Solutions: Chapter 8

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Section Review 8.1
Part A Completion
1. compound 6. low
2. molecular 7. high
3. nonmetals 8. atoms
4. diatomic
5. molecular formula
9. structure
Part B True-False
10. ST 12. AT 14. ST
11. NT 13. NT
Part C Matching
15. e 17. d 19. c
16. a 18. b
Part D Questions and Problems
20. molecular compound
21. a. 4 carbon atoms, 10 hydrogen atoms
b. 6 carbon atoms, 5 hydrogen atoms,
1 fluorine atom
22. a. molecule d. atom
b. atom
c. molecule
e. molecule
Section Review 8.2
Part A Completion
1. stable electron
2. covalent
3. shared
4. single
5. unshared pairs
6. double/triple
7. coordinate covalent bond
8. Energy
9. bond dissociation energy
10. resonance structure
Part B True-False
11. NT 13. AT 15. ST
12. NT 14. AT 16. NT
Part C Matching
17. e 19. b 21. d
18. a 20. c
Part D Questions and Problems
22. a.
b.
c.
6-9Br 6-9Br 6
H 6 C 6 6 6 N 6
H
H 6-9N
6 H
H
Section 8.3
Part A Completion
1. molecular orbitals
2. bonding orbital
3. lower
4. sigma or
5. pi or
6. three-dimensional
7. VSEPR theory
8. orbital hybridization
Part B True-False
9. AT 11. NT 13. ST
10. ST 12. AT 14. AT
Part C Matching
15. d 17. a 19. c
16. e 18. b
Part D Questions and Problems
20. Reading from left to right:
sp 3 sp 2 sp 2 sp sp sp 3
Section 8.4
Part A Completion
1. equally
2. nonpolar
3. unequally
4. polar
5. electronegativities
6. dipole interactions
7. hydrogen bond
8. electronegative
9. oxygen, nitrogen, or fluorine
Answer Key 785

Part B True-False
10. NT 12. ST 14. AT
11. AT 13. ST 15. AT
Part C Matching
16. b 18. e 20. a
17. d 19. c
Part D Questions and Problems
21. dispersion forces, dipole interactions,
hydrogen bonds
22. a. ionic
b. polar covalent bonds
c. polar covalent bonds
d. nonpolar covalent bonds
Practice Problems 8
Section 8.1
1. a. atom d. molecule
b. molecule e. atom
c. molecule
2. a. not diatomic
b. diatomic
c. diatomic
d. not diatomic
e. diatomic
3. Molecular compounds are usually composed
from two or more nonmetallic elements.
4. A molecular structure gives information
about the kinds and numbers of atoms
present in a molecule.
5. Molecular compounds tend to have lower
melting and boiling points that that of ionic
compounds
Section 8.2
1. The two atoms share a pair of electrons in
order to form a single covalent bond.
H F
2. Phosphorous needs 3 more electrons to fill
the 3p orbitals. Fluorine needs one more
electron to fill its second energy level. Since
each fluorine atom only needs one electron
and phosphorus needs 3 electrons, three
fluorine atoms are required to bond with
phosphorus.
F P
F
786 Core Teaching Resources
F
3. Nitrogen needs 3 more electrons to fill its
second energy level. Chlorine needs one
more electron to achieve a noble gas
configuration. Because each chlorine atom
needs only one electron and nitrogen needs 3
electrons, three chlorine atoms are required
to bond with nitrogen.
4. Because carbon can form four single covalent
bonds, there is an apparent shortage of atoms
with which to bond. This is a clue that a
carbon-carbon multiple bond exists in this
compound. Each carbon atom shares one
electron with one of the two hydrogen atoms.
The remaining three electrons for each
carbon atom form a triple covalent bond. The
electron dot structure is:
HCCH
5. Carbon has 4 valence electrons and each of
the oxygens has 6 valence electrons. Two
additional electrons are added to account for
the ion having a 2 charge. The carbon and
oxygen can satisfy the octet rule by having the
oxygens bonded to a central carbon. There is
one double covalent bond between a carbon
and oxygen, which can shift to any one of the
carbon-oxygen bonds giving rise to three
resonance structures.
O
C
O O
Section 8.3
Cl N Cl
Cl
2 O
C
O O
1. The four fluorine atoms are covalently
bonded to the central carbon atom. The four
shared pairs of electrons repel each other to
the corners of a tetrahedron. All four bond
angles are 109.5°.
2. The four valence electron pairs repel each
other, but the unshared pair is held closer to
the phosphorus than the three bonding pairs.
The unshared pair repels the shared pairs
more strongly. Thus, the angle between
bonds is expected to be slightly smaller than
the tetrahedral bond angle of 109.5. The
actual bond angle for NH3 , a similar
molecule, is 107.
3. Boron forms three sp2 orbitals by mixing one
2s orbital and two 2p orbitals. The three sp2 orbitals lie in the same plane, 120 apart from
one another. Each sp2 orbital overlaps with an
2
O
C
O O
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved.

© Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved.
atomic orbital of chlorine to form three
equivalent sigma bonds.
4. To form four equivalent bonds, silicon mixes
one s orbital and all three of the p orbitals.
The hybridization in SiF4 is sp3 .
5. Oxygen is the central atom in this molecule. It
has 6 valence electrons, two of which are
bonding electrons. The other 4 electrons are
unshared pairs. These 2 unshared pairs repel
the two bonding pairs and prevent F2O from
being linear. The molecule is a bent triatomic
molecule, with a bond angle of approximately
104.5°. This angle is slightly smaller than the
tetrahedral bond angle because the two
unshared pairs repel each other more
strongly than the two shared pairs.
6. Because each carbon can form single
covalent bonds with four other atoms, there
exists in this compound an apparent
shortage of atoms with which to bond. This is
a clue that CH2CF2 contains a carbon-carbon
multiple bond. The two hydrogen atoms
bond with one carbon atom while the two
fluorine atoms bond with the other carbon.
Each carbon atom has two electrons left over.
These electrons form a carbon-carbon
double covalent bond. The molecule looks
very much like the ethene molecule (C2H4 ).
H2C2H and F2C2F bond angles of 120°, the
hybridization involved in the carbon-carbon
bond is sp2 .
H F
C C
H F
7. Carbon 1 mixes one s orbital and three p
orbitals to form four sp 3 hybrid orbitals,
which form 4 sigma bonds. Carbon 2 mixes
one s and two p orbitals to form three sp 2
hybrid orbitals, which overlap with the
hybrid orbitals of the carbon and oxygen
atoms to form three equivalent sigma bonds.
The non-hybridized p carbon orbital overlaps
with an oxygen p orbital to form one pi
bonding orbital.
Section 8.4
1. a. The difference in electronegativity
between Na and O is about 2.4 and the
bond is ionic.
b. With like atoms, the difference is zero and
the bond is nonpolar covalent.
c. The electronegativity difference between
P and O is about 1.4 and the bond is polar
covalent.
2. For a bond to be classified as nonpolar
covalent, like atoms must bond, as in
diatomic molecules. Most bonds are between
unlike atoms; therefore, they must be ionic or
polar covalent.
3. Both carbon dioxide and carbon monoxide
contain polar bonds. However, the effect of
the polar bond on the polarity of the entire
molecule depends on the shape of the
molecule. In carbon monoxide, there is a
partial positive pole and a partial negative
pole. Therefore, the molecule is a dipole. In
carbon dioxide, the carbon and oxygens lie
along the same axis. The bond polarities
cancel, producing a nonpolar molecule.
4. The more electronegative atom in a covalent
bond will have the symbol and the less
electronegative atom the symbol.
a.
H
H N H δ+
δ+
δ+ δ–
b.
δ–
F
Fδ–
δ+
C
δ–
F
Fδ–
5. CaO is an ionic compound and CS2 is a polar
covalent compound. Generally, ionic
compounds have much higher melting
points than molecular compounds.
Interpreting Graphics 8
1. O 3 C 3 O ; linear; 180°; none
2.
H H
; tetrahedral; 109.5°; none
C
H H
3. O O ; trigonal planar; 120°;
S
O
O
S
O O
4. F Be F ; linear; 180°; none
O
S
O O
O
S
O O
Answer Key 787