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“Overall, I liked teaching from Atkins & Jones, Chemical Principles, 2nd Edition. I was particularly happy that I could ask students to read sections of Atkins and Jones on their own and they were reasonably able to do so. I was very excited by the order of materials in the text, the “atoms-first” approach. I still think this is a fundamentally great way to teach chemistry.”
—Dr. Nancy Doherty, U of California, Irvine (former Oxtoby user)
“Clearly superior.”
—P.S. Braterman, U of North Texas (Oxtoby user)
“The text seems to be addressed to novices, but novices who are treated as intelligent readers. I would say that a definite plus of the Atkins/Jones text, measured against the Zumdahl version we use, is the extensive Fundamentals section. Other plusses are the extra detail in such areas as atomic structure. Both books offer clear examples; the
Atkins/Jones text has more variety of styles, eg., Toolboxes and Illustrations.”
—Ronald P. Drucker, Ph.D., City College of San Francisco (Zumdahl user)
“I think that the Atkins/Jones book is superior to the Oxtoby text in the coverage of what is traditionally called descriptive chemistry. The writing style is very good as is typical of the books that Atkins has written. The tone is quite appropriate. In fact, this is one of the strongest features of the book. It is not so tightly written as Oxtoby which may be a significant advantage for most students at this level.”
—Larry C. Thompson, U of Minnesota, Duluth (Oxtoby user)
“The writing style is excellent and the tone is appealing – brisk and encouraging. I consider Atkins/Jones superior to the book used previously (Zumdahl) in that it is generally clearer and gives a better intuitive feel for the subject matter. The attempt to provide insight is evident and largely successful.”
—Mario E. Baur, UCLA (former Zumdahl user)
“We switched from Oxtoby, Gillis and Nachtrieb to Atkins/Jones because: 1) I thought the student and instructor
CD’s in A/J were more useful, 2) I liked the A/J problems a little better (more thorough), 3) AJ covers some specialized topics (materials, atmospheric chemistry, and hemoglobin) that are missing in Oxtoby, and that I include in my course,
4) the order of presentation of the material was a little better for our fall quarter course in A/J.”
—George C. Schatz, Northwestern University (former Oxtoby user)
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—Dr. Robert A. Kolodny, Armstrong Atlantic State University (former Zumdahl user)

An atoms first organization builds understanding by starting with the behavior of atoms and molecules and moving up to more complex properties and interactions.
Our first three chapters are:
1. Atoms: The Quantum World
2. Chemical Bonds
3. Molecular Shape and Structure
Fundamentals Sections preceding the first chapter offer a quick review of basic chemistry that students should know coming into the course.
• Each section contains exercises to test student comprehension
• On-line quiz of all topics contained in Fundamentals sections at www.whfreeman.com/chemicalprinciples
Integrated Media with Media Links lead students to the books web site where they will find:
• living graphs that can be manipulated
• animations of molecular level processes
• lab videos showing techniques and experiments you may not be able to perform in class
• online tools like curve plotters and graphing functions and examples of how to use them
• media based end-of-chapter exercises that enable students to solve problems using media
An Emphasis on Contemporary Chemistry that shows students how chemists propose models, test them through experimentation, and use their knowledge professionally.
• Major Techniques inter-chapters covering important experimental methods. Pages 110, 168, 270, 456, 984, and 1028
• Impact on Materials and Impact on Biology Sections addressing the burgeoning fields of materials chemistry and biochemistry. Pages 53, 257, 671, 900,1012
• Frontiers of Chemistry boxed essays explore modern chemical research and end with a How Might You Contribute features challenge students to solve global problems using chemistry. Pages 118, 255, 414, 640, 646
• Lincoln College, Oxford University
• University of Northern Colorado
Please browse through the text and consider the benefits of some of these unique features!
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C
C
B
To calculate the change in potential energy when a solid forms from gaseous ions, we consider a simple one-dimensional model of a solid.
Imagine a long line of uniformly spaced alternating cations and anions, with d the distance between their centers, the sum of the ionic radii
(Fig. 2.3). If the charge numbers of the ions have the same absolute value
( 1 and 1, or 2 and 2, for instance), then z1 z, z2 z, and
z1z2 z2. The potential energy of the central ion is calculated by summing all the Coulombic potential energy terms, with negative terms representing attractions to oppositely charged ions and positive terms representing repulsions from like-charged ions. For the interaction with ions extending in a line to the right of the central ion, the lattice energy is
1
E
P
4
4 z
0
2 e
2
4 z 2 e
0
2 d
1
0 d z 2 e 2 d ln 2
1
2 z 2 e 2
2d
1
3
1
4 z 2 e 2
3d
. . . z 2 e 2
4d
. . .
In the last step, we used the relation 1 1
2
1
3
1
4
. . . ln 2. Finally, we multiply EP by 2 to obtain the total energy arising from interactions on each side of the ion and then by the Avogadro constant, NA, to obtain an expression for the lattice energy per mole of ions. The outcome is
E
P
2 ln 2 z 2 N
A e 2
4
0 d with d rcation ranion. This energy is negative, corresponding to a net attraction. The calculation we have just performed can be extended to three-dimensional arrays of ions with different charges:
E
P
A z
1 z
2
N
A e 2
4
0 d
(2)
The factor A is a positive numerical constant called the Madelung con-
stant; its value depends on how the ions are arranged about one another.
For ions arranged in the same way as in sodium chloride, A 1.748.
The symbol x means the “absolute value” of x, the value of x without its sign; so 2 2.
FIGURE 2.3
The arrangement used to calculate the potential energy of an ion in a line of alternating cations and anions. We concentrate on one ion, the
“central” ion denoted by the vertical dotted line.
4
+ z − z d

EXAMPLE 2.1
Estimating the relative lattice energies of solids
The ionic solids NaCl and KCl form the same type of crystal structure. In which solid are the ions bound together more strongly by coulombic interactions?
S
TRATEGY
We can decide in which compound the ions bind together more strongly by taking the ratio of EP for the two compounds. The two solids have the same crystal structures, so they have the same values of the
Madelung constant A. Form the ratio of the two expressions for the potential energy, using Eq. 2.
S
OLUTION
All the constants cancel when the ratio is taken, including the ionic charges, and we are left with
E
P
(NaCl)
E
P
(KCl)
d(KCl)
d(NaCl)
Because the sodium ion has a smaller radius, d(NaCl) d(KCl), we can conclude that EP(NaCl) is larger than EP(KCl) and, therefore, that the ions are bound more strongly in NaCl than in KCl.
Self-Test 2.1A
The ionic solids CaO and KCl crystallize to form structures of the same type. In which compound are the interactions between the ions stronger?
[Answer: CaO, higher charges and smaller radii]
Self-Test 2.1B
The ionic solids KBr and KCl crystallize to form structures of the same type. In which compound are the interactions between ions stronger?
TOOLBOX 2.1
How to write the Lewis structure of a polyatomic species
Conceptual Basis
We look for ways of using all the valence electrons to complete the octets (or duplets).
of H) of each atom by placing any remaining electron pairs around the atoms. If there are not enough electron pairs, form multiple bonds.
Step 5 Represent each bonded electron pair by a line.
Procedure
We take the following steps, which are illustrated for
HCN in Fig. 2.8:
To check on the validity of a Lewis structure, verify that each atom has an octet or a duplet.
C
H
N
1
2
Step 1 Count the number of valence electrons on each atom. Divide the total number of valence electrons in the molecule by 2 to obtain the number of electron pairs.
Step 2 Predict the most likely arrangements of atoms by using common patterns and the clues indicated in the text, then write the chemical symbols of the atoms to show their layout in the molecule.
Step 3 Place one electron pair between each pair of bonded atoms.
Step 4 Complete the octet (or duplet, in the case
H
C
N
3
5 4
FIGURE 2.8
The five steps used to write the Lewis structure of HCN, as described in this Toolbox.
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EXAMPLE 2.3
Writing the Lewis structure of a molecule or an ion
Write the Lewis structures of (a) water, H2O; (b) methanal, H2CO; and
(c) the chlorite ion, ClO2 .
S
TRATEGY
We can write the Lewis structures of molecules and ions by following the steps in Toolbox 2.1 and remembering to adjust the number of electrons for ions. Therefore, we add one electron for the negative charge of ClO2 . Find the central atom by inspecting formulas. In methanal, the central atom must be C, because it has a central position and the H atoms can each form only one bond.
S
OLUTION
(a) H2O (b) H2CO
8 12
(c) ClO2
20 1 Number of valence electrons
Number of electron pairs 4 10
2 Layout of structure
3 Electron pairs between bonded atoms
H
H O H
O H
6
H
C
H
H
C O
H
O
O Cl O
O Cl O
Electrons not yet located
4 Lone pairs and multiple bonds
5 Lines to represent bonds
H O
O
H
H
C
H
H
O
O
H
O
O
Cl
Cl
O
O
Self-Test 2.5A
Write a Lewis structure for the cyanate ion, CNO . (In this case, the C atom is in the center.)
[Answer: N C O .]
Self-Test 2.5B
Write a Lewis structure for NH3.
6

EXAMPLE 2.4
Writing Lewis structures for molecules with more than one “central” atom
Write the Lewis structure for acetic acid, CH3COOH, a common example of a carboxylic acid. Acetic acid is the acid in vinegar and is formed when the ethanol in wine is oxidized. In the
∫
COOH group, both O atoms are attached to the same C atom, and one of them is bonded to the final H atom. The two C atoms are bonded to each other.
S
TRATEGY
Follow the steps in Toolbox 2.1. The formula for acetic acid suggests that it consists of two groups, with the central C atoms joined together: a CH3 ∫ group and a
∫
COOH group. We anticipate that the
CH3 ∫ group, by analogy with methane, will consist of a C atom joined to three H atoms by single bonds.
S
OLUTION
Step 1 The total number of valence electrons is
C 2 4
H 4 1
O 2 6 12
8
4
Total 24 so the molecule has 12 valence electron pairs. Step 2 The atomic arrangement in the molecule, which is suggested by the way the molecular formula is written, is shown in (9a); the linked atoms are indicated by the pale gray rectangles. Step 3 We use seven electron pairs to link neighboring atoms, as shown in (9b). Five pairs remain. Step 4 To complete the octets, we arrange electron pairs so that each atom has eight electrons: this can be achieved by adding two lone pairs to each oxygen atom and allowing the terminal oxygen atom to form a double bond to the carbon atom, as depicted in (9c). Step 5 The final Lewis structure is shown in (9d).
Self-Test 2.6A
Write a Lewis structure for the urea molecule, (NH2)2CO.
[Answer: See (10).]
Self-Test 2.6B
Write a Lewis structure for hydrazine, H2NNH2.
H
H
C
H
O
C
O H
(a)
H
H
..
C
..
H
O
..
C
..
O H
(b)
H
H
..
C
..
H
O
..
C
..
H
(c)
H O
H C C
H H
(d)
9 Acetic acid, CH
3
COOH
H
..
N
O
C
..
N
H H
10 Urea, (NH
2
)
2
CO
H
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EXERCISES
Ionic Bonds
2.1 Using data in Appendix 2D, predict which of the following pairs of ions would have the greater coulombic attraction in a solid compound: (a) K , O2 ;
(b) Ga3 , O2 ; (c) Ca2 , O2 .
2.2 Using data from Appendix 2D, predict which of the following pairs of ions would have the greater coulombic attraction in a solid compound: (a) Mg2 ,
S2 ; (b) Mg2 , Se ; (c) Mg2 , O2 .
2.3 Explain why the lattice energy of lithium chloride
(861 kJ mol
1) is greater than that of rubidium chloride (695 kJ mol
1), given that they have similar arrangements of ions in the crystal lattice.
See Appendix 2D.
2.4 Explain why the lattice energy of silver bromide
(903 kJ mol
1) is greater than that of silver iodide
(887 kJ mol
1). See Appendix 2D.
Covalent Bonds
2.35 Write the Lewis structure of (a) CCl4; (b) COCl2;
(c) ONF; (d) NF3.
2.36 Write the Lewis structure of (a) SCl2; (b) AsH3;
(c) GeCl4; (d) SnCl2.
2.37 Write the Lewis structure of (a) tetrahydridoborate ion, BH4 ; (b) hypobromite ion, BrO ; (c) amide ion,
NH2 .
2.38 Write the Lewis structure of (a) nitronium ion,
ONO ; (b) chlorite ion, ClO2 ; (c) peroxide ion, O22 ;
(d) formate ion, HCO2 .
2.39 Write the complete Lewis structure for each of the following compounds: (a) ammonium chloride;
(b) potassium phosphide; (c) sodium hypochlorite.
2.40 Write the complete Lewis structure for each of the following compounds: (a) zinc cyanide; (b) potassium tetrafluoroborate; (c) barium peroxide (the peroxide ion is O22 ).
2.41 Write the complete Lewis structure for each of the following compounds: (a) formaldehyde, HCHO, which as its aqueous solution “formalin” is used to preserve biological specimens; (b) methanol, CH3OH, the toxic compound also called wood alcohol;
(c) glycine, CH2(NH2)COOH, the simplest of the amino acids, the building blocks of proteins.
2.42 Write the Lewis structure of each of the following organic compounds: (a) ethanol, CH3CH2OH, which is also called ethyl alcohol or grain alcohol; (b) methylamine,
CH3NH2, a putrid-smelling substance formed when flesh decays; (c) formic acid, HCOOH, a component of the venom injected by ants.
2.43 Anthracene has the formula C14H10. It is similar to benzene but has three six-membered rings that share common C ∫ C bonds, as shown below.
Complete the structure by drawing in multiple bonds to satisfy the octet rule at each carbon atom.
Resonance structures are possible. Draw as many as you can find.
H H H
H H
H
C
C
C
C
H
C
C
C
C
H
C
C
C
C
H
C
C
H
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INTEGRATED EXERCISES
2.90 For a given formula, it is possible to have a number of reasonable Lewis structures where the atoms have different connectivities (a term meaning that different atoms are bonded to each other to form the various structures). Compounds with the same formula but different connectivities are known as isomers.
Hydrogen atoms in benzene can be replaced by other atoms such as the halogens. Draw Lewis structures that obey the octet rule but have different arrangements of atoms for the compound dichlorobenzene, C6H4Cl2.
2.91 A highly toxic gas that has been used in chemical warfare gives the following elemental analysis figures:
12.1% carbon, 16.2% oxygen, and 71.7% chlorine by mass, and its molar mass is 98.9 g mol
1. Write the
Lewis structure of this compound.
2.92 White phosphorus is composed of tetrahedral molecules of P4 in which each P atom is connected to three other P atoms. Draw the Lewis structure for this molecule. Does it obey the octet rule?
2.98 (a) Show that lattice energies are inversely proportional to the distance between the ions in MX (M alkali metal, X halide ion) by plotting the lattice energies of KF, KCl, and KI against the internuclear distances dM-X. The lattice energies of KF,
KCl, and KI are 826, 717, and 645 kJ mol
1, respectively.
Use the ionic radii found in Appendix 2D to calculate
dM-X. How good is the correlation? You should use a standard graphing program to make the plot that will generate an equation for the line and calculate a correlation coefficient for the fit (see the Web site for this book). (b) Estimate the lattice energy of KBr from your graph. (c) Find an experimental value for the lattice energy of KBr in the chemical literature and compare that value to that calculated in (b). How well do they agree?
2.99 (a) Show that the lattice energies of the alkali metal iodides are inversely proportional to the distances between the ions in MI
(M alkali metal) by plotting the lattice energies given below versus the internuclear distances dM-I.
Alkali metal iodide Lattice energy (kJ mol
1)
LiI
NaI
KI
RbI
CsI
759
700
645
632
601
Use the ionic radii found in Appendix 2D to calculate
dM-X. How good is the correlation? You should use a standard graphing program to make the plot that will generate an equation for the line and calculate a correlation coefficient for the fit (see the Web site for this book). (b) From the ionic radii given in Appendix 2D and the plot given in part (a), estimate the lattice energy for silver iodide. (c) Compare your results from part (b) to the experimentally determined value of 886 kJ mol
1. If they do not agree, provide an explanation for the deviation.
2.105 An important aspect of the structures of chemical compounds is whether the atoms within a molecule are equivalent or different. For example, all the hydrogen atoms in methane are equivalent: they all exist in an identical environment. On the other hand, the fluorine atoms in a molecule such as PF5 are not all equivalent.
The two fluorine atoms that lie in the axial positions are equivalent to each other but are different from the three fluorine atoms that lie in the equatorial plane. Those three fluorine atoms are all equivalent to each other. The property of the equivalence (or nonequivalence) of atoms in molecules is known as molecular symmetry. Using your knowledge of Lewis structures, predict how many different types of hydrogen atoms will be found in the following molecules: (a) C2H2; (b) C2H4; (c) C2H3Cl;
(d) cis-C2H2Cl2; (e) trans-C2H2Cl2. (f) In the molecule
C2H5Cl, the hydrogen atoms could all be different; yet by a variety of experimental techniques, only two types of hydrogen atoms are found. Propose an explanation.
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