CHAPTER 22
The Transition Elements and Coordination
Compounds
■ SOLUTIONS TO EXERCISES
22.1. a. Pentaamminechlorocobalt(III) chloride
b. Potassium aquapentacyanocobaltate(III)
c. Pentaaquahydroxoiron(III) ion
22.2. a. K4[Fe(CN)6]
b. [Co(NH3)4Cl2]Cl
c. [PtCl4]2−
22.3. a. No geometric isomers.
b.
NH3 NH3
H3N OH2 H3N NH3
Co Co
H2O NH3 H2O OH2
NH3 NH3
© 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.
, 890 Chapter 22: The Transition Elements and Coordination Compounds
22.4. a. No optical isomers.
b.
N N
O2N N NO2
N
Co Co
N N NO2
O2N
N N
22.5. The electron configuration of Ni2+ is [Ar]3d8. The distribution of electrons among the d orbitals of
Ni in [Ni(H2O)6]2+ is as follows:
Note that there is only one possible distribution of electrons, giving two unpaired electrons.
22.6. The electron configuration of the Co2+ ion is [Ar]3d7. The distribution of the d electrons in
[CoCl4]2− is as follows:
22.7. The approximate wavelength of the maximum absorption for [Fe(H2O)6]3+, which is pale purple,
is 530 nm. The approximate wavelength of the maximum absorption for [Fe(CN)6]3−, which is
red, is 500 nm. The shift is in the expected direction because CN− is a more strongly bonding
ligand than H2O. As a result, should increase, and the wavelength of the absorption should
decrease, when H2O is replaced by CN−.
© 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.
, Chapter 22: The Transition Elements and Coordination Compounds 891
■ ANSWERS TO CONCEPT CHECKS
22.1. Because silver nitrate does not precipitate AgCl from the complex, you conclude that chlorine is
not present as free chloride ions; the Cl is presumably present as a complex. Potassium ion is
likely present as K+, which would account for two of the ions present in each formula unit. The
other would be a complex ion of platinum with six chlorine atoms (a complex of Pt with six Cl−
ions). The charge on this complex ion must be 2− to counter the charges from the K+ ions, so its
formula is [PtCl6]2−. The formula of the complex, then, is K2[PtCl6].
22.2. The addition of silver nitrate to the complex precipitates AgCl equivalent to two Cl− ions per
formula unit. Since each formula unit consists of three ions, the complex appears to consist of two
Cl− ions plus a complex ion with a charge of 2+. The formula of the complex would then be
[Co(NH3)4(H2O)Br]Cl2.
22.3. a. If you were to place a mirror to the right of complex X, complex A would directly represent
what you would see in the mirror, so it is a mirror image. Note that complexes B and C are
exactly the same, and A and E are the same; in both cases, they differ only by rotation (spin
one of the complexes in each pair 180o to see this). Therefore, A and E are mirror images of
X.
b. Optical isomers are nonsuperimposable mirror images of one another. The mirror images of
molecule X, which are molecules A and E, are both nonsuperimposable mirror images, and
are optical isomers of X.
c. To answer this part, you need to rotate each of the complexes to see if they have the same
bonding arrangement in space with each of the ligands. Models A and E are the same
complex, neither of which has the same bonding arrangement as complex X, so the
complex they represent is a geometric isomer. Models B and C represent the same complex
that also has a different bonding arrangement than complex X, so the complex they
represent is also a geometric isomer. Complex D is the same complex as complex X, so it is
not a geometric isomer.
■ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
22.1. Characteristics of the transition elements that set them apart from the main-group elements are the
following: (1) The transition elements are metals with high melting points (only the 2B elements
have low melting points). Most main-group elements have low melting points. (2) Each of the
transition metals has several oxidation states (except for the Group 3B and 2B elements). Most
main-group metals have only one oxidation state in addition to zero. (3) Transition-metal
compounds are often colored, and many are paramagnetic. Most main-group compounds are
colorless and diamagnetic.
22.2. Technetium has the electron configuration [Kr] 4d55s2.
22.3. Molybdenum has the highest melting point of all elements in the fifth period because it has the
maximum number of unpaired electrons, which contributes to the strength of the metal bonding.
22.4. One reason why iron, cobalt, and nickel are similar in properties is that these elements have
similar covalent radii.
© 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.
The Transition Elements and Coordination
Compounds
■ SOLUTIONS TO EXERCISES
22.1. a. Pentaamminechlorocobalt(III) chloride
b. Potassium aquapentacyanocobaltate(III)
c. Pentaaquahydroxoiron(III) ion
22.2. a. K4[Fe(CN)6]
b. [Co(NH3)4Cl2]Cl
c. [PtCl4]2−
22.3. a. No geometric isomers.
b.
NH3 NH3
H3N OH2 H3N NH3
Co Co
H2O NH3 H2O OH2
NH3 NH3
© 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.
, 890 Chapter 22: The Transition Elements and Coordination Compounds
22.4. a. No optical isomers.
b.
N N
O2N N NO2
N
Co Co
N N NO2
O2N
N N
22.5. The electron configuration of Ni2+ is [Ar]3d8. The distribution of electrons among the d orbitals of
Ni in [Ni(H2O)6]2+ is as follows:
Note that there is only one possible distribution of electrons, giving two unpaired electrons.
22.6. The electron configuration of the Co2+ ion is [Ar]3d7. The distribution of the d electrons in
[CoCl4]2− is as follows:
22.7. The approximate wavelength of the maximum absorption for [Fe(H2O)6]3+, which is pale purple,
is 530 nm. The approximate wavelength of the maximum absorption for [Fe(CN)6]3−, which is
red, is 500 nm. The shift is in the expected direction because CN− is a more strongly bonding
ligand than H2O. As a result, should increase, and the wavelength of the absorption should
decrease, when H2O is replaced by CN−.
© 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.
, Chapter 22: The Transition Elements and Coordination Compounds 891
■ ANSWERS TO CONCEPT CHECKS
22.1. Because silver nitrate does not precipitate AgCl from the complex, you conclude that chlorine is
not present as free chloride ions; the Cl is presumably present as a complex. Potassium ion is
likely present as K+, which would account for two of the ions present in each formula unit. The
other would be a complex ion of platinum with six chlorine atoms (a complex of Pt with six Cl−
ions). The charge on this complex ion must be 2− to counter the charges from the K+ ions, so its
formula is [PtCl6]2−. The formula of the complex, then, is K2[PtCl6].
22.2. The addition of silver nitrate to the complex precipitates AgCl equivalent to two Cl− ions per
formula unit. Since each formula unit consists of three ions, the complex appears to consist of two
Cl− ions plus a complex ion with a charge of 2+. The formula of the complex would then be
[Co(NH3)4(H2O)Br]Cl2.
22.3. a. If you were to place a mirror to the right of complex X, complex A would directly represent
what you would see in the mirror, so it is a mirror image. Note that complexes B and C are
exactly the same, and A and E are the same; in both cases, they differ only by rotation (spin
one of the complexes in each pair 180o to see this). Therefore, A and E are mirror images of
X.
b. Optical isomers are nonsuperimposable mirror images of one another. The mirror images of
molecule X, which are molecules A and E, are both nonsuperimposable mirror images, and
are optical isomers of X.
c. To answer this part, you need to rotate each of the complexes to see if they have the same
bonding arrangement in space with each of the ligands. Models A and E are the same
complex, neither of which has the same bonding arrangement as complex X, so the
complex they represent is a geometric isomer. Models B and C represent the same complex
that also has a different bonding arrangement than complex X, so the complex they
represent is also a geometric isomer. Complex D is the same complex as complex X, so it is
not a geometric isomer.
■ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
22.1. Characteristics of the transition elements that set them apart from the main-group elements are the
following: (1) The transition elements are metals with high melting points (only the 2B elements
have low melting points). Most main-group elements have low melting points. (2) Each of the
transition metals has several oxidation states (except for the Group 3B and 2B elements). Most
main-group metals have only one oxidation state in addition to zero. (3) Transition-metal
compounds are often colored, and many are paramagnetic. Most main-group compounds are
colorless and diamagnetic.
22.2. Technetium has the electron configuration [Kr] 4d55s2.
22.3. Molybdenum has the highest melting point of all elements in the fifth period because it has the
maximum number of unpaired electrons, which contributes to the strength of the metal bonding.
22.4. One reason why iron, cobalt, and nickel are similar in properties is that these elements have
similar covalent radii.
© 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.
