Burdge/Overby, Chemistry: Atoms First, 2e FM | Page 13

PREFACE xiii

provides an exercise that further probes the student’s conceptual understanding of the material.
Practice Problems C are new to this edition and many employ concept and molecular art. The
regular use of the Worked Example and Practice Problems in this text will help students develop a
robust and versatile set of problem-solving skills.
Section Review. Every section of the book that contains Worked Examples and Practice Problems
ends with a Section Review. The Section Review enables the student to evaluate whether they
understand the concepts presented in the section.
Key Skills. Located between chapters, Key Skills are easy to find review modules where students
can return to refresh and hone specific skills that the authors know are vital to success in later chapters. The answers to the Key Skills can be found in the Answer Appendix in the back of the book.
REVISED PAGES

REVISED PAGES
S e C t Io N 5.5

Having determined molecular
157
Covalent Bonding and Molecules geometry, we determine overall polarity of each molecule by examining the individual bond dipoles and
their arrangement in three-dimensional space.

F
F

S
F

polyatomic ion, we use a stepwise procedure:
1. Draw a correct Lewis structure [? Chapter 6 Key Skills].
2. Count electron domains. Remember that an electron domain is a lone pair or a bond; and that a bond may be a single bond, a double
bond, or a triple bond.
3. Apply the VSEPR model to determine electron-domain geometry.
4. Consider the positions of atoms to determine molecular geometry (shape), which may or may not be the same as the electron-domain
geometry.

Determine whether or not
the individual bonds are
polar.

F

F

F

F

Ethanol

Draw the Lewis structure
S
S
Strategy Refer to the labels on the atoms (or see TableF5.3).
F
F

F

H

F

F

F

Cl
C

Cl

H

Setup There are two carbon atoms, six hydrogen atoms, and one oxygen atom, so the subscript on C
will be 2 and the subscript onelectron domains: and there willdomains: subscript on O. domains:
6 H will be 6,
5 electron be no
4 electron
Count electron domains

on the
Solution Ccentral atom
2H6O

think about to determine
Apply VSEPR It

• six bonds

• four bonds
• one lone pair

• four bonds

6 electron domains
arrange themselves in an

5 electron domains
arrange themselves in

4 electron domains
arrange themselves in

electron-domain geometry
Often the molecular formula for a compound such astrigonal bipyramid.
octahedron.
a ethanol (consisting of carbon, hydrogen, and
a tetrahedron.
oxygen) is written so that the formula more closely resembles the actual arrangement of atoms in the
Cl
F
molecule. Thus, the molecular formulaFforFethanol is commonly written as C2H5OH.
F
F
F

S

F

F

S

H

C

Cl

H

F
F
Practice Problem A t t e m p t Chloroform was used as an anesthetic for childbirth and surgery

during the nineteenth century. Write pairs on the
With no lone the molecular formulas for one of
The lone pair occupies (a) chloroform, and (b) the
With no lone pairs on acetone
Consider positions of atoms
central atom, below.
molecular
the molecular
based to determine molecular model shownthesame as the the equatorial positions, making central atom,the same as the
on the molecular
geometry is the
the molecular geometry:
geometry is
geometry

Consider the arrangement
of bonds to determine
which, if any, dipoles
cancel one another.

electron-domain geometry:
Octahedral

See-saw shaped.

electron-domain geometry:
Tetrahedral

Cl

F
C

S
F

F
The dipoles shown in red
cancel each other; those
shown in blue cancel each
other; and those shown in
green cancel each other,
SF6 is nonpolar.

Cl
H

C, H, and Cl have
electronegativities of 2.5, 2.1,
and 3.0, respectively. The
individual bonds are polar.
Bond dipoles are represented
with arrows.

F
F

S

F

F

As in SF6, the individual bonds
in SF4 are polar. The bond
dipoles are represented with
arrows.

F
F

H
F

S and F have
electronegativities of 2.5
and 4, respectively.
[? Figure 6.4, page 188]
Therefore the individual
bonds are polar and can be
represented with arrows.

Consider the examples of SF6, SF4, and CH2Cl2. We determine the molecular geometry as follows:

C

S

F
Molecular polarity is tremendously important in determining the physical
substance.
Write the molecular formula of ethanolmolecular geometry.and chemical properties of a or shape ofIndeed, molecular
based on its ball-and-stick model, shown here. or
polarity is one of the most important consequences of
To determine the geometry
a molecule

Cl

F
F

F

worked example 5.5
Molecular Shape and Polarity

F

H
F

The dipoles shown in green
cancel each other; but the
dipoles shown in red—
because they are not directly
across from each other—
do not. SF4 is polar.

Cl
H

Although the bonds are
symmetrically distributed, they
do not all have equivalent
dipoles and therefore do not
cancel each other.
CH2Cl2 is polar.

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