Chemistry part 35, Julia Burdge,2e (2009) docx

864 CHAPTER 22 Coordination Chemistry Think About It Although ligands are alphabetized in a compound's name, they do not necessarily appear in alphabetical order in the compound's formula. Figure 22 .5 Common geometries of complex ions. In each case M is a metal and L is a monodentate ligand. Strategy If you can't remember them yet, refer to Tab l es 22.4 and 22.5 for the names of ligands and anions containing metal atoms. Setup (a) There are six ligands: five NH 3 molecules and one Cl- ion. Tb.e oxidation state of cobalt is +3, making the overall charge on the complex ion +2. Therefore, there are two chloride ions as counter ions. (b) There are four ligand s: two bidentate ethylenediamines and two Cl- ions. The oxidation state of platinum is +4, making the overall charge on the complex ion +2 . Therefore, there are two nitrate ions as counter ions. Solution (a) [Co(NH 3 )sCl]CI 2 (b) [Pt(en)zCI 2 ](N0 3 )2 Practice Problem Write the formulas for (a) tris(ethylenediamine)cobalt(lII) sulfate and (b) sodium hexanitrocobaitate(III). ~ Checkpoint 22.1 Coordination Compounds 22 .1.1 22 .1.2 Select the correct name for the compound [Cu( NH 3 )4]CI 2 . a) Copperteu'aammine dichloride b) Tetraamminecopper(II) chloride c) Tetraaminedichlorocuprate(II) d) Dichlorotetraaminecopper(II) e) Tetraaminedichlorocopper(II) Select the correct name for the compound K 3 [FeF 6 ]. a) Tripotassiumironhexaft uoride b) Hexaftuorotripotassiumferrate(III) c) He xaftuoroiron(III) potassium d) Potassium hexaftuoroferrate(lII) e) Potassium ironhexaftuorate 22 .1.3 22 .1.4 Select the correct formula for pentaaminenitrocobalt(III). a) [Co(NH 3 )sN0 3 ] 3+ b) [Co(NH 3 )sN0 3 f + c) Co(NH 3 )sN03 d) [Co(NH 3 )s](N0 3 ) e) [Co(NH 3 )S] (N0 3 )2 Select the correct formula for tetraaq uodichlorochromi u rn (III) chloride. a) [Cr(H 2 O)4 CI 2]CI 3 b) [Cr(H 2 O)4 CI2]CI 2 c) [Cr(H 2 O)4 CI 2]C l d) [Cr(H 2 O)4]CI 3 e) [Cr(H 2 O)4]Cl z Structure of Coordination Compounds The geometry of a coordination compound often plays a si gnificant role in determining its proper - ties. Figure 22.5 s how s four different geometric arrangements for metal atoms with monodentate ligands. In these diagrams we See that structure and the coordination number of the meta l re l ate to each other as follow s: L M Linear Coordination Number L 2 4 6 L M L ; L L Tetrahedral Structure Linear Tetrahedral or square pla n ar Octahedra l L L L ~ / L M / /~L -M- / L L Square planar Octahedral L • SECTION 22.2 Structure of Coordination Compounds 865 In studying the geometry of coordination compounds, we sometimes find that there is more than one way to an'ange the ligands around the central atom. Such compounds in which ligands are • •• ••• ••• •••• <' •••••••••••• ••• • •• •• ••• • •••• • •• ••••••••• arranged differently, known as stereoisomers, have distinctly different physical and chemical prop- erties. Coordination compounds may exhibit two types of stereoisomerism: geometric and optical. Geometric isomers are stereoisomers that cannot be interconverted without breaking chemi- cal bonds. Geometric isomers come in pairs. We use the terms cis and trans to distinguish one ' geometric isomer of a compound from the other. Cis means that two palticular atoms (or groups of atoms) are adjacent to each other, and trans means that the atoms (or groups of atoms) are on opposite sides in the structural formula. The cis and trans isomers of coordination compounds generally have quite different colors, melting points, dipole moments, and chemical reactivities. Figure 22.6 shows the cis and trans isomers of diamrninedichloroplatinum(II). Note that although the types of bonds are the same in both isomers (two Pt-N and two Pt-Cl bonds), the spatial arrangements are different. Another example is the tetraamminedichlorocobalt(III) ion, shown in Figure 22.7. Optical isomers are nonsuperimposable mirror images. (Superimposable means that if one structure is laid over the other, the positions of all the atoms will match.) Like geometric isomers, optical isomers come in pairs. However, the optical isomers of a compound have identical physical and chemical properties, such as melting point, boiling point, dipole moment, and chemical reac- tivity toward molecules that are not themselves optical isomers. Optical isomers differ from each other, though, in their interactions with plane-polarized light, as we will see. The structural relationship between two optical isomers is analogous to the relationship between your left and right hands. If you place your left hand in front of a mirror, the image you see will look like your right hand (Figure 22.8). Your left hand and right hand are mirror images of each other. They are nonsuperimposable, however, because when you place your left hand over your right hand (with both palms facing down), they do not match. This is why a right-handed glove will not fit comfortably on your left hand. Figure 22.9 shows the cis and trans isomers of dichlorobis(ethylenediamine)cobalt(III) ion and the mirror image of each. Careful examination reveals that the trans isomer and its mirror Minor image of left hand Left hand In genera l, stereoisomers are compounds that are made up of the same types and numbers of atoms, bonded together in the same sequence, but with different spatial arrangements. Figure 22.6 The (a) cis and (b) trans isomer s of diamminedichloro- platinum(II). Note that the two Cl atoms are adjacent to each other in the cis isomer and diagonally across from each other in the trans isomer. Figure 22.7 The (a) cis and (b) trans iso mer s of tetraamminedichloro- cobalt(III) ion , [Co(NH 3)4 CI 2t . The ion has only two geometric isomers. Figure 22.8 A left hand and its . . II1lrror Image. , 866 CHAPTER 22 Coordination Chemistry Figure 22.9 The (a) cis and (b) trans isomers of dichlorobis(ethylene- diamine )cobalt(III) ion and their mirror images. If you could rotate the mirror image in (b) 90° clockwise about the vertical position and place the ion over the trans isomer, you would find that the two are superimposable. No matter how you rotate the cis isomer and its , mirror image in (a), however, you cannot superimpose one on the other. Light source Fixed polarizer Mirror (a) CI ~co ~CI Mirror (b) image are superimposable, but the cis isomer and its mirror image are not. Thus, the cis isomer and its mirror image are optical isomers. Optical isomers are described as chiral (from the Greek word for "hand") because, like your left and right hands, chiral molecules are nonsuperimposable. Isomers that are superimposable with their mirror images are said to be achiral. Chiral molecules playa vital role in enzyme reac- tions in biological systems. Many drug molecules are chiral, although only one of a pair of chiral isomers is biologically effective [ ~~ Section 10.4 ]. Chiral molecules are said to be optically active because of their ability to rotate the plane of polarization of polarized light as it passes through them. Unlike ordinary light, which vibrates in all directions, plane-polarized light vibrates only in a single plane. We use a polarimeter to mea- sure the rotation of polarized light by optical isomers (Figure 22.10). A beam of unpolarized light first passes through a Polaroid sheet, called the polarizer, and then through a sample tube contain- ing a solution of an optically active, chiral compound. As the polarized light passes through the sample tube, its plane of polarization is rotated either to the right (clockwise) or to the left (coun- terclockwise). This rotation can be measured directly by turning the analyzer in the appropriate direction until minimal light transmission is achieved (Figure 22.11). If the plane of polarization is rotated to the right, the isomer is said to be dextrorotatory and the isomer is labeled d; if the rota- tion is to the left, the isomer is levorotatory and the isomer is labeled l. The d and l isomers of a chiral substance, called enantiomers, always rotate the plane of polarization by the same amount, but in opposite directions. Thus, in an equimolar mixture of two enantiomers, called a racemic mixture, the net rotation is zero. t ! Polarimeter tube Optically active substance in solution Analyzer + O· Degree scale + 90' Plane of polarization -90' • Figure 22.10 Operation of a polarimeter. Initially, the tube is filled with an achiral compound. The analyzer is rotated so that its plane of polarization is perpendicular to that of the polarize r. Under this condition, no light reaches the observer. Next, a chiral compound is placed in the tube as shown. The plane of polarization of the polarized light is rotated as it travels through the tube so that some light reaches the observer. Rotating the analyzer (either to the left or to the right) until no light reaches the observer again allows the angle of optical rotation to be measured. SECTION 22.3 Bonding in Coordination Compounds: Crystal Field Theory 86 7 . 1 ". Bonding in Coordination Compounds: Crystal Field Theory A satisfactory theory of bonding in coordination compounds must account for properties such as color and magnetism, as well as stereochemistry and bond strength. No single theory as yet does all this for us. Rather, several different approaches have been applied to transition metal com- plexes. We will consider only one of them here crystal field theory because it accounts for both the color and magnetic properties of many coordination compounds. We will begin our discussion of crystal field theory with the most straightforward case- namely, complex ions with octahedral geometry. Then we will see how it is applied to tetrahedral and square-planar complexes. Crystal Field Splitting in Octahedral Complexes Crystal field theory explains the bonding in complex ions purely in terms of electrostatic forces. In a complex ion, two types of electrostatic interaction come into play. One is the attraction between the positive metal ion and the negatively charged ligand or the negatively charged end of a polar ligand. This is the force that binds the ligands to the metal. The second type of interaction is the electrostatic repulsion between the lone pairs on the ligands and the electrons in the d orbitals of the metals. The d orbitals have different orientations [ ~~ Section 6.7], but in the absence of an exter- nal disturbance, they all have the same energy. In an octahedral complex, a central metal atom is surrounded by six lone pairs of electrons (on the six ligands), so all fi ve d orbitals experience electrostatic repulsion. The magnitude of this repulsion depends on the orientation of the d orbital that is involved. Take the di _/ orbital as an example. In Figure 22.12, we see that the lobes of this orbital point toward the comers of the octahedron along the x and y axes, where the lone-pair elec- trons are positioned. Thus, an electron residing in this orbital would experience a greater repulsion from the ligands than an electron would in the d X) , d yz ' or d xz orbitals. For this reason, the energy of the d x 2 _ / orbital is increased relative to the d X)' d yZ ' and d xz orbitals. The d z 2 orbital's energy is also greater, because its lobes are pointed at the ligands along the z axis. As a result of these metal-ligand interactions, the five d orbitals in an octahedral complex are split between two sets of energy levels: a higher level with two orbitals (d } _ i and dl ) having the same energy, and a lower level with three equal-energy orbitals (d X)' d yz , and d xz ) ' as shown in Figure 22.13. The crystal field splitting (Ll) is the energy difference between two sets of d orbitals in a metal atom when ligands are present. The magnitude of ~ depends on the metal and the nature of the ligands; it has a direct effect on the color and magnetic properties of complex ions. Figure 22.11 Polarized lense s. 0 light passes through the lenses when they are rotated so that their planes of polarization are perpendicular . • 868 CHAPTER 22 Coordination Chemistry Figure 22.12 The fi ve d orbitals in an octahedral environment. The metal atom (or ion) is at the center of the octahedron, and the six lone pairs on the donor atoms of the ligands are at the corners. • Figure 22.13 Cr ys tal field splitting between d orbitals in an oc tahed ra l complex. 650 nm 580 nm 700 nm 400nm 1 5160 nm 430 nm 490 nm Figure 22.14 A color wheel with appropriate wa veleng th s. Complementary color s, such as red and green, are on opposite sides of the wheel. z • d, y H Color . " • d ' , d 2 x y- Z • \ . Cr ys tal field splitting, ~ In Chapter 6 we learned that white light, such as sunlight, is a combination of all colors. A sub- stance appear s black if it absorbs all the visible light that strikes it. If it absorbs no visible light, it is white or colorless. An object appears green if it absorbs all light but reflects the green com- ponent. An object also look s green if it reflects all colors except red, the complementary color of gre en (Figure 22.14). What ha s been said of reflected light also applies to transmitted light (i.e., the light that passe s through the medium , such as a solution). Consider the hydrated cupric ion ([Cu(H 2 0)6]2+); it absorbs light in the orange region of the s pectrum , so a solution of CUS04 appears blue to us. Recall from Chapter 6 that when the energy of a photon is equal to the difference between the ground state and an excited state, absorption occurs as the photon strikes the atom (or ion or com- pound ), and an electron is promoted to a higher level. Using these concepts, we can . calculate the energy change in volv ed in the electron transition. The energy of a photon is given by E= hv where h repre sents Planck 's constant (6.63 X 10- 34 J . s) and v is the frequency of the radiation, which is 5.00 X 10 14 S- I for a wa velength of 600 nm. Here E = ~,so we have ~ = hv = (6.63 X 10- 34 J . s)(5.00 X 10 14 s) = 3.32 X 10 - 19 J This value is very small, but it is the energy absorbed by only one ion. If the wavelength of the photon absorbed by an ion lies outside the visible region, then the transmitted light looks the same (to us) as the incident light-white - and the ion appears colorless. The be st way to mea s ure crystal field splitting is to use spectroscopy to determine the wa velength at which light is absorbed. The [Ti(H 2 0)6] 3+ ion provides a straightforward exam- ple, because Ti 3+ ha s only one 3d electron ( Figure 22.15). The [Ti(H 2 0)6] 3+ ion absorbs light in the visible region of the spectrum (Figure 22.16). The wavelength corresponding to maximum SECTION 22.3 Bonding in Coordination Compounds: Crystal Field Theory 869 • Photon of energy hv ~ DD [JDD 400 500 [JD • DDD (a) 600 700 Wavelength (nm) (b) absorption is 498 nm [Figure 22.lS(b)]. To calculate the crystal field splitting energy, we start by writing Next, recall that ~ = hv e v =- A where e is the speed of light and A is the wavelength. Therefore, A he (6.63 X 10- 34 J . s)(3.00 X 10 8 mls) - 19 I.l. = = = 3.99 X 10 J A (498 nm)(l X 10- 9 mlnm) This is the energy required to excite one [Ti(H 2 0)6] 3+ ion. To express this energy difference in the more convenient units of kJ/mol, we write ~ = (3.99 X 10- 19 Jlion)(6.02 X 10 23 ionslmol) = 240,000 J/mol = 240 kJ/mol Aided by spectroscopic data for a number of complexes, all having the same metal ion but different ligands, chemists calculated the crystal field splitting for each ligand and established the following spectrochemical series, which is a list of ligands arranged in increasing order of their abilities to split the d orbital energy levels: 1- < Br- < CI- < OH - < F- < H 2 0 < NH 3 < en < CN - < CO These ligands are arranged in the order of increasing value of ~. CO and CN- are called strong- field ligands, because they cause a large splitting of the d orbital energy levels. The halide ions and hydroxide ion are weak-field ligands, because they split the d orbitals to a lesser extent. Magnetic Properties The magnitude of the crystal field splitting also determines the magnetic properties of a com- plex ion. The [Ti(H 2 0)6f + ion, having only one d electron, is always paramagnetic. However, for an ion with several d electrons, the situation is less immediately clear. Consider, for example, the octahedral complexes [FeF 6 ]3- and [Fe(CN)6]3- (Figure 22.17). The electron configuration of Fe 3+ is [Ar]3d s , and there are two possible ways to distribute the five d electrons among th e d orbitals. According to Hund's rule [I ~~ Section 6. 8], maximum stability is reached when the Figure 22.15 (a) The process of photon absorption, and (b) a graph of the absorption spectrum of [Ti(H 2 0)6l H . The energy of the incoming photon is equal to the crystal field splitting. The maximum absorption peak in the visible region occurs at 498 nm . Figure 22.16 Colors of some of the first-row transition metal ions in solutio n. From left to right: TiH, Cr 3 +, Mn 2+, Fe H , C0 2 + , Ni 2+ , Cu2+. The 3+ -+ Sc and yO ions are colorless. • 870 CHAPTER 22 Coordination Chemistry Figure 22.17 Energy-level diagrams for the Fe 3+ ion and for the [FeF 6 ]3- and [Fe(CN)6]3- complex • IOns. • Figure 22.18 Orbital diagrams for the high-spin and low-spin octahedral complexes corresponding to the electron configurations of (a) d 4 , (b) d 5 , (c) d 6 , and (d) d 7 1 1 11 11 11 11 11 1 11 11 11 1 d xy fly , d Xl d 2 2 d 2 X - y Z IHIHI1 1 d xy d yZ d xz electrons are placed in five separate orbitals with parallel spins. This arrangement can be achieved only at a cost, however, because two of the five electrons must be promoted to the higher-energy d i- i and di orbitals. No such energy investment is needed if all five electrons enter the dX}' d yz ' and d xz orbitals. According to Pauli's exclusion principle [ ~~ Section 6.8], there will be only one unpaired electron present in this case. Figure 22 .1 8 shows the distribution of electrons among d orbitals that results in low- and high-spin complexes. The actual arrangement of the electrons is determined by the amount of 1 d 2 2 x -y d z' smaller D. d 2 ' d z' / 1 1 1 1 x -y. 1 1 1 d 4 larger D. d ry d Yl d Xl H 1 1 (a) 1 1 d 2 2 d ' x - y z· d ' , d 2 / 1 1 1 1 1 x y l 1 1 1 d 5 d\ y d Yl d xz H H 1 (b) 1 1 d X 2 -l d z' d 2 ' d z' / H 1 1 1 1 x -y . H 1 1 d 6 d X) , d yZ d xz H H H (c) 1 1 1 d 2 2 X -y d 2 l d ' , d ' / H 1~ ~ 7 1 1 x y. z· H H 1 d xy d yZ d xz H H H (d) SECTION 22.3 Bonding in Coordination Compounds: Crystal Field Theory 871 stability gained by having maximum parallel spins versus the investment in energy required to promote electrons to higher d orbitals. Because F- is a weak-field ligand, the five d electrons enter five separate d orbitals with parallel spins to create a high-spin complex. The cyanide ion is a strong-field ligand, though, so it is energetically preferable for all five electrons to be in the lower orbitals, thus forming a low-spin complex. High-spin complexes are more paramagnetic than low- spin complexes. The actual number of unpaired electrons (or spins) in a complex ion can be found by mag- netic measurements, and in general, experimental findings support predictions based on crystal field splitting. However, a distinction between low- and high-spin complexes can be made only if the metal ion contains more than three and fewer than eight d electrons, as shown in Figure 22.18. Sample Problem 22.4 shows how to determine the number of spins in an octahedral complex. Predict the number of unpaired spins in the [Cr(en)3f + ion. Strategy The magnetic properties of a complex ion depend on the strength of the ligands. Strong- field ligands, which cause a high degree of splitting among the d orbital energy levels, result in low- spin complexes. Weak-field ligands, which cause only a small degree of splitting among the d orbital energy levels, result in high-spin complexes. Setup The electron configuration of Cr 2+ is [Ar]3d 4; and en is a strong-field ligand. Solution Because en is a strong-field ligand, we expect [Cr(en)3f + to be a low-spin complex. According to Figure 22.18, all four electrons will be placed in the lower-energy d orbitals (d X) , d yz , and d xz ) and there will be a total of two unpaired spins. Practice Problem How many unpaired spins are in [Mn(H 2 0)6f +? (Hint: H 2 0 is a weak-field ligand.) Tetrahedral and Square-Planar Complexes So far we have concentrated on octahedral complexes. The splitting of the d orbital energy levels in tetrahedral and square-planar complexes, though, can also be accounted for satisfac- torily by the crystal field theory. In fact, the splitting pattern for a tetrahedral ion is just the reverse of that for octahedral complexes. In this case, the d xy> d yz ' and d xz orbitals are more closely directed at the ligands and therefore have more energy than the d x 2 -l and d z 2 orbitals (Figure 22.19). Most tetrahedral complexes are high-spin complexes. Presumably, the tet- rahedral arrangement reduces the magnitude of the metal-ligand interactions, resulting in a smaller Ll value. This is a reasonable assumption because the number of ligands is smaller in a tetrahedral complex. As Figure 22.20 shows, the splitting pattern for square-planar complexes is the most compli- cated. The d x 2 -i orbital possesses the highest energy (as in the octahedral case), and the d X) , orbital is the next highest. However, the relative placement of the di and the d xz and d yZ orbitals cannot be determined simply by inspection and must be calculated. /0 D D / d ry d yZ d xz / / [I 1(, Crystal field splitting L '-= ==,'== -'=~~ J , , , , "0 Think About It It is easy to draw the wrong conclusion regarding high- and low-spin complexes. Remember that the term high spin refers to the number of spins (unpaired electrons), not to the energy levels of the d orbitals. The greater the energy gap between the lower-energy and higher-energy d orbitals, the greater the chance that the complex will be low spin. • . Figure 22.19 Crystal field splitting between d orbitals in a tetrahedral complex. 872 CHAPTER 22 Coordination Chemistry Figure 22.20 Energy-level diagram for a square-planar complex. Be ca use there are more than two energy levels, we cannot defi ne crystal field splitting as we can for octahedral and tetrahedral complexes. • 22.3.1 Bonding in Coordination Compounds: Crystal Field Theory How many unpaired spins would you expect the [Mn( CO )6 l 2+ ion to have? a) 0 b)1 c) 2 d) 3 e) 5 22.3.2 Which of the following metal ions can p ote ntia ll y form both l ow -spin and high-sp in co mple xes? (Select all that apply.) a) Ti 2 + b) Cu+ c) Fe 2+ d) Ni 2+ e) Cr H Reactions of Coordination Compounds Complex ions undergo liga nd exchange (or substitution) reactions in solution. The rates of these reactions vary widel y, depending on the nature of the metal ion and the ligands. In studying ligand exchange reactions, it is often u sef ul to distinguish between the stability of a complex ion and its tendency to react, which we call kinetic lability. Stability in this context is a thermodynamic property, which is measured in terms of the species' formation constant K f [ ~~ Section 17.5] . For exampl e, we s ay that the complex ion tetracyanonickelate(II) is stable because it has a large fo rmation constant (K f = 1 X 10 30 ): By using cy anide ions labeled with the radioactive isotope carbon-14, chemists have shown that [Ni (CN)4]2- undergoes ligand exchange very rapidly in solution. The following equilibrium is established almost as soon as the species are mixed: where the asterisk denotes a 14 C atom. Complexes like the tetracyanonickelate(II) ion are termed labile complexes because they undergo rapid ligand exchange reactions. Thus, a thermodynami- cally stable species (i.e., one that h as a large formation constant) is not necessarily unreactive. A complex that is thermodynamically unstable in acidic solution is [Co(NH 3 )6 ] 3+ . The equi- librium constant for the following reac ti on is about 1 X 10 20 : When equilibrium is reached, the concentration of the [Co(NH 3 )6] 3+ ion is very low. This reaction requires several days to complet e, however, because the [Co(NH 3 )6 ] 3+ ion is so inert. This is an example of an inert complex -a complex ion that undergoes very slow exchange reactions (on the order of hours or even day s) . It shows that a thermodynamically unstable species is not necessarily chemica ll y reactive. The rate of reaction is determined by the energy of activation, which is high in this case. Most compl ex ions containing Co 3+, Cr 3+, a nd Pt 2+ are kinetically inert. Because they exchange ligands very slowl y, they are easy to study in solution. As a result, our knowledge of the SECTION 22.S Applications of Coordination Compounds 873 bonding, structure, and isomerism of coordination compounds has come largely from studies of these compounds. Applications of Coordination Compounds Coordination compounds are found in living systems and have many uses in the home, in industry, and in medicine. We briefly describe a few examples in this section. Metallurgy The extraction of silver and gold by the formation of cyanide complexes and the purification of nickel by converting the metal to the gaseous compound Ni(CO)4 are typical examples of the use of coordination compounds in metallurgical processes. Chelation Therapy Earlier we mentioned that chelation therapy is used in the treatment of lead poisoning. Other met- als, such as arsenic and mercury, can also be removed using chelating agents. Chemotherapy Several platinum-containing coordination compounds, including cisplatin [Pt(NH3)2 CI2J and carboplatin [Pt(NH3 MOCO) 2 C4H 6 J, can effectively inhibit the growth of cancerous cells. The mechanism for the action of cisplatin is the chelation of DNA, the molecule that contains the genetic code. During cell division, the double-stranded DNA unwinds into two single stra nd s, which must be accurately copied in order for the new cells to be identical to their parent cell. X-ray studies show that cisplatin binds to DNA by forming cross-links in which the two chlorides on cisplatin are replaced by nitrogen atoms in the adjacent guanine bases on the same strand of the DNA. (Guanine is one of the four bases in DNA [ ~~ Section 10.6, Figure 10.1SJ .) This causes a bend in the double-stranded structure at the binding site. It is believed that this struc- tural distortion is a key factor in inhibiting replication. The damaged cell is then destroyed by the body's immune system. Because the binding of cisplatin to DNA requires both Cl atoms to be on the same side of the complex, the trans isomer of the compound is totally ineffective as an anticancer drug. Cisplatin Chemical Analysis Although EDTA has a great affinity for a large number of metal ions (especially 2+ and 3+ ion s), other chelates are more selective in binding. Dimethylglyoxime, for example, forms an insoluble brick-red solid with Ni2+ and an insoluble bright-yellow solid with Pd 2 +. These characteristic colors are used in qualitative analysis to identify nickel and palladium. Furthermore, the quanti- ties of ions present can be determined by gravimetric analysis [ ~~ Section 4.6J as follows: To a solution containing Ni 2 + ions, say, we add an excess of dimethylglyoxime reagent, and a brick-red precipitate forms. The precipitate is then filtered, dried, and weighed. Knowing the formula of the complex (Figure 22.21), we can readily calculate the amount of nickel present in the original solution. , Figure 22.21 Structure of nickel dimethylglyoxime. Note that the overall structure is stabilized by hydrogen bonds. [...]... that would otherwise catalyze the oxidation reactions that cause food to spoil EDTA is a common preservative in a wide variety of consumer products Bringing Chemistry to life The Coordination Chemistry of Oxygen Transport ~ Multimedia Organic and Biochemistry-oxygen binding in hemoglobin Because of its central function as an oxygen carrier for metabolic processes, hemoglobin is probably the most studied... wets the mineral particles, which are then carried to the top in the froth, while the gangue settles to the bottom The froth is skimmed off, allowed to collapse, and dried to recover the mineral particles Another physical separation process makes use of the magnetic properties of certain minerals Ferromagnetic metals are strongly attracted to magnets The mineral magnetite (Fe304), in particular, can... tissues, where it delivers the oxygen molecules to myoglobin Myoglobin, which is made up of only one subunit, stores oxygen for metabolic processes in the muscle The porphine molecule forms an important part of the hemoglobin structure Upon coordination to a metal , the H+ ions that are bonded to two of the four nitrogen atoms in porphine are displaced Complexes derived from porphine are called porphyrins,... the protein The sixth ligand is a water molecule, which binds to the Fe 2 + ion on the other side of the ring to complete the octahedral complex This hemoglobin molecule is called deoxyhemoglobin and imparts a bluish tinge to venous blood The water ligand can be replaced readily by molecular oxygen (in the lungs) to form red oxyhemoglobin, which is found in arterial blood Each subunit contains a heme... b) Give the systematic name for cisplatin c) Write the formula for the compound potassium hexachloroplatinate(IV) [ ~ Sample Problem 22.3] [ ~ Sample Problem 22.2] • , 875 876 CHAPTER 22 Coordination Chemistry CHAPTER SUMMARY • Section 22.1 • Coordination compounds contain coordinate covalent bonds between a metal ion (often a transition metal ion) and two or more polar molecules or ions • • Enantiomers... The rate at which ligand exchange occurs is a measure of a complex 's kinetic lability and does not necessarily correspond directly to the complex's thermodynamic stability Section 22.5 • Coordination chemistry is important in many biological, medical , and industrial processes !(EY WORDS Chelating agent, 861 Dextrorotatory, 866 Levorotatory, 866 Racemic mixture, 866 Coordination compound, 858 Donor... histidine residues, a cysteine residue, and a methionine residue in a tetrahedral configuration How does the crystal field splitting (Ll) change between these two oxidation states? 878 CHAPTER 22 Coordination Chemistry Section 22.4: Reactions of Coordination Compounds 22.51 What are the oxidation states of Fe and Ti in the ore ilmenite (FeTi0 3)? (Hint: Look up the ionization energies of Fe and Ti in Table... be chemically reactive and a thermodynamically unstable species may be unreactive Problems 22.40 • Oxalic acid (H Z 20 4) is so metimes used to clean rust stains C from sinks and bathtubs Explain the chemistry underlying this cleaning action 22.41 The [Fe(CN)6]3- complex is more labile than the [Fe(CN)6]4complex Suggest an experiment that would prove that [Fe(CN)6f- is a labile complex 22.42 Aqueous... whereas deoxyhemoglobin is purple Show that the difference in color can be accounted for qualitatively on the basis of high-spin and low-spin complexes (Hint: Oz is a strong-field ligand See the Bringing Chemistry to Life box on page 874.) 22.62 Hydrated Mn2+ ions are practically colorless (see Figure 22.16) even though they possess five 3d electrons Explain (Hint: Electronic transitions in which there... distinguish between cis-Pt(NH3)2CI2 and trans-Pt(NH 3 )zCI 2 22.66 879 Use the information in Table 19.1 (page 767) to calculate the equilibrium constant for the reaction (b) Based on your results in part (a), explain why most Cu(I) compounds are insoluble 22.74 Consider the following two ligand exchange reactions: [Co(H 20 )6]3+ + 6NH3 ( How many geometric isomers can the following square-planar complex . structure is stabilized by hydrogen bonds. 874 CHAPTER 22 Coordination Chemistry , . ~ Multimedia Organic and Biochemistry-oxygen binding in hemoglobin. Detergents The cleansing action. is a common preservative in a wide variety of consumer products. Bringing Chemistry to life The Coordination Chemistry of Oxygen Transport Because of its central function as an oxygen. coordination compounds must account for properties such as color and magnetism, as well as stereochemistry and bond strength. No single theory as yet does all this for us. Rather, several different