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v4.1

a

Acknowledgments

The Princeton Review would like to give special thanks to Nick Leonardi for his hand in revising and updating this title to better prepare students for the AP Chemistry Exam. Nick would like, in turn, to thank the AP Chemistry teacher community from the College Board forums, and his parents, who, despite not knowing chemistry, have contributed more to his success than anybody else.

Contents

Cover

Title Page

Copyright

Acknowledgments

Register Your Book Online!

Part I: Using This Book to Improve Your AP Score

Preview Activity: Your Knowledge, Your Expectations

Your Guide to Using This Book

How to Begin

Part II: Practice Test 1

Practice Test 1: Answers and Explanations

Part III: About the AP Chemistry Exam

The Structure of the AP Chemistry Exam

Overview of Content Topics

How AP Exams Are Used

Other Resources

Designing Your Study Plan

Part IV: Test-Taking Strategies for the AP Chemistry Exam

1 How to Approach Multiple-Choice Questions

2 How to Approach Free-Response Questions

Part V: Content Review for the AP Chemistry Exam

3 Big Idea #1: Atoms, Elements, and the Building Blocks of Matter

The Periodic Table

Moles

Coulomb’s Law

Photoelectron Spectroscopy

Predicting Ionic Charges

Periodic Trends

Chapter 3 Questions

Chapter 3 Answers and Explanations

4 Big Idea #2: Bonding and Phases

Bonds Overview

Ionic Bonds

Metallic Bonds

Covalent Bonds

Intermolecular Forces

Vapor Pressure

Lewis Dot Structures

Kinetic Molecular Theory

Maxwell-Boltzmann Diagrams

Effusion

The Ideal Gas Equation

Dalton’s Law

Deviations from Ideal Behavior

Density

Solution Separation

Chapter 4 Questions

Chapter 4 Answers and Explanations

5 Big Idea #3: Chemical Reactions, Energy Changes, and Redox Reactions

Types of Reactions

Chemical Equations

Gravimetric Analysis

Enthalpy

Energy Diagrams

Catalysts and Energy Diagrams

Oxidation States

Oxidation-Reduction Reactions

Galvanic Cells

Electrolytic Cells

Chapter 5 Questions

Chapter 5 Answers and Explanations

6 Big Idea #4: Chemical Reactions and Their Rates

Rate Law Using Initial Concentrations

Rate Law Using Concentration and Time

Collision Theory

Beer’s Law

Reaction Mechanisms

Catalysts

Chapter 6 Questions

Chapter 6 Answers and Explanations

7 Big Idea #5: Laws of Thermodynamics and Changes in Matter

Heat and Temperature

State Functions

Enthalpy of Formation, ΔHº_f

Bond Energy

Hess’s Law

Enthalpy of Solution

Thermodynamics of Phase Change

Calorimetry

Heating Curves

Entropy

Gibbs Free Energy

Free Energy Change, ΔG

ΔG, ΔH, and ΔS

Voltage and Favorability

Chapter 7 Questions

Chapter 7 Answers and Explanations

8 Big Idea #6: Equilibrium, Acids and Bases, Titrations, and Solubility

The Equilibrium Constant, K_eq

Le Châtelier’s Principle

Changes in the Equilibrium Constant

The Reaction Quotient, Q

Solubility

Acids and Bases Definitions

pH

Acid Strengths

Neutralization Reactions

Buffers

Indicators

Titration

Chapter 8 Questions

Chapter 8 Answers and Explanations

9 Laboratory Overview

Part VI: Practice Test 2

Practice Test 2: Answers and Explanations

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Part I

Using This Book to Improve Your AP Score

PREVIEW ACTIVITY: YOUR KNOWLEDGE, YOUR EXPECTATIONS

Your route to a high score on the AP Chemistry Exam depends a lot on how you plan to use this book. Please respond to the following questions.

YOUR GUIDE TO USING THIS BOOK

This book is organized to provide as much—or as little—support as you need, so you can use this book in whatever way will be most helpful to improving your score on the AP Chemistry Exam.

Once you have nailed down your strengths and weaknesses with regard to this exam, you can focus your test preparation, build a study plan, and be efficient with your time.

You may choose to use some parts of this book over others, or you may work through the entire book. This will depend on your needs and how much time you have. Let’s now look how to make this determination.

HOW TO BEGIN

Part II

Practice Test 1

Click here to download the PDF version.

AP^® Chemistry Exam

SECTION I: Multiple-Choice Questions

DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.

Instructions

Section I of this examination contains 60 multiple-choice questions. Fill in only the ovals for numbers 1 through 60 on your answer sheet.

CALCULATORS MAY NOT BE USED IN THIS PART OF THE EXAMINATION.

Indicate all of your answers to the multiple-choice questions on the answer sheet. No credit will be given for anything written in this exam booklet, but you may use the booklet for notes or scratch work. After you have decided which of the suggested answers is best, completely fill in the corresponding oval on the answer sheet. Give only one answer to each question. If you change an answer, be sure that the previous mark is erased completely. Here is a sample question and answer.

Sample Question

Chicago is a

(A) state

(B) city

(C) country

(D) continent

Sample Answer

Use your time effectively, working as quickly as you can without losing accuracy. Do not spend too much time on any one question. Go on to other questions and come back to the ones you have not answered if you have time. It is not expected that everyone will know the answers to all the multiple-choice questions.

About Guessing

Many candidates wonder whether or not to guess the answers to questions about which they are not certain. Multiple-choice scores are based on the number of questions answered correctly. Points are not deducted for incorrect answers, and no points are awarded for unanswered questions. Because points are not deducted for incorrect answers, you are encouraged to answer all multiple-choice questions. On any questions you do not know the answer to, you should eliminate as many choices as you can, and then select the best answer among the remaining choices.

CHEMISTRY

SECTION I

Time—1 hour and 30 minutes

INFORMATION IN THE TABLE BELOW AND ON THE FOLLOWING PAGES MAY BE USEFUL IN ANSWERING THE QUESTIONS IN THE SECTION OF THE EXAMINATION

Use the following information to answer questions 3-6

A student titrates some 1.0 M HCl into 20.0 mL of methylamine (CH₃NH₂), a weak base which only accepts a single proton. The following titration curve results:

Use the following information to answer questions 14-16

The radius of atoms and ions is typically measured in Angstroms (Å), which is equivalent to 1 * 10^-10 m. Below is a table of information for three different elements.

Element	Atomic Radius (Å)	Ionic Radius (Å)
Ne	0.38	N/A
P	0.98	1.00
Zn	1.42	1.35

Use the following information to answer questions 22-24

10.0 g each of three different gases are present in three glass containers of identical volume, as shown below. The temperature of all three flasks is held constant at 298 K.

Questions 32-36 refer to the following.

Questions 45-48 refer to the following.

Questions 54-56 refer to the following.

END OF SECTION I

CHEMISTRY

SECTION II

7 Questions

(Total time—105 minutes)

INFORMATION IN THE TABLE BELOW AND ON THE FOLLOWING PAGES MAY BE USEFUL IN ANSWERING THE QUESTIONS IN THE SECTION OF THE EXAMINATION

YOU MAY USE YOUR CALCULATOR FOR THIS SECTION

Directions: Questions 1-3 are long free-response questions that require about 23 minutes each to answer and are worth 10 points each. Questions 4-7 are short free-response questions that require about 9 minutes each to answer and are worth 4 points each.

Write your response in the space provided following each question. Examples and equations may be included in your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures.

STOP

END OF EXAM

Practice Test 1:

Answers and Explanations

PRACTICE TEST 1: MULTIPLE-CHOICE ANSWER KEY

SECTION I—MULTIPLE-CHOICE ANSWERS AND EXPLANATIONS

Section II—Free-Response Answers and Explanations

K_a =

Part III

About the AP Chemistry Exam

THE STRUCTURE OF THE AP CHEMISTRY EXAM

The AP Chemistry Exam is a three-hour-long, two-section test that attempts to cover the material you would learn in a college first-year chemistry course. The first section is a 90-minute, 60-question multiple-choice section. As of 2015, the second section is 105 minutes and consists of three long-form free-response questions and four short-form free-response questions.

The multiple-choice section is scored by a computer, and the free-response questions are scored by a committee of high school and college teachers. The free-response questions are graded according to a standard set at the beginning of the grading period by the chief faculty consultants. Inevitably, the grading of Section II is never as consistent or accurate as the grading of Section I.

Take a look at the following data on how students scored on the May 2016 test:

Score (Meaning)	Percentage of Test Takers	Equivalent Grade in a first-year college course
5 (extremely qualified)	9.2%	A
4 (well qualified)	16.1%	A–, B+, B
3 (qualified)	28.1%	B–, C+, C
2 (possibly qualified)	24.9%	C–
1 (no recommendation)	21.7%	D

OVERVIEW OF CONTENT TOPICS

The concepts of the AP Chemistry Exam are broken down into six major themes defined by the College Board as the Big Ideas. Rather than learning multiple disparate topics (as has been done in the past), these Big Ideas interconnect principles within these topics that describe fundamental chemical phenomena. The six Big Ideas are as follows:

This book has been arranged to teach chemistry topics grouped to each of these Big Ideas.

HOW AP EXAMS ARE USED

Different colleges use AP Exam scores in different ways, so it is important that you go to a particular college’s web site to determine how it uses AP Exam scores. The three items below represent the main ways in which AP Exam scores can be used.

OTHER RESOURCES

There are many resources available to help you improve your score on the AP Chemistry Exam, not the least of which are your teachers. If you are taking an AP class, you may be able to get extra attention from your teacher, such as obtaining feedback on your free-response questions. If you are not in an AP course, reach out to a teacher who teaches chemistry, and ask if the teacher will review your free-response questions or otherwise help you with content.

Another wonderful resource is AP Students, the official site of the AP Exams. The scope of the information at this site is quite broad and includes:

The AP Students home page address is: https://apstudent.collegeboard.org.

Finally, The Princeton Review offers tutoring and small group instruction. Our expert instructors can help you refine your strategic approach and add to your content knowledge. For more information, call 1-800-2REVIEW.

DESIGNING YOUR STUDY PLAN

As part of the Introduction, you identified some areas of potential improvement. Let’s now delve further into your performance on Test 1, with the goal of developing a study plan appropriate to your needs and time commitment.

Read the answers and explanations associated with the multiple-choice questions (starting at this page). After you have done so, respond to the following:

Part IV

Test-Taking Strategies for the AP Chemistry Exam

Chapter 1

How to Approach Multiple-Choice Questions

THE BASICS

Section I of the test is composed of 60 multiple-choice questions, for which you are allotted 90 minutes. This part is worth 50% of your total score.

For this section, you will be given a periodic table of the elements along with a sheet that lists common chemistry formulas, and you may NOT use a calculator. The College Board says that this is because the new scientific calculators not only program and graph but also store information—and they are afraid you’ll use this function to cheat!

On the multiple-choice section, you receive 1 point for a correct answer. There is no penalty for leaving a question blank or getting a question wrong.

PACING

According to the College Board, the multiple-choice section of the AP Chemistry Exam covers more material than any individual student is expected to know. Nobody is expected to get a perfect or even near perfect score. What does that mean to you?

Use the Two-Pass System

Go through the multiple-choice section twice. The first time, do all the questions that you can get answers to immediately. That is, do the questions with little or no math and questions on chemistry topics in which you are well versed. Skip questions on topics that make you uncomfortable. Also, you want to skip the ones that look like number crunchers (even without a calculator, you may still be expected to crunch a few numbers). Circle the questions that you skip in your test booklet so you can find them easily during the second pass. Once you’ve done all the questions that come easily to you, go back and pick out the tough ones that you have the best shot at.

In general, the questions near the end of the section are tougher than the questions near the beginning. You should keep that in mind, but be aware that each person’s experience will be different. If you can do acid-base questions in your sleep, but you’d rather have your teeth drilled than draw a Lewis diagram, you may find questions near the end of the section easier than questions near the beginning.

That’s why the Two-Pass System is so handy. By using it, you make sure you get to see all the questions you can get right, instead of running out of time because you got bogged down on questions you couldn’t do earlier in the test.

Don’t Turn a Question into a Crusade!

Most people don’t run out of time on standardized tests because they work too slowly. Instead, they run out of time because they spend half the test wrestling with two or three particular questions.

You should never spend more than a minute or two on any question. If a question doesn’t involve calculation, then either you know the answer, you can make an educated guess, or you don’t know the answer. Figure out where you stand on a question, make a decision, and move on.

Any question that requires more than two minutes worth of calculations probably isn’t worth doing. Remember, skipping a question early in the section is a good thing if it means that you’ll have time to get two correct answers later on.

GUESSING

You get one point for every correct answer on the multiple-choice section. Guessing randomly neither helps you nor hurts you. Educated guessing, however, will help you.

Use Process of Elimination (POE) to Find Wrong Answers

There is a fundamental weakness to a multiple-choice test. The test makers must show you the right answer, along with three wrong answers. Sometimes seeing the right answer is all you need. Other times you may not know the right answer, but you may be able to identify one or two of the answers that are clearly wrong. Here is where you should use Process of Elimination (POE) to take an educated guess.

Look at this hypothetical question.

You should have no idea what the correct answer is because two of these compounds are made up, but you do know something about the obviously wrong answers. You know that sodium chloride, (C), and carbon dioxide, (D), do not turn water purple. So, using POE, you have a 50 percent chance at guessing the correct answer. Now the odds are in your favor. Now you should guess.

Guess and Move On

Remember that you’re guessing. Pondering the possible differences between brobogdium rabelide and diblythium perjuvenide is a waste of time. Once you’ve taken POE as far as it will go, pick your favorite letter and move on.

The multiple-choice section is the exact opposite of the free-response section. It’s scored by a machine. There’s no partial credit. The computer doesn’t know, or care if you know, why an answer is correct. All the computer cares about is whether you blackened in the correct oval on your score sheet. You get the same number of points for picking (B) because you know that (A) is wrong and that (B) is a nicer letter than (C) or (D) as you would get for picking (B) because you fully understood the subtleties of an electrochemical process.

ABOUT CALCULATORS

You will NOT be allowed to use a calculator on this section. That shouldn’t worry you. All it means is that there won’t be any questions in the section that you’ll need a calculator to solve.

Most of the calculation problems will have fairly user-friendly numbers—that is, numbers with only a couple of significant digits, or things like “11.2 liters of gas at STP” or “160 grams of oxygen” or “a temperature increase from 27°C to 127°C.” Sometimes these user-friendly numbers will actually point you toward the proper steps to take in your calculations.

Don’t be afraid to make rough estimates as you do your calculations. Sometimes knowing that an answer is closer to 50 than to 500 will enable you to pick the correct answer on a multiple-choice test (if the answer choices are far enough apart). Once again, the rule against calculators works in your favor because the College Board will not expect you to do very precise calculations by hand.

REFLECT

Respond to the following questions:

Chapter 2

How to Approach Free-Response Questions

OVERVIEW OF THE FREE-RESPONSE SECTION

Section II is composed of seven free-response questions. You will be given 105 minutes to complete this section, which is worth 50 percent of your total score.

The first three free-response questions are longer, and are worth 10 points each. The College Board recommends you take about 22 minutes per question on these. The last four questions are much shorter, and only worth 4 points each. For these, the College Board recommends you take about 9 minutes each.

The suggested times are just that—suggestions. Your results may vary based on how comfortable you are with the topic being tested within each question. They do give you a good guideline, though. You probably shouldn’t be spending more than 20 minutes on any of the first three questions. Don’t be afraid to cut yourself off and come back to questions later if time allows.

CRACKING THE MATH PROBLEMS

You want to show the graders that you understand the math behind these chemistry problems, so here are some suggestions.

Show Every Step of Your Calculations on Paper

This section is the opposite of multiple-choice. You don’t just get full credit for writing the correct answer. You get most of your points on this section for showing the process that got you to the answer. The graders give you partial credit when you show them that you know what you’re doing. So even if you can do a calculation in your head, you should set it up and show it on the page.

By showing every step, or explaining what you’re doing in words, you ensure that you’ll get all the partial credit possible, even if you screw up a calculation.

Include Units in All Your Calculations

Scientists like units in calculations. Units make scientists feel secure. You’ll get points for including them and you may lose points for leaving them out.

Remember Significant Figures

You can lose 1 point per question if your answer is off by more than one significant figure. Without getting too bent out of shape about it, try to remember that a calculation is only as accurate as the least accurate number in it.

(For a detailed explanation of significant figures, see Chapter 9.)

The Graders Will Follow Your Reasoning, Even if You’ve Made a Mistake

Often, you are asked to use the result of a previous part of a problem in a later part. If you got the wrong answer in part (a) and used it in part (c), you can still get full credit for part (c), as long as your work is correct based on the number that you used. That’s important, because it means that botching the first part of a question doesn’t necessarily sink the whole question.

Remember the Mean!

So let’s say that you could complete only parts (a) and (b) on the required equi-librium problem. That’s 4 or 5 points out of 10, tops. Are you doomed? Of course not. You’re above average. If this test is hard on you, it’s probably just as hard on everybody else. Remember, you don’t need anywhere near a perfect score to get a 5, and you can leave half the test blank and still get a 3!

STRATEGIES FOR CRACKING THE FREE-RESPONSE SECTION

This section is here to test whether you can translate chemistry into English. Most of the questions can be answered in two or three simple sentences, or with a simple diagram or two. Here are some tips for answering the seven free-response questions on Part II.

Show That You Understand the Terms Used in the Question

If you are asked why sodium and potassium have differing first ionization energies, the first thing you should do is tell them what ionization energy is. That’s probably worth the first point of partial credit. Then you should tell them how the differing structures of the atoms make for differing ionization energies. That leads to the next tip.

Take a Step-by-Step Approach

Grading these tests is hard work. Breaking a question into parts in this way makes it easier on the grader, who must match your response to a set of guidelines he or she has been given that describe how to assign partial and full credit.

Each grader scores each test based on these rough guidelines that are established at the beginning of the grading period. For instance, if a grader has 3 points for the question about ionization energies, the points might be distributed the following way:

You can get all three points for this question if the grader thinks that all three concepts are addressed implicitly in your answer, but by taking a step-by-step approach, you improve your chances of explicitly addressing the things that a grader has been instructed to look for. Once again, grading these tests is hard work; graders won’t know for sure if you understand something unless you tell them.

Write Neatly

This simple concept cannot be stressed enough: Write neatly, even if that means working at half-speed. You can’t get points for answers if the graders can’t understand them. Of course, this applies to the rest of the free-response section as well.

Graders Will Follow Your Reasoning, Even If You Make A Mistake

Just like in the multiple-choice section, you might be asked to use the result of a previous part of a problem in a later part. If you decide (incorrectly) that an endothermic reaction in part (a) is exothermic, you can still get full credit in part (c) for your wrong answer about the reaction’s favorability, as long as your answer in (c) is correct based on an exothermic reaction.

REFLECT

Respond to the following questions:

Part V

Content Review for the AP Chemistry Exam

Chapter 3

Big Idea #1: Atoms, Elements, and the Building Blocks of Matter

The chemical elements are fundamental building materials of matter, and all matter can be understood in terms of arrangements of atoms. These atoms retain their identity in chemical reactions.

THE PERIODIC TABLE

The most important tool you will use on this test is the Periodic Table of the Elements.

The Periodic Table gives you very basic but very important information about each element.

The horizontal rows of the periodic table are called periods.

The vertical columns of the periodic table are called groups.

Groups can be numbered in two ways. The old system used Roman numerals to indicate groups. The new system simply numbers the groups from 1 to 18. While it is not important to know the specific group numbers, it is important to know the names of some groups.

Group IA/1 – Alkali Metals

Group IIA/2 – Alkaline Earth Metals

Group B/3-12 – Transition Metals

Group VIIA/17 – Halogens

Group VIIIA/18 – Noble Gases

In addition, the two rows offset beneath the table are alternatively called the lanthanides and actinides, the rare Earth elements, or the inner transition metals.

The identity of an atom is determined by the number of protons contained in its nucleus. The nucleus of an atom also contains neutrons. The mass number of an atom is the sum of its neutrons and protons. Electrons have significantly less mass than protons or neutrons and do not contribute to an element’s mass.

Atoms of an element with different numbers of neutrons are called isotopes; for instance, carbon-12, which contains 6 protons and 6 neutrons, and carbon-14, which contains 6 protons and 8 neutrons, are isotopes of carbon. The molar mass given on the periodic table is the average of the mass numbers of all known isotopes weighted by their percent abundance.

The mass of various isotopes of an element can be determined by a technique called mass spectrometry. A mass spectrum of selenium looks like the following:

As you can see, the most abundant isotope of selenium has a mass of 80, but there are four other naturally occurring isotopes of selenium. The average atomic mass is the weighted average of all five isotopes of selenium shown on this spectra.

The molar mass of an element will give you a pretty good idea of the most common isotope of that element. For instance, the molar mass of carbon is 12.01 and about 99 percent of the carbon in existence is carbon-12.

MOLES

The mole is the most important concept in chemistry, serving as a bridge that connects all the different quantities that you’ll come across in chemical calculations. The coefficients in chemical reactions tell you about the reactants and products in terms of moles, so most of the stoichiometry questions you’ll see on the test will be exercises in converting between moles and grams, liters, molarities, and other units.

Moles and Molecules

Avogadro’s number describes how many atoms are in a single mole of any given element. Much like a dozen is always 12, Avogadro’s number is always 6.02 × 10²³. While it technically can be used to count anything, due to its extremely large value it is usually only used to count extraordinarily small things. Within the confines of this book, it will be used to count atoms, molecules, electrons, or ions, depending on the problem.

Moles and Grams

Moles and grams can be related using the atomic masses given in the periodic table. Atomic masses on the periodic table are given in terms of atomic mass units (amu); however, they also signify how many grams are present in one mole of an element. So, if 1 carbon atom has a mass of 12 amu, then 1 mole of carbon atoms has a mass of 12 grams.

You can use the relationship between amu and g/mol to convert between grams and moles by using the following equation:

Moles and Gases

We’ll talk more about the ideal gas equation in Chapter 4, but for now, you should know that you can use it to calculate the number of moles of a gas if you know some of the gas’s physical properties. All you need to remember at this point is that in the equation PV = nRT, n stands for moles of gas.

The equation above gives the general rule for finding the number of moles of a gas. Many gas problems will take place at STP, or standard temperature and pressure, where P = 1 atmosphere and T = 273 K. At STP, the situation is much simpler and you can convert directly between the volume of a gas and the number of moles. That’s because at STP, one mole of gas always occupies 22.4 liters.

Molarity

Molarity (M) expresses the concentration of a solution in terms of volume. It is the most widely used unit of concentration, turning up in calculations involving equilibrium, acids and bases, and electrochemistry, among others. When you see a chemical symbol in brackets on the test, that means they are talking about molarity. For instance, “[Na⁺]” is the same as “the molar concentration (molarity) of sodium ions.”

Molarity (M) =

Percent Composition

Percent composition is the percent by mass of each element that makes up a compound. It is calculated by dividing the mass of each element or component in a compound by the total molar mass for the substance.

Calculate the percent composition of each element in calcium nitrate, Ca(NO₃)₂.

To do this, you need to first separate each element and count how many atoms are present. Subscripts outside of parentheses apply to all atoms inside of those parentheses.

Calcium: 1

Nitrogen: 2

Oxygen: 6

Then, multiply the number of atoms by the atomic mass of each element.

Ca: 40.08 × 1 = 40.08

N: 14.01 × 2 = 28.02

O: 16.00 × 6 = 96.00

Adding up the masses of the individual elements will give you the atomic mass of that compound. Divide each individual mass by the total molar mass to get your percent composition.

40.08 + 28.02 + 96.00 = 164.10 g/mol

Ca: 40.08/164.10 × 100% = 24.42%

N: 28.02/164.10 × 100% = 17.07%

O: 96.00/164.10 × 100% = 58.50%

You can check your work at the end by making sure your percents add up to 100% (taking rounding into consideration).

24.42% + 17.07% + 58.50% = 99.99%. Close enough!

Empirical and Molecular Formulas

You will also need to know how to determine the empirical and molecular formulas of a compound given masses or mass percents of the components of that compound. Remember that the empirical formula represents the simplest ratio of one element to another in a compound (e.g., CH₂O), while the molecular formula represents the actual formula for the substance (e.g., C₆H₁₂O₆).

Let’s take a look at the following example:

A compound is found to contain 56.5% carbon, 7.11% hydrogen, and 36.4% phosphorus.

a) Determine the empirical formula for the compound.

We start by assuming a 100 gram sample; this allows us to convert those percentages to grams. After we have that done, each element needs to be converted to moles.

C: 56.5g C × = 4.71 mol C

H: 7.11 g Cl × = 7.04 mol H

P: 36.4 g P × = 1.18 mol P

We then divide each mole value by the lowest of the values. In this example, that would be the phosphorus. It is acceptable to round your answers if they are close (within 0.1) to a whole number.

C: = 4

H: = 6

P: = 1

Those values become subscripts, so the empirical formula for the compound is C₄H₆P.

b) If the compound has a molar mass of 170.14 g/mol, what is its molecular formula?

First, we determine the molar mass of the empirical formula.

(12.01 × 4) + (1.01 × 6) + 30.97 = 85.07 g/mol

Then, we divide that mass into the molar mass.

= 2

Finally, multiply all subscripts in the empirical formula by that value. So, the molecular formula is C₈H₁₂P₂.

Electron Configurations and the Periodic Table

The positively-charged nucleus is always pulling at the negatively-charged electrons around it, and the electrons have potential energy that increases with their distance from the nucleus. It works the same way that the gravitational potential energy of a brick on the third floor of a building is greater than the gravitational potential energy of a brick nearer to ground level.

The energy of electrons, however, is quantized. That’s important. It means that electrons can exist only at specific energy levels, separated by specific intervals. It’s kind of like if the brick in the building could be placed only on the first, second, or third floor of the building, but not in-between.

The Aufbau Principle

The Aufbau principle states that when building up the electron configuration of an atom, electrons are placed in orbitals, subshells, and shells in order of increasing energy.

The Pauli Exclusion Principle

The Pauli Exclusion Principle states that the two electrons which share an orbital cannot have the same spin. One electron must spin clockwise, and the other must spin counterclockwise.

Hund’s Rule

Hund’s Rule says that when an electron is added to a subshell, it will always occupy an empty orbital if one is available. Electrons always occupy orbitals singly if possible and pair up only if no empty orbitals are available.

Watch how the 2p subshell fills as we go from boron to neon.

COULOMB’S LAW

The amount of energy that an electron has depends on its distance from the nucleus of the atom. This can be calculated using Coulomb’s Law:

E =

E = energy

 k = Coulomb’s constant

+q = magnitude of the positive charge (nucleus)

−q = magnitude of the negative charge (electron)

r = distance between the charges

While on the exam, you will not be required to mathematically calculate the amount of energy a given electron has, you should be able to qualitatively apply Coulomb’s Law. Essentially, the greater the charge of the nucleus, the more energy an electron will have (as all electrons have the same amount of charge). Coulombic potential energy is considered to be 0 at a distance of infinity. The Coulombic potential energy for a 1s electron is lower (more negative) than that of say, a 3s electron. The amount of energy required to remove a 1s electron, thereby bringing its Coulombic potential energy to zero, will thus be greater than the amount needed to remove a 3s electron. This removal energy is called the binding energy of the electron and is always a positive value.

Quantum Theory

Max Planck figured out that electromagnetic energy is quantized. That is, for a given frequency of radiation (or light), all possible energies are multiples of a certain unit of energy, called a quantum (mathematically, that’s E = hv). So, energy changes do not occur smoothly but rather in small but specific steps.

The Bohr Model

Neils Bohr took the quantum theory and used it to predict that electrons orbit the nucleus at specific, fixed radii, like planets orbiting the Sun.

Each energy level is represented by a row on the periodic table. There are currently seven known energy levels, which correspond with n = 1 to n = 7. The closer an energy level is to an atom, the less energy electrons on that level have. While the Bohr model is not a perfect model of the atom, it serves as an excellent basis to understand atomic structure.

When atoms absorb energy in the form of electromagnetic radiation, electrons jump to higher energy levels. When electrons drop from higher to lower energy levels, atoms give off energy in the form of electromagnetic radiation.

The relationship between the change in energy level of an electron and the electromagnetic radiation absorbed or emitted is given below.

For a particular atom, the energy level changes of the electrons are always the same, so atoms can be identified by their emission and absorption spectra.

The frequency and wavelength of electromagnetic radiation are inversely proportional. Combined with the energy and electromagnetic radiation equation, we can see that higher frequencies and shorter wavelengths lead to more energy.

PHOTOELECTRON SPECTROSCOPY

If an atom is exposed to electromagnetic radiation at an energy level that exceeds the various binding energies of the electrons of that atom, the electrons can be ejected. The amount of energy necessary to do that is called the ionization energy for that electron. For the purposes of this exam, ionization energy and binding energy can be considered to be synonymous terms. When examining the spectra for electrons from a single atom or a small number of atoms, this energy is usually measured in electronvolts, eV (1 eV = 1.60 × 10^–19 Joules). If moles of atoms are studied, the unit for binding energy is usually either kJ/mol or MJ/mol.

All energy of the incoming radiation must be conserved and any of that energy that does not go into breaking the electron free from the nucleus will be converted into kinetic energy (the energy of motion) for the ejected electron. So:

Incoming Radiation Energy = Binding Energy + Kinetic Energy (of the ejected electron)

The faster an ejected electron is going, the more kinetic energy it has. Electrons that were originally further away from the nucleus require less energy to eject, and thus will be moving faster. So, by examining the speed of the ejected electrons, we can determine how far they were from the nucleus of the atom in the first place. Usually, it takes electromagnetic radiation in either the visible or ultraviolet range to cause electron emission, while radiation in the infrared range is often used to study chemical bonds. Radiation in the microwave region is used to study the shape of molecules.

Spectra

If the amount of ionization energy for all electrons ejected from a nucleus is charted, you get what is called a photoelectron spectra (PES) that looks like the following:

The y-axis describes the relative number of electrons that are ejected from a given energy level, and the x-axis shows the binding energy of those electrons. Unlike most graphs, binding energy (ionization energy) decreases going from left to right in a PES. The spectra on the previous page is for sulfur.

Each section of peaks in the PES represents a different energy level. The number of peaks in a section shows us that not all electrons at n = 2+ are located the same distance from the nucleus. With each energy level, there are subshells, which describe the shape of the space the electron can be found in (remember, we are in three-dimensional space here). The Bohr model is limited to two dimensions and does not represent the true positions of electrons due to that reason. Electrons do not orbit the nucleus as planets orbit the Sun. Instead, they are found moving about in a certain area of space (the subshell) a given distance (the energy level) away from the nucleus.

In all energy levels, the first subshell is called the s-subshell and can hold a maximum of two electrons. The second subshell is called the p-subshell and can hold a maximum of six electrons. In the spectra on the previous page, we can see the peak for the p-subshell in energy level 2 is three times higher than that of the s-subshell. The relative height of the peaks helps determine the number of electrons in that subshell.

In the area for the third energy level, the p-subshell peak is only twice as tall as the s-subshell. This indicates there are only four electrons in the p-subshell of this particular atom.

Electron Configuration

Studying the PES of elements allows scientists to understand more about the structure of the atom. In addition to the s and p subshells, two others exist: d (10 electrons max) and f (14 electrons max). The periodic table is designed so that each area is exactly the length of one particular subshell.

As you can see, the first two groups (plus helium) are in what is called the s-block. The groups on the right of the table are in the p-block, while transition metals make up the d-block. The inner transition metals below the table inhabit the f-block. The complete description of the energy level and subshell that each electron on an element inhabits is called its electron configuration.

Example 1: Determine the electron configuration for sulfur.

Sulfur has 16 electrons. The first two go into energy level 1 subshell s; this is represented by 1s². The next two go into energy level 2 subshell s—2s². Six more fill energy level p (2p⁶) then two more go into 3s (3s²) and the final four enter into 3p (3p⁴). So the final configuration is 1s²2s²2p⁶3s²3p⁴.

Using the periodic table as a reference can allow for the determination of any electron configuration. One thing to watch out for is that when entering the d-block (transition metals), the energy level drops by 1. The why behind that isn’t important right now, but you should still be able to apply that rule.

Example 2: Determine the electron configuration for nickel.

1s²2s²2p⁶3s²3p⁶3d⁸4s²

Electron configurations can also be written “shorthand” by replacing parts of them with the symbol for the noble gas at the end of the highest energy level which has been filled.

For example, the shorthand notation for Si is [Ne]3s²3p³ and the shorthand for nickel would be [Ar]4s²3d⁸.

PREDICTING IONIC CHARGES

One of the great rules of chemistry is that the most stable configurations from an energy standpoint are those in which the outermost energy level is full. For anything in the s or p blocks, that means achieving a state in which there are eight electrons in the outermost shell (2 in the s subshell and 6 in the p subshell).

Elements that are close to a full energy level, such as the halogens or those in the oxygen group, tend to gain electrons to achieve a stable configuration. An ion is an atom which has either gained or lost electrons, while the number of protons and neutrons remains constant. Halogens need only gain one electron to achieve a stable configuration, and as such, typically form ions with a charge of negative one. F^–, Cl^–, etc. Any particle with more electrons than protons is called an anion (negatively charged ion). Those elements in the oxygen group need two electrons for stability; thus, they have a charge of negative 2.

On the other end of the table, the alkali metals can most easily achieve a full valence shell by losing a single electron, rather than by gaining seven. So, they will form positively charged ions (cations), which have more protons than electrons. Alkali metals typically have a charge of +1, alkaline Earth metals of +2, and so forth. The table below gives a good overview of the common ionic charges for various groups:

In general, transition metals can form ions of varying charges. All transition metals lose electrons to form cations, but how many electrons they lose will vary depending on the compound they are in. For example, in CuBr₂, we know what the charge on the bromide is negative one, and with two bromides the total negative charge present in the compound is −2. The total compound charge must be zero for the compound to be stable, so it stands to reason that the charge on the single copper ion is +2. That compound would be called copper (II) bromide. However, in CuBr, the charge on the copper would only need to be +1 to balance the single bromide, so that compound would be called copper (I) bromide.

There are a few transition metals that only form ions with one possible charge. Two important examples are zinc, which always forms ions with a charge of +2, and silver, which always forms ions with a charge of +1.

There’s also a slightly different rule as to how transition metals lose electrons when forming those cations. Transition metals, when losing electrons, will lose their higher-level s electrons before losing any of the lower-level d electrons. Thus, while an iron atom has a configuration of [Ar]4s²3d⁶, an iron ion that has lost two electrons (Fe²⁺) would have a configuration of [Ar]3d⁶. If it were to lose further electrons, those would then come from the d-orbital.

Names and Theories

Dalton’s Elements

In the early 1800s, John Dalton presented some basic ideas about atoms that we still use today. He was the first to say that there are many different kinds of atoms, which he called elements. He said that these elements combine to form compounds and that these compounds always contain the same ratios of elements. Water (H₂O), for instance, always has two hydrogen atoms for every oxygen atom. He also said that atoms are never created or destroyed in chemical reactions.

Development of the Periodic Table

In 1869, Dmitri Mendeleev and Lothar Meyer independently proposed arranging the elements into early versions of the periodic table, based on the trends of the known elements.

Thomson’s Experiment

In the late 1800s, J. J. Thomson watched the deflection of charges in a cathode ray tube and put forth the idea that atoms are composed of positive and negative charges. The negative charges were called electrons, and Thomson guessed that they were sprinkled throughout the positively charged atom like chocolate chips sprinkled throughout a blob of cookie dough.

Millikan’s Experiment

Robert Millikan was able to calculate the charge on an electron by examining the behavior of charged oil drops in an electric field.

Rutherford’s Experiment

In the early 1900s, Ernest Rutherford fired alpha particles at gold foil and observed how they were scattered. This experiment led him to conclude that all of the positive charge in an atom was concentrated in the center and that an atom is mostly empty space. This led to the idea that an atom has a positively charged nucleus, which contains most of the atom’s mass, and that the tiny, negatively charged electrons travel around this nucleus.

The Heisenberg Uncertainty Principle

Werner Heisenberg said that it is impossible to know both the position and momentum of an electron at a particular instant. In terms of atomic structure, this means that electron orbitals do not represent specific orbits like those of planets. Instead, an electron orbital is a probability function describing the possibility that an electron will be found in a region of space.

PERIODIC TRENDS

You can make predictions about certain behavior patterns of an atom and its electrons based on the position of the atom in the periodic table. All the periodic trends can be understood in terms of three basic rules.

The atoms in the left-hand side of the periodic table are called metals. Metals give up electrons when forming bonds. Most of the elements in the table are metals. The elements in the upper right-hand portion of the table are called nonmetals. Nonmetals generally gain electrons when forming bonds. The metallic character of the elements decreases as you move from left to right across the periodic table. The elements in the borderline between metal and nonmetal, such as silicon and arsenic, are called metalloids.

Atomic Radius

The atomic radius is the approximate distance from the nucleus of an atom to its valence electrons.

Moving from Left to Right Across a Period (Li to Ne, for Instance), Atomic Radius Decreases

Moving from left to right across a period, protons are added to the nucleus, so the valence electrons are more strongly attracted to the nucleus; this decreases the atomic radius. Electrons are also being added, but they are all in the same shell at about the same distance from the nucleus, so there is not much of a shielding effect.

Moving Down a Group (Li to Cs, for Instance), Atomic Radius Increases

Moving down a group, shells of electrons are added to the nucleus. Each shell shields the more distant shells from the nucleus and the valence electrons get farther away from the nucleus. Protons are also being added, but the shielding effect of the negatively charged electron shells cancels out the added positive charge.

Cations (Positively Charged Ions) Are Smaller than Atoms

Generally, when electrons are removed from an atom to form a cation, the outer shell is lost, making the cation smaller than the atom. Also, when electrons are removed, electron–electron repulsions are reduced, allowing all of the remaining valence electrons to move closer to the nucleus.

Anions (Negatively Charged Ions) Are Larger than Atoms

When an electron is added to an atom, forming an anion, electron–electron repulsions increase, causing the valence electrons to move farther apart, which increases the radius.

Ionization Energy

Electrons are attracted to the nucleus of an atom, so it takes energy to remove an electron. The energy required to remove an electron from an atom is called the first ionization energy. Once an electron has been removed, the atom becomes a positively charged ion. The energy required to remove the next electron from the ion is called the second ionization energy, and so on.

Moving from Left to Right Across a Period, Ionization Energy Increases

Moving from left to right across a period, protons are added to the nucleus, which increases its positive charge. For this reason, the negatively charged valence electrons are more strongly attracted to the nucleus, which increases the energy required to remove them.

Moving Down a Group, Ionization Energy Decreases

Moving down a group, shells of electrons are added to the nucleus. Each inner shell shields the more distant shells from the nucleus, reducing the pull of the nucleus on the valence electrons and making them easier to remove. Protons are also being added, but the shielding effect of the negatively charged electron shells cancels out the added positive charge.

The Second Ionization Energy Is Greater than the First Ionization Energy

When an electron has been removed from an atom, electron–electron repulsion decreases and the remaining valence electrons move closer to the nucleus. This increases the attractive force between the electrons and the nucleus, increasing the ionization energy.

Electronegativity

Electronegativity refers to how strongly the nucleus of an atom attracts the electrons of other atoms in a bond. Electronegativity is affected by two factors. The smaller an atom is, the more effectively its nuclear charge will be felt past its outermost energy level and the higher its electronegativity will be. Second, the closer an element is to having a full energy level, the more likely it is to attract the necessary electrons to complete that level. In general:

The various periodic trends are summarized in the diagram below. The primary exception to these trends are the electronegativity values for the three smallest noble gases. As helium, neon, and argon do not form bonds, they have zero electronegativity. The larger noble gases, however, can form bonds under certain conditions and do follow the general trends as outlined in this section. This will be discussed in more detail in the next chapter.

CHAPTER 3 QUESTIONS

Multiple-Choice Questions

Click here to download the PDF version.

Use the PES spectra below to answer questions 1-4.

Use the following information to answer questions 18-20.

Free-Response Questions

Carbon, Oxygen, Phosphorus, and Chlorine

	Atomic radius (pm)	First Ionization Energy (kJ/mol–1)
Element 1	170	1086.5
Element 2	180	1011.8
Element 3	175	1251.2
Element 4	152	1313.9

CHAPTER 3 ANSWERS AND EXPLANATIONS

Multiple-Choice

Free-Response

Chapter 4

Big Idea #2: Bonding and Phases

Chemical and physical properties of materials can be explained by the structure and the arrangement of atoms, ions, or molecules and the forces between them.

BONDS OVERVIEW

Atoms engage in chemical reactions in order to reach a more stable, lower-energy state. This requires the transfer or sharing of electrons, a process that is called bonding. Atoms of elements are usually at their most stable when they have eight electrons in their valence shells. As a result, atoms with too many or too few electrons in their valence shells will find one another and pass the electrons around until all the atoms in the molecule have stable outer shells. Sometimes an atom will give up electrons completely to another atom, forming an ionic bond. Sometimes atoms share electrons, forming covalent bonds.

IONIC BONDS

An ionic solid is held together by the electrostatic attractions between ions that are next to one another in a lattice structure. They often occur between metals and nonmetals. In an ionic bond, electrons are not shared. Instead, the cation gives up an electron (or electrons) to the anion.

The two ions in an ionic bond are held together by electrostatic forces. In the diagram below, a sodium atom has given up its single valence electron to a chlorine atom, which has seven valence electrons and uses the electron to complete its outer shell (with eight). The two atoms are then held together by the positive and negative charges on the ions.

The electrostatic attractions that hold together the ions in the NaCl lattice are very strong, and any substance held together by ionic bonds will usually be a solid at room temperature and have very high melting and boiling points.

Two factors affect the melting points of ionic substances. The primary factor is the charge on the ions. According to Coulomb’s Law, a greater charge leads to a greater bond energy (often called lattice energy in ionic bonds), so a compound composed of ions with charges of +2 and –2 (such as MgO) will have a higher melting point than a compound composed of ions with charges of +1 and –1 (such as NaCl). If both compounds are made up of ions with equal charges, then the size of the ions must be considered. Smaller ions will have greater Coulombic attraction (remember, size is inversely proportional to bond energy), so a substance like LiF would have a greater melting point than KBr.

In an ionic solid, each electron is localized around a particular atom, so electrons do not move around the lattice; this makes ionic solids poor conductors of electricity. Ionic liquids, however, do conduct electricity because the ions themselves are free to move about in the liquid phase, although the electrons are still localized around particular atoms. Salts are held together by ionic bonds.

METALLIC BONDS

When examining metals, the sea of electrons model can be used. The positively-charged core of a metal, consisting of its nucleus and core electrons, is generally stationary, while the valence electrons on each atom do not belong to a specific atom and are very mobile. These mobile electrons explain why metals are such good conductors of electricity. The delocalized structure of a metal also explains why metals are both malleable and ductile, as deforming the metal does not change the environment immediately surrounding the metal cores.

Metals can also bond with each other to form alloys. This typically occurs when two metals are melted into their liquid phases, and are then poured together before cooling and creating the alloy. In an interstitial alloy, metal atoms with two vastly different radii combine. Steel is one such example—the much smaller carbon atoms occupy the interstices of the iron atoms. A substitutional alloy forms between atoms of similar radii. Brass is a good example, atoms of zinc are substituted for some copper atoms to create the alloy.

COVALENT BONDS

In a covalent bond, two atoms share electrons. Each atom counts the shared electrons as part of its valence shell. In this way, both atoms achieve complete outer shells.

In the diagram below, two fluorine atoms, each of which has seven valence electrons and needs one electron to complete its valence shell, form a covalent bond. Each atom donates an electron to the bond, which is considered to be part of the valence shell of both atoms.

The number of covalent bonds an atom can form is the same as the number of unpaired electrons in its valence shell.

The first covalent bond formed between two atoms is called a sigma (σ) bond. All single bonds are sigma bonds. If additional bonds between the two atoms are formed, they are called pi (π) bonds. The second bond in a double bond is a pi bond and the second and third bonds in a triple bond are also pi bonds. Double and triple bonds are stronger and shorter than single bonds, but they are not twice or triple the strength.

Network (Covalent) Bonds

In a network solid, atoms are held together in a lattice of covalent bonds. You can visualize a network solid as one big molecule. Network solids are very hard and have very high melting and boiling points.

The electrons in a network solid are localized in covalent bonds between particular atoms, so they are not free to move about the lattice. This makes network solids poor conductors of electricity.

The most commonly seen network solids are compounds of carbon (such as diamond or graphite) and silicon (SiO₂—quartz). This is because both carbon and silicon have four valence electrons, meaning they are able to form a large number of covalent bonds.

Silicon also serves as a semiconductor when it is doped with other elements. Doping is a process in which an impurity is added to an existing lattice. In a normal silicon lattice, each individual silicon atom is bonded to four other silicon atoms. When some silicon atoms are replaced with elements that have only three valence electrons (such as boron or aluminum), the neighboring silicon atoms will lack one bond apiece.

This missing bond (or “hole”) creates a positive charge in the lattice, and the hole attracts other electrons to it, increasing conductivity. Those electrons leave behind holes when they move, creating a chain reaction in which the conductivity of the silicon increases. This type of doping is called p-doping for the positively charged holes.

If an element with five valence electrons (such as phosphorus or arsenic) is used to add impurities to a silicon lattice, there is an extra valence electron that is free to move around the lattice, causing an overall negative charge that increases the conductivity of the silicon. This type of doping is called n-doping due to the free-moving negatively charged electrons.

Polarity

In the F₂ molecule shown on this page, the two fluorine atoms share the electrons equally, but that’s not usually the case in molecules. Usually, one of the atoms (the more electronegative one) will exert a stronger pull on the electrons in the bond—not enough to make the bond ionic, but enough to keep the electrons on one side of the molecule more than on the other side. This gives the molecule a dipole. That is, the side of the molecule where the electrons spend more time will be negative and the side of the molecule where the electrons spend less time will be positive.

In the water molecule above, oxygen has a higher electronegativity than hydrogen and thus will have the electrons closer to it more often. This gives the oxygen a negative dipole and each hydrogen a positive dipole.

Dipole Moment

The polarity of a molecule is measured by the dipole moment. The more polar the molecule is, the larger the dipole moment is. You will not need to calculate the strength of a dipole, but you should be familiar with the unit with which that strength is quantified. That unit is called the debye (D).

INTERMOLECULAR FORCES

Intermolecular forces (IMFs) are the forces that exist between molecules in a covalently bonded substance. These forces are what need to be broken apart in order for covalent substances to change phases. Note that when ionic substances change phase, bonds between the individual ions are actually broken. When covalent substances change phase, the bonds between the individual atoms remain in place, it is just the forces that hold the molecules to other molecules that break apart.

Dipole–Dipole Forces

Dipole–dipole forces occur between polar molecules: The positive end of one polar molecule is attracted to the negative end of another polar molecule.

Molecules with greater polarity will have greater dipole–dipole attraction, so molecules with larger dipole moments tend to have higher melting and boiling points. Dipole–dipole attractions are relatively weak, however, and these substances melt and boil at very low temperatures. Most substances held together by dipole–dipole attraction are gases or liquids at room temperature.

Hydrogen Bonds

Hydrogen bonds are a special type of dipole–dipole attraction. In a hydrogen bond, the positively charged hydrogen end of a molecule is attracted to the negatively charged end of another molecule containing an extremely electronegative element (fluorine, oxygen, or nitrogen—F, O, N).

Hydrogen bonds are much stronger than normal dipole–dipole forces because when a hydrogen atom gives up its lone electron to a bond, its positively charged nucleus is left virtually unshielded. Substances that have hydrogen bonds, such as water and ammonia, have higher melting and boiling points than substances that are held together only by other types of intermolecular forces.

Water is less dense as a solid than as a liquid because its hydrogen bonds force the molecules in ice to form a crystal structure, which keeps them farther apart than they are in the liquid form.

London Dispersion Forces

London dispersion forces occur between all molecules. These very weak attractions occur because of the random motions of electrons on atoms within molecules. At a given moment, a nonpolar molecule might have more electrons on one side than on the other, giving it an instantaneous polarity. For that fleeting instant, the molecule will act as a very weak dipole.

Since London dispersion forces depend on the random motions of electrons, molecules with more electrons will experience greater London dispersion forces. So among substances that experience only London dispersion forces, the one with more electrons will generally have higher melting and boiling points. London dispersion forces are even weaker than dipole–dipole forces, so substances that experience only London dispersion forces melt and boil at extremely low temperatures and tend to be gases at room temperature.

As molecules gain more electrons, the London dispersion forces between them start to become much more significant. Comparing the boiling point of a nonpolar substance with a large number of electrons vs. a polar substance with less electrons is difficult, and there is no simple rule to follow. For instance, water has hydrogen bonds and a boiling point of 100°C. Butane (C₄H₈) and octane (C₈H₁₈) are both completely nonpolar molecules, and while butane’s boiling point is 34°C, octane’s is 125°C. Even though octane has no permanent dipoles, it has so many electrons that its London dispersion forces are significant enough that they create greater intermolecular attractions than even the hydrogen bonds in water.

The role of London dispersion forces is often determined by comparing the molar mass of molecules. However, it is not the mass itself which affects the strength of the IMFs. Rather, it is simply that as mass (based on protons and neutrons) increases, so too do the number of electrons, as the molecule must remain electrically neutral.

Bond Strength

Ionic substances are generally solids at room temperature, and turning them into liquids (melting them), requires the bonds holding the lattice together to be broken. The amount of energy needed for that is based on the Coulombic attraction between the molecules.

Covalent substances, which are liquid at room temperature, will boil when the intermolecular forces between them are broken. For molecules with similar sizes, the following IMF ranking can help you determine the relative strength of the IMFs within the molecules.

The melting and boiling points of covalent substances are almost always lower than the melting and boiling point of ionic ones.

Metallic bonding, which often only involves one type of atom, tends to be very strong and thus metals (particularly the transition metals) tend to have high melting points. Network covalent bonding is the strongest type of bonding there is, and it is very difficult to cause substances that exhibit network covalent bonding to melt.

Bonding and Phases

The phase of a substance is directly related to the strength of its intermolecular forces. Solids have highly ordered structures where the atoms are packed tightly together, while gases have atoms spread so far apart that most of the volume is free space.

In other words, substances that exhibit weak intermolecular forces (such as London Dispersion forces) tend to be gases at room temperature. Nitrogen (N₂) is an example of this. Substances that exhibit strong intermolecular forces (such as hydrogen bonds) tend to be liquids at room temperature. A good example is water.

Ionic substances do not experience intermolecular forces. Instead, their phase is determined by the ionic bond holding the ions together in the lattice. Because ionic bonds are generally significantly stronger than intermolecular forces in covalent molecules, ionic substances are usually solid at room temperatures.

VAPOR PRESSURE

Beyond helping to determine the melting point and boiling point of covalent substances, the relative strength of the intermolecular forces in a substance can also predict several other properties of that substance. The most important of these is vapor pressure. Vapor pressure arises from the fact that the molecules inside a liquid are in constant motion. If those molecules hit the surface of the liquid with enough kinetic energy, they can escape the intermolecular forces holding them to the other molecules and transition into the gas phase.

This process is called vaporization. It is not to be confused with a liquid boiling. When a liquid boils, energy (in the form of heat) is added, increasing the kinetic energy of all of the molecules in the liquid until all of the intermolecular forces are broken. For vaporization to occur, no outside energy needs to be added. Note that there is a direct relationship between temperature and vapor pressure. The higher the temperature of a liquid, the faster the molecules are moving and the more likely they are to break free of the other molecules. So, temperature and vapor pressure are directly proportional.

If two liquids are at the same temperature, the vapor pressure is dependent primarily on the strength of the intermolecular forces within that liquid. The stronger those intermolecular forces are, the less likely it is that molecules will be able to escape the liquid, and the lower the vapor pressure for that liquid will be.

LEWIS DOT STRUCTURES

Drawing Lewis Dot Structures

At some point on the test, you’ll be asked to draw the Lewis structure for a molecule or polyatomic ion. Here’s how to do it.

Let’s find the Lewis dot structure for the ion.

Resonance Forms

When we put a double bond into the ion, we place it on any one of the oxygen atoms, as shown below.

All three resonance forms are considered to exist simultaneously, and the strength and lengths of all three bonds are the same: somewhere between the strength and length of a single bond and a double bond.

To determine the relative length and strength of a bond in a resonance structure, a bond order calculation can be used. A single bond has a bond order of 1, and a double bond has an order of 2. When resonance occurs, pick one of the bonds in the resonance structure and add up the total bond order across the resonance forms, then divide that sum by the number of resonance forms.

For example, in the carbonate ion above, the top C–O bond would have a bond order of , or 1.33. Bond order can be used to compare the length and strength of resonance bonds with pure bonds as well as other resonance bonds.

Incomplete Octets

Some atoms are stable with less than eight electrons in their outer shell. Hydrogen only requires two electrons, as does helium (although helium never forms bonds). Boron is considered to be stable with six electrons, as in the BF₃ diagram below. All other atoms involved in covalent bonding require a minimum of eight electrons to be considered stable.

Expanded Octets

In molecules that have d subshells available, the central atom can have more than eight valence electrons, but never more than twelve. This means any atom of an element from n = 3 or greater can have expanded octets, but NEVER elements in n = 2 (C, N, O, etc.). Expanded octets also explains why some noble gases can actually form bonds; the extra electrons go into the empty d-orbital.

Here are some examples.

Formal Charge

Sometimes, there is more than one valid Lewis structure for a molecule. Take CO₂; it has two valid structures as shown below. To determine the more likely structure, formal charge is used. To calculate the formal charge on atoms in a molecule, take the number of valence electrons for that atom and subtract the number of assigned electrons in the Lewis structure. When counting assigned electrons, lone pairs count as two and bonds count as one.

The total formal charge for a neutral molecule should be zero, which it is on both diagrams. Additionally, the fewer number of atoms there are with an actual formal charge, the more likely the structure will be—so the left structure is the more likely one for CO₂. For polyatomic ions, the sum of the formal charges on each atom should equal the overall charge on the ion.

Molecular Geometry

Electrons repel one another, so when atoms come together to form a molecule, the molecule will assume the shape that keeps its different electron pairs as far apart as possible. When we predict the geometries of molecules using this idea, we are using the valence shell electron-pair repulsion (VSEPR) model.

In a molecule with more than two atoms, the shape of the molecule is determined by the number of electron pairs on the central atom. The central atom forms hybrid orbitals, each of which has a standard shape. Variations on the standard shape occur depending on the number of bonding pairs and lone pairs of electrons on the central atom.

The following pages show the different hybridizations and geometries that you might see on the test.

Number of lone pairs	Geometry	Examples
0	B—A—B	BeCl2
	linear	CO2

KINETIC MOLECULAR THEORY

For ideal gases, the following assumptions can be made:

MAXWELL-BOLTZMANN DIAGRAMS

A Maxwell-Boltzmann diagram shows the range of velocities for molecules of a gas. Molecules at a given temperature are not all moving at the same velocity. When determining the temperature, we take the average velocity of all the molecules and use that in the relevant equation to calculate temperature. You do not need to know that equation (unless you are taking AP Physics!). All you need to know here is that temperature is directly proportional to kinetic energy.

The first type of Maxwell-Boltzmann diagram involves plotting the velocity distributions for the molecules of one particular gas at multiple temperatures. In the diagram below, there are three curves representing a sample of nitrogen gas at 100 K, 300 K, and 500 K.

As you can see, the higher the temperature of the gas, the larger the range is for the velocities of the individual molecules. Gases at higher temperatures have greater kinetic energy (KE), and as all the molecules in this example have the same mass, the increased KE is due to the increased velocity of the gas molecules.

Maxwell-Boltzmann diagrams are also used to show a number of different gases at the same temperature. The diagram on the next page shows helium, argon, and xenon gas, all at 300 K:

In this case, all of the gases have the same amount of total kinetic energy because they have identical temperatures. However, not all of the atoms have the same mass. If the atoms have smaller masses, they must have greater velocities in order to have a kinetic energy identical to that of atoms with greater mass. Because helium atoms have the least mass, they have the highest average velocity. Xenon atoms, which have a much greater mass, have a correspondingly lower velocity.

EFFUSION

Effusion is the rate at which a gas will escape from a container through microscopic holes in the surface of the container. For instance, even though the rubber or latex that makes up a balloon may seem solid, after the balloon is filled with a gas, it will gradually shrink over time. This is due to the fact that there are tiny holes in the surface of the balloon, through which the even tinier gas molecules can escape.

The rate at which a gas effuses from a container is dependent on the speed of the gas particles. The faster the particles are moving, the more often they hit the sides of the container, and the more likely they are to hit a hole and escape. The rate of effusion thus increases with temperature, but also, if examining gases at the same temperature, the gas with the lower molar mass will effuse first.

It is likely you have experienced this; a balloon filled with helium will deflate more rapidly than one filled with air (which is composed primarily of nitrogen and oxygen) or one filled with carbon dioxide. There is a formula that quantifies the rate at which a gas will effuse, but that is beyond the scope of the exam. As long as you understand the basic principles behind effusion, you should be able to answer any questions on this topic that may come up.

THE IDEAL GAS EQUATION

You can use the ideal gas equation to calculate any of the four variables relating to the gas, provided that you already know the other three.

You can also manipulate the ideal gas equation to figure out how changes in each of its variables affect the other variables. The following equation, often called the Combined Gas Law, can only be used when the number of moles is held constant.

You should be comfortable with the following simple relationships:

DALTON’S LAW

Dalton’s Law states that the total pressure of a mixture of gases is just the sum of all the partial pressures of the individual gases in the mixture.

You should also note that the partial pressure of a gas is directly proportional to the number of moles of that gas present in the mixture. So if 25 percent of the gas in a mixture is helium, then the partial pressure due to helium will be 25 percent of the total pressure. This concept, often represented by the variable x, is called the mole fraction.

DEVIATIONS FROM IDEAL BEHAVIOR

At low temperature and/or high pressure, gases behave in a less-than-ideal manner. That’s because the assumptions made in kinetic molecular theory become invalid under conditions where gas molecules are packed too tightly together.

Two things happen when gas molecules are packed too tightly.

Even under normal conditions, gases will still show some deviation from ideal behavior due to the IMFs between the various gas molecules. These interactions are generally minimal, but when considering the likelihood that a gas will deviate from ideal behavior, stronger IMFs will lead to more deviations. H₂O, with hydrogen bonding, would be significantly more likely to deviate from ideal behavior than CH₄, which only has LDFs.

If you are looking at gases that have similar IMF types, remember that the more electrons a gas has the more polarizable it is, and is more likely to deviate from ideal behavior. For instance, within the noble gases, which have only LDFs, argon is more likely to deviate from ideal behavior than helium, but less likely to deviate than xenon.

DENSITY

You may be asked about the density of a gas. The density of a gas is measured in the same way as the density of a liquid or solid: in mass per unit of volume.

The density of any gas sample can also be determined by combining the density equation with the Ideal Gas Law.

If D = , then V = .

Substituting that into the Ideal Gas Law:

= nRT

A little rearrangement yields:

D =

The term (m/n) describes mass per mole, which is how molar mass (MM) is measured. Thus:

D =

If you are given the density of a gas and need to find the molar mass, this can also be rewritten as:

MM =

Solutes and Solvents

There is a basic rule for remembering which solutes will dissolve in which solvents.

That means that polar or ionic solutes (such as salt) will dissolve in polar solvents (such as water). That also means that nonpolar solutes (such as oils) are best dissolved in nonpolar solvents. When an ionic substance dissolves, it breaks up into ions. That’s dissociation. Free ions in a solution are called electrolytes because they can conduct electricity.

The more ions that are present in an ionic compound, the greater the conductivity of that compound will be when those ions are dissociated. For instance, a solution of magnesium chloride will dissociate into three ions (one Mg²⁺ and two Cl^–). A solution of sodium chloride will dissociate into just two ions (Na⁺ and Cl^–). Thus, a solution of magnesium chloride will conduct electricity better than a solution of sodium chloride if both solutions have identical concentrations.

SOLUTION SEPARATION

You can use intermolecular forces and the various Coulombic attractions that occur between ions and polar molecules in order to help separate various substances out from each other. There are several ways to do this.

Paper Chromatography

Chromatography is the separation of a mixture by passing it in solution through a medium in which the components of the solution move at different rates. There are several major types of chromatography. The first is paper chromatography, in which paper is the medium through which the solution passes.

Many chemical solutions, such as the ink found in most pens, are a mixture of a number of covalent substances. Each of these substances has its own polarity value, and thus has a different affinity depending on the solvent. One of the most common paper chromatography experiments involves the separation of pigments in black ink. Black ink is usually made up of substances of several different colors, which when combined create black.

In paper chromatography, a piece of filter paper is suspended above a solvent so that the very bottom of the paper is touching the solvent. The ink in question is dotted onto a line at the bottom of the filter paper that starts out just above the solvent level. As the solvent climbs the paper, the various substances inside the ink will be attracted to the polar water molecules. The more polar the substance is, the more it will be attracted to the water molecules, and the further it will travel. You might end up with something that looks like this:

Looking at that strip, you can conclude that the ink was made of three different substances. The one that traveled the farthest with the water (the red pigment) experienced the strongest attractions and was the most polar, whereas the one that didn’t travel very far from the original starting line (the blue pigment) was the least polar. Paper chromatography is the most useful with colored substances, which is why ink is used in the above example. If there were components to the ink that had no visible color, you would not be able to see them on the filter paper, and that is one major limitation of paper chromatography.

The distance the ink travels along the paper is measured via the retention (or retardation) factor, also known as the R_f value. The R_f value is calculated as such:

The stronger the attraction between the solute and the solvent front is, the larger the R_f value will be. In the diagram above, the red pigment would have the highest R_f value.

Water is not the only solvent that can be used in polar chromatography. There are many nonpolar solvents (such as cyclohexane) that can be used instead. In the case of a nonpolar solvent, the position of the various ink components in the above diagram would have been reversed—the most nonpolar substance would travel the furthest, and the most polar substance would travel the least.

Column Chromatography

Another type of chromatography is column chromatography. In this process, a column is packed with a stationary substance. Then, the solution to be separated (the analyte) is injected into the column, where it adheres to the stationary phase. After that, another solution (called the eluent) is injected into the column. As the eluent passes through the stationary phase, the analyte molecules will be attracted to it with varying degrees of strength depending on their polarity. The more attracted certain analyte molecules are to the eluent, the faster they will travel through (“elute”) and leave the column.

The speed at which the substances move through the column can be monitored, and if there is a sufficient polarity difference between the components, they will leave the column at different times, allowing them to be separated. Generally, after collecting the eluted mixture, it can be analyzed for compositional analysis via a variety of methods.

In column chromatography, either liquids or gases can be used as the eluent, depending on the situation.

Distillation

A third method for separating solutions is distillation. Distillation takes advantage of the different boiling points of substances in order to separate them. For instance, if you have a mixture of water (BP: 100°C) and ethanol (BP: 78°C) and then heat that mixture to 85°C, the ethanol will boil but the water will not.

A condenser is a piece of glassware that consists of a smaller tube running through a larger tube. The larger tube has hose connections on it, allowing for water to be run through it. This effectively cools the inner tube and not the outer tube. At that point, the ethanol vapor, when run through the inner tube, will cool and condense back into a liquid form, which can be collected on the other side of the condenser.

A major advantage to distillation is that the solutions need not be colored at all to separate them. Keeping the flask at a constant temperature can be a challenge, which is why the temperature must be monitored closely to ensure that you are only boiling one component of the mixture at a time. The biggest disadvantage is that it cannot be used to separate a mixture that contains substances with unknown boiling points.

CHAPTER 4 QUESTIONS

Multiple-Choice Questions

Click here to download the PDF version.

Use the following information to answer questions 4-7.

Use the following information to answer questions 11-14.

2SO₂(g) + O₂(g) → 2SO₃(g)

Use the following Lewis diagrams to answer questions 22-24.

Use the following information to answer questions 25-28.

Use the following information to answer questions 29-31.

Free-Response Questions

CHAPTER 4 ANSWERS AND EXPLANATIONS

Multiple-Choice

Free-Response

Chapter 5

Big Idea #3: Chemical Reactions, Energy Changes, and Redox Reactions

Changes in matter involve the rearrangement and/or reorganization of atoms and/or the transfer of electrons.

TYPES OF REACTIONS

Reactions may be classified into several categories.

Finally, this can be represented using particulate diagrams, as shown below:

You may have heard that you need to memorize solubility rules—that is, what ions will form insoluble precipitates when you combine them. You DO NOT need to memorize most of these; for the most part, the AP Exam will provide you with them as needed. The only ones you need to know are:

That’s it! You should know how to interpret any solubility rules that you are given on the test, but you need not memorize anything beyond those two.

Ionic substances that dissolve in water do so because of the attraction of the ions to the dipoles of the water molecules exceeds the attraction of the ions to each other. Let’s take NaBr for example:

When it dissolves in water, the positive Na⁺ cations are attracted to the negative (oxygen) ends of the water molecules. The negative Br^– anions are attracted to the positive (hydrogen) ends of the water molecules. Thus, when the substance dissolves it looks like this on the particulate level:

CHEMICAL EQUATIONS

Balancing Chemical Equations

Normally, balancing a chemical equation is a trial-and-error process. You start with the most complicated-looking compound in the equation and work from there. There is, however, an old Princeton Review SAT trick that you may want to try if you see a balancing equation question on the multiple-choice section. The trick is called backsolving.

It works like this: To make a balancing equation question work in a multiple-choice format, one of the answer choices is the correct coefficient for one of the species in the reaction. So instead of starting blind in the trial-and-error process, you can insert the answer choices one by one to see which one works. You probably won’t have to try all four, and if you start in the middle and the number doesn’t work, it might be obviously too small or large, eliminating other choices before you have to try them. Let’s try it.

...NH₃ + ...O₂ → ...N₂ + ...H₂O

Here’s How to Crack It!

Start at (C) because it’s a middle number. If there are 3 NH₃’s, then there can’t be a whole number coefficient for N₂, so (C) is wrong, and so is the other odd number answer, (A).

Try (D).

If there are 4 NH₃’s, then there must be 2 N₂’s and 6 H₂O’s.

If there are 6 H₂O’s, then there must be 3 O₂’s, and the equation is balanced with lowest whole number coefficients.

Chemical Equations and Calculations

Because of the math involved, stoichiometry like this will most likely appear only as a free-response question. In those instances, expect the following: You will be given a balanced chemical equation and told that you have some number of grams (or liters of gas, or molar concentration, and so on) of reactant. Then you will be asked what number of grams (or liters of gas, or molar concentration, and so on) of products are generated.

Let’s try one.

2HBr(aq) + Zn(s) → ZnBr₂(aq) + H₂(g)

Here’s How to Crack It!

Let’s try another one using the same reaction.

2HBr(aq) + Zn(s) → ZnBr₂(aq) + H₂(g)

Here’s How to Crack It!

When you perform calculations, always include units. Including units in your calculations will help you (and the person scoring your test) keep track of what you are doing. While using the proper units in an answer is not typically worth any points in and of itself (with a few exceptions), AP graders may take off points if your units are missing or incorrect.

Combustion Analysis

When a hydrocarbon is combusted, producing carbon dioxide and water, the formula of that hydrocarbon can often be determined using a little math. One of the most important concepts in chemistry is the Law of Conservation of Mass. This states that matter can be neither created nor destroyed during a chemical reaction. Thus, when a hydrocarbon is combusted, all of the carbon in that hydrocarbon will end up in CO₂, and all of the hydrogen will end up in H₂O. With that in mind, you can often determine the empirical formula of a hydrocarbon by determining the mass of the products it forms.

Here’s How to Crack It!

If 44.0 grams of CO₂ are created, that is 1.00 mole of CO₂, and thus, 1.00 mole of C as well (as there is one mole of carbon in each mole of CO₂). Through conservation of mass, this means there was 1.00 mole of C in the original compound.

27.0 g of H₂O means 1.5 moles of H₂O were created. Because there are two hydrogen atoms in every H₂O, that means there were 3.0 moles of hydrogen in the water which was created, and thus, 3.00 moles of H in the original compound. The C:H ratio in the original compound must be 1:3, which is consistent with (B).

GRAVIMETRIC ANALYSIS

A common way to use precipitation reactions in order to make quantitative determinations about the identity of an unknown sample is through gravimetric analysis. In this process, a soluble sample of an unknown compound is mixed with another solution that will cause one of the ions from the unknown sample to fully precipitate out. If the identity of the precipitate is known, stoichiometric conversions will reveal the mass of the ion in the unknown, which allows for a mass percent calculation.

Due to the large amount of math involved, it’s easiest to demonstrate.

Example:

A 4.33 g sample of an unknown alkali hydroxide compound is dissolved completely in water. A sufficient solution of copper (II) nitrate is added to the hydroxide solution such that it will fully precipitate copper (II) hydroxide via the following reaction:

Cu²⁺ + 2OH^– → Cu(OH)₂(s)

After the precipitate is filtered and dried, its mass is found to be 3.81 g. Is the original alkali hydroxide sample most likely LiOH, NaOH, or KOH?

First, you convert the mass of the precipitate to moles:

3.81g Cu(OH)₂ × = 0.0391 mol Cu(OH)₂

Next, you need to determine the moles of hydroxide in the precipitate and convert that to grams.

0.0391 mol Cu(OH)₂ × = 1.33 g OH⁻

If there are 1.33 g of OH^– present in the precipitate, then there were also 1.33 g of OH^– in the original sample. The final calculation is to determine the mass percent of the OH^– in the original sample:

× 100% = 30.7%

To determine the identity of the sample, you have to know the mass percent of hydroxide in all of the possible compounds. To do that, you divide the mass of one mole of hydroxide by the molar mass of the compound.

× 100% = 71.0%

× 100% = 42.5%

× 100% = 30.3%

The hydroxide mass percent in KOH is closest to the mass percent determined via gravimetric analysis, so that is the most likely identity of the original unknown.

ENTHALPY

Enthalpy Change, ∆H

The enthalpy of a substance is a measure of the energy that is released or absorbed by the substance when bonds are broken and formed during a reaction.

The change in enthalpy, ∆H, that takes place over the course of a reaction can be calculated by subtracting the enthalpy of the reactants from the enthalpy of the products.

If more energy is released when the bonds in the products form than is necessary to break the bonds in the reactants, the overall reaction releases energy into the surroundings and the reaction is called exothermic. Exothermic reactions have a negative enthalpy change.

If more energy is required to break the bonds in the reactants than is released when the bonds in the products form, the overall reaction requires energy from the surroundings and is called endothermic. Endothermic reactions have a positive enthalpy change.

ENERGY DIAGRAMS

Exothermic and Endothermic Reactions

EXOTHERMIC REACTION

The diagram above shows the energy change that takes place during an exothermic reaction. The reactants start with a certain amount of energy (read the graph from left to right). For the reaction to proceed, the reactants must have enough energy to reach the transition state, where they are part of an activated complex. This is the highest point on the graph above. The amount of energy needed to reach this point is called the activation energy, E_a. At this point, all reactant bonds have been broken, but no product bonds have been formed, so this is the point in the reaction with the highest energy and lowest stability. The energy needed for the reverse reaction is shown as line E_a′.

Moving to the right past the activated complex, product bonds start to form, and we eventually reach the energy level of the products.

This diagram represents an exothermic reaction, so the products are at a lower energy level than the reactants and ∆H is negative.

The diagram below shows an endothermic reaction.

ENDOTHERMIC REACTION

In this diagram, the energy of the products is greater than the energy of the reactants, so ∆H is positive.

Reaction diagrams can be read in both directions, so the reverse reaction for an exothermic reaction is endothermic and vice versa.

CATALYSTS AND ENERGY DIAGRAMS

A catalyst speeds up a reaction by providing the reactants with an alternate pathway that has a lower activation energy, as shown in the diagram above.

Notice that the only difference between the catalyzed reaction and the uncatalyzed reaction is that the energy of the activated complex is lower for the catalyzed reaction. A catalyst lowers the activation energy, but it has no effect on the energy of the reactants, the energy of the products, or ∆H for the reaction.

Also note that a catalyst lowers the activation energy for both the forward and the reverse reaction, so it has no effect on the equilibrium conditions.

OXIDATION STATES

In many chemical reactions, electrons are transferred between the reactants. In order to determine which reactants are gaining electrons and which are losing electrons, we use something called an oxidation state. There are several rules when it comes to assigning oxidation states.

Rule 6 is only applied for covalent compounds to determine the oxidation state on an element which does not fall under the first five rules. Not that many nonmetals with low electronegativity values (such as carbon, nitrogen, sulfur, and phosphorus) have frequently varying oxidation states.

Example 1: Determine the oxidation state on every element in H₂SO₄.

We know that H is +1 and O is −2, and that the total charge on the compound must be 0. Thus:

+1(2) + S + −2(4) = 0.

Solve for S, and you will find S = +6.

Example 2: Determine the oxidation state on every element in the phosphate ion, PO₄³⁻.

P + −2(4) = −3

P = +5

OXIDATION-REDUCTION REACTIONS

In an oxidation-reduction (or redox, for short) reaction, electrons are exchanged by the reactants, and the oxidation states of some of the reactants are changed over the course of the reaction. Look at the following reaction:

Fe + 2HCl → FeCl₂ + H₂

The oxidation state of Fe changes from 0 to +2.

The oxidation state of H changes from +1 to 0.

In the reaction above, H was reduced.

In the reaction above, Fe was oxidized.

Oxidation and reduction go hand in hand. If one atom is losing electrons, another atom must be gaining them.

An oxidation-reduction reaction can be written as two half-reactions: one for the reduction and one for the oxidation. For example, the reaction

Fe + 2HCl → FeCl₂ + H₂

can be written as

Fe → Fe²⁺ + 2e⁻ Oxidation
2H⁺ + 2e⁻ → H₂  Reduction

Redox Titrations

A titration involves the slow addition of a solution at a known concentration to another solution of unknown concentration in order to determine the concentration of the unknown solution. To determine the endpoint of a titration, a color change is often used. Titrations are frequently used in acid-base reactions (more on that in Chapter 8), but redox reactions can also be used in titration situations.

Potassium permanganate, KMnO₄, is frequently used in redox titrations. The manganese ion has an oxidation state of +7 in the permanganate ion (MnO₄^–), and a solution containing permanganate ions is a deep purple color. The manganese in MnO₄^– reduces to Mn²⁺ (thus changing its oxidation state to +2) when mixed with compounds that can be oxidized. Mn²⁺ is a colorless ion.

Frequently, when potassium permanganate is titrated into a colorless solution that contains a compound that can be oxidized, the end of the titration is marked by the solution turning pink. Initially, the permanganate ions take electrons from the oxidized compound and reduce to Mn^2+. However, once the compound that is being oxidized runs out, there are no electrons left for the MnO₄^– to take, and thus any extra permanganate ion that is added remains unreduced and retains its purple color, which when diluted sufficiently appears pink.

Example:

A 50.0 mL solution of sodium oxalate, Na₂C₂O₄ is poured into an Erlenmeyer flask. An acidified solution of 0.135 M potassium permanganate is titrated into the flask while the solution is swirled on a stir station and heated. The solution in the flask is originally colorless, but turns pink after 14.56 mL of potassium permanganate is added. Given the following half-reactions, what is the concentration of the oxalate solution?

Reduction: 8H⁺ + MnO₄^– + 5e^– → Mn²⁺ + 4H₂O(l)

Oxidation: C₂O₄^2– → 2CO₂(g) + 2e^–

The first step to solving this problem is to balance the half-reactions. In this case, the oxidation half-reaction must be multiplied by 5 in order to make the electrons balance with those in the reduction half-reaction, and the reduction half-reaction must be multiplied by 2. Thus, the full reaction (which omits the sodium and potassium spectator atoms) is:

16H⁺ + 5C₂O₄² + 2MnO₄^– → 2Mn²⁺ + 8H₂O(l) + 5CO₂(g)

So, for every mole of permanganate that reacts, five moles of oxalate are required. The number of moles of permanganate can be easily determined:

M =

0.135 M MnO₄^–=

n = 0.00197 MnO₄^–

That can then be converted to moles of oxalate:

0.00197 MnO₄^– × = 0.00493 mol C₂O₄^2–

Finally, the concentration of the oxalate solution can be determined:

= 0.0986 M = [C₂O₄^2–] = [Na₂C₂O₄^2–]

You will not be expected to memorize any of the various color changes that can occur throughout a redox titration. However, you should understand the concept of using a color change to determine an endpoint if you are provided with the colors of the various compounds that are present during a titration.

Reduction Potentials

Every half-reaction has an electric potential, or voltage, associated with it. You will be given the necessary values for the standard reduction potential of half-reactions for any question in which they are required. Potentials are always given as reduction half-reactions, but you can read them in reverse and flip the sign on the voltage to get oxidation potentials.

Look at the reduction potential for Zn²⁺.

Zn²⁺ + 2e⁻ → Zn(s)  E ° = – 0.76 V

Read the reduction half-reaction in reverse and change the sign on the voltage to get the oxidation potential for Zn.

Zn(s) → Zn²⁺ + 2e⁻  E ° = +0.76 V

The larger the potential for a half-reaction, the more likely it is to occur.

F₂(g) + 2e⁻ → 2F^–  E ° = +2.87 V

F₂(g) has a very large reduction potential, so it is likely to gain electrons and be reduced.

Li(s) → Li⁺ + e⁻  E ° = +3.05 V

Li(s) has a very large oxidation potential, making it very likely to lose electrons and be oxidized.

You can calculate the potential of a redox reaction if you know the potentials for the two half-reactions that constitute it. There are two important things to remember when calculating the potential of a redox reaction.

Let’s look at the following reaction:

Zn + 2Ag⁺ → Zn²⁺ + 2Ag(s)

The two half-reactions are:

Oxidation:	Zn → Zn2+ + 2e−	E ° = +0.76 V
Reduction:	Ag+ + e− → Ag(s)	E ° = +0.80 V

E = E_oxidation + E_reduction

E = 0.76 V + 0.80 V = 1.56 V

Notice that we ignored that silver has a coefficient of 2 in the balanced equation.

The relative reduction strengths of two different metals can also be determined qualitatively. In the above reaction, if Zn(s) were placed in a solution containing Ag⁺ ions, the silver ions have a high enough reduction potential that they would take electrons from the zinc and start forming solid silver, which would precipitate out on the surface of the zinc.

However, if Ag(s) were placed in a solution containing Zn²⁺ ions, zinc does not have a high enough reduction potential to take electrons from silver and so no reaction would occur. So, when a solid metal is placed into a metallic solution and a new solid starts to form, the reduction potential of the metal in solution is greater than that of the solid metal. If no solid forms, the reduction potential of the solid metal is higher.

GALVANIC CELLS

In a galvanic cell (also called a voltaic cell), a favored redox reaction is used to generate a flow of current.

Look at the following favored redox reaction:

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)		E° = 1.10 V
Oxidation:	Zn(s) → Zn2+(aq) + 2e−	E° = 0.76 V
Reduction:	Cu2+(aq) + 2e− → Cu(s)	E° = 0.34 V

In a galvanic cell, the two half-reactions take place in separate chambers, and the electrons that are released by the oxidation reaction pass through a wire to the chamber where they are consumed in the reduction reaction. That’s how the current is created. Current is defined as the flow of electrons from one place to another.

In any electric cell (either a galvanic cell or an electrolytic cell, which we’ll discuss a little later) oxidation takes place at the electrode called the anode. Reduction takes place at the electrode called the cathode.

The salt bridge maintains the electrical neutrality in the cell. At the cathode, where reduction occurs and the solution is becoming less positively charged, the positive cations from the salt bridge solution flow into the half-cell. At the anode, where oxidation occurs and the solution is becoming more positively charged, the negative anions from the salt bridge solution flow into the half-cell. If the salt bridge were to be removed, the solutions in the half-cells would become electrically imbalanced, and the voltage of the cell would drop to 0. In the previous diagram, the salt bridge is filled with aqueous NaCl. As the cell functions, the Na⁺ cations flow to the cathode and the Cl^– anions flow to the anode.

Under standard conditions (when all concentrations are 1 M), the voltage of the cell is the same as the total voltage of the redox reaction. To determine what would happen to the cell potential under nonstandard conditions, you must first learn more about equilibrium, which will be covered in Chapter 8.

ELECTROLYTIC CELLS

In an electrolytic cell, an outside source of voltage is used to force an unfavored redox reaction to take place. Most electrolytic cells occur in aqueous solutions, which are created when a chemical dissolves in water; either the ions or the water molecules can be reduced or oxidized.

Let’s look at a solution of nickel (II) chloride as an example. To determine which substance is reduced, you must compare the reduction potential of the cation with that of water. The half-reaction with the more positive value is the one that will occur.

Ni2+ + 2e– → Ni(s)	E ° = –0.25 V
2H2O(l) + 2e– → H2(g) + 2OH–	E ° = –0.80 V

In this case, the Ni²⁺ reduction will occur. For the oxidation, the oxidation potential of the anion versus that of water must be examined. As with before, the half-reaction with the most positive value is the one that will occur. Remember that potentials are always given as reduction potentials, so you must flip the sign when you flip the equation to an oxidation.

2Cl– → Cl2(g) + 2e–	E ° = –1.36 V
2H2O(l) → O2(g) + 4H+ + 4e–	E ° = –1.23 V

So, in this case, the water itself would be reduced. When the equations are balanced for electron transfer, the net ionic equation looks like this:

2Ni²⁺ + 2H₂O(l) → 2Ni(s) + O₂(g) + 4H⁺

E = –0.25 V + –1.24V = –1.49 V

The anode and cathode in an electrolytic reaction are usually just metal bars that conduct current and do not take part in the reaction. In the above reaction, solid nickel is being created at the cathode, and oxygen gas is being evolved at the anode. The sign of your total cell potential (E) for an electrolytic reaction is always negative.

Occasionally, a current will either be run through a molten compound or pure water. In this case, you do not need to determine which redox reactions are taking place, as you will only have one choice for each.

The AN OX/RED CAT rule applies to the electrolytic cell in the same way that it applies to the galvanic cell.

Electroplating

Electrolytic cells are used for electroplating. You may see a question on the test that gives you an electrical current and asks you how much metal “plates out.”

There are roughly four steps for figuring out electrolysis problems.

For instance, if a current of 2.50 A is run through a solution of iron (III) chloride for 15 minutes, it would cause the following mass of iron to plate out:

It is particularly important to keep track of your units in an electroplating problem; there are many different conversions before you come up with your final answer. In general, as long as all of your conversions are set up correctly your final answer will have the correct units.

CHAPTER 5 QUESTIONS

Multiple-Choice Questions

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Use the following solubility rules to answer questions 1-4.

Salts containing halide anions are soluble except for those containing Ag⁺, Pb²⁺, and Hg₂²⁺.

Salts containing carbonate anions are insoluble except for those containing alkali metals or ammonium.

Use the following information to answer questions 12-15.

When heated in a closed container in the presence of a catalyst, potassium chlorate decomposes into potassium chloride and oxygen gas via the following reaction:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

Use the following information to answer questions 25-28.

A voltaic cell is created using the following half-cells:

Use the following information to answer questions 29-31.

Free-Response Questions

CHAPTER 5 ANSWERS AND EXPLANATIONS

Multiple-Choice

Free-Response