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Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 22 Transition Metal Chemistry and Coordination Compounds INTRODUCTION THE 22.1 PROPERTIES OF THE TRANSITION METALS SERIES OF ELEMENTS IN THE PERIODIC TABLE IN WHICH THE d AND f ORBITALS ARE GRADUALLY FILLED ARE CALLED THE TRANSITION ELE- 22.2 CHEMISTRY OF IRON AND COPPER MENTS. 22.3 COORDINATION COMPOUNDS THERE ARE ABOUT 50 TRANSITION ELEMENTS, AND THEY HAVE WIDELY VARYING AND FASCINATING PROPERTIES. TO 22.4 STRUCTURE OF COORDINATION COMPOUNDS PRESENT EVEN ONE INTERESTING FEATURE OF EACH TRANSITION ELEMENT IS BEYOND THE SCOPE OF THIS BOOK. WE 22.5 BONDING IN COORDINATION COMPOUNDS: CRYSTAL FIELD THEORY WILL THEREFORE LIMIT OUR DISCUSSION TO 22.6 REACTIONS OF COORDINATION COMPOUNDS THE TRANSITION ELEMENTS THAT HAVE INCOMPLETELY FILLED d ORBITALS AND TO THEIR TENDENCY TO FORM COMPLEX IONS. 22.7 APPLICATIONS OF COORDINATION COMPOUNDS 871 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 872 TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS 22.1 PROPERTIES OF THE TRANSITION METALS Transition metals typically have incompletely filled d subshells or readily give rise to ions with incompletely filled d subshells (Figure 22.1). (The Group 2B metals — Zn, Cd, and Hg — do not have this characteristic electron configuration, and so although they are sometimes called transition metals, they really do not belong in this category.) This attribute is responsible for several notable properties, including distinctive coloring, formation of paramagnetic compounds, catalytic activity, and especially a great tendency to form complex ions. In this chapter we focus on the first-row elements from scandium to copper, the most common transition metals. Table 22.1 lists some of their properties. As we read across any period from left to right, atomic numbers increase, electrons are added to the outer shell, and the nuclear charge increases by the addition of protons. In the third-period elements — sodium to argon — the outer electrons weakly shield one another from the extra nuclear charge. Consequently, atomic radii decrease rapidly from sodium to argon, and the electronegativities and ionization energies increase steadily (see Figures 8.5, 8.11, and 9.5). For the transition metals, the trends are different. Looking at Table 22.1 we see that the nuclear charge, of course, increases from scandium to copper, but electrons are being added to the inner 3d subshell. These 3d electrons shield the 4s electrons from the increasing nuclear charge somewhat more effectively than outer-shell electrons can shield one another, so the atomic radii decrease less rapidly. For the same reason, electronegativities and ionization energies increase only slightly from scandium across to copper compared with the increases from sodium to argon. Although the transition metals are less electropositive (or more electronegative) than the alkali and alkaline earth metals, their standard reduction potentials suggest that all of them except copper should react with strong acids such as hydrochloric acid to produce hydrogen gas. However, most transition metals are inert toward acids or react slowly with them because of a protective layer of oxide. A case in point is chromium: FIGURE 22.1 The transition metals (blue squares). Note that although the Group 2B elements (Zn, Cd, Hg) are described as transition metals by some chemists, neither the metals nor their ions possess incompletely filled d subshells. 1 1A 1 H 18 8A 13 3A 2 2A 14 4A 15 5A 16 6A 17 7A 2 He 3 4 5 6 7 8 9 10 Li Be B C N O F Ne 13 14 15 16 17 18 Al Si P S Cl Ar 11 12 Na Mg 3 3B 4 4B 5 5B 6 6B 7 7B 8 9 8B 10 11 1B 12 2B 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn 110 111 112 87 Back 88 89 104 105 106 107 108 109 Fr Ra Ac Rf Ha Sg Ns Hs Mt Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 22.1 TABLE 22.1 873 PROPERTIES OF THE TRANSITION METALS Electron Configurations and Other Properties of the First-Row Transition Metals Sc Electron configuration M M2 M3 Electronegativity Ionization energy (kJ/mol) First Second Third Radius (pm) M M2 M3 Standard reduction potential (V)* *The half-reaction is M2 (aq) Ti V Cr Mn Fe Co Ni Cu 4s23d1 — [Ar] 1.3 4s23d 2 3d 2 3 d1 1.5 4s23d 3 3d 3 3d 2 1.6 4s13d 5 3d 4 3d 3 1.6 4s23d 5 3d 5 3d 4 1.5 4s23d 6 3d 6 3d 5 1.8 4s23d 7 3d 7 3d 6 1.9 4s23d 8 3d 8 3d 7 1.9 4s13d10 3d 9 3d 8 1.9 631 1235 2389 658 1309 2650 650 1413 2828 652 1591 2986 717 1509 3250 759 1561 2956 760 1645 3231 736 1751 3393 745 1958 3578 162 — 81 147 90 77 134 88 74 130 85 64 135 80 66 126 77 60 125 75 64 124 69 — 128 72 — 2.08 1.63 1.2 0.74 1.18 0.44 0.28 0.25 0.34 2e 88n M(s) (except for Sc and Cr, where the ions are Sc3 and Cr3 , respectively). Despite a rather negative standard reduction potential, it is quite inert chemically because of the formation on its surfaces of chromium(III) oxide, Cr2O3. Consequently, chromium is commonly used as a protective and noncorrosive plating on other metals. On automobile bumpers and trim, chromium plating serves a decorative as well as a functional purpose. GENERAL PHYSICAL PROPERTIES Most of the transition metals have a close-packed structure (see Figure 11.29) in which each atom has a coordination number of 12. Furthermore, these elements have relatively small atomic radii. The combined effect of closest packing and small atomic size results in strong metallic bonds. Therefore, transition metals have higher densities, higher melting points and boiling points, and higher heats of fusion and vaporization than the Group 1A, 2A, and 2B metals (Table 22.2). TABLE 22.2 Physical Properties of Elements K to Zn 1A 2A TRANSITION METALS 2B K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn 235 197 162 147 134 130 135 126 125 124 128 138 63.7 838 1539 1668 1900 1875 1245 1536 1495 1453 1083 419.5 760 1440 2730 3260 3450 2665 2150 3000 2900 2730 2595 906 0.86 1.54 3.0 4.51 6.1 7.19 7.43 7.86 8.9 8.9 8.96 7.14 Atomic radius (pm) Melting point (°C) Boiling point (°C) Density (g/cm3 ) Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 874 TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS ELECTRON CONFIGURATIONS The electron configurations of the first-row transition metals were discussed in Section 7.9. Calcium has the electron configuration [Ar]4s2. From scandium across to copper, electrons are added to the 3d orbitals. Thus, the outer electron configuration of scandium is 4s23d1, that of titanium is 4s23d 2, and so on. The two exceptions are chromium and copper, whose outer electron configurations are 4s13d 5 and 4s13d10, respectively. These irregularities are the result of the extra stability associated with half-filled and completely filled 3d subshells. When the first-row transition metals form cations, electrons are removed first from the 4s orbitals and then from the 3d orbitals. (This is the opposite of the order in which orbitals are filled in neutral atoms.) For example, the outer electron configuration of Fe2 is 3d 6, not 4s23d 4. OXIDATION STATES Recall that oxides in which the metal has a high oxidation number are covalent and acidic, whereas those in which the metal has a low oxidation number are ionic and basic (see Section 15.11). FIGURE 22.2 Oxidation states of the first-row transition metals. The most stable oxidation numbers are shown in color. The zero oxidation state is encountered in some compounds, such as Ni(CO)4 and Fe(CO)5. Transition metals exhibit variable oxidation states in their compounds. Figure 22.2 shows the oxidation states from scandium to copper. Note that the common oxidation states for each element include 2, 3, or both. The 3 oxidation states are more stable at the beginning of the series, whereas toward the end the 2 oxidation states are more stable. The reason for this trend can be understood by examining the ionization energy plots in Figure 22.3. In general, the ionization energies increase gradually from left to right. However, the third ionization energy (when an electron is removed from the 3d orbital) increases more rapidly than the first and second ionization energy. Because it takes more energy to remove the third electron from the metals near the end of the row than from those near the beginning, the metals near the end tend to form M2 ions rather than M3 ions. The highest oxidation state for a transition metal is 7, for manganese (4s23d 5). For elements to the right of Mn (Fe to Cu), oxidation numbers are lower. Transition metals usually exhibit their highest oxidation states in compounds with very electronegative elements such as oxygen and fluorine — for example, V2O5, CrO3, and Mn2O7. Sc Ti V Cr Mn Fe Co Ni Cu +7 +6 +6 +5 +5 +5 +5 +4 +4 +4 +4 +4 +4 +3 +3 +3 +3 +3 +3 +3 +3 +2 +3 +6 +2 +2 +2 +2 +2 +2 +2 +1 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 22.2 Third 3000 2000 Second 1000 First 0 22.2 875 4000 Ionization energy (kJ/mol) FIGURE 22.3 Variation of the first, second, and third ionization energies for the first-row transition metals. CHEMISTRY OF IRON AND COPPER Sc Ti V Cr Mn Fe Element Co Ni Cu CHEMISTRY OF IRON AND COPPER Figure 22.4 shows the first-row transition metals. In this section we will briefly survey the chemistry of two of these elements — iron and copper — paying particular attention to their occurrence, preparation, uses, and important compounds. IRON in le ab il va tA o “N ” on rsi Ve xt Te e- After aluminum, iron is the most abundant metal in Earth’s crust (6.2 percent by mass). It is found in many ores; some of the important ones are hematite, Fe2O3; siderite, FeCO3; and magnetite, Fe3O4 (Figure 22.5). The preparation of iron in a blast furnace and steelmaking were discussed in Section 21.2. Pure iron is a gray metal and is not particularly hard. It is an essential element in living systems. Iron reacts with hydrochloric acid to give hydrogen gas: Fe(s) FIGURE 22.5 The iron ore magnetite, Fe3O4. in ble ila va tA No “ FIGURE 22.6 CuFeS2. Back Forward ” ion ers V xt -Te e H2(g) 3 Concentrated sulfuric acid oxidizes the metal to Fe , but concentrated nitric acid renders the metal “passive” by forming a thin layer of Fe3O4 over the surface. One of the best-known reactions of iron is rust formation (see Section 19.7). The two oxidation states of iron are 2 and 3. Iron(II) compounds include FeO (black), FeSO4 7H2O (green), FeCl2 (yellow), and FeS (black). In the presence of oxygen, Fe2 ions in solution are readily oxidized to Fe3 ions. Iron(III) oxide is reddish brown, and iron(III) chloride is brownish black. COPPER Copper, a rare element (6.8 10 3 percent of Earth’s crust by mass), is found in nature in the uncombined state as well as in ores such as chalcopyrite, CuFeS2 (Figure 22.6). The reddish-brown metal is obtained by roasting the ore to give Cu2S and then metallic copper: 2CuFeS2(s) Chalcopyrite, Cu2S(s) Main Menu 2H (aq) 88n Fe2 (aq) TOC 4O2(g) 88n Cu2S(s) O2(g) 88n 2Cu(l ) Study Guide TOC 2FeO(s) 3SO2(g) SO2(g) Textbook Website MHHE Website 876 TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS Scandium (Sc) Titanium (Ti) Vanadium (V) Chromium (Cr) Manganese (Mn) Iron (Fe) Nickel (Ni) Copper (Cu) Cobalt (Co) FIGURE 22.4 The first-row transition metals. Impure copper can be purified by electrolysis (see Section 20.2). After silver, which is too expensive for large-scale use, copper has the highest electrical conductivity. It is also a good thermal conductor. Copper is used in alloys, electrical cables, plumbing (pipes), and coins. Copper reacts only with hot concentrated sulfuric acid and nitric acid (see Figure 21.9). Its two important oxidation states are 1 and 2. The 1 state is less stable and disproportionates in solution: 2Cu (aq) 88n Cu(s) Cu2 (aq) All compounds of Cu(I) are diamagnetic and colorless except for Cu2O, which is red. The Cu(II) compounds are all paramagnetic and colored. The hydrated Cu2 ion is blue. Some important Cu(II) compounds are CuO (black), CuSO4 5H2O (blue), and CuS (black). 22.3 Recall that a complex ion contains a central metal ion bonded to one or more ions or molecules (see Section 16.10). Back Forward COORDINATION COMPOUNDS Transition metals have a distinct tendency to form complex ions, which in turn can combine with other ions or complex ions to form coordination compounds. A coordination compound is a neutral species containing one or more complex ions. Our understanding of the nature of coordination compounds stems from the classic work of Main Menu TOC Study Guide TOC Textbook Website MHHE Website 22.3 COORDINATION COMPOUNDS 877 Alfred Werner,† who prepared and characterized many coordination compounds. In 1893, at the age of 26, Werner proposed what is now commonly referred to as Werner’s coordination theory. Nineteenth-century chemists were puzzled by a certain class of reactions that seemed to violate valence theory. For example, the valences of the elements in cobalt(III) chloride and in ammonia seem to be completely satisfied, and yet these two substances react to form a stable compound having the formula CoCl3 6NH3. To explain this behavior, Werner postulated that most elements exhibit two types of valence: primary valence and secondary valence. In modern terminology, primary valence corresponds to the oxidation number and secondary valence to the coordination number of the element. In CoCl3 6NH3, according to Werner, cobalt has a primary valence of 3 and a secondary valence of 6. Today we use the formula [Co(NH3)6]Cl3 to indicate that the ammonia molecules and the cobalt atom form a complex ion; the chloride ions are not part of the complex but are held to it by ionic forces. Most, but not all, of the metals in coordination compounds are transition metals. The molecules or ions that surround the metal in a complex ion are called ligands (Table 22.3). The interactions between a metal atom and the ligands can be thought of † Alfred Werner (1866–1919). Swiss chemist. Werner started as an organic chemist but became interested in coordination chemistry. For his theory of coordination compounds, Werner was awarded the Nobel Prize in Chemistry in 1913. TABLE 22.3 Some Common Ligands NAME STRUCTURE Monodentate ligands HOOOH N A H S CqOS O S Cl S Q S CqNS Ammonia Carbon monoxide Chloride ion Cyanide ion SOO CqNS S Q Thiocyanate ion Bidentate ligands O H2NOCH2 O CH2 OO 2 NH S S Ethylenediamine S S S S S S Oxalate ion 2 S S O O M J C OC D G O O Polydendate ligand Forward Main Menu TOC S S S Study Guide TOC S S S Back 4 S S Ethylenediaminetetraacetate ion (EDTA) S S SOS SOS B B C C DG DG CH2 O CH2 O G D OOCH2 OCH2 OO N N D G CH2 O CH2 O MD GJ C C A A SQS O SQS O Textbook Website MHHE Website 878 TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS FIGURE 22.7 (a) Structure of metal-ethylenediamine complex. Each ethylenediamine molecule provides two N donor atoms and is therefore a bidentate ligand. (b) Simplified structure of the same complex. CH2 CH2 CH2 CH2 H2N NH2 H2N NH2 NH2 H2N CH2 CH2 (a) (b) S S as Lewis acid-base reactions. As we saw in Section 15.12, a Lewis base is a substance capable of donating one or more electron pairs. Every ligand has at least one unshared pair of valence electrons, as these examples show: O DG H H In a crystal lattice the coordination number of an atom (or ion) is defined as the number of atoms (or ions) surrounding the atom (or ion). O C O O C CH2 O CH2 N N CH2 C O O CH2 CH2 C O FIGURE 22.8 EDTA complex of lead. The complex bears a net charge of 2, since each O donor atom has one negative charge and the lead ion carries two positive charges. Note the octahedral geometry around the Pb2 ion. Back Forward O S Cl S Q S CqOS Therefore, ligands play the role of Lewis bases. On the other hand, a transition metal atom (in either its neutral or positively charged state) acts as a Lewis acid, accepting (and sharing) pairs of electrons from the Lewis bases. Thus the metal-ligand bonds are usually coordinate covalent bonds (see Section 9.9). The atom in a ligand that is bound directly to the metal atom is known as the donor atom. For example, nitrogen is the donor atom in the [Cu(NH3)4]2 complex ion. The coordination number in coordination compounds is defined as the number of donor atoms surrounding the central metal atom in a complex ion. For example, the coordination number of Ag in [Ag(NH3)2] is 2, that of Cu2 in [Cu(NH3)4]2 is 4, and that of Fe3 in [Fe(CN)6]3 is 6. The most common coordination numbers are 4 and 6, but coordination numbers such as 2 and 5 are also known. Depending on the number of donor atoms present, ligands are classified as monodentate, bidentate, or polydentate (see Table 22.3). H2O and NH3 are monodentate ligands with only one donor atom each. One bidentate ligand is ethylenediamine (sometimes abbreviated “en”): .. .. H2NOCH2OCH2ONH2 CH2 Pb O O N DAG HHH The two nitrogen atoms can coordinate with a metal atom as shown in Figure 22.7. Bidentate and polydentate ligands are also called chelating agents because of their ability to hold the metal atom like a claw (from the Greek chele, meaning “claw”). One example is ethylenediaminetetraacetate ion (EDTA), a polydentate ligand used to treat metal poisoning (Figure 22.8). Six donor atoms enable EDTA to form a very stable complex ion with lead. In this form it is removed from the blood and tissues and excreted from the body. EDTA is also used to clean up spills of radioactive metals. OXIDATION NUMBERS OF METALS IN COORDINATION COMPOUNDS Another important property of coordination compounds is the oxidation number of the central metal atom. The net charge of a complex ion is the sum of the charges on the central metal atom and its surrounding ligands. In the [PtCl6]2 ion, for example, each Main Menu TOC Study Guide TOC Textbook Website MHHE Website 22.3 COORDINATION COMPOUNDS 879 chloride ion has an oxidation number of 1, so the oxidation number of Pt must be 4. If the ligands do not bear net charges, the oxidation number of the metal is equal to the charge of the complex ion. Thus, in [Cu(NH3)4]2 each NH3 is neutral, so the oxidation number of Cu is 2. Example 22.1 deals with oxidation numbers of metals in coordination compounds. EXAMPLE 22.1 Specify the oxidation number of the central metal atom in each of the following compounds: (a) [Ru(NH3)5(H2O)]Cl2, (b) [Cr(NH3)6](NO3)3, (c) [Fe(CO)5], (d) K4[Fe(CN)6]. (a) Both NH3 and H2O are neutral species. Since each chloride ion carries a 1 charge, and there are two Cl ions, the oxidation number of Ru must be 2. (b) Each nitrate ion has a charge of 1; therefore, the cation must be [Cr(NH3)6]3 . NH3 is neutral, so the oxidation number of Cr is 3. (c) Since the CO species are neutral, the oxidation number of Fe is zero. (d) Each potassium ion has a charge of 1; therefore, the anion is [Fe(CN)6]4 . Next, we know that each cyanide group bears a charge of 1, so Fe must have an oxidation number of 2. Answer Similar problems: 22.13, 22.14. PRACTICE EXERCISE Write the oxidation numbers of the metals in the compound K[Au(OH)4]. NAMING COORDINATION COMPOUNDS Now that we have discussed the various types of ligands and the oxidation numbers of metals, our next step is to learn what to call these coordination compounds. The rules for naming coordination compounds are as follows: • • • • • Back Forward Main Menu The cation is named before the anion, as in other ionic compounds. The rule holds regardless of whether the complex ion bears a net positive or a negative charge. For example, in K3[Fe(CN)6] and [Co(NH3)4Cl2]Cl compound, we name the K and [Co(NH3)4Cl2] cations first, respectively. Within a complex ion the ligands are named first, in alphabetical order, and the metal ion is named last. The names of anionic ligands end with the letter o, whereas a neutral ligand is usually called by the name of the molecule. The exceptions are H2O (aquo), CO (carbonyl), and NH3 (ammine). Table 22.4 lists some common ligands. When several ligands of a particular kind are present, we use the Greek prefixes di, tri,- tetra,- penta-, and hexa- to name them. Thus the ligands in the cation [Co(NH3)4Cl2] are “tetraamminedichloro.” (Note that prefixes are ignored when alphabetizing ligands.) If the ligand itself contains a Greek prefix, we use the prefixes bis (2), tris (3), and tetrakis (4) to indicate the number of ligands present. For example, the ligand ethylenediamine already contains di; therefore, if two such ligands are present the name is bis(ethylenediamine). The oxidation number of the metal is written in Roman numerals following the name of the metal. For example, the Roman numeral III is used to indicate the 3 oxidation state of chromium in [Cr(NH3)4Cl2] , which is called tetraamminedichlorochromium(III) ion. TOC Study Guide TOC Textbook Website MHHE Website 880 TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS TABLE 22.4 Names of Common Ligands in Coordination Compounds LIGAND Bromide, Br Chloride, Cl Cyanide, CN Hydroxide, OH Oxide, O2 Carbonate, CO2 3 Nitrite, NO2 Oxalate, C2O2 4 Ammonia, NH3 Carbon monoxide, CO Water, H2O Ethylenediamine Ethylenediaminetetraacetate TABLE 22.5 Names of Anions Containing Metal Atoms METAL NAME OF METAL IN ANIONIC COMPLEX Aluminum Chromium Cobalt Copper Gold Iron Lead Manganese Molybdenum Nickel Silver Tin Tungsten Zinc Aluminate Chromate Cobaltate Cuprate Aurate Ferrate Plumbate Manganate Molybdate Nickelate Argentate Stannate Tungstate Zincate Similar problems: 22.15, 22.16. NAME OF LIGAND IN COORDINATION COMPOUND Bromo Chloro Cyano Hydroxo Oxo Carbonato Nitro Oxalato Ammine Carbonyl Aquo Ethylenediamine Ethylenediaminetetraacetato If the complex is an anion, its name ends in -ate. For example, in K4[Fe(CN)6] the anion [Fe(CN)6]4 is called hexacyanoferrate(II) ion. Note that the Roman numeral II indicates the oxidation state of iron. Table 22.5 gives the names of anions containing metal atoms. • The following examples deal with the nomenclature of coordination compounds. EXAMPLE 22.2 Write the systematic names of the following compounds: (a) Ni(CO)4, (b) [Co(NH3)4Cl2]Cl, (c) K3[Fe(CN)6], (d) [Cr(en)3]Cl3. (a) The CO ligands are neutral species and the nickel atom bears no net charge, so the compound is called tetracarbonylnickel(0), or, more commonly, nickel tetracarbonyl. (b) Starting with the cation, each of the two chloride ligands bears a negative charge and the ammonia molecules are neutral. Thus, the cobalt atom must have an oxidation number of 3 (to balance the chloride anion). The compound is called tetraamminedichlorocobalt(III) chloride. (c) The complex ion is the anion and it bears three negative charges. Thus, the iron atom must have an oxidation number of 3. The compound is potassium hexacyanoferrate(III). This compound is commonly called potassium ferricyanide. (d) As we noted earlier, en is the abbreviation for the ligand ethylenediamine. Since there are three en groups present and the name of the ligand already contains di, the name of the compound is tris(ethylenediamine)chromium(III) chloride. Answer PRACTICE EXERCISE What is the systematic name of [Cr(H2O)4Cl2]Cl? EXAMPLE 22.3 Write the formulas for the following compounds: (a) pentaamminechlorocobalt(III) chloride, (b) dichlorobis(ethylenediamine)platinum(IV) nitrate, (c) sodium hexanitrocobaltate(III). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 22.4 881 STRUCTURE OF COORDINATION COMPOUNDS Answer (a) The complex cation contains five NH3 groups, a chloride ion, and a cobalt ion with a 3 oxidation number. The net charge on the cation must be 2 . Therefore, the formula for the compound is [Co(NH3)5Cl]Cl2. (b) There are two chloride ions, two ethylenediamine groups, and a platinum ion with an oxidation number of 4 in the complex cation. Therefore, the formula for the compound is [Pt(en)2Cl2](NO3)2. (c) The complex anion contains six nitro groups and a cobalt ion with an oxidation number of 3. Therefore, the formula for the compound is Na3[Co(NO2)6]. Similar problems: 22.17, 22.18. PRACTICE EXERCISE Write the formula for the following compound: tris(ethylenediamine)cobalt(III) sulfate. 22.4 STRUCTURE OF COORDINATION COMPOUNDS In studying the geometry of coordination compounds, we often find that there is more than one way to arrange ligands around the central atom. Compounds rearranged in this fashion have distinctly different physical and chemical properties. Figure 22.9 shows four different geometric arrangements for metal atoms with monodentate ligands. In these diagrams we see that structure and coordination number of the metal atom relate to each other as follows: Coordination number 2 4 6 Structure Linear Tetrahedral or square planar Octahedral 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. There are two types of stereoisomers: geometric isomers and optical isomers. Coordination compounds may exhibit one or both types of isomerism. Note, however, that many coordination compounds do not have stereoisomers. GEOMETRIC ISOMERS Left: cis-tetraamminedichlorocobalt(III) chloride. Right: transtetraamminedichlorocobalt(III) chloride. Geometric isomers are stereoisomers that cannot be interconverted without breaking a chemical bond. 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 particular 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 FIGURE 22.9 Common geometries of complex ions. In each case M is a metal and L is a monodentate ligand. L L L L M L M L L M M L L L L L L L L L Linear Back Forward Main Menu Tetrahedral TOC Study Guide TOC Square planar Octahedral Textbook Website MHHE Website 882 TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS FIGURE 22.10 The (a) cis and (b) trans isomers of diamminedichloroplatinum(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. H3N Cl H3N Pt Cl H3N NH3 (b) Cl Cl Co NH3 H3N Cl Cl Co NH3 Cl (a) NH3 Cl NH3 Pt NH3 Cl Cl NH3 H3N Co Cl Co NH3 H3N NH3 NH3 H3N NH3 NH3 NH3 Cl (a) (b) (c) (d) FIGURE 22.11 The (a) cis and (b) trans isomers of tetraamminedichlorocobalt(III) ion. The structure shown in (c) can be generated by rotating that in (a), and the structure shown in (d) can be generated by rotating that in (b). The ion has only two geometric isomers, (a) [or (c)] and (b) [or (d)]. and trans isomers of coordination compounds generally have quite different colors, melting points, dipole moments, and chemical reactivities. Figure 22.10 shows the cis and trans isomers of diamminedichloroplatinum(II). Note that although the types of bonds are the same in both isomers (two PtON and two PtOCl bonds), the spatial arrangements are different. Another example is tetraamminedichlorocobalt(III) ion, shown in Figure 22.11. OPTICAL ISOMERS Mirror image of left hand Left hand Mirror FIGURE 22.12 A left hand and its mirror image, which looks the same as the right hand. Back Forward 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 reactivity toward molecules that are not optical isomers themselves. Optical isomers differ from each other 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 22.12). We say that your left hand and right hand are mirror images of each other. However, they are nonsuperimposable, because when you place your left hand over your right hand (with both palms facing down), they do not match. Figure 22.13 shows the cis and trans isomers of dichlorobis(ethylenediamine)cobalt(III) ion and their images. Careful examination reveals that the trans isomer and its mirror image are superimposable, but the cis isomer and its mirror image are not. Therefore, 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 play a vital role in enzyme reactions in biological systems. Many drug molecules are chiral. It is interesting to note that only one of a pair of chiral isomers is biologically effective. Main Menu TOC Study Guide TOC Textbook Website MHHE Website 22.5 FIGURE 22.13 The (a) cis and (b) trans isomers of dichlorobis(ethylenediamine)cobalt(III) ion and their mirror images. If you could rotate the mirror image in (b) 90° clockwise about the vertical position and place it over the trans isomer, you would find that the two are superimposable. No matter how you rotated the cis isomer and its mirror image, however, you could not superimpose one on the other. Polaroid sheets are used to make Polaroid sunglasses. 22.5 883 BONDING IN COORDINATION COMPOUNDS: CRYSTAL FIELD THEORY Mirror Cl Mirror Cl Cl Co Co Cl Cl Co Co Cl Cl (a) Cl (b) 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 measure the rotation of polarized light by optical isomers (Figure 22.14). A beam of unpolarized light first passes through a Polaroid sheet, called the polarizer, and then through a sample tube containing 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 or to the left. This rotation can be measured directly by turning the analyzer in the appropriate direction until minimal light transmission is achieved (Figure 22.15). If the plane of polarization is rotated to the right, the isomer is said to be dextrorotatory (d); it is levorotatory (l) if the rotation is to the left. The d and l isomers of a chiral substance, called enantiomers, always rotate the light 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. 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 FIGURE 22.14 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 polarizer. 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. Back Forward Main Menu + Analyzer Degree scale +90° 0° – –90° 180° Polarimeter tube Fixed polarizer Optically active substance in solution Light source TOC Plane of polarization Study Guide TOC Textbook Website MHHE Website 884 TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS FIGURE 22.15 With one Polaroid sheet over a picture, light passes through. With a second sheet of Polaroid placed over the first so that the axes of polarization of the sheets are perpendicular, little or no light passes through. If the axes of polarization of the two sheets were parallel, light would pass through. applied to transition metal complexes. 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 The name “crystal field” is associated with the theory used to explain the properties of solid, crystalline materials. The same theory is used to study coordination compounds. Back Forward 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 electrostatic repulsion between the lone pairs on the ligands and the electrons in the d orbitals of the metals. As we saw in Chapter 7, d orbitals have different orientations, but in the absence of external 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 five d orbitals experience electrostatic repulsion. The magnitude of this repulsion depends on the orientation of the d orbital that is involved. Take the dx2 y2 orbital as an example. In Figure 22.16, we see that the lobes of this orbital point toward corners of the octahedron along the x and y axes, where the lone-pair electrons are positioned. Thus, an electron residing in this orbital would experience a greater repulsion from the ligands than an electron would in, say, the dxy orbital. For this reason, the energy of the dx2 y2 orbital is increased relative to the dxy, dyz, and dxz orbitals. The dz2 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 energy levels: a higher level with two orbitals (dx2 y2 Main Menu TOC Study Guide TOC Textbook Website MHHE Website 22.5 BONDING IN COORDINATION COMPOUNDS: CRYSTAL FIELD THEORY 885 z FIGURE 22.16 The five 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. x y dz2 dyz dxy dx 2 – y 2 dxz and dz2) having the same energy and a lower level with three equal-energy orbitals (dxy, dyz, and dxy), as shown in Figure 22.17. The crystal field splitting ( ) 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. COLOR In Chapter 7 we learned that white light, such as sunlight, is a combination of all colors. A substance appears 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 component. An object also looks green if it reflects all colors except red, the complementary color of green (Figure 22.18). What has been said of reflected light also applies to transmitted light (that is, the light that passes through the medium, for example, a solution). Consider the hydrated cupric ion, [Cu(H2O)6]2 , which absorbs light in the orange region of the spectrum so that a solution of CuSO4 appears blue to us. Recall from Chapter 7 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 compound), and an FIGURE 22.17 Crystal field splitting between d orbitals in an octahedral complex. Energy dx 2 – y 2 Crystal field splitting dxy Back Forward Main Menu dz2 TOC Study Guide TOC dyz dxz Textbook Website MHHE Website 886 TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS 650 nm 800 nm 400 nm 430 nm electron is promoted to a higher level. This knowledge allows us to calculate the energy change involved in the electron transition. The energy of a photon, given by Equation (7.2), is: 580 nm 560 nm E h where h represents Planck’s constant (6.63 10 34 J s) and is the frequency of the radiation, which is 5.00 1014 /s for a wavelength of 600 nm. Here E , so we have 490 nm FIGURE 22.18 A color wheel with appropriate wavelengths. Complementary colors, such as red and green, are on opposite sides of the wheel. h (6.63 3.00 10 10 34 19 1014 /s) J s)(5.00 J (Note that this is the energy absorbed by 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 best way to measure crystal field splitting is to use spectroscopy to determine the wavelength at which light is absorbed. The [Ti(H2O)6]3 ion provides a straightforward example, because Ti3 has only one 3d electron (Figure 22.19). The [Ti(H2O)6]3 ion absorbs light in the visible region of the spectrum (Figure 22.20). The wavelength corresponding to maximum absorption is 498 nm [Figure 22.19(b)]. This information enables us to calculate the crystal field splitting as follows. We start by writing h (22.1) Also Equation (7.1) shows that c c . where c is the velocity of light and FIGURE 22.19 (a) The process of photon absorption and (b) a graph of the absorption spectrum of [Ti(H2O)6]3 . 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. dx 2 – y 2 Photon of energy hν dxy is the wavelength. Therefore dz2 dyz dx 2 – y 2 dxz dxy dz2 dyz dxz Absorption (a) 400 Back Forward Main Menu TOC 500 600 Wavelength (nm) (b) Study Guide TOC 700 Textbook Website MHHE Website 22.5 887 BONDING IN COORDINATION COMPOUNDS: CRYSTAL FIELD THEORY FIGURE 22.20 Colors of some of the first-row transition metal ions in solution. From left to right: Ti3 , Cr3 , Mn2 , Fe3 , Co2 , Ni2 , Cu2 . The Sc3 and V 5 ions are colorless. hc (6.63 34 10 3.99 10 19 J s)(3.00 108 m/s) J This is the energy required to excite one [Ti(H2O)6]3 ion. To express this energy difference in the more convenient units of kilojoules per mole, we write (3.99 10 19 J/ion)(6.02 1023 ions/mol) 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 splitting for each ligand and established a spectrochemical series, which is a list of ligands arranged in order of their abilities to split the d orbital energies: The order in the spectrochemical series is the same no matter which metal atom (or ion) is present. I Br Cl OH F H2O NH3 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 complex ion. The [Ti(H2O)6]3 ion, having only one d electron, is always paramagnetic. However, for an ion with several d electrons, the situation is less clearcut. Consider, for example, the octahedral complexes [FeF6]3 and [Fe(CN)6]3 (Figure 22.21). The electron configuration of Fe3 is [Ar]3d 5, and there are two possible ways FIGURE 22.21 Energy-level diagrams for the Fe3 ion and for the [FeF6]3 and [Fe(CN)6]3 complex ions. dx 2 – y 2 Energy dx 2 – y 2 dz2 Fe3+ ion dxy dyz dxz [FeF6 ]3 – (high spin) Back Forward Main Menu dz2 TOC Study Guide TOC dxy dyz dxz [Fe(CN)6 ] 3 – (low spin) Textbook Website MHHE Website 888 TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS to distribute the five d electrons among the d orbitals. According to Hund’s rule (see Section 7.8), maximum stability is reached when the electrons are placed in five separate orbitals with parallel spins. But this arrangement can be achieved only at a cost; two of the five electrons must be promoted to the higher-energy dx2 y2 and dz2 orbitals. No such energy investment is needed if all five electrons enter the dxy, dyz, and dxz orbitals. According to Pauli’s exclusion principle (p. 269), there will be only one unpaired electron present in this case. Figure 22.22 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 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 (see Figure 22.21). On the other hand, the cyanide ion is a strong-field ligand, so it is energetically preferable for all five electrons to be in FIGURE 22.22 Orbital diagrams for the high-spin and lowspin octahedral complexes corresponding to the electron configurations d 4, d 5, d 6, and d 7. No such distinctions can be made for d1, d 2, d 3, d 8 ,d 9, and d10. High spin Low spin dx 2 – y 2 d dx 2 – y 2 4 dxy dyz dz2 dz2 dxz dxy dyz dxz d5 d6 d7 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 22.5 The magnetic properties of a complex ion depend on the number of unpaired electrons present. BONDING IN COORDINATION COMPOUNDS: CRYSTAL FIELD THEORY 889 the lower orbitals and therefore a low-spin complex is formed. 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 magnetic 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 22.22. EXAMPLE 22.4 Predict the number of unpaired spins in the [Cr(en)3]2 ion. The electron configuration of Cr2 is [Ar]3d 4. Since en is a strong-field ligand, we expect [Cr(en)3]2 to be a low-spin complex. According to Figure 22.22, all four electrons will be placed in the lower-energy d orbitals (dxy, dyz, and dxz) and there will be a total of two unpaired spins. Answer Similar problem: 22.35. PRACTICE EXERCISE How many unpaired spins are in [Mn(H2O)6]2 ? (H2O 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 two other types of complexes — tetrahedral and square-planar — can also be accounted for satisfactorily 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 dxy, dyz, and dxz orbitals are more closely directed at the ligands and therefore have more energy than the dx2 y2 and dy2 orbitals (Figure 22.23). Most tetrahedral complexes are high-spin complexes. Presumably, the tetrahedral arrangement reduces the magnitude of metal-ligand interactions, resulting in a smaller value. This is a reasonable assumption since the number of ligands is smaller in a tetrahedral complex. As Figure 22.24 shows, the splitting pattern for square-planar complexes is the most complicated. Clearly, the dx2 y2 orbital possesses the highest energy (as in the octahedral case), and the dxy orbital the next highest. However, the relative placement of the dz2 and the dxz and dyz orbitals cannot be determined simply by inspection and must be calculated. FIGURE 22.23 Crystal field splitting between d orbitals in a tetrahedral complex. dyz Energy dxy Crystal field splitting dx 2 – y 2 Back Forward Main Menu dxz TOC Study Guide TOC dz2 Textbook Website MHHE Website TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS FIGURE 22.24 Energy-level diagram for a square-planar complex. Because there are more than two energy levels, we cannot define crystal field splitting as we can for octahedral and tetrahedral complexes. dx 2 – y 2 dxy Energy 890 dz2 dxz 22.6 dyz REACTIONS OF COORDINATION COMPOUNDS Complex ions undergo ligand exchange (or substitution) reactions in solution. The rates of these reactions vary widely, depending on the nature of the metal ion and the ligands. In studying ligand exchange reactions, it is often useful 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 Kf (see p. 677). For example, we say that the complex ion tetracyanonickelate(II) is stable because it has a large formation constant (Kf ≈ 1 1030) Ni2 4CN 34 [Ni(CN)4]2 By using cyanide 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: [Ni(CN)4]2 4*CN 34 [Ni(*CN)4]2 4CN 14 where the asterisk denotes a C atom. Complexes like the tetracyanonickelate(II) ion are termed labile complexes because they undergo rapid ligand exchange reactions. Thus a thermodynamically stable species (that is, one that has a large formation constant) is not necessarily unreactive. (In Section 13.4 we saw that the smaller the activation energy, the larger the rate constant, and hence the greater the rate.) A complex that is thermodynamically unstable in acidic solution is [Co(NH3)6]3 . The equilibrium constant for the following reaction is about 1 1020: [Co(NH3)6]3 6H 6H2O 34 [Co(H2O)6]3 6NH4 When equilibrium is reached, the concentration of the [Co(NH3)6]3 ion is very low. However, this reaction requires several days to complete because of the inertness of the [Co(NH3)6]3 ion. This is an example of an inert complex, a complex ion that undergoes very slow exchange reactions (on the order of hours or even days). It shows that a thermodynamically unstable species is not necessarily chemically reactive. The rate of reaction is determined by the energy of activation, which is high in this case. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 22.7 APPLICATIONS OF COORDINATION COMPOUNDS 891 Most complex ions containing Co3 , Cr3 , and Pt2 are kinetically inert. Because they exchange ligands very slowly, they are easy to study in solution. As a result, our knowledge of the bonding, structure, and isomerism of coordination compounds has come largely from studies of these compounds. 22.7 APPLICATIONS OF COORDINATION COMPOUNDS Coordination compounds are found in living systems and have many uses in the home, in industry, and in medicine. We describe a few examples here and in the Chemistry in Action essay on p. 892. METALLURGY The extraction of silver and gold by the formation of cyanide complexes (p. 840) and the purification of nickel (p. 810) by converting the metal to the gaseous compound Ni(CO)4 are typical examples of the use of coordination compounds in metallurgical processes. THERAPEUTIC CHELATING AGENTS Earlier we mentioned that the chelating agent EDTA is used in the treatment of lead poisoning. Certain platinum-containing compounds can effectively inhibit the growth of cancerous cells. A specific case is discussed on p. 898. CHEMICAL ANALYSIS Although EDTA has a great affinity for a large number of metal ions (especially 2 and 3 ions), other chelates are more selective in binding. For example, dimethylglyoxime, H3C H3C Nickel dimethylglyoxime. G D CPNOOH A CPNOOH forms an insoluble brick-red solid with Ni2 and an insoluble bright-yellow solid with Pd2 . These characteristic colors are used in qualitative analysis to identify nickel and palladium. Further, the quantities of ions present can be determined by gravimetric analysis (see Section 4.6) as follows: To a solution containing Ni2 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 22.25), we can readily calculate the amount of nickel present in the original solution. DETERGENTS The cleansing action of soap in hard water is hampered by the reaction of the Ca2 ions in the water with the soap molecules to form insoluble salts or curds. In the late Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 892 TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS FIGURE 22.25 Structure of nickel dimethylglyoxime. Note that the overall structure is stabilized by hydrogen bonds. O H3C HZ Z O N CH 3 N C C Ni C C N H 3C N OZ Z H CH 3 O 1940s the detergent industry introduced a “builder,” usually sodium tripolyphosphate, to circumvent this problem. The tripolyphosphate ion is an effective chelating agent that forms stable, soluble complexes with Ca2 ions. Sodium tripolyphosphate revolutionized the detergent industry. However, because phosphates are plant nutrients, waste waters containing phosphates discharged into rivers and lakes cause algae to grow, resulting in oxygen depletion. Under these conditions most or all aquatic life eventually succumbs. This process is called eutrophication. Consequently, many states have banned phosphate detergents since the 1970s, and manufacturers have reformulated their products to eliminate phosphates. Tripolyphosphate ion. Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Coordination Compounds in Living Systems Coordination compounds play many important roles in animals and plants. They are essential in the storage and transport of oxygen, as electron transfer agents, as catalysts, and in photosynthesis. Here we focus on coordination compounds containing iron and magnesium. Because of its central function as an oxygen carrier for metabolic processes, hemoglobin is probably the most studied of all the proteins. The molecule contains N N H N H N N N þ Fe þ four folded long chains called subunits. Hemoglobin carries oxygen in the blood from the lungs to the tissues, where it delivers the oxygen molecules to myoglobin. Myoglobin, which is made up of only one subunit, stores oxygen for metabolic processes in 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 N H H O N N N Fe N Porphine Fe 2 N N -porphyrin Simplified structures of the porphine molecule and the Fe2 porphyrin complex. The dashed lines represent coordinate covalent bonds. The heme group in hemoglobin. The Fe2 ion is coordinated with the nitrogen atoms of the heme group. The ligand below the porphyrin is the histidine group, which is attached to the protein. The sixth ligand is a water molecule. HN Protein Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 22.7 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Back APPLICATIONS OF COORDINATION COMPOUNDS 893 O derived from porphine are called porphyrins, and the O iron-porphyrin combination is called the heme group. O O O O The iron in the heme group has an oxidation number of 2; it is coordinated to the four nitrogen atoms in N N N N N N the porphine group and also to a nitrogen donor atom Fe Fe Fe in a ligand that is attached to the protein. The sixth 2 N N N N N N ligand is a water molecule, which binds to the Fe ion on the other side of the ring to complete the ocN N N tahedral complex. This hemoglobin molecule is called HN HN HN deoxyhemoglobin and imparts a bluish tinge to venous blood. The water ligand can be replaced read(a) (b) (c) ily by molecular oxygen to form red oxyhemoglobin Three possible ways for molecular oxygen to bind to the found in arterial blood. Each subunit contains a heme heme group in hemoglobin. The structure shown in (a) group, so each hemoglobin molecule can bind up to would have a coordination number of 7, which is considfour O2 molecules. ered unlikely for Fe(II) complexes. Although the end-on There are three possible structures for oxyhemo- arrangement in (b) seems the most reasonable, evidence globin. For a number of years, the exact arrangement points to the structure in (c) as the correct one. of the oxygen molecule relative to the porphyrin group The heme group in cywas not clear. Most experimental evidence suggests tochrome c. The ligands above and below the porthat the bond between O and Fe is bent relative to CH2 phyrin are the methionine the heme group. H3C CH2 group and histidine group S The porphyrin group is a very effective chelating of the protein, respectively. N N agent, and not surprisingly, we find it in a number of Fe biological systems. The iron-heme complex is present N N in another class of proteins, called the cytochromes. N The iron forms an octahedral complex in these proHN teins, but because both the histidine and the methionine groups are firmly bound to the metal ion, they cannot be displaced by oxygen or other ligands. Protein Instead, the cytochromes act as electron carriers, which are coupled to the oxidation of organic molewhich are essential to metabolic processes. In cycules such as the carbohydrates. tochromes, iron undergoes rapid reversible redox reThe chlorophyll molecule, which is necessary for actions: plant photosynthesis, also contains the porphyrin ring, but in this case the metal ion is Mg2 rather than Fe2 . Fe3 e 34 Fe2 Forward C C C C C C C N N C CH Mg HC C N N C C C Main Menu C TOC The porphyrin structure in chlorophyll. The dotted lines indicate the coordinate covalent bonds. The electron-delocalized portion of the molecule is shown in color. C C C Study Guide TOC Textbook Website MHHE Website 894 TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action A solution showing the green color of chlorophyll at the bottom of the container. The beam of light enters at the top and is absorbed by the chlorophyll molecules. Some of the energy of the absorbed light is emitted as fluorescence in the red region of the visible spectrum. Fluorescence is the emission of electromagnetic radiation from an atom or a molecule, particularly in the visible region, after an initial absorption of a photon. KEY EQUATION SUMMARY OF FACTS AND CONCEPTS Back • 1. Transition metals usually have incompletely filled d orbitals and a pronounced tendency to form complexes. Compounds that contain complex ions are called coordination compounds. 2. The first-row transition metals (scandium to copper) are the most common of all the transition metals; their chemistry is characteristic, in many ways, of the entire group. 3. Complex ions consist of a metal ion surrounded by ligands. The donor atoms in the ligands each contribute an electron pair to the central metal ion in a complex. 4. Coordination compounds may display geometric and/or optical isomerism. 5. Crystal field theory explains bonding in complexes in terms of electrostatic interactions. According to crystal field theory, the d orbitals are split into two higher-energy and three lower-energy orbitals in an octahedral complex. The energy difference between these two sets of d orbitals is the crystal field splitting. 6. Strong-field ligands cause a large crystal field splitting, and weak-field ligands cause a small splitting. Electron spins tend to be parallel with weak-field ligands and paired with strong-field ligands, where a greater investment of energy is required to promote electrons into the high-lying d orbitals. 7. Complex ions undergo ligand exchange reactions in solution. Forward Main Menu h (22.1) TOC Calculating crystal-field splitting. Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 895 8. Coordination compounds find application in many different areas, for example, as antidotes for metal poisoning and in chemical analysis. KEY WORDS Chelating agent, p. 878 Chiral, p. 882 Coordination compound, p. 876 Coordination number, p. 876 Crystal field splitting ( ), p. 885 Donor atom, p. 878 Enantiomers, p. 883 Geometric isomers, p. 881 Inert complex, p. 890 Labile complex, p. 890 Ligand, p. 877 Optical isomers, p. 882 Polarimeter, p. 883 Racemic mixture, p. 883 Spectrochemical series, p. 887 Stereoisomers, p. 881 QUESTIONS AND PROBLEMS PROPERTIES OF THE TRANSITION METALS Review Questions 22.1 What distinguishes a transition metal from a representative metal? 22.2 Why is zinc not considered a transition metal? 22.3 Explain why atomic radii decrease very gradually from scandium to copper. 22.4 Without referring to the text, write the ground-state electron configurations of the first-row transition metals. Explain any irregularities. 22.5 Write the electron configurations of the following ions: V5 , Cr3 , Mn2 , Fe3 , Cu2 , Sc3 , Ti4 . 22.6 Why do transition metals have more oxidation states than other elements? 22.7 Give the highest oxidation states for scandium to copper. 22.8 Why does chromium seem to be less reactive than its standard reduction potential suggests? COORDINATION COMPOUNDS Review Questions 22.9 Define the following terms: coordination compound, ligand, donor atom, coordination number, chelating agent. 22.10 Describe the interaction between a donor atom and a metal atom in terms of a Lewis acid-base reaction. 22.13 Give the oxidation numbers of the metals in the following species: (a) K3[Fe(CN)6], (b) K3[Cr(C2O4)3], (c) [Ni(CN)4]2 . 22.14 Give the oxidation numbers of the metals in the following species: (a) Na2MoO4, (b) MgWO4, (c) Fe(CO)5. 22.15 What are the systematic names for the following ions and compounds? (c) [Co(en)2Br2] (a) [Co(NH3)4Cl2] (b) Cr(NH3)3Cl3 (d) [Co(NH3)6]Cl3 22.16 What are the systematic names for the following ions and compounds? (a) [cis-Co(en)2Cl2] (b) [Pt(NH3)5Cl]Cl3 (c) [Co(NH3)5Cl]Cl2 22.17 Write the formulas for each of the following ions and compounds: (a) tetrahydroxozincate(II), (b) pentaaquochlorochromium(III) chloride, (c) tetrabromocuprate(II), (d) ethylenediaminetetraacetatoferrate(II). 22.18 Write the formulas for each of the following ions and compounds: (a) bis(ethylenediamine)dichlorochromium(III), (b) pentacarbonyliron(0), (c) potassium tetracyanocuprate(II), (d) tetraammineaquochlorocobalt(III) chloride. STRUCTURE OF COORDINATION COMPOUNDS Review Questions Problems 22.11 Complete the following statements for the complex ion [Co(en)2(H2O)CN]2 . (a) en is the abbreviation for _______. (b) The oxidation number of Co is _______. (c) The coordination number of Co is _______. (d) _______ is a bidentate ligand. 22.12 Complete the following statements for the complex ion [Cr(C2O4)2(H2O)2] . (a) The oxidation number of Cr is _______. (b) The coordination number of Cr is _______. (c) _______ is a bidentate ligand. Back Forward Main Menu TOC 22.19 Define the following terms: stereoisomers, geometric isomers, optical isomers, plane-polarized light. 22.20 Specify which of the following structures can exhibit geometric isomerism: (a) linear, (b) square-planar, (c) tetrahedral, (d) octahedral. 22.21 What determines whether a molecule is chiral? How does a polarimeter measure the chirality of a molecule? 22.22 Explain the following terms: (a) enantiomers, (b) racemic mixtures. Study Guide TOC Textbook Website MHHE Website 896 TRANSITION METAL CHEMISTRY AND COORDINATION COMPOUNDS Problems 2 22.38 A solution made by dissolving 0.875 g of Co(NH3)4Cl3 in 25.0 g of water freezes at 0.56°C. Calculate the number of moles of ions produced when 1 mole of Co(NH3)4Cl3 is dissolved in water, and suggest a structure for the complex ion present in this compound. 22.23 The complex ion [Ni(CN)2Br2] has a squareplanar geometry. Draw the structures of the geometric isomers of this complex. 22.24 How many geometric isomers are in the following species? (a) [Co(NH3)2Cl4] , (b) [Co(NH3)3Cl3] 22.25 Draw structures of all the geometric and optical isomers of each of the following cobalt complexes: (a) [Co(NH3)6]3 (b) [Co(NH3)5Cl]2 (c) [Co(C2O4)3]3 22.26 Draw structures of all the geometric and optical isomers of each of the following cobalt complexes: (a) [Co(NH3)4Cl2] , (b) [Co(en)3]3 . REACTIONS OF COORDINATION COMPOUNDS BONDING IN COORDINATION COMPOUNDS Problems Review Questions 22.41 Oxalic acid, H2C2O4, is sometimes used to clean rust stains from sinks and bathtubs. Explain the chemistry underlying this cleaning action. 22.42 The [Fe(CN)6]3 complex is more labile than the [Fe(CN)6]4 complex. Suggest an experiment that would prove that [Fe(CN)6]3 is a labile complex. 22.43 Aqueous copper(II) sulfate solution is blue in color. When aqueous potassium fluoride is added, a green precipitate is formed. When aqueous potassium chloride is added instead, a bright-green solution is formed. Explain what is happening in these two cases. 22.44 When aqueous potassium cyanide is added to a solution of copper(II) sulfate, a white precipitate, soluble in an excess of potassium cyanide, is formed. No precipitate is formed when hydrogen sulfide is bubbled through the solution at this point. Explain. 22.45 A concentrated aqueous copper(II) chloride solution is bright green in color. When diluted with water, the solution becomes light blue. Explain. 22.46 In a dilute nitric acid solution, Fe3 reacts with thiocyanate ion (SCN ) to form a dark-red complex: 22.27 Briefly describe crystal field theory. 22.28 Define the following terms: crystal field splitting, high-spin complex, low-spin complex, spectrochemical series. 22.29 What is the origin of color in a coordination compound? 22.30 Compounds containing the Sc3 ion are colorless, whereas those containing the Ti3 ion are colored. Explain. 22.31 What factors determine whether a given complex will be diamagnetic or paramagnetic? 22.32 For the same type of ligands, explain why the crystal field splitting for an octahedral complex is always greater than that for a tetrahedral complex. Problems 22.33 The [Ni(CN)4]2 ion, which has a square-planar geometry, is diamagnetic, whereas the [NiCl4]2 ion, which has a tetrahedral geometry, is paramagnetic. Show the crystal field splitting diagrams for those two complexes. 22.34 Transition metal complexes containing CN ligands are often yellow in color, whereas those containing H2O ligands are often green or blue. Explain. 22.35 Predict the number of unpaired electrons in the following complex ions: (a) [Cr(CN)6]4 , (b) [Cr(H2O)6]2 . 22.36 The absorption maximum for the complex ion [Co(NH3)6]3 occurs at 470 nm. (a) Predict the color of the complex and (b) calculate the crystal field splitting in kJ/mol. 22.37 From each of the following pairs, choose the complex that absorbs light at a longer wavelength: (a) [Co(NH3)6]2 , [Co(H2O)6]2 ; (b) [FeF6]3 , [Fe(CN)6]3 ; (c) [Cu(NH3)4]2 , [CuCl4]2 . Back Forward Main Menu TOC Review Questions 22.39 Define the terms (a) labile complex, (b) inert complex. 22.40 Explain why a thermodynamically stable species may be chemically reactive and a thermodynamically unstable species may be unreactive. [Fe(H2O)6]3 SCN 34 H2O [Fe(H2O)5NCS]2 The equilibrium concentration of [Fe(H2O)5NCS]2 may be determined by how darkly colored the solution is (measured by a spectrometer). In one such experiment, 1.0 mL of 0.20 M Fe(NO3)3 was mixed with 1.0 mL of 1.0 10 3 M KSCN and 8.0 mL of dilute HNO3. The color of the solution quantitatively indicated that the [Fe(H2O)5NCS]2 concentration was 7.3 10 5 M. Calculate the formation constant for [Fe(H2O)5NCS]2 . ADDITIONAL PROBLEMS 22.47 As we read across the first-row transition metals from left to right, the 2 oxidation state becomes more stable in comparison with the 3 state. Why is this so? Study Guide TOC Textbook Website MHHE Website 897 QUESTIONS AND PROBLEMS 22.48 Which is a stronger oxidizing agent, Mn3 or Cr3 ? Explain. 22.49 What is the systematic name for the compound [Fe(en)3][Fe(CO)4]? (Hint: The oxidation number of Fe in the complex cation is 2.) 22.50 What are the oxidation states of Fe and Ti in the ore ilmenite, FeTiO3? (Hint: Look up the ionization energies of Fe and Ti in Table 22.1; the fourth ionization energy of Ti is 4180 kJ/mol.) 22.51 A student has prepared a cobalt complex that has one of the following three structures: [Co(NH3)6]Cl3, [Co(NH3)5Cl]Cl2, or [Co(NH3)4Cl2]Cl. Explain how the student would distinguish between these possibilities by an electrical conductance experiment. At the student’s disposal are three strong electrolytes — NaCl, MgCl2, and FeCl3 — which may be used for comparison purposes. 22.52 Chemical analysis shows that hemoglobin contains 0.34 percent of Fe by mass. What is the minimum possible molar mass of hemoglobin? The actual molar mass of hemoglobin is about 65,000 g. How do you account for the discrepancy between your minimum value and the actual value? 22.53 Explain the following facts: (a) Copper and iron have several oxidation states, whereas zinc has only one. (b) Copper and iron form colored ions, whereas zinc does not. 22.54 A student in 1895 prepared three coordination compounds containing chromium, with the following properties: FORMULA COLOR (a) CrCl3 6H2O (b) CrCl3 6H2O (c) CrCl3 6H2O Cl IONS IN SOLUTION PER FORMULA UNIT Violet Light green Dark green 2CN 34 [Ag(CN)2] 2NH3 22.56 From the standard reduction potentials listed in Table Back Forward Main Menu Zn(s) TOC 2Cu2 (aq) 88n Zn2 (aq) 2Cu (aq) 22.57 Using the standard reduction potentials listed in Table 19.1 and the Handbook of Chemistry and Physics, show that the following reaction is favorable under standard-state conditions: 2Ag(s) 22.58 22.59 22.60 22.61 22.62 22.63 3 2 1 Write modern formulas for these compounds and suggest a method for confirming the number of Cl ions present in solution in each case. (Hint: Some of the compounds may exist as hydrates.) 22.55 The formation constant for the reaction Ag 2NH3 34 [Ag(NH3)2] is 1.5 107 and that for 2CN 34 [Ag(CN)2] is the reaction Ag 1.0 1021 at 25°C (see Table 16.3). Calculate the equilibrium constant and G° at 25°C for the reaction [Ag(NH3)2] 19.1 for Zn/Zn2 and Cu /Cu2 , calculate G° and the equilibrium constant for the reaction 22.64 22.65 22.66 Pt2 (aq) 88n 2Ag (aq) Pt(s) What is the equilibrium constant of this reaction at 25°C? The Co2 -porphyrin complex is more stable than the Fe2 -porphyrin complex. Why, then, is iron the metal ion in hemoglobin (and other heme-containing proteins)? What are the differences between geometric isomers and optical isomers? Oxyhemoglobin is bright red, whereas deoxyhemoglobin is purple. Show that the difference in color can be accounted for qualitatively on the basis of highspin and low-spin complexes. (Hint: O2 is a strongfield ligand; see Chemistry in Action essay on p. 892.) Hydrated Mn2 ions are practically colorless (see Figure 22.21) even though they possess five 3d electrons. Explain. (Hint: Electronic transitions in which there is a change in the number of unpaired electrons do not occur readily.) Which of the following hydrated cations are colorless: Fe2 (aq), Zn2 (aq), Cu (aq), Cu2 (aq), V5 (aq), Ca2 (aq), Co2 (aq), Sc3 (aq), Pb2 (aq)? Explain your choice. Aqueous solutions of CoCl2 are generally either light pink or blue. Low concentrations and low temperatures favor the pink form while high concentrations and high temperatures favor the blue form. Adding hydrochloric acid to a pink solution of CoCl2 causes the solution to turn blue; the pink color is restored by the addition of HgCl2. Account for these observations. Suggest a method that would allow you to distinguish between cis-Pt(NH3)2Cl2 and trans-Pt(NH3)2Cl2. You are given two solutions containing FeCl2 and FeCl3 at the same concentration. One solution is light yellow and the other one is brown. Identify these solutions based only on color. The label of a certain brand of mayonnaise lists EDTA as a food preservative. How does EDTA prevent the spoilage of mayonnaise? Answers to Practice Exercises: 22.1 K: 1; Au: 3. 22.2 Tetraaquodichlorochromium(III) Chloride. 22.3 [Co(en)3]2(SO4)3. 22.4 5. Study Guide TOC Textbook Website MHHE Website C HEMISTRY IN T HREE D IMENSIONS Anticancer Coordination Compounds Cisplatin, a bright yellow compound, is administered intravenously to cancer patients. It destroys the cancer cells’ ability to reproduce by changing the configuration of their DNA. Cisplatin binds to two sites on a strand of DNA, causing it to bend about 33° away from the rest of the strand. Cisplatin Pt N H3 33° N H3 Cis-diamminedichloroplatinum(II), [Pt(NH3)2Cl2], is one of a number of platinum coordination compounds used in the treatment of cancer. Commonly known as cisplatin, this compound has a square planar geometry and labile chloride ligands: H3N Cl GD Pt DG H3N Cl Cisplatin has the ability to block the uncontrolled division of cancerous cells that results in the growth of tumors. The mechanism for this action is the chelation of DNA (deoxyribonucleic acid), the molecule that contains the genetic code. During cell division, the double-stranded DNA separates into single-stranded regions, which must be accurately copied in order for the new cells to contain DNA identical to that of the parent cell. X-ray studies show that cisplatin binds to DNA by forming intrastrand crosslinks in which the two chlorides on cisplatin are replaced by nitrogen atoms in the adjacent guanine bases on the same strand of the DNA molecule. (Guanine is one of the four bases in DNA. See Figure 25.17.) Consequently, the double-stranded structure bends at the binding site. Scientists believe that this structural distortion is a key factor in preventing enzymes from repairing a cancer cell’s DNA, thereby inhibiting replication. The damaged cell is then destroyed by the body’s immune system. Although normal cells are similarly affected by the presence of cisplatin, for some reason they are better able to repair the damage and resume normal functioning. 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. 898 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website X-ray diffraction studies provided this model of DNA-cisplatin binding in which the two chloride ligands in cisplatin are replaced by the N atoms in adjacent guanine bases on one strand of DNA. The structure of the cisplatin-DNA adduct is partially stabilized by hydrogen bonds (red dotted lines). For simplicity, all H atoms in the DNA molecule are omitted. O P O C O O O N Pt N C O C C C C O N C C N P C O C C N N N N O C C C C O O N N C C C N O C C O The trans isomer of a dinuclear platinum complex can form intrastrand DNA crosslinks, which do not distort the molecule’s shape but disrupt its function. Here the two chloride ligands of the complex ion are replaced by the N atoms in the five-membered ring of the guanines on two different strands. The rodlike structure represents the bases held together by hydrogen bonds. Pt Pt Unfortunately, cisplatin can cause serious side effects, including severe kidney damage. Therefore, ongoing research efforts are directed toward finding related complexes that destroy cancer cells with less harm to healthy tissues. A recently synthesized series of dinuclear (having two metal atoms) platinum complexes binds to DNA in a different way. One of these chelating agents is the complex ion [{transPt(NH3)2Cl}2NH2(CH2)nNH2]2 NH3 Cl H3N GD GD Pt Pt DG DG NH2(CH2)n H2N H3N NH3 Cl 2 where n is an integer between 2 and 6. Because of its wider reach, this dinuclear complex ion can form cross-links with nitrogen atoms in guanines on both strands of a DNA molecule. It is interesting to note that the cis isomer has no effect on DNA because of its unfavorable geometry: Cl Cl NH3 H3N GD GD Pt Pt DG DG NH2(CH2)n H2N H3N NH3 2 The trans isomer does not bend DNA as cisplatin does, but it does have the ability to disrupt normal DNA function and hence can destroy cancer cells. 899 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website C HEMICAL M YSTERY Dating Paintings with Prussian Blue I Portrait of a Noblewoman. n 1995 a painting entitled Portrait of a Noblewoman was given to the Mead Art Museum in Amherst, Massachusetts by an anonymous donor. The small bustlength portrait on wood panel depicts an unidentified adolescent girl of noble birth. The background, blue “wallpaper” with a gold fleur-de-lys pattern above a wood-paneled wainscot, is characteristic of settings seen in full-length Renaissance portraits. The rich dress and French royal heraldic device suggest that the subject is a member of the royal court, if not the royal family itself. The painting was attributed to the school of the court painter, François Clouet (1522 – 1572). While in the donor ’s possession, the painting was scratched by a cat. An art conservator who was asked to repair the damage became suspicious about the blue paint used to render the girl’s hat and the wallpaper. Subsequent analysis of microscopic samples of the paint revealed that the pigment was Prussian blue (ferric ferrocyanide, Fe4[Fe(CN)6]3), a coordination compound discovered by a German dyemaker between 1704 and 1707. Prior to the discovery of Prussian blue, there were three blue pigments available to painters: azurite [Cu3(OH)2(CO3)2], smalt (a complex cobalt and arsenic compound), and ultramarine blue, which has the complex formula of CaNa7Al6Si6O24SO4. Prussian blue quickly came to be valued by painters for the intensity and transparency of its color, and it is commonly found in works painted after the early 1700s. In Prussian blue the Fe2 ion is bonded to the carbon atom of the cyanide group in an octahedral arrangement, and each Fe3 ion is bonded through the nitrogen atom of the cyanide group with a similar octahedral symmetry. Thus the cyanide group acts as a bidentate ligand as shown below Fe2+ C Fe3+ N The blue color arises from the so-called intervalence charge transfer between the metal ions. If we designate these two sites of iron as Fe2 and Fe3 , where A and B denote A B different sites defined by the ligands, we can represent the transfer of an electron from Fe2 to Fe3 as follows: h Fe2 , Fe3 88n Fe3 , Fe2 A B A B The right-hand side of the above equation has more energy than the left-hand side, and Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website the result is an energy level and a light-absorbing scheme similar to that shown in Figure 22.19. It was Prussian blue’s unique color that drew the conservator ’s suspicion and prompted an analysis that dated the painting at sometime after 1704. Additional analysis of the green pigment used for the jewels showed a mixture of Prussian blue and lemon yellow (zinc chromate, ZnCrO4), a pigment produced commercially beginning around 1850. Thus, Portrait of a Noblewoman is no longer ascribed to any 16th century painter, but it is now used by the museum to teach art historians about fakes, forgeries, and mistaken attributions. CHEMICAL CLUES 1. 2. Prussian blue can be prepared by mixing a FeCl3 solution with a K4Fe(CN)6 solution. 3. 4. 5. 6. Give the systematic name of Prussian blue. In what region of the visible spectrum does the intervalence charge transfer absorb light? How can you show that the color of Prussian blue arises from intervalence charge transfer and not from a transition within a single ion such as Fe(CN)3 or 6 Fe(CN)4 ? 6 Write the formulas and give the systematic names for ferrous ferrocyanide and ferric ferricyanide. Will intervalence charge transfer occur in these two compounds? Kinetically, the Fe(CN)4 ion is inert while the Fe(CN)3 ion is labile. Based on 6 6 this knowledge, would you expect Prussian blue to be a poisonous cyanide compound? Explain. When Prussian blue is added to a NaOH solution, an orange-brown precipitate forms. Identify the precipitate. How can the formation of Prussian blue be used to distinguish between Fe2 and Fe3 ions in solution? 901 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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