Electronic Configurations, Energy Levels, and Oxidation States of Atoms and Ions, Summaries of Geometry

Download Electronic Configurations, Energy Levels, and Oxidation States of Atoms and Ions and more Summaries Geometry in PDF only on Docsity! HW 5 Answer Key Chem 165 Spring14 - 1 - Chem 165 Homework #5 due in class Monday May 6/ 1. Give the electronic configuration for each of the following gas-phase atoms or ions (e.g., Na: 1s22s22s63s1 or [Ne]3s1). [Note that all 3+ ions of the lanthanide series are 6s05d0.] (a) Ge (b) Co2+ (c) W4+ (d) Gd3+ (e) Zr4+ (f) Se2– 2. For each atom or ion in Problem 1, draw an energy ladder and fill it up with electrons (as in the example for Na at right). [For the heavier elements, parts (c)-(f), don’t give all the core electrons, just show the valence orbitals.] Give the number of unpaired electrons in each case. Answers to 1 & 2: This could also be [Ar]3d64s1 or [Ar]3d54s2. The latter two could be lower in energy because they have five unpaired electrons vs. three for the configuration above. The only way to be sure is to look it up. Co2+ (3d7, 4F) is the ground state; Co2+ (3d64s1, 6D) is an excited state, 46,438 cm-1 (556 kJ/mol, 5.76 eV) above the ground state; Co2+ (3d34s2, 4F) is a very high excited state. 1s 2s 2p 3s Na: 1s22s22p63s1 4s 4p Ge: 1s22s22p63s23p63d104s24p2 = [Ar]3d104s24p2, two unpaired electrons 1. & 2. (a) 2s 1s 2p 3s 3p 3d (b) 2s 1s 2p 3s Co2+: 1s22s22p63s23p63d7 = [Ar]3d7 three unpaired electrons 3p 4s 3d (c) W4+: [Xe]4df145d2, two unpaired electrons 6s 5d [Xe core] 4f HW 5 Answer Key Chem 165 Spring14 - 2 - 3. (a) The omission of orbital phases from Figures 8.16 (and similar ones in Oxtoby) drives me crazy. Redraw the five parts of Figure 8.16 showing the phases of the d orbitals. See the posted lecture Figures for pictures of the d orbitals with phases. (b) The caption of Figure 8.16 is incorrect because it ignores the phases. Two of these five d orbitals interact with the ligands. For those orbitals, include the ligand orbitals in your drawing for part (a) above. Indicate the orbital phase of the ligand orbitals as appropriate for an antibonding interaction between the metal and ligand orbitals. See the posted lecture Figures for pictures of the d orbitals with phases. (c) For the other three d orbitals, their overlap with the ligand σ orbitals shown in Figure 18.6 is zero (by symmetry). Which ones are these? Explain this. The dxy, dxz, and dyz orbitals have no net overlap with the ligands shown in Figure 18.6. In each case, this because the two lobes of the d orbital overlap equally but with opposite sign with the facing ligand orbital. This is only evident if the orbital phases are drawn. 4. Problem 8.6. CoO + ½ O2 → Co3O4 The oxidation state of the cobalt in Co3O4 is +2.67. It could be that the d electrons are delocalized over multiple cobalt ions such that all the cobalt ions are +2.67 (some solids are like that). However, Co3O4 has a three-dimensional structure called the spinel structure, in which there are different kinds of cobalt sites. One site has the cobalt ion surrounded by 6 O2– in an octahedral arrangement (an octahedral site) and the other has the cobalt ion surrounded by 4 O2– in a tetrahedral arrangement (a tetrahedral site). It turns out that Co3O4 is has one Co2+ ion in a tetrahedral site and two Co3+ ions in octahedral sites. 5. Problem 8.18. (d) Gd3+: [Xe]4f7, seven unpaired electrons 6s 5d [Xe core] 4f Zr4+: [Kr], no unpaired electrons (no valence electrons) 6s(e) 5d (f) 4p 4s 3d Se2–: [Ar]4s23d104p6, no unpaired electrons [Ar core] HW 5 Answer Key Chem 165 Spring14 - 5 - Tyrosine cannot be a tridentate ligand using the phenol group because the O in the phenol is pointed away from the rest of the amino acid; it is too far away. The indole nitrogen in tryptophan (in the 5-membered ring) does not act as a ligand because the lone pair on this N is needed for the π bonding to making the bicyclic system aromatic (4n+2 = 10). This lone pair and is therefore not available for binding to a metal. 10. Draw all the isomers of dichloro-bis(ethylenediamine)cobalt(III) nitrate (Oxtoby problem 8.19b). Assume that the geometry about the Co ion is roughly octahedral, with two Cl and four N donors. Ethylenediamine is 1,2-diaminoethane. 11. Look at the Introduction of the recent review article, “Cross-Coupling Reactions Using Samarium(II) Iodide” M. Szostak, N. J. Fazakerley, D. Parmar, D. J. Procter Chem. Rev., DOI: 10.1021/cr400685rL. This article is “ASAP” – meaning “as soon as publishable” and it hasn’t been given page numbers yet, just posted to the web (at least as of the time I’m writing this). Just read the introduction and glance at the rest of the review – this is a large review, in which the authors try to cover all of the published literature in one specific area. (a) Where do the authors work? Did they get all of their training and positions at that University or have they moved around? Which is the senior author? The authors are at the University of Manchester in England. Prof. Procter is the senior author. The graduate students have done their training in Manchester but Proctor and Szostak did parts of their training in the U.S. (b) Summarize in one or two sentences what the review is about. What is a “cross coupling reaction”? This review covers the use of samarium(II) diiodide as a reagent to make carbon-carbon bonds between two different organic molecules. “Cross coupling” means coupling of two different things. (c) How many papers are cited in the review? Do you think it took a long time to write? I see 585 references. Two students in my lab and I wrote a Chemical Review article a few years ago and it was a lot of work – especially for the students. The majority of their time for more than a month, just to assemble the references and put them in order. And then we started writing. (d) Samarium is one of the lanthanide elements. Although these are called “rare earth” elements, Sm is the 40th most abundant element in the Earth’s crust. As free atoms in the gas phase, most 8. [CoCl2(H2NCH2CH2NH2)2]+ has two geometric isomers, one of which has two mirror images: Cl Co Cl H2 N N H2 H2 N N H2 + NH2 Co NH2 Cl Cl H2N H2N + H2N Co H2N Cl Cl NH2 NH2 + these two are mirror images so each one is chiral and they are different (the dashed line represents the mirror) HW 5 Answer Key Chem 165 Spring14 - 6 - of the lanthanide elements have the configuration [Xe]6s25d04fn, although a few have the configuration [Xe]6s25d14fn. The ground state of samarium has no d electrons; what is n for Sm? What do you think the configuration is for Sm2+ in its typical compounds? Sm is [Xe]6s25d04f6. I looked broadly online and couldn’t find the configuration of Sm2+, but I’m pretty sure it’s [Xe]6s25d04f4. (e) The lanthanides are all in essence in Group 3 of the Periodic Table and are most stable in the 3+ oxidation state. In fact, samarium is one of the few where simple compounds in the 2+ oxidation state can be made. (For lanthanum, for instance, attempts to make “LaCl2” would very likely make a 2:1 mixture of LaCl3 + La metal.) Therefore, is Sm2+ is a strong oxidizing agent or reducing agent? Write a very simple half reaction for the standard reduction potential of Sm2+ and give its potential in acidic aqueous solution (if you can find it – optional). Sm3+ + e– à Sm2+ E° = –1.55 V vs. SHE (Standard Hydrogen Electrode) see: http://jpkc.whut.edu.cn/web18/main/wangluo/webelements/webelements/elements/text/sm/redn.html (f) Do any of the reactions in Figure 1 look familiar to you? Which two look like one of your twenty reactions? Explain the similarity. The first two reactions under part b, the Grignard-Barbier and Reformatsky reactions, both look like the Grignard reaction. These use SmI2 instead of magnesium, but they do the same thing. They add a carbon that has a halogen (X) on it to the carbon of a carbonyl (C=O) group, to make the alcohol with a new C–C bond. 12. One of the largest collection of radioactive elements (radioactive waste) in the world is at the Hanford Nuclear Reservation, just 3.5 hours drive east of Seattle. Did you know that? Look at the following four sites or articles (the last three are posted with this problem set). http://energy.gov/em/hanford-site http://www.nytimes.com/2014/03/02/us/senator-expresses-concerns-about-nuclear-waste- tanks.html?_r=0 http://blog.seattlepi.com/seattlepolitics/2014/04/23/hanford-state-puts-heat-on-feds-over- radioactive-hot-stuff/ http://www.forbes.com/sites/jeffmcmahon/2012/08/27/whats-inside-the-suspect-nuclear- waste-tank-at-hanford/ (a) To my surprise, the article in Forbes (a business magazine) is the most scientifically detailed. What does it mean that the material in the tank “stewing from the heat of its own decay … at 110 to 135 degrees”? What decay is being referred to? These tanks are hot because of the heat of their radioactive decay. (f) Does the amount in this tank seem like a lot? How many tanks like this are there at Hanford? (see http://en.wikipedia.org/wiki/Hanford_Site). Do you think that there is cause for concern? For panic? There’s an enormous amount of material in each tank – 857,000 gallons in AY-102 alone – and there are 177 tanks in use. Huge. Yes, this has to be cause for concern – it’s an enormous amount of hazardous highly radioactive material. My understanding is that the radioactivity in the tanks is on the order of 1 Curie per Liter and tank AY-102 has over 4 million Liters. For comparison, “transuranic radioactive waste” is defined as material with elements beyond uranium that has concentrations greater than 100 microcuries per kilogram (~ 1L). HW 5 Answer Key Chem 165 Spring14 - 7 - However, I don’t think there’s cause for panic. The tanks have been there a long time and there have been a lot of small leaks. That’s not good, and it needs to be fixed. But the cleanup needs to be done carefully and thoughtfully. The materials is unbelievably hazardous and corrosive, so the process can’t be rushed and the government contractors are going to make mistakes. In my view, it’s good to keep the pressure on to make progress and keep monitoring, but the process can’t be rushed too much. The good think is that Hanford gets very little rain and the river is huge, so the dilution of the ground water into the river is enormous, something like 109. So even if something leaks into the ground water (and there hasn’t been very much of that), it’s going to be very diluted. But we still have to be very careful.