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CHEM 3400 - Modern Inorganic Chemistry
Exam #3
Page 1
100 points
NAME:
Please record your answers on these sheets in the space directly below each question. Please read and
answer all questions carefully. Partial credit will be considered, show all work (no work = 0 credit) on
these sheets only with answers listed below the question.
1. [12 pts] (i) Draw the structures or give the name of the following coordination complexes and (ii)
state if they are chiral:
(a) OC-6--tris(bipyridine)ruthenium(II) (a photosensitizer used in fuel cells). You may use a
curved line to represent the N-N link of the bidentate 2,2'-bipyridine chelate (see below):
chiral
(b) tri--carbonylbis(tricarbonyliron(0)) [carbonyl = carbon monoxide or CO] (a model compound
of the H2-evolving enzyme hydrogenase):
+
not chiral
(c) The anti-hypertension drug nitroprusside (NO ligand = nitrosyl):
(OC-6-22)-pentacyanonitrosylferrate(II); not chiral
(d) Wilkinson’s catalyst (complex is square-planar; Ph3P ligand = triphenylphosphine):
(SP-4)-chlorotris(triphenylphosphine)rhodium(I); not chiral
CHEM 3400 - Modern Inorganic Chemistry
NAME:
2. [26 pts] For the octahedral coordination complex, [Co(phen)2(CN)2]1+:
Exam #3
Page 2
100 points
(a) [4 pts] Provide its chemical name (OC-6-33-prefix not needed here but will be required in 1c
below).
dicyanobis(phenanthroline)cobalt(III)
(b) [4 pts] Give the d electron configuration and the number of unpaired electrons for the Co center
(your answer should look like 'dn, x unpaired electrons').
Co(III) = 3d6; 0 upe (strong-field ligands, high oxidation state metal)
(c) [18 pts] (i) Draw and (ii) formally name all possible configurational (stereo) and structural
(constitutional) isomers. List each structure with the appropriate prefix (e.g. OC-6-33 since this
is the only variable in the formal naming of this system) and write the prefix at the bottom of
each isomer. The formal name is not needed as it was listed in 1a (above). [Hint: There are 3
possible isomers (including one pair of enantiomers)].
Only 2L and 2M survive – 2M contains enantiomers
CHEM 3400 - Modern Inorganic Chemistry
Exam #3
Page 3
100 points
NAME:
Continuation of problem 2c
1+
N
N
Co
NC
N
CN
N
CHEM 3400 - Modern Inorganic Chemistry
Exam #3
Page 4
NAME:
100 points
3+
3. [10 pts] One mechanism for obtaining Fe from the biosphere is through chelation. The macrocyclic
ligand enterobactin or ent (shown below; O atoms in blue coordinate to Fe) is one such chelator
employed by bacteria and it has a high affinity for Fe3+. The formation constant (Kf) for the reaction
of Fe3+ and ent is 1052 (the largest known Kf for Fe3+ with a naturally occurring substance; structure
shown below; green sphere is Fe3+). (i) Write the expression for Kf for the reaction of ent with
[Fe(OH2)6]3+ (ii) Based on Kf, do you expect a favorable chelation reaction? Why or why not?
[Fe(ent)]3- complex
(i) Equation of coordination reaction (assuming ent is fully deprotonated):
[Fe(OH2)6]3+ + ent6- ↔ [Fe(ent)]3- + 6 H2O (eq. 1)
Expression of the formation constant (Kf):
Kf = [{Fe(ent)}3-][H2O]6/[{Fe(OH2)6}3+][ent3-] (eq. 2)
(ii) The high Kf suggests that the equilibrium lies almost exclusively to the right (see
eq. 1 above) and thus a very stable coordination complex. This stability is due to
the chelate effect (increase in S; 2 mol reactants versus 7 mol products)
CHEM 3400 - Modern Inorganic Chemistry
NAME:
4. [16 pts] Answer the following questions about ligands:
Exam #3
Page 5
100 points
(a) [9 pts] Place the following ligands in the appropriate category below: Cl‒, OH‒, Et3N, CO, en =
ethylenediamine, NO2‒, NH3, trpy = 2,2':6',2"-terpyridine (see board for structure), I‒.
[Note: each ligand belongs to only one class; placement of the same ligand in multiple categories
will result in negative points]
‒
‒
- & -donor: Cl , OH , I
‒
‒
-donor & -acceptor: CO, NO2 , trpy

-donor only: NH3, en, Et3N
(b) [3 pts] Order the ligands you selected in (a) from low to high ligand field strength (o)
-donor/-donor < -donor < -donor/-acceptor
(c) [4 pts] Which ligand class (from 4a) results in a destabilization of the t2g orbitals in an octahedral
coordination complex? What spin-state, i.e., high- or low-spin, is favored for this ligand class?
-donor/-donor destabilize t2g
high-spin is favored
CHEM 3400 - Modern Inorganic Chemistry
Exam #3
Page 6
NAME:
100 points
5. [10 pts; 2 pts/ea] State the d electron configuration and number of unpaired electrons for the
following complexes (e.g.,: 'dn, x unpaired electrons')
(a) [Fe(CN)6]3–
Fe(III) = 3d5; 1 upe (high oxidation state metal, strong field ligands)
(b) [PtCl6]2
Pt(IV) = 5d6; 0 upe (high oxidation state metal, 3rd row TM)
(c) [MnO4]1
Mn(VII) = 3d0; 0 upe
(d) [Cr(OH2)5Cl]1+
Cr(II) = 3d4; 4 upe (weak field ligands, low oxidation state metal)
(e) [Au(H2O)6]2+
Au(II) = 5d9; 1 upe
6. [4 pts] Which one of the octahedral complexes in question 5 will exhibit a Jahn-Teller distortion?
Cr(II) = 3d4 and Au(II) = 5d9; odd number of electrons in eg level results in
distortion and lowering of energy
7. [12 pts] Use the angular overlap model to calculate the energies of all d orbitals for a squarepyramidal and square-planar coordination complex with -only donor ligands. Draw the energy
level diagram, label the d orbitals and give the energies (in units of e).
CHEM 3400 - Modern Inorganic Chemistry
Exam #3
Page 7
NAME:
100 points
8. [10 pts] For each of the two geometries calculated in question 7, calculate the energies in terms of e
of a d8 complex, assuming that the pairing energy, P = 0.5 e (low-spin case). Predict which
geometry is preferred for d8.
The square-planar (D4h) geometry is more stable by 2 e for a d8 transition metal ion.
CHEM 3400 - Modern Inorganic Chemistry
Exam #3
Page 8
NAME:
100 points
9. [5 pts EXTRA CREDIT] Diamagnetic trans-[NiBr2(PEtPh2)2] converts to a form which is
paramagnetic. This change in magnetism is also accompanied by a color change from orange to blue.
Suggest a reason for this observation and state how many unpaired electrons you would expect in the
paramagnetic form.
The change in color and spin-state is due to conversion from square-planar to
tetrahedral geometry. The arrangement of the d-orbitals in Td symmetry result in 2
unpaired electrons (S =1) for the Ni(II) = 3d8 ion (see below).
Br
R3P
Br
Ni
Br
PR3
Ni
R3P
PR3
Br
x2-y2
xy
xz
x2-y2
xy
yz
z2
(tetrahedral = Td)
2
z
xz
yz
(square-planar = D4h)
END OF EXAM
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