Download Problem Set 5_Chem165_Spring14

HW 5 Answer Key
Chem 165 Spring14
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.]
(b) Co2+
(a) Ge
(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.
3s
2s
Na: 1s22s2 2p63s1
Answers to 1 & 2:
1. & 2. (a)
1s
4s
4p
3s
3p
2s
2p
3d
Ge: 1s22s22p63s23p63d104s24p2
= [Ar]3d104s24p2, two unpaired electrons
1s
(b)
2p
4s
3d
3s
3p
2s
2p
Co2+: 1s22s22p63s23p63d7 = [Ar]3d7
three unpaired electrons
1s
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.
(c)
6s
5d
4f
[Xe core]
W4+: [Xe]4df145d2, two unpaired electrons
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HW 5 Answer Key
(d)
Chem 165 Spring14
6s
5d
[Xe core]
Gd3+:
(e)
4f
[Xe]4f7,
seven unpaired electrons
6s
5d
Zr4+: [Kr], no unpaired electrons (no valence electrons)
(f)
4p
4s
3d
[Ar core]
Se2–: [Ar]4s23d104p6, no unpaired electrons
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.
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HW 5 Answer Key
Chem 165 Spring14
6. (a) Problem 8.16. The glycinate ion typically acts as a bidentate chelate ligand and forms a fivemembered ring. Could you make a structure like this with any amino acid?
O
O
M
O
O
N
H2
R
M
N
H2
Yes, this structure can be made with any amino acid:
7. For each of the following complexes, give (i) the formal oxidation state of the central metal and
(ii) the number of d electrons on the metal in this oxidation state. For example: Cu(H2O)62+ and
CuCl42- both formally contain Cu2+, d9.
Make sure to divide the complex up so that each ligand has a closed shell (octet) configuration.
(a) Cr(NH3)3Cl3
(c) Os(NH3)62+
(e) Fe(CO)5
(g)
O
(b) Ni(H2O)62+
(d) Ir(CO)(Cl)(PPh3)2
(f) Nb(NMe2)5
OH2
O
(h)
H2
N
H2 N
Cu
N
H2
NH2
3+
H2
N
Co
O
H2 N
O
OH2
N
H2
NH2
Ligand abbreviations: Me, methyl, CH3; py, pyridine, N
; Ph, phenyl, C6H5; NH3 is ammonia.
PPh3 is triphenylphosphine, :PPh3; CO is carbon monoxide, which binds through carbon, :C≡O.
8. Assuming that each ligand counts as a two-electron donor to the metal center, calculate the total
number of electrons about the metal for each compound in problem 12. Taking CuCl42- as an
example, Cu2+ is d9 and each chloride ligand contributes two electrons, so the total electron
count is 9 + 4 × 2 = 17.
12. & 13. (a) Cr(NH3)3Cl3 = Cr3+ (d3) + 3 NH3 + 3 Cl-.
Cr(NH3)3Cl3 is a 15-electron complex: 3 d electrons + 6 two-electron ligands.
(b) Ni(H2O)62+ = Ni2+ (d8) + 6 H2O
Ni(H2O)62+ is a 20-electron complex: 8 d electrons + 6 two-electron ligands.
(c) Os(NH3)62+ = Os2+ (d6) + 6 NH3
Os(NH3)62+ is an 18-electron complex: 6 d electrons + 6 two-electron ligands.
(d) Ir(CO)(Cl)(PPh3)2 = Ir+ (d8) + CO + Cl– + 2 PPh3
(PPh3, triphenylphosphine, is a neutral ligand)
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HW 5 Answer Key
Chem 165 Spring14
Ir(CO)(Cl)(PPh3)2 is a 16-electron complex: 8 d electrons + 4 two-electron ligands.
This is a famous compound, called “Vaska’s complex”
(e) Fe(CO)5 = Fe0 (d8) + 5 CO
Fe(CO)5 is an 18-electron complex: 8 d electrons + 5 two-electron ligands.
(Yes, I know that gas phase iron atoms are 4s23d6, but all metal complexes are s0dn,
so always count the number of valence electrons and put them all in the d shell.)
(f) Nb(NMe2)5 = Nb5+ (d0) + 5 NMe2Nb(NMe2)5 is a 10-electron complex: no d electrons + 5 two-electron ligands
(except that there is π bonding between each of the dimethyl amide (NMe2–) ligands and the
Nb5+ center, in addition to the σ bonds we’re counting here).
(g)
OH2
O
O
H2
N
O
O
O
:
N
H2
O:
= Cu2+ (d9) + 2 H2O + 2
Cu
NH2
OH2
This is a 21-electron complex: 9 d electrons + 6 two-electron donor ligands.
The dots indicate the lone pairs that bind to the metal, not all the lone pairs in the ligand.
(h)
NH2
H2 N
3+
H2
N
Co
H2 N
N
H2
NH2
is Co3+ (d6) and three ethylenediamine ligands, which
are neutral (like two ammonias attached together).
It is an 18-electron complex:
6 from the metal + 6 two-electron donor amine
ligands.
9. What is the common name (not the official IUPAC name) of the organic ligand bound to copper in
question 8g? How many chiral centers are there in the copper complex in 8g? Note that the “four
different groups” rule only applies to tetrahedral structures.
The acid form of this ligand is always called alanine, one of the natural amino acids. Each
copper complex contains two mono-deprotonated alanine ligands. Each of the α carbon
atoms of the amino acid are chiral. The copper may or may not be chiral, depending on the
relative chirality of the two α carbons. If both the alanines have the same configuration (for
instance, their natural L configuration), then the copper is chiral. But if one is D and the other
is L, then I don’t think the copper is chiral.
(b) Which amino acids have side chains that could act as ligands and therefore form tridentate
complexes? Why are tyrosine and tryptophan not included in this list?
The amino acids that can bind to a metal are those that have O, N or S groups in their side
chains, including:
i. Those with carboxylic acid/carboxylate groups, aspartate and glutamate;
ii. Those with basic nitrogen groups, arginine, histidine and lysine
iii. And asparagine, cysteine, glutamine, serine, and threonine.
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HW 5 Answer Key
Chem 165 Spring14
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.
8. [CoCl2(H2NCH2CH2NH2)2]+ has two geometric isomers, one of which has two mirror images:
H2
N
Cl
H2
N
+
H2N
Co
N
H2
Cl
+
NH2
Cl
Cl
Co
N
H2
H2N
NH2
Co
Cl
NH2
+
H2N
Cl
NH2
H2N
these two are mirror images so each one is chiral and
they are different (the dashed line represents the mirror)
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
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HW 5 Answer Key
Chem 165 Spring14
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-wastetanks.html?_r=0
http://blog.seattlepi.com/seattlepolitics/2014/04/23/hanford-state-puts-heat-on-feds-overradioactive-hot-stuff/
http://www.forbes.com/sites/jeffmcmahon/2012/08/27/whats-inside-the-suspect-nuclearwaste-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).
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HW 5 Answer Key
Chem 165 Spring14
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.
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