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Chapter 23

Chapter 23. The Transition Elements and Their Coordination Compounds. 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.

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Chapter 23

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  1. Chapter 23 The Transition Elements and Their Coordination Compounds

  2. 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

  3. The transition elements (d block) and inner transition elements (f block) in the periodic table. Figure 23.1

  4. The Period 4 transition metals. Figure 23.2

  5. PROBLEM: 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. Sample Problem 23.1 Writing Electron Configurations of Transition Metal Atoms and Ions 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.

  6. Horizontal trends in key atomic properties of the Period 4 elements. Figure 23.3

  7. Figure 23.4 Vertical trends in key properties within the transition elements.

  8. Mn(II) Mn(VI) Mn(VII) Mn(VII) Cr(VI) V(V) Aqueous oxoanions of transition elements. Figure 23.5 One of the most characteristic chemical properties of these elements is the occurrence of multiple oxidation states.

  9. Figure 23.6 Colors of representative compounds of the Period 4 transition metals. sodium chromate nickel(II) nitrate hexahydrate potassium ferricyanide zinc sulfate heptahydrate titanium oxide scandium oxide manganese(II) chloride tetrahydrate copper(II) sulfate pentahydrate vanadyl sulfate dihydrate cobalt(II) chloride hexahydrate

  10. PROBLEM: 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)? PLAN: Write the condensed configuration of Sm and, using Hund’s rule and the aufbau principle, place electrons into a partial orbital diagram. 6s 4f 5d Sample Problem 23.2 Finding the Number of Unpaired Electrons 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 There are 6 unpaired e- in Sm.

  11. The bright colors of chromium (VI) compounds. Figure 23.7

  12. Orbital Occupancy *Most common states in bold face.

  13. Steps in producing a black-and-white negative. Figure 23.8

  14. models wedge diagrams chemical formulas Components of a coordination compound. Figure 23.9

  15. 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.

  16. 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.

  17. Formulas and Names of Coordination Compounds continued Rules for naming complexes: 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.

  18. 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) tetrachloro-ferrate(III)? PLAN: Use the rules presented - and . SOLUTION: (a) The complex ion is [AlF6]3-. 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

  19. (c) tetraaminebromochloroplatinum(IV) chloride (d) hexaaminecobalt(III) tetrachloro-ferrate(III) 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 4 NH3 Br- Cl- Pt4+ Cl- [Pt(NH3)4BrCl]Cl2 6 NH3 Co3+ 4 Cl- Fe3+ [Co(NH3)6][Cl4Fe]3

  20. ISOMERS Same chemical formula, but different properties Constitutional (structural) isomers Stereoisomers Figure 23.10 Important types of isomerism in coordination compounds. Atoms connected differently Different spatial arrangement Coordination isomers Ligand and counter-ion exchange Linkage isomers Different donor atom Geometric (cis-trans) isomers (diastereomers) Different arrangement around metal ion Optical isomers (enantiomers) Nonsuperimposable mirror images

  21. Linkage isomers

  22. Geometric (cis-trans) isomerism. Figure 23.11

  23. Figure 23.12 Optical isomerism in an octahedral complex ion.

  24. PROBLEM: Draw all stereoisomers for each of the following and state the type of isomerism: Sample Problem 23.4 Determining the Type of Stereoisomerism (a) [Pt(NH3)2Br2] (b) [Cr(en)3]3+ (en = H2NCH2CH2NH2) 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. These are geometric isomers; they are not optical isomers since they are superimposable on their mirror images. trans cis

  25. Sample Problem 23.4 Determining the Type of Stereoisomerism continued (b) Ethylenediamine is a bidentate ligand. Cr3+ is hexacoordinated and will form an octahedral geometry. Since all of the ligands are identical, there will be no geometric isomerism possible. The mirror images are nonsuperimposable and are therefore optical isomers.

  26. Figure 23.13 Hybrid orbitals and bonding in the octahedral [Cr(NH3)6]3+ ion.

  27. Figure 23.14 Hybrid orbitals and bonding in the square planar [Ni(CN)4]2- ion.

  28. Figure 23.15 Hybrid orbitals and bonding in the tetrahedral [Zn(OH)4]2- ion.

  29. An artist’s wheel. Figure 23.16

  30. The five d-orbitals in an octahedral field of ligands. Figure 23.17

  31. Figure 23.18 Splitting of d-orbital energies by an octahedral field of ligands. D is the splitting energy

  32. The effect of ligand on splitting energy. Figure 23.19

  33. The color of [Ti(H2O)6]3+. Figure 23.20

  34. 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+ [Cr(NH3)5Cl]2+

  35. I- < Cl- < F- < OH- < H2O < SCN- < NH3 < en < NO2- < CN- < CO STRONGER FIELD WEAKER FIELD LARGER D SMALLER D LONGER  SHORTER  The spectrochemical series. Figure 23.22 • 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.

  36. PROBLEM: Rank the ions [Ti(H2O)6]3+, [Ti(NH3)6]3+,and [Ti(CN)6]3- in terms of the relative value of D 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. Sample Problem 23.5 Ranking Crystal Field Splitting Energies for Complex Ions of a Given Metal SOLUTION: The field strength according to is CN- > NH3 > H2O. So the relative values of D and energy of light absorbed will be [Ti(CN)6]3- > [Ti(NH3)6]3+ > [Ti(H2O)6]3+

  37. Figure 23.23 High-spin and low-spin complex ions of Mn2+.

  38. high spin: weak-field ligand low spin: strong-field ligand high spin: weak-field ligand low spin: strong-field ligand Orbital occupancy for high- and low-spin complexes of d4 through d7 metal ions. Figure 23.24

  39. 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. eg eg potential energy t2g t2g Sample Problem 23.6 Identifying Complex Ions as High Spin or Low Spin PLAN: The electron configuration of Fe2+ gives us information that the iron has 6d electrons. The two ligands have field strengths shown in . 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-- (high spin) no unpaired e-- (low spin)

  40. tetrahedral square planar Figure 23.25 Splitting of d-orbital energies by a tetrahedral field and a square planar field of ligands.

  41. Hemoglobin and the octahedral complex in heme. Figure B23.1

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