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The material in this booklet corresponds to Chapter 13 in Zumdahl, “Chemical Principles”,
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“Chemical Principles” pages 65-103.
What professors are saying about…
Chemical Principles: The Quest for Insight, Second Edition
By Peter Atkins and Loretta Jones
“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)
“The writing style is excellent and Atkins/Jones should be proud of this. In some sections it reads almost like a
novel. Good job! The sections I reviewed are quite superior to Zumdahl’s text in writing style.”
—Dr. Robert A. Kolodny, Armstrong Atlantic State University (former Zumdahl user)
2
and here is why…
The goal of Chemical Principles is to help students develop “chemical
insight”—the ability to see matter through chemists’ eyes, and to make
connections between chemical principles, theory, experimentation and
the world around us. Many of the new features in the second edition
have been designed with this goal in mind.
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
see: www.whfreeman.com/chemicalprinciples
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
Chemical Principles: The Quest for Insight, Second Edition
Peter Atkins • Lincoln College, Oxford University
Loretta Jones • University of Northern Colorado
Please browse through the text and consider the benefits of some of these unique features!
Order the text for your class using ISBN 0-7167-3923-2
3
CHAPTER
CHEMICAL BONDS
How do we do that?
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
z2e2
z2e2
z2e2
z2e2
...
40
d
2d
3d
4d
z2e2
1
1
1
1 ...
40d
2
3
4
2 2
ze
ln 2
40d
EP In the last step, we used the relation 1 21 13 14 . . . 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
EP 2 ln 2 z2NAe2
40d
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:
EP A z1z2NAe2
40d
(2)
The factor A is a positive numerical constant called the Madelung constant; 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.
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
The symbol x means the “absolute
value” of x, the value of x without
its sign; so 2 2.
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?
STRATEGY 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.
SOLUTION All the constants cancel when the ratio is taken, including the
ionic charges, and we are left with
EP(NaCl)
d(KCl)
EP(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:
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
To check on the validity of a Lewis structure, verify
that each atom has an octet or a duplet.
H
1
C
2
N
H
C
3
N
5
4
FIGURE 2.8 The five steps used to write the Lewis
structure of HCN, as described in this Toolbox.
5
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.
STRATEGY 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.
(c) ClO2
20
4
6
10
HOH
H
CO
H
O Cl O
HOH
H
CO
H
OClO
H
C O
H
OClO
O Cl O
2 Layout of structure
3 Electron pairs between
bonded atoms
Electrons not yet located
4 Lone pairs and
multiple bonds
HOH
Number of electron pairs
(b) H2CO
12
1 Number of valence
electrons
(a) H2O
8
SOLUTION
H
C
O
O
H
5 Lines to represent bonds
H
H
Self-Test 2.5B
6
Write a Lewis structure for NH3.
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 .]
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.
H
H
O
H
..
O
..
C
..
C
..
H
H
..
O
..
..
H
..
O
..
H
H
..
..
..
C
..
C
..
O
..
..
H
(c)
..
C
C
H
O
..
H
..
H
..
O
H
(d)
9 Acetic acid, CH3COOH
..
..
N
H
..
O
H
Write a Lewis structure for hydrazine, H2NNH2.
..
H
Self-Test 2.6B
C
(b)
C 24 8
H 41 4
O 2 6 12
Total
24
Write a Lewis structure for the urea molecule, (NH2)2CO.
[Answer: See (10).]
C
..
..
H
SOLUTION Step 1 The total number of valence electrons is
Self-Test 2.6A
O
(a)
STRATEGY 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.
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).
H
C
..
N
H
H
10 Urea, (NH2)2CO
7
EXERCISES
Ionic Bonds
Covalent 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.
2.35 Write the Lewis structure of (a) CCl4; (b) COCl2;
(c) ONF; (d) NF3.
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.
2.37 Write the Lewis structure of (a) tetrahydridoborate
ion, BH4; (b) hypobromite ion, BrO; (c) amide ion,
NH2.
2.3 Explain why the lattice energy of lithium chloride
(861 kJmol1) is greater than that of rubidium
chloride (695 kJmol1), 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 kJmol1) is greater than that of silver iodide
(887 kJmol1). See Appendix 2D.
2.36 Write the Lewis structure of (a) SCl2; (b) AsH3;
(c) GeCl4; (d) SnCl2.
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
C
8
C
C
C
C
H
C
C
C
H
H
C
C
C
C
C
C
H
H
H
H
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 gmol1. Write the
Lewis structure of this compound.
Media Link
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 kJmol1, 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?
Media Link
INTEGRATED EXERCISES
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 (kJmol1)
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 kJmol1. 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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10
11
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Chemical Principles: The Quest for Insight, Second Edition
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