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Coordination Chemistry: Complex Formation and Behavior

Explore coordination chemistry through examples of complex compounds. Learn to identify coordination spheres, ligands, and oxidation numbers of metals in complex ions. Practice naming coordination compounds.

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Coordination Chemistry: Complex Formation and Behavior

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  1. Sample Exercise 23.1Identifying the Coordination Sphere of a Complex Palladium(II) tends to form complexes with coordination number 4. A compound has the composition PdCl2· 3 NH3. (a) Write the formula for this compound that best shows the coordination structure. (b) When 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 the complex contains NH3 and Cl–. We are asked to determine (a) which 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. The electrically neutral NH3 ligands must be in the coordination sphere, if we assume four ligands bonded to the Pd(II) ion. (b) Any chlorides in the coordination sphere do not precipitate as AgCl.

  2. Sample Exercise 23.1Identifying the Coordination Sphere of a Complex Continued Solve (a) By analogy to the ammonia complexes of cobalt(III) shown in Figure 23.7, we predict that the three NH3 are ligands attached to the Pd(II) ion. The fourth ligand around Pd(II) is one chloride ion. The second chloride ion is not a ligand; it serves only as a counterion (a noncoordinating ion that balances charge) in the compound. We conclude that the formula showing the structure best is [Pd(NH3)3Cl]Cl.

  3. Sample Exercise 23.1Identifying the Coordination Sphere of a Complex Continued (b) Because only the non-ligand can react, we 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. Practice Exercise Predict the number of ions produced per formula unit in an aqueous solution of CoCl2· 6 H2O. Answer: three: [Co(H2O)6]2+ and two Cl–

  4. Sample Exercise 23.2Determining the Oxidation Number of a Metal in a Complex What is the oxidation number of the metal in [Rh(NH3)5Cl](NO3)2? Solution Analyze We are given the chemical formula of a coordination compound and asked to determine the oxidation number of its metal atom. Plan To determine the oxidation number of Rh, we need to figure out what charges are contributed by the other groups. The overall charge is zero, so the oxidation number of the metal must balance the charge due to the rest of the compound. Solve The NO3 group is the nitrate anion, which has a 1– charge. The NH3 ligands carry zero charge, and the Cl is a coordinated chloride ion, which has a 1– charge. The sum of all the charges must be zero: x + 5(0) + (–1) + 2(–1) = 0 [Rh(NH3)5Cl](NO3)2 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

  5. Sample Exercise 23.3Determining the Formula of a Complex Ion A complex ion contains a chromium(III) bound to four water molecules and to two chloride ions.What is the formula and charge of this ion? Solution Analyze 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 and charge 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 has no charge, and chloride has a 1– charge. Solve +3 + 4(0) + 2(–1) = +1 Cr(H2O)4Cl2 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 23.2. Answer: [Pt(NH3)2Br2]

  6. Sample Exercise 23.4Naming Coordination Compounds Name the compounds (a) [Cr(H2O)4Cl2]Cl, (b) K4[Ni(CN)4]. Solution Analyze We are given the chemical formulas for two coordination compounds and 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 in the text. Solve (a) The ligands are four water molecules—tetraaqua—and two chloride ions—dichloro. By assigning all the oxidation numbers we know for this molecule, we see that the oxidation number of Cr is +3: Thus, we have chromium(III). Finally, the anion is chloride. The name of the compound is (b) The complex has four cyanide ion ligands, CN–, which means tetracyano, and the oxidation state of the nickel is zero: Because the complex is an anion, the metal is indicated as nickelate(0). Putting these parts together and naming the cation first, we have +3 + 4(0) + 2(–1) + (–1) + 0 [Cr(H2O)4Cl2]Cl tetraaquadichlorochromium(III) chloride 4(+1) + 0 + 4(–1) = 0 K4[Ni(CN)4] potassium tetracyanonickelate(0)

  7. Sample Exercise 23.4Naming Coordination Compounds Continued Practice Exercise Name the compounds (a) [Mo(NH3)3Br3]NO3, (b) (NH4)2[CuBr4]. (c) Write the formula for sodium diaquabis(oxalato)ruthenate(III). Answer:(a) triamminetribromomolybdenum(IV) nitrate, (b) ammonium tetrabromocuprate(II) (c) Na[Ru(H2O)2(C2O4)2]

  8. Sample Exercise 23.5Determining the Number of Geometric Isomers The Lewis structure indicates that the CO molecule has two lone pairs of electrons. When CO binds to a transition-metal atom, it nearly always does so by using the C lone pair. 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 to determine the coordination number of the Fe and then use the coordination number to predict the geometry. We can then either make a series of drawings with ligands in different positions to determine the number of isomers or deduce the number of isomers by analogy to cases we have discussed.

  9. Sample Exercise 23.5Determining the Number of Geometric Isomers Continued 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 an octahedral geometry. Like [Co(NH3)4Cl2]+ (Figure 23.7), it has four ligands of one type and two of another. Consequently, there are two isomers possible: one with the Cl– ligands across the metal from each other, trans-[Fe(CO)4Cl2], and one with the Cl– ligands adjacent to each other, cis-[Fe(CO)4Cl2]. 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. Practice Exercise How many isomers exist for the square-planar molecule [Pt(NH3)2ClBr]? Answer: two

  10. Sample Exercise 23.6Predicting Whether a Complex Has Optical Isomers Does either cis-[Co(en)2Cl2]+ or trans-[Co(en)2Cl2]– have optical isomers? Solution Analyze We are given the chemical formula for two geometric isomers and asked to determine whether either one has optical isomers. Because en is a bidentate ligand, we know that both complexes are octahedral and both have coordination number 6. Plan(a) 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 an arc. 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 are:

  11. Sample Exercise 23.6Predicting Whether a Complex Has Optical Isomers Continued Notice that the mirror image of the isomer is identical to the original. Consequently trans-[Co(en)2Cl2]+ does not exhibit optical isomerism. The mirror image of the cis isomer cannot be superimposed on the original: Thus, the two cis structures are optical isomers (enantiomers). We say that cis-[Co(en)2Cl2]+ is a chiral complex. Practice Exercise Does the square-planar complex ion [Pt(NH3)(N3)ClBr]– have optical isomers? Explain your answer. Answer: no, because the complex is flat. This complex ion does, however, have geometric isomers (for example, the Cl and Br ligands could be cis or trans).

  12. Sample Exercise 23.7Relating Color Absorbed to Color Observed The complex ion trans-[Co(NH3)]Cl2]+ 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? Solution Analyze We need to select the system in each pair that has the greater entropy. Plan For an object that absorbs only one color from the visible spectrum, the color we see is complementary to the color absorbed. We can use the color wheel of Figure 23.25 to determine the complementary color. Solve From Figure 23.25, we see that green is complementary to red, so the complex appears green.

  13. Sample Exercise 23.7Relating Color Absorbed to Color Observed Continued Comment As noted in Section 23.2, this green complex was one of those that helped Werner establish his theory of coordination (Table 23.3). The other geometric isomer of this complex, cis-[Co(NH3)4Cl2]+, absorbs yellow light and therefore appears violet. Practice Exercise A certain transition-metal complex ion absorbs at 630 nm. Which color is this ion most likely to be—blue, yellow, green, or orange? Answer: blue

  14. Sample Exercise 23.8Using the Spectrochemical Series [Ti(H2O)6]3+, [Ti(en)3]3+, and [TiCl6]3– all absorb visible light. Which one absorbs at the shortest wavelength? Solution Analyze 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 a t2 orbital to an e orbital. The wavelength of the light absorbed is determined by the magnitude of the crystal-field splitting energy, Δ. Thus, we use the position of the ligands in the spectrochemical series to predict the relative values of Δ. The larger the splitting energy, the shorter the wavelength. Solve Of these three ligands, ethylenediamine (en) is highest in the spectrochemical series and therefore causes the largest split between the t2 and e orbitals. The larger the split, the shorter the wavelength of the light absorbed. Thus, the complex that absorbs the shortestwavelength light is [Ti(en)3]3+. Practice Exercise The absorption spectrum of a Ti(III) complex containing the ligand L, [TiL6]3–, shows a peak maximum at a wavelength intermediate between the wavelengths of the absorption maxima for [TiCl6]3– and [TiF6]3–.What can we conclude about the place of L in the spectrochemical series? Answer: It lies between Cl– and F–.

  15. Sample Exercise 23.9Predicting the Number of Unpaired Electrons in an Octahedral Complex Predict the number of unpaired electrons in high-spin and low-spin Fe3+ complexes that have a coordination number of 6. Solution Analyze We must determine how many unpaired electrons there are in the high-spin and low-spin complexes of Fe3+. Plan We need to consider how the electrons populate the d orbitals in 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 and e sets 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 electrons are unpaired, with three in the t2 orbitals and two in the e orbitals. In a low-spin complex, all five electrons reside in the t2 set, so there is one unpaired electron:

  16. Sample Exercise 23.10Populating d Orbitals in Tetrahedral and Square-Planar Complexes Nickel(II) complexes in which the metal coordination number is 4 can have either squareplanar or tetrahedral geometry. [NiCl4]2– is paramagnetic, and is diamagnetic. One of these complexes is square planar, and the other is tetrahedral. Use the relevant crystalfield splitting diagrams in the text to determine which complex has which geometry. Solution Analyze We are given two complexes containing Ni2+ and their magnetic properties. We are given two molecular geometry choices and asked to use crystal-field splitting diagrams from the text to determine which complex has which geometry. Plan We need to determine the number of d electrons in and then use Figure 23.33 for the tetrahedral complex and Figure 23.34 for the square-planar complex.

  17. Sample Exercise 23.10Populating d Orbitals in Tetrahedral and Square-Planar Complexes Continued Solve Nickel(II) has the electron configuration [Ar]3d8. Tetrahedral complexes are always high spin, and square-planar complexes are almost always low spin. Therefore, the population of the d electrons in the two geometries is The tetrahedral complex has two unpaired electrons, and the square-planar complex has none. We know from Section 23.1 that the tetrahedral complex must be paramagnetic and the square planar must be diamagnetic. Therefore, [NiCl4]2– is tetrahedral, and [Ni(CN)4]2– is square planar.

  18. Sample Exercise 23.10Populating d Orbitals in Tetrahedral and Square-Planar Complexes Continued Comment Nickel(II) forms octahedral complexes more frequently than square-planar ones, whereas heavier d8 metals tend to favor square-planar coordination. Practice Exercise How many unpaired electrons do you predict for the tetrahedral [CoCl4]2– ion? Answer: three

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