Download 2011 Midterm 1 KEY

Chem 324 Midterm 1 Fall 2011 Version 1
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Chemistry 324
Midterm 1 KEY
Wednesday, October 19, 2011
Instructor: D. J. Berg
Name: _______________________________
Answer all questions on the paper (use the back if necessary). There are
50 minutes and 50 marks so ration your time accordingly. A periodic
table is included with this examination. There are 9 pages including a
blank page for more workspace (p. 8) and a periodic table (p. 9).
1.
Draw the structures and give the metal oxidation state and dn
electron count for each of the following compounds. (3 pts each)
(a)
trans, mer-[Cr(dien)Cl2(OH2)]+ Ox. St. = 3+
dn = 3
+
NH2
OH2
HN
Cr
Cl
Cl
NH2
(b)
[Pd(µ-Cl)2]n
(c)
[(CO)3Co]2(µ-η2:η2-PhC≡CPh)
Ox. St. = 2+
Ox. St. = 0
Ph
OC
CO
OC
Co
Co
OC
CO
CO
Ph
dn = 8
dn = 9
Chem 324 Midterm 1 Fall 2011 Version 1
2.
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Somewhat surprisingly, the complex Zr(CH3)4(dppe) shows only
ONE type of metal-bonded methyl resonance (of relative
integration 12) in the 1H NMR in addition to the resonances for
dppe. Furthermore, the spectrum is not affected by cooling to -100
°C, by changes in complex concentration OR by addition of excess
dppe. What do these observations suggest about the structure of
this complex in solution? [6 pts]
The facts indicate that this is a 6-coordinate structure at all
temperatures: dppe is a bidentate phosphine and there is no
temperature or concentration effects making it unlikely that dppe is
dissociating. Fluxional 6-coordinate structures are unusual and
require high temperature when they do occur. However, if the
structure is 6-coordinate, there is no octahedral structure that
makes all 4 methyls equivalent. Thus we need to consider other
possible structures and the most common option would be a
trigonal prism. If we place the dppe ligand so that one P is in each
trigonal face then all 4 methyls are equivalent because there are
two mirror planes that inter-relate all four. This is the most likely
structure that fits the facts (mirror planes shown by dashed lines):
Chem 324 Midterm 1 Fall 2011 Version 1
3.
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Draw the ligands given by the following names or abbreviations [2
pts each]:
(a)
ox
(b)
acac
O
(c)
R or S-BINAP
O
Chem 324 Midterm 1 Fall 2011 Version 1
4.
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The ligand –SCH2CH2NH2 forms a complex of the type [Ni(κ2N,S-SCH2CH2NH2)(Br)(OH2)] that does not conduct electricity in
solution. This complex is found to exist as two optical isomers.
What does this observation tell you about the structure? Draw the
isomers to illustrate your answer. [6 pts]
This is Ni2+ and 4-coordinate so the options are square planar or
tetrahedral structures (note it does not conduct electricity so it is
presumably an intact neutral complex as written). The question is,
what does the observation of optical isomers tell us about the
structure adopted? A square planar structure might be the first
choice but if the square plane is also a mirror plane no optical
isomers are possible. That will be the case if the ligands have a
plane of symmetry as is the case here. On the other hand, a
tetrahedral structure has no mirror plane and is chiral at the metal
so optical isomers are possible. Thus we conclude that this
complex adopts a tetrahedral structure.
superimposable
H2
N
Br
H2
N
Br
Ni
S
Ni
OH2
Ni
S
H2O
Br
Ni
S
geometric isomers
(dif f erent ligands
trans to NH2)
NH
S
H2
N
H2O
HN
Br
OH2
Br
H2O
Ni
S
not superimposable
5.
What is the relationship between the following pairs of isomers? [2
pts each]
(a)
[CrCl3(OH2)3] • 3 H2O and [Cr(OH2)6]3+ 3ClHydration or ionization isomers
(b)
[Cr(NH3)6]3+[Fe(CN)6]3- and [Fe(NH3)6]3+[Cr(CN)6]3Coordination isomers
Chem 324 Midterm 1 Fall 2011 Version 1
6.
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Consider the complex [Cr(OH2)6]3+.
(a)
Show what happens to the d orbital crystal field splitting
when the z axis water ligands are both replaced by NH3
ligands. Draw a labeled splitting diagram to illustrate your
answer. [6 pts]
NH3 is a stronger field ligand than water so replacement of
the z axis waters with NH3 will cause greater repulsion in the
z direction than in the (still equivalent) x and y. This splits the
eg set with dx2-y2 lower than dz2 and it also splits the t2g set
with dxy lower than either dxz or dyz:
(b)
Discuss what would happen to this structure if the metal
oxidation state was reduced by one electron. [3 pts]
Adding an electron gives [Cr(OH2)6]2+ which is a d4 ion and
subject to a Jahn-Teller distortion (either an axial
compression or elongation).
Chem 324 Midterm 1 Fall 2011 Version 1
(c)
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What is the magnitude of the crystal field stabilization
energy for this ion in terms of Δo? [2 pts]
Cr3+ d3 so CFSE = 3(-0.4Δo) = -1.2Δo
(d)
How does this compare with the crystal field stabilization
energy of [Mo(OH2)6]3+ in absolute energy terms (i.e. not
in terms of Δo)? [2 pts]
This ion will also have three electrons in the t2g level and a
relative CFSE of -1.2Δo, however Δo is much larger for the 2nd
row of the d block so the CFSE in absolute energy terms will be
much larger for Mo3+ than for Cr3+.
Chem 324 Midterm 1 Fall 2011 Version 1
7.
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Illustrate, using a molecular orbital approach, why a π-donor
like an alkoxide (-OR) decreases Δo in a d6 complex. [6 pts]
This question comes directly from section 6 of the notes. The key
point is that the π levels of the ligand are filled, low in energy and
of the correct symmetry to overlap with the t2g metal orbitals which
house the metal d electrons. Since the ligand π levels are filled,
whatever electrons there are from the metal will now fill the t2g*
level and this is closer to the eg* level meaning that Δo is less than
in the case of pure σ-donor ligands.
Chem 324 Midterm 1 Fall 2011 Version 1
(extra page for answers)
END
(Periodic Table follows)
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Chem 324 Midterm 1 Fall 2011 Version 1
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