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FOR

DUMMIES

By Craig Gygi, Neil DeCarlo, Bruce Williams

Foreword by Stephen R. Covey

Author, The 7 Habits of Highly Effective People And The 8th Habit

WILEY

Wiley Publishing, Inc.

Six Sigma

FOR

DUMMIES

By Craig Gygi, Neil DeCarlo, Bruce Williams

Foreword by Stephen R. Covey

Author, The 7 Habits of Highly Effective People And The 8th Habit

WILEY

Wiley Publishing, Inc.

Six Sigma For Dummies9

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About the Authors

Craig Kent Gygi Began studying and applying the elements of Six Sigma well before they were formalized into today’s renowned breakthrough methodology.

As a graduate student in mechanical engineering at Brigham Young University

In the early 1990s, he integrated these cutting-edge improvement techniques

Into his coaching of student product development teams. Upon beginning his

Career in 1994 at Motorola’s Advanced Manufacturing Research Lab in Florida, he was formally introduced to the maturing Six Sigma method. It resonated deeply with his previous findings. From that time, Craig has applied, taught, and led Six Sigma in all his endeavors, including management and technical capacities at Motorola, Iomega, and General Atomics.

In 1998, Craig founded and led a software company to develop computational tools for Six Sigma practitioners. For several years, he also worked as a technical colleague of Dr. Mikel J. Harry, the original consultant of Six Sigma, co-developing and teaching new advances in its theory and application.

Mostrecently, Craig has traded his mountain home in Utah for the Sonoran

Desert of Arizona to co-found Savvi International and direct and manage its Six Sigma products, services, and tools.

A Master Black Belt, Craig has wielded Six Sigma techniques now for over 12 years, spanning projects from design to manufacturing to business

Process management. He is also an expert teacher, having instructed and

Mentored at all levels of Six Sigma, from executives to White Belts.

Neil John DeCarlo Has been a professional communicator in the continuous improvement and Six Sigma fields for more than 15 years, beginning with his work at Florida Power & Light company when it won the coveted Deming Prize

For quality. Since that time, he has authored, ghostwritten, or edited more than

150 articles and six books in association with such companies as General Electric, Dupont, Bose Corporation, McKinsey consulting, UPS, AT&T, the Six

Sigma Academy, and many others.

As a prolific author and writer, Neil’s past work has covered a range of subject matter, including Six Sigma, information technology, e-learning, knowledge management, change management, business integration, TQM, ISO, lean management, and other disciplines. He has also worked with several CEOs and consultants, including Japanese quality expert Dr. Noriaki Kano, and worked extensively with original co-architect of Six Sigma, Dr. Mikel Harry.

In addition to his writing pedigree, Neil has managed communication and

Publishing campaigns for a variety of companies and consulting firms, most notably, the Breakthrough Management Group, a Six Sigma, lean enterprise,

And performance-improvement industry leader. While not working, Neil avidly

Practices Bikram yoga and contributes to that community through his advocacy and writing.

Bruce David Williams Has been fascinated with complex systems since the

Launch of Sputnik on his third birthday. With undergraduate degrees from the

University of Colorado in Physics and Astrophysics, he entered a career in aerospace systems, where he first encountered Six Sigma after Motorola won the inaugural Baldridge Award in 1988. Later, with graduate degrees in technical management and computer science from Johns Hopkins University and the University of Colorado, and as a member of the Hubble Telescope development team, he was intrigued by how breakdowns in the smallest components could lead to colossal system failures. He entered the Six Sigma industry in the mid-1990s, when he founded a software company to pursue product life-cycle

Traceability.

Bruce has since been founder and CEO of two Six Sigma research and technology firms, and is now Chairman and CEO of Savvi International, a provider of solutions for business performance improvement using Six Sigma, lean, and

Business process management techniques. He resides in the desert foothills of North Scottsdale, Arizona, with his wife, two children, and assortment of dogs, cats, birds, and horses.

Dedication

Craig Gygi: To Jill, Ivan, and Gordon. Being part of their lives is my greatest success.

Neil DeCarlo: To Wanda Texon, who helped me believe in myself a long time ago, and who has been a constant source of support and intellectual stimulation for many years.

Bruce Williams: To Barbi, my spectacular wife of 22 years, and my amazing

Children, Hannah and Evan. Their tireless devotion, artful humor, and limitless

Thirst for knowledge truly make anything possible.

Authors’ Acknowledgments

Six Sigma per se didn’t exist twenty years ago. Miraculously, a single individual working for a large corporation in a cubicle at a nondescript office building saw something, and now — tens of thousands of individual practitioners, thousands of corporations, and hundreds of billions of dollars in savings later — we have a fully evolved system. On behalf of the entire Six Sigma movement, we’d like to acknowledge that individual: the late Bill Smith, a reliability engineer at Motorola in Arizona, who in the mid-1980s recognized a statistical correlation

Between product complexity, process capability, and system failure. We’d also

Like to acknowledge Dr. Mikel Harry, another Motorola Six Sigma pioneer, who has helped make the industry what it is today.

We’d like to thank Don L. Redinius, who contributed significant portions of

This book. We’re extremely fortunate to have had his contribution. Don is an internationally recognized business performance improvement consultant

With more than 25 years of experience who has dedicated much of his career

To Six Sigma.

In addition to writing Chapters 4 and 10, Don supplied extensive subject

Matter expertise, materials, guidance, and comments in reviewing, and in

Improving, nearly every page and figure.

Our deepest thanks to Dr. Stephen R. Covey, who honored us by writing the

Foreword for this book.

We also thank Ms. Roxanne O’Brasky, president of the International Society

Of Six Sigma Professionals, for her Afterword. Roxanne is a singular tour de force in Six Sigma and she is making the world a better place through the

Work of the society.

We owe the very existence of this book to acquisitions editor Kathy Cox and project editor Tere Stouffer Drenth, who masterfully guided the ship through the storm, avoided the reefs, and sailed us gently into port.

We’d also like to acknowledge those tens of thousands of practitioners: the

Black Belts, Green Belts, and Yellow Belts; the Champions; the Master Black Belts; and the deployment leaders. Their collective work has defined Six Sigma. They did it without a Six Sigma textbook, government study contracts, ISO standards bodies, or enterprise information systems. They made Six Sigma

Happen and evolved it to the point where this book became possible.

Publisher’s Acknowledgments

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Some of the people who helped bring this book to market include the following:

Acquisitions, Editorial, and Media Development

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In This Chapter

^ Putting all the annoying exceptions into one corral ^ Handling exceptions more easily with tips and tricks

•Exceptions seem like nature’s way of hedging its bets. As such, they’re annoying — why

Can’t nature just go ahead and Commit? But nature knows nothing of our rules, so it certainly isn’t going out of its way to annoy you. Nor is it going out of its way to make things easier for you. Either way, seeing as you have to deal with exceptions, we thought we’d corral many of them into this chapter so you can confront them more conveniently.

However, one big caveat! This chapter is at the end for a reason. Don’t get so hung up on the exceptions that you don’t master the basics properly. Many AP students lose points on the exam because they omit describing the basic principles that underlie the exceptions. It is not enough to only understand the exceptions without quoting the "normal" rules or trends.

In the field of psychology, Maslow’s hierarchy of needs declares that people need a sense of belonging. It’s a good thing hydrogen isn’t a person because it belongs nowhere on the periodic table (which we describe in Chapter 3). Although hydrogen is usually listed atop Group IA along with the alkali metals, it doesn’t really fit. Sure, hydrogen can lose an electron to form a +1 cation, just like the alkali metals, but hydrogen can also gain an electron to form hydride, H-, especially when bonding to metals. Furthermore, hydrogen doesn’t have metallic properties under normal conditions, but typically exists as the diatomic gas, H2.

These differences arise largely from the fact that hydrogen has only a single 1s orbital and lacks other, more interior orbitals that could shield the valence electrons from the positive charge of the nucleus. (See Chapter 7 for an introduction to S And other types of orbitals.) Another factor is that H+ is a bare proton with extremely compact charge density.

Hydrogen Isn’t an Alkali Metal

The Octet Rule Isn’t Always an Option

An Octet, As we explain in Chapter 5, is a full shell of eight valence electrons. The octet rule states that atoms bond with one another so as to acquire completely filled valence shells that contain eight electrons. It’s a pretty good rule. Like most pretty good rules, it has exceptions:

Atoms containing only 1s electrons simply don’t have eight slots to fill. So, hydrogen and helium obey the "duet" rule.

Certain molecules (like NO, for example) contain an odd number of valence electrons. In these cases, full octets aren’t an option. Like people born with an odd number of toes, these molecules may not be entirely happy with the situation, but they deal with it.

Atoms often attempt to fill their valence shells by covalent bonding (see Chapter 5). Each covalent bond adds a shared electron to the shell. But covalent bonding usually requires an atom to donate an electron of its own for sharing within the bond. Some atoms run out of electrons to donate, and therefore can’t engage in enough covalent bonds to fill their shell octets. Boron trifluoride, BF3, is a typical example. The central boron atom of this molecule can only engage in three B-F bonds, and ends up with only six valence electrons. This boron is said to be electron deficient. You might speculate that the fluorine atoms could pitch in a bit and donate some more electrons to boron. But fluorine is highly electronegative and greedily holds fast to its own octets. C’est la vie.

Some atoms take on more than a full octet’s worth of electrons. This is known as an Expanded octet. These atoms are said to be Hypervalent Or Hypercoordinated. The phosphorous of phosphorous pentachloride, PCl5, is an example. These kinds of situations require an atom from period (row) 3 or higher within the periodic table. The exact reasons for this restriction are still debated. Certainly, the larger atomic size of period 3 and higher atoms allows more room to accommodate the bulk of all the binding partners that distribute around the central atom’s valence shell.

Some Electron Configurations Ignore the Orbital Rules

Electrons fill orbitals from lowest energy to highest energy. This fact is true.

The progression of orbitals from lowest to highest energy is predicted by an Aufbau diagram. This isn’t always true. Some atoms possess electron configurations that deviate from the standard rules for filling orbitals from the ground up. For Aufbau’s sake, why?

Two conditions typically lead to exceptional electron configurations:

First, successive orbital energies must lie close together, as is the case with 3d and 4s orbitals, for example.

Second, shifting electrons between these energetically similar orbitals must result in a half-filled or fully filled set of identical orbitals, an energetically happy state of affairs.

Want a couple of examples? Strictly by the Aufbau, but not by the "energy" rules, chromium should have the following electron configuration:

[Ar]3d*4s2

Because shifting a single electron from 4S To the energetically similar 3D Level half-fills the 3D Set and lowers the energy, the actual configuration of chromium is

[Ar]3D54S1

For similar reasons, the configuration of copper is not the expected [Ar]3d°4s2, but instead is [Ar]3d104s’. Shifting a single electron from 4s to 3d fills the 3d set of orbitals.

Strictly speaking, the above arguments only apply to isolated atoms. Once the atom is involved in bond formation in a compound, the whole energy level scheme has to be re-examined, and apparent filling order may change again.

Flip to Chapter 5 for full details on electron configurations and Aufbau diagrams.

One Partner in Coordinate Covalent Bonds Giveth Electrons; the Other Taketh

To form a covalent bond (as we explain in Chapter 5), each bonding partner contributes one electron to a two-electron bond, right? Not always. Coordinate covalent bonds Are particularly common between transition metals and partners that possess lone pairs of electrons.

Here’s the basic idea: Transition metals have empty valence orbitals. Lone pairs Are pairs of nonbonding electrons within a single orbital. So, transition metals and lone-pair bearing molecules can engage in Lewis acid-base interactions (see Chapter 17). The lone-pair containing molecule acts as an electron donor (a Lewis base), giving both electrons to a bond with the metal, which acts as an electron acceptor (a Lewis acid). When this occurs, the resulting molecule is called a Coordination complex.

The partners that bind to the metal are called Ligands. Coordination complexes are often intensely colored and can have properties that are quite different than those of the free metal.

All Hybridized Electron Orbitals Are Created Equal

Different electron orbital types have grossly different shapes. Spherical S Orbitals look nothing like lobed P Orbitals, for example. So, if the valence shell of an atom contains both S – And P-orbital electrons, you might expect those electrons to behave differently when it comes to things like bonding, right? Wrong. If you attempt to assume such a thing, valence bond

Theory politely taps you an the shoulder to remind you that valence shell electrons occupy hybridized orbitals. These hybridized orbitals (as in sp3, sp2, and Sp Orbitals) reflect a mixture of the properties of the orbitals that make them up, and each of the orbitals is equivalent to the others in the valence shell.

Although this phenomenon represents an exception to the rules, it’s somewhat less annoying than other exceptions because hybridization allows for the nicely symmetrical orbital geometries of actual atoms within actual molecules. VSEPR theory presently clears its throat to point out that the negative charge of the electrons within the hybridized orbitals causes those equivalent orbitals to spread as far apart as possible from one another. As a result, the geometry of Sp3-hybridized methane (CH4), for example, is beautifully tetrahedral. However, this is messed up easily once total symmetry is upset. CH3I is no longer a perfect tetrahedron. Sigh.

Check out Chapter 7 for the details on VSEPR theory and hybridization.

Use Caution When Naming Compounds with Transition Metals

The thing about transition metals is that the same transition metal can form cations with different charges. Differently charged metal cations need different names, so chemists don’t get any more confused than they already are. These days, you indicate these differences by using Roman numerals within parentheses to denote the positive charge of the metal ion. An older method adds the suffixes -ous Or -ic To indicate the cation with the smaller or larger charge, respectively. For example

Cu+ = copper (I) ion or cuprous ion Cu2+ = copper (II) ion or cupric ion

Metal cations team up with nonmetal anions to form ionic compounds. What’s more, the ratio of cations to anions within each formula unit depends on the charge assumed by the fickle transition metal. The formula unit as a whole must be electrically neutral. The rules you follow to name an ionic compound must accommodate the whims of transition metals. The system of Roman numerals or suffixes applies in such situations:

CuCl = copper (I) chloride or cuprous chloride CuCl2 = copper (II) chloride or cupric chloride

Chapter 7 has the full scoop on naming ionic and other types of compounds.

You Must Memorize Polyatomic Ions

Sorry, it’s true. Not only are polyatomic ions annoying because they must be memorized, but they pop up everywhere. If you don’t memorize the polyatomic ions, you’ll waste time trying to figure out weird (and incorrect) covalent bonding arrangements when what you’re really dealing with is a straightforward ionic compound. Here they are in Table 33-1.

Liquid Water Is Denser than Ice

Kinetic molecular theory, which we discuss in Chapters 9 and 11, predicts that adding heat to a collection of particles increases the volume occupied by those particles. Heat-induced changes in volume are particularly evident at phase changes, so liquids tend to be less dense than their solid counterparts. Weird water throws a wet monkey wrench into the works. Because of H2O’s ideal hydrogen-bonding geometry, the lattice geometry of solid water (ice) is very "open" with large empty spaces at the center of a hexagonal ring of water molecules. These empty spaces lead to a lower density of solid water relative to liquid water. So, ice floats in water. Although annoying, this watery exception is quite important for biology.

No Gas Is Truly Ideal

No matter what your misty-eyed grandparents tell you, there were never halcyon Days of Old when all the gases were ideal. To be perfectly frank, not a single gas is really, truly ideal. Some gases just approach the ideal more closely than others. At very high pressures, even gases that normally behave close to the ideal cease to follow the Ideal Gas Laws that we discuss in Chapter 9.

When gases deviate from the ideal, we call them Real gases. Real gases have properties that are significantly shaped by the volumes of the gas particles and/or by interparticle forces.

Common Names for Organic Compounds Hearken Back to the Old Days

Serious study of chemistry predates modern systematic methods for naming compounds. As a result, chemists persistently address a large number of common compounds, especially organic compounds, by older, "trivial" names. This practice won’t change anytime soon. A cynical take on the situation is to observe that progress occurs one funeral at a time. A less cynical approach involves serenely accepting that which you cannot change and getting familiar with these old-fashioned names. Table 33-2 lists some important ones; head to Chapter 25 for details on organic compounds.

AG

AH

Favorable TAS

Overcomes unfavorable AH

AH

AH

Figure 21-7:

The interplay of enthalpy and entropy components of free energy.

+

Nonspontaneous Reactions:

AG

TAS

J_

AG

In This Chapter

^ Hitting the high points

^ Shaping up your skill with some practice questions

^ Discovering the answers and explanations to your questions

Chapter 7 was packed with ideas, starting with the way valence electrons distribute themselves around individual atoms, building up through the way atoms connect themselves into molecules with particular shapes and finishing with the way molecular shapes help determine the properties of different compounds. In other words, Chapter 7 was a lot to take in one bite. Here is where you chew. In the following sections we review the highlights of Chapter 7, give you practice problems to help sear important concepts into your brain, and give you full explanations for the answers to help you swallow what you’ve chewed.

Shaping Up the High Points

Before you attempt any questions in this chapter, you should review the following sections of important points from Chapter 7.

Electron dot structures and Lewis structures

Electron dot structures And Lewis structures Use dots and/or lines to represent valence electrons and/or lines to represent valence electron pairs.

Basic steps to follow when drawing a Lewis structure are

1. Add up all the valence electrons for all the atoms in the molecule.

2. Pick a "central" atom to serve as the anchor of your Lewis structure.

3. Connect the other, "outer" atoms to your central atom using single bonds only.

4. Fill the valence shells of your outer atoms. Then put any remaining electrons on the central atom.

5. Check whether the central atom now has a full valence shell.

Compounds that can engage in multiple, valid Lewis structures participate in Resonance, Based on an old idea that the molecules might actually be in equilibrium between the multiple structures. Although we now know that not to be the case, the term has stuck. Individual Resonance structures Contribute to an overall, averaged structure called a Resonance hybrid. Adjacent p-orbital electrons are commonly involved in resonance, in which they participate within delocalized N Bonds.

Valence bond theory

Valence bond theory Describes covalent bonds in terms of overlap between atomic orbitals:

IU Orbitals that overlap in a manner completely symmetric to the bond axis form Sigma bonds.

I Orbitals that overlap in a manner that is symmetric to the bond axis in only one plane form Pi bonds.

VSEPR theory

VSEPR theory Explains molecular shapes in terms of mutual repulsion between electron pairs (both bonding and nonbonding pairs) around a central atom:

IU Lone (nonbonding) pairs repel more strongly than bonding pairs.

IU Valence electrons in different kinds orbitals (like S And P Orbitals) Hybridize Into mixed, equivalent orbitals. The total number of hybridized orbitals is the same as the original number of unmixed orbitals.

I One S Orbital mixes with one P Orbital to create two identical Sp Hybrids.

Sp Hybridized centers will have linear geometry. IU One S Orbital mixes with two P Orbitals to create three identical Sp2 Hybrids.

Sp2 Hybridized centers will have trigonal planar geometry. IU One S Orbital mixes with three P Orbitals to create four identical Sp3 Hybrids.

Sp3 Hybridized centers will have tetrahedral geometry.

Bond dipoles

Individual Bond dipoles Arise from differences in electronegativity in bonded atoms. Except for homonuclear diatomic molecules (like O2, H2, and so on), all atom-atom bonds will have a at least a small bond dipole. Within a molecule, bond dipoles may or may not contribute to an overall Molecular dipole Depending on geometry:

I Bond dipoles add as vectors within molecules, building on each other or canceling each other out depending on geometry.

IU Dipole moment (□,) = amount of separated charge (Q) x distance separated (d); the equation looks like this: u, = Qd

Isomers

Isomers Are compounds made up of the same atoms, but are put together in the following different arrangements:

IU Structural isomers Have different bonds.

IU Stereoisomers Have the same bonds, but different three-dimensional arrangements of atoms.

• Geometric isomers Can occur when atoms are restricted from rotating freely about their bonds.

• Optical isomers (also called Enantiomers) Are nonsuperimposable mirror images of one another and are therefore Chiral.

Testing Your Knowledge

Now that you know your sp2′s from your sp3′s, and have your VSEPR in shape, give these practice questions a try.

1. Which of the following correctly lists the molecules in order of increasing polarity?

(A) H2O, H2S, HF, H2

(B) H2, HF, H2S, H2O

(C) H2, H2S, HF, H2O

(D) H2, H2S, H2O, HF

(E) H2S, H2O, HF, H2

2. Which of the following molecules has bond angles closest to 109.5°?

(A) BF3

(B) SiS2

(C) CCl4

(D) H2O

(E) CO2

3. Identify the incorrect set of resonance structures.

A.

-1 • • +1 . • -1 IO — N — O!

-1 • • +1 • •

! O — N — O • • | • •

I O I

"-1

• • +1 . • -1

O — N — O!

• • | • •

I O I -1 • •

B.

H

C.

H

H H H

H — C — O I

M m

H

H

HH H

‘.OL I..

HCO

D.

+1

‘, O — O — O!

+1

I O — O — O I

-1 • • • •

I O — C = O

• • I • •

‘.OL

* *-1

-1 • • • • • .-1

I O — C — O I

OCO

Mm | • •

‘.OL

* *-1

4. Group IVA elements tend to hybridize in which of the following ways?

5. Within formaldehyde, CH2O, carbon possesses which of the following traits?

I. Sp2 Hybridization

II. Three equivalent sigma bonds

III. Trigonal planar bonding geometry

(A) I only

(B) II only

(C) III only

(D) I and III only

(E) I, II and III

O

H

H

-1

O

-1

-1

E.

-1

O

Questions 6 through 10 refer to the following list of geometries:

6. Characteristic of four electron pairs, two bonding and two nonbonding

7. Typical of Sp Hybridization

8. Accounts for the nonpolarity of SiF4

9. Nitrate anion 10. PCl5

Questions 11 through 14 refer to the following molecule, and the accompanying set of choices:

O

11. Exhibit(s) Sp2 Hybridization

12. Exhibit(s) Sp3 Hybridization

13. Has a lone pair

14. Has a pi bond

Checking Your Work

Don’t play the angles. Clear up any lingering confusion by checking your answers, making sure you understand the explanations for any questions you got wrong — or the ones that you got right by guessing.

1. (D). H2 is the least polar because its sole covalent bond occurs between atoms that are identical, and which therefore have identical electronegativity. H2S and H2O are close, but H2O is slightly more polar because oxygen is slightly more electronegative than sulfur. The "bent" geometry of both H2S and H2O is key to the presence of a molecular dipole in each

Case — a linear shape would lead to canceling of the bond dipoles, resulting in nonpolar molecules. The sole covalent (nearly ionic) bond in HF is the most polar of them all.

2. (C). Bond angles of 109.5° signify a perfect tetrahedron. Many molecules approach this ideal, but perfect tetrahedral geometry typically requires that the central atom (carbon, in this case) possess four identical bonding partners (chlorines, in this case).

3. (E). The last set of resonance structures, for CO32- anion, are incorrect. Each of the structures includes an oxygen atom with an excess of valence electrons. The oxygen atom double-bonded to carbon in each structure should have only two lone pairs, not three.

4. (C). Group IVA atoms tend to engage in four covalent bonds. Because of hybridization, these bonds involve four identical Sp3 Orbitals.

5. (D). In formaldehyde, the central carbon indeed exhibits Sp2 Hybridization. This hybridization leads to a trigonal planar (though not perfectly so) molecular shape. However, the three sigma bonds are not identical, for two reasons. First, C-O bonds differ in polarity from C-H bonds. Second, the C-O bond axis includes a pi bond; double-bonding between carbon and oxygen brings those two atoms closer together than does a single bond.

6. (B). In water, for example, the two lone pairs of oxygen and the two O-H bonds lead to just such a bent shape.

7. (A). In an Sp Hybridized atom, there are only two bond axes. The electrons in these axes mutually repel to achieve the maximum separation afforded by linear geometry.

8. (D). Because of its perfect tetrahedral geometry, silicon tetrafluoride is nonpolar, despite the fact that each silicon-fluorine bond is polar.

9. (C). Nitrate displays resonance between three Sp2 Hybridized resonance structures. The hybrid structure therefore has a trigonal planar shape consistent with Sp2 Hybridization.

10. (E). This compound may have thrown you. The central phosphorous of PCl5 is "hyperva-lent," meaning that it engages in more than the three covalent bonds you’d expect from its position on the periodic table. But even if the hypervalency freaked you out, take heart: the basic principles of VSEPR still apply. Whatever the reasons for pentavalent phosphorous, the fact remains that the electrons within the five bond axes mutually repel, seeking the shape with maximum separation. That shape is trigonal bipyramidal.

11. (C). Carbon III engages in a double bond (with carbon) and in two single bonds (with hydrogen and nitrogen). This arrangement leads to three bonding axes and Sp2 Hybridization.

12. (D). Carbon I engages in four separate single bonds, one with another carbon and three with hydrogens. Four separate single bonds around a central carbon are a hallmark of Sp3 Hybridization. Nitrogen II may at first appear to be Sp2 Hybridized, because it engages in only three bonds — but the nitrogen also contains a lone pair, and is thus Sp3 Hybridized.

13. (B). Carbons I and III keep their four valence electrons occupied within various combinations of covalent bonds. Because it has one more valence electron than carbon, Nitrogen II engages in only three covalent bonds, reserving two electrons as a nonbonding "lone pair."

14. (C). Double bonds, like the carbon-carbon double bond in which Carbon III participates, include both sigma and pi bonds.

FOR

By Peter Mikulecky, Michelle Rose Gilman, and Kate Brutlag

WILEY

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AP Chemistry For Dummies®

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About the Authors

Peter Mikulecky Grew up in Milwaukee, an area of Wisconsin unique for its high human-to-cow ratio. After a breezy four-year tour in the Army, Peter earned a bachelor of science degree in biochemistry and molecular biology from the University of Wisconsin-Eau Claire, and a PhD in biological chemistry from Indiana University. With science seething in his DNA, he sought to infect others with a sense of molecular wonderment. Having taught, tutored, and mentored in classroom and laboratory environments, Peter was happy to find a home at Fusion Learning Center and Fusion Academy. There, he enjoys convincing students that biology and chemistry are, in fact, fascinating journeys, not entirely designed to inflict pain on hapless teenagers. His military training occasionally aids him in this effort. He is the author of AP Biology For Dummies And Chemistry Workbook For Dummies.

Michelle Rose Gilman Is most proud to be known as Noah’s mom. A graduate of the University of South Florida, Michelle found her niche early, and at 19, she was working with emotionally disturbed and learning-disabled students in hospital settings. At 21, she made the trek to California, where she found her passion for helping teenagers become more successful in school and life. What started as a small tutoring business in the garage of her California home quickly expanded and grew to the point where traffic control was necessary on her residential street.

Today, Michelle is the founder and CEO of Fusion Learning Center and Fusion Academy, a private school and tutoring/test prep facility in Solana Beach, California, serving more than 2,000 students per year. She is the author of ACT For Dummies, Pre-Calculus For Dummies, AP Biology For Dummies, Chemistry Workbook For Dummies, GRE For Dummies, Pre-Calculus Workbook for Dummies, And other books on self-esteem, writing, and motivational topics. Michelle has overseen dozens of programs over the last 20 years, focusing on helping kids become healthy adults. She currently specializes in motivating the unmotivatable adolescent, comforting their shell-shocked parents, and assisting her staff of 35 teachers.

Michelle lives by the following motto: There are people content with longing; I am not one of them.

Kate Brutlag Has been a full-fledged science dork since she picked up her first book on dinosaurs as a child. A native of Minnesota, Kate enjoys typical regional activities such as snow sports and cheese eating. Kate left Minnesota as a teen to study at Middlebury College in Vermont and graduated with a double major in physics and Japanese. Seeking to unite these two highly unrelated passions, she spent a year in Kyoto, Japan, on a Fulbright scholarship researching Japanese constellation lore. Kate was quickly drawn back to the pure sciences, however, and she discovered her love for education through her work at Fusion Academy, where she currently teaches upper-level sciences and Japanese. She is the author of Chemistry Workbook for Dummies.

Dedication

We would like to dedicate this book to our families and friends who supported us during the writing process. Also, to all our students who motivate us to be better teachers by pushing us to find unique and fresh ways to reach them.

Authors’ Acknowledgments

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In This Chapter

^ Getting familiar with common laboratory equipment ^ Reviewing common laboratory techniques

Chemistry is a laboratory science. Although there are such things as purely "theoretical" chemists who do all their reactions on computers, other chemists tend to make fun of these types behind their backs. Most chemistry still involves things like beakers, tubes, and interesting odors (beyond those possibly emanating from the chemists themselves). This state of affairs is not lost on the creators of the AP exam, who expect you to know your basic laboratory equipment and techniques. As you’ll know if you’re taking an AP chemistry course, lab is an important component. This chapter will help you to review the equipment and procedures that you’ve employed to create interesting odors of your own.

Gearing Up: An OverView of Common Lab Equipment

Figure 27-1 shows all of the lab equipment that an AP chemistry test-taker will need to be familiar with. You will not be directly tested on your knowledge of lab equipment, however the AP chemistry exam will often describe and/or diagram experimental setups so you will need to be able to recognize and understand the purpose of each of the pieces of equipment below.

Figure 27-1:

Common chemistry lab equipment.

Platform balance (triple beam)

Ring stand

Burner

Figure 27-1 shows you what the equipment looks like, and the list below tells you how each piece of lab equipment functions:

Balance: Used for obtaining the masses of solid and liquid samples

Beaker: A flat-bottomed, cylindrical piece of glassware used for mixing and heating compounds

Bunsen burner: Attached to a gas line and lit to provide heat for your experiments

Buret: An extremely accurate device with a stopcock at the bottom used to measure volumes of reagents

Ceramic square: Used to avoid burning the surface of your lab bench and incurring your chemistry teacher’s wrath

Clamps: Used to hold a variety of things in place, particularly test tubes Clay triangle: Used to hold a crucible while it is being heated

Condenser: Used to collect vapors by condensing them into liquid as they contact the liquid-cooled inner surface of the condenser

Crucible: A cup-shaped container capable of sustaining high temperatures. It is used to heat chemicals.

Crucible tongs: Used to handle the hot crucible

Erlenmeyer flask: Used to hold liquids. The small upper opening slows evaporation, so for some volatile liquids, a flask is a better choice than a beaker. The shape also makes it suitable for mixing and swirling liquids during a titration.

Florence flask: A type of flask, generally round-bottomed, usually suspended and heated from below. Its shape makes it easy to swirl and mix liquids inside of it.

Funnel: Used together with filter paper to filter precipitates out of solutions

Graduated cylinder: Used to precisely measure volumes

Metal spatula: Used to measure out solid substances

Mortar and pestle: Used to grind sesame seeds for cooking and chemical compounds for chemistry experiments, though we recommend using a different set for each

Pipette bulb: Used to transfer accurately measured amounts of liquid from one container to another

Rubber stoppers: Used to close flasks or test tubes to prevent evaporation of liquids or

Escape of gases

Scoopula: Another instrument used to transfer solids from one place to another

Test tube: Cylindrical open-topped piece of glassware that comes in varying sizes

Thermometer: Used to measure temperatures. Thermometers generally contain liquid mercury.

Watch glass: A piece of glassware in the shape of a large contact lens used for evaporating liquids

Wire gauze: Generally used as a surface for a beaker or flask to rest when being heated by a Bunsen burner

Following Protocol

What makes an effective laboratory "hunchback" is an intimate knowledge of two things. First, the hunchback knows the right tool for the job. Second, the hunchback knows how to use that tool. If you don’t know the first thing, you may find yourself attempting to measure the pH of a solution by placing it dropwise onto the tip of your tongue (don’t ever do this). If you know the first thing but not the second thing, you may find yourself staring in admiration at a stack of litmus paper that you have no idea how to use. To prevent both these kinds of embarrassing episodes, and to shore up your lab knowledge for the AP exam, make sure you thoroughly review this section.

Measuring mass, volume, and density

Mass and volume are by far the most frequently measured quanities in your basic chemistry lab. From mass and volume, you can calculate density. Alternately, you can use density to calculate mass or volume.

Mass: You measure mass in the laboratory using a balance. Most laboratory balances are extremely accurate, capable of measuring masses with a precision of hundredths of grams.

Volume: Techniques for measuring volume vary depending on what type of substance you are measuring, with liquids being the easiest to measure. Graduated cylinders are generally used to make approximate volume measurements for liquids. Pipettes and burets are used to measure precise volumes of liquids. Obtaining the volume of a solid can be somewhat tricky and, in some cases, impossible to do without altering the substance. If the substance does not react with water, you can submerge it in water and measure the volume of water displaced by it. Luckily, in most experiments it is mass of solid used that matters and not volume. You cannot measure the volume of a gas because gases have no definite volume. Recall that it is a fundamental property of gases that they expand to fill the entire volume available to them. However, the amount of gas can be obtained by measuring its pressure when held in a known volume.

Density: Mass and volume measurements can also be used to determine the density of a substance using the formula density=mass^volume. Density is a characteristic physical property for solids and liquids and can often be looked up in a table.

Determining melting point

Melting Is the phase transition between the solid and liquid states and is equivalent to the freezing point. Because melting point is a characteristic physical property of a solid, you can use melting point to help to determine the identity of an unknown solid or to verify that an experiment has been successful and you have isolated the desired product.

You can use several different types of apparatus for measuring melting points, but they all have the same general setup. A sample of the solid in question is placed in a thin glass tube and heated slowly from the bottom. The experimenter watches the solid through a magnifying glass until it reaches the melting point and records it. Pure crystalline solids have very distinct melting points and the phase transition between solid and liquid occurs in a few seconds, giving a very accurate melting point determination. Impurities lower the melting point of a pure solid, by an amount depending on their concentration. So it is vital to have a pure

Sample if you are using melting point to confirm an identity. Amorphous solids such as glass and plastic do not have precise melting points and so chemists make special types of melting measurements for them.

Determining boiling point and doing distillation

The boiling point of a substance is slightly more difficult to measure than the melting point. This is because it is the phase transition between a liquid and a gas and gases are difficult to see so you need a way of verifying that your liquid is at its boiling point and not merely very, very hot. One convenient phenomenon to help you in measuring a boiling point is the tendency for a liquid to maintain a constant temperature during a phase transition. All of the energy being added to the solution goes into changing its phase rather than into heating it.

Distillation is a process that can aid you not only in determining a precise boiling point for a liquid but also in purifying that liquid. Say you have a mixture of two different liquids in a flask, one with a lower boiling point than the other. If you could somehow trap the gas being released as the first liquid boils you would have isolated the pure first substance. This is the basic purpose of a distillation apparatus such as the one shown in Figure 27-2.

The flask on the left is being heated, while the one on the right is cool. Connecting the two is a piece of lab equipment called a Condenser. The inner tube in a condenser connects the first flask to the second, while cold water is run through the outer tube (this water is kept isolated and does not come into direct contact with the chemicals in the experiment). This causes the gas that is escaping from the boiling liquid in the first flask to condense in the inner tube of the condenser, which is angled downward so that it will collect in the second flask.

As the experimenter, you watch a thermometer connected to the first flask. When the first liquid begins to boil and condense in the condenser, the temperature of the liquid will remain constant until all of the lower boiling point liquid has escaped. Once it has all escaped, the condenser will collect no more liquid and the temperature in the first flask will again begin to rise. You then turn off the heat and disconnect your apparatus, having isolated a pure substance in your second flask. If your mixture contains more than two liquids or you wish to isolate a liquid with a higher boiling point, you can continue this process, switching out the second flask and isolating one substance at a time.

Dealing with solids in precipitation and filtration

In chemistry, you will often create solid precipitates in aqueous solutions. You separate the solution from the precipitate using a process called Filtration, Which involves pouring your solution/precipitate mixture through a funnel lined with filter paper. The filter paper allows the solution to pass through, but catches the precipitate. The precipitate should be washed to remove impurities.

Positive spin: Centrifugation

Centrifugation is another nifty way of separating compounds, this time according to their density. Centrifuges are quite expensive and most high school chemistry labs are not equipped with them, but you will often find reference to centrifugation in your chemistry text or on the AP exam. A centrifuge is essentially a big, extremely fast-spinning trough. A mixture in the centrifuge will separate according to density as the denser components migrate away from the center of the apparatus and the less dense migrate toward it.

Thinning out with dilutions

Rather than storing lots of different concentrations of the same type of solution, most chemistry labs will store concentrated stock solutions of each chemical. However, you don’t always want to be dealing with extremely concentrated solutions, particularly when they are strong acids and bases that have a propensity to react with skin and are generally unpleasant. Often, a dilution is required before an experiment is begun. In a dilution, a concentrated stock solution is diluted by adding water to a desired molarity. Dilutions utilize the equation Mi x Vi = Mf x Vf, where Mi And Mf are the initial and final molarities and Vi And Vf are the initial and final volumes. The amount of water added in these cases is equal to Vf—Vi.

Drying off

Because so many compounds react with water, it is often desirable to eliminate water from certain substances. Since so many substances are soluble in water, it is often very difficult to get rid of. In many cases, you can simply heat the substance you wish to rid of water to higher than 100 degrees C and wait for the water to evaporate. However, high temperatures can also cause decomposition of the substance you wish to isolate so sometimes another method is needed.

This alternative method takes advantage of substances called Desiccants That draw in water. If you leave your sample in a sealed container also containing a desiccant, over time the desiccant will absorb all of the water from your sample. Desiccants are slow acting, and your sample will generally have to be left in the container overnight to complete the drying process. Desiccants are capable of absorbing only a certain amount of water, so it is important that your sample have as much water as possible removed from it before being sealed with the desiccant. Silica gel-based desiccants are common in chemistry labs because they can be easily regenerated by heating. Most commercially available dessicants are prepared with additives so that they change color as they become more hydrated.

Making pH measurements

When doing acid/base reactions in the laboratory, you will often wish to know the pH of the solution you are dealing with. There are several ways to measure pH. Often, a material known as litmus paper is used. Litmus paper changes color as it reacts with solutions of different pH. Proper technique is to place a drop of solution onto the paper with a stir rod. When the color of the paper develops, you match its color to the pH scale that comes with the box of litmus paper. Many labs now use "pH paper" or "pH strips," that change to different colors over a range of pH values.

Many chemists prefer to use a pH indicator such as phenolphthalein instead of litmus paper, however, probably because it makes the solution itself turn pink when a pH above 9 is reached. Universal indicators can provide a range of color changes against a scale. pH meters have now become quite cheap and small, so a probe dipped in a solution can give you a direct digital readout of the pH in the solution.

Using spectroscopy/spectrometry to measure concentration

Spectroscopy Is a technique for separating light into its component colors. Spectrometry Means making measurements of the separated light. Sophisticated chemistry labs use spec-trophotometers to create a graph showing the absorption and/or emission at each wavelength of the visible spectrum. However, as is the case with centrifuges, not all high school chemistry labs are equipped with spectrophotometers. If you are asked about spectroscopy on the AP chemistry exam, you will be required to know that it relies on the separation of light into its component colors. Recall from Chapter 3 that the colors absorbed or emitted by a pure element are unique to that element and are related to its atomic structure.

In addition, you will be expected to know that a spectrometer can be used to measure the concentration of a colored substance by using Beer’s Law, and measuring the amount of light transmitted by the solution. The solution is poured into a clean cell and placed in the spectrometer. By comparing this to a cell containing a standard amount of the substance, the spectrometer can determine an unknown concentration by a simple ratio. It is important that no other light scattering (such as by a fingerprint on the cell) happens, as it will affect the result.

In This Chapter

^ Remembering the important stuff

^ Having a gas with some practice questions

^ Going through answers with explanations

■ Mye Blew through the gas laws and kinetic theory in Chapter 9, so review this condensed ▼ ▼ list of key concepts. After you think you’ve got these concepts under control, try tackling the AP Chemistry exam-style questions later in this chapter. Make sure to attempt the problems using only your cheat sheet as a reference in the same way you would use your list of formulas and constants on the real AP exam.

Doing a Quick Review

Here is a brief outline of the concepts presented in Chapter 9. Review it before attempting the practice questions and see Chapter 9 for the details of anything you’re still a little rusty on:

Kinetic theory states that as the temperature of a gas increases, so do the kinetic energies and velocities of its constituent particles.

*»»T Ideal gases are those for which

• Particle motions are random.

• Particles undergo only translational motion and do not vibrate, spin, or twist.

• Collisions between particles are elastic, meaning that no kinetic energy is transformed.

• Neighboring gas particles do not exert forces (attraction or repulsion) on one another.

The average kinetic energy of a gas particle is KE = KMv2, where KE is kinetic energy, m is mass, and v is velocity.

Average particle velocity can be expressed as

I3RT u = J—Z-z—

V M

Or as

J3kT U = 4/-

V M

Where U Is the average velocity, R Is the ideal gas constant, T Is temperature, M Is the molar mass, K Is Boltzmann’s constant, and M Is the mass of an individual gas particle.

Ii The average kinetic energy per mole of a gas is KE = KRT, where R is the ideal gas constant and T is the temperature.

The Gas Laws

• Boyle’s Law: P1xV1 = P2xV2, where P1 and V1 Are the initial pressure and volume of the gas and P2 and V2 are the final

• Charles’ Law: V1-T1 = V2-T2. Here again, V1 And T1 Are the initial volume and temperature of the gas and V2 and T2 are the final.

• The Combined Gas Law:

P1 xV1 = P2 x V2 N1 x T1 n2 x T2

Where P1, V1, n1 and T1 are the initial pressure, volume, number of moles and temperature respectively and P2, V2, n2 and T2 are the final

• The Ideal Gas Law: PV = NRT, Where P Is pressure, V Is volume, N Is the number of moles, R Is the ideal gas constant and T Is temperature

• Dalton’s Law of Partial Pressures: Ptotal = P1 + P2 + P3 + . . . + Pn, where P1, P2, P3, etc. are the partial pressures of different components of the gas mixture

• Graham’s Law:

RateA – y/molar massB RateB .^molar massA

I Chemists use the expression PV

= n

RT

To determine whether or not a real gas resembles an ideal gas closely enough to use the ideal gas laws as approximations, which correlates to N Values near 1.

For gases with N Values far from 1, the more complicated expression

Is used. Values of the parameters A And B Must be known for the gas in question.

Testing Your Knowledge

The following practice problems can help you capture the wispy gas laws in your brain once and for all.

1. Which of the following accounts for the fact that gases generally do not behave ideally under high pressures?

(A) They begin to undergo rotational motion.

(B) Collisions between gas molecules become inelastic.

(C) Average molecular speeds increase.

(D) Intermolecular forces are greater when molecules are close together.

(E) Collisions with the walls of the container become more frequent.

2. Which of the following will increase the rms speed of a molecule of gas, assuming that all other variables are kept constant?

I. Increasing pressure

II. Increasing temperature

III. Decreasing molar mass

(A) I only

(B) I and II

(C) I and III

(D) II and III

(E) I, II, and III

3. If gas A effuses (goes through a small hole into a vacuum) at three times the rate of gas B, then

(A) the molar mass of gas A is three times greater than the molar mass of gas B.

(B) the molar mass of gas B is three times greater than the molar mass of gas A.

(C) the molar mass of gas A is nine times greater than the molar mass of gas B.

(D) the molar mass of gas B is nine times greater than the molar mass of gas A.

(E) the molar masses of gas A and gas B are equal.

4. A sealed vessel contains 0.50 mol of neon gas, 0.20 mol hydrogen gas, and 0.3 mol oxygen gas. The total pressure of the the gas mixture is 8.0atm. The partial pressure of oxygen is

5. Molecules of an unknown gas diffuse at three times the rate of molecules of ammonia (NH3) under the same conditions. What is the molar mass of the unknown gas?

6. A rigid 10.0L cylinder contains 2.20g H2, 36.4g N2 and 51.7g O2.

(a) What is the total pressure of the gas mixture at 30°C?

(b) If the cylinder springs a leak, and the gases begin escaping, will the ratio of hydrogen remaining in the cylinder to the other two gases increase, decrease, or stay the same?

7. Samples of Br2 and Cl2 are placed in 1L containers at the conditions indicated in the diagram in Figure 10-1.

Figure 10-1:

Conditions for Br2 and Cl2 samples.

(a) Which of the two gases has the greater kinetic energy per mole? Justify your answer.

(b) If the volume of the container holding the Cl2 were decreased to 0.25L but the temperature remained the same, what would the change in pressure be? Assume that the gas behaves ideally.

8. Two flasks are connected by a stopcock as shown in Figure 10-2. The 10.0L flask contains CO2 at a pressure of 5atm and the 5.0L flask contains CO at a pressure of 1.5atm. What will the total pressure of the system be after the stopcock is opened?

Checking Your Work

You’ve done your best. Now check your work. Make sure to read the explanations thoroughly for any questions you got wrong — or for any you got right by guessing.

1. (D). High pressure gases do not behave ideally because their molecules are too close together to neglect intermolecular forces.

2. (C). Molecular speeds are proportional to temperature and inversely proportional to mass.

3. (D). Remember that Graham’s Law states that gases effuse at a rate inversely proportional to the square root of their molar masses.

4. You are given enough information to find the mole fraction of each gas. Oxygen is 0.3 mol out of a total of 1.0mol of gas, so its mole fraction is 0.3. Multiply this by the total pressure to get a partial pressure of 2.4 for oxygen.

5. 153g mol. This is another Graham’s Law problem. You know that the ratio of effusion rates is 3:1, so the left-hand side of Graham’s Law will be 3. You can also easily calculate the molar mass of ammonia, which is 17g mol. Plug these values into the equation to get

1 ~V 17

Square both sides to get rid of the square root and then multiply both sides by 17 to get 153g mol.

6. (a) 9.83atm. Begin by converting the masses you have been given to moles by dividing by their molar masses. Make sure to get in the habit of showing your work even on simple calculations such as these.

2.20gH2 x 1molH2 = 110molH

1 2.00gH2 1-lumuin:

34.6gN21molN2

–—- x-— = 1 23mol N

1 28.0gN2 1-"muilN;

51.7gO21molO2

-i x 32 0gO2 = 1.62molO

1 32.0gO2

Next, add these values to yield the total number of moles of gas (3.95mol). Plug this and your known values for volume and temperature into the equation P = nRTVV.

3.95mo/ x 0.0821 L X,Atn}r X 303K Mo

10.0 L

P =-^m°l X K-= 9.83am

(b) Decrease. Because hydrogen has the lowest molar mass, it will effuse the fastest under Graham’s Law, so the ratio of hydrogen to the two other gases in the container, which escape more slowly, will decrease.

7. (a) They are the same. The average kinetic energy per mole of a sample of gas depends on temperature, and the two gases are at the same temperature.

(b) 0.25atm. This is a simple application of Charles’s Law where V1 = 1.0L, V2 = 0.25L and P1 = 1atm. Plug these into the equation to yield

8. 3.8atm. Begin by finding the final pressure of each of the two gases when the entire volume is made available to them using Boyle’s Law

P2 of CO2 = P1V1 vV2 = 5.0atm x 10.0L 4 15.0L = 3.3atm

P2 of CO = P1V1^V2 = 1.5atm x 5.0L 4 15.0L = 0.5atm

According to Dalton’s Law of Partial Pressures, the final pressure of this gas mixture will simply be the sum of these two final pressures.

P2

V x P1 = 0.25L x 1atm

= 0.25atm

V1 1.0 L

In This Chapter

► Understanding the ins and outs of atomic structure

► Practicing atomic and nuclear chemistry problems

► Applying your knowledge to test questions

► Getting some answers and explanations

The concepts reviewed in Chapter 3 are the most basic in all of chemistry, and they are also some of the most important and testable concepts on the AP chemistry exam. A strong understanding of the most fundamental aspects of modern chemistry is essential to your success on the exam, and the concepts presented in Chapter 3 provide the foundation for the rest of introductory chemistry. Questions that appear regularly in both multiple-choice and free-response questions involve

Periodic table trends Nuclear chemistry

The structure of the atom and nucleus

Many AP questions that do not ask directly about atomic structure nonetheless assume that you have a solid understanding of its concepts.

In this chapter, we highlight the most important points of Chapter 3 for you, and then allow you to try your hand at AP chemistry-style test questions about atomic structure. Keep your eye out for these question types and others that build off of them in the practice tests (Chapters 29 and 30) at the end of the book as well.

Reinforcing the Foundation for Atomic Structure

For the AP exam, you should be familiar with the concepts of atomic structure as well as how to use and manipulate any equations and/or constants related to atomic structure. In this section we review these important concepts covering atomic structure so that you’re prepared for some AP-style exam questions we give you later in this chapter.

Make sure that you thoroughly understand all of the basic components of the atom and how scientists arrived at their understanding of the modern atom. Burn the following major points into your brain before the exam:

Atomic structure came to be understood gradually through the work of Thompson, Rutherford and Bohr:

• Thompson’s model, called the Plum pudding Model of the atom, consisted of blobs of negative charge in a soup of positive charge.

• Rutherford shot bullets (alpha particles) at tissue paper (gold foil) and had to dodge them every once in a while as they came bouncing back at him. He thereby discovered that the majority of the mass and one of the types of charge (the positively charged protons) of the atom were concentrated at its very center in the nucleus.

• Bohr said that the negative charges in an atom (the electrons) orbited the positively charged nucleus like the planets orbit the sun.

• Later models see the electrons in atoms as cloudlike and not as particles that orbit like planets around the sun.

Proton number is what determines the identity of an atom and is equivalent to its atomic number.

If you add the number of neutrons to the number of protons in an atom, you will get its mass number, equivalent to the number of massive particles in its nucleus (nucleons, Or protons and neutrons).

All neutral atoms have equal numbers of positive and negative charges.

Atoms with more electrons than protons have excess negative charge and are called Anions.

^ Atoms with too few electrons and excess positive charge are called Cations. i Atoms can decay via three processes:

• Alpha decay Involves the emission of a helium nucleus, resulting in a daughter atom with a decrease of two in the atomic number and four in the mass number.

• Beta decay Happens through three separate methods, all of which result in an atomic number change of one.

• Gamma decay Is simply the emission of a high-energy particle of light by an excited atom and does not result in any changes in the important numbers defining an atom.

^ Nuclear fusion Involves the joining of two light elements into one heavier element.

^ Nuclear fission Involves the splitting of an enormous atom into two smaller atomic pieces.

^ Electron configurations Show the locations of the electrons in an atom, organized by shells and subshells:

• Electrons always occupy these shells and subshells from lowest energy to highest, often according to the Aufbau diagram.

• Atoms of elements in the copper and chromium families, however, are prone to "steal" an electron from their neighboring S Orbitals in order to achieve a half full or completely full D Orbital, which is the more stable state. These elements have Exceptional electron configurations, Which differ slightly from those predicted by the Aufbau diagram.

• Unruly electrons can jump to higher energy levels by absorbing energy, which they must then emit in order to reach the Ground state. They release this energy in the form of particles of light called Photons, Which have very specific Wavelengths.

Testing Your Knowledge

Now that you’ve absorbed everything you need to know about atoms, try answering some questions. Remember that you can look back at the formulas, variables, and constants, listed on the Cheat Sheet at the front of this book, but try not to look back at the rest of the text until after you check your answers.

The best way that you can practice is by going through the questions provided in this chapter in one sitting and then checking them at the end to identify which areas you still need to review.

Questions 1 through 4 refer to atoms of the following elements.

1. In the ground state, has two electrons in one (and only one) of the P Orbitals.

2. Has the largest atomic radius.

3. Has the largest value of the first ionization energy.

4. Has the smallest second ionization energy.

5. The half-life of 3H is about 12 years. How much of a 4mg sample will remain after 36 years?

(A) 0.25mg

(B) 0.5mg

(C) 1mg

(D) 2mg

(E) 4mg

6. All of the following statements about the oxygen family of elements are true Except:

(A) The electron configuration of the valence shell of the atom is ns2np4.

(B) It contains all nonmetals.

(C) The atomic radii decrease with decreasing atomic number.

(D) Electronegativity increases with decreasing atomic number.

7. Which of the following is the correct electron configuration for molybdenum?

(A) 1s22s22p63s23p64s23d104p65s24d*

(B) [Kr] 5s24d4

(C) 1s22s22p63s23p64s23d104p65s14d5

(D) [Kr] 5S14D4

8. Which of the following sequence of decays might lead to the creation of

234 Pa

91

From

238 U

92

(A) Alpha then gamma decay

(B) Alpha then beta decay

(C) Alpha decay only

(D) Beta decay only

9(a). There are three common isotopes of naturally occuring magnesium as indicated in the table

Using the information above, calculate the average atomic mass of magnesium.

9(b). A major line in the emission spectrum of magnesium corresponds to a wavelength of

518.3nm. Calculate the energy in Joules of the transition resulting in the emission of that spectral line.

The table below shows the first three ionization energies for two mystery elements.

First Ionization Second Ionization Third Ionization Energy (kJ mol’1) Energy (kJ mol’1) Energy (kJ mol’1)

Element 1 520 7300 11820

Element 2 900 1760 14850

10(a). Elements 1 and 2 are both in the second period. What are their identities? Explain your reasoning.

10(b). Write the full electron configuration for element 1 after it has been ionized once. What other element shares the same electron configuration in its neutral state?

Checking Your Work

You’ve done your best. Now check your work. Make sure to read the explanations thoroughly for any questions you got wrong — or for any you got right by guessing.

1. (D). There are three P Subshell orbitals, and an atom that has been forced to double up on electrons in one and only one P Orbital must have four electrons in its P Subshell. The electron configuration of oxygen is 1 S22S22P4, So it fits the bill.

2. (A). Atomic radius decreases as you move to the right across a period, therefore the element in the list with the lowest atomic number (in other words, the farthest to the left), will have the largest atomic radius.

3. (E). The element with the largest value of the first ionization energy should be the one that is least likely to give up an electron. Halogens are the most likely candidates for high first ion-ization energies because they are the closest to achieving noble gas configurations through gaining electrons. Fluorine is the only halogen in the group.

4. (A). The element with the lowest second ionization energy should be one that is eager to give up its second electron. This corresponds to elements in the alkaline earth metal column, which need to be ionized twice in order to achieve noble gas configurations.

5. (B). Thirty-six years is equivalent to three half lives for 3H. This means that the amount of radioactive material in the sample will have decreased by half three times. This leaves 2mg at 12 years, 1mg at 24 years, and 0.5mg at 36 years.

6. (B). The first statement is true because elements in the oxygen family always have full S Subshells and four electrons in their P Subshells by virtue of sharing the same horizontal location in the periodic table. B is not true. Several of the oxygen family’s elements fall below the "staircase" marking the division between nonmetals and metals on the periodic table, indicating that they are metals or semimetals. C is a deceptively tricky statement because, while the atomic radius of elements in the same Family Do indeed decrease with decreasing atomic number, elements in the same period have precisely the opposite relationship, so you must be careful to keep them straight. D is also true because electronegativity increases to the right in a period and toward the top of a family.

7. (C). One might expect molybdenum to have the electron configuration shown in A and B (which are equivalent). However, molybdenum, being of the chromium family, falls under the umbrella of exceptional electron configurations. Its 4d orbital will abscond with an electron, which would normally belong to the 5S Orbital in order to make it more stable. This leaves you with the electron configuration 1 S22S22P63S23P64S23D104P65S14D5, Which can also be written as [Kr]5S14D5.

8. (B). The net result in the decay of

Is a decrease of four in mass number and a decrease of one in atomic number. Because neither beta nor gamma decay result in a decrease in mass number, an alpha decay must occur, which would leave you with

To then increase the atomic number by one from the initial alpha decay,

Must undergo a beta decay.

238

92

U

To

29314

91

Pa

9(a). 24.3amu. Multiply each mass by its percent abundance in decimal form, giving you

(24.0 x 0.79) + (25.0 x.010) + (26.0 x 0.11) = 24.32amu

9(b). 3.84 x 10-19J. This problem is a direct application of the equation E = Hv, and C = Xv. Plugging the speed of light as well as the given wavelength into the second equation and solving for frequency gives you

V = c = 3.0 x 10 Ms— = 5.79 x 1014 Hz X 518.3 x 10 9 M

Take this frequency value and plug it in turn into the first equation to give you the energy value of the spectral line in question:

E = hxu= 6.63x 10 34Jsx5.79x 1019Hz = 3.84x 10 14J

10(a). Element 1 is lithium and Element 2 is berylium. It requires roughly 14 times more energy to remove the second electron from element 1 than it did to remove the first, implying that the first electron was given up relatively easily and that the atom is very reluctant to give up another. This is consistent with the alkali metals, which only need to lose one electron to achieve a noble gas configuration. Once any element achieves noble gas configuration, it is reluctant to lose it by giving up another electron, explaining the large jump to the second ion-ization energy. Because you are told that these elements are in the second period, element 1 must be lithium. As for Element 2, it requires only twice as much energy to remove the second electron as was required to remove the first, but then it requires eight times more energy to remove the third. This implies that the first and second electrons are considerably easier to remove than the third, which is consistent with an alkaline earth metal. In this case, that metal must be berylium, which is the only alkaline earth metal in the second period.

10(b). 1 S2, Helium. The general electron configuration of lithium is 1 S22S1. When lithium is ionized, however, it loses an electron, achieving the electron configuration of the noble gas helium, whose electron configuration is simply 1 S2.

In This Chapter

Making sure you remember what’s important ^ Practicing with questions

^ Getting your questions answered and explained

LEscriptive chemistry, as you may have gathered from Chapter 23, is perhaps the most memorization-intensive area covered by the AP chemistry exam. But fight the impulse

To sweep it under the rug — having an intuitive feel for periodic trends and the broad patterns of common chemical reactions is an invaluable asset. Help gel your chemical intuition by reviewing the highlights of Chapter 23 and applying your insights to the practice questions provided here.

Keeping the Important Stuff Straight

Because descriptive chemistry includes a lot of "just-have-to-know" details, this summary of ideas from Chapter 23 is one that you’ll probably want to review well and often, even more than the summaries of many other chapters. To make the task easier, the bits and pieces are grouped into categories that reflect specific reaction types or specific perioidic patterns. Focus on one category at a time, attempting to master each one before moving on to the others. That way, you won’t lose your way in the swirl of details.

Precipitation reactions

Much of the previously mentioned memorization has to do with knowing which ionic compounds are soluble in water and which are not. You need this knowledge to predict the phases of products of ionic double displacement reactions, so you can properly write net ionic equations.

Here is a summary of solubility rules and the process of formulating net ionic equations:

Precipitation reactions can occur when ionic compounds react, typically in Double displacement Reactions.

Compounds that contain the following ions are soluble: NH4+, Group IA cations, NO3-, ClO3-, ClO2-, CH3COO-.

Compounds that contain the following ions are usually soluble: Cl-, Br-, and I – (except with Ag+, Pb2+, and Hg22+); SO42- (except with Ag+, Pb2+, Hg22+,Sr2+, Ca2+, and Ba2+).

Compounds that contain the following ions are usually insoluble: OH – (except with Group IA cations, Sr2+, Ca2+, and Ba2+); CO32-, PO43-, SO32-, and S2- (except with Group IA cations or NH4+).

Net ionic equations list only those components of the reaction that change state during the reaction. Spectator ions, those ions that appear in the same fashion on both the reactant and product sides, are omitted.

Patterns of common reaction types

Beyond precipitation, it pays to know the patterns of common reaction types that you’ll likely encounter on the AP exam, including gas-releasing reactions, complete combustion, reactions of oxides, and acid-base reactions.

Here is a concise summary of the patterns you can expect from the most commonly tested reaction types:

Gas-releasing decomposition can result from heating, electrolysis, and unstable acids (especially H2CO3 and H2SO4).

I Metals and metal hydrides can release hydrogen gas when they react with water, producing a basic solution (excess hydroxide ion).

I Combustion occurs when hydrocarbons and other compounds containing carbon-hydrogen bonds, such as alcohols, react with oxygen, forming carbon dioxide and water.

I Metal oxides often react with water to produce basic solutions. I Nonmetal oxides often react with water to produce acidic solutions. I Salts and oxides that contain transition metals are often colored. I Acids and bases

• A strong acid is a weak base. A strong base is a weak acid.

• The conjugate anion of a strong acid is a Very Weak base. The conjugate cation of a strong base is a Very Weak acid.

• The conjugate acid of a weak base is a weak acid. The conjugate base of a weak acid is a weak base.

• The conjugate acid of a weak base is a Stronger Acid Than the base And the conjugate base of of a weak acid is a Stronger Base Than the acid.

• Strong acids react with strong bases to form neutral salts and water.

• Salts of weak acids or bases react in water to form weakly acidic or basic solutions.

Periodic trends and group properties

The periodic table is periodic for a reason — the elements of the table are arranged to reflect recurring patterns in the properties of the elements. Those patterns emerge from trends in the number of valence electrons held by the elements. Although learning these patterns requires a bit of initial memorization, knowing them saves you a load of memorization in the long run.

Here are the key periodic trends and the properties of important groups of elements:

Electronegativity, electron affinity, and ionization energy tend to increase upward and to the right within the main body of the periodic table.

Atomic radius and metallic character tend to increase downward and to the left within the main body of the periodic table.

Alkali metals (Group IA) Are extremely reactive, and their cations are extremely unre-active; they are the least electronegative elements; they have an oxidation number of + 1 within compounds.

Hydrogen is not a metal: In compounds with nonmetals, hydrogen has oxidation number +1; in compounds with metals, hydrogen has oxidation number -1.

Alkaline earth metals (Group IIA) Are very reactive, and their cations are very unreac-tive; they have very low electronegativity; they have an oxidation number of +2 within compounds.

Halogens (Group VIIA) Are very electronegative and range in reactivity, with the most electronegative and most reactive at the top; they form ionic compounds with metals and molecular compounds with nonmetals; they usually have an oxidation number of -1 within compounds.

Noble gases (Group VIIIA) Are extremely unreactive. The larger atoms (Kr and Xe) Can Form compounds with strongly electronegative atoms such as fluorine and oxygen.

Testing Your Knowledge

You put in the hard work of climbing to the top of Chapter 23. Now peer from the heights, seeing what you can see from the summit. Don’t foget to make full use of the periodic table as you work these questions. The more you use it, the quicker you will find your way around it.

1. Which of the following statements are true about the combustion of propane, C3H8?

(I) Oxygen, O2, is produced.

(II) Total moles of products outnumber total moles of reactants.

(III) Products can react to form carbonic acid, H2CO3.

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) II and III only

2. Which element has the greatest electronegativity?

3. Which element has the most metallic character?

(A) Si

(B) Pb

(C) Fr

(D) K

(E) Mg

4. Which element has the lowest ionization energy?

(A) H

(B) Li

(C) Ne

(D) Cs

(E) Ra

5. Which of the following pairs of solutions produce a precipitate if mixed?

6. Which of the following reactants are least likely to yield a gaseous product?

(A) LiClO3 and heat

(B) Ca(OH)2 and heat

(C) NaH and H2O

(D) K and H+

(E) Li and H2O

7. Which mixture(s) will produce a solution with pH < 7?

8. Which of the following manipulations increase the solubility of calcium carbonate, CaCO3(s), In aqueous solution?

Questions 9 through 13 refer to the following set of choices. Pick the Best Answer in each case.

(A) Transition metals

(B) Strong acids conjugates

(C) Weak acids

(D) Salts

(E) Hydrides

9. Are weak bases 10. Often form colored compounds. 11. Result from neutralization reactions

12. Release hydrogen gas in the presence of wate

13. Are conjugates of weak base

14. Which of the following represent(s) a net ionic equation?

(I) Zn(s) + HCl(a<7) — ZnCl2(aq) + H2(g)

(II) CaCO3(s) + 2H+(aq) + 2Cl-(aq) — Ca2+(aq) + 2Cl-(aq) + H2O(l) + CO2(g)

(III) Ag+(a< ) + Cl-(a< ) — AgCl(s)

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) II and III only

Questions 15 through 16 refer to the diagram of an electrolytic cell in Figure 24-1:

- +

Figure 24-1:

Electolytic cell.

N ri

15. Electrons travel out of the liquid along this wire:

(A) 1

(B) 2

(C) 1 and 2 simultaneously

(D) Neither 1 nor 2

(E) Alternates between 1 and 2

16. If molten LiCl is the liquid indicated by 5, then the following is true:

(I) Solid metal may accumulate at 3.

(II) Cl2(g) may bubble from 4.

(III) Li2(g) may bubble from 3.

(A) I only

(B) II only

(C) III only

(D) I and II only

(E) II and III only

1

2

3

4

5

Checking Your Work

If you feel like a few of your answers may have been less than perfect, check them against the answers and explanations below. If you’re certain that you nailed the questions, check your answers anyway, if only to revel in your perfection.

1. (E). Combustion of a hydrocarbon (like propane) involves a combination reaction with oxygen gas, yielding carbon dioxide and water products. Here’s the balanced equation for the combustion of propane:

C3H8(g) + 5O2(g) —3CO2(g) + 4H2O(I)

From the balanced equation, it’s clear that oxygen is consumed, not produced. Moreover, the moles of products (3 + 4 = 7) outnumber the moles of reactants (1 + 5 = 6). Finally, the carbon dioxide and water products can undergo a further reaction to form carbonic acid:

CO2(g) + H2O(I) — H2CO3(ag)

2. (D). Overall, electronegativity increases upward and to the right on the periodic table.

3. (C). Overall, metallic character increases downward and to the left on the periodic table.

4. (D). Overall, ionization energy increases upward and to the right on the periodic table (just like electronegativity).

5. (E). HBr and Ca(OH)2 are a strong acid and a strong base, respectively, that undergo a neutralization reaction to produce water and the soluble salt, CaBr2. Although the nitrate salt of lead is soluble (like virtually all nitrate salts), most other lead salts (like PbI2) are insoluble. Although ammonium phosphate is soluble (like virtually all ammonium salts), only Group IA and ammonium salts of phosphate are soluble — other phosphate salts (like Mg3(PO4)2) are insoluble.

6. (B). Chlorate (ClO3-) salts tend to decompose with heat, yielding oxygen gas. Ionic compounds containing hydroxide (OH-) do not tend to decompose with heat. Metal hydrides (like NaH) tend to produce hydrogen gas when they react with water. Reactive metals (like Group IA and Group IIA metals) tend to react with acid (H+) to produce hydrogen gas. Reactive metals also tend to react with water to produce hydrogen gas.

7. (A). Nonmetal oxides (like P4O10) tend to react with water to produce acidic (pH < 7) solutions:

P4O10(g) + 6H2O(I) — 12H+(a<7) + 4PO43"(a<7)

Metal oxides (like MgO) tend to react with water to produce basic (pH > 7) solutions: MgO(s) + H2O(I) — Mg2+(a<7) + 2OH"(a<7)

Metal hydrides (like SrH2) tend to react with water, releasing hydrogen gas and producing a basic solution:

SrH2(s) + 2H2O(I) — Sr2*(aq) + 2OH"(ag) + 2H2(g)

8. (B). Decreasing temperature usually decreases the solubility of solid solutes. Adding CO32-anion by means of sodium carbonate decreases the solubility of CaCO3 due to the common ion effect, in which the added carbonate shifts the equilibrium toward undissolved CaCO3.

Decreasing pH increases CaCO3 solubility because the added H+ undergoes a secondary reaction with CO32-, shifting the overall equilibrium toward dissolved CaCO3:

CaCO3(s) + H2O(I) — Ca2+(ag) + CO32′(aq)

H*(aq) + CO32"(ag) — HCO3-(ag)

9. (B). Conjugates of strong acids are Very Weak bases.

10. (A). Transition metals are notorious for forming intensely colored coordination complexes.

11. (D). In a neutralization reaction, acid and base reactants react to form a salt and water.

12. (E). Hydrides (especially metal hydrides) tend to react with water to produce hydrogen gas, H2.

13. (C). Weak acids are the conjugate acids Of Weak bases.

14. (C). To generate a net ionic equation, list dissolved (soluble) compounds as dissociated ions. Then, cancel any spectator ions, which are those ions that appear in identical form on both sides of the reaction equation. What remains is the net ionic equation. For example:

AgNO3(ag) + NaCl(a<7) — NaNO3(ag) + AgCl(s)

Ag+(ag) + NO3"(a<7) + Na*(aq) + Cl~(aq) — Na*(aq) + NO3"(ag) + AgCl(s) Ag+(ag) + Cl"(a<7) — AgCl(s)

15. (B). The positive and negative signs at the top of the figure indicate the polarity of the voltage source for the electrolytic cell. Because electrons have negative charge, they migrate out of the liquid toward the positive pole along 2, and into the liquid away from the negative pole along 1.

16. (D). If the liquid indicated by 5 is molten LiCl, then the salt is dissociated into Li+ cations and Cl – anions. Molten salts can be decomposed by electrolysis in this kind of cell. The electrode at 3 acts as a cathode, delivering electrons to Na+( ) cations, which deposit there as Na(s) metal. The electrode at 4 acts as an anode, accepting electrons from Cl-(l) anions, which react with one another to form Cl2(g) at the electrode.