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Problem Set 10 Solutions 1. For each of the following, provide: i) The correct name of the entire compound ii) The oxidation state of the transition metal iii) The number of d electrons on the transition metal iv) The crystal-field energy level diagram for the d orbitals v) Whether the complex is high-spin or low-spin vi) The total number of unpaired electrons a) (NH4)[Cr(H2O)6](SO4)2 (3 pts) Ammonium hexaaquachromium(III) sulfate Oxidation state of Cr is +3 Cr3+ will have 3 d-electrons: With three electrons, high-spin and low-spin is the same. All 3 electrons are unpaired. b) K4[Os(CN)6] (3 pts) Potassium hexacyanoosmate(II) Oxidation state of Os is +2 Os2+ will have 6 d-electrons: With strong-field CN– ligands, expect this to be low-spin. There are no unpaired electrons. c) [Ni(NH3)4(H2O)2](NO3)2 (3 pts) Tetraamminediaquanickel(II) nitrate Oxidation state of Ni is +2 Ni2+ will have 8 d-electrons: With 8 electrons, high-spin and low-spin is the same. There are two unpaired electrons. d) Mo(CO)6 (3 pts) Hexacarbonylmolybdenum(0) Oxidation state of molybdenum is zero Mo(0) will have 6 d-electrons: With the strong-field ligand CO, we expect to have a low-spin configuration. There are no unpaired electrons. 2. Ruthenium can form complexes with chloride and water ligands with varying stoichiometries. Three such complexes are: [RuCl2(H2O)4]+ [RuCl3(H2O)3] [RuCl4(H2O)2]– a) What is the oxidation state of ruthenium in each of these complexes? (All of the complexes have the same oxidation state.) The oxidation state is +3. (1 pts) b) All of these complexes are low-spin. How many unpaired electrons are found in each of these complexes? d5 low spin, so only 1 unpaired electron (1 pts) c) These three complexes appear colored red, yellow, and green (but not necessarily in that order). Match each complex with its apparent color. First of all, note that Cl– is a weaker-field ligand then H2O. Then the rest is easy: (3 pts) d) [RuCl2(H2O)4]+ is yellow [RuCl3(H2O)3] is red (it absorbs high-energy violet) (it absorbs medium-energy green) [RuCl4(H2O)2]– is green (it absorbs low-energy red) Could any of these compounds exhibit geometrical isomerism? Show all the possible geometric isomers for each of these compounds. All could exhibit geometrical isomerism: (3 pts) [RuCl2(H2O)4]+ Cl Cl H2 O H2 O [RuCl4(H2O)2]– [RuCl3(H2O)3] Cl Ru OH2 Cl H2 O H2 O Cl Ru Cl Cl H2 O Cl Ru fac cis cis OH2 OH2 OH2 Cl Cl Cl H2 O OH2 H2 O Ru H2 O Cl H2 O Cl Ru OH2 trans Cl H2 O Ru OH2 Cl OH2 mer Cl Cl trans Cl 3. Draw all the unique isomers for the complex [Co(NH3)3(H2O)ClBr]+, which has cobalt in an octahedral geometry. Identify whether each isomer is chiral or achiral. Predict the number of unpaired electrons expected for this complex. (3 pts) NH3 H3 N H3 N NH3 Br Co Cl OH2 Br Cl NH3 Co NH3 OH2 chiral chiral NH3 H3 N Cl Br Co OH2 achiral NH3 NH3 H3 N Br OH2 Co Cl achiral NH3 NH3 H3 N H2 O Cl Co Br achiral NH3 This complex should have Co3+, which is usually low-spin, with no unpaired electrons. (1pt) 4. Consider the following two chromium-containing compounds: Compound A: K3[Cr(CN)6] Compound B: [Cr(en)Cl(H2O)3]Cl2 en = ethylenediamine, H2NCH2CH2NH2 a) Provide a systematic name for each compound. (2 pts) Compound A: potassium hexacyanochromate(III) Compound B: triaquachloro(ethylenediamine)chromium(III) chloride b) One of these compounds is red, the other is yellow. Choose which is which and explain your choice using the language of crystal field theory. (2 pts) Compound A is yellow, and Compound B is red. The substance that appears yellow absorbs violet light, and the substance that appears red absorbs green light. Violet light has a higher energy than green light, so the substance that appears yellow must absorb light of a higher energy than the substance that appears red. So the substance that appears yellow must have a larger ∆. Compound A should have a larger ∆ than Compound B because the cyanide ligand is a stronger field ligand than ethylenediamine, chloride, or water. c) Draw all unique geometric isomers of the complex [Cr(en)Cl(H2O)3]2+. Indicate whether each one is chiral or achiral. (2 pts) N N OH 2 Cl Achiral N OH 2 N OH 2 OH 2 Cl OH 2 OH2 Achiral