CHAPTER 9
Ionic and Covalent Bonding
■ SOLUTIONS TO EXERCISES
.
9.1. The Lewis symbol for oxygen is : O . . .
. . and the Lewis symbol for magnesium is Mg . The
magnesium atom loses two electrons, and the oxygen atom accepts two electrons. You can
represent this electron transfer as follows:
9.2. The electron configuration of the Ca atom is [Ar]4s2. By losing two electrons, the atom assumes a
2+ charge and the argon configuration, [Ar]. The Lewis symbol is Ca2+. The S atom has the
configuration [Ne]3s23p4. By gaining two electrons, the atom assumes a 2− charge and the argon
configuration [Ne]3s23p6 and is the same as [Ar]. The Lewis symbol is
.. 2-
:S:
..
9.3. The electron configuration of lead (Pb) is [Xe]4f145d106s26p2. The electron configuration of Pb2+
is [Xe]4f145d106s2.
9.4. The electron configuration of manganese (Z = 25) is [Ar]3d54s2. To find the ion configuration,
remove first the 4s electrons, then the 3d electrons. In this case, only two electrons need to be
removed. The electron configuration of Mn2+ is [Ar]3d5.
9.5. S2− has a larger radius than S. The anion has more electrons than the atom. The electron-electron
repulsion is greater; hence, the valence orbitals expand. The anion radius is larger than the atomic
radius.
9.6. The ionic radii increase down any column because of the addition of electron shells. All of these
ions are from the Group 2A family; therefore, Mg2+ < Ca2+ < Sr2+.
9.7. Cl−, Ca2+, and P3− are isoelectronic with an electron configuration equivalent to [Ar]. In an
isoelectronic sequence, the ionic radius decreases with increasing nuclear charge. Therefore, in
order of increasing ionic radius, we have Ca2+, Cl−, and P3−.
9.8. The absolute values of the electronegativity differences are C–O, 1.0; C–S, 0.0; and H–Br, 0.7.
Therefore, C–O is the most polar bond.
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, Chapter 9: Ionic and Covalent Bonding 291
9.9. First, calculate the total number of valence electrons. C has four, Cl has seven, and F has seven.
The total number is 4 + (2 7) + (2 7) = 32. The expected skeleton consists of a carbon atom
surrounded by Cl and F atoms. Distribute the electron pairs to the surrounding atoms to satisfy
the octet rule. All 32 electrons (16 pairs) are accounted for.
Cl
F C F
Cl
9.10. The total number of electrons in CO2 is 4 + (2 6) = 16. Because carbon is more electropositive
than oxygen, it is expected to be the central atom. Distribute the electrons to the surrounding
atoms to satisfy the octet rule.
O C O
All sixteen electrons have been used, but note that there are only four electrons on carbon. This is
four electrons short of a complete octet, which suggests the existence of double bonds. Move a
pair of electrons from each oxygen to the carbon-oxygen bonds.
O C O or O C O
9.11. a. There are (3 1) + 6 = 9 valence electrons in H3O. The H3O+ ion has one less electron than
is provided by the neutral atoms because the charge on the ion is 1+. Hence, there are eight
valence electrons in H3O+. The electron-dot formula is
b. Cl has seven valence electrons, and O has six valence electrons. The total number of
valence electrons from the neutral atoms is 7 + (2 6) = 19. The charge on the ClO2− is 1−,
which provides one more electron than the neutral atoms. This makes a total of 20 valence
electrons. The electron-dot formula for ClO2− is
-
O Cl O
9.12. The resonance formulas for NO3− are
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, 292 Chapter 9: Ionic and Covalent Bonding
9.13. The number of valence electrons in SF4 is 6 + (4 7) = 34. The skeleton structure is a sulfur atom
surrounded by fluorine atoms. After the electron pairs are placed on the F atoms to satisfy the
octet rule, two electrons remain.
F F
S
F F
These additional two electrons are put on the sulfur atom because it had d orbitals and, therefore,
can expand its octet.
F F
S
F F
9.14. Be has two valence electrons, and Cl has seven valence electrons. The total number of valence
electrons is 2 + (2 7) = 16 in the BeCl2 molecule. Be, a Group 2A element, can have fewer than
eight electrons around it. The electron-dot formula of BeCl2 is
Cl Be Cl
9.15. The total number of electrons in H3PO4 is 3 + 5 + 24 = 32. Assume a skeleton structure in which
the phosphorus atom is surrounded by the more electronegative four oxygen atoms. The hydrogen
atoms are then attached to the oxygen atoms. Distribute the electron pairs to the surrounding
atoms to satisfy the octet rule. If you assume all single bonds (structure on the left), the formal
charge on the phosphorus is 1+ and the formal charge on the top oxygen is 1−. Using the
principle of forming a double bond with a pair of electrons on the atom with the negative formal
charge, you obtain the structure on the right. The formal charge on all oxygens in this structure is
zero; the formal charge on phosphorus is 5 − 5 = 0. This is the better structure.
O O
H O P O H H O P O H
O O
H H
9.16. The bond length can be predicted by adding the covalent radii of the two atoms. For O–H, we
have 66 pm + 31 pm = 97 pm.
9.17. As the bond order increases, the bond length decreases. Since the C=O is a double bond, we
would expect it to be the shorter one, 123 pm.
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, Chapter 9: Ionic and Covalent Bonding 293
9.18.
H H O
C C + 3 O2 2 O C O + 2 H H
H H
One C=C bond, four C–H bonds, and three O=O bonds are broken. There are four C=O bonds
and four O–H bonds formed.
H = {[614 + (4 413) + (3 498)] − [(4 804) + (4 463)]} kJ = −1308 kJ
■ ANSWERS TO CONCEPT CHECKS
9.1. a. The 2+ ions are common in transition elements, but it is the outer s electrons that are lost to
form these ions in compounds. Iron, whose configuration is [Ar]3d64s2, would be expected
to lose two 4s electrons to give the configuration [Ar]3d6 for the Fe2+ ion in compounds.
The configuration given in the problem is for an excited state; you would not expect to see
it in compounds.
b. Nitrogen, whose ground-state atomic configuration is [He]2s22p3, would be expected to
form an anion with a noble-gas configuration by gaining three electrons. This would give
the anion N3− with the configuration [He]2s22p6. You would not expect to see the anion N2−
in compounds.
c. The zinc atom has the ground-state configuration [Ar]3d104s2. The element is often
considered to be a transition element. In any case, you would expect the atom to form
compounds by losing its 4s electrons to give Zn2+ with the pseudo-noble-gas configuration
[Ar]3d10. This is the ion configuration given in the problem.
d. The configuration of the ground-state sodium atom is [He]2s22p63s1. You would expect the
atom to lose one electron to give the Na+ ion with the noble-gas configuration [He]2s22p6.
You would not expect to see compounds with the Na2+ ion.
e. The ground state of the calcium atom is [Ne]3s23p64s2. You would expect the atom to lose
its two outer electrons to give Ca2+ with the noble-gas configuration [Ne]3s23p6, which is
the configuration given in the problem.
9.2. a. There are two basic points to consider in assessing the validity of each of the formulas
given in the problem. One is whether the formula has the correct skeleton structure. You
expect the F atoms to be bonded to the central N atoms because the F atoms are more
electronegative. The second point is the number of dots in the formula. This should equal
the total number of electrons in the valence shell of the atoms (five for each nitrogen atom
and seven for each fluorine atom), which is (2 5) + (2 7) = 24, or twelve pairs. The
number showing in the formula here is thirteen, which is incorrect.
b. This formula has the correct skeleton structure and the correct number of dots. All of the
atoms have octets, so the formula would appear to be correct. As a final check, however,
you might try drawing the formula beginning with the skeleton structure. In drawing an
electron-dot formula, after connecting atoms by single bonds (a single electron pair), you
would place electron pairs around the outer atoms (the F atoms in this formula) to give
octets. After doing that, you would have used up nine electron pairs (three for the single
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Ionic and Covalent Bonding
■ SOLUTIONS TO EXERCISES
.
9.1. The Lewis symbol for oxygen is : O . . .
. . and the Lewis symbol for magnesium is Mg . The
magnesium atom loses two electrons, and the oxygen atom accepts two electrons. You can
represent this electron transfer as follows:
9.2. The electron configuration of the Ca atom is [Ar]4s2. By losing two electrons, the atom assumes a
2+ charge and the argon configuration, [Ar]. The Lewis symbol is Ca2+. The S atom has the
configuration [Ne]3s23p4. By gaining two electrons, the atom assumes a 2− charge and the argon
configuration [Ne]3s23p6 and is the same as [Ar]. The Lewis symbol is
.. 2-
:S:
..
9.3. The electron configuration of lead (Pb) is [Xe]4f145d106s26p2. The electron configuration of Pb2+
is [Xe]4f145d106s2.
9.4. The electron configuration of manganese (Z = 25) is [Ar]3d54s2. To find the ion configuration,
remove first the 4s electrons, then the 3d electrons. In this case, only two electrons need to be
removed. The electron configuration of Mn2+ is [Ar]3d5.
9.5. S2− has a larger radius than S. The anion has more electrons than the atom. The electron-electron
repulsion is greater; hence, the valence orbitals expand. The anion radius is larger than the atomic
radius.
9.6. The ionic radii increase down any column because of the addition of electron shells. All of these
ions are from the Group 2A family; therefore, Mg2+ < Ca2+ < Sr2+.
9.7. Cl−, Ca2+, and P3− are isoelectronic with an electron configuration equivalent to [Ar]. In an
isoelectronic sequence, the ionic radius decreases with increasing nuclear charge. Therefore, in
order of increasing ionic radius, we have Ca2+, Cl−, and P3−.
9.8. The absolute values of the electronegativity differences are C–O, 1.0; C–S, 0.0; and H–Br, 0.7.
Therefore, C–O is the most polar bond.
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, Chapter 9: Ionic and Covalent Bonding 291
9.9. First, calculate the total number of valence electrons. C has four, Cl has seven, and F has seven.
The total number is 4 + (2 7) + (2 7) = 32. The expected skeleton consists of a carbon atom
surrounded by Cl and F atoms. Distribute the electron pairs to the surrounding atoms to satisfy
the octet rule. All 32 electrons (16 pairs) are accounted for.
Cl
F C F
Cl
9.10. The total number of electrons in CO2 is 4 + (2 6) = 16. Because carbon is more electropositive
than oxygen, it is expected to be the central atom. Distribute the electrons to the surrounding
atoms to satisfy the octet rule.
O C O
All sixteen electrons have been used, but note that there are only four electrons on carbon. This is
four electrons short of a complete octet, which suggests the existence of double bonds. Move a
pair of electrons from each oxygen to the carbon-oxygen bonds.
O C O or O C O
9.11. a. There are (3 1) + 6 = 9 valence electrons in H3O. The H3O+ ion has one less electron than
is provided by the neutral atoms because the charge on the ion is 1+. Hence, there are eight
valence electrons in H3O+. The electron-dot formula is
b. Cl has seven valence electrons, and O has six valence electrons. The total number of
valence electrons from the neutral atoms is 7 + (2 6) = 19. The charge on the ClO2− is 1−,
which provides one more electron than the neutral atoms. This makes a total of 20 valence
electrons. The electron-dot formula for ClO2− is
-
O Cl O
9.12. The resonance formulas for NO3− are
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, 292 Chapter 9: Ionic and Covalent Bonding
9.13. The number of valence electrons in SF4 is 6 + (4 7) = 34. The skeleton structure is a sulfur atom
surrounded by fluorine atoms. After the electron pairs are placed on the F atoms to satisfy the
octet rule, two electrons remain.
F F
S
F F
These additional two electrons are put on the sulfur atom because it had d orbitals and, therefore,
can expand its octet.
F F
S
F F
9.14. Be has two valence electrons, and Cl has seven valence electrons. The total number of valence
electrons is 2 + (2 7) = 16 in the BeCl2 molecule. Be, a Group 2A element, can have fewer than
eight electrons around it. The electron-dot formula of BeCl2 is
Cl Be Cl
9.15. The total number of electrons in H3PO4 is 3 + 5 + 24 = 32. Assume a skeleton structure in which
the phosphorus atom is surrounded by the more electronegative four oxygen atoms. The hydrogen
atoms are then attached to the oxygen atoms. Distribute the electron pairs to the surrounding
atoms to satisfy the octet rule. If you assume all single bonds (structure on the left), the formal
charge on the phosphorus is 1+ and the formal charge on the top oxygen is 1−. Using the
principle of forming a double bond with a pair of electrons on the atom with the negative formal
charge, you obtain the structure on the right. The formal charge on all oxygens in this structure is
zero; the formal charge on phosphorus is 5 − 5 = 0. This is the better structure.
O O
H O P O H H O P O H
O O
H H
9.16. The bond length can be predicted by adding the covalent radii of the two atoms. For O–H, we
have 66 pm + 31 pm = 97 pm.
9.17. As the bond order increases, the bond length decreases. Since the C=O is a double bond, we
would expect it to be the shorter one, 123 pm.
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, Chapter 9: Ionic and Covalent Bonding 293
9.18.
H H O
C C + 3 O2 2 O C O + 2 H H
H H
One C=C bond, four C–H bonds, and three O=O bonds are broken. There are four C=O bonds
and four O–H bonds formed.
H = {[614 + (4 413) + (3 498)] − [(4 804) + (4 463)]} kJ = −1308 kJ
■ ANSWERS TO CONCEPT CHECKS
9.1. a. The 2+ ions are common in transition elements, but it is the outer s electrons that are lost to
form these ions in compounds. Iron, whose configuration is [Ar]3d64s2, would be expected
to lose two 4s electrons to give the configuration [Ar]3d6 for the Fe2+ ion in compounds.
The configuration given in the problem is for an excited state; you would not expect to see
it in compounds.
b. Nitrogen, whose ground-state atomic configuration is [He]2s22p3, would be expected to
form an anion with a noble-gas configuration by gaining three electrons. This would give
the anion N3− with the configuration [He]2s22p6. You would not expect to see the anion N2−
in compounds.
c. The zinc atom has the ground-state configuration [Ar]3d104s2. The element is often
considered to be a transition element. In any case, you would expect the atom to form
compounds by losing its 4s electrons to give Zn2+ with the pseudo-noble-gas configuration
[Ar]3d10. This is the ion configuration given in the problem.
d. The configuration of the ground-state sodium atom is [He]2s22p63s1. You would expect the
atom to lose one electron to give the Na+ ion with the noble-gas configuration [He]2s22p6.
You would not expect to see compounds with the Na2+ ion.
e. The ground state of the calcium atom is [Ne]3s23p64s2. You would expect the atom to lose
its two outer electrons to give Ca2+ with the noble-gas configuration [Ne]3s23p6, which is
the configuration given in the problem.
9.2. a. There are two basic points to consider in assessing the validity of each of the formulas
given in the problem. One is whether the formula has the correct skeleton structure. You
expect the F atoms to be bonded to the central N atoms because the F atoms are more
electronegative. The second point is the number of dots in the formula. This should equal
the total number of electrons in the valence shell of the atoms (five for each nitrogen atom
and seven for each fluorine atom), which is (2 5) + (2 7) = 24, or twelve pairs. The
number showing in the formula here is thirteen, which is incorrect.
b. This formula has the correct skeleton structure and the correct number of dots. All of the
atoms have octets, so the formula would appear to be correct. As a final check, however,
you might try drawing the formula beginning with the skeleton structure. In drawing an
electron-dot formula, after connecting atoms by single bonds (a single electron pair), you
would place electron pairs around the outer atoms (the F atoms in this formula) to give
octets. After doing that, you would have used up nine electron pairs (three for the single
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
