sample exercise 24.1 identifying the coordination sphere

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Chemistry: The Central Science, Eleventh EditionBy Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. MurphyWith contributions from Patrick Woodward

Sample Exercise 24.1 Identifying the Coordination Sphere of a Complex

Palladium(II) tends to form complexes with a coordination number of 4. One such compound was originally formulated as PdCl2 · 3 NH3. (a) Suggest the appropriate coordination compound formulation for this compound. (b) Suppose an aqueous solution of the compound is treated with excess AgNO3(aq). How many moles of AgCl(s) are formed per mole of PdCl2 · 3 NH3?

SolutionAnalyze: We are given the coordination number of Pd(II) and a chemical formula indicating that NH3 and Cl– are the potential ligands. We are asked to determine (a) what ligands are attached to Pd(II) in the compound and (b) how the compound behaves toward AgNO3 in aqueous solution.Plan: (a) Because of their charge, the Cl– ions can be either in the coordination sphere, where they are bonded directly to the metal, or outside the coordination sphere, where they are bonded ionically to the complex. Because the NH3 ligands are neutral, they must be in the coordination sphere. (b) The chlorides that are in the coordination sphere will not be precipitated as AgCl.Solve:(a) By analogy to the ammonia complexes of cobalt(III), we predict that the three NH3 groups of PdCl2 · 3NH3 serve as ligands attached to the Pd(II) ion. The fourth ligand around Pd(II) is one of the chloride ions.The second chloride ion is not a ligand; it serves only as an anion in this ionic compound. We conclude thatthe correct formulation is [Pd(NH3)3Cl]Cl. (b) The chloride ion that serves as a ligand will not be precipitated as AgCl(s) followingthe reaction with AgNO3(aq). Thus, only the single “free” Cl– can react. We therefore expect to produce 1 mol of AgCl(s) per mole of complex. The balanced equation is [Pd(NH3)3Cl]Cl(aq) + AgNO3(aq) → [Pd(NH3)3Cl]NO3(aq) + AgCl(s). This is a metathesis reaction (Section 4.2) in which one of the cations is the [Pd(NH3)3Cl]+ complex ion.




Sample Exercise 24.1 Identifying the Coordination Sphere of a Complex

Predict the number of ions produced per formula unit in an aqueous solution of CoCl2· 6 H2O.Answer: three (the complex ion, [Co(H2O)6]2+, and two chloride ions)

Practice Exercise




Sample Exercise 24.2 Determining the Oxidation Number of a Metal in a Complex

What is the oxidation number of the central metal in [Rh(NH3)5Cl](NO3)2?

What is the charge of the complex formed by a platinum(II) metal ion surrounded by two ammonia molecules and two bromide ions?Answer: zero

Practice Exercise

SolutionAnalyze: We are given the chemical formula of a coordination compound, and we are asked to determine the oxidation number of its metal atom.Plan: To determine the oxidation number of the Rh atom, we need to figure out what charges are contributed by the other groups in the substance. The overall charge is zero, so the oxidation number of the metal must balance the charge that is due to the rest of the compound.Solve: The NO3 group is the nitrate anion, which has a 1– charge, NO3

–. The NH3 ligands are neutral, and the Cl is a coordinated chloride ion, which has a 1– charge, Cl–. The sum of all the charges must be zero.

The oxidation number of rhodium, x, must therefore be +3.




Sample Exercise 24.3 Determining the Formula of a Complex IonA complex ion contains a chromium(III) bound to four water molecules and two chloride ions. What is its formula?

Write the formula for the complex described in the Practice Exercise accompanying Sample Exercise 24.2.Answer: [Pt(NH3)2Br2]

Practice Exercise

SolutionAnalyze: We are given a metal, its oxidation number, and the number of ligands of each kind in a complex ion containing the metal, and we are asked to write the chemical formula of the ion.Plan: We write the metal first, then the ligands. We can use the charges of the metal ion and ligands to determine the charge of the complex ion. The oxidation state of the metal is +3, water is neutral, and chloride has a 1– charge.Solve:

The charge on the ion is 1+, [Cr(H2O)4Cl2]+.




Sample Exercise 24.4 Naming Coordination CompoundName the following compounds: (a) [Cr(H2O)4Cl2]Cl, (b) K4[Ni(CN)4].

Name the following compounds: (a) [Mo(NH3)3Br3]NO3, (b) (NH4)2[CuBr4]. (c) Write the formula for sodiumdiaquabis(oxalato)ruthenate(III).Answers: (a) triamminetribromomolybdenum(IV) nitrate, (b) ammonium tetrabromocuprate(II) (c) Na[Ru(H2O)2(C2O4)2]

Practice Exercise

SolutionAnalyze: We are given the chemical formulas for two coordination compounds, and we are assigned the task of naming them.Plan: To name the complexes, we need to determine the ligands in the complexes, the names of the ligands, and the oxidation state of the metal ion. We then put the information together following the rules listed previously.Solve: (a) As ligands, there are four watermolecules, which are indicated as tetraaqua,and two chloride ions, indicated as dichloro.The oxidation state of Cr is +3.Thus, we have chromium(III). Finally, the anionis chloride. Putting these parts together, thename of the compound is(b) The complex has four cyanide ions, CN–,as ligands, which we indicate as tetracyano.The oxidation state of the nickel is zero.Becausethe complex is an anion, the metal is indicatedas nickelate(0). Putting these partstogether andnaming the cation first, we have




Sample Exercise 24.5 Determining the Number of Geometric Isomers

SolutionAnalyze: We are given the name of a complex containing only monodentate ligands, and we must determine the number of isomers the complex can form.Plan: We can count the number of ligands, thereby determining the coordination number of the Fe in the complex and then use the coordination number to predict the geometry of the complex. We can then either make a series of drawings with ligands in different positions to determine the number of isomers, or we can deduce the number of isomers by analogy to cases we have discussed.Solve: The name indicates that the complex has four carbonyl (CO) ligands and two chloro (Cl–) ligands, so its formula is Fe(CO)4Cl2. The complex therefore has a coordination number of 6, and we can assume that it has an octahedral geometry. Like [Co(NH3)4Cl2 ]+ (Figure 24.1), it has four ligands of one type and two of another. Consequently, it possesses two isomers: one with the Cl– ligands across the metal from each other (trans-Fe(CO)4Cl2) and one with the Cl– ligands adjacent (cis-Fe(CO)4Cl2).In principle, the CO ligand could exhibit linkage isomerism by binding to a metal atom via the lone pair on the O atom. When bonded this way, a CO ligand is called an isocarbonyl ligand. Metal isocarbonyl complexes are extremely rare, and we do not normally have to consider the possibility that CO will bind in this way.Comment: It is easy to overestimate the number of geometric isomers. Sometimes different orientations of a single isomer are incorrectly thought to be different isomers. If two structures can be rotated so that they are equivalent, they are not isomers of each other. The problem of identifying isomers is compounded by the difficulty we often have in visualizing three-dimensional molecules from their two-dimensional representations. It is sometimes easier to determine the number of isomers if we use three-dimensional models.

The Lewis structure of the CO molecule indicates that the molecule has a lone pair of electrons on the C atom and one on the O atom . When CO binds to a transition-metal atom, it nearly always does so by using the lone pair on the C atom. How many geometric isomers are there for tetracarbonyldichloroiron(II)?




Sample Exercise 24.5 Determining the Number of Geometric Isomers

How many isomers exist for square-planar [Pt(NH3)2ClBr]?Answer: two

Practice Exercise




Sample Exercise 24.6 Predicting Whether a Complex Has Optical IsomersDoes either cis- or trans-[Co(en)2Cl2]+ have optical isomers?

SolutionAnalyze: We are given the chemical formula for two structural isomers, and we are asked to determine whether either one has optical isomers. The en ligand is a bidentate ligand, so the complexes are six-coordinate and octahedral.Plan: We need to sketch the structures of the cis and trans isomers and their mirror images. We can draw the en ligand as two N atoms connected by a line, as is done in Figure 24.20. If the mirror image cannot be superimposed on the original structure, the complex and its mirror image are optical isomers.Solve: The trans isomer of [Co(en)2Cl2]+ and its mirror image is

where the dashed vertical line represents a mirror. Notice that the mirror image of the trans isomer is identical to the original. Consequently trans-[Co(en)2Cl2 ]+ does not exhibit optical isomerism.The mirror image of the cis isomer of [Co(en)2Cl2]+, however, cannot be superimposed on the original

Thus, the two cis structures are optical isomers (enantiomers): cis-[Co(en)2Cl2 ]+ is a chiral complex.




Sample Exercise 24.6 Predicting Whether a Complex Has Optical Isomers

Does the square-planar complex ion [Pt(NH3)(N3)ClBr]– have optical isomers?Answer: no

Practice Exercise




Sample Exercise 24.7 Relating Color Absorbed to Color ObservedThe complex ion trans-[Co(NH3)4Cl2]+ absorbs light primarily in the red region of the visible spectrum (the most intense absorption is at 680 nm). What is the color of the complex?

The [Cr(H2O)6 ]2+ ion has an absorption band at about 630 nm. Which of the following colors—sky blue, yellow, green, or deep red—is most likely to describe this ion?Answer: sky blue

Practice Exercise

SolutionAnalyze: We need to relate the color absorbed by a complex (red) to the color observed for the complex.Plan: The color observed for a substance is complementary to the color it absorbs. We can use the color wheel of Figure 24.24 to determine the complementary color.Solve: From Figure 24.24, we see that green is complementary to red, so the complex appears green.Comment: As noted in Section 24.1, in the text discussing Table 24.1, this green complexwas one of those that helped Werner establish his theory of coordination. The other geometric isomer of this complex, cis-[Co(NH3)4Cl2]+, absorbs yellow light and therefore appears violet.




Sample Exercise 24.8 Using thenSpectrochemical SeriesWhich of the following complexes of Ti3+ exhibits the shortest wavelength absorption in the visible spectrum: [Ti(H2O)6 ]3+, [Ti(en)3]3+, or [TiCl6]3–?

The absorption spectrum of [Ti(NCS)6]3– shows a band that lies intermediate in wavelength between those for [TiCl6]3– and [TiF6]3–. What can we conclude about the place of NCS– in the spectrochemical series?Answer: It lies between Cl– and F–; that is, Cl– < NCS– < F–.

Practice Exercise

SolutionAnalyze: We are given three octahedral complexes, each containing Ti in the +3 oxidation state. We need to predict which complex absorbs the shortest wavelength of visible light.Plan: Ti(III) is a d1 ion, so we anticipate that the absorption is due to a d-d transition in which the 3d electron is excited from the lower-energy t2 set to the higher-energy e set. The wavelength of the light absorbed is determined by the magnitude of the energy difference, Δ. Thus, we use the position of the ligands in the spectrochemical series to predict the relative values of . The larger the energy, the shorter the wavelength (Equation 24.7). Solve: Of the three ligands involved—H2O, en, and Cl–—we see that ethylenediamine (en) is highest in the spectrochemical series and will therefore cause the largest splitting, Δ, of the t2 and e sets of orbitals. The larger the splitting, the shorter the wavelength of the light absorbed. Thus, the complex that absorbs the shortestwavelength light is [Ti(en)3]3+.




Sample Exercise 24.9 Predicting the Number of Unpaired Electrons in an Octahedral Complex

Predict the number of unpaired electrons in six-coordinate high-spin and low-spin complexes of Fe3+ .

For which d electron configurations in octahedral complexes is it possible to distinguish between high-spin and low-spin arrangements?Answer: d4, d5, d6, d7

Practice Exercise

SolutionAnalyze: We must determine how many unpaired electrons there are in the highspin and low-spin complexes of the metal ion Fe3+.Plan: We need to consider how the electrons populate d orbitals for Fe3+ when the metal is in an octahedral complex. There are two possibilities: one giving a high-spin complex and the other giving a low-spin complex. The electron configuration of Fe3+ gives us the number of d electrons. We then determine how these electrons populate the t2 set and e set of d orbitals. In the high-spin case, the energy difference between the t2 and e orbitals is small, and the complex has the maximum number of unpaired electrons. In the low-spin case, the energy difference between the t2 and e orbitals is large, causing the t2 orbitals to be filled before any electrons occupy the e orbitals.Solve: Fe3+ is a d5 ion. In a high-spin complex, all five of these electrons are unpaired, with three in the t2orbitals and two in the e orbitals. In a low-spin complex, all five electrons reside in the t2 set of d orbitals, so there is one unpaired electron:




Sample Exercise 24.10 Populating d Orbitals in Tetrahedral and Square-Planar Complexes

Four-coordinate nickel(II) complexes exhibit both square-planar and tetrahedral geometries. The tetrahedral ones, such as [NiCl4]2–, are paramagnetic; the square-planar ones, such as [Ni(CN)4]2–, are diamagnetic. Show how the d electrons of nickel(II) populate the d orbitals in the appropriate crystal-field splitting diagram in each case.SolutionAnalyze: We are given two complexes containing Ni2+, a tetrahedral one and a square-planar one. We are asked to use the appropriate crystal-field diagrams to describe how the d electrons populate the d orbitals in each case.Plan: We need to first determine the number of d electrons possessed by Ni2+ and then use Figure 24.34 for the tetrahedral complex and Figure 24.35 for the squareplanar complex.Solve: Nickel(II) has an electron configuration of [Ar]3d8. The population of the d electrons in the two geometries is

Comment: Notice that the tetrahedral complex is paramagnetic with two unpaired electrons, whereas the square-planar complex is diamagnetic. Nickel(II) forms octahedral complexes more frequently than square-planar ones, whereas heavier d8 metals tend to favor square planar coordination.




Sample Exercise 24.10 Populating d Orbitals in Tetrahedral and Square-Planar Complexes

How many unpaired electrons do you predict for the tetrahedral [CoCl4]2– ion?Answer: three

Practice Exercise




Sample Integrative Exercise Putting Concepts Together

The oxalate ion has the Lewis structure shown in Table 24.2. (a) Show the geometrical structure of the complex formed by coordination of oxalate to cobalt(II), forming [Co(C2O4)(H2O)4]. (b)Write the formula for the salt formed upon coordination of three oxalate ions to Co(II), assuming that the charge-balancing cation is Na+. (c) Sketch all the possible geometric isomers for the cobalt complex formed in part (b). Are any ofthese isomers chiral? Explain. (d) The equilibrium constant for the formation of the cobalt(II) complex produced by coordination of three oxalate anions, as in part (b), is 5.0 × 109. By comparison, the formation constant for formation of the cobalt(II) complex with three molecules of ortho-phenanthroline (Table 24.2) is 9 × 1019.From these results, what conclusions can you draw regarding the relative Lewis base properties of the two ligands toward cobalt(II)? (e) Using the approach described in Sample Exercise 17.14, calculate the concentration of free aqueous Co(II) ion in a solution initially containing 0.040 M oxalate ion and 0.0010 M Co2+(aq) .Solution(a) The complex formed by coordination of one oxalate ion is octahedral:




Sample Integrative Exercise Putting Concepts Together Solution (Continued)(b) Because the oxalate ion has a charge of 2–, the net charge of a complex with three oxalate anions and one Co2+ ion is 4–. Therefore, the coordination compound has the formula Na4[Co(C2O4)3].(c) There is only one geometric isomer. The complex is chiral, however, in the same way as the [Co(en)3]3+

complex, shown in Figure 24.20(b). These two mirror images are not superimposable, so there are two enantiomers:

(d) The ortho-phenanthroline ligand is bidentate, like the oxalate ligand, so they both exhibit the chelate effect. Thus, we can conclude that ortho-phenanthroline is a stronger Lewis base toward Co2+ than oxalate. This conclusion is consistent with what we learned about bases in Section 16.7, namely that nitrogen bases are generally stronger than oxygen bases. (Recall, for example, that NH3 is a stronger base than H2O.)




Sample Integrative Exercise Putting Concepts Together Solution (Continued)(e) The equilibrium we must consider involves three moles of oxalate ion (represented as Ox2– ).

The formation-constant expression is

Because Kf is so large, we can assume that essentially all of the Co2+ is converted to the oxalato complex. Under that assumption, the final concentration of [Co(Ox)3]3– is 0.0010 M and that of oxalate ion is [Ox2–] = (0.040) – 3(0.0010) = 0.037 M (three Ox2– ions react with each Co2+ ion).We then have

Inserting these values into the equilibrium-constant expression, we have

Solving for x, we obtain 4 × 10-9 M. From this, we can see that the oxalate has complexed all but a tiny fraction of the Co2+ present in solution.

sample exercise 24.1 identifying the coordination sphere

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