Chem 32
3.32
Solutions to Section 3.1 – 3.5 Homework Problems
Argon (Ar) is in Group 8A, so it has eight valence electrons and satisfies the octet rule.
3.34 Hydrogen does not behave like the other Group 1A elements, because it has only one
empty space in its valence shell. Hydrogen readily forms covalent bonds, while the other Group
1A elements make ions exclusively.
3.36 Choice a is correct. Of these three elements, carbon forms the most bonds, based on its
position on the periodic table. (Carbon makes 4 bonds, nitrogen makes 3, and bromine makes 1).
Therefore, we expect carbon to be the central atom. Note that you don’t need to draw a Lewis
structure to solve this problem.
3.38 When you draw dot structures, try to remember the typical bonding patterns for each
group of elements:
Hydrogen: makes 1 bond.
Group 4A: these atoms make 4 bonds and have no non-bonding electron pairs.
Group 5A: these atoms make 3 bonds and have 1 non-bonding electron pair.
Group 6A: these atoms make 2 bonds and have 2 non-bonding electron pairs.
Group 7A: these atoms make 1 bond and have 3 non-bonding electron pairs.
H
a) H
b)
P
(Phosphorus is in Group 5A, so it needs three more electrons
and makes three bonds, one bond to each hydrogen atom.)
H
Br
O
Br
(Oxygen is in Group 6A, so it needs two more electrons and
makes two bonds. Bromine is in Group 7A, so it needs only
one more electron. Therefore, the oxygen must be between
the two bromines.)
F
c)
F
Si
F
F
(Silicon is in Group 4A, so it needs four more electrons and
makes four bonds. Fluorine is in Group 7A, so it needs only
one more electrons. Therefore, the silicon atom must be in
the center and must bond to all four fluorine atoms.)
H
d) H
N
F
(Nitrogen is in Group 5A, so it needs three more electrons and
makes three bonds. Hydrogen and fluorine need only one more
electron, so the nitrogen atom must be in the center and must
bond to all three of the other atoms.)
e)
f)
Br
Cl
N
C
O
(Bromine needs only one electron and can only make one bond.
Nitrogen needs three electrons and must make three bonds.
Oxygen needs two electrons and must make two bonds. Since
nitrogen makes the most bonds, it must be in the center. We must
draw a double bond between N and O to allow those atoms to
make the correct numbers of bonds.)
N
(Chlorine needs only one electron and can only make one bond.
Carbon needs four electrons and must make four bonds.
Nitrogen needs three electrons and must make three bonds. Since
carbon makes the most bonds, it must be in the center. We must
draw a triple bond between C and H to allow those atoms to make
the correct numbers of bonds.)
Cl
(Carbon needs four electrons and must make four bonds. Oxygen
needs two electrons and must make two bonds. Each chlorine
needs only one electron and can only make one bond. Since
carbon makes the most bonds, it must be in the center. We must
draw a double bond between C and O to allow those atoms to
make the correct numbers of bonds.)
O
g) Cl
C
3.40 H2 is stable, but Li2 is not. Hydrogen only needs two electrons to fill its valence shell
(shell 1). For lithium, though, the valence shell is shell #2, which can hold 8 electrons.
3.42 The correct answer is shown below. The key is to keep the normal bonding requirements
of these atoms in mind. Carbon must make four bonds (because it needs four more electrons),
oxygen must make two bonds (because it needs two more electrons), and hydrogen must make
one bond (because it needs one electron). A double bond counts as two bonds.
H
O
O
H
H
O
C
C
C
C
O
H
3.44 Structure #3 is correct, because it satisfies the normal bonding requirements for all of the
atoms. Each carbon atom forms four bonds (with no non-bonding electrons), the nitrogen atom
makes three bonds (with one non-bonding electron pair), and the hydrogen atoms make one
bond.
3.48 Element X makes two bonds and has two non-bonding electron pairs, so it must come
from Group 6A. Element Q makes four bonds and has no non-bonding electrons, so it must
come from Group 4A. Since the problem tells you that both elements are in Period 2 (elements 3
through 10), X must be oxygen and Q must be carbon.
Another way to work this out is to divide up the bonding electrons between the atoms, as
shown below. Each X ends up with 6 electrons (2 bonding electrons plus 4 nonbonding
electrons), so X must be in Group 6A. Q ends up with 4 electrons, so Q must be in Group 4A.
..
..
:X::Q::X:
Dividing up the
two sets of bonding
electrons…
..
:X:
:Q:
..
:X:
…shows us that X has six valence electrons
and Q has four.
3.50 In any covalent bond, the atom with the higher electronegativity is negatively charged,
and the atom with the lower electronegativity is positively charged. Electronegativities are listed
in Table 3.2 on page 3-24.
a) Since both iodine atoms have the same electronegativity (2.5), the atoms share the
bonding electrons equally. Therefore, neither atom has any sort of electrical charge.
b) The electronegativities of C and N are 2.5 and 3.0, respectively, so nitrogen has a
negative charge.
c) The electronegativities of S and F are 2.5 and 4.0, respectively, so fluorine has a
negative charge.
3.52
a) SO3
b) Cl2O
3.54
a) phosphorus tribromide
c) carbon tetrachloride
b) sulfur hexafluoride