Download Chemistry 324 Midterm 2 Name: KEY

Chem 324 Fall 09 Midterm 2 Ver. 1
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Chemistry 324
Midterm 2
Friday, November 20, 2009
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 AND Orgel
diagrams (p. 8) are included with this examination. There are 9 pages including the
periodic table.
1.
Sketch how the following sets of orbitals split in the specified crystal field.
[2 pts each]
a) p orbitals in an octahedral field
not split in an octahedral field (all axes equivalent)
b) p orbitals in a square planar field (square plane = xy plane)
px, py degenerate
pz lowest in energy
c) p orbitals in a distorted square planar field where the ligands on the xaxis are stretched relative to those on the y axis
py highest, px in the middle, pz lowest in energy
Chem 324 Fall 09 Midterm 2 Ver. 1
2.
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For each pair of complexes indicate which will have the larger d orbital
splitting and explain why in a few words. [2 pts each]
a) [Cr(en)3]3+ OR [CrCl6]3en complex because amines are higher in the spectrochemical series than
Clb) Cr(CO)6 OR W(CO)6
W is in the third row and ∆oct increases down the row
c) [CuCl4]2- OR trans-[CuCl4(H2O)2]2[CuCl4]2- is d9 and tetrahedral while trans-[CuCl4(H2O)2]2- is octahedral
and since ∆tet is 4/9 ∆oct the latter will have a larger d splitting
3.
Explain why solid CrF3 is octahedral at the Cr ion with a single Cr-F bond
length of 1.90 Å while MnF3 is distorted 6-coordinate at the Mn ion and has
three different Mn-F bond distances (1.79, 1.91 and 2.09 Å). [4 pts]
Mn3+ is a d4 Jahn-Teller ion while Cr3+ is a d3 ion. The Jahn-Teller ion will
distort in a way that removes the degeneracy of the partially filled (eg)
orbital. In this case, the distortion occurs in all three directions but the effect
is the same: the eg orbital pair splits into two different energy levels and the
lone electron resides in the lower of the two. Cr3+ has one electron in each
t2g orbital and no electrons in the eg set so there is no energetic advantage to
distortion in this case.
Chem 324 Fall 09 Midterm 2 Ver. 1
4.
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a) Determine the Russell-Saunders ground states for the following free
ions [2 pts each]:
(i) p4
ml
+1 0 -1
This is the highest S and L value arrangement so it will be the ground
state:
ML = Σml = 1 = P term
MS = Σms = 1; S = 1 and 2S+1 = 3 triplet
SO 3P term is the ground state
(ii) f6
ml
+3 +2 +1
0
-1
-2
-3
This is the highest S and L value arrangement so it will be the ground
state:
ML = Σml = 3 = F term
MS = Σms = 3; S = 3 and 2S+1 = 7 septet
SO 7F term is the ground state
b) What phenomenon is responsible for Russell-Saunders coupling? [2
pts]
electron-electron repulsion
c) What other phenomena lead to further splitting of the ground (and
excited) state terms in actual compounds? [2 pts]
1. crystal (or ligand) field effects
2. spin-orbit coupling
Chem 324 Fall 09 Midterm 2 Ver. 1
5.
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a) The complex [Cr(ox)3]3- (ox = bidentate C2O42-) shows two weak bands in
the visible at 23,600 and 17,400 cm-1 plus a third band in the UV at
40,000 cm-1. Assign the transitions responsible for each band; include
the spin multiplicity of all terms in your answer. [4 pts]
Cr3+ is d3, octahedral. The spin multiplicity will be 2S+1 where S= 3/2 (3
unpaired spins) to give a quartet (2S+1 = 4) state. The possible
transitions can simply be read off the Orgel diagram provided for this
ion and field. The ground state will therefore be 4A2g with transitions to
all three higher energy terms:
4
A2g → 4T2g
4
A2g → 4T1g
4
A2g → 4T1g
(from P)
17,400 cm-1
23,600 cm-1
40,000 cm-1
b) In view of the Laporte (symmetry) selection rule, explain why we see
these bands at all. [2 pts]
Transitions normally require a parity change (g→u or u→g) but since
all d derived terms in a centrosymmetric octahedral geometry are g, this
can’t happen and the d-d transitions should not be observed. However,
asymmetric vibrations result in instants where the ion is not octahedral
in symmetry and as a result, these transition do occur, albeit very weakly.
Chem 324 Fall 09 Midterm 2 Ver. 1
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c) Explain why the ferricyanide ion [Fe(CN)6]3- displays TWO INTENSE
absorptions, one in the visible and one in the UV while the ferrocyanide
ion [Fe(CN)6]4- only shows ONE INTENSE band in the UV. Be sure to
address the nature of the transition and to include a simple energy level
diagram to aid your explanation. [4 pts]
These CANNOT be d→d transitions because they are intense not weak.
They must therefore be charge transfer transitions of some type. Cyanide
is a stable anion and similar to halogens except that it has electron
density in the CN pi bond. Since N is electronegative, the orbital energy
of the CN pi system will be lower than the d orbital energies so a LMCT
transition is most probable. The Fe centre in these two ions is d5 in
ferricyanide but d6 in ferrocyanide. Since CN- is a strong field ligand,
these will be low spin complexes. If this is true, then transitions could
occur from the CN pi
Fe2+
Fe3+
orbitals to either the t2g or eg
set for ferricyanide but only
to the eg set for ferrocyanide.
The eg set is higher in energy
so the latter transition falls
in the UV but the transition
to t2g is low enough in energy
to fall in the visible. In
addition, the higher charge
on iron in ferricyanide
lowers the metal orbital
energy and this helps shift
the lower energy transition
further into the visible.
6.
a) Sketch a figure illustrating how χ varies with 1/T for a paramagnetic
substance. [1 pt]
b) Sketch a figure illustrating how μ varies with 1/T for a paramagnetic
substance. [1 pt]
(a)
(b)


1/T
1/T
Chem 324 Fall 09 Midterm 2 Ver. 1
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c) Calculate the spin-only magnetic moment for a tetrahedral d8 ion such as
[NiCl4]2- [2 pts]
s = 2[S(S+1)]1/2 S = 1 (two unpaired e-)
s = 2[1(1+1)]1/2 = 2.83 B.M.
d) The observed moment for [NiCl4]2- is 3.3 B.M. Explain why this moment
is significantly higher than the value predicted by the spin-only formula
making specific reference to the electronic structure of the metal ion. [3
pts]
This is due to spin-orbit coupling. The ORGEL diagram for d8
tetrahedral shows a T1 ground state. Degenerate ground states such as
these usually show significant orbital angular momentum contributions
to the magnetic moment resulting in an observed  value that is greater
than the spin-only moment.
Chem 324 Fall 09 Midterm 2 Ver. 1
7.
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Hg2+ forms very strong complexes with CN- ion. The first four log K values
are: 18.00, 16.70, 3.83 and 2.98.
a) Write the expression for β, the overall equilibrium constant, in terms of
[Hg2+], [CN-] and [{Hg(CN)4}2-]. [3 pts]
[{Hg(CN)4}2-] / [Hg2+][CN-]4
b) What is the overall logβ value for formation of {Hg(CN)4}2-? [1 pt]
log= logK1 +logK2 +logK3 + logK3 = 18.00 + 16.70 + 3.83 + 2.98 =
41.51
c) Provide a rationale for the sharp drop off going from logK2 to logK3. [2
pts]
After two CN- have been added, the complex [Hg(CN)2] is neutral, so the
electrostatic attraction of adding a third ligand is much diminished
relative to adding the first or second.
Hg2+ prefers a linear geometry so adding more than two ligands results
in a less favourable bonding geometry.
Steric effects are NOT important with only three or four small ligands
and a large metal ion like Hg2+.
Chem 324 Fall 09 Midterm 2 Ver. 1
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d) Despite the fact that CN- is sometimes regarded as a ‘pseudo-halogen’,
the logK1 and logK2 values for the equivalent complexation of Hg2+ by
Cl- are MUCH smaller than for CN-. Explain why this might be so. [3
pts]
This is just a reflection of the hard-soft acid-base theory which
states that soft metal ions like Hg2+ prefer to bond most strongly
with soft (polarizable) ligands such as CN- in preference to hard
ligands such as Cl-. Cyanide is a soft ligand mainly by virtue of
the fact that it has a pi bonding system and is capable of acting
as a pi acceptor.
END
ORGEL DIAGRAMS
Chem 324 Fall 09 Midterm 2 Ver. 1
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