Chapter 23 Transition Metals and Coordination Chemistry

Transcription

1 Lecture Presentation Chapter 23 and Coordination Chemistry James F. Kirby Quinnipiac University Hamden, CT

2 Color Catalysts Magnets 23.1 The 1036 Why are of Interest? Biological roles Coordination compounds (metals bonded to molecules and ions)

3 Minerals Most metals, including transition metals, are found in solid inorganic compounds known as minerals. Minerals are named by common, not chemical, names. Most transition metals range from +1 to +4 oxidation state in minerals.

4 Metallurgy The science and technology of extracting metals from their natural sources and preparing them for practical use Steps often involved: 1)Mining 2)Concentrating the ore 3)Reducing the ore to free metal 4)Purifying the metal 5)Mixing it with other elements to modify its properties (making an alloy a solid mixture)

5 Properties of the First Row First row means period 4. Periods 5 and 6 have similar trends in properties.

6 Atomic Radius As one goes from left to right, a decrease, then an increase, is seen in the radius of transition metals. On the one hand, increasing effective nuclear charge tends to make atoms smaller. On the other hand, the strongest (and, therefore, shortest) metallic bonds are found in the center of the transition metals. Periods 5 and 6 are about the same size due to the lanthanide contraction the effect of 4f electrons on effective nuclear charge.

7 Metal Characteristics Partially occupied d sublevels lead to the possibility of 1)multiple oxidation states; 2)colored compounds; 3)magnetic properties.

8 Oxidation States For the period 4 transition elements, when cations are formed, they lose the 4s electrons first; all (except Sc) form a +2 cation (have a +2 oxidation state). from Sc to Mn, the maximum oxidation state is the sum of 4s and 3d electrons. after Mn, the maximum oxidation number decreases, until Zn, which is ONLY +2.

9 Magnetism Electrons possess spin, causing a magnetic moment. When all electrons are spin-paired, the moments cancel each other out: this is a diamagnetic solid. With unpaired electron(s), the substance is called paramagnetic. In these substances, the adjacent atoms don t affect each other. In three other types of magnetism, the atoms affect each other: ferromagnetic, antiferromagnetic, and ferrimagnetic. (These become paramagnetic at higher temperatures.)

10 Ferromagnetism In ferromagnetic substances, the unpaired spins influence each other to align in the same direction, thereby exhibiting strong attractions to an external magnetic field. Such species are permanent magnets. Elements: Fe, Co, Ni; also many alloys

11 Antiferromagnetism Antiferromagnetic substances have unpaired spins on adjacent atoms that align in opposing directions. These magnetic fields tend to cancel each other. Examples element: Cr; alloys: FeMn; transition metal oxides: Fe 2 O 3, LaFeO 3, MnO

12 Ferrimagnetism Ferrimagnetic substances have spins that align opposite each other, but the spins are not equal, so there is a net magnetic field. This can occur because magnetic centers have different numbers of unpaired electrons; more sites align in one direction than the other; both of these conditions apply. Examples are NiMnO 3, Y 3 Fe 5 O 12, and Fe 3 O 4.

13 23.2 -Metal Complexes 1040 Complexes Commonly, transition metals can have molecules or ions that bond to them, called ligands. These give rise to complex ions or coordination compounds. Many colors are observed in transition metal complexes. Ligands act as Lewis bases, donating a pair of electrons to form the ligand metal bond. Four of the most common ligands:

14 Alfred Werner s Theory on Metal Complexes Many compounds exist combining CoCl 3 and NH 3. Their nature was explained by Alfred Werner in The oxidation number of a metal is +3 in each compound. However, the number of atoms bonded to the metal is different. He called this the coordination number.

15 Werner s Theory The key to solving this problem is the number of ions produced in solution per formula unit: along with ONE cation, the rest would tell how many Cl ions are NOT connected directly to the metal. Precipitation of AgCl confirmed amount of free Cl. Writing the formula: the brackets show the complex; counterions are written after.

16 The Metal Ligand Bond The reaction between a metal and a ligand is a reaction between a Lewis acid (the metal) and a Lewis base (the ligand). The new complex has distinct physical and chemical properties (e.g., color, reduction potential).

17 Sample Exercise 23.1 Identifying the Coordination Sphere of a Complex Palladium(II) tends to form complexes with coordination number 4. A compound has the composition PdCl 2 3 NH 3. (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 AgNO 3 (aq), how many moles of AgCl(s) are formed per mole of PdCl 2 3 NH 3? Solution Analyze We are given the coordination number of Pd(II) and a chemical formula indicating that the complex contains NH 3 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 AgNO 3 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 NH 3 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. Solve (a) By analogy to the ammonia complexes of cobalt(iii) shown in Figure 23.7, we predict that the three NH 3 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(NH 3 ) 3 Cl]Cl.

18 Sample Exercise 23.1 Identifying the Coordination Sphere of a Complex Continued (b) Because only the non-ligand Cl can react, we expect to produce 1 mol of AgCl(s) per mole of complex. The balanced equation is [Pd(NH 3 ) 3 Cl]Cl(aq) + AgNO 3 (aq) [Pd(NH 3 ) 3 Cl]NO 3 (aq) + AgCl(s) This is a metathesis reaction (Section 4.2) in which one of the cations is the [Pd(NH 3 ) 3 Cl] + complex ion. Practice Exercise 1 When the compound RhCl 3 4 NH 3 is dissolved in water and treated with excess AgNO 3 (aq) one mole of AgCl(s) is formed for every mole of RhCl 3 4 NH 3. What is the correct way to write the formula of this compound? (a) [Rh(NH 3 ) 4 Cl 3 ], (b) [Rh(NH 3 ) 4 Cl 2 ]Cl, (c) [Rh(NH 3 ) 4 Cl]Cl 2, (d) [Rh(NH 3 ) 4 ]Cl 3, (e) [RhCl 3 ](NH 3 ) 4. Practice Exercise 2 Predict the number of ions produced per formula unit when the compound CoCl 2 6 H 2 O dissolves in water to form an aqueous solution.

19 Sample Exercise 23.2 Determining the Oxidation Number of a Metal in a Complex What is the oxidation number of the metal in [Rh(NH 3 ) 5 Cl](NO 3 ) 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 NO 3 group is the nitrate anion, which has a 1 charge. The NH 3 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: The oxidation number of rhodium, x, must therefore be +3.

20 Sample Exercise 23.2 Determining the Oxidation Number of a Metal in a Complex Continued Practice Exercise 1 In which of the following compounds does the transition-metal have the highest oxidation number? (a) [Co(NH 3 )4Cl 2 ], (b) K 2 [PtCl 6 ], (c) Rb 3 [MoO 3 F 3 ], (d) Na[Ag(CN) 2 ], (e) K4[Mn(CN) 6 ]. Practice Exercise 2 What is the charge of the complex formed by a platinum(ii) metal ion surrounded by two ammonia molecules and two bromide ions?

21 23.3 Common Ligands in Coordination Chemistry 1045 Coordination Numbers The coordination number of a metal depends upon the size of the metal and the size of the ligands. Iron(III) can bind to 6 fluorides but only 4 chlorides (larger). The most common coordination numbers are 4 and 6. They correspond to common geometries: tetrahedral or square planar; octahedral.

22 Common Ligands The table shown contains some ligands commonly found in complexes. Monodentate ligands coordinate to one site on the metal, bidentate to two sites.

23 Bidentate and polydentate ligands are also called chelating agents. There are many transition metals that are vital to human life. Chelates Several of these are bound to chelating agents.

24 Chelates in Biological Systems The porphine molecule is the basis for many important biological metal chelates, becoming a porphyrin ring. The iron in hemoglobin carries O 2 and CO 2 through the blood. It contains heme units. Chlorophylls also have metals bound to porphine units.

25 23.4 Nomenclature and Isomerism in Coordination Chemistry 1050 Nomenclature Rules for Coordination Chemistry 1. In naming complexes that are salts, the name of the cation is given before the name of the anion. 2. In naming complex ions or molecules, the ligands are named before the metal. Ligands are listed in alphabetical order, regardless of their charges.

26 Nomenclature Rules 3. The names of anionic ligands end in the letter o, but electrically neutral ligands ordinarily bear the name of the molecules (exceptions: ammonia, water, CO).

27 Nomenclature Rules 4. Greek prefixes (di-, tri-, tetra-, etc.) are used to indicate the number of each kind of ligand when more than one is present. If the ligand contains a Greek prefix or is polydentate, the prefixes bis-, tris-, tetrakis-, etc. are used and the ligand name is placed in parentheses. 5. If the complex is an anion, its name ends in -ate. 6. The oxidation number of the metal is given in parentheses in Roman numerals following the name of the metal.

28 Nomenclature Examples [Ni(NH 3 ) 6 ]Br 2 = hexaamminenickel(ii) bromide Na 2 [MoOCl 4 ] = sodium tetrachlorooxomolybdate(iv) [Co(en) 2 (H 2 O)(CN)]Cl 2 = aquacyanobis(ethylenediamine)cobalt(iii) chloride

29 Isomers Isomers have the same molecular formula but a different arrangement of atoms. There are two main subgroupings: structural isomers (same molecular formula but different connections of atoms) and stereoisomers (same connections of atoms, but different three-dimensional orientations).

30 Linkage Isomers In linkage isomers the ligand is bound to the metal by a different atom. For example, nitrite can bind via the N or via an O.

31 Coordination Sphere Isomers Coordination sphere isomers differ in what ligands are bound to the metal and which fall outside the coordination sphere. For example, CrCl 3 (H 2 O) 6 exists as [Cr(H 2 O) 6 ]Cl 3, [Cr(H 2 O) 5 Cl]Cl 2 H 2 O, or [Cr(H 2 O) 4 Cl 2 ]Cl 2H 2 O.

32 Stereoisomers Same chemical bonds but different spatial arrangements Two types: Geometric isomers Optical isomers

33 Geometric Isomers In geometric isomers, the arrangement of the atoms is different, but the same bonds exist on the complex. For example, chlorine atoms can be adjacent to each other (cis) or opposite each other (trans); found in square planar or octahedral complexes, not tetrahedral. They have different physical properties and, often, different chemical reactivity!

34 Optical Isomers Optical isomers, or enantiomers, are mirror images of one another that don t superimpose on each other. They are said to be chiral. Their properties differ from each other only when in contact with other chiral substances.

35 Optical Isomers Enantiomers are distinguished from each other by the way they rotate plane-polarized light. Substances that rotate plane-polarized light to the right are dextrorotatory. Substances that rotate plane-polarized light to the left are levorotatory. A mixture of the two is called a racemic mixture.

36 Sample Exercise 23.3 Naming Coordination Compounds Name the compounds (a) [Cr(H 2 O) 4 Cl 2 ]Cl, (b) K 4 [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 tetraaquadichlorochromium(iii) chloride.

37 Sample Exercise 23.3 Naming Coordination Compounds Continued (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 potassium tetracyanonickelate(0). Practice Exercise 1 What is the name of the compound [Rh(NH 3 ) 4 Cl 2 ]Cl? (a) Rhodium(III) tetraamminedichloro chloride, (b) Tetraammoniadichlororhodium(III) chloride, (c) Tetraamminedichlororhodium(III) chloride, (d) Tetraamminetrichlororhodium(III), (e) Tetraamminedichlororhodium(II) chloride. Practice Exercise 2 Name the compounds (a) [Mo(NH 3 ) 3 Br 3 ]NO 3, (b) (NH 4 ) 2 [CuBr 4 ]. (c) Write the formula for sodium diaquabis(oxalato)ruthenate(iii).

38 Sample Exercise 23.4 Determining 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.

39 Sample Exercise 23.4 Determining 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) 4 Cl 2. The complex therefore has a coordination number of 6, and we can assume an octahedral geometry. Like [Co(NH 3 ) 4 Cl 2 ] + (Figure 23.8), 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) 4 Cl 2 ], and one with the Cl ligands adjacent to each other, cis-[fe(co) 4 Cl 2 ]. 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.

40 Sample Exercise 23.4 Determining the Number of Geometric Isomers Continued Practice Exercise 1 Which of the following molecules does not have a geometric isomer? Practice Exercise 2 How many isomers exist for the square-planar molecule [Pt(NH 3 ) 2 ClBr]?

41 Sample Exercise 23.5 Predicting Whether a Complex Has Optical Isomers Does either cis-[co(en) 2 Cl 2 ] + or trans-[co(en) 2 Cl 2 ] 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 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) 2 Cl 2 ] + and its mirror image are: Notice that the mirror image of the isomer is identical to the original. Consequently trans-[co(en) 2 Cl 2 ] + does not exhibit optical isomerism.

42 Sample Exercise 23.5 Predicting Whether a Complex Has Optical Isomers Continued 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) 2 Cl 2 ] + is a chiral complex. Practice Exercise 1 Which of the following complexes has optical isomers? (a) Tetrahedral [CdBr 2 Cl 2 ] 2, (b) Octahedral [CoCl 4 (en)] 2, (c) Octahedral [Co(NH 3 ) 4 Cl 2 ] 2+, (d) Tetrahedral [Co(NH 3 )BrClI]. Practice Exercise 2 Does the square-planar complex ion [Pt(NH 3 )(N 3 )ClBr] have optical isomers? Explain your answer.

43 23.5 Color and Magnetism in Coordination Chemistry 1057 Color Color depends on the metal AND the ligands.

44 Color Two ways we see color in a complex: Object reflects that color of light. Object transmits all colors EXCEPT the complementary color (as is seen in an absorption spectrum).

45 Sample Exercise 23.6 Relating Color Absorbed to Color Observed The complex ion trans-[co(nh 3 ) 4 Cl 2 ] + 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 ion? Solution Analyze We need to relate the color absorbed by a complex (red) to the color observed for the complex. 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 to determine the complementary color. Solve From Figure 23.25, we see that green is complementary to red, so the complex appears green.

46 Sample Exercise 23.6 Relating 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(nh 3 ) 4 Cl 2 ] +, absorbs yellow light and therefore appears violet.

47 Sample Exercise 23.6 Relating Color Absorbed to Color Observed Continued Practice Exercise 1 A solution containing a certain transition-metal complex ion has the absorption spectrum shown here. What color would you expect a solution containing this ion to be? (a) violet, (b) blue, (c) green, (d) orange, (e) red. Practice Exercise 2 A certain transition-metal complex ion absorbs at 695 nm. Which color is this ion most likely to be blue, yellow, green, or red?

48 23.6 Crystal-Field Theory 1059 Crystal-Field Theory As was mentioned earlier, ligands are Lewis bases that are attracted to a Lewis acid (the metal). But d electrons on the metal would repel the ligand. In crystal-field theory, the approaching ligand is considered to be a point charge repelled by the electrons in a metal s d-orbitals.

49 Crystal-Field Theory Therefore, the d orbitals on a metal in a complex would not be degenerate. Those that point toward ligands would be higher in energy than those that do not.

50 Crystal-Field Theory The energy difference between the orbitals is called the crystal-field splitting energy. This energy gap between d orbitals corresponds to the energy emitted or absorbed as a photon.

51 Crystal-Field Theory The spectrochemical series ranks ligands in order of their ability to increase the energy gap between d orbitals. (This is a variation known as ligand-field theory.)

52 Crystal-Field Theory Numbers of unpaired electrons can differ depending upon the order in which orbitals are filled. Stronger ligand fields result in greater splitting of orbitals; this is a high-field but low-spin case. Weaker ligand fields result in lower splitting of orbitals; this is a low-field but high-spin case.

53 Crystal-Field Theory Octahedral complexes differ from tetrahedral and square planar complexes because the ligands approach directly on the x-, y-, and z-axes only for octahedral complexes. (Last slide was octahedral.)

54 Sample Exercise 23.7 The Spectrochemical Series, Crystal Field Splitting, Color, and Magnetism The compound hexaamminecobalt(iii) chloride is diamagnetic and orange in color with a single absorption peak in its visible absorption spectrum. (a) What is the electron configuration of the cobalt(iii) ion? (b) Is [Co(NH 3 ) 6 ] 3+ a high-spin complex or a low-spin complex? (c) Estimate the wavelength where you expect the absorption of light to reach a maximum? (d) What color and magnetic behavior would you predict for the complex ion [Co(en) 3 ] 3+? Solution Analyze We are given the color and magnetic behavior of an octahedral complex containing Co with a +3 oxidation number. We need to use this information to determine its electron configuration, its spin state (low-spin or high-spin), and the color of light it absorbs. In part (d), we must use the spectrochemical series to predict how its properties will change if NH 3 is replaced by ethylenediamine (en). Plan (a) From the oxidation number and the periodic table we can determine the number of valence electrons for Co(III) and from that we can determine the electron configuration. (b) The magnetic behavior can be used to determine whether this compound is a low-spin or high-spin complex. (c) Since there is a single peak in the visible absorption spectrum the color of the compound should be complementary to the color of light that is absorbed most strongly. (d) Ethylenediamine is a stronger field ligand than NH 3 so we expect a larger Δ for [Co(en) 3 ] 3+ than for [Co(NH 3 ) 6 ] 3+.

55 Sample Exercise 23.7 The Spectrochemical Series, Crystal Field Splitting, Color, and Magnetism Continued Solve (a) Co has an electron configuration of [Ar]4s 2 3d 7 and Co 3+ has three fewer electrons than Co. Because transition-metal ions always lose their valence shell s electrons, the electron configuration of Co 3+ is [Ar]3d 6. (b) There are six valence electrons in the d orbitals. The filling of the t 2 and e orbitals for both high-spin and low-spin complexes is shown below. Because the compound is diamagnetic we know all of the electrons must be paired up, which allows us to determine that [Co(NH 3 ) 6 ] 3+ is a low-spin complex. (c) We are told that the compound is orange and has a single absorption peak in the visible region of the spectrum. The compound must therefore absorb the complementary color of orange, which is blue. The blue region of the spectrum ranges from approximately 430 nm to 490 nm. As an estimate we assume that the complex ion absorbs somewhere in the middle of the blue region, near 460 nm.

56 Sample Exercise 23.7 The Spectrochemical Series, Crystal Field Splitting, Color, and Magnetism Continued (d) Ethylenediammine is higher in the spectrochemical series than ammonia. Therefore, we expect a larger Δ for [Co(en) 3 ] 3+. Because Δ was already greater than the spin-pairing energy for [Co(NH 3 ) 6 ] 3+, we expect [Co(en) 3 ] 3+ to be a low-spin complex as well, with a d 6 configuration, so it will also be diamagnetic. The wavelength at which the complex absorbs light will shift to higher energy. If we assume a shift in the absorption maximum from blue to violet, the color of the complex will become yellow. Practice Exercise 1 Which of the following octahedral complex ions will have the fewest number of unpaired electrons? (a) [Cr(H 2 O) 6 ] 3+, (b) [V(H 2 O) 6 ] 3+, (c) [FeF 6 ] 3, (d) [RhCl 6 ] 3, (e) [Ni(NH 3 ) 6 ] 2+.

57 Sample Exercise 23.7 The Spectrochemical Series, Crystal Field Splitting, Color, and Magnetism Continued Practice Exercise 2 Consider the colors of the ammonia complexes of Co 3+ given in Table Based on the change in color would you expect [Co(NH 3 ) 5 Cl] 2+ to have a larger or smaller value of Δ than [Co(NH 3 ) 6 ] 3+? Is this prediction consistent with the spectrochemical series?

58 Sample Exercise 23.8 Populating d Orbitals in Tetrahedral and Square-Planar Complexes Nickel(II) complexes in which the metal coordination number is 4 can have either square-planar or tetrahedral geometry. [NiCl 4 ] 2 is paramagnetic, and [Ni(CN) 4 ] 2 is diamagnetic. One of these complexes is square planar, and the other is tetrahedral. Use the relevant crystal-field splitting diagrams in the text to determine which complex has which geometry. Solution Analyze We are given two complexes containing Ni 2+ 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 Ni 2+ and then use Figure for the tetrahedral complex and Figure for the square-planar complex.

59 Sample Exercise 23.8 Populating d Orbitals in Tetrahedral and Square-Planar Complexes Continued Solve Nickel(II) has the electron configuration [Ar]3d 8. 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, [Ni(CN) 4 ] 2 is tetrahedral, and [Ni(CN) 4 ] 2 is square planar. Comment Nickel(II) forms octahedral complexes more frequently than square-planar ones, whereas d 8 metals from periods 5 and 6 tend to favor square-planar coordination.

60 Sample Exercise 23.8 Populating d Orbitals in Tetrahedral and Square-Planar Complexes Continued Practice Exercise 1 How many unpaired electrons do you predict for the tetrahedral [MnCl] 4 2 ion? (a) 1, (b) 2, (c) 3, (d) 4, (e) 5. Practice Exercise 2 Are there any diamagnetic tetrahedral complexes containing transition metal ions with partially filled d orbitals? If so what electron count(s) leads to diamagnetism?

61 Sample Integrative Exercise Putting Concepts Together The oxalate ion has the Lewis structure shown in Table (a) Show the geometry of the complex formed when this ion complexes with cobalt(ii) to form [Co(C 2 O 4 )(H 2 O) 4 ]. (b) Write the formula for the salt formed when three oxalate ions complex with 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 , and the equilibrium constant for formation of the cobalt(ii) complex with three molecules of ortho-phenanthroline (Table 23.4) is 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 M oxalate (aq) and M Co 2+ (aq).

62 Sample Integrative Exercise Putting Concepts Together Continued Solution (a) The complex formed by coordination of one oxalate ion is octahedral: (b) Because the oxalate ion has a charge of 2, the net charge of a complex with three oxalate anions and one Co 2+ ion is 4. Therefore, the coordination compound has the formula Na 4 [Co(C 2 O 4 ) 3 ].

63 Sample Integrative Exercise Putting Concepts Together Continued (c) There is only one geometric isomer. The complex is chiral, however, in the same way the [Co(en) 3 ] 3+ complex is chiral (Figure 23.22). The two mirror images are not superimposable, so there are two enantiomers:

64 Sample Integrative Exercise Putting Concepts Together Continued (d) The ortho-phenanthroline ligand is bidentate, like the oxalate ligand, so they both exhibit the chelate effect. Thus, we conclude that ortho-phenanthroline is a stronger Lewis base toward Co 2+ than oxalate. This conclusion is consistent with what we learned about bases in Section 16.7, that nitrogen bases are generally stronger than oxygen bases. (Recall, for example, that NH 3 is a stronger base than H 2 O.) (e) The equilibrium we must consider involves 3 mol of oxalate ion (represented as Ox 2 ). Co 2+ (aq) + 3 Ox 2 (aq) 3Co(Ox) 3 ] 4 (aq) The formation-constant expression is Because K f is so large, we can assume that essentially all the Co 2+ is converted to the oxalato complex. Under that assumption, the final concentration of [Co(Ox) 3 ] 4 is M and that of oxalate ion is [Ox 2 ] = (0.040) 3(0.0010) = M (three Ox 2 ions react with each Co 2+ ion). We then have [Co 2+ ] = xm, [Ox 2 ] M, [[Co(Ox) 3 ] 4 ] M

65 Sample Integrative Exercise Putting Concepts Together Continued Inserting these values into the equilibrium-constant expression, we have Solving for x, we obtain M. From this, we see that the oxalate has complexed all but a tiny fraction of the Co 2+ in solution.