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Chemistry 350
Spring 2012
Exam #3, April 27, 2012
50 minutes
CCM
100 points on 4 pages + a useful page 5
1. In discussing solid-state materials, we discussed band theory. Draw a depiction of filling of bands for
the following types of compounds. Be sure to label the band gap (with approximate energy) and bands
appropriately. (9 pts)
a) an insulator
b) a conductor
E = 0 eV
vb = valence band
cb=conduction band
E = band gap > 4eV
c) semi-conductor
E < 4eV
2. A little potpourri of transition metal questions like those on the problem set… (12 pts)
a) Determine the number of valence electrons on the central metal atom in the following complex using
either of the methods we learned about in class. You must state which method you are using and show
your work. So that there is less confusion, the answer for [Ni(OH2)6]2+ would be 20 electrons.
Ionic: Cr(+2) 4
4 PR3 8
2 NO 2
16
Neutral:
Cr
4 PR3
2 NO
4+
6
8
6
-4
16
[Cr(PR3)4(NO)2]4+ (linear NO)
b) Assuming the 18 electron rule applies, identify the second-row transition metal: [M(CO)7]
18-(7x2) = 4, so Zirconium, Zr
c) What charge, z, would be necessary for the following complex to obey the 18-electron rule?
[Re(CO)3(SiMe3)3]z
n: 18 – 7 – (3x2) – (3x1) = 2, so need 2 more electrons… z=2i: 18 – (3x2) – (3x2) = 6 electrons on Re. That is 1+. So, 1+ + (3x1-) = 2-
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Chem 350, Spring 2012 Exam 3
CCM, Page 2
3. Consider the following useful redox diagrams. (24 pts)
a) Write a balanced equation for the VO2+/V3+ couple in acid.
e- + 2H+ + VO2+  V3+ + H2O
b) Which vanadium species is the best oxidizing agent?
(circle one) VO2+
VO2+
V3+
V2+
c) Which vanadium species is the best reducing agent?
(circle one) VO2+
VO2+
V3+
V2+
d) Which vanadium species is most stable in acid?
(circle one) VO2+
VO2+
V3+
V2+
e) Fill in the blanks appropriately…
Using the _Pourbaix___________ diagram
at right, it can easily be determined that the
only stable oxidation state for vanadium in
basic solution is __V5+___ . It can also be
seen that the species one would expect to
find in bog water, which has pH=2 and a
potential of 0.4 V, is __VO2+___.
_1.10_ V
a) Assuming standard conditions and based on the data on the useful
page, label the cathode and anode in this cell and give the E0cell in
the boxes provided. Cu2+ + Zn = Cu + Zn2+
Ecell = Ecat-Ean = 0.34 – (-0.76) = 1.10 V
b) If water is added to the Zn/Zn2+ cell, the voltage of the cell will
(circle one)
increase
decrease
stay the same.
c) Briefly explain your reasoning for part g.
The voltage will change with addition of water. This is mathematically obvious from the Nernst equation.
Diluting the Zn2+ solution will decrease Q and, therefore, increase Ecell. The system is
attempting to create more Zn2+, basically.
_cath__ode
__an_ode
/24
Chem 350, Spring 2012 Exam 3
CCM, Page 3
4. a)Complete the following table and b) answer the question below. (29 pts)
correct name
oxidation state of
central metal atom
number of valence
electrons on central
metal atom
[Fe(CO)6]0
[Co(en)2(NH3)2]3+
[CrF2Br2Cl2]4-
below
below
below
0
3+
2+
neutral: 8 + 2(6) =20
ionic: 8 + 2(6) = 20
n: 9 + 2(4)+2(2) -3 =18
i: 6 + 2(4) + 2(2) = 18
n:6+2(1)+2(1)+2(1)+4=16
i: 4+2(2)+2(2)+2(2)=16
20
18
16
8
6
4
[Ar]4s23d6
[Ar]3d6
[Ar]3d4
octahedral
octahedral
octahedral
1
2
2
(e.g. [Ni(OH2)6]2+ has 20)
number of d
electrons
(e.g. [Ni(OH2)6]2+ has 8)
electron configuration
of central metal ion
likely molecular
geometry around
central atom
spin, S =
(at this point, answer as if all
complexes are “high spin” octahedral
)
Chiral? (circle)
YES
NO
YES
NO
YES
NO
There is a little more room here… so give the proper names for…
[Fe(CO)6]0 :_hexacarbonyliron(0)__________________
[Co(en)2(NH3)2]3+:_diamminebis(ethylenediamine)cobalt(III)
[CrF2Br2Cl2]4- :__dibromodichlorodifluorochromate(II)
b) At least one of the above compounds is chiral. Draw two enantiomers here and label them
appropriately.
isomer of [CrF2Br2Cl2]
4-
F
4-
Cl
F
F
Cl
Cr
Cr
Cl
4- e.g. all cis
F
Br
Br
Cl
Br
Br


/29
Chem 350, Spring 2012 Exam 3
CCM, Page 4
5. Shown is a portion of an engineering diagram for a common industrial process. (10 pts)
a) Name the process
Haber-Bosch Process
b) Write a balanced chemical equation for the overall process.
N2 (g) + 3 H2 (g)  2 NH3 (g)
c) Comment on three portions of the engineering that help the
reaction occur in a more commercially viable way (there are
more than 3).
 Couple with other processes
 High temperature
 High pressure of N2 and H2
 Use of catalyst
 Recycle unreacted gases
 Liquify NH3
6. In the space provided, draw all of the possible stereoisomers of [TiBrClFI]0. Be sure to consider both
possible coordination geometries. (5 points)
Br
Br
F
Ti
I
F
Ti
Ti
Cl Cl
F
Br Br
I
I
F
F
Ti
Cl
I
Cl
Ti
Cl
I
Br
7. In class we same the demonstration of the reaction of [Ni(OH2)6]Cl2 with ethylenediamine, en. With
the addition of en, the aqueous solution turned from green to blue to violet. (11 points)
a) Based on these observations, comment on whether [Ni(OH2)6]2+ is inert or labile. Explain your
reasoning.
Labile. The substitution reaction occurs quickly, as observable with the color change. Inert species do
not react quickly… if at all.
b) Explain why you observed multiple color changes, not a single color change.
The substitution of the hexacoordinated nickel does not occur in one step. One could imagine first
substituting two waters with one en, then repeating that process, and then a third and final time. Only the
initial change and the final color change are obvious to us on observation and the likelihood is that there is
a statistical distribution of [Ni(en)n(OH2)6-2n]2+ from the beginning until 3 equivalents of en are added.
c) We also discussed the order of addition. If we had first mixed another reagent, NH3 with
the[Ni(OH2)6]2+, we would see a different blue-colored solution. If we then added en, we would achieve
the same violet solution above. If we add NH3 to the violet solution it stays violet, i.e. there is no
reaction. Briefly explain why. The en ligand is bidentate, that is it has two lone pairs that can bind to the
metal. Higher denticity ligands (here bidentate) are more likely to stay bound than lowe denticity (here
monodentate) ligands; this is called the chelate effect. If you add the en first, it stays bound. Second, it
replaces the NH3.
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Chem 350, Spring 2012 Exam 3
CCM, Page 5
Ligand
Ionic Method
NO (bent)
NO (linear)
=O, =S, =NR
CR
N
5-C5H5 (Cp)
6-C6H6 (benzene)
7-C7H7
(cycloheptatrienyl)
3-C3H5 (allyl)
RCCR
2 (NO–)
2 (NO+)
4 (O2–, S2–, RN2–)
3 (neutral)
6 (N3–)
6 ( C5H5 –)
6
6 ( C7H7+)
Neutral
Method
1 (O=N•)
3 (O=N• + lp)
2
3
3
5
6
7
4 (C3H5–)
2 or 4
3
2 or 4
Brief Spectrochemical Series:
I- < H2O < NH3 < CN-,CO
en = ethylenediamine H2NCH2CH2NH2
= bipy
N
N