CHEM 3400 - Modern Inorganic Chemistry Exam
NAME :
Page 1
100 points
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 H
2
-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; Ph
3
P ligand = triphenylphosphine):
(SP-4)-chlorotris(triphenylphosphine)rhodium(I); not chiral

CHEM 3400 - Modern Inorganic Chemistry Exam
NAME :
2.
[26 pts] For the octahedral coordination complex, [Co(phen)
2
(CN)
2
] 1+ :
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 ' d n , x
unpaired electrons').
Co(III) = 3 d
6 ; 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
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Page 3
100 points
Continuation of problem 2c

CHEM 3400 - Modern Inorganic Chemistry Exam
Page 4
NAME :
3.
[10 pts] One mechanism for obtaining Fe 3+
100 points 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 Fe 3+ of Fe 3+ and ent is 10 52 (the largest known K f shown below; green sphere is Fe 3+
[Fe(OH
2
)
6
] 3+ (ii) Based on
K f
for Fe 3+
. The formation constant (
K f
) for the reaction
with a naturally occurring substance; structure
). (i) Write the expression for
K f
for the reaction of ent with
, do you expect a favorable chelation reaction? Why or why not?
[Fe(ent)]
3complex
(i) Equation of coordination reaction (assuming ent is fully deprotonated):
[Fe(OH
2
)
6
] 3+ + ent 6↔ [Fe(ent)] 3 + 6 H
2
O (eq. 1)
Expression of the formation constant ( K f
):
K f
= [{Fe(ent)}
3-
][H
2
O]
6
/[{Fe(OH
2
)
6
}
3+
][ent
3-
] (eq. 2)
(ii) The high K f
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 Exam
NAME :
Page 5
100 points
4.
[16 pts] Answer the following questions about ligands:
(a) [9 pts] Place the following ligands in the appropriate category below: Cl ‒ ethylenediamine, NO
2
‒ , NH
3
, OH ‒ , Et
3
N, CO, en =
, 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, NO
2
‒
, trpy

 -donor only: NH
3
, en, Et
3
N
(b) [3 pts] Order the ligands you selected in (a) from low to high ligand field strength (  o
)


< 
< 

(c) [4 pts] Which ligand class (from 4a) results in a destabilization of the t
2g
orbitals in an octahedral coordination complex? What spin-state, i.e., high- or low-spin, is favored for this ligand class?


t
2g

CHEM 3400 - Modern Inorganic Chemistry Exam
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Page 6
100 points
5.
[10 pts; 2 pts/ea] State the d
electron configuration and number of unpaired electrons for the following complexes (e.g.,: ' d n , x unpaired electrons')
(a) [Fe(CN)
6
] 3–
Fe(III) = 3 d
5 ; 1 upe (high oxidation state metal, strong field ligands)
(b) [PtCl
6
] 2 
Pt(IV) = 5 d
6 ; 0 upe (high oxidation state metal, 3 rd row TM)
(c) [MnO
4
] 1 
Mn(VII) = 3 d
0 ; 0 upe
(d) [Cr(OH
2
)
5
Cl] 1+
Cr(II) = 3 d
4 ; 4 upe (weak field ligands, low oxidation state metal)
(e) [Au(H
2
O)
6
] 2+
Au(II) = 5 d
9 ; 1 upe
6.
[4 pts] Which one of the octahedral complexes in question 5 will exhibit a Jahn-Teller distortion?
Cr(II) = 3 d
4 and Au(II) = 5 d
9 ; odd number of electrons in e distortion and lowering of energy g
level results in
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
NAME :
Page 7
100 points
8.
[10 pts] For each of the two geometries calculated in question 7, calculate the energies in terms of e
 of a d
8 complex, assuming that the pairing energy, P = 0.5 e

(low-spin case). Predict which geometry is preferred for d
8 .
The square-planar ( D
4h
) geometry is more stable by 2 e

for a d
8 transition metal ion.

CHEM 3400 - Modern Inorganic Chemistry Exam
NAME :
Page 8
100 points
9.
[5 pts EXTRA CREDIT ] Diamagnetic trans
-[NiBr
2
(PEtPh
2
)
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 T unpaired electrons ( S =1) for the Ni(II) = 3 d
8 d
ion (see below).
symmetry result in 2
Br
R
3
P
Br
Ni
Br
PR
3
R
3
P
Ni
Br
PR
3 x
2
-y
2 xy xz yz xy z
2 xz yz
D
4h
x
2
-y
2 z
2
T d
END OF EXAM