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Chemistry, Student Solutions Manual Chapter 18 Chapter 18 The Transition Metals Solutions to Problems in Chapter 18 18.1 Oxidation states are determined by applying the rules given in Section 16.1 (Chapter 16). The procedure can be simplified when a polyatomic ion of known charge is present: (a) CO32– has a charge of –2, so Mn must be +2 to give overall neutrality. (b) Cl (more electronegative) is –1, so Mo must be +5 to give overall neutrality. (c) Na is +1, so VO4 has overall charge –3; each O is –2, so V is +5. (d) O is –2, so Au must be +3 to give overall neutrality. (e) H2O has zero charge, and SO42– has a charge of –2, so Fe must be +3 to give overall neutrality. 18.3 Use the periodic table to locate and identify elements from their valence configurations: (a) There are six valence electrons, so the element is in Column 6; 3d orbital is filling: Cr (b) There are full s and d blocks, so the element is in Column 12; 4d has just filled: Cd (c) There are 11 valence electrons, so the element is in Column 11; 3d orbital is full: Cu 18.5 Use the periodic table to locate a transition metal and determine the principal quantum numbers of its valence electrons. Then remove electrons to give the appropriate cation, remembering that when cations form, the valence s electrons are the first ones removed: (a) 3d5; (b) 5d6; (c) 3d8; and (d) 4d4. 18.7 The properties of transition metals vary regularly with their valence configurations, so predict relative properties based on locations in the d block: (a) Pd is in Column 10, Cd is in Column 12, both in the n = 4 row. Beyond the middle of the d block, melting point decreases with Z because electrons are placed in antibonding orbitals, so Pd melts at higher temperature. (b) Cu and Au are both in Column 11, but Au has a higher molar mass, so Au has higher density. (c) Cr is in Column 6, Co is in Column 9, both in the n = 3 row. IE1 increases with Z across a row because Zeff increases, so Co has the higher IE1. © John Wiley and Sons Canada, Ltd. 344 Chemistry, Student Solutions Manual Chapter 18 18.9 Use the charges of ligands and ions to determine the oxidation states of transition metals in coordination complexes. Because s electrons are always removed first, the count of d electrons is given by the number of valence electrons — the oxidation state: (a) Each Cl is –1, and NH3 is neutral, so Ru has oxidation state +2. Ru is in Column 8 (eight valence electrons), giving d6. (b) Each I is –1, and en is neutral, so Cr has oxidation state +3. Cr is in Column 6 (six valence electrons), giving d3. (c) Each Cl is –1, and trimethylphosphine is neutral, so Pd has oxidation state +2. Pd is in Column 10 (10 valence electrons), giving d8. (d) Each Cl is –1, and NH3 is neutral, so Ir has oxidation state +3. Ir is in Column 9 (nine valence electrons), giving d6. (e) CO is a neutral molecule, so Ni has oxidation state 0. Ni is in Column 10 (10 valence electrons), giving s2 d8. 18.11 Compounds that contain coordination complexes are named following the six rules stated in your textbook: name the cation first, name ligands in alphabetical order, name the metal, add “o” for anions, use Greek prefixes, add “-ate” for anionic complexes, give the oxidation number: (a) Hexaammineruthenium(II) chloride (b) trans-bis(Ethylenediamine)diiodochromium(III) iodide (c) cis-Dichlorobis(trimethylphosphine)palladium(II) (d) fac-Triamminetrichloroiridium(III) (e) Tetracarbonylnickel(0) 18.13 The structure of a metal complex usually is octahedral (six ligands), tetrahedral (four ligands), or square planar (four ligands): (a) Six NH3 in an octahedron around the central Ru: (b) Two I at opposite ends of one axis, two en in a square plane around the central Cr: © John Wiley and Sons Canada, Ltd. 345 Chemistry, Student Solutions Manual Chapter 18 (c) Square planar arrangement about Pd, with two Cl adjacent to each other: (d) Octahedral arrangement around Ir, with three Cl in a triangular face: (e) Tetrahedral arrangement of C≡O about a central Ni: 18.15 The name of a complex contains the information needed to determine its chemical formula. Determine the charge of the complex from charges on the ligands and the oxidation number: (a) cis-[Co(NH3)4ClNO2]+; (b) [PtNH3Cl3]–; (c) trans-[Cu(en)2(H2O)2]2+; (d) [FeCl4]–. 18.17 (a) cis-Tetraamminechloronitrocobalt(III) has six ligands and octahedral geometry. The cis indicates that the chlorine and nitro ligands will have a 90º angle between them: © John Wiley and Sons Canada, Ltd. 346 Chemistry, Student Solutions Manual Chapter 18 (b) In amminetrichloroplatinate(II), platinum has eight d electrons in its valence shell, resulting in square planar geometry: (c) In trans-diaquabis(ethylenediamine)copper(II), ethylenediamine is a bidentate ligand, giving a coordination number of 6 and octahedral geometry. The trans term means that the water ligands are opposite each other on the complex: (d) In tetrachloroferrate(III), iron(III) is a 3d metal with five valence electrons, resulting in tetrahedral geometry: 18.19 The crystal field diagram for weak and strong octahedral fields is always the same, with the populations changing depending on how many d electrons must be accommodated. The valence configuration provides information about the number of d electrons: (a) Ti2+ (Column 4, two electrons) is d2; (b) Cr3+ (Column 6, three electrons) is d3. For these two ions, the low-field and high-field configurations are the same: © John Wiley and Sons Canada, Ltd. 347 Chemistry, Student Solutions Manual Chapter 18 (c) Mn2+ (Column 7, two electrons) is d5. (d) Fe3+ (Column 8, three electrons) is d5. These two ions have identical diagrams, with high-spin when the splitting is small and low-spin when the splitting is large: 18.21 The magnetic properties of a complex are determined by its number of d electrons and the extent of its crystal field splitting energy: (a) Ir3+ (Column 9, three electrons) is d6, and NH3 generates relatively large splitting. All the electrons will be paired, making the complex diamagnetic. (b) Cr2+ (Column 6, two electrons) is d4, and water generates relatively small splitting. The complex is paramagnetic with four unpaired electrons. (c) Pt2+ (Column 10, two electrons) is d8, so regardless of the splitting energy, this square planar complex is paramagnetic with two unpaired electrons. (d) Pd has d10 configuration, so all orbitals are filled and this complex is diamagnetic. 18.23 The colours of transition metal complexes are generally determined by d–d transitions, but Zr4+ (Column 4, four electrons) has all its valence electrons removed, so there is no valence electron that can undergo a transition involving the absorption of visible light. 18.25 (a) The coordination number is the number of bonds formed between metal and ligands. The en ligand is bidentate, so three of them form six bonds, CN = 6. (b) The oxidation number is determined by examining the net charge and correcting for charges on all species other than the transition metal. Here, en is neutral. Each Cl– anion contributes –1 charge, so the oxidation number of Fe is +3. Fe is in Column 8 of the periodic table, so its valence configuration is 3d5. (c) Coordination number 6 means octahedral geometry. (d) With an odd number of electrons, it is not possible for all of them to be paired, so this complex is paramagnetic. (e) The complex is low-spin, so all but one of the electrons is paired. 18.27 Complexes are high-spin and paramagnetic when the crystal field splitting is small, and they become low-spin if the crystal field splitting is larger than the pairing © John Wiley and Sons Canada, Ltd. 348 Chemistry, Student Solutions Manual Chapter 18 energy. According to the spectrochemical series, CN– produces greater splitting than H2O. Replacing the H2O ligands with CN– ligands increases the splitting, to the point where it takes less energy to pair the electrons than to promote them to the less stable d orbitals. Cr is in Column 6 of the periodic table, so Cr2+ has d4 valence configuration. There are four unpaired electrons in the high-spin configuration but only two in the low-spin configuration: 18.29 The wavelength of light that is absorbed provides a measure of the crystal field splitting energy: (6.626 034 J s)(2.998 108 m/s)(6.022 10 23 mol1 ) 103 kJ 465 m 1 J 257 kJ/mol E hcN A The complex absorbs visible light around 465 nm, in the blue. The colour of the complex will be the complementary colour to blue. Consult Table 18-5 to determine that this colour is orange. 18.31 The structural difference between haemoglobin and myoglobin is that the former has four subunits, whereas the latter has just one. As a consequence, haemoglobin has much more complex cooperative chemical behaviour than myoglobin does. 18.33 An iron ion bonded to four sulphur atoms from cysteines is in a tetrahedral environment, so the splitting pattern is the 2–3 pattern characteristic of tetrahedral complexes. The iron cation loses one electron and is converted from Fe2+ to Fe3+: 18.35 Consult your textbook for the chemical reactions of various metallurgical processes: (a) CuFeS2 + 3 O2 → CuO + FeO + 2 SO2 (b) Si + O2 + CaO → CaSiO3 (c) TiCl4 + 4 Na → 4 NaCl + Ti © John Wiley and Sons Canada, Ltd. 349 Chemistry, Student Solutions Manual Chapter 18 18.37 This is a standard stoichiometry problem. Begin by analyzing the chemistry. The reactants are Cu2S and air and the given products are Cu metal and SO2 gas. Balanced reaction: Cu2S + O2 → 2 Cu + SO2. Start by computing the number of moles of Cu2S, and then use the appropriate mass–mole–mass and p–V–T calculations to determine the amounts of the products: mCu2S 103 kg 2.37% 6 5.60 10 kg 1.327 g 1 kg 100% nCu2S 4 1.327 106 g 8.34 3 mol nSO2 159.2 g/mol 2 mol Cu 6 mCu 8.34 103 mol (63.55 g/mol) 1.06 g 1 mol Cu S 2 1 bar ptotal 755 Torr 1.006 bar 750.06 Torr VSO2 nRT (8.34 103 mol)(0.08314 L bar mol1 K 1 )(273.15 23.5 K) 2.04 5 L p 1.006 bar 18.39 Standard free energy changes are calculated using standard free energies of formation, found in Appendix D. Greaction Gproducts Greactants ZnO (s) + C (s) → Zn (s) + CO (g) Greaction = [1 mol(–137.2 kJ/mol) + 0] – [1 mol(–320.5 kJ/mol) + 0] = 183.3 kJ ZnO (s) + CO (g) → Zn (s) + CO2 (g) Greaction = [1 mol(–394.4 kJ/mol) + 1 mol(0)] – [1 mol(–320.5 kJ/mol) + 1 mol(–137.2 kJ/mol)] = 63.3 kJ 18.41 The coinage metals are those that have been used since antiquity for coins: copper, silver, and gold. All are in Column 11 of the periodic table. They are characterized by high electrical conductivity, good ductility, and low chemical reactivity, in particular resistance to oxidation. Hence, they are used for money (a vanishing use in technologically advanced countries), for electrical wire, for jewellery, and for other decorative objects. See your text for special examples of uses for compounds of these elements. © John Wiley and Sons Canada, Ltd. 350 Chemistry, Student Solutions Manual Chapter 18 18.43 Titanium is used as an engineering metal because of its relatively low density, high bond strength, resistance to corrosion, and ability to withstand high temperatures, all of which make it a favoured structural material. 18.45 The number of possible isomers of a complex is determined by its geometry and the number of ligands of each type: 18.47 Bidentate ligands form complexes with two links. Each Fe ion forms an octahedral complex with three oxalate anions. When oxalate is applied to rust, it complexes and dissolves the Fe3+ cations. Here is a sketch showing the ligand–metal orientations: © John Wiley and Sons Canada, Ltd. 351 Chemistry, Student Solutions Manual Chapter 18 18.49 To determine electron configurations, start from the position of the element in the periodic table and remove s electrons preferentially. (a) Cr is in Column 6, configuration [Ar] 4s1 3d5; Cr2+ is [Ar] 3d4; Cr3+ is [Ar] 3d3 (b) V is in Column 5, configuration [Ar] 4s2 3d3; V2+ is [Ar] 3d3; V3+ is [Ar] 3d2; V4+ is [Ar] 3d1; V5+ is [Ar] (c) Ti is in Column 4, configuration [Ar] 4s2 3d2; Ti2+ is [Ar] 3d2; Ti4+ is [Ar] 18.51 Compounds that contain coordination complexes are named following the six rules stated in your textbook: name the cation first, name ligands in alphabetical order, name the metal, add “o” for anions, use Greek prefixes, add “-ate” for anionic complexes, list the oxidation number: (a) cis-tetraaquadichlorochromium(III) chloride; (b) bromopentacarbonylmanganese(I); (c) cisdiamminedichloroplatinum(II). 18.53 Superoxide dismutase catalyzes the conversion of superoxide into molecular oxygen and hydrogen peroxide. This reaction occurs at a metal site that contains one Zn2+ ion and one Cu2+ ion, linked by a histidine ligand that bonds to both metal ions. An O atom binds to the Zn2+ ion: 18.55 Cu(II) in water forms an aqua complex. Addition of fluoride produces the insoluble green salt, CuF2, whereas addition of chloride produces the bright green tetrachlorocopper(II) complex, [CuCl4]2–: Cu2+ (aq) + 2 F– (aq) → CuF2 (s) Cu2+ (aq) + 4 Cl– (aq) → [CuCl4]2– (aq) 18.57 Silver (Column 11 of the periodic table) has a filled set of d orbitals, making the neutral metal difficult to oxidize. This gives silver good resistance to corrosion and makes it suitable for jewellery. Vanadium (Column 5), with d3 configuration, is readily oxidized, so it corrodes rapidly and is unsuited to jewellery. © John Wiley and Sons Canada, Ltd. 352 Chemistry, Student Solutions Manual Chapter 18 18.59 The mer isomer of an octahedral complex has three like-ligands arranged in a meridian plane: 18.61 Silver tarnish is silver sulphide, Ag2S, formed by reaction with trace amounts of H2S in the atmosphere: 4 Ag (s) + 2 H2S (g) + O2 (g) → 2 Ag2S (s) + 2 H2O (g) 18.63 In these complexes, chromium is in the +3 oxidation state. From its location in Column 6 of the periodic table, we deduce that is has d3 valence electron configuration. The three electrons occupy the three different t2g orbitals, regardless of the magnitude of the splitting energy. 18.65 Orbital sketches show that the two orbitals experience quite different electron– electron repulsion in a square planar environment: 18.67 (a) The charge on a complex is the charge on the transition metal cation, modified by any charges on the ligands. Ammonia and water are neutral, so when L is either of these, n+ = 3+; but chloride is –1, so for L = Cl–, n+ = 2+. (b) The colour of each complex is the colour that is complementary to the one corresponding to the wavelength of light that it absorbs. Consult Table 18-5 of your textbook to determine these: L = Cl–, red to purple; L = H2O, orange; and L = NH3, yellow-orange. (c) The wavelength of light that is absorbed provides a measure of the crystal field splitting energy: E © John Wiley and Sons Canada, Ltd. hcN A 353 Chemistry, Student Solutions Manual Chapter 18 L = Cl–: (6.626 1034 J s)(2.998 108 m/s)(6.022 10 23 mol1 ) 103 kJ 515 9 m 1 J 232 kJ/mol L = H2O: (6.626 1034 J s)(2.998 108 m/s)(6.022 10 23 mol1 ) 103 kJ 9 m 1 J 249 kJ/mol L = NH3: (6.626 1034 J s)(2.998 108 m/s)(6.022 10 23 mol1 ) 103 kJ 9 m 1 J 257 kJ/mol The trend matches exactly the positions of these three ligands in the spectrochemical series: Cl– < H2O < NH3. 18.69 Tetracarbonylnickel(0), [Ni(CO)4], has tetrahedral geometry since Ni is a first-row transition element. Because of the strong field ligand, CO, ∆ is large, which will result in the compound absorbing UV light and appearing colourless. Tetracyanozinc(II), [Zn(CN)4]2–, has tetrahedral geometry (Zn is a first-row transition metal) where Zn has a +2 charge (d10). This complex is colourless because there are no possible d–d transitions. 18.71 Brass is an alloy of zinc and copper, and superoxide dismutase contains zinc and copper ions in its reaction centre. 18.73 Four-coordinate complexes may be either tetrahedral or square planar. The splitting patterns for these two geometries show that the d8 configuration can have zero spin in the square planar case but not in the tetrahedral case. Thus, the magnetic behaviour indicates that [Ni(CN)4]2– is square planar and [NiCl4]2– is tetrahedral: 18.75 When ferritin is neither empty nor filled to capacity, the protein has the capacity to provide iron as needed for haemoglobin synthesis, or to store iron if an excess is absorbed by the body. © John Wiley and Sons Canada, Ltd. 354 Chemistry, Student Solutions Manual Chapter 18 18.77 Visible spectroscopy is useful when a compound has an energy gap between the highest occupied and lowest unoccupied orbital that matches the energy of visible light. Because Zn2+ has d10 configuration, its d orbitals are completely filled, and the lowest unoccupied orbital is quite high in energy. In contrast, Co2+ has d7 configuration, giving this cation unfilled d orbitals. Consequently, metalloproteins that contain Co2+ absorb visible light, making it possible to study them with visible spectroscopy. 18.79 Use standard reduction potentials from Appendix F to determine which metals can be displaced by Zn. Any metal with a less negative standard potential can be displaced: Zn 2 2 e € Zn E 0.7618 V The following are a few examples of metals that can be displaced by Zn: Co 2 2 e € Co E 0.28 V Cu 2 2 e € Cu E 0.3419 V Fe2 2 e € Fe E 0.447 V Ni 2 2 e € Ni E 0.257 V When combined with the Zn half-reaction, the overall cell voltage is positive and the process is spontaneous. For example: Zn Co2 € Co Zn 2 E (0.28 V) (0.7618 V) 0.48 V 18.81 The colours of substances depend on what wavelengths of light they absorb. According to Table 18-5, blue colour results when a substance absorbs 610 nm light (196 kJ/mol), and red colour results when a substance absorbs 500 nm light oxyhaemoglobin, suggesting that O2 causes larger energy-level splitting than H2O. The hypothesis could be tested by replacing the H2O ligand in deoxyhaemoglobin by a ligand that causes smaller splitting, such as Cl–, which should push the colour toward the green. In addition, CO, which generates the largest energy-level splitting, should push the colour to red-orange or orange. 18.83 The densities of transition metals increase with Z across each row of the d block, up to about the middle of the block, as illustrated by Figure 18-2b in your textbook. Thus, the upper-left-most transition metal in the periodic table, scandium, would be the best replacement for Al. The density of scandium is only about 10% greater than that of aluminum. The figure shows that osmium, in the middle of the 5d block, is the densest transition metal. Its density is actually twice that of lead. © John Wiley and Sons Canada, Ltd. 355 Chemistry, Student Solutions Manual Chapter 18 18.85 The chloride ions that form silver chloride precipitate are not bound to the complex and should not be included in the chemical formula of the complex ion: PtCl4•2NH3 is [PtCl4(NH3)2], diamminetetrachloroplatinum(IV), cis and trans: PtCl4•3NH3 is [PtCl3(NH3)3]Cl, the complex ion being triamminetrichloroplatinum(IV), mer and fac: PtCl4•4NH3 is [PtCl2(NH3)4]Cl2, the complex ion being tetraamminedichloroplatinum(IV), cis and trans: PtCl4•5NH3 is [PtCl(NH3)5]Cl3, the complex ion being pentaamminechloroplatinum(IV), and PtCl4•6NH3 is [Pt(NH3)6]Cl4, the complex ion being hexaammineplatinum(IV): © John Wiley and Sons Canada, Ltd. 356