Download fourth midterm examination

NAME:_____________________________________ Chem 51, fall, 2005, 4th. midterm
1) (45 points) The molecular formulae of three substances are given below. Provide the
three-dimensional structure of the complex portion of each substance. If isomers are
possible, provide the structures of all isomers. Also provide the IUPAC name of the
compound and the hybridization of the central transition metal atom.
a) [Co(NH3)4(C2O4)]Cl
hybridization of the Co:__d2sp3_____
name:__tetraämmineoxalatocobalt(III) chloride____
d6 configuration, octahedral coordination. The plane of symmetry containing to Co,
the 2 C's, and the 4 O's rules out optical isomerism.
O
O NH
3
NH3
O
H3N NH
O
3
b) K2[CoCl2F2]
hybridization of the Co:___sp3_____
name:__potassium dichlorodifluorocobaltate(II)__
d7 configuration. Halides are monodentate and the coordination number is 4.
Tetracoordinate complexes with only halides or pseudohalides as ligands usually
have a tetrahedral geometry. No isomers.
F
Cl
F
Cl
c) [Pd(NO2)2(NO)2]
hybridization of the Pd:___dsp2____
(Pd, Z = 46)
name:___dinitrodinitrosylpalladium(II) Pd(II) is d8 so expect square planar.
O
O
N
O
O
O
O
O
N Pd N
O
N
O
trans or E
N
N Pd N O
N
O
O
cis or Z
2) (15 points) The picture below shows a single crystal of the mineral corundum, Al2O3.
What can be inferred from the picture about the crystal structure of Al2O3?
We expect closest packing of the oxide anions. The hexagonal shape of the crystal shows
that the oxide lattice exhibits hexagonal closest packing.
3) (15 points) Crystalline sulfur is soluble in organic solvents at melts at 115C. When
liquid sulfur is rapidly cooled, it converts into a rubbery, insoluble form which
decomposes before melting when heated. What can be inferred from these data about the
molecular structure of the rubbery allotrope (form) of sulfur?
The insolubility and high melting point indicates that the allotrope is a network solid, i.e.
a long network of sulfur-sulfur covalent bonds. The rubbery consistency suggests a
polymer. The chains are long but do not extend throughout the solid.
4) (15 points) The mineral rutile is a colorless oxide of titanium. Which spectroscopic
method can be used to determine the location of deposits of rutile on the surface of Mars
from satellite surveillance? Why is your method appropriate for this purpose?
Infrared spectroscopy is the method of choice. All molecules transmit and receive
radiation. Polyatomic titanium oxide will have at least one vibrational mode (actually
more than one) with a changing dipole moment and the substance will therefore absorb
and transmit infrared radiation at wavelengths characteristic of the material.
5) (xx points) Cotton and Wilkinson note in their authoritative treatise Advanced
Inorganic Chemistry that "the electronic structure of manganese(II) compounds is of
unusual interest as they exemplify important fundamental principles very clearly."
a) In their discussion of the chemistry of Mn(II) they report that "tetrahedral
complexes of Mn(II) are yellow-green and the coloration is more intense than that
of the octahedral complexes." What is the basis for the difference in the intensity?
Octahedral complexes possess a center of symmetry and a greatly reduced molar
extinction coefficient. Tetrahedral complexes lack a center of symmetry and have
normal, i.e. much larger, extinction coefficients.
b) They also report that "the  value for MnBr4-2 would be expected to be less than
that for Mn(H2O)6+2." Provide an argument for this conclusion.
We shall apply our model. A ligand functions as such by virtue of its lone pair.
Bromide, being the conjugate base of a strong acid, is expected to be a very weak
base and hence a very weak ligand. Bromide is a weaker ligand than water and
interacts less with the Mn(II). This results is a reduced destabilization of the eg
antibonding MO with a resulting reduction in the t2g-eg (HOMO-LUMO) gap. In
addition, with the same ligand, the value of  is less with tetrahedral complexes
since there are only 4 M-L bonds instead of 6.
c) The paramagnetic susceptibility and the value of magnetic dipole moment are
strikingly different for Mn(H2O)6+2 and MnBr4-2.
i) Predict the number of unpaired electrons in the two cases.
Mn(II) is d5. We have a strong field case with water as a ligand and a weak
field case with bromide as a ligand.
octahedral complex
tetrahedral complex
eg
____ ____
t2g __ __ ___
one unpaired electron
t2 ___ ___ ___
e
___ ____
5 unpaired electrons
ii) For each of the two cases, predict the number of electrons in bonding (B),
non-bonding (NB), and anti-bonding (AB) molecular orbitals.
Mn(H2O)6+2 B:__12___
MnBr4-2
B:__8___
NB:__5___
NB:__2___
AB:__0____
AB:__3____
iii) Compare the values of max in the two cases. That is, in which case is max
greater? Briefly provide the basis for your answer.
 is greater for the octahedral aqua complex so the frequency of the absorbed
light will be greater. Wavelength is inversely proportional to frequency so the
max will be great for the tetrahedral bromo complex. N.B. max refers to the
spectrum and hence the absorbed light.