779 CHAPTER TWENTY-ONE TRANSITION METALS AND COORDINATION CHEMISTRY For Review 1. Chromium ([Ar]:4s 1 3d 5 ) and copper [Ar]:4s 1 3d 10 ) have electron configurations which are different from that predicted from the periodic table. Other exceptions to the predicted filling order are transition metal ions. These all lose the s electrons before they lose the d electrons. In neutral atoms, the ns and (n1)d orbitals are very close in energy, with the ns orbitals slightly lower in energy. However, for transition metal ions, there is an apparent shifting of energies between the ns and (n1)d orbitals. For transition metal ions, the energy of the (n1)d orbitals are significantly less than that of the ns electrons. So when transition metal ions form, the highest energy electrons are removed, which are the ns electrons. For example, Mn 2+ has the electron configuration [Ar]:4s 0 3d 5 and not [Ar]:4s 2 3d 3 . Most transition metals have unfilled d orbitals, which creates a large number of other electrons that can be removed. Stable ions of the representative metals are determined by how many s and p valence electrons can be removed. In general, representative metals lose all of the s and p valence electrons to form their stable ions. Transition metals generally lose the s electron(s) to form +1 and +2 ions, but they can also lose some (or all) of the d electrons to form other oxidation states as well. 2. a. Coordination compound: a compound composed of a complex ion (see b) and counter- ions (see c) sufficient to give no net charge. b. Complex ion: a charged species consisting of a metal ion surrounded by ligands (see e). c. Counterions: anions or cations that balance the charge on a complex ion in a coordina- tion compound. d. Coordination number: the number of bonds formed between the metal ion and the ligands (see e) in a complex ion. e. Ligand: Species that donates a pair of electrons to form a covalent bond to a metal ion. Ligands act as Lewis bases (electron pair donors). f. Chelate: Ligand that can form more than one bond to a metal ion. g. Bidentate: Ligand that forms two bonds to a metal ion. Because transition metals form bonds to species that donate lone pairs of electrons, transition metals are Lewis acids (electron pair acceptors). The Lewis bases in coordination com- pounds are the ligands, all of which have an unshared pair of electrons to donate. The coordinate covalent bond between the ligand and the transition metal just indicates that both electrons in the bond originally came from one of the atoms in the bond. Here, the electrons in the bond come from the ligand.
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779
CHAPTER TWENTY-ONE
TRANSITION METALS AND COORDINATION CHEMISTRY
For Review
1. Chromium ([Ar]:4s13d5) and copper [Ar]:4s13d10) have electron configurations which are
different from that predicted from the periodic table. Other exceptions to the predicted filling
order are transition metal ions. These all lose the s electrons before they lose the d electrons.
In neutral atoms, the ns and (n1)d orbitals are very close in energy, with the ns orbitals
slightly lower in energy. However, for transition metal ions, there is an apparent shifting of
energies between the ns and (n1)d orbitals. For transition metal ions, the energy of the (n1)d
orbitals are significantly less than that of the ns electrons. So when transition metal ions form,
the highest energy electrons are removed, which are the ns electrons. For example, Mn2+ has
the electron configuration [Ar]:4s03d5 and not [Ar]:4s23d3.
Most transition metals have unfilled d orbitals, which creates a large number of other
electrons that can be removed. Stable ions of the representative metals are determined by how
many s and p valence electrons can be removed. In general, representative metals lose all of
the s and p valence electrons to form their stable ions. Transition metals generally lose the s
electron(s) to form +1 and +2 ions, but they can also lose some (or all) of the d electrons to
form other oxidation states as well.
2. a. Coordination compound: a compound composed of a complex ion (see b) and counter-
ions (see c) sufficient to give no net charge.
b. Complex ion: a charged species consisting of a metal ion surrounded by ligands (see e).
c. Counterions: anions or cations that balance the charge on a complex ion in a coordina-
tion compound.
d. Coordination number: the number of bonds formed between the metal ion and the ligands
(see e) in a complex ion.
e. Ligand: Species that donates a pair of electrons to form a covalent bond to a metal ion.
Ligands act as Lewis bases (electron pair donors).
f. Chelate: Ligand that can form more than one bond to a metal ion.
g. Bidentate: Ligand that forms two bonds to a metal ion.
Because transition metals form bonds to species that donate lone pairs of electrons, transition
metals are Lewis acids (electron pair acceptors). The Lewis bases in coordination com-
pounds are the ligands, all of which have an unshared pair of electrons to donate. The
coordinate covalent bond between the ligand and the transition metal just indicates that both
electrons in the bond originally came from one of the atoms in the bond. Here, the electrons
in the bond come from the ligand.
780 CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
3. Linear geometry (180 bond angles) is observed when the coordination number is 2.
Tetrahedral geometry (109.5 bond angles) or square planar geometry (90 bond angles) is
observed when the coordination number is 4. Octahedral geometry (90 bond angles) is
observed when the coordination number is 6.
For the following complex ions, see Table 21.13 if you don’t know the formula, the charge,
or the number of bonds the ligands form.
a. Ag(CN)2−; Ag+: [Kr]4d10 b. Cu(H2O)4
+; Cu+: [Ar]3d10
c. Mn(C2O4)2−; Mn2+: [Ar]3d5 d. Pt(NH3)42+; Pt2+: [Xe]4f145d8
e. Fe(EDTA)−; Fe3+: [Ar]3d5; Note: EDTA has an overall 4 charge and is a six coordinate
ligand. f. Co(Cl)6
4−; Co2+: [Ar]3d7 g. Cr(en)3
3+ where en = ethylenediane (NH2CH2CH2NH2); Cr3+: [Ar]3d3
4. See section 21.3 for a nice summary of the nomenclature rules.
a. The correct name is tetraamminecopper(II) chloride. The complex ion is named
incorrectly in several ways.
b. The correct name is bis(ethylenediamine)nickel(II) sulfate. The ethylenediamine ligands
are neutral and sulfate has a 2 charge. Therefore, Ni2+ is present, not Ni4+.
c. The correct name is potassium diaquatetrachlorochromate(III). Because the complex ion
is an anion, the –ate suffix ending is added to the name of the metal. Also, the ligands
were not in alphabetical order (a in aqua comes before c in chloro).
d. The correct name is sodium tetracyanooxalatocobaltate(II). The only error is that tetra
should be omitted in front of sodium. That four sodium ions are needed to balance charge
is deduced from the name of the complex ion.
5. a. Isomers: species with the same formulas but different properties; they are different com-
pounds. See the text for examples of the following types of isomers.
b. Structural isomers: isomers that have one or more bonds that are different.
c. Steroisomers: isomers that contain the same bonds but differ in how the atoms are
arranged in space.
d. Coordination isomers: structural isomers that differ in the atoms that make up the
complex ion.
e. Linkage isomers: structural isomers that differ in how one or more ligands are attached to
the transition metal.
f. Geometric isomers: (cis-trans isomerism); steroisomers that differ in the positions of
atoms with respect to a rigid ring, bond, or each other.
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY 781
g. Optical isomers: steroisomers tbat are nonsuperimposable mirror images of each other;
that is, they are different in the same way that our left and right hands are different.
The trans form of Cr(en)Cl2 is not optically active, but the the cis form is optically active. See
Figure 21.17 for illustrations showing the cis and trans forms for a similar compound; shown
also is the optical activity of the cis form. The only difference between the complex in this
question, and the complex in Figure 21.17, is that Cr2+ has replaced Co2+. Note that not all cis
isomers are optically active. For example, the cis isomer of Cr(NH3)4Cl2 is not optically
active because the mirror image is superimposable (prove it to yourself).
In Figure 21.17, a plane of symmetry exists through the square planar orientation of the two
en ligands. Other planes of symmetry also exist in the trans isomer. In the cis isomer in
Figure 21.17, no plane of symmetry exists, so this cis form is optically active (as we know).
6. The crystal field model focuses on the energies of the d orbitals and what happens to the
energies of these d orbitals as negative point charges (the ligands) approach (and repel) the
electrons in the d orbitals. For octahedral geometry, six ligands are bonded to the metal ion.
Because of the different orientations of the d orbitals, not all d orbitals are affected the same
when six negative point charges (ligands) approach the metal ion along the x, y, and z axis. It
turns out that the dxy, dxz, and dzy orbitals are all destabilized by the same amount from the
octahedrally arranged point charges, as are the 22 yxd
−and 2z
d orbitals. These are the two sets
that the d orbitals split into. The dxy, dxz, and dyz set is called the t2g set, while the 22 yxd
−and
2zd set is called the eg set.
Another major point for the octahedral crystal field diagram is that the eg set of orbitals is
destabilized more than the t2g set. This is because the t2g orbital set (dxy, dxz, and dzy) points
between the point charges while the eg orbital set ( 22 yxd
−and 2z
d ) points directly at the point
charges. Hence, there is more destabilization in the eg orbital set, and they are at a higher
energy.
a. Weak field ligand: ligand that will give complex ions with the maximum number of
unpaired electrons.
b. Strong-field ligand: ligand that will give complex ions with the minimum number of
unpaired electrons.
c. Low-spin complex: complex ion with a minimum number of unpaired electrons (low-spin
= strong-field).
d. High-spin complex: complex ion with a maximum number of unpaired electrons (high-
spin = weak-field).
In both cobalt complex ions, Co3+ exists which is a d6 ion (6 d electrons are present). The
difference in magnetic properties is that Co(NH3)63+ is a strong-field (low-spin) complex
having a relatively large , while CoF63− is a weak-field (high-spin) complex having a
relatively small . The electron configurations for Co3+ in a strong field vs. a weak field is
782 CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
shown in Figure 21.22. The strong-field d6 ion is diamagnetic because all electrons are paired.
This is the diagram for Co(NH3)63+. The weak field d6 ion is paramagnetic because it has
unpaired electrons (4 total). This is the diagram for CoF63−.
Looking at Figure 21.22, d1, d2, and d3 metal ions would all have the same number of
unpaired electrons. This won’t happen again until we get all the way up to d8, d9, and d10
metal ions (prove it to yourself that d4, d5, d6, and d7 metal ions have a different d orbital
electron configurations depending on a strong-field or a weak-field). V3+ is a d2 ion (2
unpaired electrons in the t2g set). It has the same diagram no matter how strong the field
strength. The same is true for the d8 Ni2+ ion (filled t2g set and half-filled eg set). However,
Ru2+, a d6 ion, will have different diagrams depending on a strong-field or a weak-field. If a
weak-field is present, then there are four unpaired electrons. In the strong-field case, all six d
electrons are in the t2g set and all are paired (no unpaired electrons).
7. The valence d electrons for the metal ion in the complex ion are placed into the octahedral
crystal field diagram. If electrons are all paired, then the complex is predicted to be
diamagnetic. If there are unpaired electrons, then the complex is predicted to be
paramagnetic.
Color results by the absorption of specific wavelengths of light. The d-orbital splitting, , is
on the order of the energies of visible light. The complex ion absorbs the wavelength of light
that has energy equal to the d-orbital splitting, . The color we detect for the substance is not
the color of light absorbed. We detect (see) the complementary color to that color of light
absorbed. See Table 21.16 for observed colors of substances given the color of light
absorbed.
From Table 21.16, if a complex appears yellow then it absorbs blue light on the order of
~450 nm. Therefore, Cr(NH3)63+ absorbs blue light.
The spectrochemical series places ligands in order of their ability to split the d-orbitals. The
strongest field ligands (large ) are on one side of the series with the weakest field ligands
(small ) on the other side. The series was developed from studies of the light absorbed by
many octahedral complexes. From the color of light absorbed, one can determine the d-orbital
splitting. Strong-field ligands absorb higher energy light (violet light, for example, with
~400 nm), while weak-field ligands absorb lower energy light (red light, for example, with
~650 nm).
The higher the charge on the metal ion, the larger the d-orlbital splitting. Thus, the Co3+
complex ion [Co(NH3)63+], would absorb higher energy (shorter wavelength) light than a Co2+
complex ion (assuming the ligands are the same).
Cu2+: [Ar]3d9; Cu+: [Ar]3d10; Cu(II) is d9 and Cu(I) is d10. Color is a result of the electron
transfer between split d orbitals. This cannot occur for the filled d orbitals in Cu(I). Cd2+,
like Cu+, is also d10. We would not expect Cd(NH3)4Cl2 to be colored because the d orbitals
are filled in this Cd2+ complex.
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY 783
Sc3+ has no electrons in d orbitals. Ti3+ and V3+ have d electrons present. Color of transition
metal complexes results from electron transfer between split d orbitals. If no d electrons are
present, no electron transfer can occur, and the compounds are not colored.
8. The crystal field diagrams are different because the geometries of where the ligands point is
different. The tetrahedrally oriented ligands point differently in relationship to the d-orbitals
than do the octahedrally oriented ligands. Plus, we have more ligands in an octahedral
complex.
See Figure 21.27 for the tetrahedral crystal field diagram. Notice that the orbitals are reverse
of that in the octahedral crystal field diagram. The degenerate 2zd and 22 yx
d−
are at a lower
energy than the degenerate dxy, dxz, and dyz orbitals. Again, the reason for this is that
tetrahedral ligands are oriented differently than octahedral field ligands so the interactions
with specifically oriented d-orbitals are different. Also notice that the difference in magnitude
of the d-orbital splitting for the two geometries. The d-orbital splitting in tetrahedral
complexes is less than one-half the d-orbital splitting in octahedral complexes. There are no
known ligands powerful enough to produce the strong-field case, hence all tetrahedral
complexes are weak-field or high spin.
See Figure 21.28 for the descriptions of the square planar and linear crystal field diagrams.
Each is unique which is not surprising. The ligands for any specific geometry will point
differently relative to the orientations of the five d-orbitals. Different interactions result
giving different crystal field diagrams.
9. Each hemoglobin molecule can bind four O2 molecules. It is an Fe2+ ion in hemoglobin that
binds an individual O2 molecule, and each hemoglobin molecule has four of these Fe2+
binding sites. The Fe2+ ion at the binding site is six-coordinate. Five of the coordination sites
come from nitrogens in the hemoglobin molecule. The sixth site is available to attach an O2
molecule. When the O2 molecule is released, H2O takes up the sixth position around the Fe2+
ion. O2 is a strong field ligand, unlike H2O, so in the lungs, O2 readily replaces the H2O
ligand. With four sites, each hemoglobin molecule has a total of four O2 molecules attached
when saturated with O2 from the lungs. In the cells, O2 is released by the hemoglobin and the
O2 site is replaced by H2O. The oxygen binding is pH dependent, so changes in pH in the
cells as compared to blood, causes the release of O2 (see Exercise 21.72). Once the O2 is
released and replaced by H2O, the hemoglobin molecules return to the lungs to replenish with
the O2.
CN− and CO form much stronger complexes with Fe(II) than O2. Thus, O2 is not transported
by hemoglobin in the presence of CN− or CO because the binding sites prefer the toxic CN−
and CO ligands.
10. The definitions follow. See section 21.8 for examples.
a. Roasting: converting sulfide minerals to oxides by heating in air below their melting
points.
784 CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
b. Smelting: reducing metal ions to the free metal.
c. Flotation: separation of mineral particles in an ore from the unwanted impurities. This
process depends on the greater wetability of the mineral particles as compared to the
unwanted impurities.
d. Leaching: the extraction of metals from ores using aqueous chemical solutions.
e. Gangue: the impurities (such as clay, sand, or rock) in an ore.
Advantages of hydrometallurgy: cheap energy cost; less air pollution; Disadvantages of
hydrometallurgy: chemicals used in hydrometallurgy are expensive and sometimes toxic.
In zone refining, a bar of impure metal travels through a heater. The impurities present are
more soluble in the molten metal than in the solid metal. As the molten zone moves down a
metal, the impurities are swept along with the liquid, leaving behind relatively pure metal.
Questions
5. Fe2O3(s) + 6 H2C2O4(aq) → 2 Fe(C2O4)3
3−(aq) + 3 H2O(l) + 6 H+(aq); The oxalate anion
forms a soluble complex ion with iron in rust (Fe2O3), which allows rust stains to be removed.
6. Only the Cr3+ ion can form four different compounds with H2O ligands and Cl− ions. The Cr2+
ion could form only three different compounds while the Cr4+ ion could form five different
compounds.
The Cl− ions that form precipitates with Ag+ are the counter ions, not the ligands in the
complex ion. The four compounds and mol AgCl precipitate that would form with 1 mol of
compound are:
Compound mol AgCl(s)
[Cr(H2O)6]Cl3 3 mol
[Cr(H2O)5Cl]Cl2 2 mol
[Cr(H2O)4Cl2]Cl 1 mol
[Cr(H2O)3Cl3] 0 mol
7.
mirror
trans
(mirror image is
superimposable)
cis
The mirror image of the cis
isomer is also superimposable.
Co
Cl
Cl
H3N
H3N
NH3
NH3
Co
Cl
NH3
Cl
NH3
Co
Cl
NH3
H3N
H3N
Cl
NH3
NH3
NH3
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY 785
No; both the trans or the cis forms of Co(NH3)4Cl2
+ have mirror images that are superim-
posable. For the cis form, the mirror image only needs a 90 rotation to produce the original
structure. Hence, neither the trans nor cis forms are optically active.
8. The transition metal ion must form octahedral complex ions; only with the octahedral
geometry are two different arrangements of d electrons possible in the split d orbitals. These
two arrangements depend on whether a weak field or strong field is present. For four
unpaired electrons, the two possible weak field cases are for transition metal ions with 3d4 or
3d6 electron configurations:
d4 d6
Of these two, only d6 ions have no unpaired electron in the strong field case.
Therefore, the transition metal ion has a 3d6 arrangement of electrons. Two possible metal
ions that are 3d6 are Fe2+ and Co3+. Thus, one of these ions is present in the four coordination
compounds and each of these complex ions has a coordination number of 6.
The colors of the compounds are related to the magnitude of (the d-orbital splitting value).
The weak field compounds will have the smallest , so the of light absorbed will be
longest. Using Table 21.16, the green solution (absorbs 650 nm light) and the blue solution
(absorbs 600 nm light) absorb the longest wavelength light; these solutions contain the
complex ions which are the weak field cases with four unpaired electrons. The red solution
(absorbs 490 nm light) and yellow solution (absorbs 450 nm light) contain the two strong
field case complex ions because they absorb the shortest wavelength (highest energy) light.
These complex ions are diamagnetic.
9. a. CoCl42−; Co2+: 4s03d7; All tetrahedral complexes are a weak field (high-spin).
small
CoCl42− is an example of a weak-field case
having three unpaired electrons.
b. Co(CN)63−: Co3+ : 4s03d6; Because CN− is a strong-field ligand, Co(CN)6
3− will be a
strong-field case (low-spin case).
large
small small
786 CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
large
CN− is a strong field ligand so Co(CN)63− will be
a low-spin case having zero unpaired electrons.
10. a. The coordination compound has the formula [Co(H2O)6]Cl2. The complex ion is
Co(H2O)62+ and the counter ions are the Cl− ions. The geometry would be octahedral and
the electron configuration of Co2+ is [Ar]3d7.
b. The coordination compound is Na3[Ag(S2O3)2].The compound consists of Na+ counter-
ions and the Ag(S2O3)23− complex ion. The complex ion is linear and the electron
configuration of Ag+ is: [Kr]4d10.
c. The two coordination compounds are Pt(NH3)2Cl2 and K2PtCl4. For Pt(NH3)2Cl2 we need
four ligands for a square planar geometry. Since only four species are attached to Pt, then
there are no counterions. The complex is Pt(NH3)2Cl2. Because chlorines each have a 1
charge, the platinum must be +2. The electron configuration for Pt2+ is: [Xe] 4f145d8. Note
that there are two possible arrangements for the Cl− and NH3 ligands. From the name of
the compounds, the cis isomer is the one discussed in the problem.
cis trans
For the K2PtCl4 coordination compound, K+ are the counterions and PtCl42− is the square
planar complex ion. Platinum is also in the +2 oxidation state with a [Xe]4f145d8 electron
configuration.
d. The reactant coordination compound is [Cu(NH3)4]Cl2. The complex ion is Cu(NH3)42+,
and the counterions are Cl− ions. The complex ion is tetrahedral (given in the question)
and the electron configuration of Cu2+ is: [Ar]3d9. The product coordination compound is
[Cu(NH3)4]Cl. The complex ion is Cu(NH3)4+ with Cl− counterions. The complex ion is
tetrahedral, and the electron configuration of Cu+ is: [Ar]3d10.
11. At high altitudes, the oxygen content of air is lower, so less oxyhemoglobin is formed which
diminishes the transport of oxygen in the blood. A serious illness called high-altitude sickness
can result from the decrease of O2 in the blood. High-altitude acclimatization is the
phenomenom that occurs in the human body in response to the lower amounts of
oxyhemoglobin in the blood. This response is to produce more hemoglobin, and, hence,
increase the oxyhemoglobin in the blood. High-altitude acclimatization takes several weeks
to take hold for people moving from lower altitudes to higher altitudes.
12. Metals are easily oxidized by oxygen and other substances to form the metal cations. Because
of this, metals are found in nature combined with nonmetals such as oxygen, sulfur, and the
Pt
Cl NH3
Cl NH3
Pt
Cl NH3
H3N Cl
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY 787
halogens. These compounds are called ores. To recover and use the metals, we must separate
them from their ores and reduce the metal ions. Then, because most metals are unsuitable for
use in the pure state, we must form alloys with the metals in order to form materials having
desirable properties.
Exercises
Transition Metals and Coordination Compounds
13. a. Ni: [Ar]4s23d8 b. Cd: [Kr]5s24d10
c. Zr: [Kr]5s24d2 d. Os: [Xe]6s24f145d6
14. Transition metal ions lose the s electrons before the d electrons.
a. Ni2+: [Ar]3d8 b. Cd2+: [Kr]4d10
c. Zr3+: [Kr]4d1; Zr4+: [Kr] d. Os2+: [Xe]4f145d6; Os3+: [Xe]4f145d5
15. Transition metal ions lose the s electrons before the d electrons.
a. Ti: [Ar]4s23d2 b. Re: [Xe]6s24f145d5 c. Ir: [Xe]6s24f145d7
Ti2+: [Ar]3d2 Re2+: [Xe]4f145d5 Ir2+: [Xe]4f145d7
Ti4+: [Ar] or [Ne]3s23p6 Re3+: [Xe]4f145d4 Ir3+: [Xe]4f145d6
16. Cr and Cu are exceptions to the normal filling order of electrons.
a. Cr: [Ar]4s13d5 b. Cu: [Ar]4s13d10 c. V: [Ar]4s23d3
Cr2+: [Ar]3d4 Cu+: [Ar]3d10 V2+: [Ar]3d3
Cr3+: [Ar]3d3 Cu2+: [Ar]3d9 V3+: [Ar]3d2
17. a. With K+ and CN− ions present, iron has a +3 charge. Fe3+: [Ar]3d5
b. With a Cl− ion and neutral NH3 molecules present, silver has a +1 charge. Ag+:
[Kr] 4d10
c. With Br− ions and neutral H2O molecules present, nickel has a +2 charge. Ni2+:
[Ar]3d8
d. With NO2− ions, an I− ion, and neutral H2O molecules present, chromium has a +3 charge.
Cr3+: [Ar]3d3
18. a. With NH4+ ions, Cl− ions, and neutral H2O molecules present, iron has a +2 charge.
Fe2+: [Ar]3d6
788 CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
b. With I− ions and neutral NH3 and NH2CH2CH2NH2 molecules present, cobalt has a +2
charge. Co2+: [Ar]3d7
c. With Na+ and F− ions present, tantalum has a +5 charge. Ta5+: [Xe]4f14 (expected)
d. Each platinum complex ion must have an overall charge if the two complex ions are
counterions to each. Knowing that platinum forms +2 and +4 charged ions, we can
deduce that the six coordinate complex ion has a +4 charged platinum ion and the four
coordinate complex ion has a +2 charged ion. With I− ions and neutral NH3 molecules
present, the two complex ions are [Pt(NH3)4I2]2+ and [PtI4]2− .
Pt2+: [Xe]4f145d8; Pt4+: [Xe]4f145d6
19. a. molybdenum(IV) sulfide; molybdenum(VI) oxide
b. MoS2, +4; MoO3, +6; (NH4)2Mo2O7, +6; (NH4)6Mo7O244 H2O, +6
20. Fe2O3: iron has a +3 oxidation state; Fe3O4: iron has a +8/3 oxidation state. The three iron
ions in Fe3O4 must have a total charge of +8. The only combination that works is to have two
Fe3+ ions and one Fe2+ ion per formula unit. This makes sense from the other formula for
magnetite, FeO Fe2O3. FeO has an Fe2+ ion and Fe2O3 has two Fe3+ ions.
21. The lanthanide elements are located just before the 5d transition metals. The lanthanide
contraction is the steady decrease in the atomic radii of the lanthanide elements when going
from left to right across the periodic table. As a result of the lanthanide contraction, the sizes
of the 4d and 5d elements are very similar (see the following Exercise). This leads to a
greater similarity in the chemistry of the 4d and 5d elements in a given vertical group.
22. Size also decreases going across a period. Sc & Ti and Y & Zr are adjacent elements. There
are 14 elements (the lanthanides) between La and Hf, making Hf considerably smaller.
23. CoCl2(s) + 6 H2O(g) ⇌ CoCl26 H2O(s); If rain were imminent, there would be a lot of
water vapor in the air causing the reaction to shift to the right. The indicator would take on
the color of CoCl26 H2O, pink.
24. H+ + OH- → H2O; Sodium hydroxide (NaOH) will react with the H+ on the product side of
the reaction. This effectively removes H+ from the equilibrium, which will shift the reaction
to the right to produce more H+ and CrO42−. As more CrO4
2− is produced, the solution turns
yellow.
25. Test tube 1: added Cl− reacts with Ag+ to form a silver chloride precipitate. The net ionic
equation is Ag+(aq) + Cl-(aq) → AgCl(s). Test tube 2: added NH3 reacts with Ag+ ions to
form the soluble complex ion Ag(NH3)2+. As this complex ion forms, Ag+ is removed from
solution, which causes the AgCl(s) to dissolve. When enough NH3 is added, all of the silver
chloride precipitate will dissolve. The equation is AgCl(s) + 2 NH3(aq) → Ag(NH3)2+(aq) +
Cl− (aq). Test tube 3: added H+ reacts with the weak base NH3 to form NH4+. As NH3 is
removed from the Ag(NH3)2+ complex ion equilibrium, Ag+ ions are released to the solution
which can then react with Cl− to reform AgCl(s). The equations are Ag(NH3)2+(aq) + 2
27. Because each compound contains an octahedral complex ion, the formulas for the compounds
are [Co(NH3)6]I3, [Pt(NH3)4I2]I2, Na2[PtI6] and [Cr(NH3)4I2]I. Note that in some cases, the I−
ions are ligands bound to the transition metal ion as required for a coordination number of 6,
while in other cases the I− ions are counterions required to balance the charge of the complex
ion. The AgNO3 solution will only precipitate the I− counterions and will not precipitate the
I− ligands. Therefore, 3 moles of AgI will precipitate per mole of [Co(NH3)6]I3, 2 moles of
AgI will precipitate per mole of [Pt(NH3)4I2]I2, 0 moles of AgI will precipitate per mole of
Na2[PtI6], and l mole of AgI will precipitate per mole of [Cr(NH3)4I2]I.
28. BaCl2 gives no precipitate, so SO42− must be in the coordination sphere (BaSO4 is insoluble).
A precipitate with AgNO3 means the Cl− is not in the coordination sphere. Because there are
only four ammonia molecules in the coordination sphere, SO42− must be acting as a bidentate
ligand. The structure is:
29. To determine the oxidation state of the metal, you must know the charges of the various
common ligands (see Table 21.13 of the text). a. pentaamminechlororuthenium(III) ion b. hexacyanoferrate(II) ion c. tris(ethylenediamine)manganese(II) ion d. pentaamminenitrocobalt(III) ion 30. a. tetracyanonicklate(II) ion b. tetraamminedichlorochromium(III) ion c. tris(oxalato)ferrate(III) ion d. tetraaquadithiocyanatocobalt(III) ion
31. a. hexaamminecobalt(II) chloride b. hexaaquacobalt(III) iodide
c. potassium tetrachloroplatinate(II) d. potassium hexachloroplatinate(II)
e. pentaamminechlorocobalt(III) chloride f. triamminetrinitrocobalt(III)
32. a. pentaaquabromochromium(III) bromide b. sodium hexacyanocobaltate(III)
c. bis(ethylenediamine)dinitroiron(III) chloride d. tetraamminediiodoplatinum(IV)
tetraiodoplatinate(II)
Co
NH3
NH3
H3N
H3N
O
S
O
OO
+
Cl-
790 CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
cis trans
Ir
NH3
Cl
ClH3N
ClH3NIr
Cl
Cl
NH3H3N
ClH3N
33. a. K2[CoCl4] b. [Pt(H2O)(CO)3]Br2
c. Na3[Fe(CN)2(C2O4)2] d. [Cr(NH3)3Cl(H2NCH2CH2NH2)]I2
34. a. FeCl4− b. [Ru(NH3)5H2O]3+
c. [Cr(CO)4(OH)2]+ d. [Pt(NH3)Cl3] −
35. a.
Note: C2O42− is a bidentate ligand. Bidentate ligands bond to the metal at two positions
that are 90° apart from each other in octahedral complexes. Bidentate ligands do not
bond to the metal at positions 180° apart.
b.
c.
Co
OH2
OH2
O
OO
OC
C
O
O
C
C
O
O
transcis
O
C
O
CoO
CC
O
O
O
H2O
H2O
OC
O
cis
2+ 2+
trans
Pt
NH3
IH3N
H3N
I
NH3
Pt
I
IH3N
H3N
NH3
NH3
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY 791
Cr
I
N
NH3
N
I
H3N
+ + +
Cr
NH3
N
I
N
H3N
ICr
I
N
NH3
N
H3N
I
d.
Note: is an abbreviation for the bidentate ligand ethylenediamine
(H2NCH2CH2NH2).
36. a. b.
c. d.
e.
37.
M = transition metal ion
2+
H2C
H2C
OH2
OH2
H2N
H2N
CH2
CH2NH2
NH2
Cu
M
H2N
CH2
CO
O
NN
H2C
H2C
H2N
H2N
Pt
Cl
Cl H2C
H2C
H2N
H2N
Co
NH2 CH2
CH2NH2
Cl
Cl
Co
NH3
NH3
H3N Cl
H3N NO2
Co
Cl
ONO
H3N NH3
H3N NH3
792 CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
38. M = transition metal ion
39. Linkage isomers differ in the way the ligand bonds to the metal. SCN- can bond through the
sulfur or through the nitrogen atom. NO2− can bond through the nitrogen or through the
oxygen atom. OCN− can bond through the oxygen or through the nitrogen atom. N3-,
NH2CH2CH2NH2 and I− are not capable of linkage isomerism.
40.
41. Similar to the molecules discussed in Figures 21.16 and 21.17 of the text, Cr(acac)3 and cis-
Cr(acac)2(H2O)2 are optically active. The mirror images of these two complexes are
nonsuperimposable. There is a plane of symmetry in trans-Cr(acac)2(H2O)2, so it is not
optically active. A molecule with a plane of symmetry is never optically active because the
mirror images are always superimposable. A plane of symmetry is a plane through a molecule
where one side reflects the other side of the molecule.
OC
H2C H2N
Cu
NH2 CH2
CO
O
O
OC
H2C H2N
Cu
O C
CH2NH2
O O
and
HS CH2
CHHS
M
CH2OH
HO CH2
CHHS
M
CH2SH
HO CH2
CH
CH2HS
M SH
Co
NO2
NO2
H3N
H3N
NH3
NH3
Co
NO2
NO2
H3N
H3N
NH3
NH3
Co
NO2
NH3
O2N
H3N
NH3
NH3
Co
ONO
ONO
H3N
H3N
NH3
NH3
Co
ONO
NH3
ONO
H3N
NH3
NH3
Co
ONO
NO2
H3N
H3N
NH3
NH3
Co
ONO
NH3
O2N
H3N
NH3
NH3
CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY 793
42. There are five geometrical isomers (labeled i-v). Only isomer v, where the CN-, Br- and H2O
ligands are cis to each other, is optically active. The nonsuperimposable mirror image is
shown for isomer v.
Bonding, Color, and Magnetism in Coordination Compounds
43. a. Fe2+: [Ar]3d6
b. Fe3+: [Ar]3d5 c. Ni2+: [Ar]3d8
44. a. Zn2+: [Ar]3d10
Pt
CN
CNBr
Br OH2
OH2
Pt
OH2
OH2
Br
Br CN
CNPt
Br
BrH2O
H2O CN
CN
i ii iii
mirror
Pt
CN
OH2
Br
Br CN
OH2
Pt
OH2
OH2
NC
Br CN
BrPt
H2O
NC Br
Br
iv v
optically active
mirror image of v(nonsuperimposable)
NC
H2O
High spin, small Low spin, large
High spin, small
794 CHAPTER 21 TRANSITION METALS AND COORDINATION CHEMISTRY
b. Co2+: [Ar]3d7
c. Ti3+: [Ar]3d1
45. Because fluorine has a 1 charge as a ligand, chromium has a +2 oxidation state in CrF64−.
The electron configuration of Cr2+ is: [Ar]3d4. For four unpaired electrons, this must be a
weak-field (high-spin) case where the splitting of the d-orbitals is small and the number of
unpaired electrons is maximized. The crystal field diagram for this ion is:
46. NH3 and H2O are neutral ligands, so the oxidation states of the metals are Co3+ and Fe2+.
Both have six d electrons ([Ar]3d6). To explain the magnetic properties, we must have a
strong-field for Co(NH3)63+ and a weak-field for Fe(H2O)6
2+.
Co3+: [Ar]3d6 Fe2+: [Ar]3d6
Only this splitting of d-orbitals gives a diamagnetic Co(NH3)63+ (no unpaired electrons) and a