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Chapter 17

Chapter 17. The Colorful Chemistry of Transition Metals. Lewis Acids and Bases. A Lewis base is a substance that donates a pair of electrons in a chemical reaction. A Lewis acid is a substance that accepts a pair of electrons in a chemical reaction. Figure 17.1. Another Lewis Example.

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Chapter 17

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  1. Chapter 17 The Colorful Chemistry of Transition Metals

  2. Lewis Acids and Bases • A Lewis base is a substance that donates a pair of electrons in a chemical reaction. • A Lewis acid is a substance that accepts a pair of electrons in a chemical reaction. Figure 17.1

  3. Another Lewis Example

  4. Complex Ions • A complex ion is an ionic species consisting of a metal ion bonded to one or more Lewis bases. • A coordinate bond forms when one anion or molecule donates a pair of electrons to another ion or molecule to form a covalent bond. • A ligand is a Lewis base bonded to the central metal ion of a complex ion. • The inner coordination sphere of a metal consists of the ligands that are bound directly to the metal via coordinate covalent bonds.

  5. Complex Ions Dissolved metal ions are Lewis Acids, and form aqueous complexes with Lewis Bases

  6. More Metal Complexes

  7. Terms • Counter ions are the ions that balance the electrical charges of complex ions in coordination compounds (e.g. in basic solution OH- ions could act as counter ions Zn(NH3)4(OH)2 • Coordination compounds are made up of one or more complex ions; typically no more than 6 ligands surround the metal. • The coordination number (CN) of a metal ion identifies the number of electron pairs surrounding it in a complex.

  8. Which ions are the counter ions in the following: Na2[Zn(CN)4] [Co(NH3)4Cl2]NO3

  9. Bonding in Zn(NH3)42+ Ion

  10. Are any of these NOT Lewis bases? Would you expect H2O or NH3 to be the stronger Lewis Base?

  11. Complex-Ion Equilibria Cu2+(aq) + 4 NH3(aq) <==> Cu(NH3)42+(aq) Figure 17.5

  12. 17.26. When a strong base is added to a solution of CuSO4, which is pale blue, a precipitate forms and the solution above the precipitate is colorless. When ammonia is added, the precipitate dissolves and the solution turns a deep navy blue. Use appropriate chemical equations to explain why the observed changes occurred.

  13. Formation Reactions The equilibrium constants associated with complexation are called formation constants Kf = [Cu(NH3)42+]/[Cu2+][NH3]4 = 5.0E+13

  14. Metal ions as Lewis Acid also promote hydrolysis of water, and the formation of H3O+ Metal cations (Fe3+, Cr3+ and Al3+)with large positive charges are more likely to cause hydrolysis.

  15. 17.45. Sketch the titration curve (pH versus volume of 0.50 M NaOH) for a 25 mL sample of 0.5 M FeCl3 (Ka=3E-3)

  16. 17.5. Solubility of Ionic Compounds Minerals in contact with ground water will dissolve to some extent. CaCO3(s) ↔ Ca2+(aq) + CO32-(aq) Write the equilibrium expression for this dissolution.

  17. Solubility Equilibria CaCO3(s) ↔ Ca2+(aq) + CO32-(aq) K = [Ca2+(aq)][CO32-(aq)] = Ksp The equilibrium expression is called the solubility product (sp), because it involves only products of the concentrations of the dissolved species and NOT the solid. If Ksp is known, the solubility (at equilibrium) of the solid can be calculated

  18. Solubilities of Ionic Compounds The solubility-product constant (Ksp) is an equilibrium constant that describes the formation of a saturated solution of a slightly soluble salt. Mg(OH)2(s) <==> Mg2+(aq) + 2 OH-(aq) Ksp = [Mg2+][OH-]2 = 5.6 x 10-12

  19. Kspvalues of some common salts HgS(s) Ksp = [Hg2+][S2-] = 4.0 x 10-53 Fe(OH)3(s) Ksp = [Fe3+][OH-]3 = 2.8 x 10-39 AgI(s) Ksp = [Ag1+][I1-] = 8.5 x 10-17 CaCO3(s) Ksp = [Ca2+][CO32-] = 9.8 x 10-9 CaSO4(s) Ksp = [Ca2+][SO42-] = 4.9 x 10-5 Ag2SO3(s) Ksp = [Ag1+]2[SO32-] = 1.2 x 10-5 NaCl(s) Ksp = [Na1+][Cl1-] = 6.2

  20. Solubility Problem 17.58. What are the equilibrium concentrations of Pb2+ and F– in a saturated solution of lead fluoride if the Ksp value of PbF2 is 3.2 × 10–8?

  21. Solubility Problem 2.75 grams of BaF2 (FW=175.3) is placed in enough water to make 1.00 L at 25°C. After equilibrium has been established…the F- concentration equal 0.0150 M, what is the Ksp for BaF2.

  22. Solubility Problem 50 mg of PbSO4 (FW=303.3) is placed in 250 mL of pure water; What percentage of the solid dissolves? Ksp(PbSO4)=1.8E-8 (Table A5.4)

  23. Solubility Problem Calculate the pH of a saturated solution of zinc hydroxide, Ksp = 4.0E-17

  24. Solubility Problem 120 Calculate the solubility of silver chloride in seawater with a chloride concentration of 0.547 M. Ksp(AgCl) = 1.8E-10

  25. Terms • A monodentate ligand is a species that forms only a single coordinate bond to a metal ion in a complex. • A polydentate ligand is a species that can form more than one coordinate bond per molecule. • Chelation is the interaction of a metal with a polydentate ligand (chelating agent). (e.g. Google “chelation therapy”)

  26. Chelation-Examples (a) bidentate chelation Ni2+(aq) by ethylenediamine (b) 3 ethylenediamime molecules bind to Ni2+ tridentate chelation by diethylenetriamine.

  27. A Hexadentate Ligand Ethylenediamine- tetraacetic acid (EDTA) forms 6 stable, coordinate bonds with many metals

  28. Ligands and Complex Colors NiCl2 [Ni(NH3)]Cl2

  29. Complex Ions of Nickel • Ni(H2O)62+ is equivalent to Ni2+(aq) and is green in solution. Ni2+(aq)+ 6NH3(aq) <==> Ni(NH3)62+(aq) Kf = 5 x 108 Ni2+(aq)+ 3en <==> Ni(en)32+(aq) Kf = 1.1 x 1018 en=ethylenediamine • The chelate effect is the greater affinity of metal ions for polydentate ligands compared to the corresponding monodentate ligands.

  30. Crystal Field Theory…why transition metal complexes are colored • Crystal field splitting is the separation of a set of d-orbitals into subsets with different energies as a result of interactions between electrons in those orbitals and pairs of electrons in ligands surrounding the orbitals. • Crystal field splitting energy () is the difference in energy between subsets of d-orbitals split by interactions in a crystal field.

  31. d-orbitals in an octahedral field

  32. Light (photons) Can Promote Electrons in a Complex Ion’s Orbitals

  33. Fig. 17.11: Crystal field spitting of d-orbital energies that results from the interaction with p-orbitals in an octahedral geometry. The difference in energy is called the crystal field splitting energy (D). An example of this is the substitution of Cr3+ for Al3+ in the octahedral holes of the silicate structure of beryl. The Be2+ sits in tetrahedral holes. Be3CrxAl2-xSi6O18 Review: See chapter 7 for shapes of atomic orbitals.

  34. The Color of Compounds • Our eyes perceive the transmitted colors of complex ions. • [Cu(NH3)4]2+ absorbs much of the yellow, orange, and red wavelengths of light, so we see the complex as being navy blue.

  35. Color Wheel

  36. Ni(H2O)62+ --> Ni(NH3)62+ --> Ni(en)32+Green Blue Violet • Ni(H2O)62+ is green because it absorbs the color red on the opposite side of the color wheel. • Ni(NH3)62+ is blue because it absorbs the color opposite blue, which is orange. • Ni(en)32+ is violet because it absorbs colors centered on the complement of violet, which is yellow (en is the ethlyenediamine ligand).

  37. Crystal Field Splitting Energy •  must change for the three nickel compounds. • H2O as a ligand must yield the smallest  because the red light, which is lowest in energy, is absorbed by the nickel complex. • NH3 must cause a little larger  than water because a higher energy of orange light is absorbed. • The ligand en must cause the largest , because its complex absorbs yellow light, which is highest in energy of the three examined.

  38. The Spectrochemical Series A list of ligands ordered by their abilities to increase , the crystal field splitting energy of the d-orbitals.

  39. Square Planar Crystal Field Splitting

  40. How many d-electrons does Cu2+ have? Fig. 17.15: The mineral turquoise, CuAl6(PO4)4(OH)84H2O, has Cu2+ ions occupying irregular shaped octahedral holes. Consequently, the crystal field spitting produces four energy levels among the d-orbitals.

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