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Chapter 1
77
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b.
oxygen (atomic number
=
8; 2 core, 6 valence):
1
s
2
2
s
2
2
p
4
sulfur (atomic number
=
16; 10 core, 6 valence):
1
s
2
2
s
2
2
p
6
3
s
2
3
p
4
c.
nitrogen (atomic number
=
7; 2 core, 5 valence):
1
s
2
2
s
2
2
p
3
phosphorus (atomic number
=
15; 10 core, 5 valence):
1
s
2
2
s
2
2
p
6
3
s
2
3
p
3
d.
magnesium (atomic number
=
12; 10 core, 2 valence):
1
s
2
2
s
2
2
p
6
3
s
2
calcium (atomic number
=
20; 18 core, 2 valence):
1
s
2
2
s
2
2
p
6
3
s
2
3
p
6
4
s
2
7.
a.
Potassium is in the first column of the periodic table; therefore, like lithium and sodium that are also in
the first column, potassium has one valence electron.
b.
It occupies a
4
s
orbital.
8.
The polarity of a bond can be determined by the difference in the electronegativities (given in Table 1.3 on
page 10 of the text) of the atoms sharing the bonding electrons. The greater the difference in electronega-
tivity, the more polar the bond.
a.
Cl
¬
CH
3
b.
H
¬
OH
c.
H
¬
F
d.
Cl
¬
CH
3
9.
The electronegativity differences in the four listed compounds are as follows:
KCl 3.0
-
0.8
=
2.2
LiBr 2.8
-
1.0
=
1.8
NaI 2.5
-
0.9
=
1.6
Cl
2
3.0
-
3.0
=
0
a.
KCl has the most polar bond because its two bonded atoms have the greatest differences
in electronegativity.
b.
Cl
2
has the least polar bond because the two chlorine atoms share the bonding electrons equally.
10.
Solved in the text.
11.
To answer this question, compare the electronegativities of the two atoms sharing the bonding electrons
using Table 1.3 on page 10 of the text.
a.
HO H
δ δ
− +
b.
F Br
δ δ
+ −
c.
H
3
C NH
2
δ δ
− +
d.
H
3
C Cl
δ δ
+ −
e.
HO Br
δ δ
− +
f.
H
3
C Li
δ δ
+ −
g.
I Cl
δ δ
− +
h.
H
2
N OH
δ δ
+ −
(Notice that if the two atoms being compared are in the same row of the periodic table, the atom farther to
the right is the more electronegative atom; if the atoms being compared are in the same column, the one
closer to the top of the column is the more electronegative atom.)
12.
Solved in the text.
13.
The dipole moment is the magnitude of the charge times the distance between the charges. Because fluo-
rine is more electronegative than Cl, the charge on H and F in HF is larger than the charge on H and Cl in
HCl. The larger charge on F compared to the charge on Cl is more than enough to make up for the fact that
H
¬
F
is a shorter bond than
H
¬
Cl.