Metais de Transição e Compostos de Coordenação

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Chapter 25 Transition Metals and Coordination Compounds 517 (a) Ni; Ni²⁺ The noble gas that precedes Ni is Ar. Ni is in the fourth period; so the orbitals we use are 4s and 3d, and Ni has 10 more electrons than Ar does. Ni Ni will lose electrons from the 4s and then from the 3d. Ni²⁺ (b) Mn; The noble gas that precedes Mn is Ar. Mn is in the fourth period; so the orbitals we use are 4s and 3d, and Mn has seven more electrons than Ar does. Mn Mn will lose electrons from the 4s and then from the 3d. (c) The noble gas that precedes Y is Kr. Y is in the fifth period; the orbitals we use are 5s and 4d, and Y has three more electrons than Kr does. Y Y will lose electrons from the 5s and then from the 4d. (d) Ta; Ta²⁺ The noble gas that precedes Ta is Xe. Ta is in the sixth period; so the orbitals we use are 6s, 5d, and 4f, and Ta has 19 more electrons than Xe does. Ta Ta will lose electrons from the 6s and then from the 5d. Ta²⁺ 25.19 (a) V Highest oxidation state = +5. V = Because V is to the left of Mn, it can lose all of the 4s and 3d electrons; so the highest oxidation state is +5 (b) Re Highest oxidation state = +7. Re = Re can lose all of the 6s and 5d electrons, the highest oxidation state is +7. (c) Pd Highest oxidation state = +4. Pd = [Kr]4d¹⁰. Metal ions with a electron configuration exhibit a square planar geometry. Pd can lose 4 electrons from the 5s and 4d orbitals, so the highest oxidation state is +4. Coordination Compounds 25.21 (a) [Cr(H₂O)₆]³ H₂O is neutral, so Cr has an oxidation state of +3. Six H₂O molecules are attached to each Cr, so the coordination number is 6. (b) NH₃ is neutral, and has charge of The sum of the oxidation state of Co and the charge of chloride ion = -1.x + (3(-1)) = = +2; therefore, the oxidation state of Co is +2. The three NH₃ molecules and the three ions are bound directly to the Co atom; therefore, the coordination number is 6. (c) [Cu(CN)₄]²⁻ CN has a charge of 1-. The sum of the oxidation state of Cu and the charge of the cyanide ion = (4(-1)) = 2-, x = +2; therefore, the oxidation state of Cu is +2. The four cyanide ions are directly bound to the Cu atom; therefore, the coordination number is 4. (d) [Ag(NH₃)₂] is neutral, so Ag has an oxidation number of +1. Two NH₃ molecules are attached to each Ag atom, so the coordination number is 2. 25.23 (a) is hexaaquachromium(III) ion. is a complex cation. Name the ligand: is aqua. Name the metal ion: Cr³⁺ is chromium(III). Name the complex ion by adding the prefixes to indicate the number of each ligand, followed by the name of each ligand and the name of the metal ion: hexaaquachromium(III) ion. (b) [Cu(CN)₄]²⁻ is tetracyanocuprate(II) ion. [Cu(CN)₄] is a complex anion. Name the ligand: CN⁻ is cyano. Copyright © 2017 Pearson Education, Inc.