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Chem 64 Winter 2003 Exam2 -- Brief Solutions
1. a. OC-B(CF3)3 is expected to be more thermally stable. The C6F5 group is much
bigger than CF3, so there will be more steric hindrance in this adduct (it will be
unfavorable to pack 3 big C6F5 groups around a pyramidal B in the adduct).
b. Hard-soft. The soft Au(I) ion preferentially binds the softer halides in the order
Cl<Br<I.
2. a. Consider the octahedral complexes. Ni(II) is d8 with electron configuration
(t2g)6(eg)2. Cu(II) is d9 with (t2g)6(eg)3. The latter has a degeneracy in the eg MO which
is predicted to result in a structurally significant Jahn-Teller distortion. Hence, Nioctahedral and Cu-distorted.
b. Sterics; the longer W-P bond is needed to reduce ligand-ligand repulsion with the
bulkier P(t-Bu)3. Apparently this is even worse for the smaller Cr center (consistent with
a steric argument) and explains the non-existence of that complex. An additional
problem could be the generally weaker Cr-P (vs W-P) bonds, so that even a long Cr-P
bond might be too weak to be profitable.
c. Since 1 does not conduct electricity, it must be neutral. A possible structure is 5coordinate, as shown (it can’t be a 4-coordinate cation with chloride anion, or other ionic
forms). To make it approximately C2v, put the chlorides equatorial. Salt 2 has Td anion
[NiCl4]2– and approximately C2v cation [Ni(PNP)Cl][Cl]. These structures plus a balanced
equation are shown below. Note: this was the intended answer, but some students
found creative ways to get other answers that also made sense.
3 Ni(PNP)Cl2
[Ni(PNP)Cl] 2[NiCl 4] + PNP
complex 1
Me
PPh2
N Ni Cl
Cl
PPh2
salt 2
+
anion
cation
PPh2
Me
N Ni
Cl
Cl
2–
Cl Ni Cl
Cl
PPh2
2
3. a. Across the row, Zeff of the metal ion is increasing, which pulls the ligands in
closer to the metal, thus decreasing M-L bond length, or covalent radius.
Similarly, the Zeff for M(III) is bigger than for M(II), with the same trends.
Note in both cases that the d-electrons are placed in the essentially nonbonding t2g MO
and therefore have little structural effect.
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b. Complex A is octahedral, d8, (t2g)6(eg)2 with 2 unpaired electrons, hence paramagnetic.
Complex B is square planar with the electron configuration shown, hence diamagnetic.
t-Bu
t-Bu
A
H2O
O
O
Ni
O
O
OH2
t-Bu
t-Bu
eg
t2g
t-Bu
t-Bu
B
t-Bu
O
O
Ni
O
O
t-Bu
d x2 –y2
d xy
d z2
d xz, dyz
4. As described in class, this double-humped curve reflects the CFSE of the metal ions.
To see what the curve for low-spin ions would look like, calculate the CFSE for these
[this was partially done in class, but you could do it here: CFSE = (# of t2g electrons)(
2/5Δo) – (# of eg electrons)(3/5Δo) where I have used the convention that CFSE is
positive]. Results below, and shown on the graph (not to scale -- I have not tried to
make this quantitatively accurate!)
Qualitatively, the important points are results are the same for d0 to d3 and d8 to d10, but
higher for low-spin ions for d4 to d7.
CFSE
# of d electrons
0
1
2
3
4
5
6
7
8
9
10
CFSE (high-spin)
0
2/5 Δo
4/5 Δo
6/5 Δo
3/5 Δo
0
2/5 Δo
4/5 Δo
6/5 Δo
3/5 Δo
0
CFSE (low-spin)
0
2/5 Δo
4/5 Δo
6/5 Δo
8/5 Δo
10/5 Δo
12/5 Δo
9/5 Δo
6/5 Δo
3/5 Δo
0
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5. a. Here's the MO diagram for [Re2Cl8]2– and related compounds with M-M multiple
bonds, as cut-n-pasted from the lecture notes. [Tc2Cl8]2– is isoelectronic, with two Tc(III)
d4 centers yielding a quadruple bond (bond order 4). Adding one more electron as in
[Re2Cl8]3– will occupy the antibonding δ* MO, reducing the bond order to 3.5.
complex
[Tc2Cl8]3–
[Tc2Cl8]2–
Tc-Tc bond order
3.5
4
σ∗
π*
δ*
Tc
Tc
4
d4
d
δ
π
σ
b. This should be surprising, since usually higher bond order correlates with
shorter bond length.
Some possible explanations: (a) In [Tc2Cl8]2–, it’s formally Tc(III). Adding an electron to
form [Tc2Cl8]3– makes it formally Tc(2.5). The Tc-Tc bond length may be controlled by
two important factors: bond order and Coulombic repulsion between the formally
positively charged Tc ions. The bond order is clearly greater in the 2- ion, but not by
much; it also involves a relatively weak delta-bond. It’s possible that the higher formal
charge on Tc in the 2- ion leads to increased Coulombic Tc-Tc repulsion, which
outweighs the greater bond order and lengthens the bond.
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(b) Or, this change in oxidation state contracts the d-orbitals, which leads to poorer
overlap and a longer bond.
6. See the diagram below for the symmetry of the metal orbitals, straight from the
character table except for s, which as usual is not included in the table but is the first
row.
To make the MO scheme, note that there are 3 sets of A1' symmetry orbitals which will
yield 3 MO's; ditto for the E' set. In these cases the 3 MO's include bonding,
antibonding and "in-between"; I have labeled the latter as weakly antibonding but their
true nature is debatable and will depend on the relative energies of the M and L orbitals.
As long as you have them in the middle, that's the important part. The metal A2'' and E''
orbitals have no symmetry partners on the ligands so they remain nonbonding.
The 3 ligand orbitals contribute 6 electrons, which fill up the 2 lowest-energy bonding
MO's -- the next 10 will go into the box with the frontier MOs. Note that their relative
energies are not necessarily as drawn.
A1 '
E' antibonding
A2''
E'
p
A1 '
s
A2'' nonbonding (nb)
A1 '
E'
Energy
A1' E'
E''
d
nb
E''
weakly
antibonding
A1 '
E'
E' bonding
A1 '
M
ML3
L3
ligand
SALC's
5
7. a. This is Fe(III), d5, so it could be either high-spin, (t2g)3(eg)2 with 5 unpaired electrons,
or low-spin, (t2g)5, with one unpaired electron. The magnetic moments would then be
[n(n+2)]0.5 BM or 5.9 BM vs 1.7 BM. Since experiment says 5.8 BM, it must be highspin.
b. i. The change in magnetic moment is due to spin crossover; some of the complex is
becoming low-spin (which presumably involves other structural and electronic changes).
Not all, or the moment would be 1.7 BM. How much? Let x be the amount high-spin;
then 1-x is the amount low-spin, and the overall moment is a weighted average:
x(5.8 BM) + (1-x)(1.7 BM) = 4.04 BM
5.8x + 1.7 -1.7x = 4.04
4.1x = 2.34, so x = 0.57.
It's 57% high-spin and 43% low-spin at the low temperature.
ii. Changing to low-spin must affect Δo as the M-L bond lengths change; its magnitude
affects the color.
8. a. Co(II) is d7. The octahedral complex could be high-spin or low-spin; pa is
approximately like NH3 or en which are roughly in the middle of the series so this is a
tough call; assuming that since Co is a 1st-row metal it's high spin we get (t2g)5(eg)2 with
CFSE of 4/5Δo (or 9/5Δo for low-spin, see also #4 above). The tetrahedral complex, like
almost all the others, must be high-spin, (e)4(t2)3, with CFSE of 6/5Δt.
b. Δo is about twice as large as Δt, so the octahedral complex has larger CFSE and
should be thermodynamically favored.
c. Despite this prediction, the tetrahedral complex is favored. Why? This must be due to
entropic effects -- as you can see in the equilibrium one octahedral particle is making
one tetrahedral one plus 2 free amines; the increased disorder favors the right-hand side
of this equation and apparently overcomes the improved CFSE for the octahedral
complex.
d. With tn, which presumably acts as a chelate, the equilibrium is now
[Co(tn) 3]2+
[Co(tn) 2]2+ + tn
In comparison to pa it's now one particle going to two instead of three, so this is not
quite as favored entropically. You might also notice that the tn bite angle is a bit smaller
than the ideal tetrahedral angle, so the tetrahedral complex is likely to be destabilized.