Download 2010 Midterm 2 KEY

Chem 324 Midterm 2
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Chemistry 324
Midterm 2 Version 1
Friday, November 19, 2010
Instructor: D. J. Berg
Name: KEY
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 the
periodic table.
1 (a)
What is the Russell-Saunders ground state term of an f 3 ion? [2 pts]
S = 3/2 so 2S+1 = 4 and ML = Σml = 6 therefore L = 6 or an I term.
This is therefore an 4I term.
(b) What is the total degeneracy of a 4F term? [2 pts]
Total degeneracy = total number of microstates = (2S+1)(2L+1) =
(4)(2x3+1) = 28
(c) Is it possible to have a 3G term arising from a d2 configuration? Explain.
[1 pt]
No, because a G term has L = 4 and the only way this can happen is to
have the two d electrons in the ml = +2 orbital and if this is true they
MUST be antiparallel in spin so S = 0 (a singlet state)
Chem 324 Midterm 2
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(d) Indicate the number of different energy levels and their degeneracy for
a 4F term in an octahedral field. [2 pts]
An octahedral field splits an F term into two triply degenerate 4T terms
and one 4A term (3 total levels).
(e) What must the ground state value of L be for any half filled set of
orbitals? Show how you arrived at this answer.[2 pts]
L = 0 (an S term) because ML = Σml = 0 if all ml states are singly occupied.
(f)
Determine the Russell-Saunders ground state term of a p4 ion. [2 pts]
The microstate with the highest values of ML and S will be (1+, 1-, 0+, -1+)
which has S = 1 and therefore 2S+1 = 3. The maximum value of ML arises
from the term with the same maximum value of L so this is ML = 1 and L =
1 (a P term). The ground state term is therefore a 3P.
(g) Determine how the ground state term for the ion in part (f) above splits in
a square pyramidal ligand environment (crystal field) and indicate the
number of electronic transitions you expect assuming there are no higher
energy terms of the same multiplicity as the ground state. [3 pts]
A P term will split in the same way as the p orbitals in a given symmetry.
In a SqPy geometry, the z-axis is unique but the x and y axes are
equivalent. We therefore expect the 3P term to split into a 3A and a 3E set of
states. The only possible electronic transition that does not violate the spin
rule is between these two levels since there is no higher energy term with
the same multiplicity in this case.
Chem 324 Midterm 2
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2
The isoelectronic species VO43- (amber), CrO42- (yellow-orange) and MnO4(purple) all show intense transitions in the visible at 445, 525 and 640 nm,
respectively.
(a)
Identify the type of electronic transition responsible for this absorption
band. [2 pts]
Must be LMCT as it is intense and the metal has no d electrons in all three
complexes (d0).
(b)
Rationalize the trend in wavelength using a schematic energy level
diagram to explain your argument. [4 pts]
The transitions involved here are from low energy filled oxygen lone pair
orbitals to the empty (since these are d0 ions) e orbitals of the tetrahedral
metal centre. Additional transitions are also possible to the t2 set but these
would occur at shorter wavelengths (most probably in the UV). The energy
of this transition follows the order Mn7+ < Cr6+ < V5+ (opposite order to
wavelengths since larger wavelengths are lower energy). This makes sense
because a higher charge on the metal centre results in lower energy for all
metal orbitals and this brings the empty metal orbitals closer to the oxygen
lone pair energy and decreases the energy of the LMCT transition.
V5+
e
Cr6+
e
Mn7+
e
empty metal e
filled O lone pairs
Chem 324 Midterm 2
3
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The compound trans-Fe(o-phen)2(NCS)2 has a magnetic moment of 5.0
BM at room temperature but it drops steadily as the temperature is
lowered so that it is only 0.65 by 80K and approaches 0.0 BM near 0K.
(a) Calculate the approximate number of unpaired electrons in this complex
at room temperature based on the spin only formula for the magnetic
moment. [2 pts]
Using µ = 2[S(S+1)]1/2, the best fit is for four unpaired spins (S = 2) which
gives a value of µ = 4.9 BM
(b) Assuming this system is magnetically dilute (there is a great separation
between molecules in the solid state), rationalize the observed change in
the magnetic moment with temperature. The complex also shows large
changes in its UV/Vis spectrum as a function of temperature.
HINT: consider what the moment is telling you about the number of
unpaired spins. [4 pts]
If the system is magnetically dilute it means that intermolecular
interactions such as ferromagnetic or antiferromagnetic ordering can be
ignored. The moments observed indicate that there are 4 unpaired
electrons at room temperature but this decreases to zero unpaired electrons
at near 0K (that is, it becomes diamagnetic). This is an Fe2+ system which
makes it d6. We can rationalize the observed spins if it is high spin
octahedral at room temperature but switches to low spin octahedral at low
T. We conclude that the ligand set here is finely balanced between high and
low spin such that thermal changes are enough to cause a change in spin
state. (This is an example of what is called a spin-crossover system). The
change in UV/Vis is completely consistent with a change in the electronic
ground state of a single molecule and inconsistent with any long range
magnetic ordering such as antiferromagnetic coupling.
Chem 324 Midterm 2
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(c) Illustrate the temperature dependence of χ for a paramagnetic and a
ferromagnetic substance. Include a χ vs. T plot as part of your answer. [3
pts]
This is the complete figure given in the notes; this question just required
paramagnetic and ferromagnetic behaviour.

Tc
TN
ferromagnetic
paramagnetic
anti-ferromagnetic
diamagnetic
T
(d) What happens if you heat a ferromagnet above its Curie temperature in
the absence of a magnetic field? [1 pt]
It demagnetizes and becomes a simple paramagnetic material with no
permanent field.
Chem 324 Midterm 2
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(a) Define the ‘chelate effect’ and discuss the thermodynamic factors
responsible for it. [4 pts]
Chelate effect: observation that chelating ligands form more stable
complexes (higher β) than do an equivalent set of similar monodentate
ligands.
The chelate effect is entropically driven: consider displacement of water by
a chelate ligand and a set of monodentate ligands. In the case of a
bidentate chelate ligand, 6 waters are displaced by 3 chelates for a net
increase in entropy. In the case of a monodentate donor, 6 waters are
displaced by 6 donors for no net entropy change. If we assume that the
donor properties of the monodentate ligand and the chelate donors are
essentially the same, then ΔHo is the same in both cases because it depends
only on the number of donors and their bond strengths.
M(H2O)6 + 3 en → M(en)3 + 6 H2O ΔSo > 0
M(H2O)6 + 6 NH3 → M(NH3)6 + 6 H2O ΔSo = 0
(b)
Write the mathematical equation describing the formation constant K of
[Co(EDTA)]2- from EDTA4- and [Co(H2O)6]2+ in water. [3 pts]
(Note:
H4EDTA
is
ethylenediamine
tetraacetic
acid,
(HO2CCH2)2NCH2CH2N(CH2CO2H)2)
K (or β) = [Co(EDTA)2-] / [EDTA4-][Co(H2O)62+]
Chem 324 Midterm 2
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(c) Explain why the value of log K for formation of M(EDTA)n- varies with
metal ion M in the following order: Co2+ (logK = 16.1) < V3+ (25.9) < Co3+
(36.0). [3 pts]
Higher charge leads to greater binding so we expect Co3+ to be more
strongly bound than Co2+.
Co3+ is smaller than V 3+ so we expect it to bind more strongly because it
has a greater charge/size (Q/r) ratio.
5 (a)
Which high spin dn counts display the greatest crystal field stabilization
energy for the first row transition metals (i.e., which are most stabilized)
and how large is this value in terms of Δo? [3 pts]
The greatest CFSE occur at d3 and d8 counts and in both cases the
stabilization is -1.2Δo
Chem 324 Midterm 2
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(b) How would you expect the water exchange rate to compare between the
octahedral, high spin d5 ions [Mn(H2O)6]2+ and [Fe(H2O)6]3+? [2 pts]
Both ions have NO electronic effects (CFSE = 0) so this exchange rate will
depend on charge/size (Q/r) ratio. Since Mn2+ has a smaller charge than
Fe3+, it will have the weakest binding to water and exchange the fastest.
(c) Distinguish between an Ia and an A mechanism of ligand exchange. Use a
diagram to explain your answer if that helps. [3 pts]
Ia is concerted and goes through a higher coordinate transition state.
A is step-wise and goes through a higher coordination number
intermediate. Suitable diagrams to illustrate this are given in Section 15 of
the notes.
(d) How much is the rate of water substitution by an incoming ligand L
influenced by the concentration of L for typical first row, d-block
octahedral [M(H2O)6]2+ complexes? Explain your rationale. [2 pts]
Typical first row octahedral metals undergo dissociative substitution of
water (usually by an Id mechanism) so the rate should not be influenced by
the concentration of the incoming ligand L.
END (periodic table attached)
Chem 324 Midterm 2
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