Download Exam 2010 - Chemistry

Chemistry 3211
Final Examination
April 20, 2010
Prof. C. Kozak
Do not open your exam booklet until you are instructed to do so.
Write your name and student number in the space provided.
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The last 3 pages (pages 10, 11 and 12) of the exam consist of a Periodic Table and
Tanabe-Sugano diagrams. Remove these 3 pages from the booklet if you wish, but
DO NOT TAKE THEM FROM THE EXAM.
Do not remove any other pages from the exam.
Please read the whole paper before you begin.
Please answer all of the questions in the space provided.
If you write on the back of the page, please indicate this.
The total possible marks assigned to each question are given below and in square
brackets, [ ], next to each question. There is a total of 90 marks. BUDGET YOUR TIME
ACCORDINGLY!
You have 2.5 hours (150 minutes) to complete your Exam – Good Luck!
Name:
Student Number:
Question
number
Possible
Marks
1
10
2-4
18
5, 6
14
7, 8
13
9, 10
10
11, 12
11
13
14
Marks Obtained
Bonus
TOTAL
/90
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1. Provide the missing information in the table below
a) Cr(acac)3
Name:
d-electron count =
Structure:
b) mer-Trihydridotris(triphenylphospine)
iridium(III)
Formula:
Structure:
d-electron count =
c) (Ph3P)AuCl
Name:
Structure:
d-electron count =
d) Pentaamminedinitrogenruthenium(II)
chloride
Formula:
Structure:
d-electron count =
e) trans-[Cr(en)2Cl2]
Name:
Structure:
d-electron count =
[10 marks]
2
2. Define the following three sets of terms, use examples where appropriate.
(a)
Chelate effect vs. macrocycle effect
[2 marks]
(b)
Nephelauxetic series vs. spectrochemical series
[2 marks]
(c)
Oxidative addition vs. reductive elimination
[2 marks]
3. Both H- and P(C6H5)3 are ligands of similar field strength and high in the spectrochemical series.
Explain how each ligand’s specific bonding abilities explain their high position in the spectrochemical
series. Your answer must include the orbital interactions that account for the strength of each ligand.
[6 marks]
4. Crystal Field Theory (answer both parts)
(a) The complex [NiCl2(PPh3)2] is paramagnetic, but [PdCl2(PPh3)2] is diamagnetic. What does this say
about their structures/geometries?
[3 marks]
(b) In the following solution equilibria, will the products or the reactants be favoured? Give your
reasoning in each case:
(i) AgF(aq) + LiI(aq)
AgI(s) + LiF(aq)
(ii) 2 Fe(OCN)3(aq) + 3 Fe(SCN)2(aq)
2 Fe(SCN)3(aq) + 3 Fe(OCN)2(aq)
[3 marks]
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5. (a) The following is a general trend for the Trans Effect of some selected simple ligands:
Strong Trans effect
Weak Trans effect
CO ~ CN- ~ C2H4 > PR3 > CH3- > SCN- > I- > Br- > Cl- > NH3 > H2O
Predict the geometries of the products obtained from the following reactions by labelling the
complexes as cis, trans, fac or mer in the spaces provided.
[RhCl(OH2)5]2+ + Cl- → _________ [RhCl2(OH2)4]+ + H2O
[3 marks]
_________ [RhCl2(OH2)4]+ + Cl- → _________ [RhCl3(OH2)3] + H2O
_________ [RhCl3(OH2)3] + Cl- → _________ [RhCl4(OH2)2]- + H2O
(b) Predict the reaction outcomes yielding products A and B. If there are conflicting products,
identify which ligand is the better leaving group for that compound.
[PtCl3(NH3)]- + SCN- → (A)
(A) + SCN- → (B)
[3 marks]
6. Rationalize the following observations, answer both parts:
(a) Mo(CO)6 exhibits absorptions in the IR spectrum at 2004 cm-1 but upon reaction with excess
PPh3 forms Mo(CO)3(PPh3)3, which displays absorptions at 1934 and 1835 cm-1. Reaction with
pyridine forms Mo(CO)3(Py)3, which displays absorptions at 1888 and 1746 cm-1. Explain the
observed carbonyl frequencies for all three complexes.
[4 marks]
(b) The IR spectrum of dicobalt octacarbonyl, Co2(CO)8, in solution shows bands at 2069 and 2042
cm-1, but in the solid state at 2031 and 1857 cm-1. Explain by proposing possible geometries for
the compound.
[4 marks]
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7. On the basis of the 18-electron rule for each of the following, identify the 1st row transition metal,
M and its oxidation state, or the overall charge on the complex, x:
(i) [M(PF3)2(NO)2]+
(contains linear M-N-O)
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(ii) [(η -C5H5)V(η -C3H5)(dmpe)]x
(dmpe = 1,2-bis(dimethylphosphino)ethane)
(iii) [(η5-C5H5)M(CO)3]2
(assume a single M-M bond)
[3 marks]
8. Redox reactions in coordination chemistry (answer both parts)
(a) Explain the trends for the following observed rate constants for the self exchange redox
reactions:
[Ru(NH3)6]3+ + [Ru(NH3)6]2+ k = 104 L mol-1 s-1
[Co(NH3)6]3+ + [Ru(NH3)6]2+ k = 10-2 L mol-1 s-1
[Co(NH3)6]3+ + [Co(NH3)6]2+ k = 10-6 L mol-1 s-1
[4 marks]
(b) The redox reaction of [Co(NH3)5X]2+ with [Cr(OH2)6]2+ in water has rate constants of 6.0 × 105
when X = Cl- and 3.0 × 106 when X = I-.
i) Show the reaction mechanism.
ii) Why do the rate constants differ depending on the nature of X?
[6 marks]
5
9. The reaction below in liquid ammonia proceeds 1000 times faster when a catalytic amount of KNH2
(< 5% vs. [Cr]) is present. Suggest an explanation for this rate increase compared to when no KNH2 is
present.
[Cr(NH3)5Cl]2+ + NH3(l) → [Cr(NH3)6]3+ + Cl[4 marks]
10. Ionic radii generally decrease left to right across a period. However, for the M2+ transition metal
ions the following trend is observed for six-coordinate metal oxides:
Mn2+ (80 pm), Fe2+ (74 pm), Co2+ (72 pm), Ni2+ (69 pm), Cu2+ (72 pm), Zn2+ (74 pm).
Provide an explanation for the observed deviation from the expected trend.
[6 marks]
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11. Given the following spectroscopic data, answer all parts below.
ν1 (cm 1)
ν2 (cm 1)
[Ni(H2O)6]2+
8,500
14,600
2+
[Ni(en)3]
11,000
18,500
2+
[Ni(NH3)6]
10,700
17,500
−
−
ν3 (cm 1)
25,300
30,000
28,200
−
Molar absorbtivity, ε, for all bands is in the range of 10-100 M-1 cm-1.
(i) Assign bands ν1, ν2 and ν3 using term symbols from the Tanabe-Sugano Diagram.
(ii) Determine Δo for each complex ion.
(iii) Place the three ligands in a spectrochemical series of weakest to strongest field strength.
[5 marks]
12. Spectroscopic properties of metal complexes
(a) The ligand-to-metal charge transfer bands increase in energy in the series: [CoI4]- < [CoBr4]- <
[CoCl4]-. Explain. Tip: use relative orbital energy level diagrams to illustrate your reasoning.
[3 marks]
(b) Explain why [MnF6]4- is colourless and exhibits only very weak intensity bands in its UV-vis
spectrum, whereas [CoF6]3- is coloured but exhibits only a single band in the visible region.
[3 marks]
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13. Molecular Orbital Theory, answer both parts:
(a) For a transition metal complex with octahedral geometry,
(i) Draw the symmetry adapted linear combinations (LCAOs) of the ligand group orbitals
(LGOs) that form bonding interactions with metal s, p and d orbitals.
(ii) Draw the metal orbitals and provide their octahedral symmetry labels (i.e. a1g, eg, etc.)
[7 marks]
(b) Molecular oxygen, O2, is a relatively strong field σ-donor/π-acceptor ligand.
(i) Construct a complete molecular orbital diagram for O2, including atomic orbitals and their
relative energies. Fully label the orbitals (e.g. px, σ, π, etc.), HOMO, and LUMO.
(ii) Using orbital overlap representation, show the metal-ligand bonding between O2 and a
transition metal. Why is its bonding different to that exhibited by CO?
(iii) Do you expect the O-O bond length in O2 to be more or less influenced than the C-O bond
length in CO by backbonding from a metal? Explain.
[7 marks]
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Bonus marks [2 marks each]:
1.
From where does the metal vanadium get its name? What is an important use for vanadium?
2.
From where does copper get its name? What was one of copper’s earliest uses?
3.
To what does the “Goldschmidt Classification” refer? What are the two classes?
4.
Name two scientists who worked on the “Manhattan Project” together only to become bitter
ideological rivals.
A final thought…
Stars like the Sun are converting hydrogen and helium into carbon, oxygen, nitrogen, neon,
magnesium, silicon and iron, but nothing heavier than iron (atomic number 26). Red Giants make
most of the other elements (some say as heavy as cadmium, atomic number 48). The heaviest
atomic nuclei (including gold, uranium and iodine) can only be made by the death of a star, that is,
a supernova.
Because the human body needs elements heavier than iron, such as copper, zinc and selenium, we
know these must have been created by a red giant at some time in the past. Because we also need
iodine, part of us must have once been scattered into the cosmos by a supernova.
We are, therefore, truly made of star dust!
THE END
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