Ch24 Sample Exercise

7/29/2019 Ch24 Sample Exercise

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Copyright 2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458All rights reserved.

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 originallyformulated as PdCl2 3 NH3. (a) Suggest the appropriate coordination compound formulation for thiscompound. (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?

Solution

Analyze: We are given the coordination number of Pd(II) and a chemical formula indicating that NH3 andCl are the potential ligands. We are asked to determine(a) what ligands are attached to Pd(II) in thecompound 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 thecomplex. Because the NH3 ligands are neutral, they must be in the coordination sphere. (b)The chloridesthat 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 1mol 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 ofthe cations is the [Pd(NH3)3Cl]

+complex ion.


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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


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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 moleculesand two bromide ions?Answer: zero

Practice Exercise

Solution

Analyze: We are given the chemical formula of a coordination compound, and we are asked to determinethe oxidation number of its metal atom.Plan:To determine the oxidation number of the Rh atom, we need to figure out what charges arecontributed by the other groups in the substance. The overall charge is zero, so the oxidation number of themetal 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, andthe 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.


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Sample Exercise 24.3 Determining the Formula of a Complex Ion

A complex ion contains a chromium(III) bound to four water molecules and two chloride ions. What is itsformula?

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

Practice Exercise

Solution

Analyze: We are given a metal, its oxidation number, and the number of ligands of each kind in a complexion 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 todetermine the charge of the complex ion. The oxidation state of the metal is +3, water is neutral, andchloride has a 1 charge.Solve:

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


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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

Solution

Analyze: 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, andthe 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 anion

is 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


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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 mustdetermine 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 thecomplex and then use the coordination number to predict the geometry of the complex. We can then eithermake a series of drawings with ligands in different positions to determine the number of isomers, or we candeduce 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, soits formula is Fe(CO)4Cl2. The complex therefore has a coordination number of 6, and we can assume that ithas an octahedral geometry. Like [Co(NH3)4Cl2 ]

+(Figure 24.1), it has four ligands of one type and two ofanother. 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 onthe O atom. When bonded this way, a CO ligand is called an isocarbonyl ligand. Metal isocarbonylcomplexes 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 ofa single isomer are incorrectly thought to be different isomers. If two structures can be rotated so that theyare equivalent, they are not isomers of each other. The problem of identifying isomers is compounded by thedifficulty we often have in visualizing three-dimensional molecules from their two-dimensionalrepresentations. It is sometimes easier to determine the number of isomers if we use three-dimensionalmodels.

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


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Sample Exercise 24.5 Determining the Number of Geometric Isomers

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

Practice Exercise


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Sample Exercise 24.6 Predicting Whether a Complex Has Optical Isomers

Does eithercis- or trans-[Co(en)2Cl2]+have optical isomers?

Solution

Analyze: 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 drawthe en ligand as two N atoms connected by a line, as is done in Figure 24.20. If the mirror image cannot besuperimposed 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 isidentical 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.


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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


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Sample Exercise 24.7 Relating Color Absorbed to Color Observed

The complex iontrans-[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 colorssky blue,

yellow, green, or deep redis most likely to describe this ion?Answer: sky blue

Practice Exercise

Solution

Analyze: 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 colorwheel 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 thiscomplex,cis-[Co(NH3)4Cl2]

+, absorbs yellow light and therefore appears violet.


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Copyright 2009 by Pearson Education, Inc.Upper Saddle River, New Jersey 07458

All rights reserved.


Sample Exercise 24.8 Using thenSpectrochemical Series

Which 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


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Sample Exercise 24.9 Predicting the Number of Unpaired Electrons in anOctahedral Complex

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

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

Practice Exercise

Solution

Analyze: We must determine how many unpaired electrons there are in the highspin and low-spincomplexes of the metal ion Fe3+.Plan: We need to consider how the electrons populatedorbitals for Fe3+ when the metal is in an octahedralcomplex. There are two possibilities: one giving a high-spin complex and the other giving a low-spincomplex. The electron configuration of Fe3+gives us the number ofdelectrons. We then determine howthese electrons populate thet2 set andeset ofdorbitals. In the high-spin case, the energy differencebetween thet2 andeorbitals is small, and the complex has the maximum number of unpaired electrons. Inthe low-spin case, the energy difference between thet2 andeorbitals is large, causing thet2orbitals to befilled before any electrons occupy theeorbitals.Solve: Fe3+is ad5 ion. In a high-spin complex, all five of these electrons are unpaired, with three in thet2orbitals and two in theeorbitals. In a low-spin complex, all five electrons reside in thet2 set ofdorbitals, sothere is one unpaired electron:


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Sample Exercise 24.10 Populating d Orbitals in Tetrahedral and Square-PlanarComplexes

Four-coordinate nickel(II) complexes exhibit both square-planar and tetrahedral geometries. The tetrahedralones, such as [NiCl4]

2, are paramagnetic; the square-planar ones, such as [Ni(CN)4]2, are diamagnetic. Show

how thedelectrons of nickel(II) populate thedorbitals in the appropriate crystal-field splitting diagram in

each case.Solution

Analyze: We are given two complexes containing Ni2+, a tetrahedral one and a square-planar one. We areasked to use the appropriate crystal-field diagrams to describe how thedelectrons populate thedorbitals ineach case.Plan: We need to first determine the number ofdelectrons possessed by Ni2+and then use Figure 24.34 forthe tetrahedral complex and Figure 24.35 for the squareplanar complex.

Solve: Nickel(II) has an electron configuration of [Ar]3d8. The population of thedelectrons in the twogeometries is

Comment: Notice that the tetrahedral complex is paramagnetic with two unpaired electrons, whereas thesquare-planar complex is diamagnetic. Nickel(II) forms octahedral complexes more frequently than square-planar ones, whereas heavierd8 metals tend to favor square planar coordination.


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Sample Exercise 24.10 Populating d Orbitals in Tetrahedral and Square-PlanarComplexes

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

Answer: three

Practice Exercise


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Sample Integrative Exercise Putting Concepts Together

The oxalate ion has the Lewis structure shown in Table 24.2. (a) Show the geometrical structure of thecomplex 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 isNa+. (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) complexproduced by coordination of three oxalate anions, as in part (b), is 5.0 109. By comparison, the formationconstant for formation of the cobalt(II) complex with three molecules ofortho-phenanthroline (Table 24.2) is9 1019.From these results, what conclusions can you draw regarding the relative Lewis base properties of the twoligands 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 0.040M oxalate ion and 0.0010MCo2+(aq) .

Solution

(a)The complex formed by coordination of one oxalate ion is octahedral:


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Solution (Continued)

(b) Because the oxalate ion has a charge of 2, the net charge of a complex with three oxalate anions andone 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 twoenantiomers:

(d)Theortho-phenanthroline ligand is bidentate, like the oxalate ligand, so they both exhibit the chelate

effect. Thus, we can conclude thatortho-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 basesare generally stronger than oxygen bases. (Recall, for example, that NH3 is a stronger base than H2O.)


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Chemistry: The Central Science, Eleventh EditionBy Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J . MurphyWithcontributions fromPatrickWoodward


Solution (Continued)

(e)The equilibrium we must consider involves three moles of oxalate ion (represented as Ox2 ).

The formation-constant expression is

BecauseKfis 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 0.0010M and that of oxalate ion is [Ox

2

] =(0.040) 3(0.0010) =0.037M (three Ox2 ions react with each Co2+ion).We then have

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

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

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Ch24 Sample Exercise