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Chem 64
Winter 2003
NAME: ____________________
3rd (final) EXAM
• THIS EXAM IS WORTH 100 POINTS
AND CONTAINS 9 QUESTIONS
• THEY ARE NOT EQUALLY WEIGHTED!
• YOU SHOULD ATTEMPT ALL QUESTIONS
AND ALLOCATE YOUR TIME ACCORDINGLY
• IF YOU DON'T KNOW THE ANSWER, GUESS!
• ALL BOOKS AND PAPERS OTHER THAN TABLES
WHICH HAVE BEEN HANDED OUT SHOULD BE PLACED
ON THE FLOOR DURING THIS EXAM
• IF YOU DO WORK ANYWHERE OTHER THAN THE SPACE PROVIDED FOR
EACH QUESTION, INDICATE CLEARLY WHERE IT IS LOCATED
• A PERIODIC TABLE, SPECTROCHEMICAL SERIES, MAGNETISM INFO,
AND TANABE-SUGANO DIAGRAMS
ARE ATTACHED AT THE END OF THE EXAM
• PLEASE PRINT YOUR NAME BOTH ON THIS PAGE AND ON PAGE 2
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NAME: ____________________
FOR GRADING USE ONLY
Question 1 (13 pts)......................…__________
Question 2 (8 pts)........................__________
Question 3 (8 pts)........................__________
Question 4 (10 pts).........................__________
Question 5 (12 pts) ......................…__________
Question 6 (12 pts) ........................__________
Question 7 (12 pts) ........................__________
Question 8 (12 pts) ........................__________
Question 9 (13 pts) ........................__________
TOTAL (100 pts)......................===========
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1. (13 pts) Consider data for the two ligand substitution reactions shown below.
SnPh3
1
L
Pt
SnPh3
L
+ L*
SnPh3
L = PMe2Ph
Pt
L* + L
SnPh3
SiMePh2
2
Ph2MeSi
Pt
L
Reaction number
∆H‡ (kJ/mol)
∆S‡ (J/mol•K)
L
1
24
–131
L + L*
SiMePh2
Ph2MeSi
Pt
L* + L
L
2
118
120
a. (4 pts) Suggest mechanisms for both reactions. Briefly explain your answers.
b. (3 pts) The SnPh3 group is a π -acceptor. Why might this be important in your
proposed mechanism for reaction 1?
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c. (6 pts, 3 each) This table shows X-ray structural data for the complex in reaction 2
and for analogous complexes with Me and Cl ligands. [Note: L = PMe2Ph]
Complex
cis-PtL2(SiMe2Ph)2
cis-PtL2(Me)2
cis-PtL2(Cl)2
Pt-P bond length (Å)
2.359
2.284
2.258
i. Circle your choice for the trans influence of the SiMe2Ph group, relative to Me, Cl,
and other ligands with which you are familiar. Briefly explain your answer.
The trans influence of SiMe2Ph is
high
medium
ii. Why might this be important in your proposed mechanism for reaction 2?
low
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2. (8 pts, 4 each) a. For water exchange on complex 1, ∆S ‡ = –34 J/mol•K and ∆V ‡ =
–4.7 cm3/mol. Propose a mechanism for the ligand substitution and briefly explain your
answer.
2+
N
N Cu
H2
H 2O
1
2+
N
NH2
N
H2
N Cu
Me2
H 2O
NMe2
N
Me2
2
b. Water exchange in complex 2 is 105 times slower than in 1. Explain this observation
in terms of the mechanisms of these ligand substitutions.
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3. (8 pts, 4 each) The figure below shows the electronic spectra of the cis and trans
isomers of approximately octahedral [Co(en)2F2]+ (en = H2NCH2CH2NH2). There are 2
curves, labeled A and B. Fill in the box to match the spectrum with the isomer and
explain your answer. [Note: the y-axis shows ε values.]
cis
trans
b. Electronic spectra of [RhCl6]3– and [Rh(NH3)6]3+ are shown in the figure. Fill in the
box to match the spectrum to the complex and explain your answer.
[RhCl6]3–
[Rh(NH3)6]3+
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4. (10 pts) a. (4 pts) The complex [Cr(H2O)6]2+ reacts with [Co(NH3)5Cl]2+ 108 times faster
than it does with [Co(NH3)6]3+. Explain this observation.
b. (6 pts) The substitution of azide (N3–) for triflate (OSO2CF3) in the reaction below is
very slow in acidic aqueous solution.
[Co(NH3)5(OSO2CF3)] 2+
NaN3
[Co(NH3)5(N3)] 2+ very slow
However, when 1.0 M N3– in the presence of 0.10 M OH– is used, the ligand substitution
reaction is much faster, and the products are the cobalt azide complex (10%) and the
cobalt hydroxide complex (90%).
NaN3
[Co(NH3)5(OSO2CF3)] 2+
[Co(NH3)5(N3)] 2+ much faster
–
OH
[Co(NH3)5(OH)]2+
Explain these observations in terms of mechanisms for the ligand substitutions. Don't
forget to explain why the reaction in acid solution is slow.
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5. (12 pts) a. (4 pts) Both [Fe(CN)6]4– and [Fe(CN)6]3– are toxic because they dissociate
cyanide, CN–. Circle the one you predict will be more poisonous, and briefly explain
your answer.
[Fe(CN)6]4–
[Fe(CN)6]3–
b. (8 pts) The gemstone ruby consists of Cr(III) impurities in octahedral sites in Al2O3.
The electronic spectrum of ruby contains some very narrow (peak width ~ 10 cm–1),
low-intensity lines.
Identify (as specifically as possible) the electronic transitions responsible for the
narrow peaks in the ruby spectrum, and explain why the peaks are unusually narrow in
comparison to "normal" ones.
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6. (12 pts) [Fe(CN)6 ] 3– and [Co(EDTA)]2 – react slowly to yield [Fe(CN)6]4 – and
[Co(EDTA)] – . In this reaction (1), the cyano-bridged intermediate [(NC)5Fe-CNCo(EDTA)]5– is formed within milliseconds. The rates of the analogous reactions 2 and 3
are very similar to each other and to that of 1.
Suggest mechanisms for these electron-transfer reactions and explain your answers.
1
NC
NC
Fe
CN
CN
CN
5–
CN
3–
CN
NC
[Co(EDTA)] 2– [Co(EDTA)] –
[Fe(CN) 6]4–
NC
Fe
C
N
CN
CN
EDTA
[Co(EDTA)]
intermediate
2–
CN
2
NC
NC
Fe
CN
CN
3
NC
NC
Fe
CN
CN
–
O2C
–O
[Co(EDTA)]
2–
2–
CN
[Co(EDTA)] 2–
N
CO2–
CO2–
py = N
bipy = N
[Co(EDTA)]–
[Fe(CN) 5(bipy)]3–
N
2C
N
[Co(EDTA)] –
[Fe(CN) 5(py)]3–
N
N
N
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7. (12 pts) Phosphines PR3 can act as π-acceptors using the P-C σ* MO as an acceptor
orbital. Using this model, explain
a. why the P-C bond shortens on oxidation of the Co (1 ---> 2).
b. why the Co-P bond lengthens on oxidation of the Co (1 ---> 2).
Be sure to consider both σ- and π-bonding effects.
number
Compound
Co-P bond length (Å)
P-C bond length (Å)
1
[CpCo(PEt3)2]
2.218±0.001
1.846±0.003
2
[CpCo(PEt3)2][BF4]
2.230±0.001
1.829±0.003
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8. (12 pts) The isoelectronic ions [VO4]3–, [CrO4]2–, and [MnO4]– all show intense charge
transfer peaks in their UV-vis spectra. The wavelengths of these transitions increase
in this series, with [MnO4]– having its charge transfer absorption at longest wavelength.
Explain these observations in terms of a suitable MO diagram for these complexes; be
sure to identify the specific transitions responsible for the charge transfer peaks.
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9. (13 pts) a. (4 pts) Consider a linear ML2 complex (the z-axis is the molecular axis).
According to crystal field theory, the energies of the M atom's d orbitals in such a
complex split into 3 different groups. Fill in the boxes with the appropriate orbitals and
explain your answers.
Energy
d-orbitals
b. (4 pts) The complex [NiO2 ] 2– , which contains two O2– ligands, is linear and
paramagnetic. Its absorption spectrum shows two peaks attributed to d-d transitions at
9000 and 16000 cm–1 . Use your CFT result above to explain the origin of these
transitions in terms of ground state and excited state electron configurations.
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c. (5 pts) The extinction coefficient, ε, for the peaks in part b is about 5 M–1 cm–1.
Compare the magnitude of ε to that for peaks in the spectra of generic octahedral
complexes (circle your choice) and explain qualitatively why ε is similar (or different) to
that in octahedral complexes.
In comparison to ε values for octahedral complexes, this ε value is
much smaller
about the same
much bigger
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PERIODIC TABLE OF THE ELEMENTS
1
2
H
He
1.0079
4.00260
3
Li
4
5
Be
B
6.941 9.01218
11
12
Na
24.305
20
K
Ca
39.0983
37
Rb
85.4678
55
Cs
132.905
87
Fr
(223)
Al
21
Sc
40.08 44.9559
38
39
Sr
87.62
Y
88.9059
56
Ba
57
La
137.33 138.906
88
Ra
89
Ac
226.025 227.028
22
Ti
23
V
47.88
40
24
Cr
25
Mn
26
Fe
27
Co
50.9415 51.996 54.9380 55.847 58.9332
41
42
43
44
45
Zr
Nb
Mo
Tc
Ru
Rh
91.224
92.9064
95.94
(98)
101.07
102.906
73
72
Hf
178.49
104
74
W
Ta
75
Re
180.948 183.85 186.207
105
106
107
76
77
Os
Ir
190.2
192.22
108
28
29
Ni
Cu
58.69
46
63.546
47
Zn
Pd
Ag
106.42 107.868
78
Pt
79
Au
S
As
72.59 74.9216
50
51
Cd
In
Sn
Sb
112.41
114.82
118.71
121.75
80
81
Tl
200.59 204.383
F
O
P
Ge
69.72
49
9
10
Ne
12.011 14.0067 15.9994 18.9984 20.179
16
14
15
17
18
Si
Ga
Hg
195.08 196.967
N
26.9815 28.0855 30.9738
30
31
32
33
65.39
48
8
7
C
10.81
13
Mg
22.9898
19
6
82
Cl
32.06
34
Ar
35.453 39.948
35
36
Se
Br
Kr
78.96
52
79.904
53
83.80
54
Te
I
Xe
127.60 126.905 131.29
83
84
85
86
Pb
Bi
Po
At
Rn
207.2
208.980
(209)
(210)
(222)
109
Unq Unp Unh Uns Uno Une
(261)
(262)
58
Ce
(263)
59
Pr
(262)
60
Nd
140.12 140.908 144.24
90
91
92
Th
Pa
U
232.038 231.036 238.029
(265)
61
Pm
(266)
62
63
64
Tb
66
Dy
67
Ho
68
Tm
70
Yb
71
150.36
94
151.96
95
157.25 158.925 162.50
96
97
98
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
(237)
(244)
(243)
(247)
(247)
(251)
(252)
(257)
(258)
(259)
(260)
164.930
99
Er
69
Eu
(145)
93
Gd
65
Sm
167.26 168.934 173.04
100
101
102
Lu
174.967
103
SPECTROCHEMICAL SERIES
Increasing ∆ (Fixed Metal)
I– < Br– < Cl– ~ *SCN– ~ N3– < F– < urea < OH– < CH3CO2– < ox < H2O < *NCS– < EDTA <
NH3 ~ py < en ~ tren < o-phen < *NO2– < H– ~ CH3– < CN– ~ CO
Note: * denotes the ligating atom
Increasing ∆ (Fixed Ligand)
Mn2+ < Co2+ ~ Ni2+ ~ Fe2+ < V2+ < Fe3+ < Cr3+ < V3+ < Co3+ < Mn4+ < Mo3+ < Rh3+ < Ru3+ <
Ir3+ < Re4+ < Pt4+
Spin-only magnetic moment formula
µs = 2 [S(S+1)]0.5 B.M. = [n(n+2)]0.5 B.M. (Bohr magnetons)
Ligand abbreviations
Cp = η5-C5H5
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