3·5H2O, and mix to reduce the iodine and produce a clear, colorless
solution. Acidified solutions air-oxidize much faster, which is why the solution is acidified
here just before it is used.)
 Few drops of household bleach.
Procedure:
SAFETY NOTE
Wear your safety goggles.

Mix 1 mL of 6 M aqueous sulfuric acid, (HO)2SO2 with 20 mL of 20% aqueous potassium
iodide, KI, solution in the flask. Swirl to mix and have the groups record their observations.
 Add a few drops of household bleach to the flask, swirl to mix the contents, and have the
groups record their observations. The solution should now appear dark red-orange, due to the
presence of iodine, I2 [as I3–(aq) in this solution].
Anticipated results:
 These photos show a flask containing acidified iodide solution on the left and, on the right, the
flask after a few drops of bleach are added and the flask swirled to mix the reactants:
Disposal:
 Wash the solution down the drain with copious amounts of water.
Follow-up discussion:
 Use Consider This 6.81 to initiate discussion of the results of this activity.
Follow-up activities:
 Worked Example 6.82. Balancing redox equations by the half-reactions method.
 Check This 6.83. Balancing redox equations by the half-reactions method.
 Worked Example 6.84. Balancing redox equations in basic solution.
 Check This 6.85. Balancing redox equation in basic solution.
 End of chapter problems 6.56 through 6.60.
Consider This 6.81. What is the reaction between bleach and iodide?
Goal:
Conclude that the reaction between bleach and iodide in acidic solution is a redox reaction
producing iodine and possibly chloride.
ACS Chemistry FROG
49
Chapter 6
Chemical Reactions
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to complete this activity. Then, the
instructor can lead the discussion, summarizing answers on the chalkboard or an overhead
transparency or have students do so.
 This activity could also be assigned as a homework problem and then discussed at the next
class session, either in small groups initially or open class discussion.
 This activity could be conducted as an open class discussion.
Instructor notes:
 Be sure the class agrees on the observations from Investigate This 6.80 before doing this
activity.
Students should reason and conclude:
(a) When clear, colorless acid solution was added to the clear, colorless potassium iodide
solution, no change was observed, so we conclude that acid and iodide ion do not react with
one another. When bleach was added to the acidified iodide solution, the color of the solution
changed to dark red-orange indicating that a reaction had occurred between bleach and acidic
potassium iodide solution.
(b) We know that iodine is a highly colored solid and its solutions are probably also highly
colored, so it seems reasonable to conclude that one of the reaction products is elemental
iodine, I2. Since production of iodine (ON = 0) from iodide (ON = –1) is a reduction, this
must be a reduction-oxidation reaction.
(c) The chlorine atom in the hypochlorite ion has an ON = +1. Since iodide is oxidized, the
chlorine atom in the hypochlorite ion is probably reduced. A likely product would be the
chloride ion, Cl– (ON = –1). Chloride ion in solution is colorless, so no evidence for its
formation could be observed in the investigation. Further analysis would be needed. Another
possible product would be elemental chlorine, Cl2 (ON = 0), but chlorine is a gas and there is
no evidence of gas formation in the reaction system. [Even if elemental chlorine were
formed, it would react rapidly with iodide to produce iodine and chloride, so it would never
be observed.]
Follow-up discussion:
 Investigate This 6.80 and Consider This 6.81 are designed to be the lead in for you to
introduce the half-reactions method for balancing oxidation-reduction reactions. This is the
reaction that is used in the Worked Example 6.82 after the procedure is presented in Table 6.7.
Follow-up activities:
 Worked Example 6.82. Balancing redox equations by the half-reactions method.
 Check This 6.83. Balancing redox equations by the half-reactions method.
 Worked Example 6.84. Balancing redox equations in basic solution.
 Check This 6.85. Balancing redox equation in basic solution.
 End of chapter problems 6.56 through 6.60.
Section 6.11. Reduction-Oxidation Reactions of Carbon-containing Molecules
Learning Objectives for Section 6.11.
 Classify a chemical change as an oxidation-reduction reaction based on known reactants and
products, or from experimental observations on the change.
 Identify the molecules or ions that are reduced and oxidized and their respective products in
an oxidation-reduction reaction.
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ACS Chemistry FROG
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Chapter 6


Assign oxidation numbers to all the atoms in a given molecule or ion.
Balance a given oxidation-reduction reaction in acidic or basic solution by inspection, the
oxidation-number method, or the half-reactions method.
 Use your knowledge of the oxidation numbers of atoms in various molecules or ions to predict
possible reduced or oxidized products from a reaction.
 Use oxidation numbers to show that a given reaction is an internal oxidation-reduction
Investigate This 6.86. Does methanal react with silver ion?
Goal
Conclude that methanal reacts to reduce silver ion.
Set-up time:
 2-3 hours (includes test tube cleaning).
Time for activity:
 5-10 minutes (including discussion).
Materials:
 One CLEAN 200-mm test tube. Use this procedure to clean the test tube. Place about 2-3 mL
of concentrated nitric acid in the test tube, clamp the tube upright in a boiling water bath in a
hood, and cover the mouth with a small watch glass. Let the test tube heat for about 2 hours,
then remove from the bath and allow the tube and contents to cool. Discard the cooled acid
with copious amounts of water down the drain and thoroughly rinse the test tube with tap
water followed by a couple of rinses with distilled or deionized water.
NOTE: This same procedure can be used to remove the mirror and reclean the test tube after the
activity.
Reagents:
 Concentrated nitric acid, HNO3 (to clean the test tube).
 1 g silver nitrate, AgNO3(s) for Tollens reagent (see directions below for preparation).
 1 g sodium hydroxide, NaOH(s) for Tollens reagent (see directions below for preparation).
 Concentrated ammonia solution, NH3(aq), diluted 1:10 with water. Use for Tollens reagent
(see directions below for preparation).
 5-10% aqueous methanal (formaldehyde). Alternatively, a solution of glucose can be used.
Procedure:
SAFETY NOTES
Wear your safety goggles.
The basic silver Tollens solution is caustic and can harm
clothes and skin. Handle with protective gloves.
Tollens reagent is unstable and, on standing, can form
explosive silver fulminate.
 Preparation of Tollens reagent:
WARNING: DO NOT PREPARE MORE OF THE MIXTURE THAN YOU WILL USE. The
stock solutions used to prepare Tollens reagent are stable, but the mixture is unstable, and, on
standing, can form explosive silver fulminate.
(A) 1 g of silver nitrate, AgNO3, dissolved in 10 mL water.
(B) 1 g sodium hydroxide, NaOH, dissolved in 10 mL water.
(C) Ammonia solution made by diluting concentrated ammonia 1:10 with water.
ACS Chemistry FROG
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Chapter 6
Chemical Reactions
Just before doing the activity, mix 2 mL each of solutions (A) and (B) and then add solution
(C) dropwise, while swirling the test tube, until the precipitate of hydrous silver oxide just
dissolves. This solution should be essentially clear and colorless.
NOTE: You can do this preparation in the test tube to be used for the activity. Or, if you want to
do more than one reaction or test more than one reducing agent, mix a larger batch and use
aliquots.
 Add to the test tube, containing 4 mL of the Tollens reagent a few drops of an aqueous
solution of methanal (formaldehyde), H2CO.
 Swirl the test tube to mix the reagents and continue to swirl for at least a minute or two as
silver metal deposits on the inside wall of the test tube forming a silver mirror.
Alternative Activity:
 One instructor has suggested substituting the Benedict’s test for reducing sugars for the silver
mirror (Tollens) test and using the EDTA version of Benedict's solution (J. Chem. Ed. 1994,
71, 345). The reducing sugar test demonstrates that glucose, but not sucrose, reduces
copper(II) ions (blue solution) to copper(I) oxide (reddish-brown solid). After recognizing that
copper(II) ions were reduced, students realized that the only difference between the two
solutions was the sugar, and that the glucose must have been the reducing agent. The
instructor used the opportunity to introduce the intramolecular aldehyde-alcohol reaction that
leads to the closed form of glucose that many of the students are familiar with from biology
(see end of chapter problem 6.40), and we have been using in this text. After comparing the
structures of glucose and sucrose, it was pointed out why the sucrose rings, unlike the glucose
rings, cannot open up and therefore why sucrose cannot reduce the copper(II) ions. This
instructor felt that it was a good end of chapter activity because it brought together many of
the topics, including formal charge, nucleophile-electrophile reactions, and oxidation
numbers. [Benedict’s test is introduced and analyzed in Chapter 10, Section 10.8. If this
alternative is used, students can be referred back to what has been done here instead of
redoing it. You will have to introduce a bit more chemistry, if you use the alternative. As
noted, the formation of the internal hemiacetal is introduced in the end of chapter problems,
so, if you are planning to use those problems, this alternative activity could fit in well. If you
would prefer not to introduce more complexity at this stage, then this alternative is a poor
choice.]
Disposal:
 When the activity with 4 mL is complete, pour the liquid contents of the test tube into about
150 mL of 0.1 M hydrochloric acid, HCl, in a 600 mL beaker. The caustic solution is
destroyed by reaction with the acid and any residual silver ion precipitates as silver chloride.
The silver chloride can be collected and disposed of with other heavy metals.
Follow-up discussion:
 Use Consider This 6.87 to initiate discussion of this activity.
Follow-up activities:
 Consider This 6.88. What oxidation numbers are available for carbon?
 Check This 6.89. Balancing the methanal-silver ion reaction equation.
 Check This 6.90. Oxidation numbers for carbons in glucose fermentation.
 Check This 6.91. Net glycolysis reaction.
 Check This 6.92. Reduction-oxidation of glucose.
 End of chapter problems 6.61 through 6.64.
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Consider This 6.87. What is the reaction of methanal with silver ion?
Goal:
Conclude, based on the results of Investigate This 6.86, that the reaction is a redox reaction and
one of the products is silver metal.
Classroom options:
 This activity can be conducted as an open class discussion begun as the silver mirror begins to
form in Investigate This 6.86.
Instructor notes:
 Be sure the class agrees on the results of Investigate This 6.86. If you use reducing sugars
instead of methanal for the Tollens test, you will either need to amend some of the discussion
following this activity, or simply not inform the class that you used a different reducing agent.
[If you opt to do the alternative activity described above in Investigate This 6.86, you will
need to change the discussion here to focus on the reduction of copper(II) to copper(I).]
Students should reason and conclude:
Silver metal forms on the test tube wall, so one product of the reaction is Ag(s), ON = 0.
Since the silver is in the original solution as Ag(I), reduction of the silver has occurred, and
this is a reduction-oxidation reaction. The other product of the reaction, presumably an
oxidized species formed from methanal, cannot be directly identified from the observations
in Investigate This 6.86. [See the reasoning in Consider This 6.88.]
Follow-up discussion:
 Focus on figuring out what the oxidation product from methanal must be.
Follow-up activities:
 Consider This 6.88. What oxidation numbers are available for carbon?
 Check This 6.89. Balancing the methanal-silver ion reaction equation.
 Check This 6.90. Oxidation numbers for carbons in glucose fermentation.
 Check This 6.91. Net glycolysis reaction.
 Check This 6.92. Reduction-oxidation of glucose.
 End of chapter problems 6.61 through 6.64.
Consider This 6.88. What oxidation numbers are available for carbon?
Goal:
Conclude that oxidation numbers of 0, +1, +2, +3, and +4 are possible for carbon and that
methanoic (formic) acid is the most likely methanal oxidation product with silver(I) as the
oxidizing agent.
Classroom options:
 Allow 3-5 minutes for students, working in small groups, to complete this activity. Then, the
groups can share their answers with the class to begin the discussion.
 This activity could also be assigned as a homework problem and then discussed at the next
class session, either in small groups initially or open class discussion.
 This activity could be conducted as an open class discussion.
Instructor notes:
 For part (a), suggest that students draw Lewis structures before determining the oxidation
numbers for each of the atoms in the compounds.
Students should reason and conclude:
ACS Chemistry FROG
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Chapter 6
Chemical Reactions
(a)
H
0
H
0
H0
0
C
H
-2
+1
+1
C
O
H
0
H
0
H0
H0
-2
-2
O
0
H
O
C+2
H
0
0
H
C+3
-2 +1
O
H
-2
O
+4
C
-2
O
The oxidation numbers shown in red are derived from the rules in Table 6.5. In each case, the
oxidation numbers for H and/or O are assigned first and then the oxidation number for C is
assigned to make the sum of oxidation numbers zero in each case.
(b) Yes, there is a correlation between the number of bonding pairs of electrons between
carbon and oxygen and the oxidation number of carbon. The number of bonding pairs of
electrons between carbon and oxygen gives the oxidation number of carbon. Another way to
state this is to say that the oxidation number of carbon is equal to the number of bonds carbon
makes to oxygen.
(c) The oxidation product from methanal could be either methanoic (formic) acid or carbon
dioxide. Since there is no evidence for the formation of a gas (no bubbles) in the reaction, the
oxidized product is likely to be methanoic acid.
Follow-up discussion:
 Use the discussion to review how equation (6.74) is obtained from equation (6.73).
Follow-up activities:
 Check This 6.89. Balancing the methanal-silver ion reaction equation.
 Check This 6.90. Oxidation numbers for carbons in glucose fermentation.
 Check This 6.91. Net glycolysis reaction.
 Check This 6.92. Reduction-oxidation of glucose.
 End of chapter problems 6.61 through 6.64.
Section 6.13. Extension: Titration
Learning Objectives for Section 6.13.
 Explain how a titration is carried out and used to determine the amount of a reactant present in
a solution of unknown concentration.
 Use titration data to determine the amount of a reactant present in a solution of unknown
concentration.
Investigate This 6.93. How can you analyze an acid-base reaction?
Goal:
Note the color of an acid-base indicator as increasing amounts of hydroxide are added to equal
volumes of ethanoic (acetic) acid solution.
Set-up time:
 5-10 minutes (assuming reagents are made prior to activity).
Time for activity:
 5-10 minutes.
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Chapter 6
Materials:
 24-wll microtiter plate.
 Overhead projector.
Reagents:
 0.10 M aqueous sodium hydroxide, NaOH. Dissolve 0.40 g NaOH(s) in 100. mL water.
 0.10 M aqueous ethanoic (acetic) acid, CH3C(O)OH. Add 5.7 mL glacial acetic acid to about
500 mL water and bring the volume to 1000. mL.
NOTE: These solutions should be made up reasonably accurately. The results depend on the
solutions being the same concentration, so that 0.8 mL of the base does not react with all the
acid in 1.0 mL of acid, but 1.0 mL of the base does. The activity is a titration.
 Bromocresol green acid-base indicator. Dissolve about 40 mg of solid bromocresol green
indicator in 6 mL of 0.01 M NaOH solution and dilute to 100 mL with water.
Procedure:
SAFETY NOTE
Wear your safety goggles.

Add two drops of bromocresol green acid-base indicator solution to about 20 mL of the 0.10
M aqueous solution of ethanoic acid and swirl to mix well.
 Use a calibrated-stem pipet to add 1.0 mL of this solution to each of eight wells in a 24-well
microtiter plate.
 Leave the solution in the first well as it is.
 Add 0.20 mL of 0.10 M aqueous sodium hydroxide, NaOH solutions to the second well, 0.40
mL to the third, 0.60 mL to the fourth, and so on to 1.40 mL to the eighth well.
 Place the plate on an overhead projector so students can view the plate and the colors of the
solutions in the wells.
 Have students record the color in each of the wells and discuss the colors and differences in
color from one well to the next, until the class agrees on these descriptions.
Anticipated Results:
 The identity (names) of the colors assigned by the class may be somewhat different than those
in this table, but there should be a definite greenish hue for wells two through five and the
same blue color in wells six through eight.
Added Base
Color
none
Yellow
0.20 mL
Light green
0.40 mL
Green
0.60 mL
Turquoise
0.80 mL
Blue green
1.00 mL
Blue
1.20 mL
Blue
1.40 mL
Blue
ACS Chemistry FROG
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Chapter 6
Chemical Reactions
NOTE: It is wise to check these solutions before doing the activity. Equal volumes of the acid
and base (with indicator) should be blue, but solutions that have less than an equivalent of base
should show a definite green hue (mixture of yellow and blue). The indicator in the acid is
yellow. Addition of some base begins to turn the solution green and then blue at equivalence
and beyond. It is OK to dilute the samples as is done in this activity, since the intermediate
solutions are buffered and dilution has little effect on the pH. Since the solutions are being
viewed through columns of liquid containing the same number of moles of indicator, the
dilution doesn’t affect the intensity of the observed color. If students are uncomfortable with the
changing volumes, add enough water to each sample to make the volumes all 2.4 mL. The
activity can be done on a larger scale, if desired, but, if viewed from the side, the final volumes
should all be the same.
 This photograph shows the results, but the camera favors the blue a bit, so well five looks
almost the same blue as wells six through eight. When the solutions are correctly prepared, the
difference is easy to detect.
Disposal:
 Discard solutions down the drain with plenty of water. Rinse plates well with water.
Follow-up discussion:
 Use Consider This 6.94 to initiate discussion of the results of this activity.
Follow-up activities:
 Check This 6.95. Acid-base equivalence in Investigate This 6.93.
 Worked Example 6.96. An acid-base titration.
 Check This 6.97. An acid-base titration.
 End of chapter problems 6.65 and 6.66
Consider This 6.94. How can you follow an acid-base reaction?
Goal:
Conclude and explain, based on the results from Investigate This 6.93, that the stoichiometry of
an acid-base reaction can be followed using an acid-base indicator.
Classroom options:
 This activity can be conducted as an open class discussion immediately following Investigate
This 6.93.
Instructor notes:
 Students should be viewing the results form Investigate This 6.93 as they carry out this
activity.
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Chapter 6
Students should reason and conclude:
The color pattern and a photograph are shown in the Results for Investigate This 6.93. As
increasing amounts of hydroxide ion are added to the acid solutions, the color changes from
yellow through a series of green and blue-green solutions to a blue color that is the same for
the last three additions of base. The constant blue color is obtained when 1.00 mL or more of
the 0.010 M hydroxide solution is added to 1.0 mL of the acid solution. It seems likely that
the green color is a mixture of the original yellow and the final blue. The color gets more
blue as more base is added, until the final blue is obtained. There is a correlation of the color
with amount of added base.
Follow-up discussion:
 Show or refer students to equation (6.79), which shows the reaction of hydroxide ion with
ethanoic acid.
 Recall Table 6.2 to show that hydroxide ion is a stronger base than ethanoate ion.
 Make sure students understand that if more moles of hydroxide ions than ethanoic acid are
mixed, reaction (6.79) occurs until all the ethanoic acid is used up. The resulting solution will
contain high concentrations of both hydroxide and ethanoate anions.
Follow-up activities:
 Check This 6.95. Acid-base equivalence in Investigate This 6.93.
 Worked Example 6.96. An acid-base titration.
 Check This 6.97. An acid-base titration.
 End of chapter problems 6.65 and 6.66
Investigate This 6.98. What other titrations are possible?
Goal:
Determine the volume of aqueous iodine solution required to react with (titrate) a given volume
of aqueous ascorbic acid.
Set-up time:
 5-10 minutes (assuming reagents are prepared).
Time for activity:
 10-15 minutes.
Materials:
 Two small test tubes per group.
 Three thin-stem plastic pipets per group.
Reagents:
 0.0050 M ascorbic acid, C6H8O6. Dissolve 0.22 g ascorbic acid in 250 mL of water. Dispense
in labeled thin-stem plastic pipets.
 0.0050 M iodine (actually triiodide) solution. Dissolve 1.25 g of potassium iodide, KI, in 100
mL of water. To this solution, add 0.127 g of iodine, I2, and stir to dissolve (which may take
some time). Store the triiodide solution in a glass container. Iodine is absorbed by plastic, so a
glass container is essential. Iodine vapor will react with just about any kind of cap used to seal
the bottle, so it cannot be stored indefinitely. Dispense in labeled thin-stem pipets that are
filled as short a time before use as possible. The pipets will be stained by the iodine, but can
be used again for iodine solutions, if desired.
Procedure:
SAFETY NOTE
Wear your safety goggles
ACS Chemistry FROG
57
Chapter 6
Chemical Reactions

Students can work in small groups to do this investigation and to discuss and analyze the
results. The student who does the actual manipulations in each part should wear safety
goggles.
 Each group will use two small test tubes and three, labeled thin-stem plastic pipets containing,
respectively, water, 0.0050 M aqueous ascorbic acid (vitamin C) solution, and 0.0050 M
aqueous iodine, I2 [I3–(aq)], solution.
(a)
 In one test tube, mix 20 drops of water and 3 drops of iodine solution. Students should record
their observations on the appearance of the reagents and mixture.
 In the second test tube, mix 20 drops of ascorbic acid solution and 3 drops of iodine solution.
Students should record their observations on the appearance of the reagents and mixture.
(b)
 Continue to mix iodine solution, 2-3 drops at a time, with the ascorbic acid solution while
keeping a count of the total number of drops added.
 Stop adding at the first sign that the ascorbic acid has all reacted. Record the total number of
drops required.
Anticipated results:
(a) Addition of 3 drops of the red-orange solution of iodine (triiodide) to 20 drops of water
should give a light yellow-orange solution. Addition of 3 drops of the red-orange solution of
iodine (triiodide) to 20 drops of the clear, colorless ascorbic acid solution should give a clear
colorless solution. The ascorbic acid reduces iodine to iodide.
(b) Iodine (triiodide) and ascorbic acid react in a 1:1 mole ratio. Assuming that the students
use 20 drops of 0.0050 M ascorbic acid, it should require about 20 drops of 0.0050 M iodine
to titrate the ascorbic acid. The end point in the titration is signaled by the solution becoming
colored (yellow), when all the ascorbic acid has reacted.
Disposal:
 Discard solutions in the test tubes down the drain, rinsing with copious amounts of water.
Follow-up discussion:
 Since the technique, as presented, is crude (to make it fast) the drop ratio will probably not be
exactly 20:20. You can help students interpret the data by suggesting that the molar
equivalencies have to be a simple ratio of whole numbers; a little variation is expected
because of the experimental uncertainties. Alternately, you can do the titration on a larger
scale in the standard way with a buret. The titration is fast and can be done to 1-5% precision
in a couple of minutes.
 Use Consider This 6.99 to initiate discussion of this activity.
Follow-up activities:
 Check This 6.100. Ascorbic acid oxidation.
 End of chapter problems 6.65 and 6.66.
Consider This 6.99. What is the stoichiometry of the iodine-ascorbic acid reaction?
Goal:
Conclude, based on the results from Investigate This 6.98, that iodine and ascorbic acid react in
a one-to-one stoichiometric ratio.
Classroom options:
 This activity can be conducted as an open-class discussion after the groups report their results
from Investigate This 6.98.
58
ACS Chemistry FROG
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Chapter 6
Instructor notes:
 Discuss the results from Investigate This 6.98, being sure the class agrees on the observations
and on a class “average” ratio of drops of iodine solution to drops of ascorbic acid solution
required for complete reaction of the ascorbic acid.
Students should reason and conclude:
(a) When 3 drops of the red-orange iodine solution is added to 20 drops of water, the
resulting solution is light yellow-orange. When 3 drops of the red-orange iodine solution is
added to 20 drops of clear, colorless ascorbic acid solution, the resulting solution is clear and
colorless. The iodine must have reacted to produce a product that is not colored. On
possibility is that it is reduced to iodide ion.
(b) About 20 drops of the iodine solution were required to react with all the ascorbic acid in
20 drops of its solution. We knew all the ascorbic acid had reacted, because addition of
another drop or two of iodine caused the clear, colorless solution to take on a light yelloworange color indicating that iodine was now left unreacted in the solution. The concentrations
of the two reagents, iodine and ascorbic acid, are the same and equal volumes (number of
drops) react, so they must react in a 1:1 stoichiometric ratio.
Follow-up activities:
 Check This 6.100. Ascorbic acid oxidation.
 End of chapter problems 6.65 and 6.66.
ACS Chemistry FROG
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Chapter 6
Chemical Reactions
Solutions for Chapter 6 Check This Activities
Check This 6.10. Stoichiometric calculation for calcium oxalate formation
(a)/(b) Test tube #1: (sample calculation)
 1L 
mol Ca2+ = (0.10M)(0.5mL) 
= 5.0  10-5 mol
 1000 mL 
 1L 
mol C2O42- = (0.10M)(4.5mL) 
= 3.5  10-4 mol
 1000 mL 
limiting reactant is Ca2+: 5.0  10-5 mol of CaC2O4(s) is formed.
Test tube #2:
mol Ca2+ = 1.5  10-4 mol
mol C2O42- = 3.5  10-4 mol
limiting reactant is Ca2+: 1.5  10-4 mol of CaC2O4(s) is formed.
Test tube #3:
mol Ca2+ = 2.5  10-4 mol
mol C2O42- = 2.5  10-4 mol
mol Ca2+ = mol C2O42- = mol CaC2O4(s) = 2.5  10-4 mol
stoichiometric reaction: all the Ca2+(aq) and C2O42-(aq) are used up.
Test tube #4:
mol Ca2+ = 3.5  10-4 mol
mol C2O42- = 1.5  10-4 mol
limiting reactant is C2O42-: 1.5  10-4 mol of CaC2O4(s) is formed.
(c) The results in part (b) and from Worked Example 6.9 indicate that the least precipitate
should be formed in test tubes #1 and #5 and the most in test tube #3. This is what we said the
bar graph sketched in Consider This 6.7 should look like.
(d) The sample with the largest amount of precipitate contains stoichiometrically equal
quantities of Ca2+(aq) and C2O42-(aq) ions, which are essentially all used up in the reaction. The
other four test tubes have less of one or the other of the reactants and these limit the amount of
precipitate that can form to less than that formed in test tube #3.
Check This 6.11. Nickel–dimethylglyoxime reaction stoichiometry
(a) The stoichiometry is 1 mol of Ni2+(aq) to 2 mol of C4H8N2O2(alc). Test tube #4 has the
largest amount of precipitate formed and we know that the sample closest to the stoichiometric
ratio of reactants will form the largest amount of product. Comparing the number of moles of
Ni2+(aq) to the number of moles of C4H8N2O2(alc) in this sample, we see that the ratio is
roughly 1:2 (= 7.50:13.1 = 1.00:1.75). The precipitate is Ni(dmg)2.
(b) To find the number of moles of precipitate formed in each sample, we use the limiting
reagent in each case, assuming that the molar ratio of Ni2+ to C4H8N2O2 is 1:2 in the precipitate.
In samples 1, 2, and 3, Ni2+ is the limiting reagent because the Ni2+ to C4H8N2O2 ratio is less
than 1:2 in these samples. In these samples the number of moles of precipitate are equal to the
number of moles of Ni2+ that react: mol precipitate = 2.55  10–4, 4.89  10–4, and 5.98  10–4,
respectively, in samples 1, 2, and 3. In samples 4, 5, and 6 C4H8N2O2 is the limiting reagent
because the Ni2+ to C4H8N2O2 ratio is greater than 1:2 in these samples. In these samples the
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ACS Chemistry FROG
Chemical Reactions
Chapter 6
number of moles of precipitate are equal to the half the number of moles of C4H8N2O2 that
react: mol precipitate = 6.55  10–4, 5.45  10–4, and 4.10  10–4, respectively, in samples 4, 5,
and 6.
(c) The average total number of moles of the two reactants is 20.7  10–4 mol.
(d) To prepare a sample that would give the exact stoichiometric ratio of Ni2+ to C4H8N2O2, we
need one that contains a 1:2 ratio of moles of these reactants in amounts that equal a total of
20.7  10–4 mol. One-third of the moles in the mixture should be Ni2+ and the other two thirds
C4H8N2O2. A mixture that contains 6.90  10–4 [= (20.7  10–4)/3] mol Ni2+ and 13.8  10–4 mol
C4H8N2O2 will give 6.90  10–4 mol of precipitate. The precipitate is surely an electrically
neutral substance, so it is likely that hydrogen ions, H+, are lost from dmg in the process of
reacting with Ni2+ cation. We will neglect this factor and assume that the precipitate is
Ni(C4H8N2O2)2, which has a molar mass of 291 g·mol–1. The mass of precipitate formed will be:
mass ppt = (6.90  10–4 mol)(291 g·mol–1) = 0.201 g
This mass is a bit larger than the largest mass observed in the study reported here. This is the
result we expect, because this is the product of the reaction with stoichiometrically equivalent
amounts of the reactants.
Check This 6.15. [H3O+(aq)] in 0.1 M ethanoic acid solution
(a) The pH of 0.1 M ethanoic acid in Investigate This 6.13 is shown above as 2.6. This pH
corresponds to [H3O+(aq)] = 10–pH M = 10–2.6 M = 3  10-3 M, which is less than the 0.1 M
concentration of ethanoic acid. Ethanoic acid does not donate all its protons to water to form
hydronium ions. We, therefore, classify ethanoic acid as a weak acid.
(b) The molecular level diagrams from the Web Companion, Chapter 6, Section 6.4, page 2, are
reproduced here:
For hydrochloric acid, we said that all of the protons from HCl are transferred to water
molecules to form H3O+ and Cl– ions. In the left-hand diagram here, all the HA molecules have
transferred their proton to water to form H3O+ and A– ions. Thus, this diagram represents the
result of a strong acid interaction with water. In part (a), we found that only a small fraction of
ethanoic acid molecules transfer their protons to water to form H3O+ and ethanoate ions. This is
like the right-hand diagram here, where only one-fifth of the HA molecules have transferred
their proton to water. This diagram represents the result of a weak acid interaction with water.
Check This 6.16. Fraction of proton transfer between ethanoic acid and water
In Check This 6.15(a) we found that [H3O+(aq)] = 3  10-3 M = 0.003 M in a solution to which
enough ethanoic acid is added to make its concentration 0.1 M. The fraction of ethanoic acid
that transfers a proton to water is (0.003 M)/(0.1 M) = 0.03 = 3/100 = 3%.
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Check This 6.18. [OH–(aq)] in an aqueous sodium amide solution
In this solution [H3O+(aq)] = 10–pH M = 10–13 M. Therefore, [OH–(aq)] = (10–14 M2)/[H3O+(aq)]
= (10–14 M2)/(10–13 M) = 10–1 M = 0.1 M. Enough sodium amide was added to the solution to
make its concentration 0.1 M. If all the added NH2– accepted protons from water, as represented
by reaction (6.12), an equivalent amount of OH–(aq) would be formed, that is, [OH–(aq)] would
be 0.1 M. This is the concentration we calculated from the pH of the solution, so the result is
consistent with reaction (6.12). We conclude that NH2– is a strong Brønsted-Lowry base.
Check This 6.19. Writing Lewis acid–base reactions
(6.11): CH3C(O)O–H(aq) + H2O:(l)  H2O–H+(aq) + CH3C(O)O:–(aq)
(6.12): HO–H(l) + :NH2–(aq)  H–NH2(aq) + :OH–(aq)
(6.13): HO–H(l) + :NH3(aq)  H–NH3+(aq) + :OH–(aq)
Check This 6.21. Relative basicities of chloride and hydroxide ions
(a) The pH of 0.1 M NaCl is essentially the same as the pH of water. The pH is not affected by
the addition of NaCl, so [:OH–(aq)]  10–7 M compared to [:Cl–(aq)] = 0.1 M. The :Cl–(aq) is
present in higher concentration in the sodium chloride solution.
(b) Because [:OH–(aq)] << [:Cl–(aq)], we know that :OH–(aq) wins the competition. The
equilibrium position of reaction (6.16) lies far to the reactant side as written. Thus, :OH–(aq) is
the stronger Lewis base.
(c) H–Cl(aq) is the stronger acid. If :OH–(aq) is a stronger Lewis base than :Cl–(aq), then H–
Cl(aq) must be a stronger Lewis acid than H–OH(l).
Check This 6.23. Relative basicities of ethanoate and hydroxide ions
(a) In Investigate This 6.13, we found that a solution of sodium ethanoate (acetate) is more
basic (higher pH) than pure water. Thus, reaction (6.18) must proceed to some extent to produce
more :OH–(aq) than is formed in water alone.
(b) The reasoning in this part of the problem shows that CH3C(O)O:– is a stronger Lewis base
than H2O:. Therefore, CH3C(O)OH is a weaker Lewis acid than H3O+, which is also the
conclusion in Worked Example 6.22.
(c) We use the measured pH (= 8.6) to find [H3O+(aq)] and then the relationship between
[H3O+(aq)] and [:OH-(aq)] to find [:OH-(aq)]. Combining the steps, we get:
[:OH-(aq)] = (10–14 M2)/(10–pH M) = (10–14 M2)/(10–8.6 M) = 4  10-6 M.
(d) The concentration of unreacted ethanoate ion is the concentration we began with minus the
concentration of ethanoate ion formed: [CH3C(O)O:–(aq)] = (0.1 M) – (4  10-6 M)  0.1 M.
Almost all of the ethanoate ion remains unreacted: [CH3C(O)O:–(aq)] >> [:OH-(aq)].
(e) We conclude that :OH–(aq) is a stronger Lewis base than CH3C(O)O:–(aq), because the
concentration of CH3C(O)O:–(aq) is greater than the concentration of :OH–(aq) in this solution
where they are competing for protons from water.
Check This 6.25. pH of an aminium (ammonia-like) chloride solution
The pH of a solution of RNH3Cl will be close to the pH of water, but somewhat more acidic
(lower pH than water), because Table 6.2 shows that RNH3+ ion is a bit stronger acid than water
and chloride is a weaker base than water and will not affect the pH.
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Check This 6.26. Relative Lewis base strengths
(a) The charge density models shown in the Web Companion, Chapter 6, Section 6.5, page 1,
are reproduced here:
The charge density models shown here are the acids formed by addition of a proton to each of
the Lewis bases in expression (6.24). Electrons on atoms with high electronegativity are held
more tightly. Thus, they are less available for donating to form a bond with a proton with the
result that the proton is more easily transferred to another Lewis base. This effect is shown in
these Lewis acids as we see the increasingly positive (blue color) hydrogen ends of these
molecules as period two atomic center goes from carbon to fluorine. The more positive the
hydrogen, the easier it is to transfer it, which means that the corresponding Lewis base is
weaker (less able to hold the proton). Expression (6.24) shows that fluoride ion, F–, is the
weakest Lewis base in this group and the methide ion, CH3–, is the strongest base of the group,
just as these charge density models predict.
(b) Based on the electronegativities of the third period atoms, we would predict this order of
basicities: (most basic) :PH2– > :SH– > :Cl– (least basic). This is because electrons on atoms
with high electronegativity are held more tightly and are less available for donating to form a
bond with a proton. Acidities for the conjugate Lewis acids go in the opposite order: (least
acidic) H–PH2 < H–SH < H–Cl (most acidic). The Lewis bases :PH2– and :SH– are included in
Table 6.2. Their relative basicities are in the order we predicted on the basis of
electronegativities, so we have some confidence that our prediction for the relative basicity of
PH2– is also correct.
Check This 6.29. Relative Lewis base strengths
We predict :PH3 to be a weaker base than :NH3 because the phosphorus atom is larger than the
nitrogen atom. The phosphorus valence electrons occupy more space than those in nitrogen and
are, therefore, more stable and less able to act as proton acceptors.
Check This 6.32. Relative basicities of ethanoate, ethoxide, and hydroxide anions
(a) The rankings in Table 6.2 are consistent with the conclusions about the relative basicities. If
water and ethanol have the same acid strength, the ethoxide anion and hydroxide anion will
have the same base strength and they are adjacent in Table 6.2. Ethanoate (acetate) ion is further
down the list in Table 6.2 and is a weaker base than both the ethoxide and hydroxide ions.
(b) Lewis structures for hydroxide, ethoxide, and ethanoate anions are:
ACS Chemistry FROG
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Chapter 6
Chemical Reactions
H
H
O
H
H
H
C
C
H
H
H
O
C
H
H
O
C
H
O
C
O
C
O
H
There are two equivalent Lewis structures for the ethanoate anion. This means that electrons on
the oxygen atoms are delocalized through its pi orbitals. Since the electrons are delocalized,
they are more stable and less able to act as proton acceptors, which explains why the ethanoate
ion is a weaker base than the hydroxide and ethoxide ions.
Check This 6.33. Relative basicities of oxyanions and acidities of oxyacids
(a) Cyclohexanoxide anion and the ethoxide anion do not contain any pi bonds, so there is no
delocalization of pi electrons and they should be about equivalent in basicity. The phenoxide (or
phenolate) anion and ethanoate anion contain pi bonds so there can be delocalization in both
anions. The charge density models for cyclohexanoxide and phenoxide in the Web Companion,
Chapter 6, Section 6.5, page 2, clearly show the delocalization of the negative charge (red color)
into the pi-bonded ring in the phenoxide. The pi electrons of the weaker bases (phenoxide anion
and ethanoate anion) are delocalized, which stabilizes the anions and makes them weaker bases
because they are less able to act as proton acceptors. Table 6.2 shows that both phenoxide
(phenolate) anion and ethanoate anion are less basic than ethoxide anion (and, by analogy,
cyclohexanoxide anion as well).
(b) Lewis structures for the reaction of carbonic acid transferring a proton to water are:
O
C
HO
O
OH
H
H
O
H
C
HO
O
O
O
O
C
HO
H
H
and
O
HO
C
O
(c) The relative acidities of carbonic acid and ethanoic acid should be similar, because their
anions probably have about the same basicity. Both anions have pi electron delocalization over
two oxygen atoms bonded to a carbon atom. This prediction is consistent with the information
in Table 6.2, where the acids (and bases) are listed next to each other.
Check This 6.35. Acid–base properties of bleach solutions
The hypochlorite ion is a Lewis base that reacts with water to form hypochlorous acid and
hydroxide ion, so it would make bleach solutions basic.
Check This 6.36. Another way to predict relative strengths of oxyacids and oxyanions
(a) For hypochlorous acid, there are zero oxygen atoms doubly bonded to chlorine. For chlorous
acid, there is one oxygen atom doubly bonded to chlorine. For chloric acid, there are two oxygen
atoms doubly bonded to chlorine. For perchloric acid, there are three oxygen atoms double
bonded to chlorine. There is a correlation between these numbers and the relative strengths of the
oxyacids: the more oxygen atoms doubly bonded to the central atom, the stronger the acid.
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Chapter 6
Conversely, the more oxygen atoms doubly bonded to the central atom, the weaker the conjugate
base.
(b) Hydrogen sulfate has two oxygen atoms doubly bonded to the sulfur and hydrogen sulfite has
one. Using the rule in part (a), we predict that hydrogen sulfate will be a weaker Lewis base than
hydrogen sulfite. This is also what the delocalization of electrons in these ions predicts, because
there are three equivalent Lewis structures for the hydrogen sulfate anion and only two (less
delocalization and less stable) for the hydrogen sulfite anion.
Check This 6.38. Relative acid and base strengths
(a) (weakest base) NO3– < NO2– < CH3CO2–  HOCO2– < OH– (strongest base). All central
atoms are from the same period in the periodic table. NO3– has greatest amount of delocalization
(three equivalent Lewis structures) and is the weakest base. OH– has no delocalization so it will
be the strongest base of this group. CH3CO2–, HOCO2–, and NO2– all have the same amount of
delocalization (two equivalent Lewis structures). However, nitrogen is more electronegative
than carbon, so the negative charge on the electrons in NO2– will be somewhat less than in
CH3CO2– and HOCO2–. Thus, the nitrite, NO2–, ion will be a somewhat weaker base than
CH3CO2– and HOCO2–, which will have nearly the same base strength.
(b) (strongest acid) HOSO3– > HOSO2– > (HO)2PO2–  HOPHO2– (weakest acid). All central
atoms are from the same period (third) in the periodic table. The sulfate ion formed from
HOSO3– has the greatest amount of delocalization (four equivalent Lewis structures), making it
the weakest base and, therefore, HOSO3– the strongest acid of the group. The conjugate bases
formed by the other acids have the same degree of delocalization (three equivalent Lewis
structures), but the central atoms are different. Sulfur is somewhat more electronegative than
phosphorus so the negative charge on the electrons in the conjugate base sulfite will be
somewhat less than in the phosphorus bases, and sulfite will be a weaker base. Conversely, the
hydrogen sulfite anion will be a stronger acid than the phosphorus acids. The phosphorus acids
will have about the same acid strength. (Note that the H bonded directly to the P in HOPHO2– is
not acidic, that is, it does not get transferred to water in aqueous solutions.)
(c) (strongest acid) (HO)2SO2 > (HO)3PO > (HO)2CO (weakest acid). The conjugate base
formed by (HO)2SO2 has the greatest amount of electron delocalization (three equivalent Lewis
structures) compared to the bases formed by the other acids in this list. Although the conjugate
bases formed by (HO)3PO and (HO)2CO have the same degree of electron delocalization (two
equivalent Lewis structures), the phosphorus atom is larger than the carbon atom, which adds to
the delocalization in the conjugate base from (HO)3PO and makes it a stronger acid than
(HO)2CO.
Check This 6.40. Predicting the direction of reaction for acid–base reactions
Use a solution of sodium hydroxide. Table 6.2 shows that OH– is a stronger base than phenol, so
the hydroxide ion will win the competition for H+ from phenol. The phenolate anion, which is
more soluble in water than phenol, will be formed.
Check This 6.46. Keeping the shine in your hair
EDTA in shampoo complexes hard water ions (Mg2+, Fe3+, and Ca2+) in the water and prevents
them from reacting with the soaps and/or detergents in the shampoo [as you observed in
Investigate This 6.43(b)]. Without the EDTA, the ions in hard water would form soap scum that
ACS Chemistry FROG
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Chapter 6
Chemical Reactions
would tend to stick to your hair and you would have to use more shampoo to try to wash it off.
Thus, EDTA makes your shampoo work more effectively and you will use less shampoo.
Check This 6.47. Names and structures of Pt(NH3)2Cl2
Recall the introduction of the prefixes cis- and trans- in Chapter 5, Section 5.9, page 326, and
review the definitions, if necessary. The name cis-Pt(NH3)2Cl2 tells us that the NH3 ligands are
located next to each other (as are the Cl– ligands) in the structure, while in trans-Pt(NH3)2Cl2,
they are located across from each other. Check the structures in Figure 6.8 to be sure this
nomenclature makes sense to you.
Check This 6.49. The porphine ring system in chlorophyll and heme
There are 22  electrons in the porphine ring. There are 20  electrons in chlorophyll, and 22 
electrons in the heme. (Note that the alternating double and single bonds around the porphine
ring system has one fewer double bonds in chlorophyll because the five-membered ring at the
lower left of the structure in Figure 6.10(a) is missing a double bond that is present in the
uninterrupted  electron structure of porphine.)
Check This 6.52. Centers of positive and negative charge in molecules
(a)/(b) Electronegative atoms such as oxygen always have nonbonding electrons that make
them centers for high electron density. Oxygen atoms, in these scenarios, act as Lewis bases.
Carbon atoms that are bonded to these oxygen atoms will have positive centers and thus act as
Lewis acids. The Lewis structures for all the molecules, with partial charges on second period
atoms labeled, are given here:

H
O

(a)

O

C 
 O H

  H
H

O C H
H
(b) CO2, H2O, (HO)3PO, CH3C(O)OH, HOCH2CH2OH, and H2N(CH2)6NH2



O C O

H  HH
O C C O 
H 
H H


H
O
H 

 O
H
O P O
 
H
O

 H


O 
H3C C O 
H


H H H H H H H
H


 N C C C C C C N 
H
H H H H H H H


Hydrogen atoms bonded to the oxygen atoms are also positive centers and, in some of these
molecules, a proton can be transferred to water in aqueous solution, that is, the molecule will be
a Brønsted-Lowry acid.
[NOTE to Instructors: However, if an oxygen is attached to a pi orbital network ring, such as
found in salicylic acid, it is a less negative center because its electron density is spread out
66
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Chapter 6
through the pi orbital network. In organic chemistry courses, this is usually shown in terms of
quininoid Lewis structures with a double bond to the oxygen atom, which gives it a positive
charge in the non-ionized form. But, based upon electronegativity and electron density
presented in this section, we label the phenolic oxygen as "-".]
Check This 6.53. More nucleophile-electrophile reactions
(a) The functional group that is formed as a product is an ester.
H
O
C
H
H
O
O
O
O
O
O
H
H
H
O
CH3
C
O
O
H
C
O
CH3
H
O
H
CH3
methyl salicylate (an ester)
(b) The nucleophile-electrophile reaction of carbon dioxide and water is shown as the first step
in this reaction and the rearrangement that produces the final product from the intermediate is the
second step.
O
O
O
C
H
H
O
H
C
O
O
H
C
O
O
H
H
O
Check This 6.54. A condensation reaction
The first step, the nucleophile-electrophile reaction, in reaction (6.35) is:
H
O
O
C
O
C
+
O
H
H
– H
O C
H
H
H
C
H
H
O
O
O
C
O
H
C
O
O
H
CH2CH2OH
Check This 6.55. Polyamide condensation polymers
(a) The charge density models in the Web Companion, Chapter 6, Section 6.7, page 1 are:
The nucleophile –with high negative charge density (red) at the nitrogen– is the amine and the
electrophile –with high positive charge density (blue) at the carboxylic acid carbon– is the
carboxylic acid. The red arrow is our usual representation of the movement/rearrangement of
electrons in reactions.
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Chapter 6
Chemical Reactions
(b) The reaction to form nylon-66 can be represented as:
O
HO
O
H
H
CCH2CH2CH2CH2C OH
H NCH2CH2CH2CH2CH2CH2N
many of each of these
O
O
CCH2CH2CH2CH2C
H
H
H
NCH2CH2CH2CH2CH2CH2N
n
nH2O
nylon-66
The polymer is called nylon-66 because there are six carbon atoms from each dicarboxylic acid
molecule (two acid groups in the same molecule) and six carbon atoms from each diamine
molecule (two amine groups in the same molecule). The first step in this reaction is the one
represented in the Web Companion, Chapter 6, Section 6.7, page 1 [although the loss of a water
molecule when the amide bond, –C(O)-NH–, is formed is not shown in the Companion as it is
for the ester bond formation in equation (6.35)]. Since one end of the initial product still has an
amine group and the other end a carboxylic acid group, further reaction occurs to add more
molecules of the reactants (producing a water molecule for each amide bond formed). A final
product molecule of, nylon-66, incorporates n molecules of the dicarboxylic acid plus n
molecules of the diamine, and produces n molecules of water.
(c) The reaction, actually two sequential reactions, represented in Figure 1.30 are related to the
reactions discussed in this section, especially the amine-carboxylic acid reaction in this activity.
The reactions in Figure 1.30 are the reactions of amino acids to form a chain of amino acids
connected by amide bonds, which is the primary structure of protein molecules. The difference
between the amino acid condensation reactions and the ones in this section is that each amino
acid has both a nucleophilic site, the lone pair of electrons on the amine nitrogen, and an
electrophilic reactive site (the positively polarized carbon of the carboxylic acid). Look, again,
at Figure 1.30 and see how the chain can be extended at either end by nucleophile-electrophile
reactions to form further amide bonds with more amino acids.
Check This 6.56. Water as a nucleophile
The hydrolysis reaction for this amide can be represented as:
O
H2 O
CH3CH2C
O
H
NCH2CH2CH2 CH3
H
CH3 CH2 C
H
NCH2 CH2 CH2 CH3
O
H
H
O
CH3 CH2 C
OH
H
NCH2 CH2CH2CH3
O
CH3 CH2C
H
O
H
NCH2 CH2CH2CH3
H
Check This 6.57. Polymeric structure of DNA
Phosphate ester bonds link the polymer repeat units together. Since each of the phosphate
groups forms an ester link with two or the ribose sugars (one from each monomer), these are
often called phosphodiester links. or bonds. The repeating unit is:
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Chapter 6
Base
O
O
O
O
P
OH
Check This 6.60. Using formal charge
(a) The three Lewis structures we wrote for the cyanate ion in Consider This 6.59(a) are:
0
0
–1
–1 0
0
–2 0
+1
N
C
O
N
O
N
O
C
C
The two structures on the left have one atom with a –1 formal charge, which is the least we can
have, since this is a –1 ion. The negative formal charge is on an electronegative atom in each
case, which is where we expect negative charge to be. The structure on the right involves
multiple atoms with formal charge (other than zero), one of which is –2. The most stable species
have as few atoms with formal charge as possible. This structure is much higher in energy than
the other two and we can neglect it when considering the stability of the cyanate ion. The two
Lewis structures we wrote for the isocyanate ion in Consider This 6.59(a) are:
–1 +1 –1
C N O
–2 +1
C
N
0
O
These structures have multiple atoms with formal charge, one of which is –2. Both of these
structures will be of higher energy than the two lower energy structures for cyanate, so we
conclude that the isocyanate ion is less stable (higher energy) that the cyanate. (Indeed, under
appropriate conditions, isocyanates rearrange to form cyanates.)
NOTE: Since the oxygen atom is more electronegative than the nitrogen atom, the cyanate
structure on the left will have a somewhat lower energy than the one in the center. Thus, these
two Lewis structures are not strictly equivalent, but they are probably close enough in energy to
suggest that the pi electrons will be delocalized over all three atoms (although likely to have
somewhat higher density at the oxygen end compared to the nitrogen end). Delocalization will
lower the energy of this ion. All five of these Lewis structures predict that both the cyanate and
isocyanate ion are linear, because the central atom has two bonding sigma orbitals to the
flanking atoms.
(b) Table 6.3 includes these correlations. When there is one bond to an oxygen atom (and six
nonbonding electrons), the atom has a –1 formal charge. When there are two bonds to an
oxygen atom (and four nonbonding electrons), the atom has a zero formal charge When there
are three bonds to an oxygen atom (and two nonbonding electrons), the atom has a +1 formal
charge.
(c) The formal charges on the atoms in the chlorite ion structure written in this activity (on the
left here) and in the structures you wrote in Consider This 6.34(a) (on the right here) are:
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Chapter 6
Chemical Reactions
–1
–1
O Cl O
+1
0
0 –1
–1 0
O
Cl O
O
0
Cl O
Since this is a –1 ion, at least one of the atoms in the ion must have a –1 formal charge. The
structure on the left, however, has a non-zero formal charge on all three atoms, while the ones
on the right have zero formal charge on two of the atoms. Stable species usually have as many
atoms with zero formal charge as possible. The structures on the right (the ones we wrote
previously) satisfy this criterion, so we conclude that this structure with lower formal charge
and delocalized pi electrons is the more stable structure for the chlorite ion.
NOTE 1: Just as we divided up the pi electrons among the bonds in molecules with delocalized
pi orbitals (see Chapter 5, Section 5.7, pages 312-313) to get effective bond orders, we can
divide up the formal charge in these structures with more than one equivalent Lewis structure.
In the case of the chlorite ion, the “average” formal charge on each of the oxygen atoms is –1/2
in the delocalized structure represented by the two structures on the right here. This spreading of
the charge among the atoms is a favorable factor.
NOTE 2: Some other texts, including organic textbooks, use the Lewis structure written on the
left here, that preserves the octet of electrons on all possible atoms in the structure. It is clear
from the behavior of third period elements in other compounds that they do not necessarily obey
the octet rule (which only seems to apply to C, N, O, and F – atoms too small to accommodate
more than four sigma bonding orbitals), so accommodating more than an octet in these oxyacids
is not strange. In “Lewis Structures in General Chemistry: Agreement Between Electron
Density Calculations and Lewis Structures,” J. Chem. Educ. 2001 78, 981, Gordon H. Purser
found that calculations of electron densities in the bonds between O and central P, S, and Cl
atoms were consistent with the bond orders predicted from Lewis structures that minimize
formal charges. That is, the calculations are consistent with the delocalization represented by the
structures shown on the right above. These results further reinforce the idea that delocalization
of electron waves has a powerful energy lowering (stabilizing) effect on molecules and ions.
(d) The Lewis structures and formal charges for the chlorate and perchlorate anions in which the
chlorine atom is constrained to have only an octet of electrons are:
+2
–1
O
Cl
O
O
–1
–1
O
–1
+3
–1
O Cl
O
O
–1
–1
We chose to correlate base strength with the structures written in Consider This 6.34, rather than
those here, because there is an inverse correlation between the number of equivalent Lewis
structures (hence the amount of delocalization of the electrons) and the base strength that allows
us to make predictions about base strengths. There is no such correlation with these structures.
If the way we visualize molecules and ions does not help in predicting the observable properties
of the compounds, then it is not a very useful visualization. See, also, NOTE 2 in part (c) for
more reason not to use the structures written here.
Check This 6.65. Balancing simple reduction–oxidation reaction expressions
(a)
Mg2+(aq) + Cu(s)  Mg(s) + Cu2+(aq)
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Chapter 6
We see that there are the same number of magnesium and of copper species on each side of the
equation, so the atoms (mass) balance. The net charge on the left is +2 and on the right is also
+2, so the charge balances. The reaction expression is balanced as written.
(b)
Fe3+(aq) + Sn2+(aq)  Fe2+(aq) + Sn4+(aq)
We see that there are the same number of iron and of tin species on each side of the equation, so
the atoms (mass) balance, as written. The net charge on the left is +5 and on the right is +6, so
the charge does not balance. Each Sn2+ loses two electrons to form Sn4+ and each Fe3+ gains one
electron to form Fe2+, so, we have enough electrons to reduce two Fe3+ for each Sn2+ oxidized.
Thus, to balance the equation we write:
2Fe3+(aq) + Sn2+(aq)  2Fe2+(aq) + Sn4+(aq)
This equation is still balanced in atoms and now the net charge on each side is the same, +8, so
the equation is balanced.
(c)
Zn(s) + H+(aq)  Zn2+(aq) + H2(g)
The zinc species on each side of the equation balance, but there are two H atoms on the right
and only one on the left. To balance the hydrogen atoms we write:
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
In this equation, the atoms balance and the net charge on each side is +2, so the equation is
balanced.
Check This 6.69. Oxidation numbers for elements and monatomic ions
(a) Fe(s) is an elemental atom, all of which have a zero charge and an oxidation number of 0.
Fe3+(aq) is a monatomic ion, all of which have oxidation numbers equal to their charge, so it has
an oxidation number of +3. Chlorine, Cl2(g), is an elemental species, so the atoms in a molecule
of the element have oxidation numbers of 0. The chloride ion, Cl–(aq), is a monatomic ion, all
of which have oxidation numbers equal to their charge, so it has an oxidation number of –1.
(b) The reasoning for all the molecules and ions in this part is the same as in part (a). V4+(aq)
has an oxidation number of +4. H–(g) has an oxidation number of –1. P4(s) (an element) has an
oxidation number of 0. N2(g) (an element) has an oxidation number of 0. Na+(aq) has an
oxidation number of +1.
Check This 6.70. The procedure and the definition of oxidation number
Since the inner shell electrons never take part in electron transfer reactions and always remain in
their atomic core, there is the same number of core electrons in an atom of an element and in
any ion of that element. Thus, when we take the difference between the number of electrons in
the atom and the number in the ion, the number of core electrons cancels out of the calculation
and only the change in number of valence electrons needs to be accounted for.
Check This 6.72. Assigning oxidation numbers
Lewis structures with electrons assigned according to the procedure for determining oxidation
numbers are shown here with the oxidation number of each atom labeled. Check in each case
that the sum of the oxidation numbers equals the overall charge on the molecule or ion.
ACS Chemistry FROG
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–2
+1
H
–2
+1
H
O
+1
H
–3
+1
H
N
+1
H
–2 O –2
+5
O P O
+1
H
O –2
H +1
H +1
3–
–2
–2 O –2
+5
O P O
+1
H
–2
+1
0
O
Cl
Cl
0
Cl
O –2
0
H0
0
0 0
C
C H
H
0
H
0
H
H
0
0
H
–2
+1
C
O
0
H
0
–2
+1
H
H
O
0
H
–2
+3
C
O
+1
H
Check This 6.73. Reduction–oxidation reactions
(a) In reaction 6.45, Cu(s) (ON = 0) is oxidized to Cu2+(aq) (ON = +2). The nitrogen atom in
NO3–(aq) (ON = +5) is reduced to an ON = 4 in NO2(g).
0
+5
+2
+4
Cu(s) + H+(aq) + NO3–(aq)  Cu2+(aq) + NO2(g)
(b) The oxidation numbers for the atoms in equation (6.48) are:
+1 –1
+1 –1
+1 –1+
–1
HCl(aq) + H2 O(l)
H3 O (aq) + Cl–(aq)
This is not an oxidation-reduction reaction. The oxidation numbers for each atom in both the
reactants and products are the same.
(c) The oxidation numbers for the atoms in equation (6.49) are:
+1+
+1 -1
-1
+1 -2
0
H (aq) + H2O2 (aq) + I– (aq)
H2O(aq) + I2 (aq)
Reaction (6.49) is an oxidation-reduction reaction. The oxidation numbers of the oxygen and
iodine have changed.
Check This 6.75. Oxidation number for oxygen in peroxides
For the bonds between oxygen and hydrogen atoms in H–O–O–H, the oxygen atoms will be
assigned the two bonding electrons because oxygen is more electronegative than hydrogen. The
two oxygen atoms will each be assigned one electron from the electron-pair bond between them.
Each oxygen atom will have seven valence electrons (its four nonbonding electrons plus two
from the O–H bond plus one from the O–O bond) and, therefore, a –1 oxidation number (+6
charge on the atomic core minus 7 valence electrons).
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Chapter 6
Check This 6.77. Assigning oxidation numbers
The oxidation numbers assigned using the rules in Table 6.5 are the same as those from Check
This 6.72. The notes here indicate how the rules in the table are applied in each case.
H2O: Apply the first parts of rules 3 and 4 to get ON = –2 for O and ON = +1 for each of the
two Hs. Rule 5 is then satisfied.
NH3: Apply the first part of rule 4 to get ON = +1 for each of the three Hs and then rule 5 to get
ON = –3 for the N, in order to have the sum of ONs = 0.
(HO)3PO: Apply the first parts of rules 3 and 4 to get ON = –2 for each of the four Os and ON =
+1 for the three Hs. The sum of the ONs thus assigned is –5 (= –8 + 3), so, by rule 5, the ON for
the P must be +5.
PO43–: Apply the first part of rule 4 to get ON = –2 for each of the four Os. The sum of the ONs
thus assigned is –8, so, by rule 6, the ON for the P must be +5.
HOCl: Apply the first parts of rules 3 and 4 to get ON = –2 for O and ON = +1 for H. The sum
of the ONs thus assigned is –1 (= –2 + 1), so, by rule 5, the ON for the Cl must be +1.
Cl2: Apply rule 1 to get ON = 0 for both Cls.
C2H6: Apply the second part of rule 4 to get ON = 0 for all six Hs. Since the Cs are equivalent,
applying rule 5 gives ON = 0 for both Cs.
CH3OH: Apply the first parts of rules 3 and 4 to get ON = –2 for O and ON = +1 for the H
bonded to O. Apply the second part of rule 4 to get ON = 0 for the three Hs bonded to C. The
sum of the ONs thus assigned is –1 (= –2 + 1 + 0), so, by rule 5, the ON for the C must be +1.
HC(O)OH: Apply the first parts of rules 3 and 4 to get ON = –2 for the two Os and ON = +1 for
the H bonded to O. Apply the second part of rule 4 to get ON = 0 for the H bonded to C. The
sum of the ONs thus assigned is –3 (= –4 + 1 + 0), so, by rule 5, the ON for the C must be +3.
Check This 6.79. Balancing redox equations by the oxidation-number method
(a) For the N in NO, ON = +2 (using either the definition of ON or the rules in Table 6.5). The
unbalanced and balanced equations (showing the gain and loss of electrons and assuming at the
start that hydronium ion is a reactant and water a product) for this reaction are:
gain 2 electrons
Cu(s) + H+ (aq) + NO3– (aq)  Cu2+ (aq) + NO(g) + H2 O(l)
lose 3 electrons
gain 6 electrons
3Cu(s) + 8H+ (aq) + 2NO3– (aq)  3Cu2+ (aq) + 2NO(g) + 4H2 O(l)
lose 6 electrons
(b) The unbalanced equation with the gain and loss of electrons shown is:
gain 1 electron
H2 O2 (aq) + I– (aq)  H2O(l) + I2(aq)
lose 1 electron
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Chemical Reactions
Each O gains one electron and each I loses one electron in the reaction. Thus the gain and loss
of electrons are balanced, but the atoms are not. In order to balance O, there must be two H2O
molecules in the products. In order to balance I, there must be two I– in the reactants. If we
make these changes, the gain and loss of electrons will still balance, but we will need two H (as
H+) on the reactant side to balance Hs. The final balanced equation with gain and loss of
electrons show is:
gain 2 electrons
H2 O2 (aq) + 2I– (aq) + 2H+ (aq)  2H2 O(l) + I2 (aq)
lose 2 electrons
(c) Equation (6.57) is already balanced in atoms and in overall charge with a net –4 charge on
each side of the reaction. To show the balance in terms of the oxidation number changes
requires us to assign oxidation numbers to the sulfur atoms in the reactants and products. There
are two ways to do this. (1) Using the definition of oxidation numbers we assign all electrons
from S–O bonds to the Os, which will all have ON = –2, and divide the electrons in S–S bonds
equally between the two Ss. For the thiosulfate ion reactant, this gives the central S an ON = +5
and the other S and oxidation number of –1. The sum of the ONs is –2 (= –6 +5 –1), which is
the charge on the ion, as it must be. Similarly, the Ss in the product that are bonded to O have
ON = +5 and the two Ss in the center of the ion have ON = 0. Thus, there is a oxidation of one S
in thiosulfate from –1 to 0 in the reaction. Since two thiosulfates react, the overall oxidation
change is the loss of two electrons by the two Ss. These electrons are gained by I2 as it is
reduced to two I–. (2) Using the rules in Table 6.5, we assign all the Os an ON = –2. For the
thiosulfate reactant, the application of rule 6 requires that the two Ss have a combined ON = +4.
(Note that this combined ON is the net of the ONs for the individual Ss we assigned from the
definition of ON.) A similar argument for the product ion gives the four Ss a combined
ON = +10. Since two thiosulfate ions react, their four Ss have an overall ON = +8, so the Ss are
oxidized, and they lose two electrons going from reactants to products. The result, showing the
gain and loss of electrons in the balanced reaction is:
gain 2 electrons
2-
O
2
O
S
O
O
S
+ I2(aq)
O
(aq)
S
O
2-
O
S
S
S
O
O
+ 2I-(aq)
(aq)
lose 2 electrons
Check This 6.83. Balancing redox equations by the half-reactions method
(a) The oxidation numbers for Fe and Cr are different in the reactants and products, indicating an
oxidation-reduction reaction occurred.
(1) Write half reactions with appropriate stoichiometric coefficients for redox species:
Fe2+(aq)  Fe3+(aq) + e–
Cr2O72–(aq) + 6e–  2Cr3+(aq)
+
(2) Add waters and H :
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Chapter 6
Fe2+(aq)  Fe3+(aq) + e–
Cr2O72–(aq) + 14H+(aq) + 6e–  2Cr3+(aq) + 7H2O(l)
(3) Balance electrons:
6Fe2+(aq)  6Fe3+(aq) + 6e–
Cr2O72–(aq) + 14H+(aq) + 6e–  2Cr3+(aq) + 7H2O(l)
(4) Add half reactions together:
6Fe2+(aq) + Cr2O72–(aq) + 14H+(aq)  6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)
(b) The Os in the hypochlorite ion and water have ON = –2 and the Cl in the hypochlorite ion
has ON = +1. The products, O2 and Cl2, are elements whose atoms have ON = 0. The O is
oxidized and the Cl is reduced, so this is a reduction-oxidation reaction.
(1) Write half reactions with appropriate stoichiometric coefficients for redox species:
2H2O(l)  O2(g) + 4e–
2OCl–(aq) + 2e-  Cl2(g)
(2) Add waters and H+:
2H2O(l)  O2(g) + 4H+(aq) + 4e– 2OCl–(aq) + 4H+(aq) + 2e-  Cl2(g) + 2H2O(l)
(3) Balance electrons:
2H2O(l)  O2(g) + 4H+(aq) + 4e– 4OCl–(aq) + 8H+(aq) + 4e-  2Cl2(g) + 4H2O(l)
(4) Add half reactions together and cancel redundant species:
4OCl–(aq) + 4H+(aq)  2Cl2(g) + O2(g) + 2H2O(l)
(c) Follow the procedure we have been using.
(1) Write half reactions with appropriate stoichiometric coefficients for redox species:
2I–(aq)  I2(aq) + 2e–
2IO3–(aq) + 10e–  I2(aq)
+
(2) Add waters and H :
2I–(aq)  I2(aq) + 2e–
2IO3–(aq) + 12H+(aq) + 10e–  I2(aq) + 6H2O(l)
(3) Balance electrons:
10I–(aq)  5I2(aq) + 10e–
2IO3–(aq) + 12H+(aq) + 10e–  I2(aq) + 6H2O(l)
(4) Add half reactions together and cancel redundant species:
10I–(aq) + 2IO3–(aq) + 12H+(aq)  6I2(aq) + 6H2O(l)
(4a) Divide by 2 to reduce stoichiometric coefficients to their lowest possible values:
5I–(aq) + IO3–(aq) + 6H+(aq)  3I2(aq) + 3H2O(l)
Check This 6.85. Balancing redox equations in basic solution
(a) After reaction (6.69) is balanced by the oxidation number method (for neutral or acidic
solution), the conversion to reaction in basic solution is done just as it is in Worked Example
6.84 where the reaction was balanced by the half-reaction method. First we write the reaction
showing the species that are oxidized and reduced, their products, and the electron change per
atom undergoing the change:
gain 1 electron
Cl2 (g) + Cl 2(g)  OCl– (aq) + Cl– (aq)
lose 1 electron
The electron changes balance here, but the equation is not balanced in atoms or overall charge.
When we balance the Cl (keeping the electron changes balanced) and add the requisite amounts
of water and hydronium ion to balance the O, we get:
ACS Chemistry FROG
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Chapter 6
Chemical Reactions
gain 2 electrons
Cl2 (g) + Cl 2(g) + 2H2 O(l)  2OCl –(aq) + 2Cl–(aq) + 4H+ (aq)
lose 2 electrons
When we collect the Cl2(g) together, we will get equation (6.70) in Worked Example 6.84 and
the rest of the problem proceeds just as it does there. [You may divide through by 2 at this point,
rather than after the OH–(aq) is added, but the final result, equation (6.71) will be the same.]
(b) No method is suggested, so we may balance this reaction (in neutral or acidic solution) by
whichever method we prefer; we’ll use the half-reaction method here:
(1) Write half reactions with appropriate stoichiometric coefficients for redox species:
Mn2+(aq)  MnO2(s) + 2e–
H2O2(aq) + 2e–  2H2O(l)
(2) Add waters and H+:
Mn2+(aq) + 2H2O(l)  MnO2(s) + 4H+(aq) + 2e– H2O2(aq) + 2H+(aq) + 2e– 
2H2O(l)
(3) Add half reactions together and cancel redundant species:
H2O2(aq) + Mn2+(aq)  MnO2(s) + 2H+(aq)
(4) Add two OH–(aq) to each side and “react” them with H+(aq) to form water:
H2O2(aq) + Mn2+(aq) + 2OH–(aq)  MnO2(s) + 2H2O(l)
Check This 6.89. Balancing the methanal-silver ion reaction equation
(a) No method is suggested, so we may balance this reaction (in neutral or acidic solution) by
whichever method we prefer; we’ll use the oxidation-number method. This reaction is
complicated by the fact that the oxidation number change in the methanal to methanoic acid
reaction involves two atoms: O goes from ON = +2 to ON = +3 and one H goes from ON = 0 to
ON = +1. Therefore, this change is a loss of two electrons by the molecule, but not (in the way
we are counting oxidation numbers) both from the same atom:
gain 1 electron
Ag+ (aq) + CH2O(aq)  Ag(s) + HC(O)OH(aq)
lose 2 electrons
We’ll combine balancing the oxidation and reduction with the addition of water and hydronium
ion to balance the O and H (in that order). Add an H2O(l) to the reactants to balance O and two
H+(aq) to the products to balance H:
gain 2 electrons
2Ag+ (aq) + CH2O(aq) +H2 O(l)  2Ag(s) + HC(O)OH(aq) + 2H+ (aq)
lose 2 electrons
Add two OH–(aq) to each side, “react” them with H+(aq) to form water, and cancel redundant
water to get equation (6.74):
2Ag+(aq) + CH2O(aq) + 2OH–(aq)  2Ag(s) + HC(O)OH(aq) + H2O(l)
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(b) Methanoic acid should react in basic solution to transfer a proton to hydroxide ion to form
water and methanoate anion:
HC(O)OH(aq) + OH–(aq)  HC(O)O–(aq) + H2O(l)
(c) Adding the reaction in part (b) to the final reaction in part (a) gives the overall reaction of
methanal with silver ion in basic solution:
2Ag+(aq) + CH2O(aq) + 3OH–(aq)  2Ag(s) + HC(O)O–(aq) + 2H2O(l)
Check This 6.90. Oxidation numbers for carbons in glucose fermentation
H
H
OH
H
C+1
H
+1
C
H
O
H
0
H
+1
C
+4
OH
O
C
O
H
+2
C+1
C
+1 +1C
C
HO
C
H
OH
H
HO
H
OH
Check This 6.91. Net glycolysis reaction
(a) For equation (6.76), shown here, there are 6 Cs, 6 Os, and 12 Hs on each side of the
equation, so atoms are balanced. None of the nonbonding electron pairs on O are shown,
because all Os have two pairs and we know there are the same number of Os in the reactants
and products, so these electrons are the same on each side. Counting all the electron pair bonds
(lines) in the reactant glucose, we find that there are 24 pairs of electrons (48 electrons)
represented in the reactants. Each pyruvic acid product molecule has 11 electron pair bonds
represented, for a total of 22 pairs of electrons (44 electrons) in the two molecules, and the
remaining four electrons are accounted for as the 4e– lost by the glucose as it is oxidized.
H
O
H
C
H
H
C
H
O
H
O
C
C
C
O
H
O
H
O
C
C
O
H
2H
C
H
O
H
H
H
(aq)
+ 4H+ (aq) + 4e–
C
O
H
(aq)
Equation (6.76)
H
For equation (6.77), we show only the central part of the NAD+ and NADH molecules because
this is where the reduction-oxidation reaction occurs. In this condensed structure, each
intersection of a line or end of a line represents a C and we see that 7 Cs are shown on both
sides of the reaction as well as one N on both sides. Only one H is shown bonded to the ring, but
there are three more at each of the vertices to which nothing is attached in these drawings. Thus,
five Hs (including H+) are represented on each side of the equation and it is balanced in atoms.
Counting up the two electron bonds on the left, we have 12 electron pairs, 24 electrons, in the
ACS Chemistry FROG
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Chemical Reactions
molecular structure. There are two more that are used to reduce the ring, for a total of 26
electrons on the left. On the right, there are, again, 12 electron pair bonds plus a nonbonding
pair, for a total of 26 electrons, so the equation is balanced in electrons.
H
H
H
+ H+(aq) + 2e–
+
N
(aq)
N
(aq)
Equation (6.77)
(b) Combining equations (6.76) and (6.76) – taken twice – we get:
C6H12O6(aq) + 2NAD+(aq)  2C3H4O3(aq) + 2NADH(aq) + 2H+(aq)
Check This 6.92. Reduction–oxidation of glucose
(a) Combining equations (6.78) and the reverse of (6.77) – both taken twice to give the
reduction of two pyruvic acid molecules gives:
2C3H4O3(aq) + 2NADH(aq) + 2H+(aq)  2CH3CH2OH(aq) + 2CO2(g) + 2NAD+(aq)
(b) The addition of the equation from Check This 6.91(b) and the equation here in part (a) gives
the overall glucose fermentation reaction (6.75):
C6H12O6(aq) + 2NAD+(aq)  2C3H4O3(aq) + 2NADH(aq) + 2H+(aq)
2C3H4O3(aq) + 2NADH(aq) + 2H+(aq)  2CH3CH2OH(aq) + 2CO2(g) + 2NAD+(aq)
C6H12O6(aq)  2CH3CH2OH(aq) + 2CO2(g)
(c) The NADH and NAD+ cancel out between the two equations, because the NAD+ required
for glycolysis, the first reaction, is regenerated by the fermentation reaction. Recycling of
product NADH by reactions like fermentation keeps the reduction of glucose going, even
though the cell contains only a small amount of NAD+ and NADH. (Glycolysis and
fermentation are an example of coupled processes, which we will meet again in Chapter 7,
Section 7.9, page 477.)
Check This 6.95. Acid–base equivalence in Investigate This 6.93
Well #5 in Investigate This 6.93 is where the number of moles of hydroxide is equal to the
number of moles of ethanoic acid.
(0.001L)(0.1 M ethanoic acid) = (0.001 L)(0.1 M sodium hydroxide)
0.0001 mol ethanoic acid = 0.0001 mol sodium hydroxide
Well #5 is the first one in the pattern that is blue. The wells with more hydroxide are also blue,
but the ones with less hydroxide are green or greenish-blue.
Check This 6.97. An acid–base titration
(a) If two drops of phenolphthalein are added to 50.0 mL of an aqueous solution of
ethylenediamine, the solution will be red (or reddish-purple). Ethylenediamine is a Lewis base
with one pair of nonbonded electrons on each of the two nitrogen atoms and, like ammonia, can
accept protons from water to give hydroxide ion as a product. Phenolphthalein is red in basic
solutions with a pH of about 9 or greater.
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Chapter 6
(b) The color change is red to colorless. When the ethylenediamine has all reacted with added
hydronium ion (from the hydrochloric acid), there will be no base left in the solution to accept
protons from water, so the solution will no contain a high enough concentration of hydroxide
ion to make the phenolphthalein red. It will change to its colorless form in low pH solutions.
Reaction (6.80) shows that two moles of hydronium ion react with each mole of
ethylenediamine. The number of moles of hydronium ion added to make the indicator change
color, that is, to react with all the ethylenediamine is:
mol H+(aq) = (.0250 L)(0.500 mol·L–1) = 0.0125 mol
The number of moles of ethylenediamine that have reacted is:
 1 mol en 
1
mol en(aq) = {mol H+(aq)} 
= (0.0125 mol)   = 0.00625 mol
 2 mol H + 
2
The concentration of ethylenediamine in the original solution is:
0.00625 mol
[en(aq)] =
= 0.125 M
0.0500 L
Check This 6.100. Ascorbic acid oxidation
[NOTE: There is an error in equation (6.82) in the textbook. The structure of dehydroascorbic
acid should have a single bond between the carbons within the oval, as here:
H
O
H
H
O
C
O C
C
C O H
H
C C
H
H O
O H (aq)
H O
H
H
O
C
O C
C
C O H
H
C C
H
(aq)
O
O
2H+ (aq) + 2e–
The answer here will assume the correct structure, since the problem is incorrect without this
change.]
(a) The two carbon atoms in the ovals in equation (6.82) are oxidized from +1 to +2 oxidation
numbers. If we count the number of electron pairs within the oval on the left (reactant) side of
the equation, we find that there are 10 pairs (6 pairs in bonds and 4 nonbonded pairs). On the
right (product) side of the equation, there are 9 pairs (5 pairs in bonds and 4 nonbonded pairs).
The “lost” electrons are the 2e–.
(b) Using line formulas for ascorbic and dehydroascorbic acids, the balanced reaction with
iodine is:
C6H8O6(aq) + I2(aq)  C6H6O6(aq) + 2I–(aq) + 2H+(aq)
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Chapter 6
Chemical Reactions
Solutions for Chapter 6 End-of-Chapter Problems
Problem 6.1.
(a) When water is boiled in a teakettle, no chemical reaction occurs. There is a physical change
from H2O(l) to H2O(g).
(b) When boiling water is poured into a bowl containing a package of instant oatmeal and
stirred to make a hot cereal breakfast, various components of the cereal dissolve or are rehydrated (the oatmeal had been previously prepared and the chemical changes that make
oatmeal from oat flakes had occurred and then water was removed to produce the instant
product). The changes occurring when the product is re-hydrated are probably reasonably
classified as physical, not chemical changes.
(c) When a glass of a carbonated soft drink is left overnight and tastes “flat” the next morning,
CO2(g) dissolved in the soft drink as CO2(aq) has escaped to the atmosphere. You might think
of this as a physical reaction, but it can also be considered a chemical reaction, if some of the
dissolved carbon dioxide was present as carbonic acid, (HO)2CO(aq), which is a different
chemical species.
(d) When a glass of ice cubes and water is left overnight and there are no ice cubes in the water
the next morning, no chemical reaction has occurred. The ice cubes melted, H2O(s)  H2O(l),
but no new chemical species formed.
Problem 6.2.
(a) When a match is dropped on the floor, no chemical change (reaction) occurs. The match is
still the match.
(b) When a match is struck and used to start a barbeque, several chemical reactions occur. There
is a chemical reaction that uses the energy of the friction of the match head to initiate the
reaction that lights the match (which continues to burn because the energy released in the
burning sustains the reactions). Once the match is burning, the flame is used to start the
chemical reaction in the charcoal, propane, or other fuel used in the barbeque. In both cases, the
products of the change, carbon dioxide, water, ashes, and so on, are different chemical species
than were present in the match and fuel.
(c) When a piece of paper is folded to make a paper airplane, no chemical change (reaction)
occurs. The paper is still the paper.
(d) When a piece of paper is torn into many small pieces to make confetti, you can argue that no
chemical change (reaction) has occurred because the little pieces are still paper. On the other
hand, tearing the paper apart breaks a large number on intermolecular (not covalent) bonds
between the cellulose polymer molecules that make up the paper, so many of these molecules
are not making the same bonding interactions they were previously. The distinction between
chemical change and physical change is sometimes quite fuzzy and probably not worth making.
Problem 6.3.
(a) When an acorn buried and forgotten by a squirrel grows into an oak tree, a very large
number of chemical reactions have occurred. Many compounds not originally present in the
acorn are now present in the oak tree. Other compounds that were originally present in the acorn
(glucose, for example) are also present in the oak tree, but they are not the same molecules as
were in the acorn and there are, of course, a great many more of them in the tree.
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(b) When a bottle of milk left too long in the refrigerator turns sour, chemical reactions have
occurred. New compounds (some of which taste and/or smell bad) have been formed, so souring
of milk is a chemical reaction.
(c) When equal volumes of solutions of blue food coloring and yellow food coloring are mixed
and the resulting solution is green, no chemical reaction has occurred to produce the color
change. Mixing two different colored solutions results in a mixed solution that retains the
characteristics of the molecules in the original solutions. In this case, we know that a mixture of
blue and yellow colors produces green. In principle, we could use physical methods (that is,
ones that do not required changing the colored molecules to different molecules) to separate the
mixture into its original components.
(d) When a few drops of bromocresol green solution are added to 20 mL of a colorless 0.1 M
solution of sodium acetate, the resulting solution is blue. When 10 mL of a colorless solution of
0.1 M hydrochloric acid are added, the mixture is green. When a further 10 mL of the
hydrochloric acid are added, the mixture is yellow. Chemical reactions occurring in these
solutions are responsible for the color changes. Bromocresol green is an acid-base indicator, a
substance whose color depends upon the pH of the solution it is in. It is blue in a higher pH
solution, yellow in a low pH solution, and, in a solution at some intermediate pH, both the blue
and yellow forms are present, yielding a green color [compare with the dyes in part (c)]. The
acetate ion in sodium acetate solution is a base that gives the solution high enough pH to put the
indicator entirely in its blue form. Added hydrochloric acid reacts with some of the acetate and
reduces the pH to an intermediate value where the indicator is present in both its blue and
yellow forms and the solution appears green. Further addition of acid reacts with all the acetate
and lowers the pH still further, which changes the indicator entirely to its yellow form. A
chemical reaction between the hydronium ion in the solution and the indicator is responsible for
the different colors of the indicator.
Problem 6.4.
When a few small pieces of dry ice (solid carbon dioxide, –78 C) are dropped into a tall glass
cylinder containing a red solution of dilute ammonia to which has been added a few drops of
phenol red acid-base indicator solution, bubbles of gas are rapidly evolved, a white fog forms
above the solution, and, after a short time, the solution in the cylinder turns yellow. A number of
physical and chemical reactions are occurring in this system. Solid carbon dioxide is subliming,
that is going from solid to gas, CO2(g)  CO2(g), which accounts for the bubbles that are
formed. This process is usually classified as a physical change. When the cold CO2(g) reaches
the surface and leaves the liquid, it cools the gaseous molecules, including the water vapor
above the surface. Some of the water vapor condenses to tiny droplets that we observe as a fog
above the surface and spilling out of the cylinder and falling toward the ground as the fog is
carried down by the cold dense CO2(g) (and the cooled air) spilling out of the cylinder as well.
Some of the CO2(g) dissolves in the solution forming some carbonic acid, (HO)2CO(aq), which
can react with the base, ammonia: (HO)2CO(aq) + NH3(aq)  HOCO2–(aq) + NH4+(aq). This
reaction lowers the pH of the solution changing the phenol red acid-base indicator from its red
(basic) form to its yellow (acidic) form. These physical and chemical changes explain the
observations described in the first sentence.
Problem 6.5.
(a) NaCl is a water-soluble ionic compound composed of 1+ and 1– ions, Na+ and Cl–.
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(b) CaCO3 is a water insoluble ionic compound because both the cation, Ca2+, and anion, CO32–,
are multiply charged.
(c) Na2CO3 is probably a water-soluble ionic compound, because alkali metal compounds are
almost all soluble, even if the anions are multiply charged.
(d) BaCl2 is probably a water-soluble ionic compound, because most chloride compounds are
soluble, even if the cations are multiply charged.
(e) BaSO4 is a water insoluble ionic compound because both the cation, Ba2+, and anion, SO42–,
are multiply charged.
Problem 6.6.
Mixing 25.0 mL of 0.100 M sodium sulfate, Na2SO4, solution with 25.0 mL of 0.200 M barium
chloride, BaCl2 solution results in the formation of a white precipitate. The precipitate must be,
BaSO4(s), since it is the only possible ionic product with a multiply charged cation and anion.
The net ionic reaction equation for its formation is:
Ba2+(aq) + SO42–(aq)  BaSO4(s)
This equation shows that equal numbers of moles of Ba2+(aq) and SO42–(aq) react to give the
product. To find the number of moles of product formed (and from that, the mass of product
formed), we need to calculate the number of moles of each ion present in the original mixture:
mol Ba2+(aq) = (0.0250 L)(0.200 M) = 0.00500 mol Ba2+(aq)
mol SO42–(aq) = (0.0250 L)(0.100 M) = 0.00250 mol SO42–(aq)
Since there are fewer moles of the sulfate ion, SO42–(aq) is the limiting reactant and we are
limited, by the stoichiometry of the reaction equation to formation of 0.00250 mol BaSO4(s).
The molar mass of BaSO4 is 233.4 g, so the mass of product is:
 233.4 g 
mass BaSO4(s) = (0.00250 mol BaSO4) 
= 0.584 g BaSO4(s)
 1 mol BaSO4 
Problem 6.7.
(a) All of the iron in the 0.264 g of precipitated Fe2O3 came from the 2.998 sample of powdered
tablets, which was only part of the original 22.131 g of 20 powdered tablets. To determine the
average mass of iron in one tablet, we need to find out how much iron is in the sample of 20
tablets and divide this amount by 20 to find the average per tablet. To find the mass of iron in
the 20-tablet sample, we first calculate the mass of iron in the 2.998 g sample that was analyzed:
 1 mol Fe 2O 3   2 mol Fe   55.85 g Fe 
0.264 g Fe2O3 = (0.264 g Fe2O3) 


 159.7 g Fe 2O 3   1 mol Fe 2O 3   1 mol Fe 
= 0.185 g Fe
Next we need to find the mass of Fe in the total mass of powdered tablets and divide this mass
by 20 to find the average mass of Fe per tablet. We get the total mass of Fe (in 20 tablets) using
the factor by which the total sample is larger than the analyzed sample.
 22.131 g for 20 tablets  
1

mass Fe per tablet = (0.185 g Fe) 



 2.998 g analyzed sample  20 tablets 
= 0.0683 g·tablet–1
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(b) The mass percent of Fe in a tablet is equal to the mass percent of iron in any of the samples,
so we can go back to the analyzed sample that contained 0.185 g Fe in 2.998 g of sample:
 0.185 g Fe 
mass % Fe = 
100% = 6.17% Fe
 2.998 g tablets 
(c) In this analysis, we assume that all chemical conversions are quantitative, that is all iron is
converted to Fe3+, which is all precipitated, and which all appears in the final product, Fe2O3.
We assume that the stoichiometry of the product is correct and that it is really only this product
with no leftover water or other contaminants. We assume that the 20 tablets chosen accurately
represent the entire batch of tablets (which may be many thousands or millions) being tested,
that the tablets are relatively uniform in composition (so the average is meaningful), and that the
sample taken for analysis has the same composition as the bulk sample from which it came.
Sampling problems are some of the most important in analyses like this, often adding more
uncertainly than the chemistry itself.
Problem 6.8.
We are to analyze a mixture containing only potassium sulfate, K2SO4, and ammonium sulfate,
(NH4)2SO4, to determine the mass percent of K2SO4 in the mixture. A 0.649 g sample of the
mixture is dissolved in water and then treated with excess barium nitrate, Ba(NO3)2, solution to
precipitate all of the sulfate ion as barium sulfate, BaSO4(s).
(a) The net ionic equation for the precipitation reaction is:
Ba2+(aq) + SO42–(aq)  BaSO4(s)
(b) The mass of precipitate formed is 0.977 g, so the number of moles of sulfate ion in the
original sample is:
 1 mol BaSO4   1 mol SO2–

4
0.977 g BaSO4(s) = (0.977 g BaSO4(s)) 
 233.4 g BaSO4   1 mol BaSO4 
= 0.00419 mol SO42–
(c) To find the mass percent of K2SO4 in the original sample, we first find the mass of K2SO4 in
the original sample. To do this, let w = mass of K2SO4 in the sample. Then we can write the
mass of (NH4)2SO4 in the sample as (0.649 – w) g. Use these masses to express the number of
moles of sulfate present in each compound and set their sum equal to the total number of moles
of sulfate from part (b). There is one mole of sulfate in one mole of each compound, so we
have:
mol K2SO4 + mol (NH4)2SO4 = total mol SO42– = 0.00419 mol
 1 mol K 2SO 4 
 1 mol (NH 4 )2 SO 4 
(w g) 
+ [(0.649 – w) g] 

 = 0.00419 mol
 174.3 g 

132.1 g
(0.00574)·w mol + 0.00491 mol – (0.00757)·w mol = 0.00419 mol
– (0.00183)·w mol = – 0.00072 mol
w = 0.393 g K2SO4
The mass percent K2SO4 in the sample is:
 0.393 g K 2SO 4 
mass percent K2SO4 = 
100% = 60.6 %
 0.649 g sample 
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Problem 6.9.
(a) The total volume of each sample in Figure 6.1 (six units of volume) is proportional to the
total number of moles of cation plus anion, because the cation- and anion-containing solutions
have the same molarity. The mole fraction of cation in each solution is the ratio of the volume
of cation added to the total volume. For the five samples, the mole fractions of cation are: 0.17,
0.33, 0.50, 0.67, and 0.83.
(b) Count the number of cation-anion pairs at the bottom of each container in Figure 6.1 to get a
measure that is proportional to the amount of precipitate formed. For the five samples, the
results are 2, 4, 6, 4, and 2. The plot of amount of precipitate as a function of the mole fraction
of cation is:
(c) The plot in part (b) has obvious straight-line segments (shown on the graph) that meet at the
point corresponding to the largest amount of precipitate formed. This intersection point occurs
when the mole fraction of cation is 0.5. The corresponding mole fraction of anion is also 0.5.
That is, the straight lines intersect where the mole fractions of the reactants are in the same ratio
as the stoichiometry of their reaction with one another.
(d) For an unknown precipitation stoichiometry, you could perform a similar series of reactions
in which you measure the mass of precipitate formed as you keep the total number of moles of
the cation and anion constant while varying their ratio. A plot of these data will probably yield a
graph similar to the one in part (b) with the intersection of the two straight-line portions of the
data occurring at the mole fraction of cation corresponding to the fraction of cations in the
precipitate. For example, in a 1:2 ionic compound, such as PbI2, the fraction of cations in the
solid is one-third (one in three of the ions in the compound is a cation), so the lines would
intersect at about 0.33 mole fraction of cation.
Problem 6.10.
The mole fractions of nickel cation in the six samples in Check This 6.11 are obtained by
dividing the number of moles of nickel cation by the total number of moles of nickel plus
dimethylglyoxime in each sample. The results are shown in this table (where all molar
quantities are mol  104)
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Chapter 6
Sample
2+
mol Ni (aq)
mol dmg
total mol
mol fraction Ni2+(aq)
mass ppt, g
1
2
3
4
5
6
2.55
18.2
20.8
0.123
0.074
4.89
15.8
20.7
0.236
0.141
5.98
14.8
20.8
0.288
0.173
7.50
13.1
20.6
0.364
0.189
9.89
10.9
20.8
0.475
0.158
12.6
8.19
20.8
0.606
0.118
The plot of the mass of precipitate formed as a function of the cation mole fractions is:
We see that the two straight-line segments intersect at a nickel cation mole fraction of 0.33; one
third of the ions in the solid ionic compound are nickel cation, Ni2+. The other two thirds are
dimethylglyoxime anions, so the ionic compound must be Ni(dmg)2. Note that this result also
tells us that each dimethylglyoxime anion must have a charge of –1 in order to balance the
cationic charge. When you analyzed the data in Check This 6.11, you probably focused on the
sample that produced the largest amount of precipitate and found that the molar ratio of nickel
to dimethylglyoxime in this sample solution is 7.50:13.1 = 1.00:1.75 and concluded (assuming
simple small whole numbers of each ion) that the cation to anion ratio in the solid product is
1:2. The graphical method confirms this analysis and provides even more convincing evidence
for this ratio.
Problem 6.11.
Lewis (or partial Lewis) structures are drawn here for several ions or molecules, in order to help
predict which will be Lewis bases. To act as a Lewis base, an ion or molecule must that a
nonbonding pair of electrons that can bond to a proton.
(a) Ammonia, :NH3, has a nonbonding electron pair on the nitrogen atomic core and is a Lewis
base.
(b) A proton (hydrogen atomic core), H+, has no nonbonding electron pairs and cannot be a
Lewis base.
(c) The ammonium ion, NH4+, has no nonbonding electron pairs and cannot be a Lewis base.
(d) The hydroxide ion,
and is a Lewis base.
OH
, has three nonbonding electron pairs on the oxygen atomic core
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(e) The cyanide ion, C N , has a nonbonding electron pair on both a carbon and nitrogen
atomic core and is a Lewis base. Because the carbon is less electronegative, it has the greater
ability to share its electron pair, so a proton will bond there to form HCN.
(f) Methyl amine, CH3NH2 , like ammonia, has a nonbonding electron pair on the nitrogen
atomic core and is a Lewis base.
(g) Although the oxygen atomic core in the hydronium ion, H3O+, has an unshared electron pair,
the species has an overall positive charge, which will repel another proton. The hydronium ion
shows no Lewis base properties.
(h) The methide ion,
Lewis base.
CH 3
, has a nonbonding electron pair on a carbon atomic core and is a
Problem 6.12.
(a) All Lewis bases have a pair of non-bonding electrons capable of forming a covalent bond
with a Lewis acid.
(b) Lewis acids can share a pair of non-bonding electrons from a Lewis base to form a covalent
bond. Protons are Lewis acids (just as they are Brønsted-Lowry acids), but the array of ions and
molecules that act as Lewis acids is much broader than this, as this chapter tries of show.
Problem 6.13.
From the point of view of the Brønsted-Lowry acid-base model, we might write the reaction in
this problem like this:
H
H O H
H
N H
H
H
H O
H
H N H
H
The red color of one of the hydrogen atomic cores draws attention to proton transfer from the
Brønsted-Lowry acid, H3O+ (hydronium ion), to the Brønsted-Lowry base, NH3 (ammonia). As
in all acid-base reactions, there are two acid-base pairs, H3O+/H2O and NH4+/NH3, that differ by
the presence of a hydrogen atom core in the acid that is not present in the base.
From the point of view of the Lewis acid-base model, we might write the reaction like this:
H
H O H
H
N H
H
H
H O
H
H N H
H
Here, the color focuses our attention on electron pairs that bond a hydrogen core to another
atomic core in the Lewis acids and that are nonbonding electron pairs in the Lewis bases. The
identification of the Lewis acids and bases and conjugate Lewis acid-base pairs is identical to
that for the Brønsted-Lowry model above, as it always is when it is a proton that is transferred to
the Lewis acid.
Problem 6.14.
The reaction of lactic acid, CH3CH(OH)C(O)OH, transferring its acidic proton to water in a
Brønsted-Lowry acid-base reaction, with the acid-base pairs denoted, is:
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Chapter 6
CH3CH(OH)C(O)OH + H2O  CH3CH(OH)C(O)O– + H3O+
acid 1 base 2
base 1
acid 2
We write the reaction with reversible equilibrium arrows, because we are told that the only a
fraction of the lactic acid molecules transfer their protons to water. Evidently the reaction can
go in either direction and comes to a balance point (equilibrium) when there are substantial
amounts of all the species present in the solution.
Problem 6.15.
The concentration of hydronium ion in a solution of pH 1.3 is:
[H3O+(aq)] = 10–pH = 10–1.3 = 0.05 M
Since the pH of a 0.05 M aqueous solution of perchloric acid, HOClO3, is 1.3, it looks as though
the acid has transferred its proton completely to the water:
HOClO3(aq) + H2O(aq)  H3O+(aq) + ClO4–(aq)
The perchlorate ion, ClO4–(aq), must be quite a weak base (similar to the chloride ion), since it
gives up all its protons to water. Therefore, if sodium perchlorate, NaClO4, is dissolved in
water, the solution pH will be the same as the water before the salt was added, because the weak
perchlorate base does not react with water to produce hydroxide ion and perchloric acid. Note
the position of the perchlorate ion in Table 6.2.
Problem 6.16.
A 0.1 M solution of sodium cyanide, NaCN, in water has a pH of 11, so the cyanide ion, CN–
(aq), must be reacting as a base to produce a basic solution containing some OH–(aq):
CN–(aq) + H2O(aq)  HCN(aq) + OH–(aq)
The concentration of hydronium ion in a solution of pH 11 is:
[H3O+(aq)] = 10–pH = 10–11 M
Rearranging equation 6.14 and substituting this value for [H3O+(aq)], we find:
[OH–(aq)] =
10 –14 M 2
10 –14 M 2
=
= 10–3 M
+
–11
10 M
[H 3O (aq)]
If the chemical reaction written above were to go to completion, a 0.1 M solution of sodium
cyanide would have [OH–(aq)] = 10–1 M. It appears that only about 1% of the cyanide ion reacts
as shown, so we should write the equation with double equilibrium arrows to signify that the
reaction reaches a balance point (equilibrium) when there are substantial amounts of all the
species present in the solution. This conclusion tells us that hydrogen cyanide gas, HCN,
dissolved in water should transfer some of its protons to the water molecules forming some
H3O+(aq). Thus, the pH of a 0.1 M aqueous solution of HCN, will be lower, more acidic, than
the water before the HCN was dissolved. Since, some HCN(aq) forms in aqueous cyanide
solutions, we know that the acid is not a strong acid that transfers all its protons to water. The
pH will be higher than 1 (the value for a 0.1 M solution of a strong acid), but lower than 7 (the
pH of pure water). An intermediate value like 4 would be reasonable estimate based on the acidbase concepts developed so far. [The mathematical analyses from Chapter 9 give a pH of 5.]
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Problem 6.17.
(a) From its position in Table 6.2, we see that 2,4-dinitrophenol is only a little stronger acid than
ethanoic (acetic) acid. Thus, the anionic (more water soluble), basic form of 2,4-dinitrophenol
readily accepts protons from hydronium ion to produce its electrically neutral (less water
soluble), acidic form. This is just the property we want in a molecule added to interrupt the
formation of ATP (by reducing the pH difference across membranes), so we can study how this
interruption affects other metabolic processes.
(b) Our bodies need a continuous supply of ATP to keep the processes of life going. If the rate
of production of ATP goes down, we compensate by metabolizing more carbohydrate and fat
molecules to try to maintain the difference in pH across membranes that is required for ATP
synthesis. The rate of production of ATP can be slowed down by the presence of 2,4dinitrophenol, which reduces the pH difference across membranes. In this case, the body
metabolizes more carbohydrate and fat than it usually would and the result is the loss of weight,
since these are the fuel molecules the body uses to store energy. This treatment can obviously be
carried too far with too large a decrease in ATP production and inability of the body to sustain
itself.
Problem 6.18.
(a) The reaction between boric acid and water with the Lewis structures written out is:
OH
OH
HO
B
OH
+
2 O H
H
HO
B OH
OH
+
H
O H
H
The boron atom is only surrounded by three electron pairs, so it can share an electron pair from
a water molecule. Boric acid is the Lewis acid, and water is the Lewis base on the left-hand side
of this equation. Note that you might think of this reaction occurring in two steps:
HO
HO
OH H
B O +
OH H
OH
+
B
O H
OH
H
HO
OH H
B O
OH H
OH
O H
H
HO
B OH
OH
+
H
O H
H
The first step is the Lewis acid-base reaction between the electron deficient boron atomic center
(see Problem 5.58), a Lewis acid, and one of the non-bonding electron pairs on a water
molecule to form a species that has three bonds to an oxygen core. This species would have
properties similar to a hydronium ion, in particular, it could transfer a proton to water to form a
hydronium cation and the B(OH)4– anion. These reactions are reversible and solutions of boric
acid are only weakly acidic (which is why you may be familiar with its use in dilute solutions as
an eyewash solution), so the equilibrium balance must lie far to the reactant side.
(b) Writing the formula for boric acid as H3BO3 is misleading, because it might suggest that the
three hydrogen atomic centers are directly bonded to the boron atomic center. The formula,
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Chapter 6
B(OH)3, makes the connectivity explicit and shows that this structure is related to the oxyacids
(although its acidic reaction is quite different). Conventionally, oxyacids, like sulfuric,
(HO)2SO2, are written with the acidic hydrogen centers on the left by analogy with simpler
acids like hydrochloric, HCl. Writing boric acid as B(OH)3 is a subtle way of indicating that it
does not act like other oxyacids, that is, that the hydrogen cores on this species are not acidic.
(c) Polyprotic acids, like sulfuric, (HO)2SO2, and phosphoric, (HO)3PO, are all oxyacids that
can donate more than one proton to a Lewis (or Brønsted-Lowry) base, because they have more
than one –OH group attached to the central atomic core. Boric acid, B(OH)3, is an interesting
apparent exception to this observation. The difference is that the central boron atomic core is
electron deficient and itself acts as a Lewis acid [see part (a)], to add an OH– group from a water
molecule as the proton from the water molecule is transferred to another water molecule to form
a single hydronium ion for each B(OH)3 that reacts, that is, boric acid is monoprotic.
Problem 6.19.
Based on these experimental observations, we are to rank these bases from weakest to strongest:
CH3C(O)O– (ethanoate or acetate), NO3– (nitrate), OH– (hydroxide).
(i) 0.1 M CH3C(O)OH has pH = 2.9
(ii) 0.1 M HNO3 has a pH = 1.0
(iii)0.1 M NaNO3 has a pH = water in which it was dissolved
(iv) 0.1 M CH3C(O)O–Na+ (sodium ethanoate) has a pH = 8.9
(v) 0.1 M NaOH has a pH = 13.0
The ranking is (weakest) NO3– < CH3C(O)O– < OH– (strongest)
Observation (ii) shows that nitric acid, HNO3(aq), transfers its protons completely to water,
since the pH corresponds to [H3O+(aq)] = 0.1 M, which is the concentration of the nitric acid
solution. In addition, observation (iii) shows that nitrate anion does not attract enough protons
from water to form HNO3(aq) and OH–(aq) and change the pH of the water in which NaNO3 is
dissolved. Thus, nitrate anion, NO3–(aq), must be a very weak base.
Observation (i) shows that ethanoic acid, CH3C(O)OH(aq), does not transfers all its protons
completely to water, since the pH corresponds to [H3O+(aq)] = 10–2.9 ≈ 0.001 M, which is less
than the concentration of the ethanoic acid solution. In addition, observation (iv) shows that
ethanoate anion attracts enough protons from water to form CH3C(O)OH(aq) and OH–(aq) and
makes the pH of the water in which CH3COO-Na+ is dissolved somewhat basic, pH 8.9. Thus,
ethanoate anion, , must be a stronger base than nitrate ion, which has no effect on the pH.
To compare the ethanoate and hydroxide anions, we consider the reaction that makes the
solution basic in observation (iv):
CH3COO-(aq) + H2O(aq)  CH3C(O)OH(aq) + OH–(aq)
If this reaction went to completion as written, the solution would have [OH–(aq)] = 0.1 M, with
a pH of 13. But the pH is only 8.9, showing that this reaction precedes only a small amount as
written. Since the weaker base predominates in equilibria like this, the ethanoate anion is a
weaker base than the hydroxide anion, which completes the ordering we were asked to find,
Problem 6.20.
(a) NH3 is a stronger Lewis base than PH3. The central atomic center in each molecule is in
group V of the periodic table. We know that basicity decreases going down a column of the
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table, because the size of the central atomic core increases and the charge density of the
nonbonding electrons becomes less concentrated.
(b) HP2– is a stronger Lewis base than S2–. The atomic center in each ion is in the third period of
the periodic table, so these atomic cores are about the same size. P is less electronegative than S,
so holds its nonbonding electrons less strongly, which makes them are more available to bond
with a Lewis acid and makes HP2– the stronger Lewis base.
(c) It’s difficult to predict whether HP2– or O2– is the stronger Lewis base. Because of size
considerations, we know that O2– is a stronger Lewis base than S2–. But from part (b) we know
that HP2– is a stronger Lewis base than S2–. Thus, we see that O2– and HP2– are both stronger
bases than S2–, but our directional models don’t give us any quantitative information about how
much stronger each one is, so our models are inadequate to make this prediction.
(d) CH3O– is a stronger Lewis base than CH3S–. The nonbonding electrons in each ion are on
atomic centers that are in group VI of the periodic table. We know that basicity decreases going
down a column of the table, because the size of the central atomic core increases and the charge
density of the nonbonding electrons becomes less concentrated.
Problem 6.21.
When equal volumes of 0.1 M aqueous solutions of sodium hydroxide, NaOH, and ammonium
chloride, NH4Cl, are mixed this net acid-base reaction can occur:
NH4+(aq) + OH–(aq)  NH3(aq) + H2O(aq)
Since equal volumes of solutions of the same molarity are mixed, the number of moles of the
reactants, NH4+(aq) and OH–(aq), are stoichiometrically equivalent and they would react
completely to use one another up, if this reaction were to go to completion. The Lewis bases in
this reaction are OH–(aq) and NH3(aq). The strongest base that can exist in aqueous solution is
the hydroxide ion, so NH3(aq) is the weaker base and it will be the predominant Lewis base in
the solution, that is, the reaction will lie far toward the product side. Water is a Lewis acid in
this reaction system and it will be the predominant Lewis acid in the solution. As we have said
in the text, it is a bit unfair to count water as the predominant Lewis acid (or base) in aqueous
solution, because its concentration overwhelms all other species, regardless of whether it is the
weaker acid or base. In this case, the comparison is somewhat valid, because water is on the
product side and it is the weaker acid compared to the ammonium ion, NH4+(aq.
Problem 6.22.
The reactions that describe the solution that results when equal volumes of 0.10 M NH3 and
0.10 M HCl are mixed can be written as:
HCl(aq) + H2O(aq)  H3O+(aq) + Cl– (aq)
H3O+(aq) + NH3(aq)  NH4+(aq) + H2O(aq)
We know that the chloride anion is such a weak base that HCl(aq) completely transfers its
proton to water, as shown in the first equation. Since equal volumes of solutions of the same
molarity are mixed, the number of moles of the reactants, H3O+(aq) and NH3(aq), are
stoichiometrically equivalent and they would react completely to use one another up, if the
second reaction were to go to completion. The Lewis acids in this reaction are H3O+(aq) and
NH4+(aq). The strongest acid that can exist in aqueous solution is the hydronium ion, so
NH4+(aq) is the weaker acid and it will be the predominant Lewis acid in the solution, that is,
the reaction will lie far toward the product side. Thus, the predominant ions in the solution will
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be the Cl– (aq) anion from the first acid-base reaction and the NH4+(aq) cation from the second.
There will, of course, be hydronium and hydroxide ions in the solution, but their concentrations
will be much lower.
Problem 6.23.
For two oxyanions with the same amount of electron delocalization, such as (HO)2AsO2– and
(HO)2PO2–, the anion with the larger central atom is more stable (the ion is larger and the
electrons can occupy a larger volume) and does not accept protons as well. Since atomic size
increases down a family of the periodic table, we expect the dihydrogen arsenate anion to be a
somewhat weaker base than the dihydrogen phosphate anion. Conversely, we expect arsenic
acid, (HO)3AsO, to be a somewhat stronger acid (more transfer of a proton to water) than
phosphoric acid, (HO)3PO. Experimentally, we find that these two bases (acids) are almost
equal in their basicity (acidity); the size difference between a phosphorus atom and an arsenic
atom is small.
Problem 6.24.
(a) The Lewis structures for CH3C(O)O– and CH3C(O)S– are:
H
O
H C C
H O
and
H O
H C C
H O
H
S
H C C
H O
and
H
S
H C C
H O
These ions have delocalized pi electrons spread over the three atoms of the carboxyl, O–C–O,
and thiocarboxyl, S–C–O, groups.
(b) The CH3C(O)O– (ethanoate) anion is likely to be the stronger base, because the
delocalization of the electrons is likely to be enhanced by the somewhat larger sulfur atomic
center in the CH3C(O)S– (thioethanoate) anion, which would stabilize thioethanoate more than
ethanoate. This added stability would make CH3C(O)S– less likely to react with a proton donor
and hence make it the weaker base. [Quantitatively, CH3C(O)S– (pKb = 10.67) is about 26 times
less basic than CH3C(O)O– (pKb = 9.25).]
Problem 6.25.
6.25. (a) A 0.1 M solution of potassium amide, KNH2, in water has a pH of 13. Write a
chemical equation for the acid-base reaction that occurs in this solution. What is the
predominant Brønsted-Lowry acid and base in the solution? How would you characterize the
base strengths of the bases in this solution. Explain your responses.
(b)
A 0.1 M solution of ethylamine, CH3CH2NH2, in water has a pH of about 11. Write a
chemical equation for the acid-base reaction that occurs in this solution. What is the
predominant Brønsted-Lowry acid and base in the solution? How would you characterize the
base strengths of the bases in this solution. Explain your responses.
(c)
How are the reactions you write and your interpretations of the experimental evidence
in parts (a) and (b) the same and different? Explain clearly.
Answer to 6.25:
(a) A 0.1 M solution of potassium amide, KNH2, in water has a pH of 13. In order to form a
strongly basic, pH 13, solution, the amide ion must react with water to accept a proton:
NH2–(aq) + H2O(l)  NH3(aq) + OH–(aq)
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In Check This 6.18, you calculated that the concentration of hydroxide ion in this solution is 0.1
M, which is the concentration of amide ion that would be present if reaction had not occurred.
We can conclude that essentially all of the amide ion has reacted and that reaction equation
(6.12), which shows the reaction going to completion, better represents the situation in this
solution:
NH2–(aq) + H2O(l)  NH3(aq) + OH–(aq)
The predominant acid and base in this solution are ammonia, NH3(aq), and hydroxide ion, OH–
(aq), respectively. Since the weaker acid and base predominate in a solution, we can conclude
that the amide ion, NH2–(aq), is a stronger base than the hydroxide ion.
(b) A 0.1 M solution of ethylamine, CH3CH2NH2, in water has a pH of about 11. In order to
form a basic, pH 11, solution, the ethylamine must react with water to accept a proton:
CH3CH2NH2(aq) + H2O(l)  CH3CH2NH3+(aq) + OH–(aq)
In this pH 11 solution, [H3O+(aq)] = 10–11 M, and, from equation (6.14), we get:
10 –14 M 2
10 –14 M 2
[OH (aq)] =
=
= 10–3 M
+
–11
10 M
[H 3O (aq)]
–
This concentration of hydroxide ion is 100 times less than the concentration of ethylamine in the
solution. We can conclude that the reaction we wrote occurs to only a small extent, so the
predominant acid and base in this solution are water, H2O(l), and ethylamine, CH3CH2NH2(aq),
respectively. Since the weaker acid and base predominate in a solution, we can conclude that the
hydroxide ion is a stronger base than ethylamine.
(c) The reactions we wrote in parts (a) and (b) are essentially the same, a base accepts a proton
transferred from water to form a solution with excess hydroxide ion compared to pure water.
The great difference is in the extent of the proton transfer. In part (a), the transfer is “complete,”
with virtually none of the amide ion left unreacted; the amide ion is a strong base in water. In
part (b), the transfer is incomplete and a great deal of the ethylamine is left unreacted;
ethylamine is a weak base in water.
Problem 6.26.
(a) Consider the Lewis acid-base reactions occurring in these reaction steps for the formation of
carbonic acid, H2CO3 [(HO)2CO], from water and carbon dioxide:
H
O
H O
H O
H O
C
H O C
H O C
O
O
O
The water is acting as an electron-pair donor, so it is the Lewis base. The C=O bonds in carbon
dioxide are polarized with the more electronegative oxygen atomic cores pulling electron
density away from the carbon. Thus, a positive center is created at the carbon core, which acts
as an electron-pair acceptor, so it is the Lewis acid (as represented in the reaction by the arrow
showing the attraction of the electron pair to this atomic center). As the new C–O bond forms,
there has to be a redistribution of electrons, because only four electron pairs can be
accommodated around the carbon core. Therefore, as the small red arrow indicates, one of the
pairs of electrons forming the pi bond between carbon and oxygen moves to an oxygen. This
oxygen now has three nonbonding electron pairs, which gives it properties similar to the oxygen
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core in hydroxide ion, OH–, a strong base. The oxygen core just bonded to the carbon has three
sigma bonds to other atomic cores, just as in the hydronium ion, H3O+, a strong acid. Thus, the
intermediate molecular structure is set up for the electron pair on the basic oxygen to donate a
pair of electrons to the acidic hydrogen atomic core on the other, as represented by the small red
arrow on the center structure. The result is the formation of the carbonic acid structure at the
right, in which all atomic centers have their conventional number of bonding and nonbonding
sigma electron pairs.
(b) We have previously written the overall chemical equation for this reaction as:
H2O(l) + CO2(g)  H2CO3(aq)
The structural equation in part (a) gives more information about electron arrangement, making it
easier to identify the electron pair donors and acceptors. The formula equation does give not
give information about which atoms are linked to which nor any indication which are the Lewis
acids and bases in this reaction.
Problem 6.27.
(a) The chemical equation H2O + HCO3–  H3O+ + CO32– can be rewritten using the
representations given in Table 6.2 and identifying all Lewis acids and bases in the reaction and
the conjugate pairs of Lewis acids and bases (red is used to label nonbonding electron pairs in
Lewis bases and corresponding bonding pairs in Lewis bases):
conjugate pair
Lewis acid
H2O
H OCO2Π
Lewis base
H+
H2O
Lewis acid
Lewis base
OCO22Π
conjugate pair
(b) The hydronium ion, H3O+(aq), is the strongest acid that can exist in aqueous solution, so the
weaker acid, HOCO22–(aq), will be the predominant acid in this reaction. The equilibrium will
lie mostly toward the reactant side where the weaker Lewis acid (and weaker Lewis base) is
found. The same conclusion can be drawn comparing the Lewis bases, water and carbonate
anion. Sodium carbonate, Na2CO3 (often called washing soda), dissolves in water to form a
basic solution, showing that the carbonate anion wins the competition with water molecules for
protons and is the stronger base and will not predominate in the reaction we wrote.
Problem 6.28.
Since aluminum is in the boron family of elements, group III, its bonding characteristics will
probably be similar to boron’s. In boron compounds, the boron atomic center is often electron
deficient, because the atom has only three valence electrons, and, in Problem 6.18, we found
that boric acid, B(OH)3, reacts with water to form the B(OH)4– anion (and hydronium). It seems
likely that Al(OH)3(s) might dissolve in basic solution by reacting with a hydroxide ion to form
a soluble anion:
Al(OH)3(s) + OH–(aq)  Al(OH)4–(aq)
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mol fraction en
A640
A560
0.000
0.083
0.167
0.250
0.333
0.417
0.500
0.583
0.667
0.750
0.833
0.917
0.044
0.095
0.150
0.200
0.256
0301
0.001
0.026
0.050
0.075
0.107
0.146
0.261
0.187
0.307
0.411
0.312
0.208
0.105
Absorbance
Problem 6.29.
(a) The independent variable in a continuous variations analysis is the mole fraction one of the
reactants is of the total number of moles of reactants added. If we suspect that reactant A will
react with more than one mole of reactant B, then, the graphs are easier to interpret if we plot
the mole fraction of reactant B. For the copper-ethylenediamine complex, it’s likely that more
than one ethylenediamine (en) will react with each copper (Cu) ion, so we’ll use the mole
fraction of en in each sample as the independent variable. Since the molarities of the reactant
solutions are equal, there is the same total number of moles of Cu plus en in each sample. The
fraction that is en is just the ratio of the volume of en added divided by the total volume, 12 mL
in each case. This table gives the mole fraction of en in each sample and their absorbances at
640 nm (red light) and 560 nm (green light). The graph shows the two absorbances, A, plotted
as a function of the mole fraction of en with blue squares for A640 and magenta circles for A560.
[These are actual data from an experiment on the formation of Cu(II)-ethylenediamine
complexes for use as an analytical method for determining
Cu(II)
concentrations.]
G
are
Qraphics
uickTim
needed
e™
decom
toand
see
pressor
athis
picture.
0.450
0.400
0.350
0.300
0.250
0.200
0.150
0.100
0.050
0.000 0.200 0.400 0.600 0.800 1.000
0.000 mol fractionen
The maximum absorbance at 640 nm is reached at a lower mole fraction of en than for the
absorbance at 560 nm. The problem states that copper (II) solutions initially become a deep blue
when en is added. A solution that absorbs red light (640 nm) will appear blue, since blue light is
being transmitted. Note that there is some light absorption with no en added and this is
consistent with the original light blue copper (II) solution. At higher mole fractions of en, the
absorbance at 640 nm decreases, but the absorbance at 560 nm (green light) continues to grow.
If green light is absorbed by the solution, the colors that are transmitted will be violet and red,
which accounts for the magenta color of solutions with more en added. At even higher mole
fractions of en, the absorbance at 560 nm also decreases.
If we extrapolate both the increasing and decreasing parts of the A640 plot, it looks like the lines
would meet at about a mole fraction of 0.5, at an absorbance of about 0.35. At a mole fraction of
0.5, the number of moles of Cu and en are equal. We conclude that a 1:1 complex, Cu(en)2+(aq),
is a deep blue.
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Chapter 6
The peak in the 560 nm absorbance is at an en fraction of 0.67, that is two-thirds en and onethird copper ion [twice as much en as Cu(II)]. We can conclude that increasing amounts of en
produced a second complex with a 1:2 ratio of copper (II) to en, Cu(en)22+(aq). When more en is
added, the absorbance at 560 nm decreases linearly with decreasing number of moles of
copper(II) in solution. This is exactly what you expect, if no complexes with more than two en
are formed. Once there are two or more en molecules for every copper(II), all the copper(II) two
is complexed and the absorbance of the solution depends only on how many moles of copper(II)
are present. When the fraction of en is 1.00 (no copper(II) in the solution), the absorbance is
zero, as the extrapolation of the plot shows.
(b) The color (and the absorbance) of these solutions are only dependent on the ratio of en to
Cu. Thus, if the 6:6 solution had been mixed to make it uniform before it was spilled, spilling
some would make no difference to the absorbance. The students could have used the sample (as
long as there was enough left to measure in their spectrophotometer). The estimated 640 nm
absorbance of this sample was given above. You can get the estimated 560 nm absorbance by
interpolating between the points on either side of the missing point. Drawing a smooth curve
through the points gives an estimated absorbance of 0.22.
(c) A model of the 1:1 complex has one ethylenediamine attached through the lone electron
pairs on its two nitrogen atomic cores at two positions on the metal ion. The 2:1 complex has
two ligands (ethylenediamine molecules) bonded by their nitrogen lone electron pairs at four
positions around the central metal ion. The geometry of the nitrogen atomic cores around the
Cu(II) central atomic center might be either square planar or tetrahedral. [The evidence is that
the complex has the nitrogen atomic cores arranged in a square about the Cu(II) ion with water
molecules occupying positions above and below the plane of the square, so that the bonding
about the Cu(II) core is octahedral.]
Problem 6.30.
The light blue solution first formed when a 4.0 M aqueous solution of ethylenediamine, en, is
added slowly to a 200 mL of a stirred 0.10 M aqueous solution of nickel(II) ion is probably the
color of a complex between the Ni(II) ion and one en, Ni(en)2+. As more en solution is added
the solution changes from light blue to purple, which suggests that further Ni(II)-en complexes
are formed with two or more en ligands per metal ion. The color stops changing after 15 mL of
the en solution has been added. We can calculate the ratio of moles of en to moles of Ni(II)
when the color stops changing to get an indication how many en ligand molecules complex to
the Ni(II). The number of moles of en added is:
mol en = (4.0 M)(0.015 L) = 0.060 mol
The number of moles of Ni(II) in the original solution is:
Mol Ni(II) = (0.10 M)(0.200 L) = 0.020 mol
The ratio en:Ni(II) is 3:1 when the color stops changing, so we can conclude that the purple
complex is probably Ni(en)32+.
Problem 6.31.
(a) This is a representation of the structure of BAL (British AntiLewisite with the H atoms
omitted) that focuses on the relationship of the three group VI atomic cores (each of which has
two pairs of nonbonding electrons) to one another. All three of these Lewis base centers are
“reaching” toward you from the plane defined by the three carbon cores.
ACS Chemistry FROG
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Chapter 6
Chemical Reactions
C
O
C
C
S
S
(b) The nonbonding electron pairs in the molecule (and your model) are in position to form
bonds to a single Lewis acid central metal ion, such as Pb2+. Just as ethylenediamine can form
two bonds in a metal ion complex and EDTA can form six, BAL can form three, so two of these
molecules could complex with and successfully sequester potentially toxic Pb2+ ions, preventing
them from being incorporated into bone and blood. The sulfur atomic centers are weaker bases
than the oxygen center and probably will transfer their protons to water (forming BAL2–) as they
bind to the metal ion. A likely complex with Pb(II) would be Pb(BAL)2–, a soluble complex that
normal excretion processes taking place in the kidney will eliminate from the body.
Problem 6.32.
(a) Brønsted-Lowry acid-base reactions occur when a positively polarized proton donor (B-L
acid) and a negatively polarized proton acceptor (B-L base) are attracted to one another and get
close enough for the proton to be transferred from the acid to the base. The polarized molecules
charges may carry full ionic charges or partial charges caused by electronegativity differences
within the molecules. Lewis acid-base reactions shift the focus from proton exchange to
electron-pair donors (Lewis bases), which are negatively polarized, and electron-pair acceptors
(Lewis acids), which are positively polarized. When these positive and negative centers are
attracted close enough, an electron pair from the Lewis base forms an electron-pair bond with
the positive center on the Lewis acid.
(b) Brønsted-Lowry acid-base definitions are particularly straightforward to apply to aqueous
proton transfer systems. The particular usefulness of Lewis acid-base definitions is extension of
the concept of acid-base reactions to many organic reactions and to complex-ion reactions.
Problem 6.33.
(a) If we analyze this possible reaction, CH3C(O)OH + NH3  CH3C(O)NH2 + H2O, as an
electrophile-nucleophile reaction, the electrophile is CH3C(O)OH. The carbonyl carbon is
positively polarized, because the two electronegative oxygen atoms draw bonding electron
density away from the carbon. Thus, the carbonyl carbon attracts electron density from another
molecule and is an electrophile.
(b) The ammonia molecule, NH3, has an unshared pair of electrons that can interact with a
positive center in another molecule, so it is a nucleophile.
(c) These reactants, a carboxylic acid and a base (ammonia) can undergo an acid-base reaction,
CH3C(O)OH + NH3  CH3C(O)O– + NH4+, instead of an electrophile-nucleophile reaction.
The marginal note on page 391 suggests that electrophile-nucleophile reactions are often slow
and we have observed that acid-base reactions (proton transfers) are quite rapid. Thus, it seems
most likely that the acid-base reaction will be observed. [This is true.]
Problem 6.34.
(a) The structural formula of the tripeptide that is formed when three glycine molecules,
H2NCH2C(O)OH, are linked together by amide bonds structure is:
H O
H O
H
O
H N C C N C C N C C
OH
H H
H H
H H
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(b) The reaction equation for formation of this tripeptide is:
H
O
3H N C C
OH
H H
H O
H O
H
O
2 H2 O
H N C C N C C N C C
OH
2
H H
H H
H H
Problem 6.35.
The animation in the Web Companion, Chapter 6, Section 6.7, page 2, shows the (somewhat
hypothetical) formation of an amide from a carboxylic acid and an amine that we can compare
to the reaction sequence shown in equation (6.34) for formation of an ester from a carboxylic
acid and an alcohol. The animation represents the attraction of the electrophilic carbonyl carbon
for the electron pair on the nucleophilic nitrogen of the amine as a developing bond when the
molecules approach one another. This is the attraction represented in equation (6.34) by the long
curved arrow from the electron pair on the alcohol oxygen toward carbonyl carbon in the acid.
(By convention, these kinds of arrows go from electrons to the atomic core where they end up.
This is because electrons are moving rapidly and the much heavier cores move quite slowly, so
rearrangements are likely to involve mostly electronic movement.) As the molecules get closer
and the new bond in the animation gets more prominent, note that the double bond from the
carbon to the carbonyl oxygen fades to become a single bond. This is represented in equation
(6.34) by the small curved arrow in the left-hand structure that shows one of the pairs of
electrons in the C=O double bond moving onto the oxygen. The result in the animation is a
structure like the central structure in the equation: a nitrogen in the animation replaces the
oxygen that comes from the alcohol in the equation. The animation next shows the transfer of a
proton from the nitrogen (now bonded to the original carbonyl carbon) to the oxygen that was
the –OH oxygen in the carboxylic acid. The proton transfer occurs from an ammonium-like
nitrogen species (sigma bonded to four other atoms) to the oxygen. Similarly, equation (6.34)
represents such a proton transfer by the two lower curved arrows in the central structure, which
show a proton transfer from a hydronium-like oxygen species (sigma bonded to three other
atoms) to the oxygen that was the –OH oxygen in the carboxylic acid. The animation follows
this transfer with the loss of a water molecule, which takes with it the electron pair that was
bonding it to the original carbonyl carbon. As the carbon-oxygen bond extends and breaks, note
that the double bond from reforms between the carbonyl carbon and the originally doublebonded oxygen. This concerted bond breaking and formation are represented in equation (6.34)
by the upper pair of curved arrows in the central structure. The reaction is now complete with
the formation of an amide (in the animation) or an ester (in the equation) and a molecule of
water in both cases. These representations are quite parallel and hopefully reinforce and clarify
the molecular changes that occur in these electrophile-nucleophile reactions. Both
representations are probably incorrect in some details, but they provide pictures that help us
understand how these reactions can occur.
Problem 6.36.
The nucleophile, the electrophile, and any spectator ions are identified in each of these balanced
equations.
(a) NaOH + CH3–I  NaI + CH3–OH
The oxygen (electron pair) of hydroxide (HO–) is the nucleophile. The positively polarized
carbon atom of methyl iodide (CH3—I) is the electrophile. The sodium cation (Na+) is a
spectator ion.
ACS Chemistry FROG
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Chapter 6
Chemical Reactions
(b) (CH3)3N + CH3–I  (CH3)4N+ + I–
The nitrogen (electron pair) of trimethyl amine ((CH3)3N) is the nucleophile. The positively
polarized carbon atom of methyl iodide (CH3—I) is the electrophile. There are no spectator
ions.
(c) (CH3)3C–Br + CH3CH2OH  (CH3)3C–O–CH2CH3 + HBr
The oxygen (electron pair) of ethanol (CH3CH2OH) is the nucleophile. The central (tertiary)
carbon positively polarized of tertiary butyl bromide ((CH3)3C—Br) is the electrophile. There
are no spectator ions.
(d) NH3 + CH3C(O)OCH2CH3  CH3C(O)NH2 + CH3CH2OH
The nitrogen (electron pair) of ammonia (NH3) is the nucleophile. The positively polarized
carbonyl carbon of ethyl acetate (CH3C(O)OCH2CH3) is the electrophile. There are no spectator
ions.
(e) CH3NH2 + CH3C(O)Cl  CH3C(O)NHCH3 + HCl
The nitrogen (electron pair) of methylamine (CH3NH2) is the nucleophile. The positively
polarized carbonyl carbon of acetyl chloride (CH3C(O)Cl) is the electrophile. There are no
spectator ions.
Problem 6.37.
The formulas of the alcohol and the acid from which each ester is made are shown:
(a)
(b)
(c)
(d)
(e)
ester
CH3C(O)OCH2CH2CH(CH3)2
CH3CH2CH2C(O)OCH2CH3
CH3C(O)O(CH2)7CH3
C6H5C(O)OCH3
CH3C(O)OCH2C6H5
alcohol
HOCH2CH2CH(CH3)2
HOCH2CH3
HO(CH2)7CH3
HOCH3
HOCH2C6H5
acid
CH3C(O)OH
CH3CH2CH2C(O)OH
CH3C(O)OH
C6H5C(O)OH
CH3C(O)OH
Problem 6.38.
The formula of the ester that is formed by reaction of each of these acid and alcohol
combinations is shown.
(a) CH3(CH2)2C(O)OH + CH3(CH2)4OH  CH3(CH2)2C(O)O(CH2)4CH3 + H2O
(b) CH3C(O)OH + CH3CH2OH  CH3C(O)OCH2CH3 + H2O
(c) CH3C(O)OH + (CH2)3CH(CH2)2OH  CH3C(O)O(CH2)2CH(CH3)2 + H2O
(d) CH3CH2CH2C(O)OH + CH3OH  CH3CH2CH2C(O)OCH3 + H2O
(e) HC(O)OH + CH3CH2OH  HC(O)OHCH2CH3 + H2O
Problem 6.39.
The reactions of an alcohol with a carboxylic acid, reaction (6.34), and an alcohol with an
aldehyde, this problem, are shown here for comparison.
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H
H
H O
H C C
+
O H
H
H
H3 C O
O
H
C C H
H H
C
C C
H
O
H
O H
H
C H
H C
H
H H
H O
O
H
H O
H
CH2 CH3
H O
C C
+ H O
H
O
H
H C H
H C H
H
H O
H3 C O C CH2 CH3
H3 C O C CH2 CH3
H
H
In both cases, an electrophilic carbon atom doubly-bonded to oxygen reacts with a nucleophilic
nonbonding electron pair on the alcohol oxygen atom to form a new sigma bond between the
carbon and oxygen atoms. The electron rearrangement accompanying this interaction displaces
one of the electron pairs in the carbon-oxygen double bond onto the oxygen. In the
intermediates shown, the oxygen atom with three nonbonding electron pairs has a –1 formal
charge and the oxygen atom bonded to three other atoms has a +1 formal charge. In the second
reaction, these formal charges are reduced to zero by transfer of a proton from the oxygen with
the positive formal charge and an acidic proton to the oxygen with the negative formal charge
acting as a base to accept the proton. The result is the hemiacetal product shown. An exactly
analogous rearrangement can take place in the intermediate in the first reaction. The result
would be:
OH
H3 C
C OH
OCH 2CH 3
Experimentally, we find that molecules (intermediates) with carbon atoms sigma bonded to four
other groups, two of which are –OH groups, are unstable and readily lose water to form a
carbonyl double bond between the carbon atom and the remaining oxygen atom. These are the
products shown in the first reaction, which may actually proceed through this second
intermediate to the final product. Thus, the difference between reactions of alcohols with
carboxylic acids compared to those with aldehydes and ketones is a result of the intermediate
structure with its two –OH groups from the carboxylic acid. This reaction goes on to eliminate a
water molecule, a fate that is not possible for the product from the aldehyde or ketone reaction.
Problem 6.40.
(a) This intramolecular reaction sequence between an alcohol and aldehyde functional group in
the same molecule is modeled after the intermolecular sequence shown in Problem 6.39. To
simplify the pictures, we use skeletal structures with only the atoms that enter into the reaction
shown explicitly.
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H O
H
H
O
C
O
O
O
O H
C
C
H
H
(b) Skeletal structures for the product from part (a) and for glucose are:
OH
O
O
OH
HO
OH
HO
OH
The similarities between these structures are: (1) the six-membered ring with an oxygen atom as
one of the ring atoms, (2) the –OH group bonded to one of the ring carbon atoms adjacent to the
ring oxygen atom, and (3) the non-ring, sixth carbon bonded to the other ring carbon adjacent to
the ring oxygen.
Problem 6.41.
(a) The glucose ring carbon atom denoted by the arrow in this structure is bonded to two
electronegative oxygen atoms:
OH
O
HO
HO
OH
OH
This carbon atom acts very much like a carbon atom doubly bonded to an oxygen atom (which
it is in the linear form of the glucose molecule). Thus, this electrophilic carbon (Lewis acid) is
likely to be the site of reaction between this glucose molecule and an electron pair from another.
(b) One of the nonbonding electron pairs on the oxygen atom that is highlighted in the reaction
shown will be the pair to interact with the electrophilic center (carbon) in the other molecule to
form the product shown in the problem. The highlighted oxygen ends up bonded to this
electrophilic carbon:
OH
OH
O
HO
HO
OH
O
OH
OH
HO
O
OH
HO
OH
OH
HO
HO
O
O
OH HO
OH
H 2O
OH
The oxygen atomic core (and all its valence electrons) originally bonded to the electrophilic
carbon ends up in the water product. Loss of this oxygen and formation of water is represented
very schematically by the dashed ellipse.
Problem 6.42.
(a) Lewis structures for nitrite anion and isoamyl nitrite are:
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N
O
Chapter 6
N
O
O
O
H2
C
CH3
CH
C
CH3
H2
(b) The reaction between the nucleophilic nitrite anion and the electrophilic carbon center
(bonded to the electronegative bromine atom) is:
electrophile
spectator ion
Na+
N
O
O
Br
C
H2
H2
C
CH3
CH
C
CH3
H2
N
O
O
H2
C
CH3
CH
C
CH3
H2
Na+
BrΠ
nucleophile
(c) The nitrogen atomic center, with its unshared electron pair, as well as the oxygen atomic
centers of the nitrite anion, are possible nucleophiles.
(d) Reaction of the other nucleophilic site in the nitrite anion with 3-methyl-1-bromobutane is
represented by this reaction that forms 3-methyl-1-nitrobutane.
Na +
N
O
O
Br
C
H2
H2
C
CH3
C
H2
CH
O
CH3
N
O
H2
C
CH3
C
H2
CH
CH3
Na +
BrΠ
(e) There is reason to expect the oxygen atoms in the nitrite ion to be less nucleophilic than the
oxygen atom in the hydroxide. In the Lewis structure shown above, the right-hand oxygen core
is sigma bonded to another atomic core and has three pairs of unshared electrons, just like the
hydroxide ion. However, the pi electrons in the nitrite anion are delocalized and spread over all
three atomic centers:
N
O
O
O
N
O
Delocalization gives each of the oxygen atomic centers an electron density intermediate
between the extremes represented by either Lewis structure, thus making the nitrite anion
oxygen centers less nucleophilic than the oxygen in the hydroxide ion.
Problem 6.43.
(a) The stoichiometry of the reaction of glycerol, OHCH2CH(OH)CH2OH, two carboxylic
acids, R1C(O)OH and R2C(O)OH, choline, HOCH2CH2N(CH3)3+, and phosphoric acid to
produce phosphatidyl choline can be represented as:
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O
O
C
C
OH
R2
HO
O
OH
HO
OH
OH
O
R1
O
C
O
R2
C
OH
HO
N
O
P
O
O
P
O
N
4H2O
O
O
(b) All four of the bonds formed between the units are ester bonds, bonds formed by loss of
water between and an acid and an alcohol in an electrophile-nucleophile reaction. The
electrophilic carbon atoms in the two carboxylic acids and the electrophilic phosphorus atom in
the dihydrogen phosphate ion (which reacts to form a phosphodiester) are highlighted in blue.
The nucleophilic alcohol oxygen atoms are highlighted in red.
Problem 6.44.
(a) When the four pairs of sigma bonding electrons in NH4+ are divided equally between the
atomic cores they bond, the N has four valence electrons and a core charge of +5, so its formal
charge is +1. Each of the H atomic cores has one valence electron and a core charge of +1, so its
formal charge is 0. The sum of the formal charges, +1 + 0 + 0 + 0 + 0 = +1, is the charge on the
ion, as it must be.
(b) Breaking apart a Lewis structure for NO3– helps to find the formal charges on each atom:
–
O
N
O
O
N
O
O
O
The topmost O core is surrounded by six valence electrons and has a 6+ core charge, so its
formal charge is 0. Both of the lower O cores are surrounded by seven valence electrons and
have a 6+ core charge, so they each have a formal charge of –1. The N core is surrounded by
four valence electrons and has a 5+ core charge, so its formal charge is +1. The sum of the
formal charges, 0 + 1– 1– 1 = –1, is the charge on the ion.
[Recall that there are three energy equivalent Lewis structures for the nitrate ion:
O
O N
O
O
O N
O
O
O N
O
Note that it does not matter which of the three equivalent Lewis structures we use to find the
formal charges. The result is that two O atoms have –1 formal charge and the third has 0 formal
charge. Since all the Os are equivalent, we can “average” the formal charge on each to give all
three a –2/3 formal charge. The sum of the formal charges is then 1 – 2/3 – 2/3 – 2/3 = –1, as it
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must be. This is a more realistic way to think about the formal charge in a molecule or ion with
delocalized electrons.]
(c) Breaking apart the Lewis structure for NH3 gives:
H
H N H
H
H
N
H
The N core is surrounded by five valence electrons and has a 5+ core charge, so its formal
charge is 0. Each of the H atomic cores has one valence electron and a core charge of +1, so its
formal charge is 0. The sum of the formal charges is 0 for this electrically neutral molecule.
(d) Breaking apart the Lewis structure for H2O gives:
H
H O
H
O
H
The O core is surrounded by six valence electrons and has a 6+ core charge, so its formal charge
is 0. Each of the H atomic cores has one valence electron and a core charge of +1, so its formal
charge is 0. The sum of the formal charges is 0 for this electrically neutral molecule.
Problem 6.45.
Lewis structures with formal charges for each atom are shown for five anions:
[NOTE: Most of the solution for Problem 6.46 is also included here, so as to pull together the
ideas of formal charge and electron delocalization that spreads formal charge more uniformly
over a molecule or ion.]
(a) Three energy equivalent Lewis structures are possible for the carbonate anion, CO32–:
-1
2Π
0
O
0
O C
-1
O
2Π
O C
O
O
2Π
O C
O
O
The formal charge on each atom in any one of these structures is like that shown for the first
one. However, as we argued in the solution to Problem 6.44(b), these may be a bit misleading,
because delocalization of the pi electrons makes all the O atoms equivalent and each will have a
formal charge of –2/3. (In two out of three of the structures, a particular O has a –1 formal
charge and in the other structure it has a 0 formal charge, thus averaging to –2/3.) The sum of
the formal charges, 0 – 2/3 – 2/3 – 2/3 = –2, is the charge on the anion.
(b) Two energy equivalent Lewis structures are possible for the nitrite anion, NO2–:
–
N –1
O
O
0
–
0
O
N
O
The formal charge on each atom in either of these structures is like that shown for the first one.
However, delocalization of the pi electrons makes both the O atoms equivalent and each will
have a formal charge of –1/2. (In half of the structures, a particular O has a –1 formal charge and
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in the other structure it has a 0 formal charge, thus averaging to –1/2.) The sum of the formal
charges, 0 – 1/2 – 1/2 = –1, is the charge on the anion.
(c) Four energy equivalent Lewis structures are possible for the phosphate anion, PO43–:
0
-1
O
0
3Π
O
-1
O P O
O
3Π
O P O
O
-1
O
3Π
O
O P O
O P O
O
O
3Π
The formal charge on each atom in any one of these structures is like that shown for the first
one. However, delocalization of the pi electrons makes all the O atoms equivalent and each will
have a formal charge of –3/4. (In three out of four of the structures, a particular O has a –1
formal charge and in the fourth structure it has a 0 formal charge, thus averaging to –3/4.) The
sum of the formal charges, 0 – 3/4 – 3/4 – 3/4 – 3/4 = –3, is the charge on the anion.
(d) Three energy equivalent Lewis structures are possible for the hydrogen phosphate anion,
HPO42–:
0
OH 0
0
O P O
-1 O
2Π
-1
OH
O P O
O
2Π
OH
O P O
O
2Π
The formal charge on each atom in any one of these structures is like that shown for the first
one. (The formal charge on the H is 0, but, for simplicity, is not shown.) However,
delocalization of the pi electrons makes all the O atoms not bonded to hydrogen equivalent and
each will have a formal charge of –2/3. (In two out of three of the structures, a particular O has a
–1 formal charge and in the other structure it has a 0 formal charge, thus averaging to –2/3.) The
sum of the formal charges, 0 + 0 + 0 – 2/3 – 2/3 – 2/3 = –2, is the charge on the anion [just as in
the case of the carbonate anion in part (a)].
(e) Two energy equivalent Lewis structures are possible for the dihydrogen phosphate anion,
H2PO4–:
0
OH 0
0
HO P O
-1 O
0
Π
OH
HO P O
O
Π
The formal charge on each atom in either of these structures is like that shown for the first one.
(The formal charge on each H is 0, but, for simplicity, is not shown.) However, delocalization of
the pi electrons makes both the O atoms not bonded to hydrogen equivalent and each will have a
formal charge of –1/2. (In half of the structures, a particular O has a –1 formal charge and in the
other structure it has a 0 formal charge, thus averaging to –1/2.) The sum of the formal charges,
0 + 0 + 0 + 0 + 0 – 1/2 – 1/2 = –1, is the charge on the anion [just as in the case of the nitrite
anion in part (b)].
Problem 6.46.
[NOTE: This problem asks if there is more than one possible way to write Lewis structures for
any of the anions in the preceding Problem 6.45, and if so, calculate the formal charges for each
atom in each structure and explain which structure(s) are most favored. In the solution for
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Problem 6.45, energy-equivalent structures for each anion are given for the Lewis structure with
the least possible formal charge on the atoms (which, according to our rule in the textbook, is
the structure of lowest energy). The consequence of electron delocalization, the spreading of
formal charge more uniformly over the anions, is explained as well. The remainder of a solution
to the problem here is to examine other Lewis structures that might be possible and explain why
they are not favored.]
If we assume that the best Lewis structures for molecules with only second period atoms have
octets of electrons (four pairs) on each second row atom, then the only way to write Lewis
structures for CO32– and NO2– are the ones in Problem 6.45. In each case, energy equivalent
structures were written with the charges on different oxygen atoms, but the structures are the
same as far as formal charges are concerned. The oxygen atoms with three nonbonding electron
pairs have a formal charge of –1 and all the other atoms have a formal charge of zero. This is
the best we can do for formal charge; some atoms in an ion must bear formal charges and the
minimum number of formal charges is present in the structures written.
If we do not assume octets on all second period atoms, we could write these Lewis structures for
CO32– and NO2–:
-1
-1
O
2Π
+1
–
-1 N –1
O C +1
O
O
-1
O
In each case, the oxygen atoms with three nonbonding electron pairs have a formal charge of –1
and the central atoms have a +1 formal charge. The net charge on each ion is still what it should
be, but the amount of formal charge on the atoms is greater than for the structures in Problem
6.45. In addition, these are the only Lewis structures possible with these electron arrangements,
so there is no electron delocalization. These new structures are energetically unfavorable
compared to the ones written previously.
For the ions with a central phosphate atom, we are not restricted to an octet of electrons on the
phosphate atom, as you saw in the structures written in the solution for Problem 6.45. In each
case, energy equivalent structures were written with the charges on different oxygen atoms, but
the structures are the same as far as formal charges are concerned. The oxygen atoms with three
nonbonding electron pairs have a formal charge of –1 and all the other atoms have a formal
charge of zero. This is the best we can do for formal charge; some atoms in an ion must bear
formal charges and the minimum number of formal charges is present in the structures written.
If we assume that the phosphorus atoms should have octets, we can draw these structures:
-1
-1
O
3Π
-1
O P+1 O
O
-1
0
OH -1
O P+1O
2Π
-1
-1 O
0
OH -1
HO P+1O
Π
0
-1 O
In each case, all the oxygen atoms with three nonbonding electron pairs have a formal charge
of –1 and the central atoms have a +1 formal charge. The net charge on each ion is still what it
should be, but the amount of formal charge on the atoms is greater than for the structures in
Problem 6.45. In addition, these are the only Lewis structures possible with these electron
arrangements, so there is no electron delocalization. These new structures are energetically
unfavorable compared to the ones written previously.
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Problem 6.47.
Lewis structures for ozone and nitric acid can be broken apart to show how the formal charges
are determined.
O
O
O +1 -1
O
0
O
O
The core charge on O is 6+, which is balanced by the six valence electrons on the left-hand O
for a formal charge of 0. The central O has only five valence electrons and a formal charge
of +1. The right-hand O has seven valence electrons and a formal charge of –1. The pi electrons
are delocalized in the ozone molecule, so the end O atoms each have an average formal charge
of –1/2. (See the solutions to Problems 6.44 and 6.45.)
O
H O N O
0
H
0
O
O 0
+1
N
O –1
The core charge on O is 6+, which is balanced by the six valence electrons on the left-hand and
center O for a formal charge of 0 on these atoms. The right-hand O has seven valence electrons
and a formal charge of –1. The N with a core charge of +5 and four valence electrons has a
formal charge of +1. The H with a core charge of +1 and one valence electron has a formal
charge of 0. The pi electrons are delocalized in the nitric acid molecule, so the two O atoms not
bonded to H each have an average formal charge of –1/2.
Problem 6.48.
(a) You can tell that a reaction occurs by observing that a solid forms when solutions of
Sn2+(aq) and Bi3+(aq) are mixed:
3Sn2+(aq) + 2Bi3+(aq)  3Sn4+(aq) + 2Bi3+(s)
If you determine that it is a metallic solid, Bi(s), that precipitates, you can conclude that the
reaction is a reduction-oxidation reaction, because the only way to get the metal from the cation
is by reduction (addition of electrons).
(b) The element Sn, is oxidized from Sn2+ to Sn4+. Each Sn2+ cation loses two electrons in the
reaction.
(c) The element Bi is reduced from Bi3+ to Bi0. Each Bi3+ cation gains three electrons in the
reaction.
(d) Each Sn2+ cation loses two electrons in the reaction.
(e) Each Bi3+ cation gains three electrons in the reaction.
Problem 6.49.
(a) The evolution of hydrogen gas and the disappearance of the solid metallic sodium are
indications that a chemical reaction occurs when a piece of sodium metal, Na(s), is placed in
water. The resulting solution is basic, since it turns pink when phenolphthalein, an acid-base
indicator is added. Since addition of phenolphthalein to pure water does not produce a pink
color, we can conclude that another product of the reaction is hydroxide anion, OH–(aq). Some
cation must be formed to balance the charge from OH–(aq), and a logical candidate is the
sodium cation, Na+(aq), which also accounts for the disappearance of the sodium metal. The
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transformation of Na(s) to Na+(aq, is an oxidation reaction, Na0  Na+. Some other element in
the system must be reduced and it is logical to think of the H in water molecules, which has an
oxidation number +1, being reduced to the H in molecular hydrogen, H2(g), which has an
oxidation number of 0.
(b) The net ionic equation for the reaction described and analyzed in part (a) is:
2Na(s) + 2H2O(l)  2Na+(aq) + H2(g) + 2OH–(aq)
(c) The element sodium, Na(s), is the reactant that is oxidized, as shown in part (a).
(d) Hydrogen atomic cores from two water molecules are reduced to H2(g), (also producing two
hydroxide anions).
Problem 6.50.
(a) The change Zn  Zn2+ is an oxidation, because the oxidation number increases going from
Zn (the element with oxidation number 0) to Zn2+ (monatomic ion with oxidation number +2).
(b) The change S2–  S is a reduction, because the oxidation number decreases going from S2–
(monatomic ion with oxidation number –2) to S (the element with oxidation number 0).
(c) The change Fe3+  Fe2+ is a reduction, because the oxidation number decreases going from
Fe3+ (monatomic ion with oxidation number +3) to Fe2+ (monatomic ion with oxidation number
+2).
(d) The change Ag+  Ag is a reduction, because the oxidation number decreases going from
Ag+ (monatomic ion with oxidation number +1) to Ag (the element with oxidation number 0).
Problem 6.51.
(a) Balanced oxidation-reduction reactions are written for each of these observations:
(i) Mg metal reacts with Zn2+(aq) to yield Mg2+(aq) and Zn metal.
Mg(s) + Zn2+(aq)  Mg2+(aq) + Zn(s)
(ii) Zn metal reacts with Cu2+(aq) to yield Zn2+(aq) and Cu metal.
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
(iii) Mg metal reacts with Cu2+(aq) to yield Mg2+(aq) and Cu metal.
Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)
(b) Mg(s) is oxidized in reactions with both Zn2+(aq) and Cu2+(aq), observations (i) and (iii).
Thus, we conclude that Mg(s) is the most likely metal (among these three) to undergo an
oxidation,
(c) Cu2+(aq) is reduced in reactions with both Zn(s) and Mg(s), observations (ii) and (iii). Thus,
we conclude that Cu2+(aq)) is the most likely metal cation (among these three) to undergo a
reduction,
Problem 6.52.
For these reactions we (i) assign oxidation numbers to each atom, (ii) identify the element that is
oxidized, (iii) identify the element that is reduced.
(a) V2O5 + 2H2  V2O3 + 2H2O
(i) In the reactants, V = +5, O = –2, and H = 0. In the products, V = +3, O = –2, H = +1, and
O = –2
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(ii) H goes from 0  +1, an increase in oxidation number, so it is oxidized.
(iii) V goes from +5  +3, a decrease in oxidation number, so it is reduced.
(b) 2K + Br2  2K+ + 2Br(i) In the reactants, K = 0, and Br = 0. In the products, K = +1 and Br = –1.
(ii) K goes from 0  +1, an increase in oxidation number, so it is oxidized.
(iii) Br goes from 0  –1, a decrease in oxidation number, so it is reduced.
(c) N2 + 3H2  2NH3
(i) In the reactants, N = 0, and H = 0. In the products, N = –3 and H = +1.
(ii) H goes from 0  +1, an increase in oxidation number, so it is oxidized.
(iii) N goes from 0  –3, a decrease in oxidation number, so it is reduced.
Problem 6.53.
(a) The oxidation number for the sulfur atom in sulfur trioxide, SO3, is +6. as we can see by
breaking apart the molecule and assigning all bonding electrons to the most electronegative
atom and then finding the charge on the atomic core plus valence electrons for each core:
O Π2
O
O
S
O
Π2
O
S +6 Π2
O
Note that the structure here has formal charges of 0 on all the atoms. It is possible to write a
Lewis structure with an octet of electrons on the S, but this would give formal charges of –1 on
two O atoms and +2 on the S atom, which is unfavorable relative to this one. No matter which
structure you start with, all the valence electrons end up on the O atoms, just as here, when you
apply the rules to get the oxidation numbers in the molecule.
(b) The oxidation number for the sulfur atom in the sulfite ion, SO32– is +4. Let’s apply the
alternative set of rules for assigning oxidation numbers to get this value. The ON for each of the
three O atoms is –2. The total for the ONs has to sum to the charge on the ion, so we have:
ion charge = –2 = 3(–2) + ON(sulfur);
 ON(sulfur) = +4
The oxidation number for the sulfur atom in the sulfate ion, SO42– is +6:
ion charge = –2 = 4(–2) + ON(sulfur);
 ON(sulfur) = +6
(c) The dissolution of sulfur trioxide in water to give an acidic solution is not a redox reaction.
This means that the S in the acid produced has to have the same oxidation number, +6, as the S
in sulfur trioxide. The S in the sulfate ion has an oxidation number of +6, so the acid produced
must be sulfuric:
SO3(g) + H2O(l)  H2SO4(aq)
Problem 6.54.
(a) The oxidation number is –3 for the N in NH3. N is more electronegative than H, so all the
bonding electrons (plus its nonbonding pair) are assigned to N, which gives it eight electrons
(charge –8) and core charge of +5. Thus, the oxidation number is –3 (= 5 – 8).
(b) The oxidation number is +2 for the N in NO. Because this molecule has an odd number of
valence electrons (11), no Lewis structure can be written with all the electrons paired and in
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octets on these second row atoms. Since O is more electronegative than N, it will take all the
electrons it can (eight of the 11) in the imaginary separation of the atoms we use to assign
oxidation numbers. This leaves N with three valence electrons (charge –3) and core charge of
+5, for an oxidation number of +2. The same result is obtained if you apply the alternative
method for assigning oxidation numbers.
(c) The oxidation number is +3 for the N in NO2–. The Lewis structure(s) for the nitrite anion is
shown in the solutions for Problems 6.42, 6.45, and 6.46. In all the structures, the N atom has a
nonbonding pair of electrons and the others are shared with the O atoms. In the imaginary
separation of he atoms we use to assign oxidation numbers, all the electrons, except the
unshared pair on N will go to O. This leaves N with two valence electrons (charge –2) and core
charge of +5, for an oxidation number of +3. The same result is obtained if you apply the
alternative method for assigning oxidation numbers.
(d) The oxidation number is +5 for the N in NO3–. The Lewis structure(s) for the nitrate ion is
shown in the solution for Problem 6.44. All the valence electrons on the N are shared with O
atoms, so in the imaginary separation of he atoms we use to assign oxidation numbers, all the
electrons go to O. This leaves N with no valence electrons and core charge of +5, for an
oxidation number of +5. The same result is obtained if you apply the alternative method for
assigning oxidation numbers.
(e) The oxidation number for N in some common molecules and ions, parts (a) through (d),
ranges from –3 to +5. The location of nitrogen near the middle of the second period of the
periodic table helps to explain this wide range of possible oxidation numbers. The atomic core
can be assigned a maximum of eight electrons (the maximum for a second-period element)
when it is bonded to less electronegative atoms in molecules. This gives N an oxidation number
of –3. When an N atomic core shares all its valence electrons with more electronegative atoms
in a molecule, all of the valence electrons can be assigned to the other atoms, leaving N with
none and an oxidation number of +5. Intermediate values are possible, as you see in parts (b)
and (c). [What is the oxidation number of N in H2NOH, NO2, and H2NNH2, a few more
common N-containing molecules.]
Problem 6.55.
To find the oxidation number of the metal in a metal-ion complex, we take advantage of the fact
that metals (Lewis acids) have low electronegativities and, in the ligands (Lewis bases), the
atom bonded to the metal has a high electronegativity. Thus, when the bonding electrons are
partitioned between the metal and the ligand to assign oxidation numbers, the ligand will get all
the electrons. The ligands will end up with the charge they have as free molecules or ions and
the oxidation number of the metal will be whatever is required to give the correct charge on the
complex.
(a) The free ligands in PtCl42– are chloride ions, Cl–, so Pt has to have an oxidation number of
+2 to give a complex with a 2– charge.
(b) The free ligands in Cu(NH3)42+ are ammonia molecules, NH3, so Cu has to have an
oxidation number of +2 to give a complex with a 2+ charge.
(c) The free ligands in Fe(CN)63– are cyanide anions, CN–, so Fe has to have an oxidation
number of +3 to give a complex with a 3– charge.
(d) The free ligands in MoS42– are sulfide anions, S2–, so Mo has to have an oxidation number of
+6 to give a complex with a 2– charge.
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(e) The free ligands in Zn(OH)42– are hydroxide anions, OH–, so Zn has to have an oxidation
number of +2 to give a complex with a 2– charge.
Problem 6.56.
[NOTE: Both the oxidation-number and half-reaction methods for balancing oxidationreduction reactions are presented in the text with Worked Examples and Check This problems
for practice. For the reactions here, we give the reasoning for determining the reactant that is
reduced and the one that is oxidized, and then give the balanced redox reaction but do not show
the balancing in detail.]
(a)
MnO4–(aq) + SO32–(aq) + H+(aq)  Mn2+(aq) + SO42–(aq) + H2O
Mn in the MnO4–(aq) reactant has ON = +7 and Mn in the Mn2+(aq) product has ON = +2, so
the Mn is reduced and gains 5 electrons in the reaction. S in the SO32–(aq) reactant has ON = +4
and S in the SO42–(aq) product has ON = +6, so the S is oxidized and loses 2 electrons in the
reaction. We require two Mn reactants for every five S reactants, in order to balance the electron
transfer. The balanced redox equation is:
2MnO4-(aq) + 5SO32-(aq) + 6H+(aq)  2Mn2+(aq) + 5SO42-(aq) + 3H2O
(b)
NO3–(aq) + Zn(s) + H+(aq)  NH4+(aq) + Zn2+(aq) + H2O
N in the NO3–(aq) reactant has ON = +5 and N in the NH4+(aq) product has ON = –3, so the N is
reduced and gains 8 electrons in the reaction. Zn in the Zn(s) reactant has ON = 0 and Zn in the
Zn2+(aq) product has ON = +2, so the Zn is oxidized and loses 2 electrons in the reaction. We
require one N reactant for every four Zn reactants, in order to balance the electron transfer. The
balanced redox equation is:
NO3-(aq) + 4Zn(s) + 10H+(aq)  NH4+(aq) + 4Zn2+(aq) + 3H2O
(c)
Cl2(g) + OH–(aq)  ClO3–(aq) + Cl–(aq) + H2O
This reaction is a bit complicated to analyze because some of the reactant chlorine, Cl2(g), is
oxidized and some of it is reduced, [This is called disproportionation.] To make reaction easier
to analyze, we will rewrite it with the two chlorine molecules shown as reactants. We will
assume that the first is reduced and the second oxidized in the reaction:
Cl2(g) + Cl2(g) + OH–(aq)  ClO3–(aq) + Cl–(aq) + H2O
Cl in the first Cl2(g) reactant has ON = 0 and Cl in the Cl–(aq) product has ON = –1, so the Cl is
reduced and gains 1 electron in the reaction. Cl in the second Cl2(g) reactant has ON = 0 and Cl
in the ClO3–(aq) product has ON = +5, so the Cl is oxidized and loses 5 electrons in the reaction.
Since each of the Cl2(g) reactants has two Cl atoms to gain or lose electrons, we might consider
multiplying the electron gains and losses by two, but this is not necessary in this case, because it
would not change the ratio of gain to loss. We require five of the first Cl reactant for every one
of the second Cl reactant, in order to balance the electron transfer. The balanced redox equation
is:
5Cl2(g) + Cl2(g) + 12OH–(aq)  2ClO3–(aq) + 10Cl–(aq) + 6H2O
Now we can combine the Cl2(g) reactants, since there is not really any way to differentiate
them, and then divide through the entire equation by two, to reduce the coefficients to their
lowest possible integer values (as we conventionally show balanced equations).
3Cl2(g) + 6OH-(aq)  ClO3-(aq) + 5Cl-(aq) + 3H2O
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Problem 6.57.
(a) In a zinc-air cell, oxygen gas, O2, is converted in aqueous solution to hydroxide anion, OH–
and zinc metal is converted to zinc oxide, ZnO(s), during discharge of the cell. In this cell, zinc
metal is being oxidized, going from an oxidation state of 0 in Zn(s) to +2 in ZnO(s). Oxygen gas
is being reduced, going from an oxidation state of zero in O2(g) to –2 in ZnO(s) (and OH–).
(b) The net (overall) reaction in this cell is 2Zn(s) + O2(g)  2ZnO(s). You might, however,
wonder what happened to the hydroxide anion that is said to have been formed. We can break
the process into two half reactions, whose sum is the net reaction just written:
O2(g) + 2H2O(l) + 4e–  4OH–(aq)
2Zn(s) + 4HO–(aq)  2ZnO(s) + 2H2O(l) + 4e–
(c) For each mole of zinc undergoing reaction, two moles of electrons have to be transferred
from the zinc to oxygen in order to oxidize the zinc metal to zinc(II) ion. You can also see this
stoichiometry in the half reactions where four moles of electrons are transferred for every two
moles of zinc metal reacted, or, again, two moles of electrons per mole of zinc.
Problem 6.58.
[NOTE: Both the oxidation-number and half-reaction methods for balancing oxidationreduction reactions are presented in the text with Worked Examples and Check This problems
for practice. For the reactions here, we give the reasoning for determining the reactant that is
reduced and the one that is oxidized, and then give the balanced redox reaction but do not show
the balancing in detail. Both methods give the same result, as they must. Preference for one or
the other is a matter of taste.]
(a)
S2–(aq) + NO3–(aq)  NO2(g) + S8(s) (S8 rings are the common form of elemental
sulfur.)
N in the NO3–(aq) reactant has ON = +5 and N in the NO2(g) product has ON = +4, so the N is
reduced and gains 1 electron in the reaction. S in the S2–(aq) reactant has ON = –2 and S in the
S8(s) product has ON = 0, so the S is oxidized and loses 2 electrons per sulfur atom in the
reaction. Since eight sulfide anions have to be oxidized to form one S8(s), we require 16 N
reactants for every eight S reactants, in order to balance the electron transfer. The balanced
redox equation is:
8S2–(aq) + 16NO3–(aq) + 32H+(aq)  8NO2(g) + S8(s) + 16H2O
(b)
Hg2+(aq) + NO2–(aq)  Hg(l) + NO3–(aq)
Hg in the Hg2+(aq) reactant has ON = +2 and Hg in the Hg(l) product has ON = 0, so the Hg is
reduced and gains 2 electrons in the reaction. N in the NO2– (aq) reactant has ON = +3 and N in
the NO3–(aq) product has ON = +5, so the N is oxidized and loses 2 electrons in the reaction.
We require one Hg reactant for every N reactant, in order to balance the electron transfer. The
balanced redox equation is:
Hg2+(aq) + NO2–(aq) + H2O  Hg(l) + NO3–(aq) + 2H+(aq)
(c)
Fe2+(aq) + NO3–(aq)  Fe3+(aq) + NO(g)
N in the NO3–(aq) reactant has ON = +5 and N in the NO(g) product has ON = +2, so the N is
reduced and gains 3 electrons in the reaction. Fe in the Fe2+(aq) reactant has ON = +2 and Fe in
the Fe3+(aq) product has ON = +3, so the Fe is oxidized and loses 1 electron in the reaction. We
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require one N reactant for every three Fe reactants, in order to balance the electron transfer. The
balanced redox equation is:
3Fe2+(aq) + NO3–(aq) + 4H+(aq)  3Fe3+(aq) + NO(g) + 2H2O
Problem 6.59.
[NOTE: For the reactions here, we give the reasoning for determining the reactant that is
reduced and the one that is oxidized, and then give the balanced redox reaction in acidic
solution followed by the balanced reaction in basic solution but do not show the balancing or
addition of OH–(aq) in detail.]
(a)
S2–(aq) + I2(s)  SO42–(aq) + I–(aq)
I in the I2(s) reactant has ON = 0 and I in the I–(aq) product has ON = –1, so the I is reduced and
gains 2 electrons per I2(s) in the reaction. S in the S2–(aq) reactant has ON = –2 and Fe in the
SO42–(aq) product has ON = +6, so the S is oxidized and loses 8 electrons in the reaction. We
require four I2(s) reactant for every S2–(aq) reactant, in order to balance the electron transfer.
The balanced redox equations (in acid —imagined— and in base) are:
S2–(aq) + 4I2(s) + 4H2O(l)  SO42–(aq) + 8I–(aq) + 8H+(aq)
S2–(aq) + 4I2(s) + 8OH-(aq)  SO42–(aq) + 8I–(aq) + 4H2O(l)
(b)
MnO4–(aq) + C2O42–(aq)  MnO2(s) + CO2(g)
Mn in the MnO4–(aq) reactant has ON = +7 and Mn in the MnO2(s) product has ON = +4, so the
Mn is reduced and gains 3 electrons in the reaction. C in the C2O42–(aq) reactant has ON = +3
and C in the CO2(g) product has ON = +4, so the C is oxidized and loses 2 electrons per
C2O42-(aq) in the reaction. We require two MnO4–(aq) reactants for every three C2O42–(aq)
reactants, in order to balance the electron transfer. The balanced redox equations (in acid —
imagined— and in base) are:
2MnO4–(aq) + 3C2O42–(aq) + 8H+(aq)  2MnO2(s) + 6CO2(g) + 4H2O(l)
2MnO4–(aq) + 3C2O42–(aq) + 4H2O(l)  2MnO2(s) + 6CO2(g) + 8OH-(aq)
(c)
Bi(OH)3(s) + Sn(OH)3–(aq)  Bi(s) + Sn(OH)62–(aq)
Bi in the Bi(OH)3(s) reactant has ON = +3 and Bi in the Bi(s) product has ON = 0, so the Bi is
reduced and gains 3 electrons in the reaction. Sn in the Sn(OH)3–(aq) reactant has ON = +2 and
Sn in the Sn(OH)62–(aq) product has ON = +4, so the Sn is oxidized and loses 2 electrons in the
reaction. We require two Bi(OH)3(s) reactants for every three Sn(OH)3–(aq) reactants, in order
to balance the electron transfer. The balanced redox equations (in acid —imagined— and in
base) are:
2Bi(OH)3(s) + 3Sn(OH)3–(aq) + 3H2O(l)  2Bi(s) + 3Sn(OH)62–(aq) + 3H+(aq)
2Bi(OH)3(s) + 3Sn(OH)3–(aq) + 3OH-(aq)  2Bi(s) + 3Sn(OH)62–(aq)
Problem 6.60.
(a) The reaction that occurs when solid cobalt(II) sulfide dissolves in nitric acid can be
represented as:
CoS(s) + NO3–(aq)  Co2+(aq) + NO(g) + S8(s) (S8 rings are the common form of
elemental sulfur.)
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N in the NO3–(aq) reactant has ON = +5 and N in the NO(g) product has ON = +2, so the N is
reduced and gains 3 electrons in the reaction. S in the CoS(s) reactant has ON = –2 and S in the
S8(s) product has ON = 0, so the S is oxidized and loses 2 electrons per sulfur atom in the
reaction.
(b) If we imagine that elemental sulfur is S(s), then we require 2 NO3–(aq) reactants for every
three CoS(s) reactants, in order to balance the electron transfer. The balanced redox equation
and its eight-fold multiple to account for S8(s) are:
3CoS(s) + 2NO3–(aq) + 8H+(aq)  3Co2+(aq) + 2NO(g) + 3S(s) + 4H2O
24CoS(s) + 16NO3–(aq) + 64H+(aq)  24Co2+(aq) + 16NO(g) + 3S8(s) + 32H2O
Problem 6.61.
To determine whether the conversion of lactic acid, CH3CH(OH)COOH, to pyruvic acid,
CH3C(O)COOH, is an oxidation or a reduction reaction we need to focus attention on the part of
the molecule that changes, the bonding about the central carbon atomic core. In lactic acid, this
carbon is bonded to two other carbon atoms, a hydrogen atom, and an oxygen atom. Using our
rules for oxidation numbers, we assign ON = +1 to this carbon. In pyruvic acid, this carbon is
bonded to two other carbons and doubly bonded to an oxygen atom. Using our rules for
oxidation numbers, we assign ON = +2 to this carbon. Since the oxidation number has
increased, this change is an oxidation.
Problem 6.62.
The reaction of interest, ethanol in hard cider reacting with oxygen from the air to form ethanoic
acid, is an oxidation of ethanol to ethanoic acid:
CH3CH2OH(aq)  CH3C(O)OH(aq)
The ethanol has to gain an oxygen and lose two hydrogen atomic cores to undergo this
oxidation. If an oxide ion is added on the left and two hydronium ions are added on the right, we
can write the reaction as a half reaction:
CH3CH2OH(aq) + O2–(aq)  CH3C(O)OH(aq) + H+(aq) + 4e–
The problem states that the oxidant is oxygen from the air, which is reduced in the reaction, as
we can represent in another half reaction:
O2(g) + 4e–  2O2–(aq)
The sum of this oxidation and reduction (after canceling an oxide anion from each side is) is:
CH3CH2OH(aq) + O2(g)  CH3C(O)OH(aq) + 2H+(aq) + O2–(aq)
The two hydronium ions and the oxide ion will, of course, react to form water. The balanced net
reaction for making cider vinegar is:
CH3CH2OH(aq) + O2(g)  CH3C(O)OH(aq) + 2H2O(aq)
Problem 6.63.
(a) Three separate equations showing the formation of carbon dioxide, carbon monoxide, and
free carbon particles (plus water) when glucose in wood burns with oxygen gas from the air are:
C6H12O6 + 6O2  6CO2 + 6H2O
C6H12O6 + 3O2  6CO + 6H2O
C6H12O6  6C + 6H2O
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(b) All three reactions can take place at the same time. The first reaction produces nontoxic
carbon dioxide and water. The second reaction uses less oxygen, and produces toxic carbon
monoxide. In poorly ventilated areas the reaction can use up oxygen faster than it is replaced,
thus tipping the balance of the reactions toward those that require less oxygen. Thus, more
carbon monoxide (and soot) would be produced and the lack of ventilation would permit its
concentration to build up, creating a dangerous situation.
Problem 6.64.
[NOTE: Based on the rules for oxidation numbers that we have adopted in the text, we identify
each of these reactions (incomplete equations) as an oxidation, a reduction, or neither with
respect to the carbon atom on the right in each structure. This identification is not universally
accepted. Another convention that assigns and oxidation number of +1 to all H atoms and –2 to
all O atoms in a carbon-containing molecule leads to different interpretations in some cases.
This convention is sometimes useful for looking at the overall oxidation state of a molecule, but
often not useful for looking at the oxidation state of single atoms in a molecule.]
(a)
H2 C CH2
O
H3 C C
OH
The C in the reactant is bonded to two H atoms and doubly bonded to another carbon atom
(which counts as bonding to two C atoms in our convention), which gives ON = 0. The C in the
product is singly bonded to another C and an O atom, and doubly bonded to an O atom (which
counts as bonding to two O atoms), which gives ON = +3. This is an oxidation of this carbon.
The alternative convention can be applied to each carbon individually. In the reactant, both
carbons are the same, =CH2. If the H atoms are assigned ON = +1, the C must have ON = –2. In
the product, the right-hand C is part of a –C(O)OH group in which the C has to have ON = +3 in
order to give the group zero charge. This is an oxidation of this carbon, just as in the previous
paragraph. If we take account of the other C in the reactant and product, sum of the C oxidation
numbers in the reactant is –4. In the product, the left-hand C is part of a –CH3 group in which
the C has an ON = –3. The sum of the C oxidation numbers for the product is 0 [= (=3) + (–3)].
Again, we conclude that the reaction is an oxidation, but the focus has shifted from an
individual carbon atom to the whole molecule.
(b)
HO
H2 C CH 2
OH
H2 C CH 2
The C in the reactant is bonded to two H atoms, a C atom, and O atom, which gives ON = +1.
The C in the product is singly bonded to two H atoms and doubly bonded to another C atom
(which counts as bonding to two C atoms), which gives ON = 0. This is a reduction of this
carbon.
(c)
O
H3 C C
H
H2 C CH 2
The C in the reactant is bonded to an H atom, a C atom, and doubly bonded to an O atom
(which counts as bonding to two O atoms), which gives ON = +2. The C in the product is singly
bonded to two H atoms and doubly bonded to another C atom (which counts as bonding to two
C atoms), which gives ON = 0. This is a reduction of this carbon.
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OH
H3 C CH2
H2 C CH2
(d)
The C in the reactant is bonded to two H atoms and doubly bonded to another carbon atom
(which counts as bonding to two C atoms in our convention), which gives ON = 0. The C in the
product is singly bonded to another C, two H atoms, and an O atom, which gives ON = +1. This
is an oxidation of this carbon.
Problem 6.65.
(a) The oxidation number of manganese in MnO4–(aq), Mn2+(aq), MnO2(s), and MnO42–(aq) is,
respectively, +7, +2, +4, and +6. If MnO4–(aq) is used to oxidize other substances, it must itself
be reduced as electrons are transferred from the substance oxidized. The latter three manganese
species are product of these reactions under various conditions and, in each case, the oxidation
number of the manganese goes down, that is, the manganese is reduced, as it must be when
permanganate, ON = +7 is used as an oxidant.
(b) The reduction half-reaction for each of the reactions of permanganate anion acting as an
oxidizing agent is:
MnO4–(aq) + 8H+(aq) + 5e–  Mn2+(aq) + 4H2O(aq)
MnO4–(aq) + 4H2O(aq) + 3e–  MnO2(s) + 4OH–(aq)
MnO4–(aq) + e–  MnO42–(aq)
(c) The color change from intense pink of the MnO4–(aq) anion to colorless (or almost colorless)
Mn2+(aq) cation (formed as the reduction product in the titrated solution) can used as a visual
indicator of the endpoint of a permanganate oxidation in acidic solutions. Generally the reaction
proceeds until one drop of the intense pink titrant [MnO4–(aq)] changes the colorless solution
being titrated to pale pink. In neutral or slightly basic solution, formation of brown solid MnO2
could serve to signal that oxidation is taking place, although it is difficult to determine when the
solid stops forming and the brown solid obscures the disappearance of the intense pink color.
The color change in highly alkaline solutions may provide an endpoint, as a drop of the intense
pink titrant changes the pale green color of the solution being titrated [pale green because of the
reduction product MnO42–(aq)]. Of course, these visual color methods will not work if the
material being oxidized undergoes a color change itself.
Problem 6.66.
(a) Each chromium atom in the dichromate anion has ON = +6 each.
O
O
O Cr O Cr O
O
O
Oxygen is more electronegative than any metal so, all the electrons will be assigned to the O
atoms when oxidation numbers are assigned to the atoms in dichromate. Each oxygen will have
8 electrons for an oxidation number of –2 (= 6 – 8). One way to get the oxidation number of the
Cr atoms is to set the sum of the oxidation numbers to the charge on the ion:
ion charge = –2 = 7(–2) + 2ONCr;
 ONCr = +6
Another way to find ONCr is to use the information in the problem that the chromium atom has
six valence electrons, which gives the atom a core charge of +6. If, when we assign electrons to
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get the oxidation numbers, the Cr atoms are left with no valence electrons (the O atoms have
taken them all), each has an oxidation number of +6 (= 6 – 0).
(b) Each Cr in the dichromate anion has an ON = +6 and ends up as a Cr3+ cation, with an
ON = +3. Each Cr gains three electrons in the reaction. The dichromate must gain six electrons,
in order to reduce both Cr atoms. To balance this gain of electrons, the iron, oxidizing from
ON = +2 to ON = +3, must lose six electrons; so there must be six irons in the balanced
equation:
gain 6 electrons
Cr2O72Π(aq) + 6Fe 2+(aq)
2Cr3+(aq) + 6Fe3+(aq)
lose 6 electrons
This equation is balanced in electron transfers, but not in atoms. Seven oxygen atoms are
needed on the right; add seven water molecules. This also adds 14 hydrogen atoms on the right;
add 14 hydronium cations on the left to balance the hydrogen:
Cr2O72–(aq) + 6Fe2+(aq) + 14H+(aq)  2Cr3+(aq) + 6Fe3+(aq) + 7H2O(aq)
(c) This balanced equation gives the stoichiometric relationship between dichromate and
iron(II): 1 mole of dichromate reacts with 6 moles of iron(II). The titration used 0.03457 L
(34.57 mL) of 0.0100 M dichromate; the number of moles of dichromate is:
(0.03457 L)(0.0100 mol·L–1) = 3.457  10–4 mol dichromate
The number of moles of iron(II) in the sample is:
 6 mol iron(II) 
(3.457  10–4 mol dichromate)
= 2.07  10–3 mol iron(II)
1 mol dichromate 
The mass of iron that is equivalent to this number of moles is:
55.85 g iron 
(2.07  10–3 mol iron(II)) 
= 0.116 g iron
1 mol iron 
The mass percent of iron in the ore sample is:
 0.116 g iron 
 0.178 g ore  100% = 65.2% iron in the ore
Problem 6.67.
Apparently, in the presence of hydroxide ion, OH–(aq), dichromate ion reacts to produce
chromate ion:
Cr2O72–(aq) + OH–(aq)  CrO42–(aq)
orange
yellow
The oxidation number of the chromium atoms in dichromate is +6 (See Problem 6.66).
Likewise, the oxidation number of chromium in chromate ion is +6. This reaction is not an
oxidation-reduction reaction. To balance the reaction, two chromate ions are produced for each
dichromate that reacts:
Cr2O72–(aq) + OH–(aq)  2CrO42–(aq) + H+(aq)
The hydronium ion product is required to balance the equation in terms of hydrogen. This
equation is not, however, a good representation of the chemistry of the system. In basic solution,
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the hydronium immediately reacts with hydroxide to produce water. To balance the equation
“chemically,” add a hydroxide on each side and let it react with the hydronium on the right:
Cr2O72–(aq) + 2OH–(aq)  2CrO42–(aq) + H2O(l)
This reaction is not an oxidation-reduction and certainly not a precipitation. It must be a Lewis
acid-base reaction. The Lewis base (nucleophilic nonbonding electron pair on hydroxide)
attracts the Lewis acid (electrophilic chromium in dichromate):
O
O
O Cr O Cr O
O
O
(aq)
O H
(aq)
O
O Cr O
O
(aq)
O
H O Cr O
O
(aq)
In basic solution, the hydrogen (proton) on the hydrogen chromate ion will react with base
(another Lewis acid-base reaction) to form water. Adding that reaction to this one gives the
previous balanced reaction equation. The reaction is readily reversed by adding acid, which
converts the chromate to dichromate.
Problem 6.68.
The responses here are based on the data in this table:
Metal Ion
Li+
Ca2+
Mg2+
Al3+
Ratio of
Ionic Charge/Ionic Radius
1.5
2.1
3.1
6.7
(a) The process of a metal ion interacting with water involves the metal ion acting as a Lewis
acid and water acting as a Lewis base. The strength of this interaction varies with concentration
of the charge on the Lewis acid. If the charge is spread out over a large ion, the interaction is
decreased. Metal ions that are strong Lewis acids have a high charge to radius ratio because this
means the charge will be concentrated on a small ion. In this list, the Al3+ ion has the highest
charge to radius ratio and will be the strongest Lewis acid. Indeed, Al3+ compounds are often
used in reactions in which a strong Lewis acid is required, for example, as a catalyst,
(b) Magnesium and calcium are both members of group II on the periodic table. Both ions have
the same charge, 2+. However, the calcium ion (period 4), is larger than the magnesium ion
(period 3), which reduces the ratio of charge to radius, as shown in the table.
(c) Sodium ion, Na+, does not act as a Lewis acid with water. This means the cation must have
a low charge density, low charge to radius ratio. Sodium and lithium are both in group I on the
periodic table and both form +1 ions. Since Na is in period 3 and Li in period 2, Na+ will have a
larger radius and a lower charge to radius ratio than 1.5, the value for Li+ in the above table. We
might predict a value close to 1. (The actual value given in the literature is 1.0.)
Problem 6.69.
(a) As molten nylon polymer flows through the very fine holes in the spinnerets shown in the
chapter opening illustration the molecules represented in the Web Companion, Chapter 6,
Section 6.7, page 2, are forced to come out more or less lined up so the long molecules can get
through the opening (rather like a thread passing through the eye of a needle). Thus the
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molecules are lying side-by-side and hydrogen bonds are readily formed between adjacent
molecules, which help to account for the strength of nylon fibers.
(b) Spiders produce several different relatively simple protein polymers that are mixed in
various ratios to produce more complex polymers with different properties, because this is a
more efficient use of their DNA and it allows them to store these ingredients in easily usable
liquid suspensions. If each complex polymer had to be coded separately by its DNA, the spider
would require many very long sequences of DNA that were highly repetitive from one coding
gene to another. By synthesizing several smaller polymers that can be mixed and matched to
make the final filament, a much smaller amount of DNA is required. Also, if much longer
polymers were synthesized, they would probably be a good deal less soluble and would
probably have to be made as needed (a relatively slow process), instead of stored for quick use.
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