Chemistry, 8th Edition Jill Robinson Solution Manual

Type:

Solution Manual

Resource:

Chemistry

Edition:

8th Edition

Author(s):

Jill Robinson John E. McMurry Robert C. Fay

CHAPTER 1 CHEMICAL TOOLS: EXPERIMENTATION AND MEASUREMENT Chapter Learning Objectives Section 1.1

Identify the steps in the scientific method.

Differentiate between a qualitative and a quantitative measurement.

Section 1.2

Write numbers in scientific notation and use prefixes for multiples of SI units.

Section 1.3

Convert between different prefixes used in mass measurements.

Section 1.4

Convert between different prefixes used in length measurements.

Section 1.5

Convert between common units of temperature measurement.

Section 1.6

Convert between SI and metric units of volume.

Convert between different prefixes used in volume measurements.

Section 1.7

Calculate mass, volume, or density using the formula for density.

Predict whether a substance will float or sink in another substance based on density.

Section 1.8

Calculate the kinetic energy of a moving object.

Convert between common energy units.

Section 1.9

Specify the number of significant figures in a measurement.

Evaluate the level of accuracy and precision in a data set.

Report a measurement to the appropriate number of significant figures.

Section 1.10

Report the result of a mathematical calculation to the correct number of significant figures.

Section 1.11

Convert from one unit to another using conversion factors.

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Chapter 1—Chemical Tools: Experimentation and Measurement

Lecture Outline Introduction 1.1. The Scientific Method: Nanoparticle Catalysts for Fuel Cells A. Scientific method – systematic approach to research 1. Observations – qualitative descriptions or quantitative measurements 2. Hypothesis – possible explanation of the observations 3. Experimentation – testing of the hypothesis 4. Theory – consistent explanation of known observations and logical interpretations of experimental results; has predictive power and can be modified or replaced B. Scientific method applied to PdCu nanoparticles 1. Observation – the size, shape, and composition of PdCu nanoparticles affect catalytic activity 2. Hypothesis – PdCu nanoparticles with ordered structures will have superior catalytic activity to those with random arrangements of palladium and copper atoms 3. Experimentation – controlled synthesis techniques for the PdCu nanoparticles were developed and catalytic activity tested 4. Theory – improved design of nanoparticle catalysts for fuel cells and other applications 1.2. Measurements: SI Units and Scientific Notation1 A. International System (SI) of Units – seven base units (and units derived from them) that suffice for all scientific measurements (textbook Table 1.1) B. Common prefixes used to modify SI units (textbook Table 1.2) 1. giga (G); factor = 109 2. mega (M); factor = 106 3. kilo (k); factor = 103 4. deci (d); factor = 10–1 5. centi (c); factor = 10–2 6. milli (m); factor = 10–3 7. micro (); factor = 10–6 8. nano (n); factor = 10–9 C. Scientific notation – exponential format for very large or very small numbers (textbook Appendix A) D. All measurements of physical quantities contain a 1. number 2. unit label 1.3. Mass and Its Measurement2 A. Mass (SI unit = kg) – amount of matter in an object 1. Measured on a balance by comparison to another mass 2. A standard reference exists B. Matter – anything with physical dimensions: mass and volume C. Weight – pull of gravity on an object 1.4. Length and Its Measurement3 A. Meter 1. Standard SI unit of length 2. Distance traveled by light in a vacuum in 1/299,792,458th of a second

Test Item Questions: Multiple Choice 4 – 10, Algorithmic 1 – 13, Short Answer 3 – 4 Test Item File Questions: Multiple Choice 11 – 15, Algorithmic 14 – 20, Short Answer 5 3 Test Item File Questions: Multiple Choice 17 – 20, Algorithmic 21 – 26, Short Answer 6 1 2

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Chapter 1—Chemical Tools: Experimentation and Measurement 1.5. Temperature and Its Measurement4 A. Common unit – Celsius degree (oC) B. Scientific unit – Kelvin 1. Kelvin has an absolute zero – the lowest temperature that can ever be reached.

C. Celsius and Kelvin scales – 100 degrees between the freezing point and boiling point of water Temperature in K = temperature in C + 273.15 Temperature in C = temperature in K − 273.15 D. Fahrenheit scale – 180 between the freezing point and boiling point of water 2. 180 F encompasses the same range as 100 C

1 °C 

180 °C 9 =  °F 100 °F 5

E. Conversion from Celsius to Fahrenheit 1. Size correction 2. Followed by zero-point correction F. Conversion from Fahrenheit to Celsius 1. Zero-point correction 2. Followed by a size correction o

æ 9 oF ö F = ç o ´ oC÷ + 32 o F è5 C ø C =

5 C  ( F − 32 F ) 9 F

1.6. Derived Units: Volume and Its Measurement5 A. Derived quantities – quantities expressed in terms of one or more of the seven base units (See textbook Table 1.3.)

B. Volume – amount of space occupied by an object 1. Measured in SI units by the cubic meter (m 3) 2. Commonly used measurements: a. Cubic decimeter (dm3) = metric liter (L) b. Cubic centimeter (cm3) = metric milliliter (mL) 1.7. Derived Units: Density and Its Measurement6 A. Density – intensive property relating the mass of an object to its volume 1. Expressed in units of g/cm3 or g/mL 2. Temperature dependent property 3. Useful property – allows conversion of an object’s volume to its mass 1.8. Derived Units: Energy and Its Measurement7 A. Energy 1. Capacity to do work or supply heat 2. Energy = work + heat

Test Item File Questions: Multiple Choice 21 – 27, Algorithmic 27 – 29, Short Answer 7 Test Item File Questions: Multiple Choice 28, Algorithmic 30 – 36, Short Answer 8 – 9 6 Test Item File Questions: Multiple Choice 30 – 39, Algorithmic 38 – 41, Short Answer 10 – 11 7 Test Item File Questions: Multiple Choice 40 – 42, Algorithmic 43 – 45, Short Answer 12 – 13 4 5

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Chapter 1—Chemical Tools: Experimentation and Measurement B. Kinetic energy 1. Energy of motion 2. EK = ½mv2 C. Potential energy = stored energy D. Joule 1. SI unit for energy 2. 1 J = 1(kg  m2)/s2 E. Calorie 1. Amount of energy necessary to raise the temperature of 1 g of water by 1 C 2. 1 cal = 4.184 J 3. Nutritional calorie (Calorie) 4. 1 Cal = 1000 cal = 1 kcal = 4.184 kJ 1.9. Accuracy, Precision, and Significant Figures in Measurement8 A. Accuracy – how close a measurement is to the true value B. Precision – how well a number of independent measurements agree with one another 1. To indicate the precision of a measurements, use all the digits known with certainty, plus one additional estimated digit, 2. Number of digits in the measurement = number of significant figures C. Determining the number of significant figures – see rules in textbook D. Using scientific notation – helpful in expressing the correct number of significant figures E. Exact numbers – infinite number of significant figures 1.10. Significant Figures in Calculations9 A. In carrying out a multiplication or division, the answer can’t have more significant figures than either of the original numbers.

B. In carrying out an addition or subtraction, the answer can’t have more digits to the right of the decimal point than either of the original numbers.

C. Rules for rounding off numbers 1. See text 2. Doing calculations a. Use all figures, significant or not b. Only round off final answer 1.11. Converting from One Unit to Another10 A. Dimensional analysis method 1. Quantity described in one unit converted into an equivalent quantity described in a different unit 2. Conversion factor used to express relationship between units

Conversion factor =

equivalent quantity starting quantity

Starting quantity  conversion factor = equivalent quantity 3. Units a. Can be multiplied and divided b. Note that addition and subtraction can only be used with numbers having the same units

Test Item File Questions: Algorithmic 46 – 54, Short Answer 14 – 15 Test Item File Questions: Multiple Choice 44 – 47, Algorithmic 55 – 62, Short Answer 16 10 Test Item File Questions: Multiple Choice 51 – 55, Algorithmic 63 – 67, Short Answer 17 8 9

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Chapter 1—Chemical Tools: Experimentation and Measurement 4. Equation a. Set up so that all unwanted units cancel b. Leave only desired units c. Correct answer obtained only if equation set up so unwanted units cancel 5. Method: a. Start with given information, including units b. Identify information needed, including units c. Find relationship between known information and unknown answer d. Plan a strategy to get from known information to unknown answer e. Solve i. Use conversion factor that allows units in given information to cancel ii. Continue conversion factor use to cancel the previous unit until desired unit remains f. Consider whether answer obtained is reasonable (Refer to Ballpark Check in textbook.)

6. Number of significant figures in answer determined from given information Inquiry – What are the unique properties of nanoscale materials?11

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Test Item File Questions: Algorithmic 68 – 69 ..

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Chapter 1—Chemical Tools: Experimentation and Measurement

CHAPTER 1 CHEMICAL TOOLS: EXPERIMENTATION AND MEASUREMENT Teaching Tips, Points of Emphasis, and Common Misconceptions Section 1.1

Chemistry is an experimental science. Challenge students to sharpen their skills of observation and measurement, relating laboratory work to concepts discussed in lecture.

Section 1.1

A theory provides the best explanation based on available information. Additional data can result in modification or replacement of the theory.

Section 1.2

All systems of measurement are based on arbitrary standards. The SI system is as arbitrary as the English system, but the relationship of one unit to another is more systematic and easier to remember in the metric system.

Section 1.2

For very large or very small quantities, commonly used prefixes represent amounts that vary by three powers of 10, for example, milli, micro, nano, and pico (10–3, 10–6, 10–9, and 10–12, respectively).

Section 1.6

Remind students to raise both the measured number and unit to the appropriate power when dealing with areas and volumes. For example, (3 cm)3 = 33 cm3 = 27 cm3, not 3 cm3 or 27 cm.

Section 1.7

Density is sometimes confused with mass. A kg of feathers weighs the same as a kg of gold, but gold occupies less space because it has a greater density, the ratio of mass to volume.

Section 1.7

Because students are very familiar with water they tend to use it as a standard when relating the density of a solid to the density of a liquid. Most solids sink in their liquids; water is unusual in that ice floats atop liquid water.

Section 1.9

Significant figures tend to be abstract to students until they realize that significant figures are a shortcut to expressing the limitations of laboratory measurements.

Section 1.9

Only measured quantities may contain a measured error, which limits the number of significant figures. In addition to counted objects (7 days = 1 week), numbers related by definition (1 in = 2.54 cm) are exact numbers with an infinite number of significant figures.

Section 1.10

The answer given to a series of calculations involving measured quantities is limited by the least precise measurement.

Section 1.10

Students can avoid round-off errors by always using one more significant figure for constants or molar masses than required by measured data and rounding off only the final answer and not intermediate ones.

Section 1.10

Based on calculator displays, students often report answers to problems that contain too many or too few significant figures. Remind students that calculators are incapable of distinguishing between measured and exact quantities. One consequence is that calculators do not record a zero at the end of numbers even when the zero may be a significant figure.

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Chapter 1—Chemical Tools: Experimentation and Measurement Section 1.11

Mathematical problem solving is more difficult for most students than understanding chemistry concepts. Knowing the goal of a series of calculations and where to start is not trivial for many students. Students need to be taught a system, such as writing the units of the desired answer on the right side of the paper, writing a measurement or conversion factor containing the same units on the left, and using appropriate relationships to cancel all but the desired units.

Lecture/Laboratory Demonstration References Section 1.2

Charles W. Earley, “A Simple Demonstration for Introducing the Metric System to Introductory Chemistry Classes,” J. Chem. Educ., Vol. 76, 1999, 1215–1216.

Section 1.7

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Sugar in a Can of Soft Drink: A Density Excercise,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2, (American Chemical Society, Washington, DC, 1988) pp. 126–127. Regular and diet soft drinks are identified by differences in their densities.

Section 1.7

David A. Franz and David Speckhard, “Densities and Miscibilities of Liquids and Liquid Mixtures,” J. Chem. Educ., Vol. 68, 1991, 594. Using a graduated cylinder and electronic balance, the densities of water, dichloromethane, and petroleum ether are determined. A heavier mass of water is shown to form a layer on top of the more dense dichloromethane, but below the less dense petroleum ether. Mixtures of the miscible organic solvents are prepared with average densities above or below the density of water.

Literature References Section 1.2

Mark L. Campbell, “Having Fun with the Metric System,” J. Chem. Educ., Vol. 68, 1991, 1048. A word game.

Section 1.7

Kenneth E. Kolbe and Doris K. Kolbe, “Method for Separating or Identifying Plastics,” J. Chem. Educ., Vol. 68, 1991, 348. A practical use of density.

Section 1.9

Charles J. Guare, “Error, Precision and Uncertainty,” J. Chem. Educ., Vol. 68, 1991, 649-652.

Section 1.9

Richard S. Treptow, “Precision and Accuracy in Measurements: A Tale of Four Graduated Cylinders,” J. Chem. Educ., Vol. 75, 1998, 992-995.

Section 1.9

Arden P. Zipp, “A Simple but Effective Demonstration for Illustrating Significant Figure Rules When Making Measurements and Doing Calculations,” J. Chem. Educ., Vol. 69, 1992, 291.

Section 1.9

Ben Ruekberg, “A Joke Based on Significant Figures,” J. Chem. Educ., Vol. 71, 1994, 306.

Section 1.9

Kenton B. Abel and William M. Hemmerlin, “Significant Figures,” J. Chem. Educ., Vol. 67, 1990, 213.

Section 1.9

H. Graden Kirksey and Paul Krause, “Significant Figures, A Classroom Demonstration,” J. Chem. Educ., Vol. 69, 1992, 497-498.

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Chapter 1—Chemical Tools: Experimentation and Measurement

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 1.11

Worked Example 1.8; Unit conversions using significant figures

Section 1.11

Worked Example 1.9: Unit conversions with squared and cubed units

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CHAPTER 2 ATOMS, MOLECULES, AND IONS Chapter Learning Objectives Section 2.1

Correlate the atomic symbol with the element name.

Section 2.2

Identify the location of metals, nonmetals, and semimetals on the periodic table.

Indicate the atomic number, group number, and period number for an element whose position in the periodic table is given.

Identify groups as main group, transition metal group, or inner transition metal group.

Section 2.3

Specify the location and give examples of elements in the alkali metal, alkaline earth metal, halogen, and noble gas groups.

Classify an element as a metal, nonmetal, or semimetal using its properties.

Section 2.4

Determine the mass of the products in a reaction using the law of mass conservation.

Section 2.5

Demonstrate the law of multiple proportions using mass composition of two compounds of the same elements.

Use the law of multiple proportions to find the chemical formula by comparing the mass ratio to a similar composition compound whose formula is known.

Section 2.6

Describe Thomson’s cathode-ray experiment and what it contributed to the current model of atomic structure.

Describe Millikan’s oil drop experiment and what it contributed to the current model of atomic structure.

Section 2.7

Describe Rutherford’s gold foil experiment and what it contributed to the current model of atomic structure.

Calculate the number of atoms in a sample given the size of the atom and the size of the sample.

Section 2.8

Determine the mass number, atomic number, and number of protons, neutrons, and electrons from an isotope symbol.

Section 2.9

Calculate atomic weight given the natural abundance and mass of each isotope.

Convert between grams and numbers of moles using molar mass; convert between moles and number or atoms using Avogadro’s number.

Section 2.10

Use data from a mass spectrum to calculate the atomic weight of an element.

Section 2.11

Classify matter as a mixture, pure substance, element, or compound.

Convert between structural formulas, ball-and-stick models, and chemical formulas.

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Chapter 2—Atoms, Molecules, and Ions Section 2.12

Classify bonds as ionic or covalent.

Determine the number of electrons and protons from chemical symbol and charge.

Match the molecular representation of an ionic compound with its chemical formula.

Section 2.13

Convert between name and formula for ionic compounds.

Convert between name and formula for molecular compounds.

Lecture Outline 2.1. Chemistry and the Elements1 A. Element – fundamental substance that can’t be chemically changed or broken down into anything simpler 1. Chemical symbol – represents a specific element a. Capitalize first letter b. Lower case used for second letter if present B. Periodic table – tabular organization of all 118 elements 2.2. Elements and the Periodic Table2 A. Creation of the periodic table – ideal example of how scientific theory comes into being 1. Separate observations 2. Organization of data in ways that make sense 3. Consistent hypothesis emerges a. Explains known facts b. Makes predictions about unknown phenomena B. Mendeleev’s hypothesis about organizing known chemical information – meets criteria for a good hypothesis 1. Listed the known elements by atomic weight 2. Grouped them together according to their chemical reactivity 3. Was able to predict the properties of unknown elements – eka–aluminum, eka–silicon C. Periodic table – grid of the elements arranged in 7 horizontal rows and 18 vertical columns 1. Periods – seven horizontal rows in the periodic table 2. Groups – 18 vertical columns in the periodic table a. Groups numbered 1 → 18 (1A → 8A and 1B → 8B) b. Actually have 32 groups – lanthanides (14 elements after lanthanum) and actinides (14 elements after actinium) not included in the group numbers c. Elements in a given group have similar chemical properties.

D. The periodic table of the elements is the most important organizing principle of chemistry.

1. Regular progression in size of the seven periods – reflects a similar regularity in atomic structure 2. Main Group (or Representative) Elements – Groups 1, 2, 13 – 18; (two larger groups on the left and six larger groups on the right of the table) 3. Transition Metal Elements – Groups 3 – 12; (10 smaller groups in the middle of the table) 4. Inner Transition Metal (or Rare Earth) Elements (14 groups shown separately at the bottom of the table)

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Test Item File Questions: Multiple Choice 1 – 3, Algorithmic 1 – 5, Short Answer 1 – 2 Test Item File Questions: Multiple Choice 4 – 7, Algorithmic 10, Short Answer 3 – 5 10

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Chapter 2—Atoms, Molecules, and Ions 2.3. Some Common Groups of Elements and Their Properties3 A. Property – any characteristic used to describe or identify matter 1. Intensive properties – sample size–independent properties 2. Extensive properties – sample size–dependent properties 3. Physical properties – characteristics not involving chemical change of the sample 4. Chemical properties – properties that do change chemical makeup of the sample B. Elements within a group have similar chemical properties.

1. Group 1A – Alkali metals a. Lustrous, silvery metals b. React rapidly with water to form highly alkaline products 2. Group 2A – Alkaline earth metals a. Lustrous, silvery metals b. Less reactive than alkali metals 3. Group 7A – Halogens a. Corrosive, nonmetallic elements b. Salt formers 4. Group 8A – Noble gases: gases with low reactivity C. Three major categories of elements 1. Metals a. Largest category of elements b. Found on the left side of the periodic table (left of the heavy zigzag line) i. Solids (except mercury) ii. Malleable iii. Ductile – can be drawn into thin wires without breaking iv. Conduct heat and electricity 2. Nonmetals a. Relatively small number b. Found on the right side of the periodic table (right of the heavy zigzag line) i. Gases, liquids, or solids ii. Brightly colored iii. Brittle solids iv. Poor conductors of heat and electricity 3. Semimetals (metalloids) a. Elements adjacent to the zigzag boundary between metals and nonmetals b. Properties fall between metals and nonmetals i. Brittle ii. Poor conductors of heat and electricity 2.4. Observations Supporting Atomic Theory: The Conservation of Mass and the Law of Definite Proportions4 A. Element – a substance that cannot be further broken down B. Compound – combination of different elements a. Chemical formula – lists the symbols of its constituent elements and uses subscripts to indicate the number of atoms of each.

C. Chemical equation – the standard format for describing a chemical transformation or reaction in which the reactant substances undergoing change are written on the left, the product substances being formed are written on the right, and an arrow is drawn between them to indicate the direction of the chemical transformation.

D. Law of Mass Conservation – Mass is neither created nor destroyed in chemical reactions.

E. Law of Definite Proportions – Different samples of a pure chemical substance always contain the same proportion of elements by mass; elements do not combine chemically in random proportion.

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Test Item File Questions: Multiple Choice 8 – 15, Algorithmic 11 – 34, Short Answer 6 – 8, Test Item File Questions: Multiple Choice 47 – 50, Algorithmic 35 – 39, Short Answer 9

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Chapter 2—Atoms, Molecules, and Ions 2.5. The Law of Multiple Proportions and Dalton’s Atomic Theory5 A. Law of Multiple Proportions – If two elements combine in different ways to form different substances, the mass ratios are small, whole-number ratios of each other.

B. Dalton’s Atomic Theory 1. Elements made of tiny particles called atoms 2. Each element characterized by the mass of its atoms a. Atoms of the same element – identical masses b. Atoms of different elements – different masses 3. Atoms combine in small, whole-number ratios – form new substances, called compounds 4. Chemical reactions only rearrange the way that atoms are combined; atoms themselves are not changed.

2.6. Atomic Structure: Electrons6 A. Thomson – found that cathode rays consist of tiny, negatively charged particles called electrons 1. Electrons are emitted from electrodes made of two thin pieces of metal.

2. Many different metals may be used to make electrodes – all contain electrons.

3. Cathode rays can be deflected by bringing either a magnet or an electrically charged plate near the tube. Deflection depends on the a. strength of the deflecting magnetic or electric field. b. size of the negative charge on the electron. c. mass of the electron.

4. Charge-to-mass ratio, e/m of the electron = 1.758 819  108 C/g B. Millikan determined that the charge on a drop of oil was always a small, whole-number multiple of e. (See Fig. 2.4 in the textbook.)

1. e = 1.602 177  10–19 C 2. Knowing the values for e/m and e for an electron, m can be calculated. a. m = 9.109 390  10–28 g 2.7. Atomic Structure: Protons and Neutrons7 A. Rutherford directed a beam of alpha particles at a thin gold foil.

1. Alpha () particles a. 7000 times more massive than electrons b. Have a positive charge twice the magnitude of, but opposite in sign to, the charge on an electron 2. Beam of  particles a. Most pass through the thin gold foil b. A few deflected at large angles B. Nuclear model of the atom 1. Nucleus a. A tiny central core in an atom where the mass of the atom is concentrated b. Contains the atom’s positive charges 2. Electrons move in space a relatively large distance away from the nucleus.

C. Nucleus composed of two kinds of particles 1. Protons a. Mass = 1.672 623  10–24 g b. Positively (+) charged c. Number of protons = number of electrons in a neutral atom 2. Neutrons a. Neutron mass  proton mass b. Charge = 0 2.8. Atomic Numbers8 Test Item File Questions: Multiple Choice 54 – 62, Algorithmic 40, Short Answer 10 Test Item File Questions: Multiple Choice 69 – 70, Short Answer 11 7 Test Item File Questions: Multiple Choice 71 – 74, Algorithmic 41, Short Answer 12 5 6

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Chapter 2—Atoms, Molecules, and Ions A. B. C. D.

Elements differ from one another according to the number of protons in their atoms.

Atomic number (Z) = the number of protons in an atom = the number of electrons in an atom Most nuclei also contain neutrons.

Mass number (A) = number of protons + number of neutrons in an atom (whole number count of protons and neutrons in the nucleus of a particular isotope) E. Isotopes – atoms with identical atomic numbers but different mass numbers 1. Mass number written as left superscript 2. Atomic number (Z) written as left subscript (The atomic number is sometimes left off because all atoms of an element always contain the same number of protons.)

3. Number of neutrons in an isotope calculated from A – Z 4. Number of neutrons in an atom has little effect on chemical properties of the atom.

2.9. Atomic Weights and the Mole9 A. Unified atomic mass unit (u) previously the atomic mass unit (amu) 1. Exactly 1/12th the mass of an atom of 126 C 2. 1 u = 1.660 539  10–24 g 3. Unit also called a dalton (Da) B. Atomic mass 1. Mass of an atom in atomic mass units 2. Numerically close to the atom’s mass number C. Atomic weight values – weighted averages for the naturally occurring mixtures of isotopes 1. Average atomic mass of an element (atomic weight) = (mass of each isotope  the fraction of the isotope) a.  used for the term “the sum of” 2. Use atomic weights to count number of atoms by weighing a sample of the element.

D. One mole of any element is the amount whose mass in grams is numerically equal to its atomic weight.

1. Mole (mol) = 6.022  1023 of anything a. Avogadro’s number (abbreviated NA) b. 1 mol N atoms = 6.022  1023 atoms of N c. Likewise, 1 mol electrons = 6.022  1023 electrons 2. Importance of the atomic weight – provides a relationship between numbers of atoms and masses of atoms a. Allows the counting of a large number of atoms by weighing a sample of a substance E. Molar mass of an element 1. One mole of any element has a mass equal to its atomic weight in grams.

2. Mass of 6.022  1023 atoms of an element 3. Serves as a conversion factor between numbers of atoms and mass 2.10 Measuring Atomic Weight: Mass Spectrometry A. Most common method used for measurement of atomic weight B. Method 1. Vaporized sample ionized by bombardment with high-energy electrons.

2. Rapidly moving ions pass through a curved magnetic field a. Pathway followed by ions is mass dependent. b. Lighter ions deflected to a greater degree than heavier ions

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Test Item File Questions: Multiple Choice 75 – 91, Algorithmic 42 – 50, Short Answer 13 – 18 Test Item File Questions: Multiple Choice 92 – 98, Algorithmic 51 – 55, Short Answer 19 – 22

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Chapter 2—Atoms, Molecules, and Ions 3. Ions separated by mass enter a detector assembly 4. A mass spectrum, which is a plot of ion intensity versus the mass to charge ratio, is generated.

2.11. Mixtures and Chemical Compounds; Molecules and Covalent Bonds 10 A. Different kinds of matter on earth classified as either pure substances or mixtures – textbook Figure 2.12 B. Mixtures – blends of substances involving no chemical changes C. Pure substance 1. Elements 2. Compounds D. Chemical compounds – pure substance formed from the combination of atoms of two or more different elements 1. Have constant atomic composition 2. Composition indicated by a chemical formula a. Lists the symbols of individual constituent elements b. Number of each atom is given by subscript 3. Formed when atoms undergo chemical combination in a specific manner 4. Transformation from elements to compound happens by one type of chemical reaction E. Chemical bonds – connections that join atoms together in a compound 1. Formed by atom’s electrons 2. Classified as: a. Covalent bonds – occur between two nonmetals b. Ionic bonds – occur between a metal and a nonmetal F. Covalent bond – two atoms share electrons 1. Molecule – unit of matter that results when two or more atoms joined by covalent bonds 2. Structural formula a. Shows specific connections between atoms b. Contains more information than the chemical formula 3. Some elements exist as molecules: H2, O2, N2, F2, Cl2, Br2, I2 2.12. Ions and Ionic Bonds11 A. Ionic Bond – complete transfer of one or more electrons from one atom to another 1. Formed between metals and nonmetals a. Metal – gives up electrons b. Nonmetal – accepts electrons 2. Ions – charged particles resulting from loss or gain of electrons a. Cation – positively charged particle resulting from loss of one or more electrons b. Anion – negatively charged particle resulting from gain of one or more electrons c. Polyatomic ions i. Charged, covalently bonded groups of atoms ii. Like charged molecules – specific numbers and kinds of atoms joined in a definite way by covalent bonds 3. Ionic solids a. Cations and anions packed together in a regular manner b. Charges cancel 2.13. Naming Chemical Compounds12 A. Binary Ionic Compounds – ionic compounds containing only two elements, a cation and an anion 1. Identify cation first, then anion a. Cation i. Same name as the element Test Item File Questions: Multiple Choice 56, Algorithmic 56 – 63, Short Answer 23 Test Item File Questions: Multiple Choice 118 – 121, Algorithmic 69 – 85, Short Answer 24 – 26 12 Test Item File Questions: Multiple Choice 122 – 140, Algorithmic 86 – 117, Short Answer 27 – 28 10 11

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Chapter 2—Atoms, Molecules, and Ions ii. Remember metals form cations b. Anion i. First part of its name from the element ii. Adds the ending ide iii. Remember nonmetals form anions 2. Common main group and transition metal ions – textbook Figures 2.17 and 2.18 a. Elements within a group often form similar kinds of ions. b. Main-group metal cations: charge = group number c. Main-group nonmetal anions: charge = group number – 8 (1A – 8A convention) d. Some metals form more than one kind of cation. i. Charge indicated by a Roman numeral in parentheses ii. Common for transition metal complexes 3. Electrical neutrality – cations and anions combine in such a manner that the overall charge on a compound is equal to zero a. Total positive charge = total negative charge b. Determine number of positive charges on the cation by counting the number of negative charges on the anion (and vice versa) 4. Formulas for ionic compounds always contain the smallest whole number ratio of cation to anion.

B. Ionic compounds containing polyatomic ions named by following rules for naming binary ionic compounds.

1. Identify cation 2. Identify anion a. Names, formulas, and charge numbers of the most common polyatomic anions found in textbook Table 2.5 b. Most names end with the suffix ite or ate c. Several pairs of ions related by presence or absence of hydrogen atom 3. Oxoanions – an atom of the same element combined with different numbers of oxygen atoms a. Learn the name and formula of the ion whose name ends with ate (including the charge on the anion). b. Add one O; add prefix per. c. Remove one O; change ate to ite. d. Remove two O’s; add prefix hypo and change ate to ite.

C. Binary molecular compounds – molecular compounds (nonmetals) containing only two elements 1. One of the elements is more cationlike – takes the name of the element 2. One of the elements is more anionlike – takes an ide ending 3. Character depends on relative positions of the two elements in the periodic table a. More cationlike – farther left in the periodic table b. More anionlike – farther right in the periodic table 4. To specify the numbers of each element present, use numerical prefixes (textbook Table 2.6) – the mono prefix not used for the atom named first Inquiry: How can measurements of oxygen and hydrogen isotopes in ice cores determine past climates?

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Chapter 2—Atoms, Molecules, and Ions

CHAPTER 2 ATOMS, MOLECULES, AND IONS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 2.1

Elements can be broken into simpler substances (protons, neutrons, electrons, etc.), but the energy required is greater than available under ordinary chemical conditions.

Section 2.2

Iodine is a silvery-gray solid that sublimes to a purple gas. The brown solution that was once used as an antiseptic medicine is known as tincture of iodine (iodine dissolved in alcohol).

Section 2.2

Because elements in the same group tend to exhibit similar chemical behavior, a group is also called a family.

Section 2.3

While elements within the same group tend to have similar chemical properties, due to its small size the first member of each main group exhibits a chemistry that differs most from the other members of the group.

Section 2.3

Hydrogen is a unique element. Although listed under group 1A, hydrogen is not a metal under ordinary conditions. In its ionic chemistry, hydrogen sometimes behaves as though it belongs in group 7A. In covalent compounds, hydrogen behaves as though it belongs between boron and carbon.

Section 2.5

Dalton did not know about isotopes – atoms of the same element that have different masses due to different numbers of neutrons. Challenge students to determine which of Dalton’s postulates are incorrect.

Section 2.7

Demonstrate the relative insignificance of an electron’s mass by calculating its percent contribution to the approximate mass of a simple atom, such as carbon-12.

Section 2.9

The use of carbon-12 to define the amu is another example of a system of measurement based on an arbitrary standard. In earlier times the amu was based on oxygen-16 and hydrogen-1.

Section 2.9

The atomic weight of an element is a weighted, not a simple, average.

Section 2.9

An effective example showing the difference between simple average and weighted average is the use of a hypothetical student’s test scores of 100, 100, 100, 60. The weighted average = 90 since there are only two different grades, 100 and 60. The simple average is 80.

Section 2.11

The element oxygen is written O2. P4 and S8 are other examples of elements that occur as polyatomic species.

Section 2.12

NaCl is a simplest formula or empirical formula. It does not reveal that the compound is made up of ions (Na+Cl–), nor that it is an extended system (NaCl)x.

Section 2.12

Polyatomic ions are also known as molecular ions because they are molecules with a charge.

Section 2.12

Some students think HCl is ionic because it dissolves in water to form H + and Cl–.

The interaction of a molecule and a solvent can substantially change the properties of the molecule.

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Chapter 2—Atoms, Molecules, and Ions

Section 2.13

In compounds, elements on the left of the periodic table tend to be relatively positive, and elements on the right tend to be relatively negative.

Lecture/Laboratory Demonstration References Section 2.2

James L. Marshall, “A Living Periodic Table,” J. Chem. Educ., Vol. 77, 2000, 979–983. Describes a portable and permanent collection of 87 elements.

Section 2.8

Arthur B. Ellis, Edward A. Adler, and Frederick H. Juergens, “Dramatizing Isotopes: Deuterated Ice Cubes Sink,” J. Chem. Educ., Vol. 67, 1990, 159-160.

Literature References Section 2.1

P.G. Nelson, “Important Elements,” J. Chem. Educ., Vol. 68, 1991, 732–737.

Section 2.1

Vivi Ringes, “Origin of the Names of Chemical Elements,” J. Chem. Educ., Vol. 66, 1989, 731-738.

Section 2.2

Milton J. Wieder, “It’s Elementary,” J. Chem. Educ., Vol. 78, 2001, 468-469.

Section 2.4

John J. Fortman, “Analogical Demonstration,” J. Chem. Educ., Vol. 69, 1992, 323–324. Discusses the law of conservation of mass and the law of multiple proportions.

Section 2.5

Doris Eckey, “A Millikan Oil Drop Analogy,” J. Chem. Educ., Vol. 73, 1996, 237–238.

Section 2.7

Mary V. Lorentz, “Bowling Balls and Beads, A Concrete Analogy to the Rutherford Experiment,” J. Chem. Educ., Vol. 65, 1988, 1082.

Section 2.7

Barrie M. Peake, “The Discovery of the Electron, Proton, and Neutron,” J. Chem.

Educ., Vol. 66, 1989, 738.

Section 2.8

William Spindel and Takanobu Ishida, “Isotope Separation,” J. Chem. Educ., Vol. 68, 1991, 312-318.

Section 2.9

Arthur M. Last and Michael J. Webb, “Using Monetary Analogies to Teach Average Atomic Mass,” J. Chem. Educ., Vol. 70, 1993, 234–235.

Section 2.9

John J. Fortman, “Pictorial Analogies IV: Relative Atomic Weights,” J. Chem.

Educ., Vol. 70, 1993, 235–236.

Section 2.9

Josefina Arce de Sanabia, “Relative Atomic Mass and the Mole: A Concrete Analogy to Help Students Understand These Abstract Concepts,” J. Chem. Educ., Vol. 70, 1993, 233–234.

Section 2.13

Gerhard Lind, “Teaching Inorganic Nomenclature: A Systematic Approach,” J. Chem. Educ., Vol. 69, 1992, 613–614.

Section 2.13

Steven J. Hawkes, “A Mnemonic for Oxy-Anions,” J. Chem. Educ., Vol. 67, 1990, 149.

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Chapter 2—Atoms, Molecules, and Ions

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 2.9

Worked Example 2.5: Calculating an atomic weight

Section 2.9

Worked Example 2.6: Converting between mass and numbers of moles and atoms

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CHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS Chapter Learning Objectives Section 3.1 and Section 3.2

Visualize bonds broken and formed in a chemical reaction and relate numbers of molecules or atoms to the balanced chemical equation.

Balance a chemical reaction, given the formulas of reactants and products.

Section 3.3 and Section 3.4

Calculate formula weight, molecular weight, and molar mass, given a chemical formula or structure.

Convert between mass, moles, and molecules or atoms of a substance.

Relate the amount (moles or mass) of reactants and products in a balanced equation using stoichiometry.

Section 3.5

Calculate percent yield of a reaction.

Section 3.6

Determine the relative amounts of atoms or molecules in the reactants and products of a balanced reaction, given a molecular representation.

Determine which reactant is limiting and calculate the theoretical yield of the product and the amount of excess reactant.

Calculate the percent yield when one reactant is limiting.

Section 3.7 and Section 3.8

Calculate the percent composition, given a chemical formula or structure.

Determine the empirical and molecular formula, given the mass percent composition and molecular weight of a compound.

Determine the empirical and molecular formula, given combustion analysis data and molecular weight.

Section 3.9

..

Determine empirical and molecular formula using data from mass spectrometry and combustion analysis.

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Chapter 3—Mass Relationships in Chemical Reactions

Lecture Outline 3.1. Representing Chemistry on Different Levels1 A. Chemical symbols – represent both a microscopic and a macroscopic level 1. Microscopic level – chemical symbols represent behavior of individual atoms and molecules 2. Macroscopic level – formulas and equations represent large-scale behavior of atoms and molecules, giving rise to observable properties a. Deal with macroscopic behavior in the laboratory b. Use microscopic models to explain macroscopic behavior 3.2. Balancing Chemical Equations2 A. Balanced equations 1. Numbers and kinds of atoms on both sides of the arrow are equal 2. Chemical equations balanced to obey law of mass conservation B. Formula unit – one unit (an atom, ion, or molecule) corresponding to a given formula C. Balancing process 1. Write unbalanced equation using correct chemical formulas for all reactants and products.

2. Use coefficients (numbers placed before each formula) to indicate number of formula units required to balance the equation. a. Coefficients can be changed b. Formulas must stay the same 3. Reduce coefficients to smallest whole numbers.

4. Check your answer – add up each atom on both sides of the arrow.

3.3. Molecular Weights and Molar Mass A. Dealing with macroscopic behavior requires that the number ratio of reactants or products from a balanced equation be converted to a mass ratio (In the laboratory we do not count numbers of reactants and products; we weigh grams of reactants and products.)

1. Number ratio determined from coefficients in the balanced equation 2. Mass ratio determined using molecular weights a. Molecular weight – sum of atomic weights of all atoms in a molecule b. Formula weight – sum of atomic weights of all atoms in one formula unit of any substance B. One mole of a compound has a mass in grams equal to the compound’s molecular weight in daltons and contains Avogadro’s number (6.022  1023) of molecules.

3.4. Stoichiometry: Relating Amounts of Reactants and Products A. Stoichiometry – refers to the chemical arithmetic used to relate amounts (mass and/or moles) of reactants and products in a chemical reaction B. Coefficients in a balanced chemical equation indicate the number of formula units or moles of each substance needed for the reaction.

C. Converting from grams to moles in a formula unit – use the molar mass of the substance as a conversion factor D. Calculating the grams of reactants needed and products produced, knowing the grams and moles of another reactant 1. Begin with the known grams of reactant, convert to moles 2. Using the balanced chemical equation, convert moles of one substance to moles of another 3. Knowing moles, convert to grams

1Test Item File Questions: Multiple Choice 1 – 3 2Test Item File Questions: Multiple Choice 8 – 13, Algorithmic 2 – 5, Short Answer 1 – 3

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Chapter 3—Mass Relationships in Chemical Reactions 3.5. Yields of Chemical Reactions3 A. Actual yield 1. Amount of product actually formed 2. Less than the amount predicted by theory B. Theoretical yield – amount of product calculated to form if the maximum amount of the reactants are converted to products 1. The theoretical yield is based on the limiting reactant as discussed in Section 3.6.

C.

Percent yield =

Actual yield 100% Theoretical yield

3.6. Reactions with Limiting Amounts of Reactants4 A. Many reactions carried out using an excess of one reactant B. Excess reactant – reactant present in more than the amount needed according to the stoichiometry 1. Only amount required by stoichiometry reacts 2. Any excess amount of the reactant acts as a spectator and plays no role in the reaction C. Limiting reactant 1. Reactant present in the smallest molar amount as required by stoichiometry 2. Determines the extent to which a chemical reaction takes place 3. Reaction stops when limiting reactant depleted D. Determination of whether a limiting amount of one reactant is present and calculation of the amount of excess reactant consumed: 1. Find moles of each reactant present.

2. Look at the coefficients in the balanced equation to see the required mole ratio of the two reactants. Use the coefficients to calculate the amount of product that could be made if all of each reactant were to react.

3. The limiting reactant is the reactant that creates the least amount of product.

4. Subtract number of moles of excess reactant consumed from the number of moles of excess reactant present. a. Convert to grams. b. This is grams of nonlimiting reactant in excess 5. Determination of the amount of product formed: a. Based on moles of limiting reactant b. Moles of limiting reactant → moles product → grams product 3.7. Percent Composition and Empirical Formulas5 A. Composition of a compound – identity and amount of elements present in a compound B. Percent composition 1. Mass percent of each element present in a compound 2. Calculation of chemical formula from percent composition: a. Find relative numbers of moles of each element in the compound. i. Assume 100 g of compound ii. Use molar masses of elements as conversion factors. b. Find ratio of the numbers of moles by dividing larger number of moles by smaller number of moles. c. Multiply subscripts by small integers in a trial-and-error procedure until whole numbers found.

3Test Item File Questions: Multiple Choice 49 – 50, Algorithmic 38 – 43, Short Answer 13 – 14 4Test Item File Questions: Multiple Choice 51 – 59, Algorithmic 44 – 52, Short Answer 15 – 17 5Test Item File Questions: Multiple Choice 67 – 70, Algorithmic 54 – 61, Short Answer 18 – 19

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Chapter 3—Mass Relationships in Chemical Reactions

C. Empirical formula – gives only ratios of atoms in a compound D. Molecular formula – gives actual numbers of atoms in a molecule 1. May be the same as the empirical formula 2. May be an integer multiple of the empirical formula: Integer Multiple =

Molecular mass Empirical formula mass

E. Can derive percent composition from chemical formula 3.8. Determining Empirical Formulas: Elemental Analysis 6 A. Combustion analysis – compound of unknown composition burned with oxygen to produce volatile combustion products CO2 and H2O 1. CO2 and H2O separated by a gas chromatograph and their masses determined 2. Used to determine empirical formula of a compound B. Determination of empirical formula of a compound from combustion reaction: 1. Perform gram-to-mole conversions to find molar amounts of C and H in CO2 and H2O. 2. Perform mole-to-gram conversions to find number of grams of C and H in original sample.

3. Subtract masses of C and H from the mass of starting sample to determine number of grams of third element.

4. Determine number of moles of third element.

5. Find mole ratios by dividing larger number of moles by smaller number of moles.

C. Determination of molecular formula of a substance: 1. Molecular mass determined from a separate experiment (e.g., using mass spectrometry) 2. Subscripts in empirical formula multiplied by ratio of molecular mass to empirical formula mass 3.9. Determining Molecular Masses: Mass Spectrometry7 A. Most common method for molecular mass determination B. Method 1. Vaporized sample ionized by bombardment with high-energy electrons 2. Some molecular ions fragment into smaller ions

6Test Item File Questions: Multiple Choice 72, 74 – 82, Algorithmic 62 – 63, Short Answer 20 7Test Item File Questions: Multiple Choice 80 – 82

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Chapter 3—Mass Relationships in Chemical Reactions 3. Rapidly-moving ions pass through a curved magnetic field a. Pathway followed by ions mass-dependent b. Lighter ions deflected more strongly than heavier ions 4. Ions separated by mass enter a detector assembly a. Signal intensities proportional to the relative number of each ion b. Mass of each ion assigned by comparison to a standard 5. Very precise – molecular weights determined to as many as seven significant figures Inquiry: How is the principle of atom economy used to minimize waste in a chemical synthesis?

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Chapter 3—Mass Relationships in Chemical Reactions

CHAPTER 3 MASS RELATIONSHIPS IN CHEMICAL REACTIONS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 3.2

Equations may be balanced using fractions. H 2(g) + ½ O2(g) → H2O(l) is a balanced equation.

Section 3.2

When balancing chemical equations involving organic compounds (compounds of carbon), balance the atoms in the order C then H then O.

Section 3.3

Emphasize using chemical formulas with units: 18.0 g H2O/mol H2O instead of 18.0 g/mol.

Just as density provides a link between mass and volume, molar mass links mass and moles, allowing interconversions between these units.

Section 3.4

Coefficients in chemical formulas or balanced chemical equations represent numbers of formula units or moles but never numbers of grams. Show that the mass ratio of C2H4 is not the same as the atomic or mole ratio.

The number of significant figures in the answer to any calculation should be limited only by the number of significant figures in measured quantities. When selecting an atomic mass for use in a calculation, use one more significant figure than is found in the least precise measured quantity involved in the calculation.

Section 3.5

Actual yield is an experimental quantity. It cannot be calculated.

Section 3.6

There are three common methods of determining limiting reactant: (1) comparing stoichiometric ratio to actual ratio of reactants; (2) determining a theoretical yield based on each reactant; and (3) determining the amount of one reactant needed to react with a given quantity of a second reactant.

The excess reactant is not necessarily the reactant present in the greatest amount.

Section 3.7

Experimental errors in the data or calculation round-off errors usually result in mole ratios that are not whole numbers.

Mole ratios frequently contain fractions close to 1/4, 1/3, or 1/2 which should be multiplied by 4, 3, and 2, respectively. Mole ratios close to whole numbers should be rounded to that number.

Section 3.8

Combustion analysis can be used to determine the empirical formula of a compound containing carbon, hydrogen, and any third element. Frequently the third element is oxygen.

When doing combustion analysis problems students frequently try to determine the amount of oxygen in a sample from the amounts of water and carbon dioxide produced. This is not possible because the compound is burned in oxygen which makes up part of the water and carbon dioxide produced.

Molecular weight (molecular mass) cannot be calculated from combustion analysis data. It requires a separate technique, such as mass spectroscopy.

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Chapter 3—Mass Relationships in Chemical Reactions

Lecture/Laboratory Demonstration References Section 3.3

Sally Solomon and Chinhyu Hur, “Measuring Avogadro’s Number on the Overhead Projector,” J. Chem. Educ., Vol. 70, 1993, 252–253. A monolayer of stearic acid on water is used to estimate Avogadro’s number.

Section 3.3

William Laurita, “Demonstrations for Nonscience Majors: Using Common Objects to Illustrate Abstract Concepts,” J. Chem. Educ., Vol. 67, 1990, 60–61. A demonstration of the measurement of Avogadro’s number.

Section 3.7

Bassam Z. Shakhashiri, “Colorful Acid-Base Indicators,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989) pp. 33–40.

Section 3.9

Nancy C. Grim and Jerry L. Saqquis, “Mass Spectrometry Analogy on the Overhead Projector,” J. Chem. Educ., Vol. 72, 1995, 930-931. Mass spectrometry principles are demonstrated on an overhead projector using nickel spheres and magnets.

Literature References Section 3.2

William C. Herndon, “On Balancing Chemical Equations: Past and Present (A Critical Review and Annotated Bibliography),” J. Chem. Educ., Vol. 74, 1997, 1359–1362.

Section 3.2

Zoltan Toth, “Balancing Chemical Equations by Inspection,” J. Chem. Educ., Vol. 74, 1997, 1363–1364.

Section 3.3

Dawn M. Wakeley and Hans de Grys, “Developing an Intuitive Approach to Moles,” J. Chem. Educ., Vol. 77, 2000, 1007–1009.

Section 3.3

Henk van Lubeck, “How to Visualize Avogadro’s Number,” J. Chem. Educ., Vol. 66, 1989, 762.

Section 3.3

Miriam Toloudis, “The Size of a Mole,” J. Chem. Educ., Vol. 73, 1996, 348.

Section 3.3

Damon Diemente, “Demonstrations of the Enormity of Avogadro’s Number,” J. Chem. Educ., Vol. 75, 1998, 1565–1566.

Section 3.3

Carmela Merlo and Kathleen E. Turner, “A Mole of M&M’s,” J. Chem. Educ., Vol. 70, 1993, 453.

Section 3.3

Sheryl Dominic, “What’s a Mole For?”, J. Chem. Educ., Vol. 73, 1996, 309.

Section 3.3

Shanthi R. Krishnan and Ann C. Howe, “The Mole Concept: Developing an Instrument to Assess Conceptual Understanding,” J. Chem. Educ., Vol. 71, 1994, 635–655.

Section 3.4

Addison Ault, “How to Say How Much: Amounts and Stoichiometry,” J. Chem.

Educ., Vol. 78, 2001, 1347–1348.

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Chapter 3—Mass Relationships in Chemical Reactions Section 3.4

Richard L. Poole, “Teaching Stoichiometry: A Two-Cycle Approach,” J. Chem.

Educ., Vol 66, 1989, 57.

Section 3.4

Karla R. Krieger, “Stoogiometry: A Cognitive Approach to Teaching Stoichiometry,” J. Chem. Educ., Vol. 74, 1997, 306–309.

Section 3.4

John J. Fortman, “Pictorial Analogies XII: Stoichiometric Calculations,” J. Chem.

Educ., Vol. 71, 1994, 571–572.

Section 3.4

John Olmsted III, “Amounts Tables as a Diagnostic Tool for Flawed Stoichiometric Reasoning,” J. Chem. Educ., Vol. 76, 1999, 52–54.

Section 3.6

Ernest F. Silversmith, “Limiting and Excess Reagents, Theoretical Yield,” J. Chem.

Educ., Vol. 62, 1985, 61.

Section 3.6

A. H. Kalantar, “Limiting Reagent Problems Made Simple for Students,” J. Chem.

Educ., Vol. 62, 1985, 106.

Section 3.6

Zoltan Toth, “Limiting Reactant: An Alternative Analogy,” J. Chem. Educ., Vol. 76, 1999, 934.

Section 3.7

George L. Gilbert, “Percentage Composition and Empirical Formula – A New View,” J. Chem. Educ., Vol. 75, 1998, 851.

Section 3.7

An easy way to remember the strategy for converting percentage composition to an empirical formula: “Percent to mass, mass to mol, divide by small, multiply ‘til whole.” Joel S. Thompson, “A Simple Rhyme for a Simple Formula,” J. Chem.

Educ., Vol. 65, 1988, 704.

Section 3.7

P. K. Thamburaj, “A Known-to-Unknown Approach to Teach About Empirical and Molecular Formulas,” J. Chem. Educ., Vol. 78, 2001, 1050–1052.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 3.4

Worked Example 3.5: Relating the masses of reactants and products

Section 3.5

Worked Example 3.6: Calculating percent yield

Section 3.6

Worked Example 3.8: Calculating the amount of product or excess reactant when one reactant is limiting

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CHAPTER 4 REACTIONS IN AQUEOUS SOLUTION Chapter Learning Objectives Section 4.1

Calculate the molarity of a solution given the mass of solute and total volume.

Calculate the amount of solute in a given volume of solution with a known molarity.

Describe the proper technique for preparing solutions of known molarity.

Section 4.2

Calculate the concentration of a solution that has been diluted.

Calculate the amount of a stock solution to use to create a known volume of a diluted solution.

Describe the proper technique for diluting solutions.

Section 4.3

Classify a substance as a strong electrolyte, weak electrolyte, or nonelectrolyte.

Calculate the concentration of ions in a strong electrolyte solution.

Section 4.4

Classify a reaction as precipitation, acid–base neutralization, or oxidation–reduction (redox) reaction.

Section 4.5

Write a net ionic equation and identify spectator ions given the molecular equation.

Section 4.6

Use solubility guidelines to predict the solubility of an ionic compound in water.

Predict whether a precipitation reaction will occur and write the ionic and net ionic equations.

Section 4.7

Convert between name and formula for an acid.

Classify acids as strong or weak based on the molecular picture of dissociation.

Write the ionic equation and net ionic equation for an acid–base neutralization reaction.

Section 4.8

Convert between moles and volume using molarity in stoichiometry calculations.

Use reaction stoichiometry to relate moles of one reactant in solution to the moles of another reactant.

Section 4.9

Determine the concentration of a solution using titration data.

Interpret molecular representations of substances in solution in a titration procedure.

Section 4.10

Assign oxidation numbers to atoms in a compound.

Section 4.11

Identify redox reactions, oxidizing agents, and reducing agents.

Section 4.12

Use the location of elements in the periodic table and activity series to predict if a redox reaction will occur.

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Chapter 4—Reactions in Aqueous Solution

Develop an activity series and predict if a redox reaction will occur, based on experimental data provided.

Section 4.13 and Section 4.14

28

Use a redox titration to determine the concentration of an oxidizing or reducing agent in solutions.

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Chapter 4—Reactions in Aqueous Solution

Lecture Outline 4.1. Solution Concentrations: Molarity1 A. Most chemical reactions carried out in solution – reactants must have considerable mobility for reaction to occur B. Solute – substance dissolved in a solvent to make a solution C. Concentration – relative amount of solute to solvent or solution D. Molarity (M) – number of moles of a solute dissolved in each liter of solution 1. Dissolve solute in enough solution to give a final solution volume of 1.00 L 2. Used as conversion factor 4.2. Diluting Concentrated Solutions2 A. Concentrated solution + Solvent → Diluted solution B. Dilution 1. Number of moles of solute remains the same 2. Only volume is changed by adding more solvent 3. A convenient relationship:

 mol   mol  Moles of solute = Mi    Vi ( L ) = M f    Vf (L) L    L  4. Note: This equation should be used only for problems that deal with dilution of a solution!

4.3. Electrolytes in Aqueous Solution3 A. Electrolytes – dissolve in water to produce ions in solutions 1. Strong Electrolytes – completely dissociate in water a. Soluble ionic compounds b. Strong acids c. 70% ionization 2. Weak Electrolytes – incompletely dissociate in water; establish dynamic equilibrium between the forward and backward reactions a. Weak acids b. Weak bases B. Nonelectrolytes – do not produce ions in aqueous solution a. Water b. Alcohols c. Sugars d. Other water-soluble organic compounds (compounds of carbon) C. Textbook Table 4.1 4.4. Types of Chemical Reactions in Aqueous Solution 4 A. Precipitation reactions – solid precipitate forms and drops out of solution B. Acid–base neutralization reactions – acid reacts with base to produce salt and water C. Oxidation–reduction (redox) reactions – transfer of electrons between reactants 4.5. Aqueous Reactions and Net Ionic Equations5 A. Molecular equations – reactant and product formulas written as if they were molecules B. Ionic equations – any reactants or products that completely dissociate in water (strong electrolytes) shown as free ions

Test Item File Questions: Multiple Choice 1 – 4, Algorithmic 1 – 4, Short Answer 1 – 4 Test Item File Questions: Multiple Choice 5 – 8, Algorithmic 5 – 8, Short Answer 5 3 Test Item File Questions: Multiple Choice 12 – 20, Algorithmic 9 – 19, Short Answer 6 – 7 4 Test Item File Questions: Multiple Choice 23 – 25, Algorithmic 20 – 22, Short Answer 8 – 9 5 Test Item File Questions: Multiple Choice 27 – 31, Algorithmic 23 – 25, Short Answer 10 – 11 1 2

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Chapter 4—Reactions in Aqueous Solution C. Spectator ions – ions not undergoing change during reaction 1. Spectator ion identity not important 2. Role is to balance charge D. Net ionic equation 1. Equation for net change taking place during reaction 2. Spectator ions not included 4.6. Precipitation Reactions and Solubility Guidelines 6 A. Solubility – amount of compound dissolved in a given amount of solvent at a given temperature 1. Low solubility – precipitate forms 2. High solubility – no precipitate will form 3. Solubilities predicted using guidelines found in textbook B. Solubility guidelines 1. Use to predict if a precipitate will form in a reaction 2. Use to prepare and isolate a specific compound by carrying out a precipitation reaction 3. Compounds that are probably soluble usually contain a. Group 1A cations, most commonly Na+ or K+ b. Ammonium ion, NH4+ c. Halides (but not F–), unless the cation is Ag+, Hg22+, or Pb2+ d. Nitrate, NO3– e. Perchlorate, ClO4– f. Acetate, CH3CO2– g. Sulfate, SO42–, unless the cation is Sr2+, Ba2+, Hg22+, or Pb2+ 4. The most soluble compounds contain cations with a +1 charge and anions with a –1 charge. a. Solubility decreases as the charges on the ions increase b. Lattice energy effect (strong ionic bonds) 4.7. Acids, Bases, and Neutralization Reactions7 A. Arrhenius acid – dissociates in water to produce H+ 1. Hydronium ion− a. H+ attached to a water molecule b. Represented as H3O+ 2. Strong acids – strong electrolytes 3. Weak acids – weak electrolytes 4. Polyprotic acids a. Acids with more than one acidic hydrogen b. Dissociate in steps B. Arrhenius base – dissociates in water to produce OH– 1. Strong bases – strong electrolytes; most metal hydroxides 2. Weak bases – weak electrolytes NH3(g) + H2O(l) → NH4+(aq) + OH–(aq) C. Common acid and bases, Table 4.3 D. Naming acids 1. Binary acids – contain hydrogen and one other element or a polyatomic anion that is not an oxoanion a. Aqueous solutions b. Use the prefix hydro followed by the first part of the name of the other element, the suffix ic, and acid.

1) When writing the formula, indicate aqueous solution by (aq).

2) HCl(aq) is hydrochloric acid but HCl(g) is hydrogen chloride.

2. Oxoacids – contain oxygen in addition to hydrogen and another element 6 7

Test Item File Questions: Multiple Choice 32 – 43, Algorithmic 26 – 38, Short Answer 12 – 13 Test Item File Questions: Multiple Choice 52 – 68, Algortihmic 39 – 47, Short Answer 14 – 18 30

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Chapter 4—Reactions in Aqueous Solution a. Yields H+ and a polyatomic oxoanion when dissolved in H2O b. Names related to the names of the corresponding oxoanions 1) Replace suffix ite with ous acid 2) Replace suffix ate with ic acid c. Textbook Table 4.4 E. Neutralization reaction – reaction between acid and base producing salt and water 1. Salt produced from cation of base and anion of acid 2. Net reaction for a strong acid + strong base: H+(aq) + OH– (aq) → H2O(l) 3. Net reaction for a weak acid + strong base: HA(aq) + OH– (aq) → H2O(l) + A– (aq) 4.8. Solution Stoichiometry8 A. The molarity is important for performing stoichiometry calculations on substances in solution B. Use molarity to calculate either the number of moles (if volume is known) or volume (if the number of moles is known) 4.9. Measuring the Concentration of a Solution: Titration9 A. Procedure for determining the concentration of a solution B. Carefully measured volume of one solution reacts with a second solution of known concentration and volume 4.10. Oxidation–Reduction (Redox) Reactions10 A. Oxidation and reduction 1. Oxidation a. Loss of one or more electrons b. Increase in oxidation number 2. Reduction a. Gain of one or more electrons b. Decrease in oxidation number 3. Two processes occur simultaneously B. Redox reaction 1. Electrons transferred from one substance to another.

2. Total number of electrons lost by substance being oxidized = total number of electrons gained by substance being reduced C. Oxidation number 1. Means of determining whether an atom is neutral, electron-rich, or electron-poor 2. Does not necessarily imply ionic charges 3. Rules for assigning oxidation number 4.11. Identifying Redox Reactions11 A. Oxidation and reduction – occur together 1. One atom loses one or more electrons (is oxidized) 2. Another atom must gain those electrons (be reduced) B. Reducing agent – substance that causes reduction to occur 1. Loses one or more electrons a. Undergoes oxidation b. Oxidation number of atom increases 2. Metals act as reducing agents

Test Item File Questions: Multiple Choice 73, 74, 76 – 80, Algorithmic 48 – 52, Short Answer 19 Test Item File Questions: Algorithmic 53 10 Test Item File Questions: Multiple Choice 83 – 88, Algorithmic 54 – 60, Short Answer 20 – 21 11 Test Item File Questions: Multiple Choice 90 – 95, Algorithmic 61, Short Answer 22 – 23 8 9

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Chapter 4—Reactions in Aqueous Solution C. Oxidizing agent – the substance that causes oxidation to occur 1. Gains one or more electrons a. Undergoes reduction b. Oxidation number of atom decreases 2. Reactive nonmetals act as oxidizing agents 4.12. The Activity Series of the Elements12 A. Simple redox process 1. Reaction of an aqueous cation with a free element to produce a different ion and a different element 2. Process dependent on ease of oxidation of the element and ease of reduction of the cation B. Activity Series – ranks elements in order of reducing ability in aqueous solution (textbook Table 4.5) 1. Any element higher in the activity series will react with the ion of any element lower in the activity series 2. Position of hydrogen – indicates which metals will react with H+(aq) to produce H2(g) 3. Most reactive metals – top of the activity series 4. Least reactive metals – bottom of the activity series 4.13. Redox Titrations13 A. Redox titration – method for determining concentration of an oxidizing or reducing agent in solution 1. Unknown must react in 100% yield 2. Color change should signal the end of reaction B. Strategy for redox titrations – textbook Figure 4.9: titration of KMnO4 with H2C2O4 1. Measure a known amount (moles) of one substance.

2. Using the mole ratio of the balanced equation, determine number of moles of a second substrate.

3. Volume of solution containing the molar amount of a second substance measured during titration 4. Determine molar concentration of the second substance.

4.14. Some Applications of Redox Reactions14 A. Redox reactions 1. Involve almost every element in the periodic table 2. Occur in a vast number of processes B. Examples of redox reactions 1. Combustion 2. Bleaching 3. Batteries 4. Metallurgy 5. Corrosion 6. Respiration Inquiry – How do sports drinks replenish the chemicals lost in sweat?

Test Item File Questions: Multiple Choice 96 – 100, Algorithmic 62 – 69, Short Answer 24 – 27 Test Item File Questions: Multiple Choice 103 – 107, Algorithmic 70 – 73 14 Test Item File Questions: Multiple Choice 12 13

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Chapter 4—Reactions in Aqueous Solution

CHAPTER 4 REACTIONS IN AQUEOUS SOLUTION Teaching Tips, Points of Emphasis, and Common Misconceptions Section 4.1

Molarity is moles of solute per liter of solution, not per liter of solvent. The volume of solvent used is not determined when making a solution of known molarity.

Section 4.2

The relationship M1  V1 = M2  V2 is convenient for dilutions only. Students tend to use it for solution stoichiometry problems, which only works if the stoichiometry is 1:1.

Section 4.3

Relating properties to structure is an important skill. Later in this text it will be shown that species that are soluble in water are either ionic or polar, like water, as indicated in the rule of thumb “Like dissolves like.” Showing that water, sucrose, methyl alcohol, and ethyl alcohol all contain –OH groups will demonstrate their “likeness.”

Section 4.3

Dissociation is the fragmentation of a neutral species into ionic species.

Section 4.3

Strong electrolytes are usually soluble ionic compounds or strong acids. Weak electrolytes are typically weak acids or weak bases. Common nonelectrolytes are water, alcohols, and sugars.

Section 4.3

Note that HCl, HBr, and HI are strong electrolytes, but HF is a weak electrolyte. This is an example of how the properties of the first element in a group (fluorine) differ from those of the other members of the group.

Section 4.5

A molecular equation shows complete formulas of reactants and products; an ionic equation shows which of these dissociate into ions; a net ionic equation shows only those ions and molecules that react.

Section 4.5

While not included in the net ionic equation, spectator ions are present in solution and must be included in the computation of the molar mass of the compound dissolved.

Section 4.6

A functional definition of insoluble is that some of the compound remains in the solid state when about 10–3 mole of the compound is placed in contact with one liter of water. It does not mean that none of the compound dissolves.

Section 4.6

Hg22+ is a polyatomic ion with Hg in the +1 oxidation state.

Section 4.7

All compounds that contain H are not acids. NH3 is a base and CH4 is neither an acid nor a base. All compounds that contain OH are not bases. SO2(OH)2 (usually written as H2SO4) is an acid.

Section 4.7

H+(aq) and H3O+(aq) are both representations of a proton dissolved in water.

Aquated proton is actually a complex mixture of hydrates, including H 9O4+.

Section 4.7

Acidity is a function of both the solute and the solvent. HCl is a strong acid in water, a weak acid in glacial acetic acid, and remains undissociated in benzene. Arrhenius theory pertains to aqueous solutions.

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Chapter 4—Reactions in Aqueous Solution Section 4.7

Acids with one, two, and three ionizable protons are called monoprotic, diprotic, and triprotic, respectively.

Section 4.7

Ammonium hydroxide and aqueous ammonia are the same substance: NH3(g) + H2O(l) ⇌ NH4+(aq) + OH–( aq). There is no evidence for the undissociated species NH4OH.

Section 4.7

Due to economic considerations NaOH and KOH are the only two common strong bases with group 1A metal ions. The others (LiOH, RbOH, CsOH) are equally strong bases.

Section 4.10

Two commonly used redox reaction memorization aids are: 1) LEO the lion goes GER (Lose Electrons Oxidation, Gain Electrons Reduction); and OIL RIG (Oxidation Involves Loss of electrons, Reduction Involves Gain of electrons).

Section 4.10

Oxidation numbers of atoms in a compound are the charges the atoms would have if the compound were ionic.

Section 4.10

Oxygen has an oxidation number of –2 except when it is bonded to itself. The oxidation number of oxygen is 0 in O2, 1/2 in O2– (superoxide), and –1 in O22– (peroxide).

Section 4.10

A useful rule of thumb: For any main group element, maximum oxidation number = group number; minimum oxidation number = group number – 8.

Section 4.11

Species to the left and bottom of the periodic table tend to lose electrons and are good reducing agents; species at the top and right of the periodic table tend to gain electrons and are good oxidizing agents.

Section 4.12

The activity series can be roughly divided into three groups based on periodic table location: 1) active metals from Groups 1A, 2A, and Al; 2) lighter transition metals (Cr, Mn, Fe, Co, Ni, Zn), Sn, and Pb; 3) Cu and heavier transition metals (Ag, Pt, Au, Hg).

Section 4.12

A memorization aid: the elements below hydrogen in the activity series form an upside-down T on the periodic table: Cu Ag Pt Au Hg

Lecture/Laboratory Demonstration References Section 4.3

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Bassam Z. Shakhashiri, “Conductivity and Extent of Dissociation of Acids in Aqueous Solution,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp. 140– 145. Universal indicator and a conductivity probe are used to explore the relative acidity and conductivity of a series of aqueous acids.

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Chapter 4—Reactions in Aqueous Solution Section 4.6

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Solubility of Some Silver Compounds,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 83–85. The solubility of a series of silver salts and complexes is explored in this demonstration.

Section 4.6

L. R. Summerlin, C. L. Borgford, and J. B. Ealy, “Name That Precipitate,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 121–123. Blue, brown, red, white, and yellow precipitates are formed when solutions containing +2 cations of mercury, nickel, lead, and barium are mixed with solutions containing carbonate, chloride, iodide, and sulfate anions. Students are asked to name and determine the formula of each precipitate and to write an ionic equation for each reaction.

Section 4.10

C. E. Ophardt, “Redox Demonstrations and Descriptive Chemistry: Part I. Metals,” J. Chem. Educ., Vol. 64, 1987, 716. Redox reactions of iron(III) with thiosulfate, iron(II) with permanganate, and tin(II) with mercury(I) are used to show how an abbreviated table of standard reduction potentials is used to predict the products of these reactions from the relative positions of the oxidizing agents and reducing agents in the table.

Section 4.10

Lee R. Summerlin and James. L. Ealy, Jr., “Oxidation States of Manganese: Mn 7+, Mn6+, Mn4+, and Mn2+,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 133–134.

Section 4.12

Bassam Z. Shakhashiri, “Reactions of Metals and Hydrochloric Acid,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983), pp. 25–26. Reactions of iron, zinc, and magnesium with hydrochloric acid to produce aqueous solutions of the +2-metal ions and hydrogen gas are shown to be exothermic. Observed temperature changes are related to enthalpies of reaction.

Section 4.12

Bassam Z. Shakhashiri, “An Activity Series: Zinc, Copper, and Silver Half Cells,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 4 (The University of Wisconsin Press, Madison, 1992), pp. 101–106.

Section 4.12

Lee R. Summerlin and James L. Ealy, Jr., “Activity Series for Some Metals,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), p. 150. An overhead projector demonstration showing hydrogen gas generated by a redox reaction.

Section 4.12

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Producing Hydrogen Gas from Calcium Metal,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 51–52.

Literature References Section 4.1

Lloyd J. McElroy, “Teaching Dilutions,” J. Chem. Educ., Vol. 73, 1996, 765–766.

Section 4.3

John J. Fortman, “Pictorial Analogies X: Solutions of Electrolytes,” J. Chem. Educ., Vol. 71, 1994, 27–28.

Section 4.3

Albert Kowalak, “When Is a Strong Electrolyte Strong?” J. Chem. Educ., Vol. 65, 1988, 607.

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Chapter 4—Reactions in Aqueous Solution

Section 4.5

Betty J. Wruck, “Reinforcing Net Ionic Equation Writing,” J. Chem. Educ., Vol. 73, 1996, 149–150.

Section 4.6

Richard A. Kjonas, “An Analogy for Solubility: Marbles and Magnets,” J. Chem.

Educ., Vol. 61, 1984, 765.

Section 4.7

H. van Lubeck, “Significance, Concentration Calculations, Weak and Strong Acids,” J. Chem. Educ., Vol. 60, 1983, 189.

Section 4.9

Ara S. Kooser, Judith L. Jenkins, and Lawrence E. Welch, “Acid-Base Indicators: A New Look at an Old Topic,” J. Chem. Educ., Vol. 78, 2001, 1504–1506.

Section 4.10

Gion Calzaferri, “Oxidation Numbers,” J. Chem. Educ., Vol. 76, 1999, 362–363.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 4.1

Worked Example 4.2: Calculating the number of moles of solute in a solution

Section 4.9

Worked Example 4.1: Determining the concentration of a solution using a titration procedure

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CHAPTER 5 PERIODICITY AND THE ELECTRONIC STRUCTURE OF ATOMS Chapter Learning Objectives Section 5.1

Label the wavelength, frequency, and amplitude in an electromagnetic wave.

Convert between wavelength and frequency of electromagnetic radiation.

Section 5.2

Calculate the energy of electromagnetic radiation in units of J/photon or kJ/mol, when given the frequency or wavelength.

Describe the photoelectric effect and explain how it supports the theory of particlelike properties of light.

Describe quantization of energy.

Calculate the frequency or wavelength of radiation needed to produce the photoelectric effect given the work function of a metal.

Section 5.3

Describe the difference between a continuous and line spectrum.

Describe how the line spectrum relates to the quantization of electron energy levels.

Compare the wavelength and frequency of different electron transitions in the Bohr model of the atom.

Use the Balmer–Rydberg equation to calculate the wavelength and energies of electronic transitions in the hydrogen atom.

Section 5.4

Calculate the wavelength of a moving object using the de Broglie equation.

Describe how the de Broglie hypothesis shows how particles can behave like waves.

Explain why the wavelength of macroscopic objects is not observed.

Section 5.5

Calculate the uncertainty in the position of the moving object if the velocity is known.

Section 5.6

Identify and write valid sets of quantum numbers that describe electrons in different types of orbitals.

Apply the orbital notation to designating the quantum numbers of electron orbitals.

Section 5.7

Identify an orbital based on its shape and describe it using a set of quantum numbers.

Locate the nodal planes in different types of orbitals and different shells.

Section 5.8

Describe the spin quantum number.

Explain how the Pauli Exclusion Principle allows sets of four quantum numbers to describe each electron in an atom.

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Chapter 5—Periodicity and the Electronic Structure of Atoms Section 5.9

Section 5.10 and Section 5.11 Section 5.12

Explain how electron shielding gives the order of subshells from lowest to highest in energy.

Assign the electron configuration to atoms in the ground state.

Draw orbital filling diagrams for the ground state of an atom and determine the number of unpaired electrons.

Identify atoms from orbital filling diagrams or electron configurations.

Describe how the shape of the Periodic Table arises from the electron configurations of the elements.

Section 5.13

Explain the periodic trend in atomic radii.

Predict the relative size of atoms based upon their position in the periodic table.

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Chapter 5—Periodicity and the Electronic Structure of Atoms

Lecture Outline 5.1. Wave Properties of Radiant Energy and the Electromagnetic Spectrum1 A. Electromagnetic radiation – forms of radiant energy (light in all its varied forms) B. Electromagnetic spectrum – a continuous range of wavelengths and frequencies of all forms of electromagnetic radiation C. Electromagnetic energy – has wavelike properties 1. Frequency () – the number of peaks (maxima) that pass by a fixed point per unit time (s–1 or Hz) 2. Wavelength () – the length from one wave maximum to the next 3. Amplitude – the height measured from the middle point between peak and trough (maximum and minimum) 4. Intensity of electromagnetic energy is proportional to the square of the wave amplitude2 D. Speed of light (c) – rate of travel of all electromagnetic radiation in a vacuum 1. c = 3.00  108 m/s 2. Wavelength  Frequency = Speed (m)   (s–1) = c (m/s) 3. Frequency and wavelength inversely related: c c  = or ν = ν  a. long ; low ν b. short ; high ν E. Wave properties 1. Diffraction – the bending of light 2. Interference – two or more waves superimpose to form a new wave F. The double slit experiment demonstrated that light consists of waves.

5.2. Particle-like Properties of Radiant Energy: The Photoelectric Effect and Planck’s Postulate2 A. The photoelectric effect can be seen when light above a minimum frequency strikes some metal surfaces resulting in electron emission.

B. Einstein postulated that light behaves like a stream of particles called photons.

1. Based on Planck’s Postulate, the energy of a photon is quantized—occurs in a discrete quantity. hc a. E = h ; E = ; h = 6.626  10–34 Js (Planck’s constant)



b. High-energy radiation – higher , shorter  c. Low-energy radiation – lower , longer  C. The work function, , is the minimum amount of energy required to cause electron emission.

1. Low intensity, high frequency light results in few electrons emitted.

2. High intensity, high frequency light results in many electrons emitted.

D. Einstein concluded that the nature of light is more complex than previously believed.

1. Light can behave as waves.

2. Light can behave as small particles—photons with a quantized or discrete amount of energy.

1 2

Test Item File Questions: Multiple Choice 1 – 6, Algorithmic 1, Short Answer 1 – 6 Test Item File Questions: Multiple Choice 12 – 15, Algorithmic 2, Short Answer 7 – 9

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Chapter 5—Periodicity and the Electronic Structure of Atoms 5.3. The Interaction of Radiant Energy with Atoms: Line Spectra3 A. Individual energetically-excited atoms emit light.

1. Provides clue to atomic makeup 2. Consists of only a few  3. Line spectrum – series of discrete lines (or wavelengths) separated by blank areas 4. Each element has its own unique line spectrum B. The Bohr Model – The hydrogen atom has a small, positively charged nucleus with electrons in specific orbitals, like planets.

1. Each orbital has a specific energy.

2. Electron energies are quantized – have only specific energy values associated with the orbital.

C. Addition of energy to an atom causes an electron to transition from a lower to a higher energy orbital.

D. When an electron falls from a higher- to lower-energy orbit, energy is released.

1. The energy of the emitted photon equals the energy difference between the levels.

2. The frequency of the emitted photon is related to this energy difference.

3. Quantization of energy in the atom arises from the fact that an electron cannot reside between orbits in the Bohr model.

E. Balmer series – pattern in visible region atomic line spectra for the hydrogen atom 1. Other series in ultraviolet and infrared regions 2. Every line in the entire hydrogen spectrum fits the generalized Balmer-Rydberg equation

1



1   1 = R  2 − 2  where n  m m n  

3. R (Rydberg constant) = 1.097 10–2 nm–1 F. The Bohr model failed to predict line spectra of other atoms.

G. The model does not provide an accurate depiction of the electron’s location.

H. A key idea of the Bohr model is quantized energy levels for the electron 5.4. Wavelike Properties of Matter: deBroglie’s Hypothesis4 A. de Broglie – matter can behave in some respects like light 1. Both light and matter wavelike as well as particle-like 2. Relationship between  of an electron or of any other particle or object of mass m moving at velocity v

m=

h h ; = v mv

B. Dual wave/particle description of light and matter 1. Wave properties of macroscopic objects have not been observed.

2. On the atomic scale, matter behaves differently. Diffraction (wave characteristic) of electrons has been observed.

C. Results of the double-slit experiment are consistent with subatomic particles such as electrons having wave properties 5.5. The Quantum Mechanical Model of the Atom: Heisenberg Uncertainty Principle5 A. Schrödinger 1. Quantum mechanical model of the atom 2. Abandoned idea of an electron as a small particle moving around the nucleus in a defined path 3. Concentrated on the electron’s wavelike properties

Test Item File Questions: Multiple Choice 16 – 19, Algorithmic 3, Short Answer 10 – 11 Test Item File Questions: Multiple Choice 28 – 32, Algorithmic 5 – 6, Short Answer 12 5 Test Item File Questions: Multiple Choice 33 – 36, 3 4

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Chapter 5—Periodicity and the Electronic Structure of Atoms B. Heisenberg Uncertainty Principle – it is impossible to know precisely where an electron is and what path it follows h (Dx)(Dmv) ³ 4p 1. Cannot know both the position and the velocity of an electron with a high degree of certainty 2. If velocity is known with a high degree of certainty, a. mv is small b. x (position of the electron) is large 3. If the exact position of the electron is known, a. x is small b. mv is large 5.6. The Quantum Mechanical Model of the Atom: Orbitals and Quantum Numbers6 A. Quantum mechanical model of atomic structure 1. Mathematical form is a wave equation 2. Wave function () or orbital – solution to wave equation. a. 2 – gives the probability of finding an electron within a given region in space b. Defines a volume of space around the nucleus where there is a high probability of finding an electron c. Says nothing about the electron’s position B. Wave function – contains a set of three variables, quantum numbers 1. Describes the energy level of an orbital 2. Defines the shape and orientation of the region in space where the electron is most likely to be found C. Principal quantum number (n) – describes the size and energy level of the orbital 1. A positive integer (n = 1, 2, 3, 4, …)

2. As the value of n increases, a. number of allowed orbitals increases b. size of the orbitals increases c. energy of the electron in the orbital increases 3. Shell – grouping of orbitals according to the principal quantum number D. Angular-momentum quantum number (l) – defines the 3-D shape of the orbital 1. Integral value from 0 to n – 1 2. Within each shell, n different shapes for orbitals 3. Subshells – grouping of orbitals according to the angular-momentum quantum number a. Referred to by letter rather than by number i. quantum number l: 0 1 2 3 4 ... ii. subshell notation: s p d f g ...

E. Magnetic quantum number (ml) – defines the spatial orientation of the orbital with respect to a standard set of coordinate axes 1. Integral values from – l to +l 2. Within each subshell (same n and same l), 2 l + 1 different spatial orientations F. Energy level of various orbitals – textbook Figure 5.12 1. Hydrogen – energy levels depend only on n 2. Multielectron atoms – energy levels depend on both n and l

6

Test Item File Questions: Multiple Choice 37 – 48, Algorithmic 7 – 12, Short Answer 13 – 18

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Chapter 5—Periodicity and the Electronic Structure of Atoms 5.7. The Shapes of Orbitals7 A. s orbitals – textbook Figure 5.13 1. Electron density plot represents 2 a. Spherical – probability of finding an electron depends only on the distance from the nucleus. The value of electron density for a 1s orbital is greatest near the nucleus and drops off as the distance from the nucleus increases 2. The radial distribution plot represents the probability of finding the electron in a thin shell at any given distance from the nucleus.

3. Differences among s orbitals in different shells a. Size increases in successively higher shells b. Electron distribution in outer s orbitals – several different regions of maximum probability separated by a node 4. Node – a surface of zero probability a. Intrinsic property of a wave – zero amplitude at node (textbook Figure 5.16) b. Can be seen in the radial probability plots for 2s and 3s orbitals 5. Radial probability plots show that on average, an electron in a 1s orbital spends more time closer to the nucleus than an electron in a 2s orbital etc.

6. Larger orbitals are higher in energy than smaller orbitals because the electron has a greater probability of being further from the nucleus and it requires energy to separate a positive and negative charge B. p orbitals – textbook Figure 5.18 1. Dumbbell shaped 2. Electron distribution concentrated in identical lobes on either side of the nucleus 3. Nodal plane cuts through the nucleus 4. Probability of finding a p electron near the nucleus is zero 5. Have different phases (mathematical signs) a. Crucial for bonding b. Only lobes with same phase interact to form covalent bonds 6. Three p orbitals oriented along the x-, y-, and z-axes. (px, py, pz) C. d and f orbitals 1. d orbitals – textbook Figure 5.19 a. Four (dxy , dxz, dyz, dx2– y2) – cloverleaf shaped i. Four lobes of maximum electron probability separated by two nodal planes through the nucleus b. dz2 – similar in shape to a pz orbital with additional donut-shaped region of electron probability in the xy plane c. Alternating lobes have different phases 2. f orbitals – eight lobes of maximum electron probability a. Three nodal planes through the nucleus 5. 8. Electron Spin and the Pauli Exclusion Principle 8 A. Fourth quantum number – ms 1. Related to property called electron spin 2. Electron behaves as if spinning around an axis a. Gives rise to a tiny magnetic field 3. Has two values: +1/2 (  ) or –1/2 (  ) 4. Independent of other quantum numbers B. Pauli Exclusion Principle 1. No two electrons in an atom can have same four quantum numbers 2. Only two electrons with opposite spins per orbital 5.9. Orbital Energy Levels in Multi-electron Atoms9 Test Item File Questions: Multiple Choice 49 – 52, Algorithmic 13 – 15, Short Answer 19 – 21 Test Item File Questions: Multiple Choice 61, Algorithmic 16 – 19 9 Test Item File Questions: Multiple Choice 62 – 65, Algorithmic 20 7 8

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Chapter 5—Periodicity and the Electronic Structure of Atoms A. Energy level of an orbital in multielectron – depends on both n and l.

1. Energy difference due to electron-electron repulsions 2. Outer-shell electrons a. Pushed farther from nucleus b. Held less tightly 3. Partially cancels electron – nucleus attractions 4. Electrons shielded from nucleus by the other electrons.

B. Effective nuclear charge, Zeff – net nuclear charge actually felt by an electron 1. Zeff = Zactual – electron shielding 2. Lower than the actual nuclear charge 3. For the same shell a. Lower value of l corresponds to higher value of Zeff b. Corresponds to lower energy for the orbital 4. Useful for explaining various chemical phenomena 5.10. Electron Configurations of Multi-electron Atoms10 A. Electron configuration – describes the orbitals occupied by electrons in an atom B. Aufbau principle 1. Fill lowest-energy orbitals first 2. Only two electrons with opposite spin per orbital 3. Follow Hund’s rule: if two or more orbitals with same energy available, one electron with parallel spin in each until all are half full C. Ground-state configuration – lowest-energy electron configuration D. Degenerate orbitals – orbitals with the same energy level E. Orbital-filling diagrams – electrons are represented by arrows 5.11. Anomalous Electron Configurations11 A. Unusual stability when subshells half-filled or filled – leads to anomalies in electron configurations B. Anomalies 1. Occur when energy differences between subshells small 2. Transfer of an electron from one subshell to another a. Lowers total energy of the atom b. Due to decrease in electron-electron repulsions c. Most anomalies are found for elements with Z > 40 and small energy differences between subshells.

5.12. Electron Configurations and the Periodic Table 12 A. Valence-shell electrons – outermost shell of electrons 1. Elements in each group of the periodic table have similar valence-shell electron configurations.

2. Least tightly held 3. Determine an element’s chemical properties B. Similar electron configurations explain why the elements in a given group have similar chemical behavior.

C. Blocks of elements in the periodic table – depends upon the valence orbitals being filled 1. s-block elements – Groups 1A and 2A (filling of an s orbital) 2. p-block elements – Groups 3A through 8A (filling of p orbitals; ns orbitals are filled) 3. d-block elements – transition metals (filling of (n – 1)d orbitals) 4. f-block elements – lanthanide and actinide elements (filling of (n – 2) f orbitals) D. Layout of the periodic table provides a useful way to remember the order of orbital filling.

5.13. Electron Configurations and Periodic Properties: Atomic Radii13 Test Item File Questions: Multiple Choice 66 – 67, Algorithmic 22 Test Item File Questions: Multiple Choice 68, Algorithmic 23 – 24, Short Answer 22 12 Test Item File Questions: Multiple Choice 69 – 74, Algorithmic 25 – 36, Short Answer 23 – 24 13 Test Item File Questions: Multiple Choice 97, Algorithmic 37 – 42, Short Answer 25 10 11

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Chapter 5—Periodicity and the Electronic Structure of Atoms A. Radius of an atom – half the distance between nuclei of two identical atoms when covalently bonded together B. Atomic radius 1. Increases down a group 2. Successively larger valence-shell orbitals occupied C. Atomic radius 3. Decreases across a period due to increase in Zeff for valence-shell electrons across a period a. Value of Zeff – dependent upon the amount of shielding felt by an electron b. Amount of shielding – depends on both the shell and subshell of other electrons c. Valence-shell electrons i. Strongly shielded by electrons in inner shells ii. Less strongly shielded by electrons in same shell – order: s > p > d > f iii. Weakly shielded by electrons in same subshell Inquiry – How does knowledge of atomic emission spectra help us build more efficient light bulbs?

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Chapter 5—Periodicity and the Electronic Structure of Atoms

CHAPTER 5 PERIODICITY AND THE ELECTRONIC STRUCTURE OF ATOMS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 5.1

Radiation refers to any form of energy that travels in all directions from a single source, such as light from the sun.

Section 5.1

Many students associate radiation only with nuclear decay processes and their negative connotations. This form of radiation represents only a small segment of the electromagnetic radiation spectrum.

Section 5.1

Wavelength can be related to colors of the spectrum ranging from 400 nm (violet) to 700 nm (red) using the acronym ROY G BIV (Red, Orange, Yellow, Green, Blue, Indigo, Violet).

Section 5.1

Students best relate some wave properties to radio waves: the frequency of a favorite radio station and loudness (amplitude).

Section 5.2

Wave-particle duality does not mean that energy is sometimes a wave and other times a particle; it means we must use both wave and particle theories to describe energy.

Section 5.2

A photon is a quantum of light energy.

Section 5.3

The Bohr model does not fit with the current understanding of atomic structure.

Section 5.3

For a single atom, transition of an electron from a higher to a lower energy level results in the emission of one photon of light with an energy that corresponds to the difference in energy of the two levels.

Section 5.6

Quantum number n is related to the size and energy of an orbital. The larger the value of n, the larger the orbital and the higher the energy of an electron occupying the orbital.

Section 5.6

Quantum number l defines orbital shape. Some students believe that the shape of an orbital shows how the electron moves about the atom. An orbital defines the probability of finding an electron in a region of space.

Section 5.6

Quantum number ml defines the different orientations an orbital of a given shape may assume. This gives the number of orbitals of a given type: s orbital (one), p orbital (three), d orbital (five), f orbital (seven).

Section 5.6

For neutral atoms 4s < 3d but for cations 3d < 4s in energy.

Section 5.7

An electron’s orbital is a wave theory concept. Since a wave exists simultaneously on both sides of a node, an electron does not “move” from one lobe of an orbital to another according to this mathematical model.

Section 5.7

The closer an electron is brought to the nucleus of an atom, the greater the energy of the photon emitted and the lower the energy of the electron.

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Chapter 5—Periodicity and the Electronic Structure of Atoms Section 5.7

The Planck equation is for a single photon or single electron energy. On a macroscopic scale it is useful to convert to energy per mole of photons or electrons by multiplying by Avogadro’s number.

Section 5.8

If a beam of atoms having one unpaired electron (such as atoms of silver) pass through a magnetic field, half are attracted to one pole and half to the other pole of the magnetic field.

Section 5.8

Electrons are said to be paired on an atom or ion if there are the same number of electrons with ms = + 1/2 as with ms = – 1/2; any electron with ms = + 1/2 without a corresponding electron with ms = – 1/2 is said to be unpaired. Experimentally, atoms and ions that have only paired electrons are repelled from a magnetic field and are called diamagnetic; atoms and ions with one or more unpaired electrons are attracted to a magnetic field and are called paramagnetic.

Section 5.10

Electron orbital-filling diagrams explicitly show the number of unpaired electrons; electron configurations do not. Students who ignore the use of orbital-filling diagrams are prone to say that all electrons are paired on an atom with an electron configuration such as 1s22s22p2.

Section 5.13

Core electron configuration does have some effect on the chemistry of an element.

Due to filling of the d orbitals, p-block elements in period 4 have an additional 10 core electrons relative to elements in period 3. This influences some properties, such as atomic radius. A similar and more dramatic effect, called the lanthanide contraction, caused by f orbital filling, occurs for period 6 d-block and p-block elements.

Lecture/Laboratory Demonstration References Section 5.3

N. C. Thomas and R. Brown, “A Spectacular Demonstration of Flame Tests,” J. Chem. Educ., Vol. 69, 1992, 326. Brightly-colored flashes of light resulting from the ignition of cellulose nitrate impregnated with salts of sodium, potassium, strontium, and barium ions demonstrate light emission from these energetically excited ions.

Literature References Section 5.2

Robert R. Perkins, “Put Body to Them,” J. Chem. Educ., Vol. 72, 1995, 151–152.

Analogies for quantized states.

Section 5.5

Bianca L. Haendler, “Presenting the Bohr Atom,” J. Chem. Educ., Vol. 59, 1982, 372–376.

Section 5.5

Oliver G. Ludwig, “On a Relation Between the Heisenberg and de Broglie Principles,” J. Chem. Educ., Vol. 70, 1993, 28.

Section 5.5

Pedro L. Muino, “Introducing the Uncertainty Principle Using Diffraction of Light Waves,” J. Chem. Educ., Vol. 77, 2000, 1025–1027.

Section 5.5

Lawrence S. Bartell, “Perspectives on the Uncertainty Principle and Quantum Reality,” J. Chem. Educ., Vol. 62, 1985, 192–196.

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Chapter 5—Periodicity and the Electronic Structure of Atoms Section 5.6

Ngai Ling Ma, “Quantum Analogies on Campus,” J. Chem. Educ., Vol. 73, 1996, 1016–1017.

Section 5.6

Mali Yin and Raymond S. Ochs, “The Mole, The Periodic Table, and Quantum Numbers: An Introductory Trio,” J. Chem. Educ., Vol. 78, 2001, 1345–1347.

Section 5.6

Geoff Raynner-Canham, “A Student's Travels, Close Dancing, Bathtubs, and the Shopping Mall: More Analogies in Teaching Introductory Chemistry,” J. Chem.

Educ., Vol. 71, 1994, 943–944.

Section 5.6

John J. Fortman, “Pictorial Analogies VII: Quantum Numbers and Orbitals,” J. Chem. Educ., Vol. 70, 1993, 649–650.

Section 5.7

Elvin Hughes, Jr., and Arnold George, “Suitable Light Sources and Spectroscopes for Student Observation of Emission Spectra in Lecture Halls,” J. Chem. Educ., Vol. 61, 1984, 908–909.

Section 5.10

Robert D. Freeman, “New Schemes for Applying the Aufbau Principle,” J. Chem.

Educ., Vol. 67, 1990, 576.

Section 5.10

James R. Hanley, III, and James R. Hanley, Jr., “A Low-Cost Classroom Demonstration of the Aufbau Principle,” J. Chem. Educ., Vol. 56, 1979, 747.

Section 5.10

Ronald J. Gillespie, James N. Spencer, and Richard S. Moog, “Demystifying Introductory Chemistry; Part 1. Electron Configurations from Experiment,” J. Chem.

Educ., Vol. 73, 1996, 617–622.

Section 5.10

Anthony Garofalo, “Housing Electrons: Relating Quantum Numbers, Energy Levels, and Electron Configurations,” J. Chem. Educ., Vol. 74, 1997, 709–719.

Section 5.10

M. Bonneau, “The Quantum Shoe Store and Electron Structure,” J. Chem. Educ., Vol. 68, 1991, 837.

Section 5.11

Peter Cann, “Ionization Energies, Parallel Spins, and the Stability of Half-Filled Shells,” J. Chem. Educ., Vol. 77, 2000, 1056–1061.

Section 5.12

Judith A. Strong, “The Periodic Table and Electron Configurations,” J. Chem. Educ., Vol. 63, 1986, 834.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 5.3

Worked Example 5.4: Relating the Bohr model and the Balmer-Rydberg equation

Section 5.6

Worked Example 5.6: Assigning quantum numbers to an orbital

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CHAPTER 6 IONIC COMPOUNDS: PERIODIC TRENDS AND BONDING THEORY Chapter Learning Objectives Section 6.1

Write the ground-state electron configuration for main-group and transition-metal ions.

Determine the number of unpaired electrons in main-group and transition-metal ions.

Section 6.2

Predict the relative size of anions, cations, and neutral atoms.

Predict the relative size of isoelectronic ions.

Section 6.3

Order elements from lowest to highest ionization energy.

Explain the periodic trend in ionization energy.

Section 6.4

Compare successive ionization energies (higher second, third, fourth, etc.) for different elements.

Identify elements based on values of successive ionization energies.

Section 6.5

Compare the value of the electron affinity for different elements.

Explain the periodic trend in electron affinity.

Section 6.6

Use the octet rule to predict charges on main-group ions, electron configurations of main-group ions and formulas for ionic compounds.

Section 6.7

Draw a Born–Haber cycle and calculate the energy change that occurs when an ionic compound is formed from its elements.

Section 6.8

Predict the relative magnitude of lattice energy given the formula or molecular representation of an ionic compound.

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49

Chapter 6—Ionic Compounds: Periodic Trends and Bonding Theory

Lecture Outline 6.1. Electron Configurations of Ions1 A. Main-group elements 1. Main-group elements lose or gain electrons, achieving a noble-gas electron configuration.

2. Main-group ions are isoelectronic to noble gases.

3. Electrons lost by a metal come from the highest-energy occupied orbital 4. Electrons gained by a nonmetal go into lowest-energy unoccupied orbital B. Transition metals – lose valence-shell s electrons first, then d electrons 6.2. Ionic Radii2 A. Cations have smaller radii than neutral atoms 1. Electrons removed from larger valence-shell orbitals 2. Increase in Zeff when electrons removed B. Anions have larger radii than neutral atoms 1. Decrease in Zeff when electrons added 2. Increase in electron–electron repulsions 6.3. Ionization Energy3 A. Ionization energy (Ei) – amount of energy required to remove the outermost electron from an isolated neutral atom in the gaseous state B. Periodic trends in ionization energy 1. Minimum Ei – group 1A alkali metals 2. Maximum Ei – group 8A noble gases 3. Ei increases across a period 4. Ei decreases down a group C. Periodicity due to electron configurations 1. Single s electron in valence shell of alkali metals feels low Zeff a. Single valence electron well shielded by the core (inner shell) electrons b. Electron loosely held c. Low energy needed to remove electron 2. Electrons in filled valence subshell of noble gas elements feel high Zeff a. Electrons tightly held b. Radius of atom shrinks c. High energy needed to remove electron 3. Increase in atomic number down a group a. Average distance of the electron from the nucleus increases as the value of n increases b. Valence-shell electrons less tightly held than core electrons c. Ei decreases D. Minor irregularities occur from left to right across a row of the periodic table 1. Group 2A elements – Ei (Be) > Ei (B) (due to electron configuration) a. 2p electron of boron shielded somewhat by 2s electrons b. Feels a smaller Zeff c. More easily removed 2. Group 6 A elements – Ei (N) > Ei (O) (due to electron configuration) a. 2p electron removed from nitrogen removed from a half-filled orbital (stable configuration) b. 2p electron removed from oxygen removed from a filled orbital i. Electrons in filled orbitals forced together and have slightly higher energy ii. Easier to remove electrons from filled orbitals

Test Item File Questions: Multiple Choice 1 – 6, Algorithmic 4 – 9, Short Answer 1 – 12 Test Item File Questions: Multiple Choice 11 – 17, Algorithmic 10, Short Answer 13 3 Test Item File Questions: Multiple Choice 32 – 40, Algorithmic 11 – 15, Short Answer 14 1 2

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Chapter 6—Ionic Compounds: Periodic Trends and Bonding Theory 6.4. Higher Ionization Energies4 A. Energy involved with the removal of two, three, or even more electrons sequentially from an atom B. Successively larger amounts of energy required for each ionization step – more difficult to remove a negatively charged electron from a positively charged ion C. Large jumps in successive ionization energies of the elements – textbook Table 6.2 1. Valence-shell electrons easily lost during ionization 2. Core electrons relatively difficult to remove D. Valence-shell electron configuration of an atom controls its chemistry.

6.5. Electron Affinity5 A. Electron affinity (Eea) – energy change when electron added to an isolated gaseous atom B. Eea usually has a negative value 1. Energy usually released when an atom adds an electron 2. More negative the value, a. greater the tendency of the atom to accept an electron b. more stable the anion C. Eea > 0 for atoms that form an unstable anion on addition of an electron D. No experimental measurements can be made.

E. Periodic trends of Eea related to electron configurations 1. Sign and magnitude of Eea due to offsetting factors a. Negative Eea due to attraction between the additional electron and the atomic nucleus b. Positive Eea due to repulsions between the additional electron and the electrons in the atom 2. Halogens have large, negative Eea a. High Zeff b. Room in valence shell for an additional electron c. High attraction between additional electron and atomic nucleus 3. Noble-gas elements have positive Eea a. Filled s and p sublevels b. Additional electron goes into the next higher shell – feels a low Zeff c. Small attraction between the additional electron and the atomic nucleus d. High electron–electron repulsions 4. Alkaline earth metals have Eea  0 a. Filled s subshell b. Added electron goes into p subshell of higher energy c. Relatively low Zeff 6.6. The Octet Rule6 A. Octet rule – main-group elements tend to undergo reactions that leave them with eight valence electrons 1. Useful for making predictions; provides insights about chemical bonds 2. “Magic number” = 8 a. Filled octet – s and p subshells both filled b. Can’t take electrons away – electrons tightly held by high Zeff c. No low-energy orbitals available to accept electrons 3. Failure of rule a. Elements on right side of periodic table in third or lower period b. More than eight electrons sometimes accommodated by larger atoms

Test Item File Questions: Multiple Choice 46 – 53, Algorithmic 16 – 21, Short Answer 15 – 16 Test Item File Questions: Multiple Choice 54 – 60, Algorithmic 22 – 23, Short Answer 17 – 18 6 Test Item File Questions: Multiple Choice 61, Algorithmic 24 – 31, Short Answer 19 4 5

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Chapter 6—Ionic Compounds: Periodic Trends and Bonding Theory B. Factors that determine cation or anion formation 1. Cations a. Electrons highly shielded from nucleus by core electrons i. Feel low Zeff ii. Easily lost b. Charge based on electronic configuration – energy required unavailable to lose another electron once next-lower noble-gas configuration achieved 2. Anions a. Electrons poorly shielded from nucleus by core electrons i. Feel high Zeff ii. Can gain electrons because of Zeff b. Charge based on electronic configuration – low-energy orbitals required for additional electrons unavailable once next-higher noble gas configuration achieved 6.7. Ionic Bonds and the Formation of Ionic Solids 7 A. Element with small Ei can transfer an electron to an element with negative Eea 1. Produces a cation and an anion 2. Cation and anion mutually attracted by electrostatic forces 3. Ionic bond created 4. Three-dimensional network of ions (ionic solid) formed a. Each cation surrounded by and attracted to many anions b. Each anion surrounded by and attracted to many cations B. Ionic bond formation leads to large gain in stability 1. Overcomes unfavorable energy change of electron transfer C. Born–Haber cycle – series of hypothetical steps each of which contributes to the overall energy change during the formation of an ionic compound 1. Net process is the sum of individual steps 2. Typical individual steps a. Sublimation of a metal atom to a gaseous atom b. Dissociation of nonmetal molecules into atoms c. Ionization of isolated metal atoms d. Formation of nonmetal anions e. Formation of solid ionic compound 6.8. Lattice Energies in Ionic Solids8 A. Lattice energy (U) – measure of electrostatic interaction energies between ions in a solid 1. A measure of the strength of a compound’s ionic bonds 2. Refers to the breakup of a crystal into individual ions a. Positive value b. When using the lattice energy in a Born–Haber cycle must change the sign because last step is the reverse of the breakup of the crystal 3. Lattice energy mostly coulombic

zz Coulomb’s law: F = k  1 22 d 4. Effect of ion size – compounds of ions with smaller sizes have greater lattice energies than compounds of ions with larger sizes 5. Effect of ion charge – compounds of ions with higher charges have greater lattice energies than compounds of ions with lower charges Inquiry – How do ionic liquids lead to more environmentally friendly processes?

7 8

Test Item File Questions: Multiple Choice 62 – 71, Algorithmic 32 Test Item File Questions: Multiple Choice 81 – 84, Algorithmic 33 – 39, Short Answer 20 52

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Chapter 6—Ionic Compounds: Periodic Trends and Bonding Theory

CHAPTER 6 IONIC COMPOUNDS: PERIODIC TRENDS AND BONDING THEORY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 6.1

Electrons are always lost from orbitals having the highest value of the principal quantum number, n, before they are lost from orbitals with lower values of n.

Section 6.2

The larger the proton-to-electron ratio, the greater the attraction of the nucleus for the valence electrons and the smaller the size of the ion.

Section 6.3

Ionization energy is defined for isolated, gaseous-state atoms. Other factors, such as lattice energy or solvation energy, must be considered when forming ions in a condensed state.

Section 6.3

Core electrons shield valence electrons effectively; electrons in the same shell shield one another ineffectively.

Section 6.3

The closer an electron is to the nucleus, the higher its ionization energy. Hence, small atoms have high ionization energies. Due to its small size, the first element in every group tends to differ significantly from the other elements in the same group.

Section 6.4

A large change in two successive ionization energies signifies that a core electron is being ionized. In ordinary chemical reactions there is not enough energy available to ionize core electrons. Hence, only valence electrons are involved in chemical reactions.

Section 6.5

Electron affinity is defined for isolated, gaseous-state atoms. Other factors, such as lattice energy or solvation energy, must be considered when forming ions in a condensed state.

Section 6.5

Just as there are successive ionization energies: Ei1, Ei2, Ei3,..., there are successive electron affinities: Eea1, Eea2, Eea3,.... While energy is generally released on addition of one electron, energy is always required to add two or more electrons to a neutral gaseous atom.

Section 6.7

An ionic compound will form when an element from the far left-hand side combines with an element from the far right-hand side of the periodic table (a notable exception being BeCl2). Compounds encountered in a general chemistry course are usually ionic if formed from reaction of a metal with a nonmetal.

Section 6.7

Formulas for ionic compounds are empirical formulas giving the relative number of cations and anions. Although the formula for sodium chloride is NaCl, molecules of NaCl do not exist under ordinary laboratory conditions.

Section 6.7

Because the sum of ionization energy and electron affinity is always positive, obtaining noble gas electron configurations upon formation of ions is not the driving force behind the formation of an ionic compound. The ions must be stabilized by strong forces arising from electrostatic attraction.

Section 6.7

The Born–Haber cycle is an example of Hess’s law, which will be introduced in Chapter 9. It is used to calculate properties that are impossible to measure directly.

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Chapter 6—Ionic Compounds: Periodic Trends and Bonding Theory Section 6.8

When an actual lattice energy is more negative than that calculated on the basis of an ionic model, some covalent bonding between ions is occurring. This typically happens when the electronegativity of the metal is high (approximately > 1.5).

Lecture/Laboratory Demonstration References Section 6.5

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Halogens Compete for Electrons,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 60–61. The relative tendency of halogens to gain or lose electrons is demonstrated.

Section 6.7

Kristin A. Johnson, Rodney Schreiner, and Jon Loring, “A Dramatic Flame Test Demonstration,” J. Chem. Educ., Vol. 78, 2001, 640–641.

Section 6.7

M.D. Alexander, “Reactions of the Alkali Metals with Water: A Novel Demonstration,” J. Chem. Educ., Vol. 69, 1992, 418. The reaction of sodium metal with water to produce an aqueous solution of sodium hydroxide and hydrogen gas is performed at the interface between paint thinner and the more dense water.

Periodically, bubbles of hydrogen gas carry the sodium metal into the organic layer, temporarily stopping the reaction. The presence of the aqueous layer is shown by a phenolphthalein indicator color change.

Section 6.8

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Producing Hydrogen Gas from Calcium Metal,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 51–52.

Literature References Section 6.

None

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 6.4

Worked Example 6.4: Higher ionization energies

Section 6.7

Worked Example 6.7: Calculating the energy change in the formation of an ionic compound

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CHAPTER 7 COVALENT BONDING AND ELECTRON-DOT STRUCTURES Chapter Learning Objectives Section 7.1

Describe the difference between an ionic and a covalent bond.

Describe the changes in energy that occur as two nuclei approach to form a covalent bond (Figure 7.2).

Section 7.2

Predict trends in bond length and bond dissociation energy based on bond order and atomic size.

Section 7.3

Rank elements by increasing value of electronegativity.

Classify bonds as nonpolar covalent, polar covalent, or ionic.

Interpret electrostatic potential maps to determine regions of high and low electron density.

Calculate the percent ionic character of a bond.

Section 7.4

Explain the difference in physical properties between ionic and covalent compounds.

Section 7.5

Draw an electron-dot structure that satisfies the octet rule.

Section 7.6

Use the five-step procedure to draw electron-dot structures for all types of molecules, including those with expanded octets and those containing multiple bonds.

Section 7.7

Draw electron-dot structures for radicals.

Section 7.8

Draw electron-dot structures for molecules with more than one central atom.

Section 7.9

Draw resonance structures and use curved arrows to depict how one structure can be converted to another.

Describe the bonding in a molecule, using the resonance hybrid model.

Section 7.10

Calculate the formal charge on each atom in an electron-dot structure.

Use the formal charge to evaluate the contribution of different resonance structures to the resonance hybrid.

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55

Chapter 7—Covalent Bonding and Electron-Dot Structures

Lecture Outline 7.1. Covalent Bonding in Molecules1 A. Formation of a covalent bond 1. Two atoms come close together, and electrostatic interactions develop a. Nuclei repel each other; electrons repel each other b. Each nucleus attracts the same electrons; electrons attract both nuclei 2. If attractive forces > repulsive forces; then covalent bond formed B. Distance between the two atoms affects the magnitude of attractive and repulsive forces.

1. Bond length – optimum point where net attractive forces maximized a. Each covalent bond has a characteristic length that leads to maximum stability. b. Bonds between the same pairs of atoms in different compounds have similar bond lengths. c. Can estimate from atomic radii 7.2. Strengths of Covalent Bonds 2 A. Formation of covalent bonds leads to lower energy.

B. Bond dissociation energy, D 1. Amount of energy necessary to break a chemical bond in an isolated gaseous-state molecule 2. Only positive values possible 3. When a bond forms, H = – D a. Bond breaking requires energy (D) b. Bond formation releases energy (– D) C. Bonds between the same pairs of atoms usually have similar bond dissociation energies.

D. Shorter bonds are typically stronger bonds.

1. Bond energies decrease on going down the column for different members of the same family bonded to a given atom – as atomic size increases, bond length increases and bond strength decreases 2. Multiple bonds are shorter and stronger than single bonds. a. Typically D (single bond) < D (double bond) < D (triple bond) 7.3. Polar Covalent Bonds: Electronegativity3 A. Ionic and purely covalent bonds – extremes of a continuous spectrum of possibilities B. Polar covalent bonds – bonding electrons attracted somewhat more strongly by one atom in a bond 1. Electrons not completely transferred 2. More electronegative atom: – ( represents partial charge formed.)

3. Less electronegative atom: + 4. The more polar the bond, the greater the bond dissociation energy.

C. Electronegativity (EN) – ability of an atom in a molecule to attract shared electrons in a bond 1. Metallic elements – low electronegativities 2. Halogens and other elements in upper right-hand corner of periodic table – high electronegativities D. Increased bond polarity leads to increased bond strength.

E. Predicting bond polarity 1. Atoms with same or similar electronegativities form nonpolar bonds 2. Atoms whose electronegativities differ by more than two (EN > 2) – form ionic bonds 3. Atoms whose electronegativities differ by less than two (EN < 2) – form polar covalent bonds F. Percent ionic character is a way to represent the extent of charge separation in a bond.

7.4. A Comparison of Ionic and Covalent Compounds 4 Test Item File Questions: Multiple Choice 1 – 3 Test Item File Questions: Multiple Choice 4 –8, 10-11, Algorithmic 1, Short Answer 1 3 Test Item File Questions: Multiple Choice 12 – 24, Algorithmic 3 – 9, Short Answer 2 – 6, 1 2

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Chapter 7—Covalent Bonding and Electron-Dot Structures A. Ionic compounds 1. High-melting solids 2. Must overcome every ionic attraction in the entire crystal B. Covalent compounds 1. Low-melting solids, liquids, or even gases 2. Bonds within individual molecule may be very strong 3. Attractive forces between different molecules (intermolecular forces) are relatively weak.

7.5. Electron-Dot Structures: The Octet Rule5 A. Electron-dot structure – atom’s valence-electron distribution in a molecule B. Sharing electrons in a bond allows atoms to achieve a complete valence shell or noble gas electron configuration.

C. Electron-dot structures for elements 1. Electrons represented by dots around element symbol a. One to each side until all four sides occupied b. Then pair dots until all valence-shell electrons used c. Note: pairing of dots does not correspond to pairing of electrons in electron configuration D. Simple molecules of second-period main-group elements – textbook Table 7.3 1. Octet rule used to determine number of covalent bonds formed by an element 2. Unpaired electrons form bonds 3. Electrons shared until each atom surrounded by four pairs of electrons (an octet) 4. Lone pairs (nonbonded pairs) – electron pairs not used in bonding 5. Bonding pairs – electrons shared in bond formation E. Multiple covalent bonds – sometimes more than one pair of electrons shared 1. Shorter and stronger than single bonds 2. Double bond – two pairs of electrons shared 3. Triple bond – three pairs of electrons shared 4. Bond order – number of electron pairs shared/2 a. Bond order = 1 i. Single bond ii. One electron pair shared (two electrons) b. Bond order = 2 i. Double bond ii. Two electron pairs shared (four electrons) c. Bond order = 3 i. Triple bond ii. Three electron pairs shared (six electrons) F. Coordinate covalent bonds 1. Two electrons (a lone pair) donated by one atom to another atom in bond formation 2. Second atom has vacant valence orbital before bonding 3. Often formed by N, O, P, and S 7.6. Procedure for Drawing Electron-Dot Structures6 A. Use lines to represent shared electrons in a covalent bond.

1. One line represents the two shared electrons of a single bond.

2. Two lines represent the four shared electrons of a double bond.

3. Three lines represent the six shared electrons of a triple bond.

Test Item File Questions: Multiple Choice 25 – 27, Algorithmic 10 – 11, Short Answer 7 – 8 Test Item File Questions: Multiple Choice 28 – 30, Short Answer 9 6 Test Item File Questions: Multiple Choice 31 – 40, Algorithmic 12 – 20, Short Answer 10 – 11 4 5

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Chapter 7—Covalent Bonding and Electron-Dot Structures B. Use dots to represent non-bonding electrons C. Guidelines.

1. Find the number of valence electrons in the molecule or ion. a. Find the sum of the valence electrons for all atoms. b. Add an additional electron for each negative charge. c. Subtract an electron for each positive charge.

2. Connect the atoms with single lines using the following guidelines. a. The central atom is usually the one with the lowest electronegativity (except hydrogen, which is never a central atom). b. Hydrogen and the halogens usually form only a single bond. c. Elements in the second row form the number of bonds given in Table 7.4 d. Elements in Group 3A often have less than eight electrons (are electron deficient). e. Elements in the third row and lower can have expanded octets.

3. Subtract the number of electrons used to make bonds from the total above. Assign remaining electrons to terminal atoms, giving each an octet (except for H).

4. Place any remaining electrons on the central atom.

5. If the central atom does not have an octet, form multiple bonds to terminal atoms.

7.7. Drawing Electron-Dot Structures for Radicals A. Radicals are compounds that have an odd number of electrons; one electron is in a half-filled orbital.

1. Very reactive.

2. Found in a large number of important processes, both commercial and physiological.

7.8. Electron-Dot Structures of Compounds Containing Only Hydrogen and Second-Row Elements7 A. Compounds of second-row elements: C, H, N, O 1. Octet rule almost always applies 2. Easy to predict number of bonds formed by each element – similar to above 3. For larger molecules containing several second-row elements, multiple electron-dot structures are often possible. Additional information about atom connectivity will be needed in order to draw these structures 7.9. Electron-Dot Structures and Resonance8 A. Sometimes more than one electron-dot structure needed for molecular description: resonance 1. These electron-dot structures do not accurately represent electronic structure 2. Limitation of Lewis theory B. Resonance hybrid – average of various possible electron-dot structures for a molecule 1. Resonance indicated by straight double-headed arrows () joining resonance structures 2. Resonance structures differ only in valence electron placement a. The number of valence electrons remains constant b. Atom connectivities do not change c. Bonding and non-bonding electrons can shift 3. Double bonds are delocalized 7.10. Formal Charges9 A. Result of electron bookkeeping 1. Formal charge = (number of valence electrons in free atom) – 1/2(number of bonding electrons) – (number of nonbonding electrons) 2. Sum of formal charges on all atoms = zero for a molecule or the overall charge for a polyatomic ion

Test Item File Questions: Multiple Choice 41 – 42, Algorithmic 21 – 22, Test Item File Questions: Multiple Choice 43 – 44, Algorithmic 23 – 25, Short Answer 12 9 Test Item File Questions: Multiple Choice 48 – 56, Algorithmic 26 – 27, Short Answer 13 – 14 7 8

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Chapter 7—Covalent Bonding and Electron-Dot Structures B. Use to evaluate relative importance of different resonance structures 1. Smaller formal charges on atoms are preferred 2. Negative formal charges reside on more electronegative atoms 3. Like formal charges should not be on adjacent atoms Inquiry – How does bond polarity affect the toxicity of organophosphate insecticides?

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Chapter 7—Covalent Bonding and Electron-Dot Structures

CHAPTER 7 COVALENT BONDING AND ELECTRON-DOT STRUCTURES Teaching Tips, Points of Emphasis, and Common Misconceptions Section 7.2

The units of bond dissociation energy (bond energy) are kJ/mol. As with ionization energy and electron affinity, bond energies are given for isolated, gaseous-state molecules.

There is a general correlation between bond length and bond strength. The shorter the bond, the stronger the bond.

Section 7.5

The Lewis structure is a helpful tool for understanding the configuration and chemistry of molecules of main-group elements. In general, Lewis structures cannot be drawn for transition-metal compounds because of the complex way that partially filled d orbitals influence the configuration and chemistry of these compounds.

Section 7.5

Once formed, there is no difference between an ordinary covalent bond and a coordinate covalent bond.

Section 7.5

Students sometimes inquire about the origin of a positive or negative charge on a molecule. Showing how a charge on an atom occurs (H → H+ + e–) and how that simple ion can combine with a neutral molecule to form a polyatomic ion may be helpful (H+ + H2O → H3O+).

Section 7.6

This general method for writing electron-dot structures works equally well for the simple molecules discussed in the previous section.

Section 7.6

Warn students that it is crucial to follow the steps for writing electron-dot structures exactly in the order given. Otherwise, an incorrect electron-dot structure may result.

Section 7.6

Except for hydrogen, terminal atoms always have complete octets.

Section 7.6

Multiple bonds are never introduced into the electron-dot structure before step 5.

Section 7.10

The best resonance structure has (a) formal charges as close to zero as possible and (b) a negative formal charge on the most electronegative element and a positive formal charge on the least electronegative element.

Lecture/Laboratory Demonstration References Section 7.

None

Literature References Section 7.1

Georgios R. Tsaparlis, “The Chemical Bond as an Atomic Tug-of-War,” J. Chem.

Educ., Vol. 61, 1984, 677.

Section 7.1

Roger L. DeKock, “The Chemical Bond,” J. Chem. Educ., Vol. 64, 1987, 934–941.

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Chapter 7—Covalent Bonding and Electron-Dot Structures Section 7.2

Raymond F. Peterson and David F. Treagust, “Grade-12 Students’ Misconceptions of Covalent Bonding and Structure,” J. Chem. Educ., Vol. 66, 1989, 459–460.

Section 7.3

Gordon Sproul, “Electronegativity and Bond Type: Predicting Bond Type,” J. Chem.

Educ., Vol. 78, 2001, 387–390.

Section 7.3

James N. Spencer, Richard S. Moog, and Ronald J. Gillespie, “Demystifying Introductory Chemistry: Part 3. Ionization Energies, Electronegativity, Polar Bonds and Partial Charges,” J. Chem. Educ., Vol. 73, 1996, 627–631.

Section 7.3

William B. Jensen, “Electronegativity from Avogadro to Pauling, Part I: Origins of the Electronegativity Concept,” J. Chem. Educ., Vol. 73, 1996, 11–20.

Section 7.5

Natalie Foote Tiernan, “Gilbert Newton Lewis and the Amazing Electron Dots,” J. Chem. Educ., Vol. 62, 1985, 569–570.

Section 7.5

Anthony N. Stranges, “Reflections on the Electron Theory of the Chemical Bond: 1900-1925,” J. Chem. Educ., Vol. 61, 1984, 185–190.

Section 7.5

William B. Jensen, “Abegg, Lewis, Langmuir, and the Octet Rule,” J. Chem. Educ., Vol. 61, 1984, 191–200.

Section 7.5

Linus Pauling, “G. N. Lewis and the Chemical Bond,” J. Chem. Educ., Vol. 61, 1984, 201–203.

Section 7.5

S. R. Logan, “The Role of Lewis Structures in Teaching Covalent Bonding,” J. Chem. Educ., Vol 78, 2001, 1457–1458.

Section 7.5

Wan-Yaacob Ahmad and Mat B. Zakaria, “Drawing Lewis Structures from Lewis Symbols: A Direct Electron Pairing Approach,” J. Chem. Educ., Vol. 76, 1999, 329–331.

Section 7.6

Gordon H. Purser, “Lewis Structures Are Models for Predicting Molecular Structure, Not Electronic Structure,” J. Chem. Educ., Vol. 76, 1999, 1013–1017.

Section 7.6

Juan Quilex Pardo, “Teaching a Model for Writing Lewis Structures,” J. Chem.

Educ., Vol. 66, 1989, 456–458.

Section 7.6

James Allen Carroll, “Drawing Lewis Structures Without Anticipating Octets,” J. Chem. Educ., Vol. 63, 1986, 28–31.

Section 7.6

Melvin E. Zandler and Erach R. Talty, “The ‘6N + 2 Rule’ for Writing Lewis Octet Structures,” J. Chem. Educ., Vol. 61, 1984, 124–127.

Section 7.6

Thomas J. Clark, “Another Procedure for Writing Lewis Structures,” J. Chem. Educ., Vol. 61, 1984, 100.

Section 7.9

Todd P. Silverstein, “The ‘Big Dog-Puppy Dog’ Analogy for Resonance,” J. Chem.

Educ., Vol. 76, 1999, 206–207.

Section 7.9

Ronald Starkey, “Resonance Analogy Using Cartoon Characters,” J. Chem. Educ., Vol. 72, 1995, 542.

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Chapter 7—Covalent Bonding and Electron-Dot Structures Section 7.9

Kenton B. Abel and William M. Hemmerlin, “Explaining Resonance–A Colorful Approach,” J. Chem. Educ., Vol. 68, 1991, 834.

Section 7.9

Francis Delvigne, “A Visual Aid for Teaching the Resonance Concept,” J. Chem.

Educ., Vol. 66, 1989, 461–462.

Section 7.10

David G. Dewit, “Using Formal Charges in Teaching Descriptive Inorganic Chemistry,” J. Chem. Educ., Vol. 71, 1994, 750–755.

Section 7.10

Packer and Sheila D. Woodgate, “Lewis Structures, Formal Charge, and Oxidation Numbers: A More User-Friendly Approach,” J. Chem. Educ., Vol. 68, 1991, 456–458.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 7.6

Worked Example 7.6: Drawing electron-dot structures for molecules with one central atom and multiple bonds

Section 7.9

Worked Example 7.10: Drawing resonance structures in organic compounds

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CHAPTER 8 COVALENT COMPOUNDS: BONDING THEORIES AND MOLECULAR STRUCTURE Chapter Learning Objectives Section 8.1

Use the VSEPR model to predict shape and bond angles in a molecule or ion with one central atom.

Use the VSEPR model to predict bond angles in a molecule with more than one central atom.

Section 8.2 Section 8.3 and Section 8.4

Describe the difference between a sigma and a pi bond in valence bond theory.

Write an electron-dot structure for a molecule and determine hybridization and bond angles.

Identify which orbitals overlap to form  and  bonds in molecules.

Section 8.5

Predict whether a given molecule has a dipole moment and draw its direction.

Interpret electrostatic potential maps of molecules.

Section 8.6

Identify the types of intermolecular forces experienced by a molecule.

Relate the strength of intermolecular forces to physical properties such as melting point and boiling point.

Sketch the hydrogen bonding that occurs between two molecules.

Section 8.7

Interpret the molecular orbital diagram for first row diatomic molecule or ion.

Section 8.8

Interpret the molecular orbital diagram for a second row diatomic molecule or ion.

Calculate the bond order and predict magnetic properties.

Section 8.9

Draw orbital overlap diagrams for molecules and describe the use of both valence bond theory and molecular orbital theory.

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Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure

Lecture Outline 8.1. Molecular Shapes: The VSEPR Model1 A. Molecular shape – determined by numbers of valence electrons around atoms B. Valence-shell electron-pair repulsion model – predicts approximate molecular shape 1. Bonded and lone electrons occupy charge clouds 2. Count the number of charge clouds surrounding atom of interest 3. Geometry based on number of charge clouds a. Charge clouds – stay as far away as possible from each other b. Textbook Table 8.1 8.2. Valence Bond Theory2 A. Easily visualized orbital picture of how electron pairs are shared in a covalent bond B. Orbital overlap – bond formation 1. Greater the overlap, stronger the bond 2. Singly-occupied valence orbital on one atom overlaps a singly occupied valence orbital on another atom C. For orbitals other than s orbitals, orbitals used in bond formation have directionality D. Each bonded atom maintains its own orbital but shares electrons 8.3. Hybridization and sp3 Hybrid Orbitals3 A. Hybrid orbitals – combination of wave functions for atomic orbitals that forms a new set of equivalent wave functions B. Combine one s with three p – form four equivalent new orbitals, each named sp3 1. Each hybrid a. Two lobes b. One larger than the other 2. Four large lobes point toward corners of a tetrahedron 3. Bonds formed with sp3 orbitals are very strong C. A tetrahedral arrangement of charge clouds always implies sp3 hybridization.

8.4. Other Kinds of Hybrid Orbitals4 A. Geometries related to hybrid orbitals used 1. Textbook Table 8.2 2. Each electron-cloud geometry has a corresponding set of hybrid orbitals B. Sigma () bond 1. Bond formed by head-on overlap 2. Has shared electrons centered about the axis between the two nuclei C. Pi () bond 1. Bond formed by side-by-side overlap 2. Has shared electrons occuping a region above and below a line connecting the two nuclei 8.5. Polar Covalent Bonds and Dipole Moments5 A. Bond dipole 1. A bond with partial positive (+) and partial negative (–) ends 2. Due to an electronegativity difference of the atoms in the bond 3. Represented by +→; indicates direction of electron displacement a. Point of arrow represents – end of dipole b. Crossed end represents + end of dipole

Test Item File Questions: Multiple Choice 1 – 10, Algorithmic 1 – 14, Short Answer 1 – 4 Test Item File Questions: Multiple Choice 43 3 Test Item File Questions: Multiple Choice 44 – 46, Algorithmic 36, Short Answer 5 4 Test Item File Questions: Multiple Choice 54 – 62, Algorithmic 37 – 38, Short Answer 6 – 7 5 Test Item File Questions: Multiple Choice 78 – 84, Algorithmic 39 – 47, Short Answer 8 – 14 1 2

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Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure B. Polar molecules – due to the net sum of individual bond polarities and lone-pair contributions in the molecule 1. Molecular dipoles a. Visible in an electrostatic potential map b. Center of mass of positive charge (nuclei) doesn’t coincide with the center of mass of negative charge (electrons). c. Results in a net molecular polarity 2. Lone pairs of electrons make substantial contributions to net molecular polarity.

3. Individual bond polarities cancel in symmetrical molecules.

C. Dipole moment () 1. Magnitude of the charge Q at either end of the molecular dipole times the distance r between the charges 2.  = Q  r D. Predict molecular polarity, see textbook Figure 8.8 8.6. Intermolecular Forces6 A. Intermolecular forces 1. Attractive forces between different molecules a. Influence the macroscopic properties of a substance b. The strength of a substance’s intermolecular forces and its temperature affect the phase of the substance 2. Often called van der Waals forces 3. Divided into categories a. London dispersion forces, dipole-dipole forces, hydrogen bonds and ion-dipole forces b. Electrostatic in origin B. London dispersion forces 1. Result from motion of electrons around atoms a. Due to instantaneous unsymmetrical electron distribution b. Creates a short-lived dipole moment (instantaneous dipole) 2. Instantaneous dipole induces a temporary dipole on neighboring atoms 3. Weak attractive forces – strength depends on molecule’s polarizability 4. Polarizability – ease of molecule’s electron cloud distorted by a nearby electric field a. Smaller molecules and lighter atoms with fewer electrons i. Relatively nonpolarizable ii. Smaller dispersion forces b. Larger molecules and heavier atoms with more electrons i. More polarizable ii. Larger dispersion forces c. Molecules with more spread-out shapes i. Maximize molecular surface area ii. Greater contact between molecules iii. Have higher dispersion forces C. Dipole–dipole force 1. Results from electrical interactions among dipoles on neighboring molecules 2. Generally weak – strength depends on sizes of the dipole moments involved 3. Significant only when molecules in close contact 4. Correlation between dipole moment and boiling point a. High dipole moment corresponds to strong intermolecular forces b. Substance must overcome intermolecular forces to boil c. Stronger intermolecular forces require higher boiling points D. Hydrogen bonds – attractive interaction between a hydrogen atom bonded to an electronegative O, N, or F atom and an unshared electron pair on another nearby electronegative atom 6

Test Item File Questions: Multiple Choice 101 – 103, Algorithmic 50 – 59, Short Answer 15 – 18

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Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure 1. Quite strong 2. Responsible for water’s remarkable properties 3. Results in higher boiling points than might be expected 4. Contributing factors a. Extraordinary electronegativity and small sizes of F, O, and N: H–F, H–O, H–N bonds are highly polar b. Small size of hydrogen allows it to be approached closely by other molecules E. Ion-dipole forces 1. Result of electrical interactions between an ion and partial charges on a polar molecule 2. Favored orientation a. + end of molecule near the anion b. – end of molecule near the cation 3. Important in aqueous solutions of ionic substances, where dipolar water molecules surround the ions 8.7. Molecular Orbital Theory: The Hydrogen Molecule7 A. Molecular orbital model 1. More complex bonding model than valence bond 2. Considers molecule as a whole rather than concentrating on individual atoms 3. Sometimes bonding description offers better agreement with experimental observations B. Atomic orbital – wave function whose square gives probability of finding the electron within a given region of space in an atom C. Molecular orbital – a wave function whose square gives probability of finding the electron within a given region of space in a molecule 1. Specific energy level 2. Specific shape 3. Occupied by a maximum of two electrons with opposite spins D. Orbital interactions in molecular orbital theory 1. Additive interaction a. For H2 – formation of egg-shaped molecular orbital b. Denoted  c. Lower in energy than the two isolated 1s orbitals d. Bonding molecular orbital (MO) i. Electrons in bonding MO spend most time in region between the two nuclei ii. Bonds atoms together 2. Subtractive interaction a. For H2 – formation of molecular orbital that has a node between the two atoms b. Denoted * c. Higher in energy than the two isolated 1s orbitals d. Antibonding molecular orbital i. Electrons in antibonding MO do not occupy central region between the nuclei ii. Don’t contribute to bonding E. Bond order = ½(number of bonding electrons – number of antibonding electrons) F. Number of molecular orbitals formed = the number of atomic orbitals combined G. Key ideas of molecular orbital theory – see textbook

7

Test Item File Questions: Multiple Choice 104 – 105

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Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure 8.8. Molecular Orbital Theory: Other Diatomic Molecules8 A. Paramagnetic – substances with unpaired electrons are attracted by magnetic fields B. Diamagnetic – substances whose electrons are all spin-paired C. Molecular orbital of O2 1. Two 2s orbitals interact to form 2s and *2s molecular orbitals 2. Two 2p orbitals on the internuclear axis a. Interact head-on b. Form 2p and *2p molecular orbitals 3. Remaining 2p orbitals perpendicular to the internuclear axis a. Interact in sideways manner b. Form 2p and *2p molecular orbitals 4. Energy levels of molecular orbitals – textbook Figure 8.21 D. Molecular orbital diagrams require some experience to generate.

8.9. Combining Valence Bond Theory and Molecular Orbital Theory 9 A. Molecules with resonance structures 1. -bond framework a. Localized b. Describe with valence bond theory 2. -bond framework a. Delocalized b. Describe with molecular orbital theory Inquiry –Which is better for human health, natural or synthetic vitamins?

8 9

Test Item File Questions: Multiple Choice 111 – 112, Short Answer 19 – 21 Test Item File Questions: Multiple Choice 127 – 129, Short Answer 22 - 23

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Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure

CHAPTER 8 COVALENT COMPOUNDS: BONDING THEORIES AND MOLECULAR STRUCTURE Teaching Tips, Points of Emphasis, and Common Misconceptions Section 8.1

A good way to introduce the tetrahedron is to draw a cube with atoms on alternating corners. Each atom will be on a face diagonal to each other atom, and the symmetry will be more evident. Likewise, an octahedron can be shown as a cube with an atom on each face.

Section 8.1

Molecular shapes are described by the relative positions of the nuclei, not the location of electron clouds (electron geometry). NH3 is trigonal pyramidal, not tetrahedral.

Section 8.1

Electron-dot structures do not imply geometry. It makes no difference whether the two pairs of nonbonded electrons on H2O are placed 90o or 180o to one another in the Lewis structure.

Section 8.1

Use models to show that a tetrahedron is sometimes viewed down its three-fold symmetry axis (for example, CHCl3) and sometimes viewed down its two-fold symmetry axis (for example, CH2Cl2).

Section 8.1

Nonbonded electrons are always placed in the trigonal plane of a trigonal bipyramid.

Section 8.1

When VSEPR is used to describe the shape of an organic molecule, it is applied to each individual carbon, nitrogen, and oxygen atom.

Section 8.2

Electron-dot structures describe a bond as a sharing of a pair of electrons. Valence bond theory explains how electrons become shared by the overlap of atomic orbitals.

Section 8.4

Hybrid orbital sets have the same geometries as electron cloud sets predicted by VSEPR. Therefore, each VSEPR geometry has a corresponding set of hybrid orbitals used by the central atom.

Section 8.5

A bond is polar if the atoms joined have different electronegativities. A molecule is polar if the bond polarities are not symmetrically arranged, which results in an uneven distribution of charges about the molecule and a net dipole moment.

Section 8.6

Intramolecular forces are the relatively strong bonding forces that hold atoms together in the molecular unit. Intermolecular forces are the relatively weak forces that hold one molecular unit to other molecular units.

Section 8.7

Valence bond theory assumes electrons are in atomic orbitals; molecular orbital theory places electrons in orbitals belonging to the entire molecule.

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Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure

Lecture/Laboratory Demonstration References Section 8.5

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Bending a Stream of Water,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), p. 91. This demonstration compares the polarities of water and hexane.

Section 8.6

William M. Hemmerlin and Kenton Abel, “Viscosity Races,” J. Chem. Educ., Vol. 68, 1991, 417.

Section 8.6

Andrew J. Rosenthal, “Demonstration of Surface Tension,” J. Chem. Educ., Vol. 78, 2001, 332–333.

Section 8.6

Todd P. Silverstein, “Polarity, Miscibility, and Surface Tension of Liquids,” J. Chem. Educ., Vol. 70, 1993, 253.

Section 8.6

Paul G. Jasian, Glenn Barnett, and David Speckhard, “Lowering the Surface Tension of Water: An Illustration of the Scientific Method,” J. Chem. Educ., Vol. 70, 1993, 251–252.

Section 8.6

Lee R. Summerlin and James L. Ealy, Jr., “Surface Tension of Water: The Magic Touch,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), p. 45.

Section 8.6

Bassam Z. Shakhashiri, “At the Water’s Edge: Surface Spreading and Surface Tension,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp. 301–304.

Literature References Section 8.1

James P. Birk and Soraya Abbassian, “Teaching VSEPR: The Plastic Egg Model,” J. Chem. Educ., Vol. 73, 1996, 636–637.

Section 8.1

Ronald J. Gillespie, James N. Spencer, and Richard S. Moog, “Demystifying Introductory Chemistry: Part 2. Bonding and Molecular Geometry Without Orbitals– The Electron Domain Model,” J. Chem. Educ., Vol. 73, 1996, 622–627.

Section 8.1

Brian W. Pfennig and Richard L. Frock, “The Use of Molecular Modeling and VSEPR Theory in the Undergraduate Curriculum to Predict the Three-Dimensional Structure of Molecules,” J. Chem. Educ., Vol. 76, 1999, 1018–1022.

Section 8.1

Ronald J. Gillespie, “Multiple Bonds and the VSEPR Model,” J. Chem. Educ., Vol. 69, 1992, 116–121.

Section 8.3

D. W. Emerson, “A Colorful Demonstration to Simulate Orbital Hybridization,” J. Chem. Educ., Vol. 65, 1988, 454.

Section 8.3

John Barbaro, “Orbital Bartending,” J. Chem. Educ., Vol. 71, 1994, 1012. An analogy for orbital hybridization.

Section 8.5

R. J. Tykodi, “Identifying Polar and Nonpolar Molecules,” J. Chem. Educ., Vol. 66, 1989, 1007–1011.

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Chapter 8—Covalent Compounds: Bonding Theories and Molecular Structure Section 8.5

Thomas H. Bindel and Timothy C. Smiley, “The Ropes: A Molecular Polarity Activity,” J. Chem. Educ., Vol. 71, 1994, 945.

Section 8.5

Carlos Furio and Ma. Luisa Calatayud, “Difficulties with the Geometry and Polarity of Molecules: Beyond Misconceptions,” J. Chem. Educ., Vol. 73, 1996, 36–41.

Section 8.5

Gordon H. Purser, “The Significance of the Bond Angle in Sulfur Dioxide,” J. Chem. Educ., Vol. 66, 1989, 710–713.

Section 8.6

Robert R. Perkins, “Put the Body to Rest,” J. Chem. Educ., Vol. 72, 1995, 151–152.

An analogical demonstration of molecular polarity and chemical and physical changes involving intramolecular and intermolecular forces.

Section 8.6

John J. Fortman, “Pictorial Analogies I: States of Matter,” J. Chem. Educ., Vol. 70, 1993, 56–57.

Section 8.6

Ponnadural Ramasami, “Students as Solids, Liquids, and Gases,” J. Chem. Educ., Vol. 76, 1999, 485.

Section 8.6

Doris R. Kimbrough and Ronald DeLorenzo, “Solving the Mystery of Fading Fingerprints with London Dispersion Forces,” J. Chem. Educ., Vol. 75, 1998, 1300–1301.

Section 8.6

C. Jayne Wilcox, “London Dispersion Forces and the Wave,” J. Chem. Educ., Vol. 75, 1998, 1300–1301.

Section 8.6

John W. Hill, “A People-and-Velcro Model for Hydrogen Bonding,” J. Chem. Educ., Vol. 67, 1990, 223.

Section 8.7

A. B. Sannigrahi and Tapas Kar, “Molecular Orbital Theory of Bond Order and Valency,” J. Chem. Educ., Vol. 65, 1988, 674–676.

Section 8.8

Albert Haim, “The Relative Energies of Molecular Orbitals for Second-Row Homonuclear Diatomic Molecules; The Effect of s-p Mixing,” J. Chem. Educ., Vol. 68, 1991, 737–738.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 8.4

Worked Example 8.4: Identifying hybridization and orbital overlap in single and double bonds

Section 8.5

Worked Example 8.6: Predicting the presence of a dipole moment

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CHAPTER 9 THERMOCHEMISTRY: CHEMICAL ENERGY Chapter Learning Objectives Section 9.1

Calculate the kinetic energy of an object in motion.

Convert between common units for energy.

Define conservation of energy.

Section 9.2

Identify state functions.

Identify the sign of heat and work.

Section 9.3

Calculate PV work.

Section 9.4

Calculate the internal energy change (E) for a reaction.

Section 9.5 and Section 9.6

Calculate the amount of heat transferred given a thermochemical equation and the amount of reactant or product which reacts.

Classify a reaction as endothermic or exothermic and give the sign of H.

Section 9.7

Calculate heat capacities, temperature changes, or heat transfer, using the equation for heat capacity (C), specific heat (c), or molar heat capacity (Cm).

Calculate enthalpy changes in a calorimetry experiment.

Section 9.8

Use Hess’s law to find H for an overall reaction, given reaction steps and their H values.

Section 9.9

Identify the standard state of elements.

Write standard enthalpy of formation reactions (Hf) for compounds from their elements.

Use values of standard heats of formation (Hf) for elements and compounds to calculate H for a reaction.

Section 9.10

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Use bond dissociation energies to estimate H for a reaction.

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Chapter 9—Thermochemistry: Chemical Energy Section 9.11 and Section 9.12

Predict the sign of the entropy change (S) given the chemical equation or a molecular diagram.

Predict the sign of G, H, S, using the equation for Gibbs free-energy change (G).

Use the Gibbs free-energy equation to calculate the temperature at which a reaction crosses over from spontaneous to nonspontaneous.

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Chapter 9—Thermochemistry: Chemical Energy

Lecture Outline 9.1. Energy and Its Conservation1 A. Energy 1. Capacity to do work or supply heat 2. Energy = work + heat B. Kinetic energy 1. Energy of motion 2. EK = 1/2mv2 C. Potential energy – stored energy D. Joule 1. SI unit for energy 2. 1 J = 1(kg  m2)/s2 E. Law of Conservation of Energy 1. Energy can be neither created nor destroyed 2. It can only be converted from one form into another F. Many forms of energy 1. Thermal energy a. Kinetic energy of molecular motion b. Temperature – measure of the kinetic energy of molecular motion 2. Heat a. Energy transferred from one object to another b. Result of a temperature difference between them 3. Chemical energy a. A type of potential energy b. Chemical bonds of molecules act as the storage medium G. First Law of Thermodynamics: The energy of the universe is constant.

9.2. Internal Energy and State Functions2 A. System – everything we focus on in an experiment (reactants and products) B. Surroundings – everything other than the system C. Internal energy – energy of the system D. System isolated from the surroundings 1. No energy transfer to surroundings 2. E = 0 3. First Law of Thermodynamics: The total internal energy of an isolated system is constant.

E. System is not isolated from surroundings 1. Energy flows to or from surroundings 2. E = Efinal – Einitial F. Energy changes measured from the point of view of the system 1. Energy flows out of system to surroundings – negative value 2. Energy flows into system from surroundings – positive value G. State function – function or property whose value depends only on the present state (condition) of the system, not on path used to arrive at that condition 1. Reversible 2. Overall change is zero if system returns to its original condition

1 2

Test Item File Questions: Multiple Choice 1, Algorithmic 1 – 3, Short Answer 1 – 7 Test Item File Questions: Multiple Choice 2, Algorithmic 4 – 8, Short Answer 8 – 12

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Chapter 9—Thermochemistry: Chemical Energy 9.3. Expansion Work3 A. Work 1. Force (f) that produces movement of an object times distance moved (d) 2. w = F  d B. Expansion work (PV work) is the most common type of work seen for chemical systems 1. Work done as result of volume change in the system 2. w = – (P  V) 3. Expansion of the system. a. System does work on surroundings b. E is negative c. w = – PV i. Work is negative ii. V is positive 4. Contraction of the system a. Surroundings do work on the system b. E is positive c. w = – PV i. Work is positive ii. V is negative 5. When there is no volume change of the system, no PV work is done.

9.4. Energy and Enthalpy4 A. Total energy change of a system 1. E = q + w (q = heat) 2. E = q + (– PV) B. Amount of heat transferred: q = E + PV C. Reactions carried out with constant volume 1. V = 0 2. No PV work done 3. qv = E D. Reactions carried out at constant pressure 1. V  0 2. Energy change due to both heat transfer and PV work 3. qp = E + PV E. Heat of reaction, H 1. Enthalpy change of a system 2. H = E + PV 3. State function – value depends only on the current state of the system 4. H = Hproducts – Hreactants 5. For many reactions, the difference between H and E is small and the two are nearly equal.

9.5. Thermodynamic Equations and the Thermodynamic Standard State5 A. Value of enthalpy change, H, reported for a reaction represents amount of heat released when reactants converted to products in molar amounts represented by coefficients of the balanced equation.

1. Actual amount of heat involved in a reaction depends on actual amounts of reactants.

2. Physical states of reactants and products must be specified.

Test Item File Questions: Multiple Choice 7 – 10, Algorithmic 9 Test Item File Questions: Multiple Choice 11 – 19, Algorithmic 10 – 12, Short Answer 13 – 17 5 Test Item File Questions: Multiple Choice 20 – 25, Algorithmic 13 – 15 3 4

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Chapter 9—Thermochemistry: Chemical Energy B. Temperature and pressure also must be reported. Thermodynamic Standard State = 298.15 K (25 C), 1 atm pressure of each gas, 1 M concentration (for solutions) 1. Allows different reactions to be compared 2. Indicated by addition of a superscript °, H C. Standard enthalpy of reaction – enthalpy change measured under standard conditions 1. H values for a given equation a. Equation balanced for number of moles of reactants and products b. All substances are in standard states. c. Physical state of each substance specified d. Refers to reaction going in direction written e. Reversing direction of reaction changes sign of H 9.6. Enthalpies of Chemical and Physical Changes6 A. Enthalpies of chemical change 1. Heats of reaction – enthalpies of chemical change 2. Endothermic reactions a. Hproducts > Hreactants b. Heat flows into system from surroundings c. H is positive 3. Exothermic reactions a. Hproducts < Hreactants b. Heat flows to surroundings from system c. H is negative B. Enthalpies of physical change 1. Heat of fusion – amount of heat required for melting a substance without changing its temperature 2. Heat of vaporization – amount of heat required for vaporizing a substance 3. Sublimation a. Direct conversion of solid to vapor without going through a liquid state b. Heat of sublimation = Heat of fusion + Heat of vaporization 9.7. Calorimetry and Heat Capacity7 A. Calorimetry – experimental technique that allows energy change associated with a chemical or physical process to be determined 1. Temperature change observed when system gains or loses energy in the form of heat 2. Performed in a calorimeter 3. For an exothermic reaction: amount of heat released by reaction = amount of heat gained by calorimeter + amount of heat gained by solution B. Heat capacity (C) q C= DT 1. Amount of heat required to raise the temperature of an object or substance a given amount 2. Extensive property C. Specific heat capacity 1. Amount of heat necessary to raise the temperature of exactly 1 g of a substance by exactly 1 C 2. q = (specific heat)  (mass of substance)  (T) D. Molar heat capacity (Cm) 1. Amount of heat necessary to raise the temperature of 1 mole of a substance by 1 C 2. qm = (Cm)  (moles of substance)  (T)

6 7

Test Item File Questions: Multiple Choice 26 – 37, Algorithmic 16 – 24, Test Item File Questions: Multiple Choice 38 – 45, Algorithmic 25 – 32, Short Answer 18 – 20

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Chapter 9—Thermochemistry: Chemical Energy 9.8. Hess’s Law8 A. Hess’s law 1. Overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for individual steps in the reaction.

2. Reactants and products in the individual steps added and subtracted like algebraic quantities in determining the overall equation B. Combine individual reactions so that their sums will be the desired reaction.

1. All reactants must appear on the left.

2. All products must appear on the right.

3. All intermediate products must appear on both the left and right so that they cancel.

4. A reaction written in the reverse of the direction given must have the sign of its H reversed.

5. If a reaction is multiplied by a coefficient, H for the reaction must be multiplied by that same coefficient.

9.9. Standard Heats of Formation9 A. Standard heat of formation 1. Enthalpy change Hof for the hypothetical formation of 1 mole of a substance in its standard state from the most stable forms of its constituent elements in their standard states 2. The most stable forms of all elements in their standard state have Hof = 0.

B. Standard enthalpy change found by subtracting the sum of the heats of formation of reactants from the sum of the heats of formation of products: o o DH o = S DH products - S DH reactants

C. Table 9.2 and Textbook Appendix B 9.10. Bond Dissociation Energies10 A. Bond dissociation energies 1. Enthalpy changes, H, for corresponding bond-breaking reactions 2. H = D = bond dissociation energy 3. Always positive 4. Always need energy to break a bond B. H = D(Reactant bonds) – D(Product bonds) C. Textbook Table 9.3 9.11. An Introduction to Entropy11 A. Spontaneous process 1. Process that proceeds on its own without continuous external influence 2. May happen quickly or slowly B. Entropy (S) 1. Amount of molecular randomness in a system 2. Larger the value of S, greater the molecular randomness 3. S = Sfinal – Sinitial 4. Sfinal > Sinitial a. S is positive b. System has become more random 5. Sfinal < Sinitial a. S is negative b. System has become less random

Test Item File Questions: Multiple Choice 46 – 48, Algorithmic 35 – 37, Short Answer 21 Test Item File Questions: Multiple Choice 58 – 62, Algorithmic 38 – 40 10 Test Item File Questions: Multiple Choice 63 – 66, Algorithmic 41 – 42 11 Test Item File Questions: Multiple Choice 67 – 74, Algorithmic 43 – 46, Short Answer 23 8 9

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Chapter 9—Thermochemistry: Chemical Energy C. Spontaneous process 1. Favored by decrease in H (negative H) 2. Favored by increase in S (positive S) D. Nonspontaneous process 1. Favored by increase in H (negative H) 2. Favored by decrease in S (positive S) 9.12. An Introduction to Free Energy12 A. Gibbs free-energy change (G); G = H – TS. B. Sign of G used as a criterion for determining spontaneity of a process 1. G negative – spontaneous 2. G positive ≠ nonspontaneous C. Temperature dependence (TS) term for G 1. Spontaneity of some processes depends on temperature 2. Low temperatures – H dominates and controls spontaneity 3. High temperatures – TS dominates and controls spontaneity D. G = 0 1. Process at equilibrium 2. Balanced between spontaneous and nonspontaneous 3. Temperature at which a nonspontaneous reaction becomes spontaneous DH T= DS Inquiry – How do we determine the energy content of biofuels?

12

Test Item File Questions: Multiple Choice 88 – 97, Algorithmic 48 – 51, Short Answer 25 – 27

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Chapter 9—Thermochemistry: Chemical Energy

CHAPTER 9 THERMOCHEMISTRY: CHEMICAL ENERGY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 9.2

Students need to relate direction of heat flow with observed temperature change.

Demonstrate that for a chemical reaction (system) performed in a beaker (surroundings), if heat energy is released by the system, the beaker absorbs the heat and becomes hotter; if heat energy is absorbed by the system, the beaker contributes the heat and becomes colder.

Section 9.2

It is important for students to understand that the amount of energy released or gained in a reaction is proportional to the amounts of reactants and products involved in the reaction.

Section 9.3

Gases do work when the molecules push against the walls of their container and the volume of the container increases. If there is no volume change, no work is done.

Section 9.4

Relate energy change to physical work: When a person does work, the person’s energy diminishes, so when a system does work, w is negative and E is negative.

Section 9.4

If the volume of the system does not change, the heat transferred to or from a system is the energy change of that system.

Section 9.4

If the pressure of the system does not change, the heat transferred to or from a system is the enthalpy change of that system.

Section 9.4

When reactions are performed in an open container, such as a coffee-cup calorimeter, the pressure (= atmospheric pressure) does not change and qP = H; in a closed container, such as a bomb calorimeter, the volume (= volume of the bomb) does not change and q qx = E. Even in an open container, unless a gas is involved in the reaction, volume changes are usually negligible and H  E.

Section 9.5

The states of reactants and products are important in determining an energy change or enthalpy change for a reaction. The energy change will be different for a reaction producing a product as a liquid (with lower energy) versus producing the same product as a gas (with higher energy).

Section 9.5

Standard conditions as defined for thermodynamics (P = 1 atm; T = 298.15 K) are different from standard conditions as defined for gas laws (P = 1 atm; T = 273.15 K).

Section 9.6

Endothermic means heat flows into a system and H > 0. Exothermic means heat flows out of a system and H < 0.

Section 9.6

Temperature does not change during a change of state; hence, Hfusion for water refers to H2O(s, 0 oC) → H2O(l, 0 oC) and Hvap for water refers to H2O(l, 100 oC) → H2O(g, 100 oC).

Section 9.7

Just as density provides a link between mass and volume, heat capacity links heat flow and temperature change, allowing the three variables to be interconverted.

Section 9.7

The amount of heat transferred and, therefore, the temperature change depend on the amount of substance involved. Specific heat is heat capacity per gram of substance

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Chapter 9—Thermochemistry: Chemical Energy and has units of J/(g  oC); molar heat capacity is heat capacity per mole of substance and has units J/(mol  oC).

Section 9.7

Specific heat and molar heat capacity are intensive properties that depend on the state of the substance.

Section 9.8

Students have already employed Hess’s law when performing Born–Haber cycle calculations.

Section 9.8

When an equation is reversed, H must be multiplied by –1. When an equation is multiplied by a coefficient, H must be multiplied by the same coefficient.

Section 9.9

Remind students to multiply the heat of formation of each substance by its coefficient in the chemical equation and also to multiply the standard heat of formation of each product by –1.

Section 9.10

Bond dissociation energies are defined for isolated, gaseous-state species. Other factors must be considered when dealing with condensed-state systems.

Section 9.10

Reactant bonds are broken and product bonds are formed: Ho = D(reactant bonds broken) – D(product bonds formed). Students should not confuse this equation for that used for heats of formation: Ho = Hof(products) – Hof(reactants), in which products and reactants appear on the opposite side of the minus sign.

Section 9.10

Electron-dot structures should be drawn for each reactant and product to see the number of each type of bond being broken and formed. These numbers must also be multiplied by the coefficients in the balanced chemical equation.

Section 9.11

Another way to introduce entropy is to use a spontaneous process for which H = 0, such as the mixing of two ideal gases.

Section 9.12

H has units of kJ, S has much smaller units of J/K. Thus, reaction spontaneity as measured by G = H –TS is usually dominated by H except at high temperatures or when the reaction results in an increase in the number of moles of gaseous products.

Lecture/Laboratory Demonstration References Section 9.6

Bassam Z. Shakhashiri, “Endothermic Reactions of Hydrated Barium Hydroxide and Ammonium Salts,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983), pp. 10–12.

Section 9.7

Bassam Z. Shakhashiri, “Boiling Water in a Paper Cup: Heat Capacity of Water,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp. 239–241.

Section 9.7

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “A Chemical Hand Warmer,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 101–102.

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79

Chapter 9—Thermochemistry: Chemical Energy Section 9.7

Bassam Z. Shakhashiri, “Chemical Cold Pack,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983), pp. 8–9.

Section 9.7

Bassam Z. Shakhashiri, “Heat of Neutralization,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983), pp. 15–16.

Section 9.7

Lee R. Summerlin and James L. Ealy, Jr., “Endothermic Reaction: Ammonium Nitrate,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), p. 65.

Literature References Section 9.7

John J. Fortman, “Analogical Demonstrations,” J. Chem. Educ., Vol. 69, 1992, 323–324.

Section 9.7

Charles M. Wynn, Sr., “Heat Flow vs. Cash Flow: A Banking Analogy,” J. Chem.

Educ., Vol. 74, 1997, 397–398.

Section 9.7

Brother Thomas McCullogh, “A Specific Heat Analogy,” J. Chem. Educ., Vol. 57, 1980, 896.

Section 9.7

Doris R. Kimbrough, “Heat Capacity, Body Temperature and Hypothermia,” J. Chem. Educ., Vol. 75, 1998, 48–49.

Section 9.11

John J. Fortman, “Pictorial Analogies III: Heat Flow, Thermodynamics, and Entropy,” J. Chem. Educ., Vol. 70, 1993, 102–103.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 9.7

Worked Example 9.6: Calculating H in a constant-pressure calorimetry experiment

Section 9.7

Worked Example 9.7: Calculating E in constant-volume calorimetry

Section 9.12

Worked Example 9.12: Predicting the signs of H, S, and G for a reaction

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CHAPTER 10 GASES: THEIR PROPERTIES AND BEHAVIOR Chapter Learning Objectives Section 10.1

Convert between different units of pressure.

Determine the pressure from the height of a liquid mercury column in a barometer or manometer.

Section 10.2

Use the individual gas laws to calculate pressure, volume, molar amount, or temperature of a gas sample when conditions change.

Section 10.3

Use the ideal gas law to calculate pressure, volume, moles of gas, or temperature of a gas sample.

Section 10.4

Calculate volumes of gases in chemical reaction using the ideal gas law.

Calculate the density or molar mass of a gas using the formula for gas density.

Section 10.5

Calculate the partial pressure, mole fraction, or amount of each gas in a mixture.

Section 10.6

Use the assumptions in the kinetic–molecular theory to predict behavior.

Calculate the average molecular speed of a gas particle at a given temperature.

Section 10.7

Describe a molecular picture of effusion and diffusion.

Use Graham’s law to estimate the relative rates of effusion of two different gases.

Section 10.8

Identify temperature and pressure conditions that cause significant deviations from ideal behavior.

Use the van der Waals equation to calculate the pressure properties of a real gas.

Section 10.9 and Section 10.10

Explain the principle of the greenhouse gas effect.

Predict if a molecule is a greenhouse gas by examining molecular vibrations.

Section 10.11

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Describe the trends in greenhouse gas concentrations over time and the measured and predicted effects of climate change.

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Chapter 10—Gases: Their Properties and Behavior

Lecture Outline 1

10.1 Gases and Gas Pressure A. Gases 1. Little attraction between constituent atoms or molecules 2. Free to move about in available volume B. Some properties of gases 1. Mixtures always homogeneous a. Very weak attraction between gas molecules b. Identity of neighbor is irrelevant 2. Compressible a. Volume contracts when pressure applied – different behavior than solids or liquids b. Particles (atoms or molecules) of a gas are far apart unlike a solid or liquid c. Only 0.10% of volume of gas occupied by molecules 3. Exert a measurable pressure on the walls of their container C. Pressure 1. Force exerted per unit area, P = F/A 2. SI unit a. Pascal (Pa) b. 1 Pa = 1 N/m2 (1 N = 1 (kgm)/s2) 3. Alternative units a. Millimeters of mercury (mm Hg) also called a torr b. Atmosphere (atm) = 760 mm Hg = 101,325 Pa c. Barr = 100,000 Pa = 1000 kPa = 0.986923 atm d. Textbook Table 10.2 D. Atmospheric pressure 1. Pressure created from the mass of atmosphere pressing down on the earth’s surface 2. Standard atmospheric pressure at sea level = 760 mm Hg = Patm E. Measuring pressure 1. Barometer a. Long thin mercury-filled tube sealed at one end and inverted into a dish of mercury b. Downward pressure of Hg in column equals outside atmospheric pressure 2. Manometer a. U-tube filled with mercury i. One end connected to the gas-filled container ii. Other end open to the atmosphere b. Pgas = Patm: liquid level in both arms is equal c. P gas > Patm i. Liquid level in the arm connected to the gas-filled cylinder will be lower ii. P gas = Patm + PHg (PHg = the difference in the heights of the two mercury columns) d. P gas < Patm i. Liquid level in the arm connected to the gas-filled cylinder will be higher ii. P gas + PHg = Patm 2 10.2. The Gas Laws A. Different gases show similar physical behavior B. Any gas can be defined by four variables – pressure, temperature, volume, and number of moles a. Relationships of variables – gas laws b. Ideal gas – behavior follows the gas laws exactly C. Boyle’s law 1Test Item File Questions: Multiple Choice 1 – 11, Algorithmic 1 – 6, Short Answer 1 – 3 2Test Item File Questions: Multiple Choice 22 – 33, Algorithmic 7 – 12, Short Answer 4 – 6

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Chapter 10—Gases: Their Properties and Behavior 1. Relationship between volume and pressure for given amount of gas at constant temperature 2. V  1/P D. Charles’ law 1. Relationship between volume and temperature for a given amount of gas at constant pressure 2. V  T (temperature expressed in Kelvins) E. Avogadro’s law 1. Relationship between volume and amount of gas at constant pressure and temperature 2. V  n (where n = number of moles of gas) 3. 1 mol of gas at 273.15 K and 1.00 atm = 22.4 L of gas – Standard Molar Volume 3 10.3 The Ideal Gas Law A. Describes how volume of gas affected by changes in pressure, temperature, and amount L  atm J 1. PV = nRT; R = gas constant = 0.08206 = 8.3145 K  atm K  mol 2. Can be rearranged in different ways to take the form of Boyle’s law, Charles’ law, and Avogadro’s law B. Standard temperature and pressure (STP) 1. T = 0 C = 273.15 K 2. P = 1 atm C. All gases deviate somewhat from ideal behavior 1. Individual gases have molar volumes at STP that are not exactly 22.4 L – Textbook Table 10.5 2. Deviations are small and typically ignored 4 10.4. Stoichiometric Relationships with Gases A. Stoichiometric calculations involving the application of the ideal gas law B. Three of four variables, P, V, T, and n known, the fourth calculated C. Includes calculations involving gas density and molar mass 5 10.5. Mixtures of Gases: Partial Pressure and Dalton’s Law A. Gas laws apply to mixtures of gases B. Dalton’s law of partial pressures 1. Ptotal = P1 + P2 + P3 + .... at constant V, T 2. P1, P2, .... refer to the pressures of individual gases in the mixture C. Partial pressure 1. Refers to the pressure an individual gas would exert if alone in the container (P1, P2, .....)

2. Total pressure – depends on total molar amount of gas present 3. Mole fraction (X) a. Number of moles of one component divided by total number of moles in the mixture moles of component b. Mole fraction ( X ) = total moles in mixture 4. P1 = X1 Ptotal

3Test Item File Questions: Multiple Choice 40 – 50, Algorithmic 13 – 25, Short Answer 7 – 9 4Test Item File Questions: Multiple Choice 52 – 64, Algorithmic 26 – 34, Short Answer 10 – 12 5Test Item File Questions: Multiple Choice 66 – 73, Algorithmic 35 – 41, Short Answer 13 – 14

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Chapter 10—Gases: Their Properties and Behavior 6

10.6. The Kinetic–Molecular Theory of Gases A. Model that explains gas behavior B. Assumptions 1. Gas particles in constant random motion 2. Most gas volume is empty space 3. Negligible attractive forces between gas molecules 4. Constant total kinetic energy of gas particles at constant T 5. Average EK  T C. Above assumptions used to explain gas laws 7 10.7. Gas Diffusion and Effusion: Graham’s Law A. Consequence of constant motion and high velocities of gas particles – gases mix rapidly 1. Diffusion – mixing of different gases a. Random molecular motion b. Frequent collisions 2. Effusion a. Gas molecules escape through a tiny hole in a membrane into a vacuum b. No collisions B. Graham’s law 1. Rate of gas effusion inversely proportional to the square root of its molar mass 1 Rate µ M 2. Two gases at same temperature and pressure:

Rate1 = Rate 2

M2 M1

3. Different rates of effusion used to separate gas mixture into separate components 8 10.8. The Behavior of Real Gases A. Ideal gas 1. Volume occupied by gas particles is negligible compared to the total gas volume – true at low pressures 2. No attractive forces between molecules – true at low pressures B. Real gas 1. Attractive forces more important at higher pressures – decreases volume of gas from that predicted by the ideal gas law 2. Molecular volume – actual gas volume at high pressure larger than predicted by the ideal gas law C. Two effects – molecular volume and intermolecular forces cancel out at intermediate pressures; at pressures . 300 atm molecular volume effect dominates D. van der Waals equation: æ an2 ö ç P + 2 ÷ V - nb = nRT V ø è 1. a = correction for intermolecular attractions 2. b = correction for molecular volume

(

)

6Test Item File Questions: Multiple Choice 78 – 86, Algorithmic 45 – 54, Short Answer 15 – 18 7Test Item File Questions: Multiple Choice 87 – 90, Algorithmic 55 – 59, Short Answer 19 – 20 8Test Item File Questions: Multiple Choice 95 – 96, Short Answer 21

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Chapter 10—Gases: Their Properties and Behavior 9

10.9. The Earth’s Atmosphere and the Greenhouse Effect A. Atmosphere has four regions – troposphere (region nearest Earth’s surface), stratosphere, mesosphere and thermosphere 1. Atmospheric pressure decreases in a predictable way 2. Temperature versus altitude is more complex B. Absorption of infrared (IR) radiation by the atmosphere resulting in an increase in planetary temperature 1. Naturally occurring phenomenon that affects climate by regulating temperature 2. Enables life as we know it on Earth C. Principle of the greenhouse effect 1. Sun emits ultraviolet (UV) and visible radiation.

2. Ozone absorbs much of the UV radiation.

3. Absorption of visible radiation results in heating of Earth’s surface.

4. IR radiation re-emitted a. Some escapes into space b. Some absorbed by greenhouse gases in the atmosphere 5. Greenhouse gases return some of the IR radiation to Earth, warming it.

10 10.10. Greenhouse Gases A. Carbon dioxide is a greenhouse gas, as it absorbs IR radiation.

B. Absorbed IR radiation causes a molecule to reach an excited vibrational state.

C. Water is the largest contributor to the natural greenhouse effect.

D. Carbon dioxide, nitrous oxide, methane, and halogen containing gases are greenhouse gases that occur due to human activities.

E. Global-warming potential (GWP) is a relative measure of how much heat a greenhouse gas traps in the atmosphere on a per mass basis.

1. Carbon dioxide set as reference of 1 2. Textbook Table 10.7 11 10.11. Climate Change A. Increasing concentrations of greenhouse gases change the thermal balance of the Earth.

1. Other factors also part of this complex process 2. Not all areas of the Earth affected the same B. Combustion of fossil fuels has increased the concentrations of greenhouse gases in the atmosphere significantly since the onset of the industrial revolution.

C. Atmospheric levels of CO2 and CH4 are currently at historically high levels D. A historical correlation is seen between high CO2 levels and high Antarctic temperatures.

E. Average global surface temperatures are increasing.

1. Arctic sea ice coverage is decreasing.

2. Causes significant impact to various ecosystems, possibly leading to species extinction 3. Sea levels are rising.

F. Climate scientists agree that warming trends are due to human activity.

Inquiry – How do inhaled anesthetics work?

9Test Item File Questions: Multiple Choice 97 – 101, Algorithmic 60, Short Answer 23 – 26 10Test Item File Questions: Multiple Choice 102 – 104, Algorithmic 61 – 62 11Test Item File Questions: Algorithmic 60

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Chapter 10—Gases: Their Properties and Behavior

CHAPTER 10 GASES: THEIR PROPERTIES AND BEHAVIOR Teaching Tips, Points of Emphasis, and Common Misconceptions Section 10.1

Atmospheric pressure is measured with a barometer; the pressure of any gas is measured with a manometer. If gas pressure > atmospheric pressure, Pgas = Patm + hmercury; if gas pressure < atmospheric pressure, Pgas = Patm – hmercury.

Section 10.2

Students tend to remember the standard molar volume while forgetting that this value holds true only under standard conditions. One mole of an ideal gas occupies 22.4 L only at 1.00 atm and 273 K.

Section 10.2

Students sometimes think that some substances are ideal gases while other substances are nonideal. All gases are ideal under certain conditions and nonideal under others.

Lecture/Laboratory Demonstration References Section 10.1

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Boiling at Reduced Pressure,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 24–25.

Section 10.2

Bassam Z. Shakhashiri, “Boyle’s Law,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985), pp. 14–19.

Section 10.2

Bassam Z. Shakhashiri, “Effect of Pressure on the Size of a Balloon,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985), pp. 12–13.

Section 10.2

Bassam Z. Shakhashiri, “Boyle’s Law and the Mass of a Textbook,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985), pp. 20–23.

Section 10.2

Rick Broniec, “Boyle’s Law and the Monster Marshmallow,” J. Chem. Educ., Vol. 59, 1982, 974.

Section 10.2

N. Szczepanski, “A Simple Demonstration of Charles’s Law,” J. Chem. Educ. 71, 1994, 433.

Section 10.2

Bassam Z. Shakhashiri, “Thermal Expansion of Gases,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985), pp. 24–27.

Section 10.2

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Charles’ Law: The Relationship Between Volume and Temperature of a Gas,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), p. 23.

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Chapter 10—Gases: Their Properties and Behavior Section 10.4

Bassam Z. Shakhashiri, “Determination of the Molecular Mass of a Volatile Liquid,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983), pp. 51–54. A boiling-water bath is used to heat a volatile liquid until it is vaporized, completely filling an Erlenmeyer flask covered by aluminum foil with a pinhole orifice. From atmospheric pressure, bath temperature, volume of the flask, and mass of recondensed vapor, the molar mass of the volatile liquid is determined.

Section 10.4

Lee R. Summerlin, and James L. Ealy, Jr., “Determining the Molecular Weight of a Gas,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 19–20. The molar mass of butane is determined by the vapor density method.

Section 10.7

Bassam Z. Shakhashiri, “Ratio of Diffusion Coefficients: The Ammonium Chloride Ring,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983), pp. 59–62. Ammonia vapor and hydrogen chloride gas are simultaneously introduced into opposite ends of dry glass tubing. A white ring of solid ammonium chloride forms inside the tube closer to the end at which the hydrogen chloride was introduced, showing an inverse relationship between the rate of diffusion and the molar mass of each reactant. No distinction is made between diffusion and effusion, and Graham’s law is erroneously used in explaining the data.

Section 10.7

Lee R. Summerlin, and James L. Ealy, Jr., “Diffusion of Gases,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 14–15. Ammonia and hydrogen chloride diffusion rates are determined.

Section 10.7

Dianne N. Epp, “Overhead Projection of Graham’s Law of Gaseous Diffusion,” J. Chem. Educ. 67, 1990, 1061.

Section 10.7

Bassam Z. Shakhashiri, “Relative Velocity of Sound Propagation: Musical Molecular Weights,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985), pp. 88–89.

Literature References Section 10.1

Joseph S. Schmuckler, “Gases and Their Behavior,” J. Chem. Educ., Vol. 57, 1980, 85.

Section 10.2

John T. Petty, “Charles’ Law of Gases: A Simple Demonstration,” J. Chem. Educ.

72, 1995, 257.

Section 10.2

R. P. Graham, “Guy-Lussac: Chemist Extraordinary,” J. Chem. Educ. 58, 1981, 789.

Section 10.2

Harold Goldwhite, “Guy-Lussac after 200 Years,” J. Chem. Educ. 55, 1978, 366–368.

Section 10.4

Andreas Madlung, “The Chemistry Behind the Air Bag,” J. Chem. Educ. 73, 1996, 347–348.

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Chapter 10—Gases: Their Properties and Behavior Section 10.4

William L. Bell, “Chemistry of Air Bags,” J. Chem. Educ. 67, 1990, 61.

Section 10.6

Dacio R. Hartwig and Romeu C. Rocha Filho, “Cinema, Flirts, Snakes, and Gases,” J. Chem. Educ. 59, 1982, 295. A kinetic molecular theory of gases analogy is presented.

Section 10.6

Reggie L. Hudson, “Toy Flying Saucers and Molecular Speeds,” J. Chem. Educ. 59, 1982, 1025–1026.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 10.2

Worked Example 10.3: Visual representation of gas laws

Section 10.3

Worked Example 10.5: Using the ideal gas law when variables change

Section 10.8

Worked Example 10.8: Calculating partial pressure

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CHAPTER 11 LIQUIDS AND PHASE CHANGES Chapter Learning Objectives Section 11.1

Match an observation of a physical phenomenon with the correct liquid property.

Predict which substance has higher viscosity or surface tension based on molecular structure.

Section 11.2

Use the Clausius–Clapeyron equation to calculate Hvap or vapor pressure at varying temperatures.

Section 11.3

Use the equation for free energy to calculate the temperature at which a phase change becomes spontaneous or S for a phase change.

Section 11.4

Draw or interpret a heating curve for a substance.

Calculate the amount of heat associated with temperature changes and the amount of heat associated with phase changes of substances.

Section 11.5

Use a phase diagram to identify the phase of a substance at a given temperature and pressure or the phase transition that occurs when the temperature and pressure are changed.

Sketch a phase diagram given the appropriate data.

Section 11. 6

Describe a liquid crystal phase based on molecular structure.

Classify the types of intermolecular forces between molecules in a liquid crystal.

Match a molecular image with different states of matter, including solid, liquid, and common liquid crystal phases.

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Chapter 11—Liquids and Phase Changes

Lecture Outline 1

11.1. Properties of Liquids A. Viscosity 1. Measure of a liquid’s resistance to flow 2. SI unit: N  s/m2 3. Ease with which molecules move around in the liquid 4. Related to intermolecular forces – stronger the forces, higher the viscosity B. Surface tension 1. Resistance of a liquid to spreading out and increasing its surface area 2. Due to difference in intermolecular forces felt by liquid surface molecules and liquid interior molecules 3. Related to intermolecular forces – stronger the forces, greater the surface tension C. Properties temperature dependence 1. Increase in temperature corresponds to increase in kinetic energy 2. Molecules with high kinetic energies more easily overcome intermolecular forces 2 11.2. Vapor Pressure and Boiling Point A. Vapor pressure – pressure exerted by vapor molecules over a liquid in a closed container 1. Dynamic equilibrium a. Number of molecules escaping liquid = number of molecules returning to liquid b. Constant total number of molecules in both liquid and vapor phases c. Individual molecules constantly passing back and forth from one phase to another B. Explained by kinetic molecular theory – increase in temperature results in higher fraction of molecules with sufficient kinetic energy to overcome surface tension and escape into the vapor C. Value of vapor pressure 1. Related to intermolecular forces and temperature 2. Smaller the intermolecular forces, higher the vapor pressure – molecules loosely held in liquid and easily escape 3. Higher the temperature, higher the kinetic energy – molecules have sufficient kinetic energy to escape the liquid D. Clausius–Clapeyron equation 1. Heat of vaporization (ΔHvap) is the amount of energy needed to convert one mole of liquid into a gas.

2. Clausius–Clapeyron relates vapor pressure of a liquid to inverse of its temperature

 −H vap  1n ( Pvap ) =   + C2 where C = a constant characteristic of each substance  RT  3.

The vapor pressure at some temperature can be calculated if the vapor pressure at another temperature and Hvap are known.

 P   H vap   1 1  ln  1  =   −   P2   R   T2 T1  4.

Knowing vapor pressure at several temperatures, can calculate Hvap of a liquid from

1Test Item File Questions: Multiple Choice 1 – 3, Algorithmic 1 – 3, Short Answer 1 – 2 2Test Item File Questions: Multiple Choice 13 – 22, Algorithmic 10 – 13, Short Answer 4, 5

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Chapter 11—Liquids and Phase Changes

H vap =

(1n P2 − 1n P1 ) R 1 1  −   T1 T2 

E. Boiling point 1. Temperature at which vapor pressure of a liquid equals the external pressure pushing on the surface, and all of the liquid can change into the vapor phase 2. Normal boiling point – external pressure = 1 atm 3. External pressure < 1 atm – liquid boils at a lower temperature 4. External pressure > 1 atm – liquid boils at a higher temperature 3 11.3. Phase Changes between Solids, Liquids, and Gases A. Phase changes (changes of state) 1. Physical form but not chemical identity of a substance changes 2. Matter in any one state can change into either of the other two a. Fusion (melting) – solid to liquid b. Freezing – liquid to solid c. Vaporization – liquid to gas d. Condensation – gas to liquid e. Sublimation – solid changes directly into a gas f. Deposition – gas changes directly into a solid B. Phase change associated with free-energy change, G 1. G = H – TS 2. Enthalpy part – heat flow associated with making or breaking intermolecular attractions holding liquids and solids together 3. Entropy part – associated with change in molecular randomness between various states 4. Solid → liquid, solid → gas, liquid → gas: H and S both positive 5. Gas → liquid, gas → solid, liquid → solid: H and S both negative 6. Knowing H and S for a phase transition, can calculate temperature at which two phases are in equilibrium. a. At equilibrium, G = 0 b. T = H/S 11.4. Energy Changes During Phase Transitions A. Heating curve – graphical display of results of adding heat to a sample (See textbook Figure 11.6.)

B. Heating solid H2O 1. Added heat increases the temperature of the solid.

2. Molar heat capacity relates the amount of added heat to the temperature change.

C. Melting solid H2O (fusion) 1. Melting point – temperature at which solid and liquid coexist in equilibrium as molecules break free from their position in the crystal and enter the liquid phase 2. Heat of fusion (Hfusion) – amount of energy required to overcome enough intermolecular forces to convert solid into liquid 3. Temperature remains constant D. Heating liquid H2O E. Vaporizing liquid H2O 1. Boiling point - temperature at which liquid and vapor coexist in equilibrium as molecules break free from the surface of the liquid and enter the gas phase 2. Heat of vaporization (Hvap) – amount of energy necessary to convert liquid into gas F. Heating H2O vapor

3Test Item File Questions: Multiple Choice 4 – 12, Algorithmic 4 – 9, 14, Short Answer 3 – 4,

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Chapter 11—Liquids and Phase Changes G. Hvap >> Hfusion 1. For vaporization, must overcome all intermolecular forces in compound 2. For fusion, must overcome fewer intermolecular forces 4 11.5. Phase Diagrams A. Change any one state of matter spontaneously into either of the other two, depending on temperature and pressure B. Phase diagram – graphical method of illustrating pressure and temperature dependencies of a pure substance in a closed system 1. Boundary line – points on this line represent pressure/temperature combinations at which two phases are in equilibrium 2. Triple point – a unique combination of pressure and temperature at which all three phases coexist in equilibrium 3. Normal melting point – melting point at 1 atm external pressue 4. Critical point – end of the vaporization curve 5. Critical temperature – temperature beyond which a gas cannot be liquefied 6. Critical pressure – pressure beyond which a liquid cannot be vaporized 7. Supercritical fluid – substance that is neither liquid nor gas a. Pressure of a gas at the critical point is so high and the molecules are so close together that it is difficult to distinguish between gas and liquid. b. Temperature of a liquid at the critical point is so high that it is hard to distinguish between liquid and gas.

8. Effect of pressure on the slope of solid/liquid boundary line depends on the relative densities of the solid and liquid phases 11.6. Liquid Crystals A. Intermediate phase 1. Exhibits properties of both liquids and solids 2. Ordered like a solid, flows like a liquid B. Organic molecules with elongated rod-like shape C. Classified by orientation pattern 1. Nematic – long axes of molecules are approximately parallel but ends do not line up 2. Smectic – long axes are parallel and ends aligned D. Have anisotropic properties which depend on the directional orientation of the molecules.

Inquiry – How is caffeine removed from coffee?

4Test Item File Questions: Multiple Choice 36 – 40, Algorithmic 15

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Chapter 11—Liquids and Phase Changes

CHAPTER 11 LIQUIDS AND PHASE CHANGES Teaching Tips, Points of Emphasis, and Common Misconceptions Section 11.2

The Clausius–Clapeyron equation is based on experimental observation relating an independent variable, temperature, and a dependent variable, vapor pressure of a particular liquid.

Section 11.3

A phase change usually refers to what occurs when a solid, liquid, or gas is converted to one of the other two states. Sometimes more than one solid state exists for a substance, such as the metallic and nonmetallic forms of solid tin, and a phase change can occur between the two.

Section 11.3

On a heating curve for a pure substance, the temperature remains constant during a phase change. Within a given state, temperature rises with a slope that depends on the heat capacity of the substance in that state.

Lecture/Laboratory Demonstration References Section 11.2

J. A. Sears and C. J. Grieve, “A Vapor Pressure Demonstration,” J. Chem. Educ., Vol. 67, 1990, 427. Hexane is introduced into a sealed Erlenmeyer flask containing hot water and fitted with a delivery tube. The hexane is vaporized by the hot water, forcing water from the flask into a graduated cylinder. As the flask cools, water is drawn back into the flask, showing that the vapor pressure of the hexane decreases with a decrease in temperature.

Section 11.2

Lee R. Summerlin and James L. Ealy, Jr., “The Effect of Pressure on Boiling Point,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), p. 21.

Section 11.5

Stephen J. Hawkes, “A Simple Experiment for the Demonstration of the Phase Diagram of Carbon Dioxide,” J. Chem. Educ., Vol. 73, 1996, 837–838.

Literature References Section 11.1

Todd P. Silverstein, “Why Do Alcoholic Beverages Have Legs?” J. Chem. Educ., Vol. 75, 1998, 723–724.

Section 11.5

Nicholas K. Kildahl, “Journey Around a Phase Diagram,” J. Chem. Educ., Vol. 71, 1994, 1052–1054.

Section 11.5

Stephen J. Hawkes, “There Is No Perceptible Inflection at the Triple Point,” J. Chem.

Educ., Vol. 76, 1999, 226.

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Chapter 11—Liquids and Phase Changes

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 11.4

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Worked Example 11.3: Calculating the amount of heat of a temperature change

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CHAPTER 12 SOLIDS AND SOLID-STATE MATERIALS Chapter Learning Objectives Section 12.1

Classify types of solids based on their chemical composition and properties.

Section 12.2

Use the Bragg equation to calculate spacing between atomic layers in a crystal.

Section 12.3 through Section 12.5

Identify the four kinds of spherical packing arrangements in crystalline solids and the three kinds of cubic unit cells.

Calculate the density of a substance, atomic radii of its atoms or molecular mass given its unit cell dimensions.

Use an image of a unit cell to determine the formula and geometry of ionic compounds.

Section 12.6

Describe the electron-sea model of metals, and explain how it accounts for their properties.

Describe the molecular orbital theory for metals (band theory), draw MO energylevel diagrams for metals, and use band theory to account for their properties.

Section 12.7

Draw MO diagrams for insulators, semiconductors, and conductors. Use band theory to explain why different substances have different conductivities.

Classify doped semiconductors as n-type or p-type.

Predict relative band-gap energies based on periodic trends in atomic size and electronegativity.

Relate the wavelength of light emitted or absorbed by a semiconductor to the band gap in its band theory energy-level diagram.

Section 12.8

Describe some semiconductor applications.

Section 12.9

Describe the appearance of a plot of electrical resistance versus temperature for a superconductor.

Section 12.10

Describe the bonding in ceramics and account for their properties in terms of the bonding model.

Describe the synthesis of ceramics, and write balanced equations for the reactions that occur in the sol–gel synthesis method.

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Chapter 12—Solids and Solid-State Materials

Lecture Outline 1

12.1. Types of Solids A. Crystalline solids 1. Solids with atoms, ions, or molecules in an ordered, long-range arrangement 2. Macroscopically – observe flat faces and sharp angles 3. Types of crystalline solids – textbook Table 12.1 a. Ionic solids i. Constituent particles are ions ii. Example: NaCl iii. Ordered into a 3-D arrangement held together by ionic bonds b. Molecular solids i. Constituent particles are molecules held together by intermolecular forces. ii. Examples: sucrose and ice c. Covalent network solids i. Atoms linked together by covalent bonds into a giant 3-D array ii. Examples: diamond and quartz d. Metallic solids i. Large arrays of metal atoms ii. Examples: Ag and Fe iii. Have metallic properties B. Amorphous solids 1. Constituent particles randomly arranged 2. No ordered long-range structure 2 12.2. Probing the Structure of Solids: X–Ray Crystallography A. Diffraction 1. Beam of electromagnetic radiation scattered by an object containing regularly spaced lines or points (e.g., atoms in a crystal) 2. Spacing must be comparable to the  of the radiation 3. Interference between two ’s passing through the same region of space at the same time a. Constructive interference i. Waves in-phase (peak-to-peak and trough-to-trough) ii. Increases intensity of wave b. Destructive interference i. Waves out-of-phase ii. Waves cancel B. Bragg analysis 1. X-rays diffracted by different layers of atoms in the crystal undergo constructive and destructive interference nl 2. Bragg equation: nl = 2d ´ sin q; d = 2 sin q a.  = known b. Sin  = angle at which incoming rays are reflected (can be measured) c. n = an integer (usually 1)

1Test Item File Questions: Multiple Choice 1 – 2, Algorithmic 1 – 10, Short Answer 2 – 3 2Test Item File Questions: Multiple Choice 3 – 4, Short Answer 4– 5

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Chapter 12—Solids and Solid-State Materials 3

12.3. The Packing of Spheres in Crystalline Solids: Unit Cells A. Particles pack together in crystals so that they can be as close together as possible to maximize intermolecular attractions.

B. Simple cubic packing – spheres in one layer sit directly on top of those in the previous layer 1. All layers identical 2. Primitive-cubic unit cell 3. Coordination number = 6: sphere touches four neighbors in the same layer, one above and one below 4. Uses only 52% of available volume C. Body-centered cubic packing – spheres in alternate layers in an a-b-a-b arrangement where spheres in b layers fit into small depressions between spheres in neighboring a layers 1. Body-centered cubic cell 2. Coordination number = 8; four neighbors above and four neighbors below 3. Occupies 68% of available volume D. Hexagonal closest-packed – noncubic unit cell with two alternating layers (a-b-a-b) 1. Hexagonal arrangement of touching spheres 2. Spheres in b layer fit into small triangular depressions between spheres in a layer.

3. Coordination number = 12: six neighbors in the same layer, three above and three below 4. Uses 74% of available volume E. Cubic closest-packed – face-centered cubic unit cell with three alternating layers, a-b-c-a-b-c 1. a-b layers identical to hexagonal closest-packed 2. Third layer offset from both a and b 3. Coordination number = 12 4. Uses 74% of available volume F. Unit cells – small repeating units found in crystals 1. Symmetrical geometries 2. Stack to minimize space 3. Fourteen different geometries a. Parallelepipeds (six-sided geometric solids whose faces are parallelograms) b. Differ in lengths of cell edges and angles between edges G. Cubic cells – all edges equal in length, and all angles 90 1. Primitive-cubic unit cell for metals a. An atom at each of the eight corners b. Each atom shared with seven other neighboring cubes that come together at the same point c. One-eighth of each corner atom belongs to a single cube 2. Body-centered cubic unit cell – additional atom in the center of the cube 3. Face-centered cubic unit cell – additional atom on each of the six cube faces a. Shared with one other neighboring cube b. One-half of each face atom belongs to a single cube 4 12.4. Structures of Some Ionic Solids A. Spheres not all the same size – anions larger than cations B. Adopt a variety of different unit cells depending on ion sizes and charges 1. Face-centered cubic – NaCl, KCl. 2. Other common ionic unit cells – textbook Figure 12.12

3Test Item File Questions: Multiple Choice 8 – 21, Algorithmic 11 – 17, Short Answer 6 – 11 4Test Item File Questions: Multiple Choice 22 – 27, Short Answer 12

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Chapter 12—Solids and Solid-State Materials 5

12.5. Structure of Some Covalent Network Solids A. Carbon 1. Allotropes a. Different structural forms of the same element b. Differ in physical and chemical properties c. More than 40 forms of carbon, most are amorphous 2. Diamond a. Each carbon atom sp3 hybridized b. Covalently bonded with tetrahedral geometries 3. Graphite a. Two-dimensional sheets of fused six-membered rings b. Each carbon sp2 hybridized 4. Fullerene a. Spherical C60 molecule b. Shape of a soccer ball B. Silica (SiO2) 1. Four single bonds between silicon and four oxygens in a covalent network structure 2. Quartz glass – result of heating silica above 1600 C then cooling the viscous liquid a. Si–O bonds reform in a random arrangement b. Amorphous solid c. Mix in additives – prepare a wide variety of glass i. Window glass – add CaCO3 and Na2CO3 ii. Colored glass – add transition metal ions iii. Borosilicate glass (Pyrex) – add B2O3; resistant to thermal shock because it doesn’t expand much on heating 6 12.6. Bonding in Metals A. Metals – malleable, ductile, and good conductors of heat and electricity 1. Malleable – can be hammered into sheets 2. Ductile – can be drawn into wires B. Can understand properties by examining bonding in metals C. Valence electrons 1. Can’t be localized in a bond between any particular pair of atoms 2. Delocalized and belong to the crystal as a whole D. Electron-sea model – crystal is a three-dimensional array of metal cations immersed in a sea of delocalized electrons free to move throughout the crystal 1. Delocalized, mobile valence electrons act as electrostatic glue holding metal cations together.

2. Simple qualitative explanation for electrical and thermal conductivity of metals a. Mobile electrons free to move from a negative electrode to a positive electrode when subjected to an electrical potential b. Mobile electrons conduct heat by carrying kinetic energy from one part of the crystal to another. c. Electrons extend in all directions, therefore, when a metallic crystal is deformed, no localized bonds are broken. d. Explains malleability and ductility

5Test Item File Questions: Multiple Choice 41, Algorithmic 18 – 20 6Test Item File Questions: Multiple Choice 43 – 49, Algorithmic 22 – 23, Short Answer 13, 15, 16

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Chapter 12—Solids and Solid-State Materials E. Molecular orbital theory for metals 1. Number of molecular orbitals formed equals number of atomic orbitals combined 2. Difference in energy between successive MOs decreases as number of atoms increases. a. MOs merge into an almost continuous band of energy levels. b. MO theory for metals is called band theory. c. Bottom half of band consists of bonding MOs. d. Top half of band consists of antibonding MOs. 3. Each electron in a metal has a discrete kinetic energy and a discrete velocity. a. Depends on the particular MO b. Increases from bottom to the top of a band 4. One-dimensional metal wire or three-dimensional crystal a. Energy levels within a band occur in degenerate pairs (textbook Figure 12.17). i. One set – electrons moving to the right ii. Other set – electrons moving to the left b. No electrical potential i. No net electric current in either direction ii. Two sets equally populated c. Electrical potential i. Electrons moving right (toward the + battery terminal) are accelerated while electrons moving left (toward the − battery terminal) are decelerated. ii. Electrons moving very slowly left, change direction iii. Number of electrons moving right > number of electrons moving left iv. Net electric current 5. Electrical potential can shift electrons from one set of energy levels to the other only if the band is partially filled. a. Electrical insulators — materials that have only completely filled bands b. Metals — materials that have partially filled bands c. Can have s and p subshells sufficiently close in energy so that the s and p bands overlap, resulting in a partially filled composite band 6. Transition metals – d band overlaps the s band a. Composite band has six MOs per metal atom. b. Maximum bonding for metals i. Six valence electrons per metal atom ii. Near group 6B c. Causes melting points of transition metals to be at a maximum at group 6B 7 12.7. Semiconductors A. Semiconductor – a material with an electrical conductivity intermediate between that of a metal and that of an insulator B. Valence band – the bonding MOs C. Conduction band – the higher-energy, antibonding MOs D. Band gap – the separation of the valence band and the conduction band in terms of energy E. Insulator 1. No vacant MOs in the valence band; large band gap 2. Can’t excite electrons within the valence band 3. Can’t excite electrons to vacant MOs in the conduction band due to the large band gap F. Metal conductors 1. No energy gap between the highest occupied and lowest unoccupied MOs 2. Electrical conductivity decreases with increasing temperature – increased vibrational motion of cations interferes with flow of electrons

7Test Item File Questions: Multiple Choice 50 – 61, Algorithmic 24 – 26, Short Answer 14, 17 – 19

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Chapter 12—Solids and Solid-State Materials G. Semiconductor – smaller band gap than insulators 1. A few electrons can jump the gap and occupy the higher-energy conduction band 2. Partially filled conduction band and valence band – unoccupied MOs in valence band result from electrons that jumped the gap 3. Electrical potential can accelerate the electrons in the partially filled bands – semiconductor conducts a small current 4. Electrical conductivity increases with increasing temperature – electrons with enough energy to jump the band gap increase H. Best criterion for distinguishing a metal from a semiconductor – temperature-dependence of electrical conductivity I. Doping – a process in which conductivity of a semiconductor is increased by adding certain impurities in small amounts 1. Add an impurity with more electrons than the semiconductor a. n-type semiconductor b. Charge carriers are negative electrons.

2. Example of n-type semiconductor: Si doped with P a. Each P atom occupies a Si position and introduces an extra electron. b. Extra electron occupies the conduction band. c. More electrons in conduction band cause the conductivity to be higher.

3. Add an impurity with fewer electrons than the semiconductor a. p-type semiconductor b. Creates positive holes in the valence band 4. Essential components in modern solid-state electronic devices.

8 12.8. Semiconductor Applications A. Diodes 1. Convert alternating current to direct current 2. Permit electron flow in only one direction 3. Constructed of a p-type semiconductor in contact with an n-type semiconductor – called a p–n junction B. Light-emitting diodes – LEDs 1. Electrons and holes combine at p–n-junction a. Energy released as light b. Wavelength of light roughly equal to band gap energy 2. The band gap and thus the color of the light emitted by an LED can be tuned by varying the semiconductor composition guided by period trends involving atomic radii and electronegativity.

3. LED comparison to incandescent lightbulbs – smaller, brighter, longer lived, and faster switching times C. Photovoltaic cells 1. Solar cells 2. p–n junctions 3. Covert sunlight into energy D. Transistors 1. Control and amplify electrical signals in integrated circuits 2. n–p–n or p–n–p junctions

8Test Item File Questions: Multiple Choice 62 – 63, Short Answer 20 – 25,

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Chapter 12—Solids and Solid-State Materials 9

12.9. Superconductors A. Superconductor 1. A material that loses all electrical resistance below a characteristic temperature, the superconducting transition temperature (Tc) 2. Below Tc, once an electric current is started, it will flow indefinitely without loss of energy.

B. 1986 – BaxLa2–xCuO4 found to have a Tc = 35 K 1. Soon after, other copper-containing oxides with even higher Tc discovered 2. Ceramics a. Nonmetallic inorganic solids b. Most are electrical insulators C. No generally accepted theory of superconductivity in ceramic superconductors – infinitely extended layers of Cu and O and the fractional oxidation number of Cu appear to play a role in the current flow D. Superconductors can levitate a magnet – cool a superconductor below Tc and lower a magnet toward it, the magnet and superconductor repel each other.

E. Present applications of conventional superconductors require the use of liquid helium (expensive and requires cryogenic equipment).

1. Can use liquid N2 with new higher-temperature superconductors 2. Abundant refrigerant that is cheaper than milk F. Problems with high-temperature superconductors 1. Brittle powders with high melting points 2. Not easily made into wires and coils needed for electrical equipment 3. Commercial applications are emerging.

G. High-temperature superconductors based on fullerene – the allotrope of carbon that contains C60 molecules 1. K3C60 – metallic conductor at room temperature; superconductor at 19.5 K, related fullerides have even higher Tc values 2. Fullerides are three-dimensional superconductors.

10 12.10. Ceramics and Composites A. Ceramics – inorganic, nonmetallic, nonmolecular solids, including both crystalline and amorphous materials, such as glasses 1. Traditional silicate ceramics – made by heating aluminosilicate clays to high temperatures 2. Advanced ceramics – materials that have high-tech engineering, electronic, and biomedical applications a. Oxide ceramics – alumina (Al2O3) b. Nonoxide ceramics – silicon carbide and silicon nitride B. Properties of ceramics – superior to those of metals 1. Higher melting points 2. Stiffer, harder, and more resistant to wear and corrosion 3. Maintain their strength at high temperatures 4. Less dense than steel – attractive lightweight, high-temperature materials for replacing metal components in aircraft, space vehicles, and automobiles C. Strong chemical bonding 1. Covalent network solid a. Highly directional covalent bonds prevent planes of atoms from sliding over one another when the solid is subjected to the stress of a load or an impact. b. Solid can’t deform to relieve the stress. c. Maintains its shape up to a point, but then the bonds give way suddenly

9Test Item File Questions: Multiple Choice 64 – 71, Algorithmic 27 – 28, Short Answer 26 10Test Item File Questions: Multiple Choice 72 – 79, Algorithmic 29 – 31, Short Answer 27 - 28

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Chapter 12—Solids and Solid-State Materials 2. Oxide ceramics – bonding largely ionic a. Behave similarly to covalent network solids 3. Metals – deform under stress because planes of metal cations slide easily in the electron sea 4. Metals dent; ceramics shatter D. Ceramic processing – the series of steps that leads from raw material to the finished ceramic object 1. Determines product strength and resistance to fracture 2. Sintering –particles of powder welded together without completely melting a. Occurs below the melting point b. Crystal grains grow larger and the material density increases as void spaces between particles disappear. c. Need high-purity, fine powders that are tightly compacted prior to sintering – impurities and remaining voids can lead to microscopic cracks E. Sol–gel method – method used for preparing high-purity, fine powders that can be tightly compacted 1. Synthesis of a metal oxide powder from a metal alkoxide (a compound derived from a metal and an alcohol) 2. Sol – colloidal dispersion consisting of extremely fine particles having a diameter of only 0.001 to 0.1 m 3. Gel – a rigid, gelatin-like material F. Ceramic composite – a hybrid material in which a ceramic powder is mixed, prior to sintering, with fibers of a second ceramic material 1. Combines advantageous properties of both components 2. Whiskers – tiny fiber-shaped particles that are very strong because they are single crystals G. Fibers and whiskers increase the strength and fracture toughness of composite materials.

1. Most of the chemical bonds are aligned along the fiber axis, giving the fibers great strength.

2. Can deflect cracks – prevents cracks from moving cleanly in one direction 3. Can bridge cracks – holds two sides of a crack together H. Composites in which the two phases are different types of materials 1. Ceramic–metal composites (cermets) 2. Ceramic–polymer composites 3. High strength-to-weight ratios I. Ceramic fibers used in composites usually made by high-temperature methods Inquiry – What are quantum dots and what controls their color?

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Chapter 12—Solids and Solid-State Materials

CHAPTER 12 SOLIDS AND SOLID-STATE MATERIALS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 12.2

The variable n in the Bragg equation is the order of diffraction. After a rainfall, double or triple rainbows are sometimes observed with each successive rainbow fainter than the previous one. The most intense spectrum is due to a first-order diffraction. The fainter spectra are due to higher orders of diffraction.

Section 12.3

Students tend to consider the center atom of a body-centered cubic cell as different from the corner atoms. All atoms in the unit cell are identical. The unit cell can be reconfigured so that the atom originally in the center is positioned on a corner and vice versa.

Section 12.3

Primitive cubic is also known as simple cubic.

Section 12.3

Atoms that belong to more than one unit cell are classified as corner, edge, or face atoms, which contribute, respectively, 1/8, 1/4, and 1/2 atom per unit cell.

Section 12.3

Visualizing three-dimensional unit cells is difficult for many students. It is helpful to introduce the unit cell using a two-dimensional analog, such as a square array and a closest packed array of pennies laid flat on a table. The number of pennies per unit cell, the coordination number of each penny, and the percent empty space can be determined.

Section 12.3

When determining the density of a metal, students sometimes use the volume of the atoms instead of the volume of the entire unit cell. The density of a substance includes the empty space between atoms.

Section 12.4

The ratio of cations to anions in a unit cell must be consistent with the empirical formula of the ionic compound, as predicted in Chapter 2.

Section 12.6

In the electron-sea model, electrons are delocalized and free to move throughout the entire metal.

Section 12.6

The number of molecular orbitals equals the number of atomic orbitals from which the molecular orbitals were formed.

Section 12.6

Metals have electrons in a partially filled band or in a filled band that overlaps with a second, empty band to give a partially filled composite band. Insulators have electrons in completely filled bands that do not overlap with empty bands.

Section 12.6

Population of the lower (bonding) portions of a band by electrons strengthens metallic bonding, whereas population of the upper (antibonding) portion of a band weakens metallic bonding.

Section 12.7

Silicon and germanium are the two most common semiconductors.

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Chapter 12—Solids and Solid-State Materials Section 12.7

A semiconductor has a band gap between its valence and conduction bands that is small enough to allow a small equilibrium thermal electron population in the conduction band, leaving holes in the valence band. These electrons and holes can then move when a potential is applied.

Section 12.7

The electrical conductivity of a metal decreases with an increase in temperature, whereas the electrical conductivity of a semiconductor increases with an increase in temperature.

Section 12.7

Doping a semiconductor enhances its electrical conductivity by increasing the population of electrons and/or positive holes in the partially filled bands.

Section 12.10

A sol is a colloid of finely divided metal hydroxide particles. Removing water from a sol converts some hydroxide groups to oxide bridges, resulting in a gel. Further removal of water yields a metal oxide ceramic powder.

Section 12.10

Composites are hybrid materials of two ceramics, a ceramic and a metal, or a ceramic and a polymer.

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Lecture/Laboratory Demonstration References None.

Literature References Section 12.1

John J. Fortman, “Pictorial Analogies II: Types of Solids,” J. Chem. Educ., Vol. 70, 1993, 57-58.

Section 12.3

Keenan E. Dungey, George Lisensky, and S. Michael Condren, “Kixium Monolayers: A Simple Alternative to the Bubble Raft Model for Close-Packed Spheres,” J. Chem. Educ., Vol. 76, 1999, 618-619.

Section 12.3

Daryl L. Ostercamp, “Close Packing of Identical Spheres,” J. Chem. Educ., Vol. 69, 1992, 162.

Section 12.3

Frank Galasso, “The Importance of Understanding Structure,” J. Chem. Educ., Vol. 70, 1993, 287-290. A discussion of the relationship between unit cells and the structure of solids.

Section 12.4

James P. Birk, “A Model to Illustrate the Brittleness of Ionic and Metallic Crystals,” J. Chem. Educ., Vol. 62, 1985, 667.

Section 12.5

Robert F. Curl and Richard E. Smalley, “Fullerenes,” Scientific American, October 1991, 54-63.

Section 12.6

Peter P. Edwards and M. J. Sienko, “On the Occurrence of Metallic Character in the Periodic Table of the Elements,” J. Chem. Educ., Vol. 60, 1983, 691-696.

Section 12.6

Joseph F. Lomax, “Conducting Midshipmen – A Classroom Activity Modeling Extended Bonding in Solids,” J. Chem. Educ., Vol. 69, 1992, 794-795.

Section 12.6

Frank Rioux, “An Ionic Model for Metallic Bonding,” J. Chem. Educ., Vol. 62, 1985, 383-384.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 12.3

Worked Example 12.3: Using unit-cell dimensions to calculate the density of a metal

Section 12.6

Worked Example 12.4: Using band theory to account for melting points

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CHAPTER 14 CHEMICAL KINETICS Chapter Learning Objectives Section 14.1

Determine the average reaction rate over a specified period of time and estimate instantaneous reaction rate from a graph of concentration versus time.

Relate the rate of consumption of any reactant to the rate of formation of any product using reaction stoichiometry.

Section 14.2 and Section 14.3

Find the reaction order with respect to each reactant, the overall reaction order, and the units of the rate constant from a given rate law.

Predict the change in reaction rate when the concentration of a reactant changes by a specified amount.

Determine the rate law using the Method of Initial Rates.

Determine the rate law and rate constant using initial rate and concentration data.

Section 14.4 through Section 14.6

Use the integrated rate law to determine the half-life and the concentrations remaining at various times for a zeroth order reaction.

Use the integrated rate law to determine the half-life and the concentrations remaining at various times for a first order reaction.

Use the integrated rate law to determine the half-life and the concentrations remaining at various times for a second order reaction.

Determine the reaction order and rate constant graphically.

Section 14.7 and Section 14.8

Interpret potential energy diagrams and suggest geometries for successful collisions and transition states using collision theory.

Use the Arrhenius equation to calculate the activation energy of a reaction given rate constants at different temperatures.

Section 14.9 and Section 14.10

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Given a reaction mechanism, write the overall reaction, identify intermediates, and determine the molecularity for each elementary step.

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Chapter 14—Chemical Kinetics Section 14.11

For mechanisms with an initial slow step (a) predict the rate law for the overall reaction or (b) propose a mechanism that agrees with the experimental rate law.

For mechanisms with an initial fast equilibrium step, predict the rate law.

Section 14.12 and Section 14.13

Use a molecular diagram for a catalyzed reaction to determine the rate law and propose a mechanism consistent with the rate law.

Identify catalysts and intermediates in the mechanism of a catalyzed reaction.

Interpret a potential energy diagram representing the mechanism in a catalyzed reaction.

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Chapter 14—Chemical Kinetics

Lecture Outline 1

14.1. Reaction Rates A. Rate of a reaction – how fast the concentration of a reactant or a product changes per unit time D concentration Rate = D time 1. Increase in the concentration of a product per unit time 2. Decrease in the concentration of a reactant per unit time 3. Units: M/s or mol/(L  s) a. Allows rate to be independent of the scale of the reaction b. Use a minus sign in calculating rate of disappearance of reactant B. Relative rates of product formation and reactant consumption 1. Depend on coefficients in the balanced equation 2. Must specify the reactant or product when quoting a rate C. Rate changes as the reaction proceeds.

1. Must specify the time 2. Reaction rates decrease as the reaction mixture runs out of reactants.

3. General rate of a reaction is defined as the rate of consumption of a reactant or formation of a product divided by its coefficient in the balanced chemical equation.

D. Plot concentration (y-axis) versus time (x-axis).

1. [concentration] and time represent vertical and horizontal sides of a right triangle.

2. Slope of the triangle hypotenuse is the average rate during that time period.

E. Instantaneous rate at time t – the slope of the tangent to a concentration-time curve at time t F. Initial rate – the instantaneous rate at the beginning of a reaction (t = 0) 2 14.2. Rate Laws and Reaction Order A. Rate law – states the dependence of the reaction rate on concentration 1. Equation shows how the rate depends on the concentration of each reactant.

2. For the reaction a A + b B → products m n a. Rate law: Rate = k  A   B b. k = proportionality constant called the rate constant B. Reaction order – determined by the values of the exponents 1. Values of exponents in the rate law must be determined by experiment; they cannot be deduced from the stoichiometry of the reaction.

2. Values of m and n indicate reaction order with respect to A and B. a. Exponent = 1: first order b. Exponent = 2: second order c. Exponent = 3: third order 3. Overall reaction order = m + n 4. Indicates how change in concentration affects rate 5. Unrelated to coefficients in the balanced equation 6. Usually small positive integers, but can be negative, zero, or even fractions a. Exponent = 1: rate depends linearly on the concentration of the corresponding reactant b. Exponent = 0: rate is independent of the concentration of the corresponding reactant c. Exponent < 1: rate decreases as concentration of the corresponding reactant increases

1Test Item File Questions: Multiple Choice 6 – 13, 29, 38, Algorithmic 1 – 8, Short Answer 1 – 3 2Test Item File Questions: Multiple Choice 30 – 36, Algorithmic 9 – 20, Short Answer 4 – 10

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14.3. Method of Initial Rates: Experimental Determination of a Rate Law A. Determination of reaction order (values of the exponents in a rate law) – measure initial rate of a reaction as a function of different sets of initial concentrations.

1. Design pairs of experiments to investigate the effect of initial concentration of a single reactant on initial rate of change.

2. If, by doubling the concentration of a reactant, the rate also doubles, then the reaction is first order with respect to that reactant.

3. If, by doubling the concentration of a reactant, the rate increases by a factor of 2 2 = 4, then the reaction is second order with respect to that reactant.

4. Use initial rates to avoid complications from the reverse reaction. a. Initial rates measure only the rate of the forward reaction. b. Only reactants and catalysts appear in the rate law.

B. Can determine the value of k from the rate law.

1. Characteristic of a reaction 2. Depends on temperature 3. Does not depend on concentration 4. Units depend on the number of concentration terms in the rate law and values of the exponents.

4 14.4. Integrated Rate Law: Zeroth-Order Reactions A. Integrated rate law 1. A concentration–time equation that allows us to calculate the concentration of a reactant at any time t or the fraction of a reactant that remains at any time t.

2. Can be used to calculate the time required for the initial concentration of a reactant to drop to any particular value or to any particular fraction of its initial concentration.

B. Integrated rate law for a zeroth-order reaction, A → products [A] = −kt + [A]o C. Rate is independent of the concentration of the reactant.

D. Plot of [A] versus time gives a straight line if zeroth order.

1. Slope = –k 2. Intercept = [A]o E. Relatively uncommon; occurs under special circumstances 5 14.5. Integrated Rate Law: First-Order Reactions A. For the reaction A → products, the integrated rate law is [A] In = − kt [A] 1. Can rearrange the equation to give ln[A]t = −kt + ln[A]o 2. Plot of ln[A]t versus time gives a straight line if the reaction is first order in A. 3. k = −(slope) B. Half-life = t1/2 1. Time required for reactant concentration to drop to one-half its initial value 0.693 t1/2 = k 2. For a first-order reaction half-life is constant.

3. Depends only on the rate constant 6 14.6. Integrated Rate Law: Second-Order Reactions

3Test Item File Questions: Multiple Choice 39 – 48, Short Answer 11 4Test Item File Questions: Multiple Choice 51 – 52, Algorithmic 21 – 23, Short Answer 12 5Test Item File Questions: Multiple Choice 61 – 72, Algorithmic 24 – 31, Short Answer 13 – 14

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Chapter 14—Chemical Kinetics A. Integrated rate law for a second-order reaction 1 1 = kt + [A]t [A]o B. Plot of 1/[A]t versus time gives a straight line if the reaction is second order.

1. Slope of line = k 2. Intercept = 1/[A]o C. Half-life for a second-order reaction – depends on both the rate constant and the initial concentration.

1 t1/2 = k[A]o 7

14.7. Reaction Rates and Temperature: The Arrhenius Equation A. Reaction rates tend to double when the temperature is increased by 10 C. B. Collision theory model: a bimolecular reaction occurs when two properly oriented reactant molecules come together in a sufficiently energetic collision.

C. For the reaction: A + BC → AB + C 1. For a single-step reaction, a new bond, A–B, develops at the same time as the old bond, B–C, breaks.

2. Nuclei pass through a configuration in which all three atoms are weakly linked together. a. A + B–C → A—B- -C → A–B + C 3. Need energy to overcome repulsions a. Comes from kinetic energy of the colliding particles b. Stored as potential energy in A—B—C. c. A—B—C has more potential energy than either the reactants or products.

4. A potential energy barrier exists that must be surmounted before reactants can be converted to products. a. Potential energy profile – plot of potential energy versus reaction progress b. Activation energy, Ea, – the height of the barrier c. Transition state (or activated complex) – configuration of atoms at the maximum in the potential energy profile 5. All the energy needed to climb the potential energy barrier must come from the kinetic energy of the colliding molecules.

D. Comparison of collision rates and reaction rates leads to experimental evidence for the idea of an activation energy barrier.

1. Only a small fraction of collisions lead to reaction. a. Very few collisions occur with a kinetic energy as large as the activation energy. b. Area under the curve to the right of Ea represents the fraction of the collision with an energy equal to or greater than the activation energy. c. Fraction of molecules having sufficient energy for reaction

f = e− Ea / RT 2. Distribution of collision energies broadens and shifts to higher energies with an increase in temperature.

3. Exponential increase in the fraction of collisions that lead to products 4. Accounts for exponential dependence of reaction rates on temperature 5. Explains why reaction rates are so much lower than collision rates.

6Test Item File Questions: Multiple Choice 73 – 80, Algorithmic 33 – 36, Short Answer 15 7Test Item File Questions: Multiple Choice 81 – 89, Algorithmic 38 – 39

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Chapter 14—Chemical Kinetics E. Orientation requirement also reduces the fraction of collisions that lead to products.

1. Reactants won’t react unless the orientation of the reaction partners is correct for formation of the transition state.

2. Steric factor, p – fraction of collisions having proper orientation F. For a bimolecular collision between A and B. 1. Collision rate = Z [A][B]; Z is a constant related to collision frequency.

2. Reaction rate < collision rate by a factor of p  f.

3. Reaction rate = p f Z [A][B] 4. Reaction rate = k[A][B] 5. Rate constant, k k = pfZ = pZe− E2 / RT G. Arrhenius equation k = Ae

- Ea /R T

where A = frequency factor = pZ 8

14.8. Using the Arrhenius Equation A. Can determine the activation energy if values of the rate constant are known at different temperatures B. Rearrange the equation to get

æ -E ö æ 1 ö ln k = çç a ÷÷ ç ÷ + ln A è R øèT ø 1. Plot of ln k versus 1/T gives a straight line.

2. slope = –Ea/R C. Can estimate the activation energy from rate constants at just two temperatures, using another form of the equation.

æ k ö æ -E ö æ 1 1 ö ln çç 2 ÷÷ = çç a ÷÷ çç - ÷÷ è k ø è R ø è T2 T1 ø 9

14.9. Reaction Mechanisms A. Reaction mechanism – the sequence of molecular events, or reaction steps, that defines the pathway from reactants to products 1. Reaction steps – involve breaking of chemical bonds and/or making of new bonds 2. Knowing reaction mechanisms allows for better control of known reactions and predictions of new reactions.

B. Elementary step – a single step in a reaction mechanism 1. Description of an individual molecular event (collisions of individual molecules) 2. Classified on the basis of molecularity a. Molecularity – the number of molecules on the reactant side of the chemical equation b. Unimolecular reaction – elementary reaction that involves a single reactant molecule c. Bimolecular reaction – elementary reaction that results from energetic collisions between two reactant molecules d. Thermolecular reaction – involves three atoms or molecules; rare 3. Reaction intermediate – a species that is formed in one step of a reaction mechanism and consumed in a subsequent step a. Does not appear in the net equation for the overall reaction b. Presence noticed only in elementary steps

8Test Item File Questions: Multiple Choice 90 – 96, Algorithmic 40, Short Answer 16 – 18 9Test Item File Questions: Multiple Choice 97 – 100, Algorithmic 42 – 43, Short Answer 19 – 20

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Chapter 14—Chemical Kinetics C. Balanced equation for an overall reaction 1. Provides no information about how the reaction occurs 2. Merely describes reaction stoichiometry 10 14.10. Rate Laws for Elementary Reactions A. Rate law for overall reaction – determined by experimentation B. Rate law for an elementary reaction – determined from its molecularity 1. Elementary reaction is an individual molecular event.

2. Contains the concentration of each reactant raised to an exponent equal to its coefficient in the chemical equation for the elementary reaction 3. Applies only to elementary reactions, not overall reactions 4. Rate of a unimolecular reaction is first order in the concentration of the reactant molecule.

5. Overall reaction order for an elementary reaction is equal to its molecularity – textbook Table 13.5 11 14.11. Rate Laws for Overall Reactions A. Experimentally observed rate law for an overall reaction depends on the reaction mechanism.

1. If the overall reaction occurs in a single elementary step: experimental rate law = rate law for the elementary step 2. If the overall reaction occurs in two or more steps: a. Rate-determining step = the slowest step in a reaction mechanism b. Limits the rate at which reactants can be converted to products c. Overall reaction can occur no faster than the speed of the rate-determining step. d. A rate law derived for a reaction for which the slowest step is not the first step of the mechanism will require combining information from two or more steps in order for it to take the form of the experimentally determined rate law in which only initial reactants appear.

B. Two criteria for an acceptable reaction mechanism 1. Elementary steps must sum to give the overall reaction.

2. Mechanism must be consistent with the observed rate law for the overall reaction.

C. Procedure used for establishing a reaction mechanism 1. Determine the overall rate law experimentally.

2. Devise a series of elementary steps.

3. Predict the rate law based on the reaction mechanism.

4. If observed and predicted rate laws agree, the proposed mechanism is a plausible pathway for the reaction.

5. Easy to disprove a mechanism; impossible to “prove” a mechanism 12 14.12. Catalysis A. Catalyst – a substance that increases the rate of a reaction without being consumed in the reaction 1. Important in chemical industry and in living organisms 2. Chemical industry – favors formation of specific products and lower reaction temperatures 3. Living organisms – enzymes (catalysts) facilitate specific reactions of crucial biological importance.

B. Accelerates the rate of a reaction by making a new and more efficient pathway available for the conversion of reactants to products 1. Speeds up reaction in two ways a. Increases the frequency factor A b. Decreases the activation energy 2. Rate constant is more sensitive to the Ea, and a catalyst usually functions by lowering the activation energy.

10Test Item File Questions: Multiple Choice 105, Algorithmic 44 – 46 11Test Item File Questions: Multiple Choice 106 – 107, Algorithmic 47, Short Answer 21 12Test Item File Questions: Multiple Choice 108 – 110, Algorithmic 48 – 50, Short Answer 22

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14.13. Homogeneous and Heterogeneous Catalysts A. Homogeneous catalyst – one that exists in the same phase as the reactants B. Heterogeneous catalyst – exists in a different phase from the reactants 1. Mechanism complex and not well understood 2. Important steps a. Adsorption of reactants onto the surface of the catalyst b. Conversion of reactants to products on the surface c. Desorption of products from the surface 3. Adsorption steps a. Chemical bonding of reactants to the highly reactive metal atoms on the surface b. Breaking or weakening of bonds in the reactants 4. Industrial chemical processes use mostly heterogeneous catalysts due to the ease of separation of the catalyst from the reaction products.

5. Used in automobile catalytic converters Inquiry – How do enzymes work?

13Test Item File Questions: Algorithmic 52 – 53, Short Answer 23 – 24

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CHAPTER 14 CHEMICAL KINETICS Teaching Tips, Points of Emphasis, and Common Misconceptions None

Lecture/Laboratory Demonstration References Section 14.1

Bassam Z. Shakhashiri, “Color Variations of the Landolt Reaction,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 4 (The University of Wisconsin Press, Madison, 1992), pp. 26–28. Three pairs of colorless solutions are mixed. After 10 seconds the three mixtures turn red, yellow, and blue, respectively.

Each clock reaction is the reduction of iodate by bisulfite. The observed colors are related to the amount of triiodide produced, which is dependent upon the relative amounts of iodate and bisulfite reacted, and whether or not starch is present.

Section 14.1

Lee R. Summerlin and James L. Ealy, “The Starch-Iodine Clock Reaction,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 107–108. The iodine clock experiment.

Section 14.1

Bassam Z. Shakhashiri, “Hydrogen Peroxide Iodine Clock: Oxidation of Potassium Iodide by Hydrogen Peroxide,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 4 (The University of Wisconsin Press, Madison, 1992), pp. 37–43. The iodine clock experiment.

Section 14.1

Xavier Creary and Karen M. Morris, “A New Twist on the Iodine Clock Reaction: Determining the Order of a Reaction,” J. Chem. Educ., Vol. 76, 1999, 530–531.

Section 14.2

Steven W. Wright, “Tick Tock, a Vitamin C Clock,” J. Chem. Educ., Vol. 79, 2002, 40A-40B. Presents an activity that uses supermarket chemicals to perform a clock reaction in which the endpoint is signaled by an abrupt change in the appearance from colorless to blue-black. This activity can be used to explore reaction kinetics and the effect of reactant concentrations on the apparent rate of reaction.

Section 14.3

Bassam Z. Shakhashiri, “Hydrolysis of 2-chloro-2-methylpropane,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 4 (The University of Wisconsin Press, Madison, 1992), pp. 56–64. The time required for the color change in this reaction is shown to increase with increasing sodium hydroxide concentration, decrease with increasing 2-chloro-2-methylpropane concentration, and increase as the ratio of water/acetone increases. These observations are related to the mechanism of 2-chloro-2-methylpropane hydrolysis.

Section 14.7

Bassam Z. Shakhashiri, “Lightsticks,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 1 (The University of Wisconsin Press, Madison, 1983), pp. 146–152. Cyalume lightsticks are used to demonstrate the effect of temperature on reaction rates.

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Bassam Z. Shakhashiri, Lloyd G. Williams, Glen E. Dirreen, and Ann Francis, “Cool-Light Chemiluminescence,” J. Chem. Educ., Vol. 58, 1981, 70–72. Cyalume lightstick chemiluminescence reactions.

Section 14.11

Jose E. Cortes-Figueroa and Deborah A. Moore, “Using CBL Technology and a Graphing Calculator to Teach the Kinetics of Consecutive First-Order Reactions,” J. Chem. Educ., Vol. 76, 1999, 635–638.

Section 14.13

A. Sae, “Hydrogen Peroxide Demo Curing the Brown-Bottle-with-Cap Syndrome,” J. Chem. Educ., Vol. 71, 1994, 433. Hydrogen peroxide solutions from three different consumer products are decomposed catalytically with iodide producing oxygen gas and water in the presence of bubble bath. The rate of oxygen production is proportional to the amount of bubbling observed. Observations are related to the reaction mechanism.

Section 14.13

Lee R. Summerlin and James L. Ealy, “Catalytic Decomposition of Hydrogen Peroxide: Foam Production,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 101–102. Oxygen gas is catalytically produced in the presence of detergent resulting in the formation of a large quantity of foam.

Section 14.13

Christine L Copper and Edward Koubeck, “An Experiment To Demonstrate How a Catalyst Affects the Rate of a Reaction,” J. Chem. Educ., Vol. 76, 1999, 1714–1715.

Describes a chemistry experiment that allows students to calculate rates of reaction, orders of reaction, and activation energies. The activity demonstrates that to increase a reaction’s rate, a catalyst need only provide any additional pathway for the reaction, not necessarily a pathway having lower activation energy.

Literature References Section 14.2

Lionello Pobliani and Mario N. Berberan-Santos, “Inflation Rates, Car Devaluation, and Chemical Kinetics,” J. Chem. Educ., Vol. 73, 1996, 950–952.

Section 14.2

Charles J. Marzzacco, “An Analogy to Help Students Understand Reaction Orders,” J. Chem. Educ., Vol. 75, 1998, 482.

Section 14.4

Francisco J. Arnaiz, “Mice in the Box for Zero-Order Kinetics,” J. Chem. Educ., Vol. 76, 1999, 1458.

Section 14.5

Edward Todd Urbansky, “Don’t Be Tricked by Your Integrated Rate Plot,” J. Chem.

Educ., Vol. 78, 2001, 921–923.

Section 14.7

Ilya A. Leenson, “The Arrhenius Law and Storage of Food,” J. Chem. Educ., Vol. 76, 1999, 504–505.

Section 14.7

Keith L. Laidler, “Just What Is a Transition State?” J. Chem. Educ., Vol. 65, 1988, 540–542.

Section 14.7

Kent W. Piepgrass, “Audience-Appropriate Analogies: Collision Theory,” J. Chem.

Educ., Vol. 75, 1998, 72.

Section 14.7

Lee A. Krug, “The Collision Theory and an American Tradition,” J. Chem. Educ., Vol. 64, 1987, 1000.

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Section 14.9

John J. Fortman, “Pictorial Analogies XIII: Kinetics and Mechanisms,” J. Chem.

Educ., Vol. 71, 1994, 848–849.

Section 14.9

Arthur M. Last, “Doing the Dishes: An Analogy for Use in Teaching Reaction Kinetics,” J. Chem. Educ., Vol. 62, 1985, 1015–1016.

Section 14.11

Richard A. Potts, “Auto Analogies,” J. Chem. Educ., Vol. 62, 1985, 579.

Section 14.11

David W. Ball, “Another Auto Analogy: Rate-Determining Steps,” J. Chem. Educ., Vol. 64, 1987, 486–487.

Section 14.12

Doris Kolb, “Catalysis,” J. Chem. Educ., Vol. 56, 1979, 743–747.

Section 14.12

Albert Haim, “Catalysis: New Reaction Pathways, Not Just a Lowering of the Activation Energy,” J. Chem. Educ., Vol. 66, 1989, 935–937.

Section 14.13

Cynthia M. Friend, “Catalysis on Surfaces,” Scientific American, April 1993, 74–79.

Section 14.13

Sir John Meurig Thomas, “Solid Acid Catalysts,” Scientific American, April 1992, 112–118.

Section 14.13

Mitch Jacoby, “Getting Auto Exhausts to Pristine,” Chem. Eng. News, January 25, 1999, 36–44.

Section 14.13

Robert J. Farrauto, Ronald M. Heck, and Barry K. Speronello, “Environmental Catalysts,” Chem. Eng. News., September 7, 1992, 34–44.

Section 14.13

Reed A. Howald, “The Fizz Keeper, a Case Study in Chemical Education, Equilibrium, and Kinetics,” J. Chem. Educ., Vol. 76, 1999, 208–209.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 14.3

Worked Example 14.3: Determining a rate law from initial rates

Section 14.5

Worked Example 14.6: Using the integrated rate law for a first-order reaction

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CHAPTER 15 CHEMICAL EQUILIBRIUM Chapter Learning Objectives Section 15.1

Describe characteristics of a reaction in chemical equilibrium.

Section 15.2

Write an equilibrium constant expression Kc for gas and solution phase reactions.

Determine a new equilibrium constant expression when reactions are manipulated or added.

Calculate the value of the equilibrium constant Kc given concentrations of reactant and products.

Calculate the equilibrium constant from a molecular representation of a reaction mixture.

Section 15.3

Write the equilibrium expression Kp in terms of partial pressures of reactants and products. Relate the equilibrium constants Kp and Kc.

Section 15.4

Write equilibrium constant expressions for reactions involving heterogeneous equilibria.

Section 15.5

Determine the extent of a reaction, given the equilibrium constant.

Predict the direction a reaction will shift to reach equilibrium given non-equilibrium concentrations.

Calculate concentrations or partial pressures or products and reactants in equilibrium.

Calculate concentrations of reactants and products in equilibrium using the quadratic equation to solve the mathematical operations.

Section 15.6 through Section 15.9

Predict the direction a reaction in equilibrium will shift when concentration of reactants or products are changed.

Predict the direction a reaction in equilibrium will shift when volume or pressure is changed.

Predict the direction a reaction in equilibrium will shift when temperature is changed.

Section 15.10

Describe the effect of a catalyst on the equilibrium constant for a reaction.

Relate the equilibrium constant to the rate of the forward and reverse reactions.

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Lecture Outline 1

15.1. The Equilibrium State A. Reversible reaction – reaction in which product molecules recombine to form reactant molecules 1. Reactants – substances on the left side of the equation 2. Products – substances on the right side of the equation 3. Use a double arrow to indicate a reaction at equilibrium.

Reactants ⇌ Products B. With time, concentrations of reactants decrease and concentrations of products increase until all concentrations level off at constant equilibrium values.

1. Rates of the forward and reverse reactions are equal.

2. State of chemical equilibrium a. Equilibrium mixture – the mixture of reactants and products in the equilibrium state b. Dynamic state – forward and reverse reactions continue at equal rates c. No net conversion of reactants to products 2 15.2. The Equilibrium Constant Kc A. The equilibrium mixture obeys an equilibrium equation (or law of mass action).

B. For the reaction: a A + b B ⇌ c C + d D [C]c [D]d Kc = [A]a [B]b C. Kc = equilibrium constant 1. Values are reported without units.

2. Temperature dependent D. Equilibrium constant expression – expression on right side of equilibrium equation E. For the reverse reaction 1. Equilibrium expression = reciprocal of the original expression 2. Equilibrium constant for reverse reaction 1 K c’ = Kc F. For a chemical equation multiplied by a common factor, n.

1. Equilibrium expression = original expression raised to the n power 2. Kc’ = (Kc)n G. Whenever two or more chemical equations are added to give the equation for an overall reaction 1. Equilibrium expression (overall reaction) = product of the individual equilibrium expressions.

2. Kc (overall) = Kc1  Kc2 3 15.3. The Equilibrium Constant Kp A. For gas-phase reactions, use partial pressures rather than molar concentrations in equilibrium equations.

B. Kp – equilibrium constant defined using partial pressures C. Relationship between Kc and Kp 1. Kp = Kc(RT)n 2. n = the sum of the coefficients of gaseous products minus the sum of the coefficients of gaseous reactants 4 15.4. Heterogeneous Equilibria 1Test Item File Questions: Multiple Choice 1, Short Answer 1 – 2 2Test Item File Questions: Multiple Choice 2 – 12, Algorithmic 1 – 5, Short Answer 3 – 4 3Test Item File Questions: Multiple Choice 31 – 45, Algorithmic 6 – 7, Short Answer 5 – 10

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Chapter 15—Chemical Equilibrium A. Homogeneous equilibria – reactants and products in a single phase B. Heterogeneous equilibria – reactants and products present in more than one phase 1. Solids – constant molar concentrations a. Calculated from the densities and molar masses b. Independent of its amount 2. Concentrations of pure solids or pure liquids not included when writing the equilibrium equation for any heterogeneous equilibrium 3. The concentrations are constant and are included in value of equilibrium constant.

5 15.5. Using the Equilibrium Constant A. Judging extent of reaction (how much reactant is converted to product) 1. Large value of Kc ( > 103) a. Products predominate over reactants. b. Reaction proceeds almost to completion.

2. Small value of Kc ( < 10–3) a. Reactants predominate over products. b. Reaction hardly proceeds.

3. Intermediate value of Kc (10–3 < Kc < 103) a. Appreciable amounts of both reactants and products present b. Both forward and reverse reactions are important.

B. Predicting the direction of reaction 1. Reaction quotient, Qc, same as the equilibrium constant expression except that concentrations are not necessarily equilibrium values.

2. Predict direction of reaction by comparing the value of Qc to Kc. a. Qc < Kc; reaction goes from left to right (products favored). b. Qc > Kc; reaction goes from right to left (reactants favored). c. Qc = Kc; reaction is at equilibrium.

C. Calculating equilibrium concentrations 1. If the equilibrium constant is known and all but one equilibrium concentration is known, calculate the unknown concentration directly from the equilibrium constant expression.

2. If the equilibrium constant is known and initial concentrations are known, calculate equilibrium concentrations using the steps outlined in textbook Figure 15.7.

6 15.6. Factors That Alter the Composition of an Equilibrium Mixture: Le Châtelier’s Principle A. Factors that alter the composition of an equilibrium mixture 1. Concentrations of reactants or products 2. Pressure and volume 3. Temperature B. Le Châtelier’s principle: If a stress is applied to a reaction mixture at equilibrium, a net reaction occurs in the direction that relieves the stress.

1. Predicts changes in equilibrium mixture composition if one of the above changes occurs 2. Reaction moves in a direction that relieves the stress.

3. A stress is a change in concentration, pressure, volume, or temperature that disturbs the equilibrium.

4Test Item File Questions: Multiple Choice 46 – 53, Algorithmic 8 – 11, Short Answer 11 5Test Item File Questions: Multiple Choice 56 – 81, Algorithmic 12 – 26, Short Answer 12 – 17 6Test Item File Questions: Multiple Choice 83, Algorithmic 27 – 29,

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15.7. Altering an Equilibrium Mixture: Changes in Concentration A. Disturb an equilibrium by adding or removing reactant or product.

1. Stress of an added reactant or product relieved by reaction in the direction that consumes the added substance. a. Add reactant – reaction shifts right (toward product) b. Add product – reaction shifts left (toward reactant) 2. Stress of removing reactant or product is relieved by reaction in the direction that replenishes the removed substance. a. Remove reactant – reaction shifts left b. Remove product – reaction shifts right B. Changes occur due to change in value of Qc. 1. Add reactant – denominator in Qc expression becomes larger. a. Qc < Kc b. To return to equilibrium, Qc must increase. i. Numerator of Qc expression must increase and the denominator must decrease. ii. Implies net conversion of reactants to products. iii. Reaction shifts right.

2. Remove reactant – denominator in Qc expression becomes smaller. a. Qc > Kc b. To return to equilibrium, Qc must decrease. i. Numerator of Qc expression must decrease and the denominator must increase. ii. Implies net conversion of products to reactants. iii. Reaction shifts left.

8 15.8. Altering an Equilibrium Mixture: Changes in Pressure and Volume A. If the number of moles of gaseous reactants in the balanced equation is different from the number of moles of gaseous products, a change in pressure (due to changing volume) changes the equilibrium mixture composition.

B. Increase in pressure (due to decrease in volume) results in reaction in the direction of fewer number of moles of gas.

C. Decrease in pressure (due to increase in volume) results in reaction in the direction of greater number of moles of gas.

D. Changes occur due to change in value of Qc. 1. Decrease volume – molarity (= n/V) increases 2. If reactant side has more moles of gas a. Increase in denominator is greater than increase in numerator. b. Qc < Kc c. To return to equilibrium, Qc must increase. i. Numerator of Qc expression must increase and the denominator must decrease. ii. Implies net conversion of reactants to products (shifts toward fewer moles of gas).

3. If product side has more moles of gas a. Increase in numerator is greater than increase in denominator. b. Qc > Kc c. To return to equilibrium, Qc must decrease. i. Denominator of Qc expression must decrease and the numerator must increase. ii. Implies net conversion of products to reactants (shifts toward fewer moles of gas).

E. If reaction involves no change in the number of moles of gas, there is no effect on equilibrium mixture composition.

7Test Item File Questions: Multiple Choice 84 – 85, Algorithmic 30 – 31, 37, Short Answer 18 – 19 8Test Item File Questions: Algorithmic 32 – 36, Short Answer 21 – 23

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Chapter 15—Chemical Equilibrium F. For a heterogeneous equilibrium 1. Volume is nearly independent of pressure.

2. Effect of pressure changes on solids and liquids can be ignored.

G. Change in pressure due to addition of an inert gas.

1. No change in the molar concentrations of reactants or products 2. No effect on composition of equilibrium mixture 9 15.9. Altering an Equilibrium Mixture: Changes in Temperature A. Disturb an equilibrium by a change in concentration, pressure, or volume.

1. Qc  Kc 2. Value of Kc remains constant with constant T. B. Change in temperature always changes the equilibrium constant 1. Depends on sign of H for the reaction 2. Exothermic reaction (H = –) – equilibrium constant decreases as the temperature increases 3. Endothermic reaction (H = +) – equilibrium constant increases as the temperature increases C. Le Châtelier’s principle – add heat to an equilibrium mixture, net reaction occurs in the direction that relieves the stress of the added heat.

1. Endothermic reaction – heat is absorbed by reaction in the forward direction (heat is written on the reactant side of the reaction). a. Equilibrium mixture contains more product than reactant. b. Kc increases with increasing temperature.

2. Exothermic reaction – heat is absorbed by reaction in the reverse direction (heat is written on the product side of the reaction). a. Equilibrium mixture contains more reactant than product. b. Kc decreases with increasing temperature.

10 15.10. The Link Between Chemical Equilibrium and Chemical Kinetics A. For the reaction: A + B ⇌ C + D 1. Rate of forward reaction = kf [A][B] 2. Rate of reverse reaction = kr [C][D] 3. At equilibrium: kf[A][B] = kr[C][D] kf [C][D] = = Kc kr [A][B] B. Relative values of kf and kr determine the composition of the equilibrium mixture.

1. kf >> kr a. Kc is very large. b. Reaction goes to completion. c. Irreversible reaction d. Reverse reaction too slow to be detected 2. kf  kr a. Kc  1 b. Reactants and products are present at equilibrium. c. Reversible reaction C. The effect of a catalyst on equilibrium.

1. A catalyst provides a new, lower energy pathway, increasing the rate of the chemical reaction.

2. Forward and reverse reactions pass through the same transition state.

3. Rates for forward and reverse reactions increase by the same factor.

4. The catalyst does not affect the composition of the equilibrium mixture.

9Test Item File Questions: Multiple Choice 95 – 97, Algorithmic 37– 40, Short Answer 24 – 25 10Test Item File Questions: Multiple Choice 102 – 109, Algorithmic 41, Short Answer 26-27

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Chapter 15—Chemical Equilibrium a. Does not alter the equilibrium constant or equilibrium concentrations b. Does not appear in the balanced chemical equation 5. Can influence choice of optimum reaction conditions Inquiry – How does high altitude oxygen transport in the body?

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CHAPTER 15 CHEMICAL EQUILIBRIUM Teaching Tips, Points of Emphasis, and Common Misconceptions Section 15.2

The subscript c in Kc means that the amounts of reactants and products are expressed as molar concentrations.

Section 15.2

Both the equilibrium equation and the value of the equilibrium constant depend on how the chemical equation is written and balanced.

Section 15.5

For an equilibrium system, the concentration of any reactant or product can never be zero. No matter how small or how large the equilibrium constant, there must always be some products and some reactants in order for an equilibrium to exist.

Section 15.5

The equilibrium constant expression = Kc only at equilibrium. The equilibrium constant expression = Qc, the reaction quotient, whether or not the system is at equilibrium.

Section 15.5

Students should apply these criteria results not by rote memorization but by analysis of what must occur when Qc  Kc.

Section 15.5

Strongly encourage using tables. Although not necessary, it is less confusing for many students to make the coefficient used with the unknown, x, the same as the coefficient in the balanced chemical equation.

Section 15.5

Mathematical tricks are not apparent to many students. Point out the value of setting up the problem to obtain a perfect square.

Section 15.5

Mathematics sometimes yields results that have no physical meaning. Students should learn to analyze answers to problems to determine whether they make sense.

Section 15.7

When an equilibrium is upset by a change in concentration, pressure and volume, or temperature, equilibrium will be restored, but the concentrations of reactants and products will be different from those in the original equilibrium mixture.

Section 15.7

If a reactant or product is added to a system at equilibrium, the reaction will occur in a direction away from that reactant or product. If a reactant or product is removed from a system at equilibrium, the reaction will occur in a direction toward that reactant or product.

Section 15.10

The addition of a catalyst equally affects the rate of the forward and of the reverse reaction; hence, a catalyst does not change the equilibrium mixture or the value of the equilibrium constant, K.

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Chapter 15—Chemical Equilibrium

Lecture/Laboratory Demonstration References Section 15.1

Lee R. Summerlin and James L. Ealy, Jr., “Equilibrium in the Gas Phase,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 60-61. Reddish-brown nitrogen dioxide is produced and placed in two test tubes where it exists in equilibrium with colorless nitrogen dioxide. One test tube is placed in a boiling water bath, causing an increase in the intensity of the color due to an increase in the intensity in the amount of NO 2. The second test tube is placed in an ice-water bath, which causes the gaseous mixture to become almost colorless due to a decrease in the amount of NO 2. When returned to room temperature, the gaseous mixtures in both test tubes exhibit the same brown color. Pressure and temperature effects on the equilibrium are related to the stoichiometry and enthalpy of the reaction.

Section 15.7

Lee R. Summerlin and James L. Ealy, Jr., “Equilibrium and Le Châtelier’s Principle,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 77–78.

Section 15.8

Lee R. Summerlin and James L. Ealy, Jr., “Equilibrium in the Gas Phase,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 85–86.

Section 15.9

Lee R. Summerlin and James L. Ealy, Jr., “Effect of Temperature Change on Equilibrium: Cobalt Complex,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 60–61.

Concentrated hydrochloric acid is added to pink [Co(H 2O)6]2+ until blue [CoCl4]2– is formed. When heated the solution turns darker blue; when cooled the solution turns pink, indicating that the reaction is endothermic. Students are asked to examine the equilibrium reaction and predict how the system will shift upon the addition of water.

Literature References Section 15.1

Vladimir E. Fainzilberg and Stewart Karp, “Chemical Equilibrium in the General Chemistry Curriculum,” J. Chem. Educ., Vol. 71, 1994, 769–770.

Section 15.1

Carl W. David, “An Elementary Discussion of Chemical Equilibrium,” J. Chem.

Educ., Vol. 65, 1988, 407–409.

Section 15.1

Jan H. Van Driel, Wobbe de Vos, and Nico Verloop, “Introducing Dynamic Equilibrium as an Explanatory Model,” J. Chem. Educ., Vol. 76, 1999, 559–561.

Section 15.1

Louise Tyson, David F. Treagust, and Robert B. Bucat, “The Complexity of Teaching and Learning Chemical Equilibrium,” J. Chem. Educ., Vol. 76, 1999, 554–558.

Section 15.1

Audrey H. Wilson, “Equilibrium: A Teaching/Learning Activity,” J. Chem. Educ., Vol 75, 1998, 1176–1177.

Section 15.1

Penelope A. Huddle, Margie W. White, and Fiona Rogers, “Simulations for Teaching Chemical Equilibrium,” J. Chem. Educ., Vol 77, 2000, 920–926.

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John Olmsted III, “Amounts Tables as a Diagnostic Tool for Flawed Stoichiometric Reasoning,” J. Chem. Educ., Vol. 76, 1999, 52–54.

Section 15.5

William R. Smith and Ronald W. Munson, “Chemical Equilibrium and Polynomial Equations: Beware of Roots,” J. Chem. Educ., Vol. 66, 1989, 489–490.

Section 15.5

E. Weltin, “Calculating Equilibrium Concentration by Iteration: Recycle Your Approximations,” J. Chem. Educ., Vol. 72, 1995, 36–38.

Section 15.6

Martin R. Feldman and Monica L. Tarver, “Fritz Haber,” J. Chem. Educ., Vol. 60, 1983, 463–464.

Section 15.8

Ron DeLorenzo, “From Chicken Breath to the Killer Lakes of Cameroon: Uniting Seven Interesting Phenomena with a Single Chemical Underpinning,” J. Chem.

Educ., Vol. 78, 2001, 191–194.

Section 15.8

Reed A. Howard, “The Fizz Keeper, a Case Study in Chemical Education,” J. Chem.

Educ., Vol. 76, 1999, 208–209.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 15.5

Worked Example 15.9: Calculating equilibrium concentrations (solving the equilibrium equation by taking the square root of both sides of the equation)

Section 15.5

Worked Example 15.12: Calculating equilibrium partial pressures from initial partial pressures

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CHAPTER 16 AQUEOUS EQUILIBRIA: ACIDS AND BASES Chapter Learning Objectives Section 16.1

Identify Arrhenius acids and bases and Brønsted–Lowry acids and bases and conjugate acid–base pairs.

Section 16.2

Rank acids in order of increasing strength given a molecular image representing dissociation in solution.

Predict the direction of a reaction based upon relative strengths of the conjugate acid–base pairs.

Section 16.3

Predict the relative strengths of binary acids (HA) and oxoacids (HnYOm) based on their chemical structure.

Section 16.4

Calculate the concentration of [H3O+] or [OH−] using the value of Kw.

Section 16.5 and Section 16.6

Section 16.7 Section 16.8 through Section 16.10

Calculate the pH of a solution given [H3O+] or [OH−] or calculate [H3O+] or [OH−] given the pH of a solution.

Calculate the pH of a strong acid or base solution.

Calculate the value for Ka of a weak acid given the initial concentration of the acid and its equilibrium pH. Given the Ka value for a weak acid and its initial concentration, calculate the pH, percent dissociation, and the concentration of all species present in solution.

Section 16.11

Calculate the pH and the concentration of all species present in a solution of a diprotic acid.

Section 16.12

Calculate the pH and equilibrium concentrations in a solution of a weak base.

Section 16.13

Relate Ka, Kb, pKa, and pKb for a conjugate acid–base pair.

Section 16.14

Predict whether a salt is acidic, basic, or neutral and calculate the pH of the salt solution.

Section 16.15

Identify the Lewis acid and Lewis base and use curved arrow notation to indicate donation of a lone pair of elections in a Lewis acid–base reaction.

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Lecture Outline 1

16.1. Acid–Base Concepts: The Brønsted–Lowry Theory A. Arrhenius theory of acids and bases 1. Acids – substances that dissociate in water to produce hydrogen ions HA (aq) ⇌ H+(aq) + A–(aq) 2. Bases – substances that dissociate in water to produce hydroxide ions MOH (aq) ⇌ M+(aq) + OH– (aq) 3. Limitations to theory a. Restricted to aqueous solutions b. Doesn’t account for the basicity of substances that don’t contain OH groups B. Brønsted–Lowry theory of acids and bases 1. Acid – any substance that can transfer a proton, H+, to another substance (proton donor) 2. Base – any substance that can accept a proton, H+, (proton acceptor) 3. Acid–base reaction – proton-transfer reactions C. Conjugate acid–base pair – chemical species whose formulas differ only by one proton 1. Conjugate base has one less proton than its acid: A – is the conjugate base of HA. 2. Conjugate acid has one more proton than its base: BH+ is the conjugate acid of B. D. Acid-dissociation equilibrium 1. Acid transfers a proton to the solvent, which acts as a base.

2. HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq) E. Base-dissociation equilibrium 1. Base accepts a proton from the solvent, which acts as an acid.

2. NH3(aq) + H2O(l) ⇌ OH–(aq) + NH4+(aq) F. All Brønsted–Lowry bases have one or more lone pairs of electrons: unshared pair of electrons are used for bonding to the proton.

G. Hydrated protons and hydronium ions 1. Hydronium ion, H3O+ –i on in which H+ is bonded to the oxygen atom of a solvent water molecule 2. Simplest hydrate of the proton – [H(H2O)]+ 3. Produces higher hydrates through hydrogen bonding with other water molecules 4. General formula – [H(H2O)n]+ 5. Symbols H+(aq) and H3O+(aq) used to represent a proton hydrated by an unspecified number of water molecules 2 16.2. Acid Strength and Base Strength A. In an acid-dissociation equilibrium, the two bases, H2O and A–, are competing for protons.

B. Direction of reaction to reach equilibrium is proton-transfer from the stronger acid to the stronger base.

C. Strong acid – completely dissociated in aqueous solution 1. Has weak conjugate base 2. H2O is a stronger base than A– D. Weak acid – partially dissociates in aqueous solution 1. Has strong conjugate base 2. A– is a stronger base than H2O E. Inverse relationship between the strength of an acid and the strength of its conjugate base, textbook Table 16.1

1Test Item File Questions: Multiple Choice 1 – 9, Algorithmic 1 – 12, Short Answer 1 2Test Item File Questions: Multiple Choice 16 – 27, Short Answer 2

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16.3. Factors That Affect Acid Strength A. The extent of dissociation of an acid, HA, is often determined by the strength and polarity of the H–A bond.

1. Weaker the H–A bond, stronger the acid 2. More polar the H–A bond, stronger the acid B. For binary acids 1. H–A bond strength decreases down a group in the periodic table. a. Increase in size of A down a group produces poorer orbital overlap and a weaker bond. b. H–A bond strength is the most important factor for determining acid strength for binary acids of elements in the same column. c. Acidity increases down a group.

2. H–A bond polarity is the most important factor for determining acid strength for binary acids of elements in the same row. a. Increase in H–A bond polarity as electronegativity of A increases b. In general, electronegativities increase from left to right across the periodic table. c. Acidity increases across a period.

C. Oxoacids, HnYOm (Y = a nonmetallic atom, n and m are integers) 1. Y is always bonded to one or more hydroxyl (OH) groups (can also be bonded to one or more oxygen atoms).

2. Dissociation of oxoacid involves breaking an O–H bond.

3. Strength of the acid increases as the O–H bond is weakened or becomes more polar. a. For oxoacids that contain the same number of OH groups and the same number of O atoms, acid strength increases with increasing electronegativity of Y. b. For oxoacids that contain the same atom Y but different numbers of oxygen atoms, acid strength increases with increasing oxidation number of Y, which increases, in turn, with an increasing number of oxygen atoms.

4 16.4. Dissociation of Water A. Water can act both as an acid and as a base.

B. Dissociation of water – one water molecule donates a proton to another water molecule.

1. H2O(l) + H2O(l) ⇌ H3O+(aq) + OH–(aq) 2. Ion-product constant for water, Kw Kw = [H3O+] [OH– ] 3. Forward and reverse reactions are rapid.

4. Position of the equilibrium lies far to the left.

5. In an aqueous solution, [H3O+][OH–] = 1.0  10–14 at 25 C. C. Relative values of H3O+ and OH– concentrations can be used to determine whether an aqueous solution is neutral, acidic, or basic.

1.0 ´10 -14 1. [H3 O+ ] = [OH ] 2. [OH- ] =

1.0 ´ 10-14 +

[H3 O ] 3. [H3O+] = [OH–], neutral solution and [H3O+] = 1.0  10–7 M 4. [H3O+] > [OH–], acid solution and [H3O+] > 1.0  10–7 M 5. [H3O+] < [OH–], basic solution and [H3O+] < 1.0  10–7M

3Test Item File Questions: Multiple Choice 28 – 32, Algorithmic 13 – 20, Short Answer 3 – 4, 23 4Test Item File Questions: Multiple Choice 33 – 40, Algorithmic 21 – 25, Short Answer 5 – 9

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16.5. The pH Scale A. pH scale – logarithmic scale used to express the hydronium ion concentration 1. pH = –log[H3O+] 2. [H3O+] = antilog(–pH) = 10–pH 3. Significant figures in a logarithm are the digits to the right of the decimal point.

4. Number to the left of the decimal point is an exact number related to the integral power of 10 in the exponential expression for [H3O+].

B. pH decreases as [H3O+] increases.

1. Acidic solutions, pH < 7 2. Basic solutions, pH > 7 3. Neutral solutions, pH = 7 6 16.6. Measuring pH A. Acid–base indicators (HIn) – substances that change color in a specific pH range 1. Weak acids 2. Have different colors in their acid (HIn) and conjugate base (In –) forms.

3. Change color over a range of  2 pH units 4. Can determine the pH of a solution to within  1 pH unit B. pH meter 1. Electronic instrument that measures a pH-dependent electrical potential of the test solution 2. Determines pH values more accurately than acid-base indicators 7 16.7. The pH in Solutions of Strong Acids and Strong Bases A. Strong monoprotic acids – 100% dissociated in aqueous solution (HClO4, HCl, HNO3) 1. Contain a single dissociable proton 2. pH = –log[H3O+] 3. [H3O+] = [A–] = initial concentration of the acid 4. [undissociated HA] = 0 B. Strong bases 1. Alkali metal hydroxides, MOH a. Water-soluble ionic solids b. Exist in aqueous solution as alkali metal cations and hydroxide anions c. Calculate pH from [OH–] 2. Alkaline earth metal hydroxides, M(OH)2 (M = Mg, Ca, Sr, Ba) a. Less soluble than alkali hydroxides, therefore lower [OH –] b. Most important strong base – lime, CaO i. Produces Ca(OH)2 when mixed with water ii. Used in steelmaking, water purification, and chemical manufacture 8 16.8. Equilibria in Solutions of Weak Acids A. Acids – partially dissociated B. Can write an acid–dissociation equilibrium equation.

HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq) C. Position of acid–dissociation equilibrium – characterized by the acid–dissociation constant, Ka [H O + ][A − ] Ka = 3 [HA] D. The larger the value of Ka, the stronger the acid.

5Test Item File Questions: Multiple Choice 55 – 66, Algorithmic 26 – 32, Short Answer 10 – 11 6Test Item File Questions: Multiple Choice 67 – 69, Short Answer 12 – 13 7Test Item File Questions: Multiple Choice 70 – 76, Algorithmic 33 – 41, Short Answer 14-15 8Test Item File Questions: Multiple Choice 77 – 83, 85, Short Answer 16 – 20

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Chapter 16—Aqueous Equilibria: Acids and Bases E. pKa = − log Ka F. Ka values can be determined from pH measurements.

9 16.9. Calculating Equilibrium Concentrations in Solutions of Weak Acids A. Equilibrium concentrations and solution pH of a weak acid can be calculated from the value of Ka. B. To solve acid–base equilibrium problems, think about the chemistry, and use the following steps:

10

16.10. Percent Dissociation in Solutions of Weak Acids [HA] dissociated  100 A. Percent dissociation = [HA] undissociated B. Useful measure of acid strength C. Value depends on the acid D. Increases with increasing value of Ka E. Increases with increasing dilution.

11 16.11. Polyprotic Acids A. Polyprotic acid – acids that contain more than one dissociable proton.

B. Dissociate in a stepwise manner: each dissociation step has its own Ka. C. Stepwise dissociation constants decrease in the order Ka1 > Ka2 > Ka3.

D. More difficult to remove a positively charged proton from a negative ion.

9Test Item File Questions: Multiple Choice 86 – 91, Algorithmic 42 – 46 10Test Item File Questions: Multiple Choice 92 – 96, Short Answer 21 – 23 11Test Item File Questions: Multiple Choice 98 – 103, Algorithmic 47 – 52,

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Chapter 16—Aqueous Equilibria: Acids and Bases E. Diprotic acid solution contains a mixture of acids: H 2A, HA–, and H2O. 1. Strongest acid – H2A 2. Principal reaction – dissociation of H2A 3. All of H3O+ comes from first dissociation step.

12 16.12. Equilibria in Solutions of Weak Bases A. Weak bases – partially dissociated.

1. Can write a base-dissociation equilibrium equation.

2. B(aq) + H2O(l) ⇌ BH+(aq) + OH–(aq) B. Position of base-dissociation equilibrium – characterized by the base–dissociation constant, Kb [BH + ][OH − ] Kb = [B] C. Amines – organic compounds that are weak bases 1. Derivatives of ammonia 2. One or more hydrogen atoms replaced by another group 3. Basicity due to lone pair of electrons on nitrogen atom: can be used for bonding to a proton D. To solve for equilibrium concentrations and solution pH, use the same procedure as for solving weak acid problems.

13 16.13. Relation Between Ka and Kb A. For a conjugate acid–base pair, can calculate either Ka or Kb from the other B. Sum of an acid–dissociation reaction and a base-dissociation reaction is the dissociation of water.

1. Equilibrium constant for the net reaction (dissociation of water) = the product of the equilibrium constants for the individual reactions.

2. Ka  Kb = Kw 3. pKa + pKb = pKw = 14.004. For Ka  Kb to remain constant, the strength of a conjugate base must decrease as the strength of the acid increases.

14 16.14. Acid–Base Properties of Salts A. pH of a salt solution is determined by acid–base properties of the constituent cations and anions.

1. In an acid–base reaction, the influence of the stronger partner is dominant.

2. Strong acid + Strong base → Neutral solution 3. Strong acid + Weak base → Acidic solution 4. Weak acid + Strong base → Basic solution B. Salts that yield neutral solutions 1. Derived from a strong base and a strong acid 2. Neither the cation nor the anion reacts with water to produce H 3O+ or OH– ions.

3. Cations that do not react with H2O a. Group 1A b. Group 2A 4. Anions that do not react with H2O: anions from strong, monoprotic acids C. Salts that yield acidic solutions 1. Derived from a weak base and a strong acid 2. Anion is neither an acid nor a base.

12Test Item File Questions: Multiple Choice 106 – 111, Algorithmic 53 – 56, Short Answer 24 13Test Item File Questions: Multiple Choice 112 – 116, Algorithmic 57 14Test Item File Questions: Multiple Choice 117 – 125, Algorithmic 58 – 69, Short Answer 25

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Chapter 16—Aqueous Equilibria: Acids and Bases 3. Cation is a weak acid a. Reacts with water (hydrolysis reaction) to produce H 3O+ b. Small, highly charged cations are also acidic.

D. Salts that yield basic solutions 1. Derived from a strong base and a weak acid 2. Cation is neither an acid nor a base.

3. Anion is a weak base: reacts with water to produce OH –.

E. Salts that contain acidic cations and basic anions 1. Derived from a weak acid and a weak base 2. Both the cation and the anion can undergo proton-transfer reactions.

3. pH depends on the relative acid strength of the cation and the base strength of the anion. a. Ka (for the cation) > Kb (for the anion); acidic solution b. Kb (for the anion) > Ka (for the cation); basic solution c. Ka (for the cation)  Kb (for the anion); neutral solution 15 16.15. Lewis Acids and Bases A. Lewis base – electron-pair donor 1. All Lewis bases are Brønsted–Lowry bases.

2. All Brønsted–Lowry bases are Lewis bases.

B. Lewis acid – electron-pair acceptor 1. More general definition than Brønsted–Lowry 2. Includes cations and neutral molecules having vacant valence orbitals that can accept a share in a pair of electrons from a Lewis base 3. Cationic Lewis acids – metal ions 4. Neutral Lewis acids – halides of group 3A elements and oxides of nonmetals Inquiry – Has the problem of acid rain been solved?

15Test Item File Questions: Multiple Choice 130 – 134, Algorithmic 70 – 73, Short Answer 26 – 27

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CHAPTER 16 AQUEOUS EQUILIBRIA: ACIDS AND BASES Teaching Tips, Points of Emphasis, and Common Misconceptions Section 16.5

Typically pH measurements fall within the range 0–14; however, the scale extends beyond those values in both directions and includes negative pHs. Sometimes students want to use the concentration of the acid to calculate the pH. It should be emphasized that only hydronium ion concentration is reflected in the pH.

Section 16.11

A polyprotic acid has two or more dissociable protons. Acetic acid, CH3CO2H, contains four protons, only one of which is dissociable. Hence, acetic acid is not a polyprotic acid.

Section 16.13

Caution students that the equilibrium constants are multiplied, and they should not confuse this principle with Hess’s law for which enthalpy changes are added.

Lecture/Laboratory Demonstration References Section 16.3

K. E. Kolb and D. Kolb, “Illustrating the Inductive Effect on Acid Strength of Carboxylic Acids,” J. Chem. Educ., Vol. 66, 1989, 1034. Substituent effects on the acidity of carboxylic acids are demonstrated from the colors formed when methyl violet indicator is added to aqueous carboxylic acid solutions.

Section 16.3

Bassam Z. Shakhashiri, “Effect of Molecular Structure on the Strength of Organic Acids in Aqueous Solutions,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp. 158–161.

Section 16.6

J. J. Fortman and K. M. Stubbs, “Demonstrations with Red-Cabbage Indicator,” J. Chem. Educ., Vol. 69, 1992, 66 Anthocyanin pigment extract from red cabbage is used as a pH indicator exhibiting colors that range from green to aqua to violet to red in the presence of various concentrations of household ammonia and white vinegar.

Section 16.14

Bassam Z. Shakhashiri, “Hydrolysis: Acidic and Basic Properties of Salts,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp. 103–108.

Section 16.14

Edward Koubek, “Acid–Base Chemistry of the Aluminum Ion in Aqueous Solution,” J. Chem. Educ., Vol. 75, 1998, 60.

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Literature References Section 16.1

Doris Kolb, “Acids and Bases,” J. Chem. Educ., Vol. 55, 1978, 459–464.

Section 16.1

George B. Kauffman, “The Brønsted–Lowry Acid–Base Concept,” J. Chem. Educ., Vol. 65, 1988, 28–31.

Section 16.1

Jamie L. Adcock, “Teaching Brønsted-Lowry Acid-Base Theory in a Direct Comprehensive Way,” J. Chem. Educ., Vol. 78, 2001, 1495–1496.

Section 16.3

Manus Monroe and Karl Abrams, “The Relative Strength of Oxyacids and Its Application,” J. Chem. Educ., Vol. 62, 1985, 41–43.

Section 16.5

Doris Kolb, “The pH Concept,” J. Chem. Educ., Vol. 56, 1979, 49–53.

Section 16.5

Stephen J. Hawkes, “Teaching the Truth about pH,” J. Chem. Educ., Vol. 71, 1994, 747–749.

Section 16.5

Addison Ault, “Do pH in Your Head,” J. Chem. Educ., Vol. 76, 1999, 936–938.

Section 16.8

Todd P. Silverstein, “Weak vs Strong Acids and Bases,” J. Chem. Educ., Vol. 77, 2000, 849–850.

Section 68

John J. Fortman, “Pictoral Analogies XI: Concentrations and Acidity of Solutions,” J. Chem. Educ., Vol. 71, 1994, 430–432.

Section 16.15

Glenn T. Seaborg, “The Research Style of Gilbert N. Lewis: Acids and Bases,” J. Chem. Educ., Vol. 61, 1984, 93–100.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 16.9

Worked Example 16.10: Calculating the pH and the equilibrium concentrations in a solution of a weak acid

Section 16.12

Worked Example 16.12: Calculating the pH and the equilibrium concentrations in a solution of a weak base

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CHAPTER 17 APPLICATIONS OF AQUEOUS EQUILIBRIA Chapter Learning Objectives Section 17.1

Write a balanced equation for a neutralization reaction and predict whether the pH after neutralization is greater than, equal to, or less than 7.

Section 17.2

Calculate the concentration of all species, pH, and percent dissociation in a solution of a weak acid when a common ion is present.

Section 17.3 and Section 17.4

Calculate the initial pH of a buffer solution and the pH after the addition of a strong acid or a strong base.

Use the Henderson–Hasselbalch equation to calculate the pH of a buffer solution.

Section 17.5 through Section 17.9

Calculate the pH at various points in a strong acid–strong base titration.

Calculate the pH at various points in a weak acid–strong base titration or weak base– strong acid titration.

Calculate the pH at various points in a diprotic acid–strong base titration.

Visualize the molecular species present during a titration and interpret titration curves.

Section 17.10 and Section 17.11

Write the equilibrium-constant expression for dissolution of an ionic compound, and calculate the value of its Ksp. Calculate equilibrium concentrations of ions and the solubility of an ionic compound from its Ksp.

Section 17.12

Predict how the presence of molecular and ionic species will affect the solubility of an ionic compound.

Calculate solubility in a solution that contains a common ion.

Use the formation constant Kf to calculate ion concentrations in a solution that contains a complex ion.

Use the formation constant Kf to calculate the solubility of an ionic compound when the cation forms a complex ion.

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Chapter 17—Applications of Aqueous Equilibria Section 17.13 through Section 17.15

Calculate the ion product IP for an ionic compound, and determine whether a precipitate will form when various solutions are mixed.

Design a scheme for separating ions in a mixture based on selective precipitation.

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Chapter 17—Applications of Aqueous Equilibria

Lecture Outline 1

17.1. Neutralization Reactions A. Strong acid–strong base 1. Net ionic equation for the neutralization reaction: H 3O+(aq) + OH–(aq) ⇌ 2 H2O(l) 2. When equal numbers of moles of acid and base are mixed together, [H3O+] = [OH–] = 1  10–7 M. 3. Reaction proceeds far to the right.

4. Equilibrium constant, Kn, for the reaction is the reciprocal of the ion-product constant for water.

1 1 = Kn = [H 3O + ][OH − ] K w 5. pH = 7.00 B. Weak acid–strong base 1. Net ionic equation for the neutralization reaction involves proton transfer from HA to the strong base, OH–.

HA(aq) + OH–(aq) ⇌ H2O(l) + A–(aq) 2. Equilibrium constant, Kn, is obtained from multiplying the equilibrium constants for the reactions that add to give the net ionic equation: Kn = Ka(1/Kw).

HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq) Ka H3O+(aq) + OH–(aq) ⇌ 2 H2O(l) 1/Kw 3. Neutralization of any weak acid by a strong base goes 100% to completion because OH– has a great affinity for protons.

4. pH > 7.00 due to the hydrolysis of A–, which behaves as a weak base.

C. Strong acid–weak base 1. Net ionic equation for the neutralization reaction involves proton transfer from the strong acid, H3O+, to the weak base, B. H3O+(aq) + B(aq) ⇌ H2O(l) + BH+(aq) 2. Equilibrium constant, Kn, is obtained from multiplying the equilibrium constants for the reactions that add to give the net ionic equation: Kn = Kb(1/Kw).

B(aq) + H2O(l) ⇌ BH+(aq) + OH–(aq) Kb H3O+(aq) + OH–(aq) ⇌ 2 H2O(l) 1/Kw 3. Neutralization of any weak base with a strong acid goes 100% to completion because H 3O+ is a powerful proton donor 4. pH < 7.00 due to the hydrolysis of BH+, which behaves as a weak acid.

D. Weak acid–weak base 1. Neutralization reaction involves proton transfer from the weak acid to the weak base.

HA(aq) + B(aq) ⇌ BH+(aq) + A–(aq) 2. Obtain Kn from Ka for the weak acid, Kb for the weak base, and 1/Kw: Kn = KaKb(1/Kw).

HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq) Ka B(aq) + H2O(l) ⇌ BH+(aq) + OH–(aq) Kb H3O+(aq) + OH–(aq) ⇌ 2 H2O(l) 1/Kw 3. Less tendency to proceed to completion than neutralizations involving strong acids or strong bases

1Test Item File Questions: Multiple Choice 1 – 4, Algorithmic 1 – 10, Short Answer 1 – 4

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Chapter 17—Applications of Aqueous Equilibria 2

17.2. The Common-Ion Effect A. Common-ion effect – the shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium B. Example of Le Châtelier’s principle C. Determination of the pH of a solution prepared from a weak acid and a salt of its conjugate base 1. Identify the acid–base properties of the various species in solution.

2. Consider the possible proton-transfer reaction these species can undergo.

3. Principal reaction – dissociation of the weak acid 4. Set up a table of concentrations of species involved in the principal reaction.

5. A– comes from 2 sources. a. Salt of the conjugate base, which provides the initial concentration of A – b. Dissociation of weak acid, which determines the change in the concentration of A – 3 17.3. Buffer Solutions A. Buffer solutions 1. Resist drastic changes in pH 2. Contain a weak acid and its conjugate base or a weak base and its conjugate acid B. Add a small amount of base to a buffer solution – acid component of the solution neutralizes the added base.

C. Add a small amount of acid to a buffer solution – base component of the solution neutralizes the added acid.

D. Important in biological systems – pH of human blood (pH = 7.4) controlled by conjugate acid– base pairs (H2CO3/HCO3–).

E. For a buffer prepared from a weak acid, HA, and its conjugate base, B. [HA] [H 3O+ ] = Ka [B] 1. Can use initial concentrations in the calculations 2. Ka is small for commonly used buffer solutions.

3. Initial concentrations are relatively large.

4. Change in concentration is generally negligible.

F. Addition of acid or base to a buffer solution 1. Must take neutralization into account before calculating [H 3O+] a. Addition of OH– HA(aq) + OH–(aq) ⇌ H2O(l) + A–(aq) b. Addition of H3O+ A–(aq) + H3O+(aq) ⇌ H2O(l) + HA(aq) 2. Neutralization alters the number of moles of acid and conjugate base.

3. Concentration ratio, [HA]/[B], changes by only a small amount.

4. [H3O+] changes by only a small amount.

G. Buffer capacity – a measure of the amount of acid or base that a buffer solution can absorb without a significant change in pH 1. Depends on how much weak acid and conjugate base is present.

2. For equal volumes of solutions, the more concentrated the solution, the greater the buffer capacity.

3. For solutions with the same concentration, increasing the volume increases the buffer capacity.

2Test Item File Questions: Multiple Choice 5 – 8, Algorithmic 11 – 17, Short Answer 5 – 8 3Test Item File Questions: Multiple Choice 9 – 21, Algorithmic 18 – 22

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Chapter 17—Applications of Aqueous Equilibria 4

17.4. The Henderson–Hasselbalch Equation A. Explains how pH affects the percent dissociation of a weak acid. [base] pH = pKa + log [acid] [base] log = pH − pK a [acid] B. Useful for determining how to prepare a buffer solution 1. Select a weak acid whose pKa is close to the desired pH of the buffer solution.

2. pKa of the weak acid should be within 1 pH units of the desired pH. 3. Adjust the [base]/[acid] ratio.

C. The pH of a buffer solution does not depend on the volume of the solution; depends only on pKa and the relative molar amounts of weak acid and conjugate base.

17.5. pH Titration Curves A. Acid–base titration – a titrant, a solution containing a known concentration of base (or acid), is slowly added from a buret to a solution containing an unknown concentration of acid (or base).

B. Equivalence point – the point at which stoichiometrically equivalent quantities of acid and base have been mixed together.

C. pH titration curve – a plot of the pH of the solution versus the volume of added titrant.

1. Use to identify the equivalence point in a titration 2. Useful in selecting a suitable indicator to signal the equivalence point 3. Calculate pH titration curves from the principles of aqueous solution equilibria 5 17.6. Strong Acid–Strong Base Titrations A. Before addition of any strong base – initial [H3O+] = concentration of strong acid B. Addition of strong base before the equivalence point 1. Decrease in [H3O+] – added base will neutralize some of the H3O+ present.

2. mmol H3O+ after neutralization = mmol H3O+ initial – mmol OH– added. [mmol H 3O+ after neutralization] [H 3O+ ] after neutralization = [total volume of acid and base] C. Addition of strong base at the equivalence point.

1. Add just enough base to neutralize all the acid initially present 2. pH = 7.00 3. Solution contains water and a salt derived from a strong base and a strong acid.

D. Addition of strong base after the equivalence point 1. Excess of OH– present 2. mmol of excess OH– = mmol of OH– added – mmol of acid initially present

[OH− ] after neutralization = [H 3O + ] =

[mmol of excess OH− ] [total volume of acid and base]

[ Kw ]

[OH − ] E. pH titration curve 1. Plot of pH versus milliliters of strong base added 2. Sharp increase in pH in region near the equivalence point a. Characteristic of the titration curve for any strong acid–strong base titration b. Can use to identify the equivalence point when the concentration of the acid is unknown

4Test Item File Questions: Multiple Choice 38 – 47, Short Answer 9 – 15 5Test Item File Questions: Multiple Choice 48 – 49, Algorithmic 23 – 24, Short Answer 16 – 18

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17.7. Weak Acid–Strong Base Titrations A. Before addition of strong base – pH calculated in the same manner as for a solution of a weak acid B. Addition of strong base before the equivalence point 1. mmol of base, A–, after neutralization = mL of strong base  [strong base] mmol of A − 2. [A − ] = total mL of acid and base 3. mmol of acid, HA, after neutralization = mmol of HA initially – mmol of A– mmol HA after neutralization [HA] after neutralization = total volume of acid and base [A − ] [HA] C. Halfway to the equivalence point 1. Exactly half of the weak acid has been neutralized.

2. [A−] = [HA] 3. pH = pKa of the weak acid D. Addition of strong base at the equivalence point 1. Basic salt solution.

2. mmol A– after neutralization = mL of strong base  [strong base] mmol of A − [A − ] = total mL of acid and base 3. Calculate pH from the method outlined in Section 16.14 in your textbook.

4. pH > 7.00 due to hydrolysis of the anion, which behaves as a weak base.

E. Addition of strong base after the equivalence point.

1. mmol OH– added = mL of strong base  [strong base] 2. mmol A– present = mmol A– at the equivalence point pH = pK a + log

3. mmol OH– present = mmol OH– added – mmol A– present mmol OH − present [OH - ] present = total mL of acid and base [H 3O + ] =

[ Kw ]

[OH − ] F. Key features of the weak acid–strong base titration curve compared to the strong acid–strong base itration curve: 1. The initial rise in pH is greater for the weak acid–strong base curve but is more level in the region midway to the equivalence point.

2. The increase in the pH near the equivalence point is smaller for the weak acid–strong base curve.

3. The pH at the equivalence point is greater than 7.

G. Can use pH range at the equivalence point to select a suitable indicator.

7 17.8. Weak Base–Strong Acid Titrations A. Before addition of strong acid – calculate pH in the same manner as a weak base.

B. Addition of strong acid before the equivalence point 1. Have a B/BH+° buffer solution 2. Observe a leveling of the titration curve in the buffer region between the start of the titration and the equivalence point.

3. pH calculated from the Henderson–Hasselbalch equation 6Test Item File Questions: Multiple Choice 60 – 66, Algorithmic 25 – 32, Short Answer 19 7Test Item File Questions: Multiple Choice 71 – 73, Algorithmic 33 – 37, Short Answer 20

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Chapter 17—Applications of Aqueous Equilibria C. Addition of strong acid at the equivalence point 1. Acidic salt solution of BH+ 2. pH calculated using the method outlined in textbook Section 15.14 3. pH < 7.00 D. Addition of strong acid after the equivalence point 1. Excess H3O+ present.

2. pH calculated from excess [H3O+] 8

17.9. Polyprotic Acid–Strong Base Titrations A. Diprotic acid – has two dissociable protons and reacts with two molar amounts of OH – B. Amino acids – both acidic and basic and can be protonated by strong acids 1. H2A+– protonated form; neutralized to HA H2A+(aq) + H2O(l) ⇌ H3O+(aq) + HA(aq) Ka1 2. HA – neutralized to A– HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq) K a2 C. Before addition of strong base 1. Calculate the pH of a diprotic acid.

2. Principal reaction – dissociation of H2A+ D. Addition of strong base before the first equivalence point 1. H2A+ is converted to HA. 2. H2A+/HA buffer solution.

3. Calculate [H2A+] and [HA] of the amino acid in the same way [HA] and [A –] are calculated in a weak acid–strong base titration.

4. Use the Henderson-Hasselbalch equation to calculate pH. E. Addition of strong base at the first equivalence point 1. All of the H2A+ is converted to HA. 2. Principal reaction – proton transfer between HA molecules 2 HA(aq) ⇌ H2A+(aq) + A–(aq) K = Ka2/Ka1 3. pH at the first equivalence point = average of pKa1 and pKa2 pKa1 + pKa2 pH = 2 a. Isoelectric point – the [HA] is at a maximum and the [H2A+] and [A–] are very small and equal b. Useful for separating mixtures of amino acids F. Addition of strong base between the first and second equivalence points 1. HA converted to A– 2. HA/ A– buffer solution 3. Calculate [HA] and [A–] of the amino acid in the same way [HA] and [A–] are calculated in a weak acid–strong base titration.

4. Use the Henderson-Hasselbalch equation to calculate pH. G. Addition of strong base at the second equivalence point 1. All of the HA is converted A–.

2. [A–] = [H2A+] initially 3. Solution of a basic salt a. Principal reaction A–(aq) + H2O(l) ⇌ HA(aq) + OH–(aq) Kb = Kw/Ka2 b. Calculate [OH–] from equilibrium equation; calculate [H3O+] and pH from [OH–] H. Addition of strong base beyond the second equivalence point – pH calculated from the excess [OH–]

8Test Item File Questions: Multiple Choice 74 – 78, Short Answer 21

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Chapter 17—Applications of Aqueous Equilibria 9

17.10. Solubility Equilibria for Ionic Compounds A. Dissolution or precipitation of a sparingly soluble ionic compound B. Principles of solubility equilibria – examine quantitative aspects of solubility and precipitation phenomena C. Solubility product constant = Ksp 1. MaXb(s) ⇌ a Mb+(aq) + b Xa–(aq) 2. Ksp = [Mb+]a[Xa–]b 10 17.11. Measuring Ksp and Calculating Solubility from Ksp A. Ksp – measured by experiment B. Temperature dependent C. Use to calculate solubility of a compound D. Can be calculated from the molar solubility of a compound 11 17.12. Factors That Affect Solubility A. Common-ion effect – solubility of a slightly soluble ionic compound is decreased by the presence of a common ion in the solution B. pH of the solution – solubility of an ionic compound increases with decreasing pH of solution if the compound contains a basic anion C. Formation of complex ions 1. Solubility of an ionic compound increases dramatically if the solution contains a Lewis base that can form a coordinate covalent bond to the metal cation.

2. Complex – ion that contains a metal cation bonded to one or more small molecules or ions.

3. Formation constant, Kf a. Equilibrium constant for complex ion formation from the hydrated metal cation b. Measures complex ion stability D. Amphoterism 1. Amphoteric oxides – soluble both in strongly acidic and in strongly basic solutions 12 17.13. Precipitation of Ionic Compounds A. Ion product (IP) – determines if a precipitate will form when solutions containing constituent ions are mixed 1. For the salt, MaXb; IP = [Mb+]a[Xa–]b 2. Defined in the same way as Ksp 3. Concentrations are initial concentrations, not equilibrium concentrations.

B. IP > Ksp 1. Supersaturated solution 2. Precipitation occurs C. IP = Ksp 1. Saturated solution 2. Equilibrium exists D. IP < Ksp 1. Unsaturated solution 2. Precipitation will not occur

9Test Item File Questions: Algorithmic 38 – 39, 10Test Item File Questions: Multiple Choice 94 – 95, Algorithmic 40 – 47, Short Answer 22 – 24 11Test Item File Questions: Multiple Choice 96 – 105, Algorithmic 48 – 51, Short Answer 26 12Test Item File Questions: Multiple Choice 117 – 123, Algorithmic 52 – 54, Short Answer 27

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Chapter 17—Applications of Aqueous Equilibria 13

17.14. Separation of Ions by Selective Precipitation A. To separate a mixture of ions in solution, add a reagent that will precipitate some ions but not others.

B. Can separate insoluble metal sulfides from the more soluble metal sulfides 1. Carry out separation in acidic solution.

2. MS(s) + 2 H3O+(aq) ⇌ M+(aq) + H2S(aq) + 2 H3O(l) 3. Kspa, solubility product in acid [M 2+ ][H 2S] Kspa = [H 3O+ ] 4. Qc > Kspa; metal sulfide precipitates out.

5. Adjust [H3O+] so that Qc > Kspa for the more insoluble metal sulfide.

6. The more insoluble metal sulfide will precipitate out from the more soluble metal sulfide.

17.15. Qualitative Analysis A. Procedure for identifying ions present in an unknown solution B. Traditional scheme of analysis for metal cations involves separation of 20 cations into five groups by selective precipitation – textbook Figure 16.18 C. Determine the presence or absence of the ions in each group with further separations and tests D. Excellent vehicle for developing laboratory skills and learning about acid–base, solubility, and complex-ion equilibria Inquiry – What is causing ocean acidification?

13Test Item File Questions: Algorithmic 55 – 56

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CHAPTER 17 APPLICATIONS OF AQUEOUS EQUILIBRIA Teaching Tips, Points of Emphasis, and Common Misconceptions Section 17.2

The value of x is negligible compared with the initial concentration and can be ignored when subtracted from or added to the initial concentration; however, by itself x is significant.

Section 17.3

A mixture of a weak acid with a salt of its conjugate base is an acidic buffer. In calculations involving an acidic buffer, the only chemical equation needed is that for the dissociation of the weak acid: HA + H2O ⇌ H3O+ + A–.

Section 17.3

A mixture of a weak base with a salt of its conjugate acid is a basic buffer. In calculations involving a basic buffer, the only chemical equation needed is that for the dissociation of the weak base: B + H2O ⇌ HB+ + OH–.

Section 17.3

Use the following approach for any neutralization reaction: (1) Determine the moles (or concentration) of each species before reaction. (2) Allow the reaction to go to completion, and determine the mole of each species “after reaction.” (3) Allow the system to equilibrate, and calculate the concentration of each species at equilibrium.

Section 17.3

This expression is the basis for the Henderson–Hasselbalch equation, which is used to calculate the pH of a buffer solution as shown in the next section.

Section 17.4

The acid and base in this expression refers to a conjugate acid/base pair, such as CH3CO2H/CH3CO2– or NH4+/NH3. CH3CO2H is the conjugate acid of CH3CO2– and CH3CO2– is the conjugate base of CH3CO2H. Similarly, NH4+ is the conjugate acid of NH3 and NH3 is the conjugate base of NH4+.

Section 17.4

This expression contains a ratio of the concentration of a conjugate base to the concentration of its conjugate acid remaining after reaction. Because the base and its conjugate acid are contained in the same volume, moles of base and conjugate acid remaining after reaction can be used instead of molarities.

Section 17.6

The change in pH seen in a titration curve is due primarily to the neutralization reaction, but it is also due to the dilution effect of mixing two solutions. Volume changes constantly during a titration.

Section 17.6

Titration curve points of particular interest are pH values (1) before any base (or acid) is added, (2) before the equivalence point (especially at the half-equivalence point), (3) at the equivalence point, and (4) beyond the equivalence point.

Section 17.7

For an acid–base titration the pH at the equivalence point will be (1) > 7 if the base is stronger than the acid, (2) = 7 if the acid and base have equal strengths, (3) < 7 if the acid is stronger than the base.

Section 17.7

The weaker the acid, the shorter is the steeply rising portion of the titration curve. If the acid is very weak, the equivalence point is difficult to detect.

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Chapter 17—Applications of Aqueous Equilibria Section 17.9

As noted earlier, the weaker the acid, the shorter is the steeply rising portion of the titration curve. If the acid is very weak, the equivalence point is difficult to detect.

Although H3PO4 is a triprotic acid, HPO42– is such a weak acid that the third equivalence point is not observed in the H3PO4/NaOH titration curve.

Section 17.10

The solubility product constant, Ksp, is for a saturated solution. Unsaturated solutions have not reached equilibrium.

Section 17.11

No ionic compound is 100% insoluble. All ionic compounds dissolve to some extent giving rise to a Ksp expression.

Section 17.11

The relative solubilities of two ionic compounds can be determined by comparing their Ksp values only if they have the same number of ions. It is possible for a 1:1 compound, such as CaCO3, to have a larger Ksp value but a lower solubility than a 1:2 compound, such as CaF2.

Section 17.12

The addition of a common ion to a solution of a sparingly soluble ionic compound lowers the solubility of that compound.

Section 17.12

The solubility of a sparingly soluble ionic compound is increased on addition of H 3O+ if the anion of the compound is the conjugate base of a weak acid.

Section 17.12

As with acid–base neutralization reactions, the best approach to doing complex ion formation problems is to allow the reaction to go to completion and then let the system equilibrate.

Section 17.12

Remember that amphoteric means able to react as an acid or as a base.

Section 17.13

Precipitation reactions involve the mixing of solutions with different ionic compounds that contain a common ion. The compound that precipitates is not one that is contained in the original solutions.

Mixing two solutions with a common ion does not necessarily result in the formation of a precipitate. Calculation of the ion product is required to determine is equilibrium has been reached, resulting in precipitation or if equilibrium has not been reached with no solid product being formed.

Mixing two solutions causes the ion concentrations to change due to dilution.

Section 17.14

164

Like oxides, which react with H+ to form H2O, sulfides become more soluble on addition of H+ due to the formation of the weak acid H 2S.

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Chapter 17—Applications of Aqueous Equilibria

Lecture/Laboratory Demonstration References Section 17.2

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Equilibrium: The Dissociation of Acetic Acid,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 160–161.

Section 17.2

Bassam Z. Shakhashiri, “Effect of Acetate Ion on the Acidity of Acetic Acid: The Common Ion Effect,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp.155–157.

Section 17.2

Lee R. Summerlin and James L. Ealy,Jr., “The Common Ion Effect: Second Demonstration,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 93–94.

Section 17.2

Lee R. Summerlin and James L. Ealy,Jr., “The Common Ion Effect: Ammonium Hydroxide and Ammonium Acetate,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), p. 95.

Section 17.3

Bassam Z. Shakhashiri, “Buffering Action of Alka-Seltzer,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp.186–187.

Section 17.3

Bassam Z. Shakhashiri, “Buffering Action and Capacity,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp.173–185.

Section 17.7

Dianne N. Epp, “Teas as Natural Indicators,” J. Chem. Educ., Vol. 70, 1993, 326.

Section 17.11

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Solubility of Some Silver Compounds,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 83–85.

Section 17.11

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Red and White Precipitates in Sodium Silicate,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), p. 131.

Section 17.12

Bassam Z. Shakhashiri, “Fizzing and Foaming: Reactions of Acids with Carbonates,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp. 96–99.

Section 17.12

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Colorful Complex Ions in Ammonia,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 75–76.

Section 17.12

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Green and Blue Copper Complexes,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 71–72.

Section 17.12

Bassam Z. Shakhashiri, “Acidic and Basic Properties of Oxides,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp. 109–113.

Section 17.13

R. A. Pacer, “Will a precipitate form? Will it dissolve?” J. Chem. Educ., Vol. 71, 1994, 69. Principles of ionic compound precipitate formation are shown in a series of four

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Chapter 17—Applications of Aqueous Equilibria colorful demonstrations: (1) Yellow PbI2 is formed only when Q > Ksp. (2) Upon addition of hydrochloric acid, green NiCO3(s) is dissolved due to the formation of the bluish-green complex ion [Ni(NH3)6]2+. (3) Kinetic effects are seen when MgC2O4(s) fails to form at room temperature from a solution having Q > Ksp but does form when the solution is heated with stirring to 90 oC. (4) Several Ag(I) precipitates and complex ions are formed when OH–, Cl–, NH3, Br–, S2O32–, I–, and S2– are successively added to a solution containing Ag+.

Section 17.14

R. J. Stolzberg, “Shedding a New Light on Solubility Calculations,” J. Chem. Educ., Vol. 65, 1988, 65. Principles of precipitate formation, fractional precipitation, and precipitate dissolution are demonstrated from observing a series of reactions involving chloride and chromate salts of silver. Scattering of a laser beam is used to allow the observation of colloidal precipitate formation.

Literature References Section 17.3

Charles L. Bering, “A Good Idea Leads to a Better Buffer,” J. Chem. Educ., Vol. 64, 1987, 803.

Section 17.3

Edwin S. Gould, “Phosphate Buffers and Telephone Poles – A Useful Analogy with Limitations,” J. Chem. Educ., Vol. 76, 1999, 1511.

Section 17.4

Henry N. Po and N. M. Senozan, “The Henderson–Hasselbalch Equation: Its History and Limitations,” J. Chem. Educ., Vol. 78, 2001, 1499–1503.

Section 17.5

Dennis Barnum, “Predicting Acid–Base Titration Curves Without Calculations,” J. Chem. Educ., Vol 76, 1999, 938–942.

Section 17.7

Ara S. Kooser, Judith L. Jenkins, and Lawrence E. Welch, “Acid–Base Indicators: A New Look at an Old Topic,” J. Chem. Educ., Vol. 78, 2001, 1504–1506.

Section 17.7

Robert C. Mebane and Thomas R. Rybolt, “Edible Acid–Base Indicators,” J. Chem.

Educ., Vol. 62, 1985, 285.

Section 17.11

Stephen J. Hawkes, “What Should We Teach Beginners about Solubility and Solubility Product?” J. Chem. Educ., Vol. 75, 1998, 1179–1181.

Section 17.11

Roy W. Clark and Judith M. Bonicamp, “The Ksp-Solubility Conundrum,” J. Chem.

Educ., Vol. 75, 1998, 1182-1185.

Section 17.11

Andres Raviolo, “Assessing Students’ Conceptual Understanding of Solubility Equilibrium,” J. Chem. Educ., Vol. 78, 2001, 629–631.

Section 17.11

Robert Perkins, “The Useless Tea Kettle,” J. Chem. Educ., Vol. 618, 1984, 383.

Section 17.12

Edward Koubek, Cole McWherter, and George L. Gilbert, “Acid–Base Chemistry of Aluminum Ion in Aqueous Solution,” J. Chem. Educ., Vol. 75, 1998, 60.

Section 17.15

Dale D. Clyde, “Swimming Pools, Hot Rods, and Qualitative Analysis,” J. Chem. Educ., Vol. 65, 1988, 911–913.

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Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 17.3

Worked Example 17.4: Calculating the pH of a buffer after the addition of OH − or H3O+

Section 17.7

Worked Example 17.8: Calculating the pH at the equivalence point in the titration of a weak acid with a strong base

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CHAPTER 18 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND SPONTANEITY Chapter Learning Objectives Section 18.1

Section 18.2 through Section 18.4

Define a spontaneous process and classify various physical processes and chemical reactions as spontaneous or nonspontaneous.

Predict the sign of ΔS for various physical processes and chemical reactions.

Determine the number of arrangements in a system and use Boltzmann’s equation to calculate the entropy.

Calculate ΔS for the expansion or compression of an ideal gas at constant temperature.

Predict how the phase of matter, or conditions such as temperature, volume, and pressure affect the entropy of a substance.

Section 18.5

Predict how atomic structure affects the entropy of a substance.

Use standard molar entropies (S°) to calculate the standard entropy of reaction (ΔS°).

Section 18.6

Describe the factors that affect spontaneity of a reaction.

Calculate values of ΔSsys, ΔSsurr, and ΔStotal for a reaction, and use these values to determine whether the reaction is spontaneous.

Section 18.7

Describe how the sign and magnitude of ΔS, ΔH, and temperature affect the value of ΔG and the spontaneity of the reaction.

Estimate the temperature at which a reaction changes between spontaneous and nonspontaneous.

Section 18.8 and Section 18.9

Calculate the standard free-energy change (ΔG°) from ΔH° and ΔS° and interpret the meaning of the sign of ΔG°.

Calculate ΔG° for a reaction from values of ΔG°f and determine if the reaction is spontaneous under standard conditions and how temperature will affect spontaneity.

Section 18.10 and Section 18.11

Calculate ΔG for a reaction under nonstandard-state conditions.

Use the relationship between ΔG° and the equilibrium constant for a reaction, and be able to calculate one from the other.

Interpret the shape of the curve in a plot of free energy versus reaction progress.

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Chapter 18—Thermodynamics: Entropy, Free Energy, and Spontaneity

Lecture Outline 1

18.1. Spontaneous Processes A. Spontaneous process – one that proceeds on its own without any external influence B. A spontaneous reaction always moves a system toward equilibrium.

1. Spontaneity of forward or reverse reaction depends on a. Temperature b. Pressure c. Composition of the reaction mixture 2. Q < K – reaction proceeds in the forward direction.

3. Q > K – reaction proceeds in the reverse direction.

C. Spontaneity of a reaction is no indication of the speed of the reaction.

2 18.2. Enthalpy, Entropy, and Spontaneous Processes A. Spontaneous reactions 1. Some are accompanied by conversion of potential energy to heat – reactions are exothermic.

2. Some spontaneous reactions are endothermic – system moves spontaneously to a state of higher potential energy by absorbing heat from the surroundings.

3. Enthalpy alone does not account for the direction of spontaneous change.

4. Second thermodynamic driving force for spontaneous change – nature’s tendency to move to a condition of maximum randomness B. Molecular systems tend to move spontaneously to a state of maximum randomness.

C. Entropy, S – molecular randomness or disorder 1. State function 2. S = Sfinal – Sinitial 3. When S is positive there is an increase in randomness or disorder of a system.

4. Examples a. Solid → liquid b. Liquid → gas c. A reaction that results in an increase in the number of gaseous molecules d. When certain solutes dissolve 5. When S is negative there is a decrease in randomness or disorder of a system.

3 18.3. Entropy and Probability A. Systems tend to a state of maximum randomness because the random state, which is more probable, can be achieved in more ways.

B. Probabilities of ordered and random states are proportional to the number of ways that the state can be achieved.

C. Boltzmann – The entropy of a particular state is related to the number of ways that the state can be achieved.

S = k ln W 1. ln W = the number of ways that the state can be achieved 2. k = 1.38  10–23 J/K D. Gas expands spontaneously because the state of greater volume is more probable.

1. For an ideal gas V DS = R ln final Vinitial

1Test Item File Questions: Multiple Choice 1 – 6, Algorithmic 1 – 2, Short Answer 1 2Test Item File Questions: Multiple Choice 7 – 11, Algorithmic 3 – 6, Short Answer 2-3 3Test Item File Questions: Multiple Choice 12 – 21, Algorithmic 7 – 10, Short Answer 4 – 8

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P=

nRT V

DS = R ln

Pinitial

Pfinal 2. The entropy of a gas increases when its pressure decreases at constant temperature.

3. The entropy of a gas decreases when the pressure increases at constant temperature.

4 18.4. Entropy and Temperature A. Entropy – also associated with molecular motion B. Random molecular motion increases as the temperature of a substance increases.

1. Increase in the average kinetic energy of the molecules a. Total energy is distributed among the individual molecules in a number of ways. b. Boltzmann – The more ways, W, that the energy can be distributed the greater the randomness of the state and the higher the entropy.

2. Plot of entropy versus temperature – at absolute zero, every substance is a solid where particles are rigidly fixed in a crystalline structure C. Third law of thermodynamics — The entropy of a perfectly ordered crystalline substance at 0 K is zero.

5 18.5. Standard Molar Entropies and Standard Entropies of Reaction A. Standard molar entropy, S 1. The entropy of 1 mole of the pure substance at 1 atm pressure and a specified temperature, usually 25 C. 2. Absolute entropies – measured with respect to an absolute reference point B. Standard entropy of reaction, S 1. S = S(products) – S(reactants) 2. Multiply the S value for each substance by the stoichiometric coefficient of that substance in the balanced equation.

3. Entropy increases whenever a molecule breaks into two or more pieces.

6 18.6. Entropy and the Second Law of Thermodynamics A. Value of G – criterion for spontaneity 1. G = H – TS 2. G > 0: nonspontaneous reaction 3. G < 0: spontaneous reaction 4. G = 0: reaction is at equilibrium B. First law of thermodynamics – in any process, spontaneous or nonspontaneous, the total energy of a system and its surroundings is constant.

1. Helps keep track of energy flow between system and the surroundings 2. Does not indicate the spontaneity of the process C. Second law of thermodynamics – in any spontaneous process, the total entropy of a system and its surroundings always increases.

1. Provides a clear-cut criterion of spontaneity 2. Direction of spontaneous change is always determined by the sign of the total entropy change.

DStotal = DSsystem + DSsurroundings a. Stotal > 0; spontaneous reaction b. Stotal < 0; nonspontaneous reaction c. Stotal = 0; reaction at equilibrium

4Test Item File Questions: Multiple Choice 22 – 23, Algorithmic 11 – 15, Short Answer 9 5Test Item File Questions: Multiple Choice 24 – 27, Algorithmic 16 – 21, Short Answer 10 – 12 6Test Item File Questions: Multiple Choice 29 – 33, Algorithmic 22 – 24, Short Answer 13 – 16

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Chapter 18—Thermodynamics: Entropy, Free Energy, and Spontaneity D. All reactions proceed spontaneously in the direction that increases the entropy of the system plus surroundings.

E. To determine Stotal, need to know Ssystem and Ssurroundings 1.  Ssystem a. The entropy of reaction b. Calculate from standard molar entropies 2. At constant pressure DH rxn DSsurroundings = T DH rxn o DStotal = DSrxn T 7 18.7. Free Energy and the Spontaneity of Chemical Reactions A. Second law restated in terms of the thermodynamic properties of the system – chemists are more interested in the system B. Free energy: G = H – TS 1. TS = part of the system’s energy that is already disordered 2. H – TS = part of the system’s energy that is still ordered and therefore free to cause spontaneous change by becoming disordered 3. State function G = H – TS C. Relationship between free energy and spontaneity 1. Stotal = S – H/T 2. –TStotal = H – TS 3. –TStotal = G 4. G and Stotal have opposite signs.

D. In any spontaneous process at constant temperature and pressure, the free energy of the system decreases.

E. Temperature acts as a weighting factor that determines the relative importance of the enthalpy and entropy changes (textbook Table 18.2).

1. When H and S have the same sign, the spontaneity is temperature dependent, as the sign of G depends on the temperature.

2. Can estimate the temperature at which G changes from positive to negative with

DG = DH - TDS

3. To estimate the temperature at which G changes from a positive to a negative value, set G = H – TS = 0. H T= S 8

18.8. Standard Free-Energy Changes for Reactions A. Standard free-energy change, G 1. Change in free energy that occurs when reactants in their standard states are converted to products in their standard states 2. G – an extensive property; refers to the number of moles indicated in the chemical equation 3. Calculate G from standard enthalpy change, H, and standard entropy change, S. G = H – TS

7Test Item File Questions: Multiple Choice 34 – 46, Algorithmic 25 – 27, Short Answer 17 – 20 8Test Item File Questions: Multiple Choice 67 – 70, Algorithmic 28, Short Answer 21

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18.9. Standard Free Energies of Formation A. Standard free energy of formation, DG f 1. Free-energy change for formation of 1 mole of the substance in its standard state from the most stable form of the constituent elements in their standard states 2. For an element in its most stable form at 25 C, DG f = 0 B.

DGf measures a substance’s thermodynamic stability with respect to its constituent elements.

1. DG f is negative – substance is stable and does not decompose to its constituent elements

2. DG f is positive – substance is thermodynamically unstable (but may not decompose to its constituent elements if the rate of decomposition is slow) C. Can use DG f to calculate standard free-energy changes for reactions

G = Gf ( products ) − Gf ( reactants )

10

18.10. Free-Energy Changes for Reactions under Nonstandard-State Conditions A. Free-energy change for a reaction when the reactants and products are present at nonstandardstate pressures and concentrations B. G = G + RT ln Q C. Q = the reaction quotient — an expression having the same form as the equilibrium constant expression (but concentrations are not necessarily at equilibrium) 18.11. Free Energy and Chemical Equilibrium11 A. When the reaction mixture is mostly reactants 1. Q << 1 2. RT ln Q << 0 3. G < 0 4. Total free energy decreases as the reaction proceeds spontaneously in the forward direction.

B. When the reaction mixture is mostly products 1. Q >> 1 2. RT ln Q > 0 3. G > 0 4. Total free energy decreases as the reaction proceeds spontaneously in the reverse direction.

C. Free-energy curve – shows how the total free energy of a reaction mixture changes as the reaction progresses (textbook Figure 18.11) 1. Goes through a minimum somewhere between pure reactants and pure products 2. System is at equilibrium D. Relationship between free energy and the equilibrium constant 1. At equilibrium a. G = 0 b. Q = K 2. G = −RT lnK E. Relationship between G and K 1. G < 0, ln K > 0 – equilibrium mixture is mainly products 2. G > 0, ln K < 0 – equilibrium mixture is mainly reactants 3. G = 0, ln K = 0 – equilibrium mixture contains comparable amounts of reactions and products

9Test Item File Questions: Multiple Choice 71 – 77, Algorithmic 29, Short Answer 22 10Test Item File Questions: Multiple Choice 78 – 84, Algorithmic 30 11Test Item File Questions: Multiple Choice 85 – 94, Algorithmic 31 – 32, Short Answer 23 – 27

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Chapter 18—Thermodynamics: Entropy, Free Energy, and Spontaneity F.

Properties of nature that determine the direction and extent of a chemical reaction 1. K determined by Grxn

2. Grxn determined by H f and S rxn Inquiry – Does the formation of highly ordered molecules violate the second law of thermodynamics?

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CHAPTER 18 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND SPONTANEITY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 18.1

Most, but not all, spontaneous chemical and physical changes are accompanied by heat transfer from system to surroundings (H is negative) and a lowering of the potential energy of the system. When H is positive for a spontaneous physical or chemical change, another driving force (entropy) is playing an important role.

Section 18.3

The value of R used in energy-related calculations is 8.314 J/K, not 0.0821 Latm mol–1 K–1.

Section 18.3

The greater the volume (and the lower the pressure) of a gas, the greater the entropy.

Section 18.4

Entropy increases with an increase in the motion of atoms, ions, and molecules. The higher the temperature, the more rapid the motion and the greater the entropy.

Section 18.5

In general, So (gas) > So (liquid) > So (solid), and So (large molecule) > So (small molecule).

Section 18.5

Remind students to multiply the standard molar entropy of each reactant and product by the appropriate coefficient from the balanced chemical equation when calculating standard entropy of a reaction.

Section 18.6

A reaction is spontaneous only if Ssystem + Ssurroundings = Stotal > 0. Ssystem is calculated from standard molar entropies. Ssystem = –H/T, where –H is calculated from standard molar enthalpies.

Section 18.7

Two types of reactions have a spontaneity that cannot be reversed by changing the temperature (those with H and S both negative or both positive), and two have a spontaneity that can be reversed by changing the temperature (those with opposite signs for H and S).

Section 18.7

According to G = H – TS, a spontaneous reaction is favored by a decrease in enthalpy and an increase in entropy. At lower temperatures the enthalpy term usually dominates; at higher temperatures the entropy term can become dominant.

Section 18.9

For an element in its most stable form at 25 C, both the standard free energy of formation and the standard enthalpy of formation are equal to zero.

Section 18.9

A substance with a negative standard free energy of formation is stable and does not decompose to its constituent elements. However, the substance may be unstable with respect to some other reaction under the same set of conditions.

Section 18.11

When all reactants and products are under standard conditions, Q = 1 and G = Go.

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Lecture/Laboratory Demonstration References Section 18.4

Brian B. Laird, “Entropy, Disorder, and Freezing,” J. Chem. Educ., Vol. 76, 1999, 1388–1390.

Literature References Section 18.1

Ruth Ben-Zvi, Judith Silberstein, and Rachel Mamiok, “A Model of Thermal Equilibrium: A Tool for the Introduction of Thermodynamics,” J. Chem. Educ., Vol. 70, 1993, 31–34.

Section 18.1

James N. Spencer, Richard S. Moog, and Ronald J. Gillespie, “Demystifying Introductory Chemistry. Part 4: An Approach to Reaction Thermodynamics Through Enthalpies, Entropies, and Free Energies of Atomization,” J. Chem. Educ., Vol. 73, 1996, 631–636.

Section 18.1

Raymond S. Ochs, “Thermodynamics and Spontaneity,” J. Chem. Educ., Vol. 73, 1996, 952–954.

Section 18.2

John J. Fortman, “Pictorial Analogies of Heat Flow, Thermodynamics, and Entropy,” J. Chem. Educ., Vol. 70, 1993, 102–103.

Section 18.2

Joseph H. Lechner, “Visualizing Entropy,” J. Chem. Educ., Vol. 76, 1999, 1382–1387.

Section 18.2

Peter Weiss, “Another Face of Entropy,” Science News, Vol. 154, 1998, 108–109.

Section 18.2

L. Glasser, “Order, Chaos, and All That!” J. Chem. Educ., Vol. 66, 1989, 997–1001.

Section 18.2

John P. Lowe, “Entropy: Conceptual Disorder,” J. Chem. Educ., Vol. 65, 1988, 403–406.

Section 18.3

Douglas K. Russell, “The Boltzmann Distribution,” J. Chem. Educ., Vol. 73, 1996, 299–300.

Section 18.3

P. G. Nelson, “Derivation of the Second Law of Thermodynamics from Boltzmann’s Distribution Law,” J. Chem. Educ., Vol. 65, 1989, 390–393.

Section 18.5

Travis Thoms, “Periodic Trends for the Entropy of Elements,” J. Chem. Educ., Vol. 72, 1995, 16.

Section 18.7

Sidney Rosen, “J. Willard Gibbs (1839-1903): A Modern Genius,” J. Chem. Educ., Vol. 60, 1983, 593–594.

Section 18.7

Robert J. Deltete and David L. Thorsell, “Josiah Willard Gibbs and Wilhelm Ostwald: A Contrast in Scientific Style,” J. Chem. Educ., Vol. 73, 1996, 289–295.

Section 18.11

Zheng Xianmin, “The Free Energy Prediction and the Principle of Le Châtelier,” J. Chem. Educ., Vol. 66, 1989, 401–402

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Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 18.2

Worked Example 18.1: Predicting the sign of S

Section 18.10

Worked Example 18.8: Calculating G for a reaction under nonstandard-state conditions

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CHAPTER 19 ELECTROCHEMISTRY Chapter Learning Objectives Section 19.1

Assign oxidation numbers and classify half-reactions as oxidation or reduction reactions.

Use the half-reaction method to balance a redox reaction in acidic solution.

Use the half-reaction method to balance a redox reaction in basic solution

Section 19.2 and Section 19.3

Draw a galvanic cell. Label the anode and cathode, indicate the direction of electron and ion flow, and write balanced equations for the electrode and overall cell reactions.

Use shorthand notation to represent a galvanic cell.

Section 19.4

Section 19.5 and Section 19.6

Interconvert between cell potential and standard free-energy change for electrochemical reactions.

Use tabulated values of standard reduction potentials for half-reactions (textbook Appendix D) to order substances by increasing strength as an oxidizing or reducing agent.

Use standard reduction potentials for half-reactions to calculate cell potential of a redox reaction and determine spontaneity.

Use standard reduction potentials for half-reactions to calculate the voltage of a galvanic cell.

Section 19.7 and Section 19.8

Calculate the cell potential under nonstandard-state conditions using the Nernst equation.

Use the cell potential to calculate solution concentration.

Section 19.9

Use standard reduction potentials to calculate the equilibrium constant.

Section 19.10 and Section 19.11

Describe batteries, and calculate values of E°, G°, and K for the cell reaction.

Describe how corrosion occurs and electrochemical methods used to prevent it.

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Chapter 19—Electrochemistry Section 19.12 through Section 19.14

Draw an electrolytic cell. Label the anode and cathode, indicate the direction of electron and ion flow, and write balanced equations for the electrode and overall cell reactions.

Predict the products formed and the electrode and overall cell reactions when an aqueous solution of an ionic compound is electrolyzed.

Relate the current, time, and amount of product produced in an electrolytic cell.

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Lecture Outline 1

19.1. Balancing Redox Reactions by the Half-Reaction Method A. Half-reaction method of balancing redox reactions – focuses on transfer of electrons B. Key – overall reaction broken into two parts (half-reactions) 1. Oxidation part of the reaction 2. Reduction part of the reaction C. Steps – textbook Figure 19.1 2 19.2. Galvanic Cells A. Galvanic cell – a spontaneous chemical reaction generates an electric current B. Electrolytic cell – an electric current drives a nonspontaneous reaction C. Redox reaction 1. Oxidation – a loss of electrons (an increase in oxidation number) 2. Reduction – a gain of electrons (a decrease in oxidation number) 3. Represent oxidation and reduction aspects of the reaction with half-reactions 4. Oxidizing agent – species that causes oxidation to occur and is itself reduced 5. Reducing agent – species that causes reduction to occur and is itself oxidized 6. If a spontaneous reaction is carried out in a beaker a. Oxidizing agent and reducing agent are in direct contact. b. Electrons are directly transferred. c. Enthalpy of reaction is lost to the surroundings for an exothermic reaction.

7. If a spontaneous reaction is carried out in a galvanic cell – chemical energy released by the reaction is converted to electrical energy D. For the reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) 1. Use a Daniell cell, a type of galvanic cell, to carry out the reaction (textbook Figure 18.2). a. Consists of two half-cells i. A beaker with a strip of Zn in a solution of ZnSO 4 ii. A beaker with a strip of Cu in a solution of CuSO4 b. Electrodes – strips of zinc and copper c. Salt bridge – a U-shaped tube that contains a gel permeated with a solution of an inert electrolyte 2. The electrons can be transferred only through the wire. a. Oxidation and reduction half-reactions occur at separate electrodes. b. Electric current flows through the wire.

3. Anode – the electrode at which oxidation takes place a. The negative (–) electrode b. Produces electrons 4. Cathode – the electrode at which reduction takes place a. The positive (+) electrode b. Consumes electrons 5. Anode and cathode half-reactions must add to give the overall cell reaction.

6. Salt bridge maintains electrical neutrality by a flow of ions. a. Anions flow through the salt bridge from the cathode to the anode compartment. b. Cations migrate through the salt bridge from the anode to the cathode compartment.

7. Electrons move through the external circuit from the anode to the cathode.

1Test Item File Questions: Multiple Choice 1 – 13, Algorithmic 1 – 15, Short Answer 1 – 5 2Test Item File Questions: Multiple Choice 14 – 23, Algorithmic 16 – 20, Short Answer 6

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19.3. Shorthand Notation for Galvanic Cells A. Single vertical line (|) represents a phase boundary.

B. Double vertical line (||) represents a salt bridge.

C. Shorthand for the anode half-cell is always written on the left of the salt-bridge symbol, followed on the right of the symbol by the shorthand for the cathode half-cell.

1. Reactants in each half-cell are written first, followed by products.

2. Electrodes are written on extreme left (anode) or extreme right (cathode).

3. Electrons move through the external circuit from left to right (anode to cathode).

4. For Zn (s) + Cu2+(aq) → Zn2+(aq) + Cu(s): Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) D. Cell involving a gas 1. Additional vertical line due to presence of additional phase 2. List the gas immediately adjacent to the appropriate electrode.

E. Detailed notation includes ion concentrations and gas pressures.

4 19.4. Cell Potentials and Free-Energy Changes for Cell Reactions A. Electromotive force (emf) – the driving force (electrical potential) that pushes the negatively charged electrons away from the anode and pulls them toward the cathode.

1. Also called cell potential (E) or cell voltage 2. Potential of a galvanic cell is a positive quantity with units of volts (V).

B. Coulomb (C) – amount of charge transferred when a current of 1 ampere (A) flows for 1 s.

1J=1C1V C. Cell potential – measured with a voltmeter 1. Gives a positive reading when the + and – terminals of the voltmeter are connected to cathode (+) and anode (–), respectively 2. Can use voltmeter–cell connections to determine which electrode is the anode and which is the cathode D. Two driving forces of a chemical reaction: cell potential, E, and free-energy change, G 1. Related by G = −nFE a. n = number of moles of electrons transferred in the reaction b. F (faraday) – the electrical charge on 1 mol of electrons c. 1 F = 96,500 C/mol e– 2. G and E have opposite signs.

3. Spontaneous reaction has a positive cell potential but negative G. E. Standard cell potential, E– the cell potential when both reactants and products are in their standard states 1. Solutes at 1 M concentration 2. Gases at a partial pressure of 1 atm 3. Solids and liquids in pure form 4. T = 25 C F. G = –nFE 5 19.5. Standard Reduction Potentials A. Standard cell potential is the sum of anode and cathode half-cell potentials. o o o Ecell = Eanode + Ecathode 1. Potential of a single electrode not measurable 2. Potential difference measured by placing a voltmeter between two electrodes 3. Set of standard half-cell potentials developed a. Arbitrary standard half-cell chosen as reference point – assigned an arbitrary potential b. Potential of all other half-cells expressed relative to the reference half-cell

3Test Item File Questions: Multiple Choice 24 – 32, Algorithmic 21 – 28, Short Answer 8 – 12 4 Test Item File Questions: Multiple Choice 40 – 48, Algorithmic 29 – 30, Short Answer 13 – 15 5 Test Item File Questions: Multiple Choice 56 – 59, Algorithmic 31 – 34, Short Answer 16 – 18

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Chapter 19—Electrochemistry B. Standard hydrogen electrode (S.H.E.) – reference half-cell 1. Corresponding half-reaction – assigned an arbitrary potential of exactly 0 V 2 H+(aq, 1 M) + 2 e– → H2(g, 1 atm) Eo = 0.00 V 2. Shorthand notation for S. H. E. H+(1 M) | H2(1 atm) | Pt(s) C. Standard potentials for half-cells determined by constructing a galvanic cell with the half-cell of interest paired with the standard hydrogen electrode 1. Standard oxidation potential – the corresponding half-cell potential for an oxidation halfreaction 2. Standard reduction potential – the corresponding half-cell potential for a reduction halfreaction 3. Whenever half-reaction is reversed, the sign of Eo must be reversed.

4. Standard oxidation potential and standard reduction potential always have the same magnitude but opposite signs.

5. Table of standard reduction potentials – textbook Table 18.1 and Appendix D D. Conventions used in constructing a table of half-cell potentials 1. Half-reactions written as reductions a. Oxidizing agents and electrons shown on reactant side b. Reducing agents shown on product side 2. Half-cell potentials are standard reduction potentials, also known as standard electrode potentials.

3. Half-reactions are listed in order of decreasing standard reduction potential. a. Strongest oxidizing agents are located in the upper left of the table. b. Strongest reducing agents are in the lower right of the table.

4. Half-reaction ordering corresponds to ordering of oxidation reactions in the activity series.

5. The more active metals at the top of the activity series have the more positive oxidation potential (more negative reduction potential).

6 19.6. Using Standard Reduction Potentials A. Table of standard reduction potentials – summarizes much chemical information 1. Two or more oxidizing or reducing agents arranged in order of increasing strength 2. Use to predict the spontaneity or nonspontaneity of thousands of redox reactions a. Combine half-reactions of interest and use o o o Ecell = Eanode + Ecathode b. May need to multiply half-reactions by some factor to ensure that electrons cancel c. Values of E for the half-reactions not multiplied by that factor B. E values independent of amount of reaction 1. G = –nFE 2. G is an extensive property (depends on amount of substance). a. When amount of substance that reacts changes, G changes by the same amount as does n, the number of electrons transferred. b. But E = –G/nF remains constant.

C. Can predict reaction spontaneity by knowing the location of oxidizing and reducing agents in the table.

1. An oxidizing agent can oxidize any reducing agent that lies below it in the table.

2. E for the overall reaction must be positive.

6Test Item File Questions: Multiple Choice 60-71, Algorithmic 35 – 42, Short Answer 19 – 20

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Chapter 19—Electrochemistry 7

19.7. Cell Potentials under Nonstandard-State Conditions: The Nernst Equation A. Cell potentials depend on temperature and on the composition of the reaction mixture.

1. G = G + RT ln Q 2. G = −nFE; G = −nFE 3. −nFE = −nFE + RT ln Q B. Nernst equation in volts at 25 C used to calculate cell potentials under nonstandard-state conditions 0.0592 E = Eo − log Q n 8 19.8. Electrochemical Determination of pH A. Important application of Nernst equation 1. Electrochemical determination of pH 2. Measured by a pH meter B. Consider a cell with a hydrogen electrode anode and a second reference electrode cathode.

1. Pt(s) | H2(1 atm) | H+(? M) || reference cathode 2. Ecell = 0.0592 pH + Eref 3. pH is a linear function of the cell potential.

E − Eref pH = cell 0.0592 C. pH of a solution measured by measuring Ecell D. Actual pH measurements use a glass electrode with a calomel electrode as the reference.

9 19.9. Standard Cell Potentials and Equilibrium Constants A. Standard free-energy change for a reaction is related to both the standard cell potential and the equilibrium constant.

1. G = −nFE 2. G = −RT ln K B. Combining the two equations: RT 2.303RT Eo = ln K = log K nF nF 0.0592 Eo = log K n C. Most common use – calculating equilibrium constants from standard cell potentials D. Equilibrium constants for redox reactions tend to be either very large or very small in comparison with equilibrium constants for acid–base reactions.

1. Positive value of E corresponds to K > 1.

2. Negative value of E corresponds to K < 1.

E. Three different ways to determine the value of an equilibrium constant K 1. K from concentration data: [C]c [D]d K= [A]a [B]b 2. K from thermochemical data: −G o ln K = RT 3. K from electrochemical data: nFE o ln K = RT 7 Test Item File Questions: Multiple Choice 74 – 85, Algorithmic 43 – 46, Short Answer 21 8 Test Item File Questions: Multiple Choice 94 – 98, Algorithmic 47, Short Answer 22 9Test Item File Questions: Multiple Choice 99 – 102, Algorithmic 48 – 49, Short Answer 23

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Chapter 19—Electrochemistry 10

19.10. Batteries A. Most important practical application of galvanic cells B. Battery features depend on application C. General features 1. Compact and lightweight 2. Physically rugged and inexpensive 3. Provide a stable source of power for relatively long periods of time D. Lead storage battery 1. Used as a reliable power source for starting automobiles for almost 100 years 2. 12 V battery – six 2 V cells connected in series 3. Anode – series of grids packed with spongy lead 4. Cathode – series of grids packed with lead dioxide, dipped into an aqueous solution of H2SO4 (38% w/w) 5. Electrode half-reactions and the overall cell reaction Anode: Pb(s) + HSO4–(aq) → PbSO4(s) + H+(aq) + 2 e–

E = 0.296 V

Cathode: PbO2(s) + 3 H+(aq) + HSO4–(aq) + 2 e– → 2 PbSO4(s) + 2 H2O(l) E = 1.628 V _____________________________________________________________________________ Overall: Pb (s) + PbO2(s) + 2 H+(aq) + 2 HSO4–(aq) → 2 PbSO4(s) + 2 H2O(l)

E = 1.924 V

6. PbSO4 adheres to the surface of the electrodes. a. Recharge by using an external source of direct current to drive the cell reaction in the reverse, nonspontaneous direction E. Dry-cell batteries (Leclanché cells) – common household batteries 1. Anode – Zn metal can 2. Cathode – inert graphite rod surrounded by a paste of solid MnO 2 and carbon black 3. Electrolyte – a moist paste of NH4Cl and ZnCl2 in starch a. Surrounds the MnO2 containing paste b. Acidic – causes corrosion of the Zn anode (Zn → Zn2+) 4. Alkaline dry cell – modified version of Leclanché a. Replace NH4Cl (acidic) with NaOH or KOH b. Electrode reactions – oxidation of zinc and reduction of manganese dioxide i. Produces ZnO due to basic conditions ii. Zinc corrodes more slowly iii. Battery has a longer life c. Produces higher power and more stable current and voltage – more efficient ion transport in the alkaline electrolyte F. Nickel–cadmium batteries – used in calculators and portable power tools 1. Rechargeable 2. Anode – cadmium metal 3. Cathode – NiO(OH) supported on nickel metal 4. Solid products of electrode reaction adhere to the surface of the electrodes – allows battery to be recharged G. Nickel–metal hydride batteries – used in consumer electronic products and in hybrid gaselectric automobiles 1. Less expensive than nickel-cadmium batteries 2. Light weight 3. Environmentally friendly

10Test Item File Questions: Multiple Choice 103 – 110, Algorithmic 50 – 51, Short Answer 24

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Chapter 19—Electrochemistry 4. Anode – metal alloy, LaNi5 5. Cathode – NiO(OH) supported on nickel metal H. Lithium and lithium-ion batteries – used in portable electronic products 1. Li has a higher Eo value for oxidation and lower atomic mass than any other metal.

2. Anode – lithium metal or lithium atoms in a graphite electrode 3. Cathode – manganese dioxide or metal sulfide that can incorporate Li+ 4. Electrolyte – lithium salt in an organic solvent 5. Used in cell phones, laptop computers, and digital cameras, power tools, headlamps, hybrid electric cars 11 19.11. Corrosion A. Corrosion – oxidative deterioration of a metal B. Well-known example of corrosion – conversion of iron to rust 1. Requires both oxygen and water 2. Involves pitting of the metal surface 3. Rust is deposited at a location physically separated from the pits.

C. Proposed mechanism for formation of rust – an electrochemical process in which iron is oxidized in one region of the surface and oxygen is reduced in another region 1. Anode region: Fe(s) → Fe2+(aq) + 2 e– E = 0.45 V 2. Cathode region: O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l) E = 1.23 V 3. Electrons flow from anode to cathode through the metal.

4. Ions migrate through the water droplets. a. Fe2+ reacts with O2 and is oxidized to Fe3+. b. Fe3+ reacts with H2O to form Fe2O3  xH2O(s) (rust).

5. Dissolved salt in water greatly increases electrolyte conductivity.

6. Explains why cars rust more rapidly when road salt is used to melt snow and ice 7. O2 able to oxidize all metals except a few – O2/H2O half-reaction lies above the Mn+/M half-reaction 8. Some metals (Al, Mg, Ti, Cr, Zn) form a metal oxide coating that adheres to the surface of the metal, protecting it from further contact and reaction with oxygen.

D. Prevention of corrosion – shield the metal surface from oxygen and moisture 1. Durable surface coating – metals such as chromium, tin, or zinc 2. Galvanizing – coating by dipping into a bath of molten zinc 3. Cathodic protection – protecting a metal from corrosion by connecting it to a second metal that is more easily oxidized 12 19.12. Electrolysis and Electrolytic Cells A. Electrolytic cell – electric current used to drive a nonspontaneous reaction B. Processes occurring in galvanic and electrolytic cells are the reverse of each other.

C. Electrolysis – the process of using an electric current to bring about chemical change D. Electrolytic cell 1. Two electrodes in an electrolyte connected to a battery or some other source of direct electric current 2. Battery – an electron pump, pushing electrons into one electrode and pulling them out of the other electrode 3. Anode – electrode where oxidation takes place a. Positive sign b. Battery pulls electrons out of it 4. Cathode – electrode where reduction takes place a. Negative sign b. Battery pushes electrons into it

11Test Item File Questions: Multiple Choice 111 – 114, Algorithmic 54 12Test Item File Questions: Multiple Choice 115 – 119, Algorithmic 52, 55

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Chapter 19—Electrochemistry E. Electrolysis of molten NaCl 1. Cathode – attracts Na+ Na+ + e– → Na(l) 2. Anode – attracts Cl– 2 Cl– → Cl2(g) + e– F. Electrolysis of aqueous NaCl 1. Electrode reactions in an aqueous solution may differ from those for a molten salt.

2. Cathode reaction can involve the reduction of Na + or the reduction of water. a. Reduction of water preferred – Eo less negative for H2O than Na+ b. 2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq) 3. Anode reaction can involve the oxidation of Cl– or the oxidation of water. a. Actual reaction – oxidation of Cl– due to overvoltage b. 2 Cl–(aq) → Cl2(g) + 2 e– 4. Overvoltage – amount of voltage needed above the calculated standard reduction (or oxidation) potential for electrolysis to occur a. Needed when the half-reaction has a substantial barrier for electron transfer (slow rate) i. Surmounts barrier ii. Reaction proceeds at satisfactory rate. b. Small overvoltage needed for solution for deposition of metals c. Large overvoltage needed for formation of O 2 or H2 d. Can’t predict – need experimental evidence if cell potentials are similar 5. Overall cell reaction 2 Cl–(aq) + 2 H2O(l) → Cl2(g) + 2 H2(g) + 2 OH–(aq) 6. Na+ is a spectator ion and reacts with the OH– to form NaOH G. Electrolysis of water 1. Electrolysis of any aqueous solution requires the presence of an electrolyte to carry the current in solution.

2. If the ions of the electrolyte are less easily oxidized and reduced than water is, then water will react at both electrodes.

3. Anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e– 4. Cathode: 4 H2O(l) + 4 e– → 2 H2(g) + 4 OH–(aq) 5. Overall cell reaction: 2 H2O(l) → O2(g) + 2 H2(g) 13 19.13. Commercial Applications of Electrolysis A. Manufacture of sodium 1. Produced commercially in a Downs cell by electrolysis of a molten mixture of NaCl and CaCl2 2. Liquid Na produced at the cylindrical steel cathode is less dense than the molten salt and thus floats to the top part of the cell, where it is drawn off into a suitable container.

B. Manufacture of chlorine and sodium hydroxide 1. Electrolysis of aqueous NaCl 2. Basis of chlor-alkali industry 3. Anode and cathode electrolysis reactions of aqueous NaCl carried out in a membrane cell 4. Membrane keeps Cl2 and OH– apart but allows a current of Na+ to flow.

C. Manufacture of aluminum 1. Hall–Heroult process 2. Electrolysis of a molten mixture of Al2O3 and cryolite (Na3AlF6) at 1000 oC in a cell with graphite electrodes 3. Success of process – use of cryolite as a solvent 4. Electrode reactions involve the formation of complex ions. a. Ions are reduced at the cathode to produce Al(l).

13Test Item File Questions: Multiple Choice 120 – 123

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Chapter 19—Electrochemistry b. Ions are oxidized at the anode to produce O2(g). i. O2(g) reacts with the graphite electrode to produce CO2(g) ii. Requires frequent replacement of the anodes 5. Largest single consumer of electricity in the U.S. 6. One mole of electrons produces only 9 g of Al. D. Electrorefining – purification of a metal by means of electrolysis E. Electroplating – coating of one metal on the surface of another using electrolysis 1. Cathode – object to be plated (carefully cleaned) 2. Electrolytic cell contains a solution of ions of the metal to be deposited.

14 19.14. Quantitative Aspects of Electrolysis A. Amount of substance produced at an electrode by electrolysis depends on the quantity of charge passed through the cell.

B. Follows directly from the stoichiometry of the reaction and the atomic mass of the product.

C. Moles of electrons passed through a cell are determined from the electric current and the time that the current flows.

1 mol e− Moles of e− = charge (C)  96,500 C D. Sequence of conversion used to calculate the mass or volume of product produced by passing a known current for a fixed period of time.

Current  time → Charge → Moles of e– → Moles of product → Grams or liters of product 1. Current in amperes (A) 2. Time in seconds (s) 3. Charge in coulombs (C) E. Think of electrons as reactants in a balanced equation, and proceed as with any other stoichiometry problem.

15 Inquiry – How do hydrogen fuel cells work?

14Test Item File Questions: Multiple Choice 124 – 129, Algorithmic 56 – 59, Short Answer 24 – 28 15Test Item File Questions: Algorithmic 58

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CHAPTER 19 ELECTROCHEMISTRY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 19.1

The most common error that students make when balancing redox reactions by the half-reaction method is to miscount charges, which results in an incorrect number of electrons used to balance the half-reaction.

Section 19.1

Balancing a redox equation by the half-reaction method will reveal that a halfreaction is an oxidation when electrons appear on the right and a reduction when electrons appear on the left of the arrow.

Section 19.1

The number of electrons used to balance the half-reaction should agree with the change in oxidation number of the element undergoing oxidation or reduction.

Electrons that are negatively charged should be added to the side of the half-reaction that has the greater positive charge.

Electrons should be added to product side of the oxidation half-reaction and to the reactant side of the reduction half-reaction.

Section 19.2

Anions flow through the salt bridge into the anode compartment while cations flow into the cathode compartment to maintain electroneutrality.

Section 19.2

The anode of a galvanic cell is negative (like an anion), and the cathode is positive (like a cation). Electrons move from the negative anode through an external circuit to the positive cathode.

Section 19.4

The driving force of a cell reaction, measured in volts and called cell voltage, cell potential, or electromotive force, is a function of the natural tendency of one substance to lose electrons and of a second substance to gain electrons. The greater these tendencies, the higher the cell voltage.

Section 19.4

When a chemical equation is balanced by the half-reaction method, the number of electrons that occur on both sides of the balanced equation (before canceling) is equal to the value of n.

Section 19.4

For a spontaneous reaction, G is negative but E is positive.

Section 19.4

Standard conditions for Eo, the standard cell potential, are identical with standard conditions for Go, Ho, and So.

Section 19.5

Every reduction half-cell reaction has a corresponding standard reduction potential, Eored. If the reaction is reversed, it becomes an oxidation half-cell with a corresponding standard oxidation half-cell potential, Eoox which has the opposite sign. For any given half-cell reaction, Eoox = –Eored.

Section 19.5

The more positive the half-cell potential, the greater the tendency for the reaction to occur as written. The more negative the half-cell potential, the greater the tendency for the reverse reaction to occur.

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Chapter 19—Electrochemistry Section 19.6

Half-cell potentials are never multiplied by a coefficient when used in the equation Eocell = Eoox + Eored.

Section 19.6

When predicting whether a reaction is spontaneous using textbook Table 19.1, some students find it useful to circle the two potential reactants and connect them with a line. If the line has a negative slope, the reaction is spontaneous; if the line has a positive slope, the reaction is nonspontaneous. If both potential reactants are on the same side of the arrows, no reaction can occur because an oxidation requires a simultaneous reduction.

Section 19.7

The general form of the Nernst equation is used to correct the standard cell potential for concentrations other than 1 M, partial pressures other than 1 atm, or a temperature other than 25 °C. This form of the Nernst equation applies only to reactions occurring at 25 °C.

Section 19.8

The Nernst equation relates the concentration of any chemical species, including H +, to cell potential. This relationship is utilized in a pH meter which measures a cell potential and displays its value in terms of pH.

Section 19.9

The more positive the standard cell potential, the larger the value of the equilibrium constant, and the farther the equilibrium lies to the right.

Section 19.10

A multicell battery is composed of galvanic cells wired in series, so the battery voltage equals the sum of the voltages of the individual cells.

Section 19.10

Although many metals react more readily than iron to form oxides, their oxides provide a protective coating against further corrosion. The corrosion product of iron (“rust”) flakes off and exposes a fresh surface to further corrosion.

Section 19.11

In an electrolytic cell a constant input of energy from an external direct current (dc) source, such as a battery, is used to drive a nonspontaneous reaction. The voltage of the battery must be more positive than the absolute value of the voltage for the electrolysis reaction.

Section 19.11

Overvoltage is additional voltage that must be applied in excess of the calculated voltage to initiate an electrolytic reaction. It is most important for those systems involving gases that impede the flow of electrons between electrode surfaces and reactants in solution.

Section 19.12

This procedure and the Hall–Heroult process discussed below are examples of commercial uses of a colligative property (freezing point depression) to enable electrolytic reactions to be carried out more economically.

Section 19.12

Originally, chlorine was a useless byproduct of this process. The search for uses of chlorine resulted in many commercial chlorinated compounds, including plastic film used to wrap food.

Section 19.13

Stress the use of this flowchart along with the dimensional analysis method of problem solving.

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Lecture/Laboratory Demonstration References Section 19.2

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Visible OxidationReduction in Electrochemical Cells,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 205–206.

Section 19.5

J. J. Fortman and R. Battino, “Determining the Metal Activity Series Using a Potato Porcupine,” J. Chem. Educ., Vol. 70, 1993, 939. Pieces of metal are inserted into a potato. A voltmeter is attached to a given metal and a copper reference electrode.

With the potato serving as a salt bridge and source of electrolytes, the observed cell potentials are used to rank the metals in an activity series.

Section 19.11

Bassam Z. Shakhashiri, “Electrolytic Cells in Series: A Red, White, and Blue Electrolysis,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 4 (The University of Wisconsin Press, Madison, 1992), pp.170–173.

Section 19.12

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Electroplating Copper,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), pp. 199–200.

Literature References Section 19.1

Marten J. ten Hoor, “Redox Balancing Without Puzzling,” J. Chem. Educ., Vol. 74, 1997, 1368–1376.

Section 19.2

Ron DeLorenzo, “Electrochemical Errors,” J. Chem. Educ., Vol. 62, 1985, 424–425.

Section 19.2

Penelope Ann Huddle, Margaret Dawn White, and Fiona Rogers, “Using a Teaching Model to Correct Known Misconceptions in Electrochemistry,” J. Chem. Educ., Vol. 77, 2000, 104–110.

Section 19.6

Richard S. Treptow, “Dental Filling Discomforts Illustrate Electrochemical Potentials of Metals,” J. Chem. Educ., Vol. 55, 1978, 189.

Section 19.10

Michael J. Smith and Colin A. Vincent, “Structure and Content of Some Primary Batteries,” J. Chem. Educ., Vol. 78, 2001, 519–521.

Section 19.10

Gerald Ondrey, Charlene Crabb, and Takeshi Kamiya, “Batteries: Full Speed Ahead,” Chem. Eng., Vol. 106 (2), 1999, 47–51.

Section 19.11

A. I. Onuchukwu, “Evaluation of Corrosion Susceptibility of a Metal: Student Corrosion Experiment II,” J. Chem. Educ., Vol. 65, 1988, 934.

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Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 19.1

Worked Example 19.1: Balancing a redox equation for a reaction in acidic solution

Section 19.6

Worked Example 19.8: Calculating the voltage of a galvanic cell under standard-state conditions

Section 19.14

Worked Example 19.13: Calculating the amount of product produced by electrolysis

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CHAPTER 20 NUCLEAR CHEMISTRY Chapter Learning Objectives Section 20.1 and Section 20.2

Write balanced nuclear reactions for the different kinds of radioactive decay.

Section 20.3

Predict the type of radioactive decay for a given isotope.

Write a balanced nuclear reaction for a radioactive decay series.

Section 20.4

Relate half-life and decay constant.

Calculate the amount of radioactive isotope remaining after a given amount of time.

Relate decay rates to decay constant, half-life, or amount remaining.

Section 20.5

Calculate the age of an object using radioisotopic dating.

Section 20.6

Calculate the mass defect and binding energy of a nucleus.

Calculate the mass and energy change of a chemical reaction.

Section 20.7

Identify key aspects of fission and fusion reactions and their role in nuclear power and weapons.

Calculate the energy change associated with a fission or fusion reaction.

Section 20.8

Write balanced equations for nuclear transmutations.

Section 20.9

Calculate the radiation intensity in a radioactive sample and convert between units of radiation intensity.

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Chapter 20—Nuclear Chemistry

Lecture Outline 20.1 Nuclear Reactions and Their Characteristics1 A. Nuclear chemistry is the study of properties and reactions of atomic nuclei B. Differences between nuclear reactions and chemical reactions 1. Nuclear reaction – a change in an atom’s nucleus; oftentimes, the identity of the atom changes.

2. Chemical reaction – a change in the way that atoms combine; the atom’s identity doesn’t change.

3. Different isotopes of an element a. Have essentially the same behavior in chemical reactions b. Have a different behavior in nuclear reactions 4. The energy changes accompanying nuclear reactions are far greater than those accompanying chemical reactions.

20.2. Radioactivity2 A. Radioactive – the spontaneous decomposition (decay) of an unstable isotope to form a more stable isotope of the same element or of another element and emission of radiation 1. Radioisotope – unstable radioactive isotope 2. Three common types of radiation with different properties: alpha (), beta (), gamma () B. Alpha () radiation – a stream of particles that consists of two protons and two neutrons 1. Repelled by a positively charged electrode 2. Attracted by a negatively charged electrode 3. A helium ion 4 He2+ =  2 4. Emission of an  particle reduces the mass number of the nucleus by four and reduces the atomic number by two.

238 U → 42 He + 23490Th 92 5. Common for heavy radioactive isotopes (radionuclides) C. Beta () radiation 1. Attracted by a positively charged electrode 2. Repelled by a negatively charged electrode 3. In nuclear reactions electrons are beta particles 0 −1 e =  4. Superscript 0 – the mass of an electron very, very small when compared with that of a proton or neutron and is therefore assigned a “mass number” of zero 5. Subscript (–1) – the electron does not have an atomic number, assigning it a value of –1 facilitates balancing nuclear reactions 6. Occurs when a neutron in the nucleus spontaneously decays into a proton plus an electron, which is then ejected.

7. Emission of a beta particle does not affect the mass number but increases the atomic number by one.

153 131 0 53 I → 54 Xe + −1e D. Gamma () radiation – electromagnetic radiation of very high energy and short wavelength 1. Not attracted by a positively charged electrode 2. Not repelled by a negatively charged electrode 3. No mass 4. High energy electromagnetic radiation

1 2

Test Item File Questions: Multiple Choice 1 – 3, Algorithmic 1 – 5, Short Answer 1 – 2 Test Item File Questions: Multiple Choice 4 – 20, Algorithmic 6 – 11, Short Answer 3 – 6

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Chapter 20—Nuclear Chemistry 5. Almost always accompanies  and  emission – mechanism for release of energy 6.  emission often not shown in nuclear equations – doesn’t change either the mass number or the atomic number of the product nucleus E. Positron emission and electron capture 1. Positron emission – conversion of a proton in the nucleus into a neutron plus an ejected positron, a particle with the same mass as an electron but opposite charge: a. Decrease in atomic number of the product nucleus 0 + +1 e =  b. No change in mass number 40 40 0 19 K → 18 Ar + 1 e c. When decay occurs the positron is annihilated upon contact with an electron releasing gamma rays 2. Electron capture – a proton in the nucleus captures an inner-shell electron, which is converted into a neutron a. No change in mass number of the product nucleus b. Atomic number decreases by one F. Balanced nuclear reaction 1. Not balanced in usual chemical sense – nuclei are not the same on both sides of the reaction 2. Sums of nucleons (nuclear particles: neutrons and protons) on both sides are equal.

3. Sums of charges on nuclei and any elementary particles on both sides are equal.

20.3. Nuclear Stability3 A. The ratio of neutrons to protons in the nucleus affects nuclear stability, see textbook Figure 20.2.

1. 3600 known isotopes found in the band of nuclear stability 2. 264 isotopes are nonradioactive 3. Others undergo spontaneous radioactive decay B. Generalizations regarding nuclear stability 1. Every element has at least one radioactive isotope.

2. Hydrogen is the only element whose most abundant isotope contains more protons than neutrons.

3. As atomic number increases, the ratio of neutrons to protons gradually increases.

4. All isotopes heavier than Bi-209 are radioactive.

C. Within the band of nuclear stability, radioactive nuclei undergo decay reactions related to the isotopic neutron/proton ratio.

1. Nuclei on the top side of the band are neutron-rich and undergo  emission which decreases the ratio.

2. Nuclei on the bottom side of the band are neutron-poor and undergo positron emission, electron-capture, or  emission which increase the ratio.

D. Many radioactive nuclei decay via a series of nuclear reactions, called a radioactive decay chain, until a stable state is achieved.

20.4. Radioactive Decay Rates4 A. Decay of radioactive nuclei – a first-order process 1. Integrated rate law – analogous to that given in Section 13.5.

N ln t = − kt N0 a. N0 is the number of radioactive nuclei initially present. b. Nt is the number of radioactive nuclei present at time t. c. k is the decay constant.

3 4

Test Item File Questions: Multiple Choice 22 – 31, Algorithmic 12 – 20, Short Answer 7 Test Item File Questions: Multiple Choice 42 – 50, Algorithmic 21 – 23, Short Answer 8

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Chapter 20—Nuclear Chemistry 2. Decay rate – characterized by half-life 0.693 ln 2 t1/ 2 = t1/ 2 = k k 3. Calculation of the ratio of remaining and initial amounts of radioactive sample Nt/N0 at any time t N   t  ln  t  = ( −ln 2 )    N0   t1/ 2  20.5. Dating with Radioisotopes5 A. Radiocarbon dating 1. Depends on the slow and constant production of radioactive carbon-14 in the upper atmosphere by neutron bombardment of nitrogen atoms 14 1 14 1 7 N + 0 n → 6 C + 1H 2. Carbon-14 reacts to form 14CO2, which mixes with 12CO2 and is taken up by plants during photosynthesis.

14 C + O2 → 14 CO2 3. Ratio of 14C to 12C in a living organism is the same as that in the atmosphere.

4. 14C/12C decreases when the organism dies because 14C undergoes radioactive decay.

14 14 0 − 6 C → 7 N + 1e 5. For 14C, t½ = 5730 y 6. Can determine how long ago the organism died by measuring amount of 14C remaining 7. As the amount of 14C diminishes, accuracy decreases with an outer limit of about 60,000 y B. Use other radionuclides for dating rocks.

1. 238U a. For 238U, t½ = 4.4  109 y b. 238U decays through a series to give 206Pb c. Measurement of 238U/206Pb ratio can determine the age of the uranium-containing rock 40 2. K a. For 40K, t½ = 1.25  109 y b. Potassium-40 decays through electron capture and positron emission 40 0 40 19 K + −1 e → 18 Ar 40 K → 18 Ar + −01e c. Compare amount of 40Ar to 40K remaining in sample 20.6. Energy Changes During Nuclear Reactions6 A. Can calculate the energy change on formation of the nucleus from isolated protons and neutrons 1. E = mc2 2. Relates energy change of a nuclear process to corresponding mass change B. Mass defect 1. Loss in mass that occurs when protons and neutrons combine to form a nucleus 2. Lost mass converted into energy known as the binding energy a. Binding energy – energy that holds the nucleons together b. Calculate from the Einstein equation c. Usually expressed on a per-nucleon basis using the electron volts as the energy unit: eV = 1.60  10−19 J C. Plot binding energy per nucleon – higher binding energy corresponds to higher stability 40 19

5 6

Test Item File Questions: Multiple Choice 59 – 65, Algorithmic 26, Short Answer 9-11, 29-30 Test Item File Questions: Multiple Choice 66 – 75, Algorithmic 27 – 28, 30, Short Answer 12 – 20

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Chapter 20—Nuclear Chemistry D. Mass and energy are interconvertible.

1. Laws of conservation of mass and conservation of energy must be combined.

2. Neither mass nor energy is conserved separately.

3. Mass and energy are interconverted in a nuclear reaction, but the combination of the two is conserved.

20.6. Nuclear Fission and Fusion7 A. Lighter and heavier elements are less stable than mid-mass elements near iron-56.

1. Heavy nuclei – gain stability and release energy if they fragment to yield midmass elements 2. Light nuclei – can gain stability and release energy if they fuse together B. Fission – the fragmentation of heavy nuclei 1. Nuclei break into fragments when struck by neutrons.

2. Doesn’t occur in exactly the same way each time 3. Uranium-235 – more than 800 different fission products a. Frequently occurring pathway 1 235 142 91 1 0 n + 92 U → 56 Ba + 36 Kr + 30 n b. Three neutrons released induce nine more fissions → 27 neutrons c. Chain reaction – a reaction that continues to occur even if the supply of neutrons from outside is cut off d. Small sample size – many neutrons escape before initiating additional fission events e. Critical mass – a sufficient amount of radioactive nuclide that allows the chain reaction to become self-sustaining 4. Using mass defect and the Einstein equation can calculate the amount of energy released during nuclear fission – use the masses of the atoms corresponding to the relevant nuclei C. Nuclear reactors.

1. Principle of a nuclear reactor – a subcritical amount of uranium fuel is placed in a containment vessel surrounded by circulating coolant, and control rods are added 2. Control rods are made of boron and cadmium. a. Absorb and regulate the flow of neutrons b. Raised and lowered to maintain fission at a controlled rate 3. Energy from the controlled fission heats the circulating coolant.

4. Heated circulating coolant produces steam to drive a turbine and produce electricity.

D. Nuclear fusion – the joining together of light nuclei 1. Releases enormous amounts of energy 2. Fusion of hydrogen nuclei – a potential power source a. Hydrogen isotopes are cheap and plentiful. b. Fusion products are nonradioactive and nonpolluting.

3. Technical problems to achieving a practical and controllable fusion method – high initiation temperature: T = 40 ≈ 106 K 20.8. Nuclear Transmutation8 A. Nuclear transmutation 1. The change of one element into another, often brought about by bombardment of an atom with a high-energy particle 2. Creates an unstable nucleus, which causes a nuclear change to occur B. All transuranium elements have been produced by bombardment reactions.

7 8

Test Item File Questions: Multiple Choice 77 – 88, Algorithmic 31 – 32, Short Answer 21 – 23 Test Item File Questions: Multiple Choice 90 – 93, Algorithmic 33 – 35, Short Answer 25 – 26

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Chapter 20—Nuclear Chemistry 20.9. Detecting and Measuring Radioactivity9 A. Detect radiation by measuring its ionizing properties B. Devices such as Geiger counters are used for measuring radiation.

C. Expression of radiation intensity depends on what is being measured.

1. Becquerel – SI unit for measuring the number of radioactive disintegrations occurring each second in a sample: Bq = 1 disintegration/s 2. Curie (Ci): 1 Ci = 3.7  1010 disintegrations/s a. Larger unit than the becquerel and more commonly used b. Often used in medicine and biochemistry 3. Gray – SI unit for measuring the amount of energy absorbed per kilogram of tissue exposed to a radiation source a. 1 Gy = 1 J/kg b. 1 rad = 0.01 Gy 4. Sievert (Sv) – SI unit that measures the amount of tissue damage caused by radiation rem (roentgen equivalent for man) = 0.01 Sv D. The effects of ionizing radiation on the human body depend on several factors.

1. Kind of radiation and its energy 2. Length of exposure 3. Whether the source is outside or inside the body a. From an external source, x-rays and  radiation are more harmful than  and  particles. b. From an internal source,  and  particles are more dangerous.

4. The average radiation dose received annually by most people is only 120 mrem. a. About 70% is from natural sources b. About 30% is from medical procedures c. Exposure to emissions from nuclear power plants and nuclear weapons fallout is barely detectable.

Inquiry – How are radioisotopes used in medicine?

9

Test Item File Questions: Multiple Choice 95 – 111, Algorithmic 36 – 40, Short Answer 27 – 28

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CHAPTER 20 NUCLEAR CHEMISTRY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 20.1

Energy changes for nuclear reactions are on the order of a million times greater than energy changes for chemical reactions.

Section 20.2

An alpha () particle ( 42 He2+ ) is a helium nucleus. The left subscript indicates the charge on the particle, and the +2 charge (right superscript) is not used in writing nuclear equations.

Section 20.2

A beta () particle ( −1e or –) is an electron.

Section 20.2

For the purpose of balancing nuclear equations, the mass number of the electron is taken to be 0 and its “atomic number” is taken to be –1.

Section 20.2

Electron capture: proton + core electron → neutron.

Section 20.2

For a nuclear equation to be balanced: sum of nuclear charge of reactants (subscripts) = sum of nuclear charge of products, and sum of nucleons of reactants (superscripts) = sum of nucleons of products. Ionic charges are ignored.

Section 20.3

Transuranium elements, those with atomic numbers higher than uranium, do not occur naturally but are produced by nuclear transmutation reactions.

Section 20.6

When a nucleus is formed from protons and neutrons, some mass (mass defect) is converted to energy (binding energy), as related by the Einstein equation, E = mc2.

Section 20.6

The variation in slope of a plot of the binding energy versus mass number shows that a typical fusion process releases far more energy than the fission processes employed in nuclear reactors.

Section 20.9

Ionizing radiation is any particle or photon with sufficient energy to remove an electron on collision with an atom or molecule, thus creating an ion.

Section 20.9

High-energy photons ( and  rays) are the most penetrating forms of radiation.

Particulate forms of radiation ( and  particles) are less penetrating.

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Lecture/Laboratory Demonstration References None

Literature References Section 20.1

Enrique A. Hughes and Anita Zalts, “Radioactivity in the Classroom,” J. Chem.

Educ., Vol. 77, 2000, 613–614.

Section 20.1

“Nuclear Chemistry: State of the Art for Teachers,” a series of articles in the October 1994 issue of the Journal of Chemical Education.

Section 20.1

S. G. Hutchinson and F. I. Hutchinson, “Radioactivity in Everyday Life,” J. Chem.

Educ., Vol. 74, 1997, 501–505.

Section 20.1

Charles H. Atwood, “Teaching Aids for Nuclear Chemistry,” J. Chem. Educ., Vol. 71, 1994, 845–847.

Section 20.2

C. Ronneau, “Radioactivity: A Natural Phenomenon,” J. Chem. Educ., Vol. 67, 1990, 736-737.

Section 20.2

Robert Suder, “Beta Decay Diagram,” J. Chem. Educ., Vol. 66, 1989, 231.

Section 20.4

Kent J. Crippin and Robert D. Cutright, “Modeling Nuclear Decay: A Point of Integration Between Chemistry and Mathematics,” J. Chem. Educ., Vol. 75, 1998, 434–436.

Section 20.4

M. W. Rowe, “Radioactive Dating: A Method for Geochronology,” J. Chem. Educ., Vol. 62, 1985, 580–584.

Section 20.4

M. W. Rowe, “Archeological Dating,” J. Chem. Educ., Vol. 63, 1986, 16–20.

Section 20.4

Ron DeLorenzo, “California Earthquakes: Predicting the Next Big One Using Radiocarbon Dating,” J. Chem. Educ., Vol. 57, 1980, 601.

Section 20.4

Kent J. Crippen and Robert D. Curtright, “Modeling Nuclear Decay: A Point of Integration Between Chemistry and Mathematics,” J. Chem. Educ., Vol. 75, 1998, 1434–1436.

Section 20.4

M. W. Rowe, “Radioactive Dating: A Method for Geochronology,” J. Chem. Educ., Vol. 62, 1985, 580–584.

Section 20.7

Ruth Lewin Sime, “Lise Meitner and the Discovery of Fission,” J. Chem. Educ., Vol. 66, 1989, 373–378.

Section 20.7

Ruth Lewin Sime, “Lise Meitner and the Discovery of Fission,” Scientific American, January 1998, 80–85.

Section 20.7

Michael Freemantle, “Ten Years After Chernobyl: Consequences Are Still Emerging,” Chem. Eng. News, April 29, 1996, 18–28.

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Chapter 20—Nuclear Chemistry Section 20.7

Gregory R. Choppin, “Aspects of Nuclear Waste Disposal of Use in Teaching Basic Chemistry,” J. Chem. Educ., Vol. 71, 1994, 826–829.

Section 20.7

Torkil H. Jensen, “Fusion – A Potential Power Source,” J. Chem. Educ., Vol. 71, 1994, 820–823.

Section 20.9

Charles H. Atwood, “How Much Radon Is Too Much?” J. Chem. Educ., Vol. 69, 1992, 351–353.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 20.4

Worked Example 20.3: Using half-life to calculate an amount remaining

Section 20.6

Worked Example 20.6: Calculating a mass defect and a binding energy

Section 20.8

Worked Example 20.9: Balancing a nuclear transmutation equation

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CHAPTER 21 TRANSITION ELEMENTS AND COORDINATION CHEMISTRY Chapter Learning Objectives Section 21.1

Write electron configurations for transition metals and predict the number of unpaired electrons in transition metal ions.

Section 21.2

Describe the trends in physical and chemical properties of transition metal elements across a period. Explain the reasons for trends.

Section 21.3

Relate the oxidizing and reducing strength of a transition metal ion to its oxidation state and location in the periodic table.

Section 21.4

Identify Lewis acids, Lewis bases, ligands, and donor atoms in a coordination complex.

Determine the formula of a coordination complex, identify the oxidation state and coordination number of the metal atom, and draw the structure of the complex.

Section 21.5

Classify a ligand as mono, bi, tri, tetra, or hexadentate based on its chemical structure.

Section 21.6

Write the systematic name of a coordination complex.

Section 21.7

Classify isomers according the scheme given in textbook Figure 21.11.

Draw diastereomers for octahedral and square planar complexes.

Section 21.8

Classify a coordination complex as chiral or achiral and draw enantiomers for chiral compounds.

Section 21.9

Relate the color of a metal complex, the wavelength of light it absorbs, and the energy difference between its ground and excited state.

Section 21.10

Draw crystal field energy level diagrams for octahedral, tetrahedral, and square complexes. Classify ligands as strong or weak-field and predict (or account for) color and magnetic properties of complexes in terms of crystal field theory.

Section 21.11

Describe the bonding in a metal complex in terms of valence bond theory, including orbital diagrams, the hybrid orbitals used by the metal atom, and the number of unpaired electrons.

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Chapter 21—Transition Elements and Coordination Chemistry

Lecture Outline 1

21.1. Electron Configurations A. 4s subshell filled first; 3d subshell filled according to Hund’s rule 1. Add one electron to each of the five 3d orbitals before adding a second electron.

2. Two exceptions: Cr and Cu a. Cr: 4s13d5 b. Cu: 4s13d10 B. Electron configurations depend on both orbital energies and electron–electron repulsions.

1. If two valence subshells have similar energies, can’t always predict the configurations 2. Exceptions from the expected orbital–filling pattern result in either half-filled or completely filled subshells.

3. For Cr and Cu: 3d and 4s have similar energies, and as a consequence, one electron shifts from 4s to 3d. a. Shift decreases the electron–electron repulsions. b. Gives either a filled or half-filled subshell C. Easy to predict electron configurations of transition metal cation 1. All valence electrons occupy d orbitals.

2. Neutral atom loses one or more electrons; Zeff increases. a. Remaining electrons more strongly attracted to the nucleus b. Orbital energies decrease c. 3d orbitals experience a steeper drop in energy with increasing Zeff than does the 4s. d. 3d orbitals in cations lower in energy than the 4s orbitals D. Groups 1B – 8B 1. Electron configuration similar to analogous elements in groups 1A, 8A a. Groups 1B, 2B: electron configuration similar to groups 1A, 2A b. Groups 3B, 8B: electron configuration similar to groups 3A, 8A 2 21.2. Properties of Transition Elements A. Properties can be understood in terms of electron configurations.

B. Metallic properties 1. Malleable, ductile, lustrous, and good conductors of heat and electricity 2. Sharing of d, as well as s, electrons gives rise to stronger metallic bonding.

3. Transition metals are harder, have higher melting and boiling points, and are more dense than the group 1A and 2A metals.

4. Melting points increase as the number of unpaired d electrons available for metallic bonding increases and then decrease as the d electrons pair up and become less available for bonding.

C. Atomic radii and densities 1. Decrease in radii with increasing atomic number a. Added d electrons only partially shield the added nuclear charge. b. Increase in radii toward the end of each series c. Due to more effective shielding and increasing electron–electron repulsion as d orbitals become doubly occupied 2. Similar radii – accounts for ability to blend together in forming alloys 3. Second and third series a. Group 4B on have nearly identical radii b. Smaller than expected size of the third-series atoms due to lanthanide contraction 4. Lanthanide contraction – the general decrease in atomic radii of the f-block lanthanide elements between the second and third transition series a. Increase in effective nuclear charge as 4f subshell is filled

1Test Item File Questions: Multiple Choice 1 – 12, Algorithmic 1 – 11, Short Answer 1 – 8 2Test Item File Questions: Multiple Choice 21 – 23, Algorithmic 12 – 17, Short Answer 9 – 10

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Chapter 21—Transition Elements and Coordination Chemistry b. Size decrease due to a larger Zeff almost exactly compensates for the expected size increase due to an added quantum shell of electrons. c. Atoms of third series have radii very similar to those of the second series.

5. Densities – inversely related to atomic radii a. Second and third series have nearly the same atomic volume. b. Third series has unusually high densities.

D. Ionization energies and standard oxidation potentials.

1. Ionization energies a. Increase left to right across a transition series b. Due to increase in Zeff and a decrease in atomic radius 2. Es for oxidation potentials of first series transition metals are positive (except Cu). a. Solid metal is oxidized to aqueous cation more readily than H 2 gas is oxidized to H+. b. M(s) + 2 H+(aq) → M2+(aq) + H2(g) c. Better reducing agents than H2 d. Can be oxidized by “nonoxidizing” acids (lack an oxidizing anion) e. Cu(s) requires a stronger oxidizing agent.

3. General trend for E correlates with the general trend in ionization energies.

4. Ease of oxidation of the metal decreases as the ionization energies increase across the first series.

3 21.3. Oxidation States of Transition Elements A. Exhibit a variety of oxidation states – have oxidation states less than their group number B. First series forms + 2 cation (except Sc) – due to loss of the two 4s electrons C. First series – can also lose 3d electrons 1. 3d and 4s energies are similar.

2. Energy required to remove the third electron is provided by the larger G of hydration of the more highly charged +3 cation.

D. Highest oxidation state for the groups 3B–7B metals is the group number.

1. Corresponds to loss or sharing of all the valence s and d electrons.

2. Later transition metals – loss or sharing of all valence electrons prohibited by the increasing Zeff value E. Transition metal ions in a high oxidation state – good oxidizing agents F. Early transition metal ions in low oxidation state – good reducing agents G. Divalent ions of the later metals – poor reducing agents because of the larger value of Z eff 4 21.4. Coordination Compounds A. Coordination compound – a compound in which a central metal ion is attached to a group of surrounding molecules or ions by coordinate covalent bonds B. Ligands – molecules or ions that surround the central metal ion in a complex C. Donor atoms – atoms that are attached directly to the metal ion D. Complex formation is a Lewis acid–base interaction in which ligands act as Lewis bases (electron pair donors) and the central metal ion behaves as a Lewis acid (electron pair acceptor).

E. Some coordination compounds are salts which contain a complex cation or anion along with enough ions of opposite charge to give a compound that is electrically neutral overall.

1. Enclose the complex ion in brackets in the formula.

2. Indicates that the complex ion is a discrete structural unit F. Metal complex – refers both to both neutral molecules and to complex ions G. Coordination number – the number of ligand donor atoms that surround a central metal ion in a complex 1. Most common coordination numbers: 4 and 6

3Test Item File Questions: Multiple Choice 50 – 55, Algorithmic 18 – 20, Short Answer 11 – 12 4Test Item File Questions: Multiple Choice 56 – 60, Algorithmic 22 – 23, Short Answer 13 – 16

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Chapter 21—Transition Elements and Coordination Chemistry 2. Depends on the metal ion’s size, charge, and electron configuration and the size and shape of the ligands H. Characteristic shape of metal complex is determined by the metal ion’s coordination number.

1. Two-coordinate complexes – linear 2. Four-coordinate complexes – tetrahedral and square planar 3. Six-coordinate complexes – octahedral I. Charge on a metal complex is equal to the charge on the metal ion plus the sum of the charges on the ligands.

5 21.5. Ligands A. All ligands are Lewis bases.

1. Have at least one unshared pair of electrons.

2. Forms a coordinate covalent bond to a metal ion 3. Classified as monodentate or polydentate a. Depends on number of ligand donor atoms that bond to the metal b. Use the electron pair of a single donor atom – monodentate c. Use electron pairs on more than one donor atom – polydentate B. Chelating agents – polydentate ligands 1. Multipoint attachment to a metal ion resembles the grasping of an object by the claws of a crab.

2. Forms a chelate ring 3. Metal chelate – a complex that contains one or more chelate rings 4. Naturally occurring chelating ligands are essential constituents of many important biomolecules.

6 21.6. Naming Coordination Compounds A. If the compound is a salt, name the cation first and then the anion.

B. In naming a complex ion or a neutral complex, name the ligands first, in alphabetical order, and then the metal.

1. Anionic ligands end in –o. (textbook Table 21.4) 2. Complex name is one word.

C. If more than one ligand of a particular type – use Greek prefixes to indicate number D. If the name of a ligand itself contains a Greek prefix, put the ligand name in parentheses and use an alternate prefix.

E. Use a Roman numeral in parentheses immediately following the name of the metal, to indicate the metal’s oxidation state.

F. In naming the metal, use the ending –ate if the metal is in an anionic complex.

7 21.7. Isomers A. Isomers 1. Compounds with the same formula but a different arrangement of constituent atoms 2. Different compounds with different physical and chemical properties B. Constitutional isomers – isomers that have different connections among their constituent atoms 1. Linkage isomers – arise when a ligand can bond to a metal through either of two different donor atoms 2. Ionization isomers – isomers that differ in the anion that is bonded to the metal ion C. Stereoisomers 1. Isomers that have the same connections among atoms but have a different arrangement of the atoms in space

5Test Item File Questions: Multiple Choice 62 – 63, Algorithmic 24 – 29, Short Answer 17 6Test Item File Questions: Multiple Choice 69 – 75, Algorithmic 30, Short Answer 18 7Test Item File Questions: Multiple Choice 76 – 77, Algorithmic 31 – 32, Short Answer 19 – 20

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Chapter 21—Transition Elements and Coordination Chemistry 2. Diastereoisomers (geometric isomers) – have different relative orientations of metal-ligand bonds a. cis isomer – identical ligands occupy adjacent corners of the square in a square planar complex b. trans isomer – identical ligands are across from one another in a square planar complex c. Have different properties i. cis–Pt(NH3)2Cl2 – polar molecule; more soluble in water ii. trans–Pt(NH3) 2Cl2 – nonpolar; two Pt–Cl and Pt–NH3 dipoles point in opposite directions and cancel out d. Square planar complexes of the type MA2B2 and MA2BC (M= metal ion; A, B, C = ligands) can exist as cis-trans isomers. e. No cis-trans isomers for four coordinate tetrahedral complexes – all four corners of a tetrahedron adjacent to one another f. Octahedral complexes of the type MA4B2 can also exist as diastereoisomers – two B ligands can be either on adjacent or on opposite corners of the octahedron g. Easy to distinguish – various bonds in cis and trans isomers point in different directions 8 21.8. Enantiomers and Molecular Handedness A. Enantiomers – molecules or ions that are nonidentical mirror images of one another 1. Differ from one another because of their handedness a. Chiral – objects that have a handedness to them b. Achiral – objects that lack handedness B. An object is not chiral if it has a symmetry plane cutting through its middle so that one half of the object is a mirror image of the other half.

C. Certain molecules and ions are chiral.

1. Tris(ethylenediamine)cobalt(III) ion, [Co(en) 3]3+ (textbook Figure 21.16) – two nonidentical mirror image forms a. “Right-handed” enantiomer – the three ethylenediamine ligands spiral to the right (clockwise) b. “Left-handed” enantiomer – the ethylenediamine ligands spiral to the left (counterclockwise) 2. [Co(NH3) 6] 3+ – achiral; has several symmetry planes D. Enantiomers have identical properties except for their reactions with other chiral substances and their effect on plane-polarized light.

1. Plane-polarized light – light in which the electric vibrations of the light wave are restricted to a single plane 2. Pass ordinary light through a polarizing filter then through a solution of one of the enantiomers a. Plane of polarization rotated either to the right or to the left b. Pass through the solution of the other enantiomer – plane of polarization rotated through an equal angle, but in the opposite direction 3. Optical isomers – another term used for enantiomers because of their effect on planepolarized light 4. Label enantiomers (+) or (–) depending on the direction of rotation of the plane of polarization 5. Racemic mixture a. 50:50 mixture of the (+) and (–) isomers that produces no net optical rotation b. Rotations produced by the individual enantiomers exactly cancel.

8Test Item File Questions: Multiple Choice 83, Algorithmic 33 – 35

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Chapter 21—Transition Elements and Coordination Chemistry 9

21.9. Color of Transition Metal Complexes A. Color of transition metal complexes depends on the identity of the metal and the ligands.

B. Metal complex can absorb light by undergoing an electronic transition from its lowest energy state (E1) to a higher energy state (E2).

1. Wavelength of absorbed light depends on the energy separation E = E2 – E1 between the two states.

2. Absorbance – the measure of the amount of light absorbed by a substance a. Absorption spectrum – plot of absorbance versus wavelength b. Color that we see is complementary to the color absorbed.

10 21.10. Crystal Field Theory A. Crystal field theory – a model that views bonding in complexes as arising from electrostatic interactions and considers the effect of ligand charges on energies of metal ion d orbitals.

1. Accounts for color and magnetic properties of transition metal complexes 2. No covalent bonds, no shared electrons, and no hybrid orbitals 3. Electrostatic interactions within an array of ions 4. Neutral, dipolar ligand – the electrostatic interactions are of the ion–dipole type B. Octahedral complexes 1. Metal positively charged 2. Ligands negatively charged a. Ligands repel each other. b. Minimize repulsion by getting as far apart from one another as possible c. Metal-ligand attractions > ligand–ligand repulsions d. Complex is more stable than the separated ions.

C. Effect of ligand charges on the energies of d orbitals 1. Negatively charged d electrons are repelled by the negatively charged ligands. a. Orbital energies are higher in the complex. b. Not all orbitals raised in energy by the same amount 2. d z 2 , d x 2 − y 2 are raised higher in energy than the d xy , d xz , d yz (point between ligands).

3. Crystal field splitting () – the energy splitting between the two sets of d orbitals (textbook Figure 21.23) D. Crystal field splitting energy,  – corresponds to ’s in the visible region of the spectrum 1. Color of complexes due to electronic transitions between the lower and higher energy sets of d orbitals 2. Size depends on the nature of the ligands a. Spectrochemical series – changes in the crystal field splitting as the ligand varies b. Weak-field ligands i. Produce a relatively small value of  ii. High-spin complexes c. Strong-field ligands i. Produce a relatively large value of  ii. Low-spin complexes E. Spin-state depends on the relative values of  and P, the spin-pairing energy.

1. Spin-pairing energy – energy needed to overcome repulsion that exists when an electron is placed into an orbital that already contains another electron.

2.  > P; low-spin arrangement has lower energy.

3.  < P; high-spin arrangement has higher energy.

F. Choice of spin-states arises only for d4–d7 complexes.

G. Tetrahedral and square planar complexes – have different energy splitting for the d orbitals (textbook Figure 21.25) 9Test Item File Questions: Algorithmic 36, Short Answer 21 10Test Item File Questions: Multiple Choice 93 – 102, Algorithmic 37 – 39, Short Answer 23 – 25

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Chapter 21—Transition Elements and Coordination Chemistry H. Energy splitting in tetrahedral complexes is just the opposite of that in octahedral complexes.

1. (tetrahedral complexes)  ½ (octahedral complexes) 2. None of the orbitals points directly at the ligands.

3. Four ligands instead of six 4.  < P 5. All complexes are high-spin.

I. Square planar complexes – two trans ligands along the z-axis are missing 1. Large energy gap between the x2–y2 orbital and the four lower-energy orbitals.

2. Most common for metal ions with d8 electronic configurations 3. Favors low-spin complexes in which all four lower-energy orbitals are filled and the higher energy x2–y2 orbital is vacant 11 21.11. Bonding in Complexes: Valence Bond Theory A. Valence bond theory 1. Bonding results when a filled ligand orbital containing a pair of electrons overlaps a vacant hybrid orbital on the metal ion.

2. Produces a coordinate covalent bond B. If you know the geometry of a complex, you know which hybrid orbitals the metal ion uses. (textbook Table 21.6) C. Magnetic behavior of transition metal complexes 1. Paramagnetic – substance containing unpaired electrons and attracted by magnetic fields 2. Diamagnetic substance – contains only paired electrons and weakly repelled by magnetic fields 3. Number of unpaired electrons in a transition metal complex can be determined by quantitative measurement of the force exerted on the complex by a magnetic field.

D. High-spin complex – one in which the d electrons are arranged according to Hund’s rule to give the maximum number of unpaired electrons E. Low-spin complex – one in which the d electrons are paired up to give a maximum number of doubly occupied d orbitals and a minimum number of unpaired electrons Inquiry – How does cisplatin kill cancer cells?

11Test Item File Questions: Multiple Choice 107 – 113, Algorithmic 40 – 41, Short Answer 27

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Chapter 21—Transition Elements and Coordination Chemistry

CHAPTER 21 TRANSITION ELEMENTS AND COORDINATION CHEMISTRY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 21.1

Anomalous electron configurations frequently occur if the atom can attain half-filled or completely filled subshells.

Section 21.1

There are no anomalous electron configurations for transition metals in a positive oxidation state. The ns orbital is vacant, and the remaining valence electrons occupy (n –1)d orbitals, where n is the period.

Section 21.2

Within any transition series, the melting points increase for the first three or four elements, then show a general decrease.

Section 21.2

Due to the lanthanide contraction, second- and third-series transition elements have similar atomic radii, so the heavier third-series elements are much denser than the lighter second-series elements.

Section 21.2

Of the first-series metals, only copper cannot be oxidized by hydrochloric acid.

Section 21.3

Transition metals tend to have many more oxidation states than main-group metals.

Section 21.3

All first series-transition metals except scandium form +2 ions.

Section 21.3

Electrons generally become more difficult to remove from left to right across a transition series. The first elements in a transition series tend to occur in relatively high oxidation states, and later elements tend to occur in low oxidation states.

Section 21.4

Students should review Lewis acid–base theory in preparation for the sections in this chapter on coordination chemistry.

Section 21.4

Ions present in a coordination compound to balance the charge of a complex ion are sometimes called counterions.

Section 21.4

Most transition metal complexes with a coordination number 4 are tetrahedral. Metal ions with eight d electrons usually form complexes that are square planar.

Section 21.5

Chelate ligands usually form five- or six-membered rings with a metal ion.

Section 21.5

Older literature named ligand water aquo instead of aqua. The ending -o implies an anionic ligand.

Section 21.5

Ammonia as a ligand has the name ammine, with two ms. If one or more of the hydrogen atoms are replaced by groups of atoms bonded through carbon, the resulting ligand is an organic amine, with one m.

Section 21.7

Isomers have the same chemical formula but different physical and chemical properties.

Section 21.7

When counting diastereoisomers, students frequently arrive at too many because they draw the same isomer in different orientations.

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Chapter 21—Transition Elements and Coordination Chemistry Section 21.9

The color of a complex corresponds to wavelengths of light that are not absorbed by the complex. The observed color is usually the complement of the color absorbed. If all wavelengths of light are absorbed, a complex appears black. If no wavelengths of light are absorbed, a complex appears white (colorless).

Section 21.10

Crystal field theory is an electrostatic model. Valence bond theory is an orbital overlap model.

Section 21.10

The stronger the interaction between a ligand and a metal ion, the greater the crystal field splitting, , and the higher the energy (and the lower the wavelength) of light absorbed by the complex.

Section 21.10

The strong-field ligands CN– and CO are toxic to humans because of their strong interactions with iron in cytochrome oxidase and hemoglobin, respectively.

Section 21.10

If  > P, the complex is predicted to be low-spin. If  < P, the complex is predicted to be high-spin.

Section 21.10

Almost all tetrahedral complexes are high-spin, and square planar complexes are low-spin. Octahedral complexes can be either low-spin or high-spin.

Section 21.11

The number of hybrid orbitals used by a metal ion in a coordination complex equals the number of atomic orbitals from which the hybrid orbitals were formed and equals the number of ligand donor atoms.

Section 21.11

Although both have a coordination number of four, the metal ions in square planar and tetrahedral complexes require different hybrid orbitals because of the differences in the orientation of the ligands.

Lecture/Laboratory Demonstration References Section 21.3

Doris Kolb, “Oxidation States of Manganese,” J. Chem. Educ., Vol. 65, 1988, 1004.

To demonstrate the numerous oxidation states attainable by manganese, several aquated ions, oxoanions, and oxides are formed in beakers positioned on an overhead projector: Mn2+(pale pink), Mn3+(rose-colored), MnO2(brown), MnO43–(blue), MnO42–(green), and MnO4–(purple).

Section 21.4

Lee R. Summerlin and James L. Ealy, Jr., “Cobalt Complexes: Changing Coordination Numbers,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 1 (American Chemical Society, Washington, DC, 1988), pp. 41–42.

Literature References Section 21.2

Peter P. Edwards and M. J. Sienko, “On the Occurrence of Metallic Character in the Periodic Table of the Elements,” J. Chem. Educ., Vol. 60, 1983, 691–696.

Section 21.2

Frank Rioux, “An Ionic Model for Metallic Bonding,” J. Chem. Educ., Vol. 62, 1985, 383–384.

Section 21.4

R. Bruce Martin, “A Stability Ruler for Metal Ion Complexes,” J. Chem. Educ., Vol. 64, 1987, 402.

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Chapter 21—Transition Elements and Coordination Chemistry Section 21.4

D. Venkataraman, Yahua Du, Scott R. Wilson, A. Hirsch, Peng Zhang, and Jeffrey S. Moore, “A Coordination Geometry Table of the d-Block Elements and Their Ions,” J. Chem. Educ., Vol. 74, 1997, 915–918.

Section 21.5

Daniel T. Haworth, “Some Linguistic Detail on Chelation,” J. Chem. Educ., Vol. 75, 1998, 47.

Section 21.5

Mark Conway, Smallwood Holoman, Ladell Jones, Ray Leenhouts, and Gerald Williamson, “Selecting and Using Chelating Agents,” Chem. Eng., Vol. 106(3), 1999, 86–90.

Section 21.5

J. Roger Hart, “EDTA-Type Chelating Agents in Everyday Consumer Products,” J. Chem. Educ., Vol. 61, 1984, 1060–1061.

Section 21.5

N. M. Senozan and R. L. Hunt, “Hemoglobin: Its Occurrence, Structure, and Adaptation,” J. Chem. Educ., Vol. 59, 1982, 173–178.

Section 21.5

Colin J. Rix, “The Biochemistry of Some Iron Porphyrin Complexes,” J. Chem.

Educ., Vol. 59, 1982, 389–392.

Section 21.5

Ei-Lchiro Ochiai, “Toxicity of Heavy Metals and Biological Defense: Principles and Applications in Bioinorganic Chemistry, Part VII,” J. Chem. Educ., Vol. 72, 1995, 479–484.

Section 21.7

John J. Fortman, “Pictorial Analogies VIII: Types of Formulas and Structural Isomers,” J. Chem. Educ., Vol. 70, 1993, 755.

Section 21.7

Hannia Lujan-Upton, “Introducing Stereochemistry to Non-science Majors,” J. Chem. Educ., Vol. 78, 2001, 475–477.

Section 21.8

Steven C. Stinson, “Chiral Drugs,” Chem. Eng. News, September 19, 1994, 38–57.

Section 21.8

Karen F. Schmidt, “Mirror Image Molecules: New Techniques Promise More Potent Drugs and Pesticides,” Science News, May 29, 1993, 348–350.

Section 21.9

Laurence Poncini and Franz L. Wimmer, “Color Classification of Coordination Compounds,” J. Chem. Educ., Vol. 64, 1987, 1001–1002.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 21.8

Worked Example 21.7: Drawing diastereoisomers and enantiomers

Section 21.10

Worked Example 21.8: Drawing crystal field energy-level diagrams for octahedral complexes

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CHAPTER 22 THE MAIN-GROUP ELEMENTS Chapter Learning Objectives Section 22.1

Use the periodic table to predict the chemical and physical properties of the maingroup elements.

Section 22.2

Compare the number and types of bonds in compounds with second-row elements and compounds with third- and higher-row elements.

Section 22.3

Write and balance a chemical equation for the reaction of ionic hydrides with water.

Classify binary hydrides and describe their bonding and properties.

Section 22.4

Use periodic trends to predict chemical and physical properties of Group 1A and 2A elements.

Write and balance reactions of alkali and alkaline earth metals with halogens, oxygen, and water.

Write electrode reactions and perform calculations in the electrolysis of molten salts to produce alkali and alkaline earth metals.

Section 22.5

Use periodic trends to predict chemical and physical properties of Group 3A elements.

Describe the structure and bonding in boron halides and diborane.

Section 22.6

Draw electron-dot structures and predict hybrid orbitals and geometry for molecules and ions in main-group compounds containing elements from Group 4A. Use the shorthand notation for a silicate anion to represent its structure and interpret the notation to find the formula and charge of a silicate anion.

Section 22.7

Write electron-dot structures and describe the structure, bonding, and properties of molecules and ions in main group compounds containing nitrogen and phosphorus.

Determine oxidation states of nitrogen and phosphorus in acids and write acid–base reactions.

Section 22.8

Classify the oxide of an element as ionic or covalent, and as acidic, basic, or amphoteric and write balanced equations for the reaction that occurs when oxides dissolve in water.

Write electron-dot structures and describe the structure, bonding, and properties of molecules in main-group compounds containing oxygen and sulfur.

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Chapter 22—The Main-Group Elements Section 22.9

Write balanced equations for the formation of halogens and reactions of their compounds.

Determine oxidation states of halogens in oxoacids and predict the relative acid strengths of oxoacids.

Section 22.10

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Draw electron-dot structures and predict geometry for molecules and ions containing noble gases.

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Chapter 22—The Main-Group Elements

Lecture Outline 1

22.1. A Review of General Properties and Periodic Trends A. Zeff increases across the periodic table.

1. Each additional valence electron does not completely shield the additional nuclear charge.

2. Atom’s electrons are more strongly attracted to the nucleus.

3. Ionization energy increases, atomic radius decreases.

4. Electronegativity increases.

5. Metallic character decreases, and nonmetallic character increases across the table.

B. Atomic radius increases from the top to the bottom of a group in the periodic table.

1. Additional shells of electrons are occupied.

2. Valence electrons are farther from the nucleus. a. Ionization energy and electronegativity generally decrease. b. Metallic character increases. c. Nonmetallic character decreases.

2 22.2. Distinctive Properties of the Second-Row Elements A. Properties of second-row elements differ markedly from those of heavier elements in the same periodic group.

B. Second-row atoms have especially small sizes and especially high electronegativities.

1. Accentuate nonmetallic behavior 2. BeO – amphoteric (other group 2A element oxides are basic) 3. Boron – forms mainly covalent molecular compounds 4. Hydrogen bonding interactions – restricted to compounds of highly electronegative second-row elements N, O, and F C. Second-row elements have only four valence orbitals – form a maximum of four covalent bonds D. Small size of second-row atoms – allows formation of multiple bonds involving  overlap of the 2p orbitals 1. 3p orbitals are more diffuse. a. Longer bond distances b. Poor  overlap 2.  bonds involving p orbitals – rare for elements of the third and higher rows 3 22.3. Group 1A: Hydrogen A. Properties of hydrogen 1. Colorless, odorless gas 2. Nonpolar, diatomic molecule 3. Weak intermolecular forces 4. Low boiling and melting points 5. Bond dissociation energy = 436 kJ/mol B. Natural occurrence 1. 75% of the mass of the universe 2. Atmospheric abundance – 0.53 ppm by volume 3. Ninth most abundant element in terms of mass in the earth’s crust and oceans 4. Third most abundant element on an atom basis C. Laboratory synthesis – reaction of dilute acid with an active metal M(s) + 2n H+(aq) → n H2(g) + Mn+(aq)

1Test Item File Questions: Multiple Choice 1 – 9, Algorithmic 1 – 9, Short Answer 1 – 2 2Test Item File Questions: Multiple Choice 43 – 44, Algorithmic 10 – 16, Short Answer 3 3Test Item File Questions: Multiple Choice 44 – 49, Algorithmic 17 – 26

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Chapter 22—The Main-Group Elements D. Hydrogen compounds 1. Hydrogen has properties similar to both the alkali metals and the halogens. a. Lose electron → H+ i. Ei = 1312 kJ/mol ii. Doesn’t completely transfer its valence electron in ordinary chemical reactions iii. Shares electron with a nonmetallic element to produce a covalent compound b. Gain electron → H– i. Eea = – 73 kJ/mol ii. Will accept an electron from an active metal to give an ionic hydride E. Binary hydride – compound that contains hydrogen and just one other element F. Ionic hydride – formed by alkali metals and heavier alkaline earth metals 1. Saltlike, high-melting, white crystalline compounds 2. Prepared by direct reaction of the elements.

3. Alkali-metal hydrides – face-centered cubic crystal structure 4. H– – good proton acceptor a. Ionic hydrides react with water to give H2 gas and OH– ions b. Redox reaction as well as acid–base reaction 5. Good reducing agents G. Covalent hydrides – hydrogen attached to another element by a covalent bond 1. Most common – hydrides of nonmetallic elements 2. Discrete, small molecules a. Relatively weak intermolecular forces b. Gases or volatile liquids at ordinary temperatures H. Metallic hydrides – lanthanide, actinide and certain d-block transition metals.

1. Interstitial hydrides – crystal lattice of metal atoms with smaller hydrogen atoms occupying holes (interstices) between larger metal atoms 2. Nonstoichiometric compounds – atomic composition can’t be expressed as a ratio of small whole numbers 3. Properties composition dependent – function of the partial pressure of H2 gas in the surroundings 4 22.4. Group 1A: Alkali Metals and Group 2A: Alkaline-Earth Metals A. General properties of alkali metals: 1. Valence-shell electron configuration – ns1 a. Common ion M+° dominates chemistry b. Lowest ionization energies of all elements c. Most powerful reducing agents in the periodic table 2. Metals a. Bright, silvery solids b. Malleable c. Good conductors of electricity d. Soft enough to cut with a knife 3. Low densities and melting points 4. Very reactive a. Must be stored under oil to prevent instantaneous reaction with oxygen and water 5. Occur in nature as salts B. Produced by reaction of their chloride salts 1. Lithium and sodium produced by electrolysis – electric current is passed through molten salt 2. Potassium, rubidium, and cesium produced by chemical reduction

4Test Item File Questions: Multiple Choice 68 – 71, Algorithmic 27 – 28, Short Answer 5 – 6

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Chapter 22—The Main-Group Elements C. Reactions of alkali metals 1. Halogens a. General reaction: 2 M(s) + X2 (g) → 2 MX(s) b. Products are halides, crystalline ionic salts c. Lower the ionization energy, greater the reactivity 2. Oxygen a. All react rapidly b. Form different kinds of products i. Li → oxide (oxidation number of O = –2) ii. Na → peroxide (oxidation number of O = –1 iii. K, Rb, Cs → superoxide (oxidation number O = –1/2) 3. Water a. General reaction: 2 M(s) + 2 H2O(l) → 2 M+(aq) + 2 OH–(aq) + H2(g) b. Reactivity increases down the group c. Produces an alkaline solution d. Redox process D. General properties of alkaline earth metals: 1. Similar to alkali metals 2. Valence-shell electron configuration – ns2 a. Common ion – M2+ 3. Soft, silvery metals 4. Higher melting points and densities than alkali metals 5. Less reactive toward oxygen and water than alkali metals 6. Occur in nature as salts E. Produced by reduction of their salts F. Reactions of alkaline earth metals 1. Same kinds of redox reactions as alkali metals 2. Less reactive than alkali metals – Ei (alkaline earth) > Ei (alkali) 3. Halogens → ionic halide salts, MX2 4. Oxygen → oxides, MO; Sr, Ba → peroxides, MO2 5. Water → metal hydroxides, M(OH)2 5 22.5. Group 3A A. Group 3A is the first of the p-block elements with valence electron configurations of ns2 np1 1. Primary oxidation state of +3 2. Heavier elements exhibit an oxidation state of +1 B. Except for boron these elements are metals.

1. Metallic character increases upon moving down the periodic table.

2. Boron a. Smaller and more electronegative than the other members of the group and forms covalent bonds rather than ionic bonds. b. Semimetal.

3. Aluminum has many commercial applictions.

4. Gallium has the largest liquid range of any metal and is used in making GaAs, a semiconductor.

5. Indium is also used in making semiconductor devices.

6. Thallium is extremely toxic and has only specialized uses.

C. Boron 1. Occurs in concentrated deposits of borate minerals 2. Obtained from BBr3 2 BBr3(g) + 3 H2(g) → 2 B(s) + 6 HBr(g)

5Test Item File Questions: Multiple Choice 79 – 91, Algorithmic 31 – 35, Short Answer 8 – 10

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Chapter 22—The Main-Group Elements 3. Crystalline boron – strong, hard, high-melting substance a. Chemically inert at room temperature b. Desirable component in high-strength composite materials D. Boron compounds 1. Boron halides – highly reactive, volatile covalent compounds a. BX3 molecules b. Behave as Lewis acids – uses vacant 2p orbitals in accepting a share in a pair of electrons from a Lewis base 2. Boron hydrides (boranes) – volatile, molecular compounds with formulas BnHm a. Simplest – diborane (B2H6) b. Two BH2 groups connected by two bridging H atoms i. Geometry around B atoms – tetrahedral ii. Bridging B–H bonds significantly longer than terminal B–H bonds c. Diborane – 12 valence electrons; electron deficient i. B atoms use sp3 hybrid orbitals to bond to four neighboring H atoms. ii. Bond formed from overlap of a boron sp3 hybrid orbital and hydrogen 1s orbital iii. Three-center, two-electron bond — joins each bridging H atom to both B atoms iv. Electron density between adjacent atoms less than in an ordinary two-center, twobond bond v. Two electrons in B–H–B bridge spread out over three atoms 6 22.6. Group 4A A. Group members include C, Si, Ge, Sn, and Pb. 1. Especially important, both in industry and in living organisms 2. Increase in metallic character down a group in the periodic table 3. Valence electron configuration: ns2 np2 4. Most common oxidation state: + 4; covalent compounds 5. +2 oxidation state a. Occurs for Sn and Pb b. Most stable for Pb c. Ionic compounds 6. No simple M4+(aq) ions for any of the group 4A elements B. Carbon 1. In uncombined form, carbon found as diamond and graphite 2. Graphene: two-dimensional array of hexagonally arranged carbon atoms that is just one atom thick a. Strong and flexible b. Superb electrical conductor 3. Oxides of carbon a. CO i. Colorless, odorless, toxic gas ii. Forms when C or hydrocarbon fuels burned in a limited supply of oxygen iii. Toxicity a. Due to its ability to bond strongly to the Fe2+ atom of hemoglobin b. Impairs the ability of hemoglobin to carry O2 to the tissues b. CO2 i. Colorless, odorless, nonpoisonous gas ii. Formed when fuels burn in an excess of O2 4. Carbonates a. Carbonic acid – forms two series of salts: carbonates and hydrogen carbonates b. CO32– removes cations from hard water

6Test Item File Questions: Multiple Choice 92 – 114, Algorithmic 36 – 40, Short Answer 11 – 15

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Chapter 22—The Main-Group Elements c. NaHCO3 – baking soda i. Reacts with acidic substances in food ii. Produces CO2 iii. Causes dough to rise 5. Hydrogen cyanide and cyanides a. HCN i. Highly toxic, volatile substance ii. CN–(aq) + H+(aq) → HCN(aq) b. CN– i. Aqueous solutions are very weakly acidic. ii. Acts as a Lewis base – bonds to transition metals through the lone pair of electrons on carbon c. CN– used to extract Au and Ag from their ores i. Produces M(CN)2– ion ii. M(CN)2– reduced by Zn to produce M iii. No longer used in U.S. due to toxicity of CN– 6. Carbides a. Binary compounds of carbon b. Carbon atom has a negative oxidation state. i. Carbides of active metals ii. Interstitial carbides of transition metals iii. Covalent network carbides C. Silicon 1. Hard, gray, semiconducting solid that melts at 1410 C a. Diamondlike structure b. Relatively poor overlap of  orbitals – no graphitelike allotrope c. Naturally found combined with oxygen (SiO2 and various silicate minerals) 2. Ultrapure silicon needed for making solid-state semiconductor devices. Purification methods: a. Conversion to SiCl4 i. Si(s) + 2 Cl2(g) → SiCl4(l) ii. Separate by fractional distillation b. Conversion of SiCl4 back to elemental silicon by reduction with hydrogen SiCl4(g) + 2 H2(g) → Si(s) + 4 HCl(g) c. Further purification – zone refining (textbook Figure 22.8) 3. Silicates – 90% of earth’s crust a. Silicon oxoanions b. Basic structural building block i. SiO4 tetrahedron ii. Occurs as simple orthosilicate ion, SiO44– c. Common minerals – anions in which two or more O atoms bridge between Si atoms to give rings, chains, layers, and extended three-dimensional structures.

4 Sharing of two oxygen atoms per SiO4 tetrahedron → cyclic anions or infinitely extended chain anions a. Repeating unit Si2O64– (textbook Figure 22.9) b. Si6O1812– – cyclic anion i. Present in beryl, Be3Al2Si6O18 ii. Emerald gemstone – 2% of Al3+ in beryl is replaced with Cr 5. Additional sharing of O atoms gives a. (Si4O116–)n – double-stranded chain anions i. Found in asbestos minerals ii. Tremolite – Ca2Mg5(Si4O11)2(OH)2

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Chapter 22—The Main-Group Elements iii. Fibrous material iv. Bonds between chains relatively weak and easily broken b. (Si4O104–) n – infinitely extended two-dimensional layer anions i. Found in clay minerals, micas, and talc ii. Talc – Mg3(OH) 2(Si4O10) iii. Mica a. Sheetlike material b. Bonds between two-dimensional layers relatively weak and easily broken 6. Silica – an infinitely extended three-dimensional structure in which the layer anions (Si4O104–)n are stacked on top of one another, and SiO4 and AlO4 tetrahedra share all four of their corners with neighboring tetrahedra a. Quartz – crystalline form of SiO2 7 22.7. Group 5A A. Group members are nitrogen, phosphorus, arsenic, antimony, and bismuth.

1. Down the periodic group – increasing atomic size, decreasing ionization energy, decreasing electronegativity, and increasing metallic character 2. Electron configuration: ns2 np3 a. Maximum oxidation state: +5 b. Minimum oxidation state: –3 c. N and P – exhibit all oxidation states between –3 and +5 d. As and Sb – most important oxidation states: +3 and +5 3. Metallic character – increases down the group a. Sb3+ and Bi3+ – found in salts b. No simple cations in compounds of N or P 4. As, Sb, and Bi found in sulfide ores and used in making various metal alloys B. Nitrogen 1. Colorless, odorless, tasteless gas 2. Makes up 78% of earth’s atmosphere 3. Separated out of liquid air by fractional distillation 4. Uses of N2 a. Gas – protective inert atmosphere in manufacturing processes b. Liquid – refrigerant c. Most important use – Haber process (manufacture of NH3) 5. N2 unreactive due to N–N triple bond a. Reactions involving N2 have a high activation energy and/or an unfavorable equilibrium constant. b. Formation of nitric oxide from N2 and O2 has an unfavorable large equilibrium constant. c. Formation of nitric oxide is more favorable at high temperatures such as in a car engine—major source of air pollution.

C. Ammonia – gateway to nitrogen chemistry 1. Starting material for industrial synthesis of other important nitrogen compounds 2. Colorless, pungent-smelling gas 3. Polar, trigonal pyramidal NH3 molecules 4. Very soluble in water (due to hydrogen bonding) 5. Easily condensed 6. Reacts with acids to yield ammonium salts – resemble alkali metal salts in their solubility D. Hydrazine (H2NNH2) – derivative of NH3 1. Replace one H atom with an NH2 group 2. Prepared by reaction of ammonia with OCl– 3. Poisonous, colorless liquid 4. Explosive in presence of air or other oxidizing agents 7Test Item File Questions: Multiple Choice 127 – 143, Algorithmic 43 – 44, Short Answer 21 – 23

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5. Used as a rocket fuel 6. Weak base and versatile reducing agent in aqueous solutions Oxides of nitrogen 1. N2O – colorless, sweet-smelling gas a. Laughing gas – small doses are mildly intoxicating b. Used as a dental anesthetic c. Propellant for dispensing whipped cream 2. NO – colorless gas a. Produced in small quantities by reacting Cu with dilute HNO 3(aq) b. Important in biological processes 3. NO2 – highly toxic, reddish brown gas that forms rapidly when NO is exposed to air a. Dimerizes to form N2O4 b. N2O4 predominates at lower temperatures and NO2 predominates at higher temperatures.

Nitric acid – one of the most important inorganic acids 1. Used in making ammonium nitrate for fertilizers 2. Used in manufacturing explosives, plastics, and dyes 3. Strong acid – 100% dissociated in water 4. Strong oxidizing agent a. Stronger oxidizing agent than H+(aq) b. Can oxidize inactive metals c. Product depends upon the nature of reducing agent and reaction conditions 5. Aqua regia – mixture of concentrated HCl and concentrated HNO3 (3:1) a. More potent oxidizing agent than HNO3 b. NO3– – serves as the oxidizing agent c. Cl– – converts Au3+ to AuCl4– Phosphorus 1. Most abundant Group 5A element 2. Found in phosphate rock, Ca3(PO4)2, and in fluorapatite, Ca5(PO4)3F 3. Apatites – phosphate minerals with the formula 3 Ca3(PO4)2  CaX2 (X– = F– or OH–) 4. Sixth most abundant element in human body a. Bones – Ca3(PO4)2 b. Tooth enamel – Ca5(PO4)3OH c. Found in DNA and RNA 5. Phosphoric acid is one of the ten most important commercial chemicals.

Exists in two common allotropic forms 1. White phosphorus – toxic, waxy, white solid existing as a. Discrete tetrahedral P4 molecules b. Low melting point c. Soluble in nonpolar solvents d. Highly reactive – due to unusual bonding e. P–O bonds are “bent,” relatively weak, and highly reactive 2. Red phosphorus a. Nontoxic b. Polymeric structure c. More stable than white phosphorus Phosphorus compounds 1. Most important oxidation states: +3 and +5 2. More likely to be found in a positive oxidation state because of its electronegativity Phosphine (PH3) – colorless, extremely poisonous gas 1. Simplest hydride of phosphorus 2. Neutral aqueous solutions 3. Poor proton acceptor 4. Easily oxidized: burns in air to form H3PO4 221

Chapter 22—The Main-Group Elements K. Phosphorous halides (PX3 or PX5) 1. Product depends on the relative amounts of reactants 2. Gases, volatile liquids, or low-melting solids L. Oxides and oxoacids of phosphorus 1. Burn P in presence of oxygen → P4O6 or P4O10 a. Molecular compounds with a tetrahedral array of P atoms b. Acidic oxides react with water → H3PO3 or H3PO4 c. P4O10 – used as a drying agent 2. H3PO3 a. Weak diprotic acid b. Only two of the three H atoms are bonded to oxygen.

3. H3PO4 – low-melting, colorless, crystalline solid a. Phosphoric acid used in the laboratory – syrupy aqueous solution (82% H3PO4 by mass) b. Manufacturing method – depends on use i. Food additive – begin with molten phosphorus ii. Fertilizers begin with Ca3(PO4)2 8 22.8. Group 6A A. Group 6A members are oxygen, sulfur, selenium, tellurium, and polonium 1. Periodic trends follow the usual patterns a. Oxygen and sulfur – typical nonmetals b. Gray selenium i. Most stable allotrope of selenium ii. Lustrous semiconducting solid c. Tellurium – semiconductor; classified as a semimetal d. Polonium i. Radioactive element that occurs in trace amounts in uranium ores ii. Silvery white metal 2. Electron configuration: ns2np4 a. Common oxidation state: –2 b. Stability of state decreases with increasing metallic character i. Oxygen – powerful oxidizing agent ii. H2Se and H2Te – reducing agents c. S, Se, and Te – less electronegative than oxygen i. Found in positive oxidation states ii. Common oxidation state: +4 and +6 3. Commercial uses of Se, Te, and Po – limited a. Se – making red-colored glass and in photocopiers b. Te – used in alloys c. Po – heat source in space equipment and a source of alpha particles in research B. Oxygen 1. Most abundant element on the planet’s surface 2. Crucial to human life 3. Oxidizing agent in metabolic “burning” of food 4. Occurrence in nature a. 23% of atmosphere – primarily as O2 b. 46% of lithosphere c. 85% of hydrosphere – in the form of H2O C. Properties of oxygen 1. Pale blue liquid and solid 2. Paramagnetic in all three phases 3. O2 – double bonded 8Test Item File Questions: Multiple Choice 151 – 165, Algorithmic 47 – 59, Short Answer 28 – 30

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a. Bond length – 121 pm b. Bond dissociation energy = 498 kJ Preparation and uses of oxygen 1. Laboratory preparation - thermal decomposition of an oxoacid salt 2. Industrial synthesis a. Fractional distillation of liquefied air b. Third ranking industrial chemical produced in the U.S. 3. Industrial uses a. Steelmaking – removes impurities from iron by forming oxides b. Sewage treatment – destroys malodorous compounds by oxidation c. Paper bleaching – oxidizes compounds that cause unwanted colors d. Inexpensive and readily available oxidizing agent Oxygen compounds 1. High electronegativity 2. Needs only two electrons to achieve an octet 3. Reacts with active metals to produce an ionic oxide – gains two electrons and achieves a noble gas configuration 4. Forms covalent oxides with nonmetals a. Achieves a noble gas configuration by sharing two additional electrons b. Forms two single bonds or one double bond i. Double bonds are formed between oxygen and other small atoms. ii. Good  overlap between oxygen p orbital and p orbitals of small atoms Acid–base properties of oxides 1. Oxidation number = –2, oxides 2. Categorize in terms of acid–base character a. Basic oxides i. Ionic ii. Formed by metals on left side of periodic table b. Acidic oxides (acid anhydrides) i. Covalent ii. Formed by nonmetals on right side of periodic table c. Amphoteric oxides i. Exhibit both acidic and basic properties ii. Formed by elements with intermediate electronegativity iii. Bonds have intermediate ionic–covalent character d. Acid–base properties and ionic–covalent character dependencies i. Position of element in the periodic table a. Acidic and covalent character increases left to right across the periodic table b. Basic and ionic character increases top to bottom down the periodic table ii. Oxidation state of element – acidic and covalent character increases with increasing oxidation state Sulfer 1. One of the ten most abundant elements on earth 2. Occurs in elemental form in underground deposits 3. Present in numerous minerals such as pyrite, galena, cinnabar, and gypsum 4. Found in natural gas and crude oil as H2S 5. Incorporated into various proteins in plants and animals 6. Used in the manufacture of sulfuric acid, a commercially important chemical Many allotropic forms 1. Rhombic sulfur – yellow crystalline solid a. Most stable form of sulfur b. Contains crown-shaped S8 rings

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Chapter 22—The Main-Group Elements c. Above 95 oC converts to monoclinic sulfur (cyclic S8 molecules packed differently in the crystal) d. Melts at 113 oC when heated at an ordinary rate 2. Above melting point, sulfur is a fluid, straw-colored liquid.

3. Between 160 C and 195 C a. Dark, reddish brown, viscous liquid b. S8 ring opens and forms S8 chains that form long polymers with more than 200,000 S atoms in the chain.

4. Above 195 C a. More fluid liquid b. Chains fragment 5. Boils at 445 C a. Pour boiling liquid into water – forms plastic sulfur b. Chains freeze in disordered array I. Hydrogen sulfide – colorless gas with strong, foul odor of rotten eggs 1. Extremely toxic 2. Prepared by treating FeS with dilute H2SO4 3. Generated in solution by hydrolysis of thioacetamide for use in qualitative analysis 4. Weak diprotic acid and mild reducing agent J. Oxides of sulfur 1. SO2 – colorless, toxic gas formed when sulfur burns in air a. Slowly oxidized in the atmosphere to SO3 b. Used for sterilizing wine and dried fruit 2. SO3 dissolves in rainwater to give sulfuric acid K. Sulfuric acid – world’s most important industrial chemical 1. Manufactured by contact process a. S burns in air to give SO2. b. SO2 is oxidized to SO3. c. SO3 reacts with water to give H2SO4.

2. Strong acid in dissociation of first proton 3. Forms two series of salts: sulfates and hydrogen sulfates 4. Oxidizing properties depend on concentration and temperature a. Dilute solutions at room temperatures – behave like HCl b. Hot, concentrated H2SO4 can oxidize metals not oxidized by H+.

9 22.9. Group 7A: The Halogens A. General properties 1. Exist as diatomic molecules 2. Too reactive to exist in nature in their elemental form 3. Valence-shell electron configuration – ns2 np5 a. Gain electrons b. Large positive ionization energies c. Large negative Eea 3. Occur in nature as salts and minerals B. Produced by oxidation of their anions C. Halogen compounds 1. Most reactive elements in the periodic table 2. Less reactive going down the periodic table a. Flourine often reacts violently b. Chlorine and bromine less reactive c. Iodine often reacts sluggishly.

3. Metals → metal halides, MXn 9Test Item File Questions: Multiple Choice 182 – 197, Short Answer 32 – 34

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Chapter 22—The Main-Group Elements D. Hydrogen halides 1. HX – behave as acids when dissolved in water 2. HCl used in a large number of commercial applications 3. HF used to etch glass; one of only a few substances that react with glass E. Oxoacids and oxoacid salts 1. Most important compounds of halogens in positive oxidation states 2. General formula – HXOn 3. Oxidation state = +1, +3, +5, or +7 (depends on value of n) 4. Acid strength increases with increasing oxidation state of the halogen a. Acidic hydrogen bonded to oxygen b. Strong oxidizing agents 10 22.10. Group 8A: Noble Gases A. General properties 1. Colorless, odorless, unreactive gases 2. Valence-shell electron configuration – ns2 np6 a. Stable configuration b. Do not form cations or anions B. Reactions of noble gases 1. No known compounds of helium or neon 2. Only one compound of argon known 3. Kr and Xe react only with fluorine 4. Lack of reactivity due to valence-shell configuration a. Large Ei b. Small Eea Inquiry – What are the barriers to a hydrogen economy?

10Test Item File Questions: Multiple Choice 202 – 203,

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CHAPTER 22 THE MAIN-GROUP ELEMENTS Teaching Tips, Points of Emphasis, and Common Misconceptions Section 22.2

Due to small size, high electronegativity, and the ability to form  bonds, the first member of each s- and p-block group differs substantially in its chemistry from the other members of the group.

Section 22.2

Compounds of period 2 elements have a maximum of four covalent bonds and an octet of electrons; elements beyond period 2 may have more than four bonds and more than eight electrons.

Section 22.2

Molecules of period 2 p-block elements contain single, double, or triple bonds; molecules of higher period elements generally contain only single bonds.

Section 22.3

The properties of molecular hydrogen compounds are best predicted if H is assigned a position in the periodic table based on its electronegativity (between B and C).

Section 22.3

Chemical species with high charge-to-radius ratios, such as H+, are unstable and tend to form new species with lower charge and/or charge dispersed over a greater volume.

Section 22.3

Like all ionic compounds, ionic hydrides are crystalline solids with high melting points. They are formed by direct reaction with only the most active metals.

Section 22.3

Like all molecular compounds which have weak intermolecular forces of attraction, covalent hydrides are gases, liquids, or low-melting solids.

Section 22.5

2c–2e terminal B–H bonds are stronger and shorter than bridging 3c–2e bonds in B2H6.

Section 22.6

Students tend to equate ionic character with high charge. On the contrary, the higher the oxidation state, the more covalent the bonds formed by an element. Bonds in SnCl4 are more covalent than those in SnCl2.

Section 22.6

Diamond is an electrical insulator, but graphite is an electrical conductor.

Section 22.6

CN– and CO have the same number of electrons and the same Lewis electron-dot structure (:CN:– and :CO:).

Section 22.6

The number of terminal (unshared) oxygen atoms equals the charge on the silicate.

Section 22.7

Due to the high strength of the nitrogen–nitrogen bond, N2, the most common molecule in Earth’s atmosphere, is practically unreactive under ordinary conditions.

An exception is its use by plants in nitrogen fixation, a process yet to be duplicated in the laboratory.

Section 22.7

Any highly exothermic reaction that generates gaseous products has the potential to occur with explosive force.

Section 22.7

When an element occurs in two extremely different oxidation states in a compound, as does nitrogen in NH4NO3, the compound may decompose violently.

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Section 22.7

The equilibrium for an exothermic reaction shifts to the left as the temperature is raised (Le Châtelier’s principle).

Section 22.7

A change in reaction conditions may result in a change in the reaction products.

Section 22.7

Note that both allotropes of period 3 phosphorus have only single bonds, whereas period 2 nitrogen has a triple bond in N2.

Section 22.7

In oxoacids, only hydrogen atoms attached to oxygen are acidic.

Section 22.8

Remember that double bonds are shorter and stronger than single bonds but longer and weaker than triple bonds formed from the same two atoms.

Section 22.8

Oxygen forms single and double (occasionally triple) bonds with other period 2 nonmetals in order to attain a complete octet. Less commonly does it form multiple bonds with period 3 and higher-period elements.

Section 22.8

The most basic oxides are ionic solids formed by the reaction of oxygen with elements at the left and bottom of the periodic table; the most acidic oxides are covalent and formed by the reaction of oxygen with elements at the right and top of the periodic table.

Section 22.8

Note that all allotropes of period 3 sulfur have only single bonds, whereas period 2 oxygen has a double bond in O2.

Section 22.8

Expecting to see a yellow solid, students frequently do not recognize finely divided sulfur, which appears white.

Section 22.8

A small segment of the population experiences a severe allergic reaction to foods treated with sulfites, the presence of which is noted on wine-bottle labels but not always on labels of dried fruits and nuts.

Section 22.9

The greater the number of oxygen atoms, the higher the oxidation state of the halogen and the stronger the oxoacid.

Lecture/Laboratory Demonstration References Section 22.1

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Halogens Compete for Electrons,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washingon DC, 1988), pp. 60–61

Section 22.1

Bassam Z. Shakhashiri, “Acidic and Basic Properties of Oxides,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 3 (The University of Wisconsin Press, Madison, 1989), pp.109–113.

Section 22.7

Bassam Z. Shakhashiri, “Preparation and Properties of Ammonia,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985), pp. 202–204. Ammonia gas is produced from the reaction of ammonium chloride with calcium hydroxide and is collected in an Erlenmeyer flask by the displacement of air. The ammonia gas turns moistened litmus paper blue, demonstrating the basic properties of ammonia dissolved in water.

When the mouth of an ammonia-filled Erlenmeyer flask is placed below the surface

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Chapter 22—The Main-Group Elements of a pool of water, water rapidly enters the flask to replace ammonia that dissolves in the water. White smoke forms when ammonia vapor is passed over the top of a beaker of concentrated hydrochloric acid forming solid ammonium chloride.

Section 22.8

Bassam Z. Shakhashiri, “Preparation and Properties of Oxygen,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985), pp.137–141. Oxygen is produced by the reaction of hydrogen peroxide potassium permanganate and collected by displacing water from an inverted Erlenmeyer flask. The oxygen is shown to support vigorous combustion when a glowing wooden ember and hot steel wool burst into flames when placed in an atmosphere of oxygen gas.

Section 22.8

Bassam Z. Shakhashiri, “Combining Volume of Oxygen with Sulfur,” Chemical Demonstrations, A Handbook for Teachers of Chemistry, Vol. 2 (The University of Wisconsin Press, Madison, 1985), pp.190–192. SO2, not SO3, produced in the combustion reaction of sulfur and oxygen.

Section 22.8

Lee R. Summerlin, Christie L. Borgford, and Julie B. Ealy, “Plastic Sulfur,” Chemical Demonstrations, A Sourcebook for Teachers, Vol. 2 (American Chemical Society, Washington, DC, 1988), p. 53.

Section 22.9

C. E. Ophardt, “Redox Demonstrations and Descriptive Chemistry: Part 2.

Halogens,” J. Chem. Educ., Vol. 64, 1987, 807. Using an abbreviated table of reduction potentials as a predictive tool, reactions of bromine and iodine in various oxidation states are demonstrated. When excess I– is reacted with BrO3– in the presence of starch, a blue color forms immediately, the final products being I 2 and Br–. When I– is reacted with excess BrO3– in the presence of starch, within minutes a blue color turns purple then yellow with the final products being Br 2 and IO3–. A second series of reactions demonstrates the instability of certain oxidation states of iodine due to reaction with oxygen and disproportionation under acidic and basic conditions.

Section 22.9

Robert C. Hansen, “An Overhead Demonstration of Some Descriptive Chemistry of the Halogens and Le Châtelier’s Principle,” J. Chem. Educ., Vol. 65, 1988, 264–265.

Literature References Section 22.1

George Gorin, “Mendeleev and Moseley: The Principal Discoverers of the Periodic Law,” J. Chem. Educ., Vol. 73, 1996, 490–493.

Section 22.1

Eric R. Scerri, “The Evolution of the Periodic System,” Scientific American, September 1998, 8–83.

Section 22.1

N. K. Goh and L. S. Chia, “Using the Learning Cycle to Introduce Periodicity,” J. Chem. Educ., Vol. 66, 1989, 747.

Section 22.1

Robert H. Goldsmith, “Metalloids,” J. Chem. Educ., Vol. 59, 1982, 526–527.

Section 22.1

Kimberley A. Waldron, Eric M. Fehringer, Amy E. Streeb, Jennifer E. Trotsky, and Joshua J. Pearson, “Screen Percentages Based on Slater Effective Nuclear Charge as a Versatile Tool for Teaching Periodic Trends,” J. Chem. Educ., Vol. 78, 2001, 635–639.

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Section 22.1

Joan Mason, “Periodic Contractions Among the Elements; Or, on Being the Right Size,” J. Chem. Educ., Vol. 65, 1988, 17–20.

Section 22.1

Gabriel Pinto, “Using Balls from Different Sports to Model the Variation of Atomic Sizes,” J. Chem. Educ., Vol. 75, 1998, 725–726.

Section 22.1

Ronald L. Rich, “Periodicity in the Acid–Base Behavior of Oxides and Hydroxides,” J. Chem. Educ., Vol. 62, 1985, 44.

Section 22.8

Foil A. Miller, “Joseph Priestly, Preeminent Amateur Chemist,” J. Chem. Educ., Vol 66, 1989, 801.

Section 22.8

Michael Laing, “The Three Forms of Molecular Oxygen,” J. Chem. Educ., Vol 66, 1989, 453–455.

Section 22.9

Miriam C. Nagel, “Herman Frasch, Sulfur King,” J. Chem. Educ., Vol. 58, 1981, 60–61.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 22.6

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Worked Example 22.4: Interpreting representations of silicate anions

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CHAPTER 23 ORGANIC AND BIOLOGICAL CHEMISTRY Chapter Learning Objectives Section 23.1

Represent the chemical structure of alkanes as condensed structures or line drawings.

Identify and draw isomers of alkanes.

Section 23.2

Identify chiral centers in a molecule and classify molecules as chiral or achiral.

Section 23.3

Identify and name functional groups in organic molecules.

Represent molecules with functional groups using line drawings.

Section 23.4

Classify a monosaccharide as an aldose or ketose and draw a typical open-chain structure.

Classify the type of isomerism in monosaccharides.

Section 23.5

Draw orbital overlap pictures to describe bonding in organic compounds.

Identify and draw cis–trans isomers.

Section 23.6

Recognize and draw the structure of triacyl glycerols.

Relate lipid structure to physical properties.

Section 23.7

Used curved arrows to draw resonance structures and specify formal charges in organic line drawings.

Section 23.8

Draw orbital overlap pictures for conjugated systems and identify localized and delocalized electrons.

Identify conjugated systems and the hybridization of atoms included in the system.

Section 23.9

Identify amino acids and draw structures for peptides.

Section 23.10

Predict if an organic compound is aromatic.

Section 23.11

Draw structures showing how different components of the nucleic acid are joined, and write the DNA sequence that is complementary to a specified strand.

Draw resonance structures to show that DNA bases are aromatic.

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Lecture Outline 1

23.1. Organic Molecules and Their Structures: Constitutional Isomers A. Hydrocarbons – molecules that contain only carbon and hydrogen B. Alkanes 1. Hydrocarbons that have only single bonds 2. Methane, CH4, is the simplest alkane C. Straight-chain alkanes – compounds with all their carbons connected in a row D. Branched-chain alkanes – those with a branching connection of carbons E. Constitutional isomers 1. Molecules that have the same molecular formula but different connections between atoms 2. Different isomers are different chemical compounds.

3. Have different structures’ different chemical and physical properties F. Condensed structures – carbon–hydrogen and carbon–carbon single bonds aren’t shown.

1. Bonds are understood.

2. The horizontal bonds between carbons aren’t shown but vertical bonds are shown.

G. A molecule can be arbitrarily shown in a great many ways.

1. Rotation around carbon–carbon single bonds is possible.

2. Conformations – a number of possible three-dimensional structures resulting from the ability of parts of a molecule to spin around a carbon–carbon single bond H. Line drawings are another shorthand representation of molecular structure.

1. Only the bonds in the carbon backbone are shown.

2. The end of each line segment represents a carbon atom.

3. Mentally supply the proper number of hydrogen atoms needed to fill out carbon’s valency.

I. Acylic hydrocarbons – open-chain hydrocarbons J. Cycloalkanes – alkanes that contain rings of carbon atoms K. Condensed structures can be awkward so line structures used 1. Structures are represented by polygons.

2. Rings larger than four carbons have a puckered shape.

23.2. Stereoisomers: Chiral Molecules A. Chiral molecules – mirror image that cannot be superimposed on the original 1. Chiral objects have handedness – no plane of symmetry 2. Achiral objects have a plane of symmetry B. Nonidentical mirror image molecules are called enantiomers 1. Type of stereoisomer 2. Same connections between atoms but different arrangements in space of the atoms 3. Same chemical and physical properties 4. Differ in reactions with other chiral molecules 5. Often called optical isomers because of their ability to rotate plane-polarized light 2 23.3. Families of Organic Molecules: Functional Groups A. Can classify organic compounds into families according to their structural features – chemical behavior of the members of a family is predictable B. Functional group – a part of a larger molecule consisting of an atom or group of atoms that has characteristic chemical behavior – a given functional group undergoes the same kinds of reactions in every molecule of which it is a part C. The chemistry of an organic molecule, regardless of its size and complexity, is determined by the functional groups it contains.

D. Most common functional groups – textbook Table 23.1

1Test Item File Questions: Multiple Choice 1 – 17, Algorithmic 1 – 2, Short Answer 1 – 9 2Test Item File Questions: Multiple Choice 28 – 43, Algorithmic 3 – 21, Short Answer 10-16

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Chapter 23—Organic and Biological Chemistry 3

23.4. Carbohydrates: A Biological Example of Isomers A. Carbohydrates are a large class of polyhydroxylated aldehydes and ketones—commonly called sugars 1. Exhibit several types of isomerism 2. Simple sugars or monosaccharides cannot be broken down into smaller molecules by hydrolysis.

3. Complex carbohydrates or polysaccharides consist of many simple sugars linked together and can be broken down by hydrolysis.

B. Monosaccharides 1. Classified as aldoses, which contain an aldehyde carbonyl group, or ketoses, which contain a ketone carbonyl group 2. The -ose suffix indicates a sugar.

3. The number of carbon atoms in the sugar is indicated by the appropriate numerical prefix 4. Often depicted as having open-chain structures 5. Exist primarily as cyclic molecules in which an –OH group near one end of the chain adds to the carbonyl group at or near the other end of the chain to form a ring a. In the  cyclic form, the –OH group at C1 is on the bottom of the ring b. In the  cyclic form, the –OH group at C1 is on the top of the ring 6. Three types of isomers possible a. Constitutional isomers, which have the same chemical formula but different atom connections such as linear and branched b. Enantiomers, which are nonidentical mirror images c. Anomers which have a –OH group on different sides of the ring C. Polysaccharides 1. Sucrose or table sugar is a common disaccharide composed of one molecule of glucose joined to one molecule of fructose with a glycosidic linkage 2. Cellulose is a fibrous polysaccharide composed of several thousand -glucose molecules joined in a chain 3. Starch is an edible polysaccharide that contains -glucose units 4 23.5. Valence Bond Theory and Orbital Overlap Pictures A. Simple representations of organic molecules include electron dot and line structures.

1. Indicate atom connections and the location of multiple bonds and electron lone pair 2. Limited in that they are do not adequately represent differences in structure or reactivity B. Valence bond theory describes a covalent bond as arising from overlapping orbitals on different atoms.

1. Hybridized and unhybridized orbitals can overlap 2. Overlap type leads to sigma, , and pi, , bonds C. Many organic molecules have a carbon atom with sp2 hybridization which leads to  bonds.

1. Energetically unfavorable rotation around the double bond 2. Geometric or cis/trans isomers are possible. a. cis arrangement has groups on the same side of the double bond. b. trans arrangement has groups on opposite sides of the double bond.

5 23.6. Lipids: A Biological Example of cis-trans Isomerism A. Lipids are naturally occurring organic molecules that dissolve in nonpolar solvents.

B. There are a great many different kinds of lipids; all contain large hydrocarbon portions 1. Animal fats are usually solid, for example, butter and lard.

2. Vegetable oils are usually liquid, for example, peanut and corn oils.

3. Both have similar structures.

3Test Item File Questions: Multiple Choice 62 – 70, 77 Algorithmic 28 – 29, Short Answer 17 4Test Item File Questions: Multiple Choice 78 – 80, Algorithmic 31 – 33 5Test Item File Questions: Multiple Choice 89 – 95, Short Answer 18 – 29

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Chapter 23—Organic and Biological Chemistry C. Fats and oils are triacylglycerols or triglycerides.

1. Esters of glycerol (1,2,3-propanetriol) with three long-chain carboxylic acids called fatty acids 2. Fatty acids are usually unbranched and have an even number of carbon atoms ranging from 12–22.

3. The three fatty acids of a given triacylglycerol need not be the same.

4. A fat or oil is a mixture of many different triacylglycerols.

D. About 40 different fatty acids occur naturally.

1. Saturated fatty acids include palmitic acid (C16) and stearic acid (C18).

2. Monounsaturated fatty acids have one double bond, for example, oleic acid.

3. Polyunsaturated fatty acids have more than one double bond, for example, linoleic acid and linolenic acid.

E. Vegetable oils generally have a higher proportion of unsaturated fatty acids than do animal fats.

1. Fats have saturated fatty acids that can line up with one another to maximize their London dispersion forces, leading to a solid.

2. Oils with unsaturated fatty acids in the cis confirmation have a geometry which disrupts the intermolecular forces, leading to liquids.

F. Oils can be processed by hydrogenation which converts unsaturated oils to saturated fats.

1. The amount of unsaturation affects the melting point.

2. Hydrogenation improves shelf-life, as double bonds are susceptible to reaction with oxygen.

3. Isomerization of the double bonds can occur, leading to trans fats.

G. Unsaturated facts are considered to be healthier.

23.7. Formal Charge and Resonance in Organic Compounds A. Formal charge and resonance play important roles in understanding structure and reactivity.

1. Most organic reactions occur between regions of high and low electron density.

2. Resonance helps predict stability and sites of high and low electron density.

B. Benzene has two resonance structures which contribute to the resonance hybrid.

1. Six electrons in the double bonds are delocalized over the six-ring carbon atoms. a. Reduces electron–electron repulsions b. Stabilizes the molecule 2. Resonance hybrid is an average of the resonance structures.

C. Common patterns that indicate resonance 1. Lone pair on an atom one  bond away from a  bond.

2. Polar  bond.

3. A positive formal charge on an atom one  bond from a  bond.

6 23.8. Conjugated Systems A. Can combine a line drawing with orbital overlap picture 1. Provides a good representation of the bonding in organic molecules.

2. Lines represent  bonds.

3. p orbitals superimposed on the line structure to show multiple bonds B. Geranial has a  system with more than two adjacent p orbitals that can overlap 1. Four sp2 hybridized carbon atoms in a row 2. The four atoms are conjugated. a. Alternating single and double bonds b. Two or more  bonds separated by  bonds c. Double bonds are linked and do not act as separate, independent double bonds C. Conjugation and resonance 1. Line drawing localizes electron in a bond.

2. In a conjugated system, the electrons are best described as being delocalized or associated with more than one or two atoms at the same time.

6Test Item File Questions: Algorithmic 34

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Chapter 23—Organic and Biological Chemistry 3. Conjugation provides a structural basis for resonance theory.

4. Resonance theory says that a  bond doesn’t fluctuate between two locations but is in both at the same time.

5. Simplified orbital picture shows electrons are localized in  bonds but delocalized in  bonds in a conjugated system. a. Overlap of the p orbitals involving the  bond and lone pair spread the four electrons over three atoms. b. Delocalization leads to resonance stabilization.

D. Conjugation and hybridization 1. According to valence bond theory atoms will adopt the hybridization that makes the molecule most stable by lowering the overall energy. a. Conjugation is stabilizing. b. An atom that is not conjugated will adopt a stable hybridization according to the VSEPR model.

7 23.9. Proteins: A Biological Example of Conjugation A. Proteins are a group of biologically important molecules 1. Make up 50% of human body’s dry weight 2. Catalyze biological reactions B. Proteins have many different biological functions.

1. Keratin is a structural protein found in skin, hair, and fingernails.

2. Hormones serve as chemical messengers, such as insulin that controls carbohydrate metabolism.

3. Enzymes are biological catalysts, such as DNA polymerase (FYI textbook Chapter 12) C. Proteins are long chains of amino acids.

1. Amino acids contain a basic amino functional group (−NH2) and an acidic functional group (−CO2H).

2. Amino acids are linked by a peptide bond involving the –NH2 group of one and the – CO2H group of another to form an amide. a. A dipeptide results when two amino acids are linked by one amide bond. b. A tripeptide results when three amino acids are linked by two amide bonds. c. Short chains of up to 50 linked amino acids are called peptides. d. Polypeptides and proteins involve longer chains.

3. Structure of peptide bond best understood in terms of conjugation and hybridization a. Nitrogen lone pair is one  bond away from a C–O  bond. b. Conjugated system c. C–N peptide bond has significant double bond character even though it is represented in a line structure as a single bond.

4. Twenty different amino acids are commonly found in proteins. a. Each referred to by a three letter code, textbook Figure 23.13 b. In all twenty, the amine nitrogen is connected to the carbon atom next to or  to the carboxylic acid group. c. The twenty amino acids differ in the nature of the group, called the side chain, attached to the  carbon. d. Nine amino acids are essential amino acids, which must be obtained from the diet as the body cannot synthesize them 5. Amino acids classified as neutral, basic, or acid depending upon the structure of the side chain a. Two acidic amino acids with an additional carboxylic acid group in the side chain b. Three basic amino acids with an additional amine group in the side chain c. Fifteen neutral amino acids i. Hydrophobic (water fearing) amino acids have nonpolar side chains. ii. Hydrophilic (water loving) amino acids have polar side chains.

7Test Item File Questions: Multiple Choice 98 – 115, Algorithmic 35 – 56, Short Answer 20, 22

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Chapter 23—Organic and Biological Chemistry 6. All noncyclic proteins have an N-terminal amino acid with a free –NH2 on one end and a C-terminal amino acid with a free –CO2H on the other end a. By convention, a protein is written with the free –NH2 on the left and the free –CO2H on the right. b. The name of the protein is indicated using the three-letter abbreviations of the component amino acids.

8 23.10. Aromatic Compounds and Molecular Orbital Theory A. Aromatic compounds 1. Class of compounds which are cyclic, planar, and have adjacent p orbitals around a ring 2. Simplest aromatic – benzene, which is a six-membered ring with alternating single and double bonds 3. Many biological and pharmaceutical agents contain an aromatic ring.

B. Benzene’s stability is a consequence of its electronic structure.

1. Conjugation and resonance explain the structure of benzene but not its stability.

2. Molecular orbital theory is more useful.

C. Molecular orbital theory says that electrons are in molecular orbitals that cover the entire molecule.

D. In benzene, six molecular orbitals are formed from the mathematical combination of the six p orbitals.

1. Three bonding molecular orbitals which are lower in energy than the atomic p orbitals 2. Three antibonding molecular orbitals which are higher in energy than the atomic p orbitals 3. Six electrons are paired in the three bonding molecular orbitals, stabilizing the molecule.

E. The number of electrons is important.

1. Four electrons in the molecular orbitals result in two unpaired electrons—a destabilizing arrangement.

2. Eight electrons in the molecular orbitals result in two electrons in an antibonding orbital— a destabilizing arrangement.

3. Hückel’s Rule says that the number of electrons in an aromatic system is 4n + 2 where n is zero or a positive integer.

4. Aromatic biological commonly have 6 or 10 electrons F. Aromatic compounds have certain characteristics.

1. Must be cyclic.

2. Must be planar so that p orbitals can overlap and mix mathematically.

3. Must have p orbitals on adjacent atoms around the ring.

4. Must have (4n + 2)  electrons in the ring.

9 23.11. Nucleic Acids: A Biological Example of Aromaticity A. Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are chemical carriers of genetic information.

1. Nucleic acids are made of nucleotides linked together in a chain.

2. Each nucleotide is composed of a nucleoside plus phosphoric acid, H 3PO4.

3. Each nucleoside is composed of an aldopentose sugar plus an amine base. a. Ribose is the sugar component of RNA. b. 2-deoxyribose is the sugar compo nent of DNA. c. Four different cyclic amine bases occur in DNA: adenine, guanine, cytosine, and thymine. d. Four different cyclic amine bases occur in RNA: adenine, guanine, cytosine, and uracil.

4. In both DNA and RNA the cyclic amine base is bonded to the C1’’ of the sugar and the phosphoric acid is bonded to the C5’’ of the sugar.

5. Nucleotides join together by forming a bond between the phosphate group of one nucleotide and the hydroxyl group on the sugar component of another.

8Test Item File Questions: Multiple Choice 158 – 163, Short Answer 21 9Test Item File Questions: Multiple Choice 164 – 175, Algorithmic 57, Short Answer 24 – 26

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Chapter 23—Organic and Biological Chemistry B. The structure of a nucleic acid depends on the sequence of individual nucleotides.

1. Abbreviations are used for each nucleotide.

2. A for adenosine, G for guanine, C for cytosine, T for thymidine, and U for uracil C. DNA consists of two polynucleotide strands coiled around each other in a double helix.

1. The sugar–phosphate backbone is on the outside of the helix.

2. Amine bases are on the inside.

3. A base on one strand is directly opposite a base on the other strand.

4. The two strands run in opposite directions.

5. Strands are held together by hydrogen bonds between opposite base pairs a. Adenine and thymine form two strong hydrogen bonds to each other but not to guanine or cytosine. b. Guanine and cytosine form two strong hydrogen bonds to each other but not to adenine or thymine.

D. DNA strands are complementary, not identical.

1. Whenever guanine occurs in one strand, cytosine occurs in the other.

2. Whenever adenine occurs in one strand, thymine occurs in the other.

E. The bases in the DNA must be planar so that they can stack in the strands.

1. Base stacking is key to DNA double-helix stability.

2. Tight stacking of bases leads to a structure with strong intermolecular attractions between adjacent bases.

F. Conjugation and resonance show that all of the DNA bases are aromatic.

Inquiry – Why do enantiomers have different biological responses?

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Chapter 23—Organic and Biological Chemistry

CHAPTER 23 ORGANIC AND BIOLOGICAL CHEMISTRY Teaching Tips, Points of Emphasis, and Common Misconceptions Section 23.1

An alkane is a hydrocarbon (containing only carbon and hydrogen atoms) in which each carbon atom is bonded to four other atoms. Because each carbon atom is bonded to its maximum number of atoms and cannot bond to additional atoms, the alkane is said to be saturated.

Section 23.1

C–H and most C–C bonds are not shown in condensed structures.

Section 23.1

An electron-dot structure does not give the shape of a molecule. The shape is defined by applying the VSEPR model to each carbon, nitrogen, and oxygen atom present in the molecule.

Section 23.1

Although cycloalkanes are usually drawn as planar polygons, most are not planar, but puckered.

Section 23.3

To emphasize the role of a functional group in the chemistry of a molecule, it is sometimes useful to represent nonreacting portions of the molecule with the letter R. For example, both CH3OH and CH3CH2OH are represented by the general formula ROH, because the chemistry of both depends primarily on the –OH group and little or not at all on the nature of R.

Literature References Section 23.1

Raymond B. Seymour, “Alkanes: Abundant, Pervasive, Important, and Essential,” J. Chem. Educ., Vol. 66, 1989, 59–63.

Section 23.1

A. Haudrechy, “A Simple Method of Drawing Stereoisomers from Complicated Symmetrical Structures,” J. Chem. Educ., Vol. 77, 2000, 864–866.

Section 23.6

Ellin Doyle, “Trans Fatty Acids,” J. Chem. Educ., Vol. 74, 1997, 1030–1032.

Media References: List of Interactive Videos available in the eText and on MasteringTM Chemistry Section 23.2

Worked Example 23.3: Identifying chiral molecules

Section 23.10

Worked Example 23.13: Predicting if an organic compound is aromatic

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