CHAPTER 8 | Chemical Bonding and Climate Change
8.2. Collect and Organize
In the periodic table shown in Figure P8.2, groups 2, 13, 14, 15, and 16 are highlighted. We are to determine
which groups have 2, 5, and 3 valence electrons.
Analyze
The number of valence electrons in an element is equal to the number of electrons in the outermost shell. For
the groups highlighted, the electron configurations for the valence electrons are
Group 2
ns2
Group 13
ns2np1
Group 14
ns2np2
Group 15
ns2np3
Group 16
ns2np4
Solve
(a) Group 2 (red) elements have 2 valence electrons.
(b) Group 15 (peach) elements have 5 valence electrons.
(c) Group 13 (green) elements have 3 valence electrons.
Think about It
The number of valence electrons for neutral atoms is related to the group number. For groups 1 and 2, the
number of valence electrons equals the group number. For the other representative (main group) elements,
groups 13–18, the number of valence electrons is 10 fewer than the group number (15–10 = 5 valence electrons
for group 15 elements).
8.4. Collect and Organize
We are given Lewis symbols for neutral N, N2+, N3–, O2–, and O2+ and asked which are correct. The number of
valence electrons for each species can be determined from the electron configuration for the species. The
number of valence electrons equals the number of electrons in the outermost shell.
Analyze
The electron configuration and number of valence electrons for each species are
N
[He]2s22p3
5 valence electrons
2+
N
[He]2s22p1
3 valence electrons
N3–
[He]2s22p6
8 valence electrons
O2–
[He]2s22p6
8 valence electrons
O2+
[He]2s22p2
4 valence electrons
Solve
None of the Lewis symbols shown is correct!
Think about It
The correct Lewis symbols for the species are
8.6. Collect and Organize
All of the elements highlighted in Figure P8.6 are in period 2. We are asked to name the element (Li, Be, B, C,
N, O, F, Ne) that is the most electronegative.
Analyze
As we move across a row in the periodic table, electronegativity increases.
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Chemical Bonding and Climate Change | 395
Solve
According to the periodic trend we might be tempted to jump to name Ne (peach) as the most electronegative
element in period 2. However, remember that electronegativity is defined as the power of an atom that is
chemically bonded to another atom to attract electrons to itself. Neon does not form compounds! Therefore,
fluorine (lilac) has the greatest electronegativity of the elements in period 2.
Think about It
The least electronegative element in this period is Li (red).
8.12. Collect and Organize
Of the four drawings of bent triatomic molecules in Figure P8.12, we are to choose the one that represents the
electron density distribution in ozone, O3.
Analyze
Differences in electron density within a molecule depend on the different pulling powers of the atoms in the
molecule for electrons (electronegativity) and on the arrangement of the atoms in space.
Solve
All of the atoms in ozone are the same and thus all have the same electronegativity. It would seem that no one
atom pulls electrons toward itself over the others, so the electron density distribution is expected to be even
throughout the ozone molecule. However, the molecule is bent and is the combination of two resonance forms:
The electrons in the double bonds, because of resonance, will result in a buildup of electron density on the
terminal oxygen atoms, reducing the electron density for the central O atom. Therefore, drawing a best
represents ozone.
Think about It
Because of the symmetrical atom arrangement (and because all the atoms are the same), drawings c and d
would not be good choices for the electron density distribution in ozone.
8.18. Collect and Organize / Analyze
We are asked to define which electrons are the valence electrons in an atom.
Solve
Valence electrons are the outermost electrons in the atom. They are in the highest n level and can be involved
in bonding.
Think about It
The core electrons lie below the valence electrons and do not become involved in forming chemical bonds.
8.32. Collect and Organize
For each atom or ion (Xe, Sr2+, Cl, and Cl–) we are to write the Lewis symbol and determine the number of
valence electrons.
Analyze
The number of valence electrons is determined by the number of electrons in the highest-lying energy levels
through the electron configuration. The Lewis symbol shows those valence electrons around the element
symbol.
Solve
Atom or
Ion
Electron
Configuration
Number of
Valence e–
Xe
[Kr]4d105s25p6
8
Sr2+
[Kr]
0
Lewis Symbol
396 | Chapter 8
Cl
[Ne]3s23p5
7
Cl–
[Ne]3s23p6
8
Think about It
Notice that the 4d electrons in Xe become “core electrons” once that shell is filled and therefore the total
number of valence electrons on Xe is 8, not 18.
8.36. Collect and Organize
For the diatomic species N2+, CS+, CN, and CO, we are to determine the total number of valence electrons.
Analyze
For each species we need to add the valence electrons for each atom. If the species is charged, we need to
reduce or increase the number of electrons as necessary to form cations or anions, respectively.
Solve
(a) 5 valence e– (N) + 5 valence e– (N) – 1 e– (positive charge) = 9 valence e–
(b) 4 valence e– (C) + 6 valence e– (S) – 1 e– (positive charge) = 9 valence e–
(c) 4 valence e– (C) + 5 valence e– (N) = 9 valence e–
(d) 4 valence e– (C) + 6 valence e– (O) = 10 valence e–
Think about It
N2+, CS+, and CN are all odd-electron species. One of the atoms will not be able to satisfy the octet rule in a
Lewis structure. These odd-electron species are called radicals.
8.39. Collect and Organize
For CO, O2, ClO–, and CN– we are to determine the number of electron pairs that are shared in each molecule.
Analyze
The Lewis structures clearly show the shared pairs as covalent bonds .
Solve
(a) CO has a triple bond between C and O, so 6 electrons or three electron pairs are shared.
(b) O2 has a double bond between the oxygen atoms, so 4 electrons or two electron pairs are shared.
(c) ClO– has a single bond between Cl and O, so 2 electrons or one electron pair is shared.
(d) CN– has a triple bond between C and N, so 6 electrons or three electron pairs are shared.
Think about It
Each bond contains one shared pair. A single bond has 2 electrons, a double bond has 4 electrons, and a triple
bond has 6 electrons.
8.54. Collect and Organize
Of the bonds listed, we are to determine which is the least polar.
Analyze
The lower the difference in electronegativity between the bonded atoms, the less polar bond. We need the
electronegativity values from Figure 8.5 in the textbook.
Chemical Bonding and Climate Change | 397
Solve
Bond
C—Se
C O
Cl—Br
O O
N—H
C—H
The O
Electronegativity
Difference
2.5 – 2.4 = 0.1
3.5 – 2.5 = 1.0
3.0 – 2.8 = 0.2
3.5 – 3.5 = 0
3.0 – 2.1 = 0.9
2.5 – 2.1 = 0.4
O bond is nonpolar and, therefore, the least polar in the list.
Think about It
The most polar bond in the list is C
atoms.
O because it has the greatest electronegativity difference between the
8.55. Collect and Organize
From the list of pairs of atoms, we are to discriminate between binary compounds with nonpolar covalent
bonds, polar covalent bonds, and ionic bonds.
Analyze
When the difference in electronegativity between the atoms is zero, the bond is nonpolar. If the
electronegativity difference is below 2.0, the bond is polar covalent. If the electronegativity difference is 2.0 or
greater, the bond is ionic.
Solve
Bond
(a) C and S
(b) C and O
(c) Al and Cl
(d) Ca and O
Binary compounds of (b) C and O
Ca and O has ionic bonds.
Electronegativity
Bond Type
Difference
2.5 – 2.5 = 0
Nonpolar covalent
3.5 – 2.5 = 1.0
Polar covalent
3.0 – 1.5 = 1.5
Polar covalent
3.5 – 1.0 = 2.5
Ionic
and (c) Al and Cl have polar covalent bonds. The binary compound of (d)
Think about It
Although both C — O and Al — Cl bonds are polar covalent, the Al — Cl bond has greater ionic character (is
more polar) than the C — O bond.
own for part b are those that give products with zero formal charges on all of the atoms.