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? Solution

Analyze: 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 · 3 NH3 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 that the correct formulation is [Pd(NH3)3Cl]Cl. (b) The chloride ion that serves as a ligand will not be precipitated as AgCl(s) following – the 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.

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

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

Solution Analyze: 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.

Practice Exercise 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

+ The charge on the ion is 1+, [Cr(H2O)4Cl2] . Practice Exercise Write the formula for the complex described in the Practice Exercise accompanying Sample Exercise 24.2. Answer: [Pt(NH3)2Br2]

Name the following compounds: (a) [Cr(H2O)4Cl2]Cl, (b) K4[Ni(CN)4]. 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, 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 water molecules, 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, the name 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 indicated as nickelate(0). Putting these partstogether and naming the cation first, we have Practice Exercise

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

Chemistry: The Central Science, Eleventh Edition Copyright ©2009 by Pearson Education, Inc. By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy Upper Saddle River, New Jersey 07458 With contributions from Patrick Woodward All rights reserved. Sample Exercise 24.5 Determining the Number of Geometric Isomers 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)? Solution Analyze: 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. Chemistry: The Central Science, Eleventh Edition Copyright ©2009 by Pearson Education, Inc. By Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, and Catherine J. Murphy Upper Saddle River, New Jersey 07458 With contributions from Patrick Woodward All rights reserved. Sample Exercise 24.5 Determining the Number of Geometric Isomers Practice Exercise

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

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.

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

Practice Exercise 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

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.

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 of these 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:

(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.)

The formation-constant expression is

2+ Because Kf is so large, we can assume that essentially all of the Co is converted to the oxalato complex. 3– 2– Under that assumption, the final concentration of [Co(Ox)3] is 0.0010 M and that of oxalate ion is [Ox ] = (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.