Download 2011 Midterm 2 KEY

Chem 324 Midterm 2 Fall 2011
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Chemistry 324
Midterm 2 Version 1
Friday, November 18, 2011
Instructor: D. J. Berg
Name: KEY
Answer all questions on the paper (use the back if necessary). There are
50 minutes and 47 marks so ration your time accordingly. A periodic
table and ORGEL diagram are included with this examination. There are
10 pages including the additional resources.
1 (a)
Explain why the energy states of a d2 ion in an octahedral field are not
adequately represented by t2g2eg0 (at least as far as electronic spectroscopy
is concerned). [4 pts]
While the t2g2eg0 designation does take crystal field effects into account it
ignores electron-electron repulsion effects (Russell-Saunders splitting).
This is not reasonable for multi-electron first row transition metals where
the crystal field is weak or moderate (it is much better if the crystal field is
strong). This is particularly important in electronic spectroscopy because
the observed transitions are best understood in terms of transitions between
Russell-Saunders states split by a smaller crystal field effect.
Chem 324 Midterm 2 Fall 2011
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(b) What is the Russell-Saunders ground state term for an f8 ion? [3 pts]
S = 3 and L = 3 so this is a 7F term.
(c) What is the highest possible L value Russell-Saunders term for an f2 ion?
[2 pts]
Highest possible L value will have both electrons in the ml = +3 orbital so it
would be L = 6 (or an I term)
(d) Determine (i) the ground state of an f7 ion and (ii) how it splits in a
trigonal planar field. [4 pts]
(i) 8S
(ii) Not split since L = 0 (S term)
Chem 324 Midterm 2 Fall 2011
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2
The electronic spectrum of [VCl4(bipy)] shows a single asymmetric band
at 21,300 cm-1 (ca. 470 nm) of moderate intensity (ε < 800 L mol-1) with a
shoulder at lower energy (17,400 cm-1 or ca. 575 nm).
(a)
How many absorption bands are expected if this complex is regarded as a
perfect Oh? Show explicitly what gives rise to the band(s) observed. [3 pts]
The electronic transitions here are far too weak to be charge transfer
transitions, despite the fact that bipy often shows such transitions. This
means they must be d→d transitions of some type. This is an octahedral V4+
(d1) complex so we expect a 2D Russell-Saunders ground state term that will
be split into a lower energy 2T2g and a higher energy 2Eg state by the
octahedral field. There are no higher lying terms for a d1 ion so the only
possible transitions are from 2T2g to 2Eg giving rise to a single transition in
the UV-visible region.
(b)
Propose an explanation for the presence of the lower energy absorption.
(There are at least two possibilities here but one will do). [3 pts]
The lower energy transition must also arise from a d→d transition due to
its weak intensity. The most likely way to get more than one transition is if
the degeneracy of the 2T2g or 2Eg levels is somehow broken. There are a
number of ways this could happen but the two most likely explanations are:
1. the actual ground state symmetry of the complex in not Oh so this will
split the T2g state into an E and an A state. This immediately gives rise to
more bands in the spectrum.
2. the excited state effectively has a single electron in the eg set of
orbitals (those that are directed right at the ligands) so this should result in
a Jahn-Teller distortion in the excited state that breaks the Eg degeneracy
and gives rise to two bands.
Chem 324 Midterm 2 Fall 2011
(c)
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Reduction of this complex with sodium metal produces the intensely green,
but somewhat unstable, complex Na+ [VCl4(bipy)]-. Suggest explanations
for the observations that it is:
(i) intensely green in colour [2 pts]
The intense colour is strongly suggestive of a charge transfer
transition. Assuming reduction occurs at the metal, then the most
likely transition is from a metal-based orbital to a bipy π* level.
(ii)
somewhat unstable [1 pt]
Reduction puts a negative charge on the complex and since this is 6coordinate, loss of Cl- could occur to generate a neutral 5-coordinate
complex, VCl3(bipy). [I was looking for any reasonable explanation
here]
3
Consider a LMCT transition when answering the following questions:
(a) What characteristic distinguishes this type of transition from a d→d or f→f
transition? [2 pts]
High molar extinction coefficient (molar absorptivity); that is, intense
colours.
(b) Explain why MnO4- is intensely purple coloured while ReO4- is not highly
coloured. [2 pts]
These are both M7+ species and the intense colour of MnO4- suggests a
charge transfer band that must be LMCT. The question is really why ReO4is not very coloured. Presumably this must mean that its LMCT band falls
in the UV rather than in the visible. This makes sense because Re is less
electronegative than Mn and as such will have higher energy metal orbitals
so the gap between the O lone pair orbitals and metal orbitals is larger and
the transition occurs at shorter wavelength.
Chem 324 Midterm 2 Fall 2011
4.
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Ni2+ is unusual in that it can form octahedral, tetrahedral or square planar
complexes depending on the ligands involved.
(a) Calculate the spin-only moment (μs) for the tetrahedral complex. [2 pts]
If it is tetrahedral, it has 2 unpaired electrons in the t2 level so S = 1. This
means the spin-only moment is:
μs = 2[S(S+1)]1/2 = 2[2]1/2 = 2.83 BM
(b) The actual moments observed for these complexes are: octahedral, 2.9
BM; tetrahedral, 4.1 BM; square planar, 0.0 BM. Rationalize the
observed moments in terms of the electronic structure for EACH of the
three geometries. [6 pts]
From basic crystal field splitting, we expect Ni2+ to have 2 unpaired
electrons in either a Td or Oh geometry but no unpaired electrons in a
square planar geometry (because the highly destabilized dx2-y2 orbital is too
high to be occupied). In order to explain why there is a deviation from the
expected spin-only moments, we need to see whether there is likely to be
any contribution from orbital angular momentum. This requires us to look
more carefully at the ground state of each geometry taking both electron
repulsion and crystal field effects into account. From the Orgel diagram
we can see that a d8 ion has a T1 ground state in Td geometry but an A2g
ground state in an Oh geometry. Orbital angular momentum is expected to
be quenched for an A2g state so it is not surprising that the Oh complex is
far closer to the expected spin-only moment. However, a significant orbital
angular momentum contribution is expected for the degenerate T1 ground
state of the Td geometry and this should result in a significant increase in μ
compared to the spin-only moment prediction and this is in fact observed
here.
Chem 324 Midterm 2 Fall 2011
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(c) The values reported in (b) above have been corrected for the diamagnetic
susceptibility. Explain:
(i) what this contribution to the magnetic susceptibility of the complex is
due to [1 pt]
circulation of electrons in filled orbitals
(ii) its sign [1 pt]
negative
(iii) its magnitude relative to the paramagnetic susceptibility in the
octahedral or tetrahedral complexes. [1 pt]
diamagnetic susceptibilities are between 1/100th and 1/10th the size of
paramagnetic susceptibilities, depending on how many paired
electrons there are in the molecule.
Chem 324 Midterm 2 Fall 2011
5.
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(a) Write the expression for the overall stability constant β3 in terms of the
concentrations of the species shown in the following equilibrium. [3 pts]
PMe2
[M(H2O)6]2+ + 3
[M(dmpe)3]2+ + 6 H2O
PMe2
(dmpe)
β = [M(dmpe)32+] / [M(OH2)62+][dmpe]3
(b) Would you expect the enthalpy contribution to be favourable (∆Ho < 0)
or unfavourable (∆Ho > 0) for the equilibrium shown in (a) above if the
metal is cobalt? Explain your answer. [2 pts]
Cobalt in a relatively low oxidation state like 2+ is a fairly soft metal
and as such should prefer binding the softer phosphines of dmpe over
the harder O of H2O. I would therefore expect ΔH0 to be favourable for
the forward direction (< 0).
Chem 324 Midterm 2 Fall 2011
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(c) Would you expect the entropy contribution to be favourable (∆So > 0)
or unfavourable (∆So < 0) for the equilibrium shown in (a) above?
Explain your answer. [2 pts]
The entropy contribution should definitely be favourable (positive)
because the total number of particles goes from 4 to 7 in the forward
direction.
(d) Why does the lattice enthalpy become less favourable going from VCl2
to MnCl2, despite the fact that Mn2+ is smaller than V2+? [3 pts]
The lattice enthalpy would be expected to be more favouarble
(exothermic) for the smaller metal ion (Mn2+) based on charge/size
arguments. However, electronic stabilization also play a role in the total
enthalpy. V2+ is a d3 system with a total CFSE of -1.2Δo if it adopts an
octahedral lattice. On the other hand Mn2+ is d5 high spin and will have
a CFSE of 0 regardless of geometry. Thus, the CFSE effects outweigh
the changes in ion size and result in a more negative lattice energy for
VCl2 than for MnCl2.
END (Orgel diagram and periodic table attached)
General Orgel Diagram for Octahedral and Tetrahedral
Complexes with Various d Counts
Chem 324 Midterm 2 Fall 2011
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Chem 324 Midterm 2 Fall 2011
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