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Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 23 The Transition Elements and Their Coordination Compounds 23-1 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Transition Elements and Their Coordination Compounds 23.1 Properties of the Transition Elements 23.2 The Inner Transition Elements 23.3 Highlights of Selected Transition Metals 23.4 Coordination Compounds 23.5 Theoretical Basis for the Bonding and Properties of Complexes 23-2 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.1 23-3 The transition elements (d block) and inner transition elements (f block) in the periodic table. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.2 23-4 The Period 4 transition metals. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 23-5 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 23.1 PROBLEM: Writing Electron Configurations of Transition Metal Atoms and Ions Write condensed electron configurations for the following: (a) Zr; (b) V3+; (c) Mo3+. (Assume that elements in higher periods behave like those in Period 4.) PLAN: The general configuration is [noble gas] ns2(n-1)dx. Recall that in ions the ns electrons are lost first. SOLUTION: (a) Zr is the second element in the 4d series: [Kr]5s24d2. (b) V is the thired element in the 3d series: [Ar]4s23d3. In forming V3+, three electrons are lost (two 4s and one 3d), so V3+ is a d2 ion: [Ar]3d10. (c) Mo lies below Cr in Group 6B(6), so we expect the same except in configuration as for Cr. Thus, Mo is [Kr]5s14d5. In forming the ion, Mo loses the one 5s and two of the 4d electrons to become a 4d3 ion: [Kr]4d3. 23-6 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.3 23-7 Horizontal trends in key atomic properties of the Period 4 elements. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.4 Vertical trends in key properties within the transition elements. 23-8 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.5 Aqueous oxoanions of transition elements. One of the most characteristic chemical properties of these elements is the occurrence of multiple oxidation states. Mn(II) Mn(VI) Mn(VII) Mn(VII) Cr(VI) V(V) 23-9 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 23-10 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 23-11 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.6 Colors of representative compounds of the Period 4 transition metals. sodium chromate titanium oxide scandium oxide vanadyl sulfate dihydrate 23-12 nickel(II) nitrate hexahydrate potassium ferricyanide manganese(II) chloride tetrahydrate cobalt(II) chloride hexahydrate zinc sulfate heptahydrate copper(II) sulfate pentahydrate Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 23-13 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 23.2 PROBLEM: PLAN: Finding the Number of Unpaired Electrons The alloy SmCo5 forms a permanent magent because both samarium and cobalt have unpaired electrons. How many unpaired electrons are in the Sm atom (Z = 62)? Write the condensed configuration of Sm and, using Hund’s rule and the aufbau principle, place electrons into a partial orbital diagram. SOLUTION: Sm is the eighth element after Xe. Two electrons go into the 6s sublevel and the remaining six electrons into the 4f (which fills before the 5d). Sm is [Xe]6s24f6 6s 4f There are 6 unpaired e- in Sm. 23-14 5d Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.7 23-15 The bright colors of chromium (VI) compounds. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Orbital Occupancy *Most common states in bold face. 23-16 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.8 23-17 Steps in producing a black-and-white negative. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.9 Components of a coordination compound. models 23-18 wedge diagrams chemical formulas Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Structures of Complex Ions: Coordination Numbers, Geometries, and Ligands •Coordination Number - the number of ligand atoms that are bonded directly to the central metal ion. The coordination number is specific for a given metal ion in a particular oxidation state and compound. •Geometry - the geometry (shape) of a complex ion depends on the coordination number and nature of the metal ion. •Donor atoms per ligand - molecules and/or anions with one or more donor atoms that each donate a lone pair of electrons to the metal ion to form a covalent bond. 23-19 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 23-20 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 23-21 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 23-22 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Formulas and Names of Coordination Compounds Rules for writing formulas: 1. The cation is written before the anion. 2. The charge of the cation(s) is balanced by the charge of the anion(s). 3. In the complex ion, neutral ligands are written before anionic ligands, and the formula for the whole ion is placed in brackets. 23-23 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Formulas and Names of Coordination Compounds Rules for naming complexes: continued 1. The cation is named before the anion. 2. Within the complex ion, the ligands are named, in alphabetical order, before the metal ion. 3. Neutral ligands generally have the molecule name, but there are a few exceptions. Anionic ligands drop the -ide and add -o after the root name. 4. A numerical prefix indicates the number of ligands of a particular type. 5. The oxidation state of the central metal ion is given by a Roman numeral (in parentheses). 6. If the complex ion is an anion we drop the ending of the metal name and add -ate. 23-24 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 23.3 Writing Names and Formulas of Coordination Compounds PROBLEM: (a) What is the systematic name of Na3[AlF6]? (b) What is the systematic name of [Co(en)2Cl2]NO3? (c) What is the formula of tetraaminebromochloroplatinum(IV) chloride? (d) What is the formula of hexaaminecobalt(III) tetrachloroferrate(III)? PLAN: Use the rules presented - SOLUTION: and . (a) The complex ion is 3-. [AlF6] Six (hexa-) fluorines (fluoro-) are the ligands - hexafluoro Aluminum is the central metal atom - aluminate Aluminum has only the +3 ion so we don’t need Roman numerals. sodium hexafluoroaluminate 23-25 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 23.3 Writing Names and Formulas of Coordination Compounds continued (b) There are two ligands, chlorine and ethylenediamine dichloro, bis(ethylenediamine) The complex is the cation and we have to use Roman numerals for the cobalt oxidation state since it has more than one - (III) The anion, nitrate, is named last. dichlorobis(ethylenediamine)cobalt(III) nitrate (c) Cl Pt4+ Cltetraaminebromochloroplatinum(IV) chloride 4 NH3 Br- [Pt(NH3)4BrCl]Cl2 (d) 6 NH3 Co3+ 4 Cl- Fe3+ hexaaminecobalt(III) tetrachloro-ferrate(III) [Co(NH3)6][Cl4Fe]3 23-26 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 23-27 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.10 Important types of isomerism in coordination compounds. ISOMERS Same chemical formula, but different properties Constitutional (structural) isomers Stereoisomers Atoms connected differently Different spatial arrangement Coordination Coordination isomers isomers Linkage Linkage isomers isomers Ligand Ligand and and counter-ion counter-ion exchange exchange Different Different donor donor atom atom 23-28 Geometric Optical Optical isomers isomers Geometric (cis(cistrans) (enantiomers) (enantiomers) trans) isomers isomers (diastereomers) (diastereomers) Nonsuperimposable Nonsuperimposable Different mirror Different mirror images images arrangement arrangement around around metal metal ion ion Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Linkage isomers 23-29 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.11 23-30 Geometric (cis-trans) isomerism. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.12 Optical isomerism in an octahedral complex ion. 23-31 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 23.4 Determining the Type of Stereoisomerism PROBLEM: Draw all stereoisomers for each of the following and state the type of isomerism: (b) [Cr(en)3]3+ (en = H2NCH2CH2NH2) (a) [Pt(NH3)2Br2] PLAN: Determine the geometry around each metal ion and the nature of the ligands. Place the ligands in as many different positions as possible. Look for cis-trans and optical isomers. SOLUTION: (a) Pt(II) forms a square planar complex and there are two pair of monodentate ligands - NH3 and Br. Br NH3 H3N Pt H3N 23-32 Pt Br trans Br H3 N Br cis These are geometric isomers; they are not optical isomers since they are superimposable on their mirror images. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 23.4 Determining the Type of Stereoisomerism (b) Ethylenediamine is a bidentate ligand. Cr3+ is hexacoordinated and will form an octahedral geometry. continued Since all of the ligands are identical, there will be no geometric isomerism possible. 3+ 3+ N N N N N N Cr Cr N The mirror images are nonsuperimposable and are therefore optical isomers. N N N N N rotate 3+ N N N Cr N N N 23-33 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.13 Hybrid orbitals and bonding in the octahedral [Cr(NH3)6]3+ ion. 23-34 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.14 Hybrid orbitals and bonding in the square planar [Ni(CN)4]2- ion. 23-35 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.15 Hybrid orbitals and bonding in the tetrahedral [Zn(OH)4]2- ion. 23-36 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.16 23-37 An artist’s wheel. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 23-38 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.17 23-39 The five d-orbitals in an octahedral field of ligands. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.18 Splitting of d-orbital energies by an octahedral field of ligands. Δ is the splitting energy 23-40 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.19 23-41 The effect of ligand on splitting energy. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.20 23-42 The color of [Ti(H2O)6]3+. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.21 Effects of the metal oxidation state and of ligand identity on color. [V(H2O)6]3+ [V(H2O)6]2+ [Cr(NH3)6]3+ 23-43 [Cr(NH3)5Cl ]2+ Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.22 The spectrochemical series. •For a given ligand, the color depends on the oxidation state of the metal ion. •For a given metal ion, the color depends on the ligand. I- < Cl- < F- < OH- < H2O < SCN- < NH3 < en < NO2- < CN- < CO WEAKER FIELD 23-44 STRONGER FIELD SMALLER Δ LARGER Δ LONGER λ SHORTER λ Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 23.5 Ranking Crystal Field Splitting Energies for Complex Ions of a Given Metal PROBLEM: Rank the ions [Ti(H2O)6]3+, [Ti(NH3)6]3+, and [Ti(CN)6]3- in terms of the relative value of Δ and of the energy of visible light absorbed. PLAN: The oxidation state of Ti is 3+ in all of the complexes so we are looking at the crystal field strength of the ligands. The stronger the ligand the greater the splitting and the higher the energy of the light absorbed. SOLUTION: The field strength according to is CN- > NH3 > H2O. So the relative values of Δ and energy of light absorbed will be [Ti(CN)6]3- > [Ti(NH3)6]3+ > [Ti(H2O)6]3+ 23-45 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.23 23-46 High-spin and low-spin complex ions of Mn2+. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.24 Orbital occupancy for high- and low-spin complexes of d4 through d7 metal ions. high spin: weak-field ligand 23-47 low spin: strong-field ligand high spin: weak-field ligand low spin: strong-field ligand Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Sample Problem 23.6 Identifying Complex Ions as High Spin or Low Spin PROBLEM: Iron (II) forms an essential complex in hemoglobin. For each of the two octahedral complex ions [Fe(H2O)6]2+ and [Fe(CN)6]4-, draw an orbital splitting diagram, predict the number of unpaired electrons, and identify the ion as low or high spin. PLAN: The electron configuration of Fe2+ gives us information that the iron has 6d electrons. The two ligands have field strengths shown in . potential energy Draw the orbital box diagrams, splitting the d orbitals into eg and t2g. Add the electrons noting that a weak-field ligand gives the maximum number of unpaired electrons and a high-spin complex and vice-versa. [Fe(CN)6]4[Fe(H2O)6]2+ SOLUTION: 4 unpaired e-eg (high spin) eg no unpaired e-(low spin) t2g t2g 23-48 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 23.25 Splitting of d-orbital energies by a tetrahedral field and a square planar field of ligands. tetrahedral square planar 23-49 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B23.1 23-50 Hemoglobin and the octahedral complex in heme. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 23-51 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure B23.2 23-52 The tetrahedral Zn2+ complex in carbonic anhydrase.