NAME:________SOLUTIONS_______ Chem 1b, spring, 2007, 3rd. midterm
1) (35 points) The molecular formulae of three substances are given below. Predict the
3D structure of the complex portion of each substance. If isomers are possible, provide
the structures of all isomers. Also provide the IUPAC name of each compound.
Transition ratios: cubic(0.732)octahedral(0.414)tetrahedral(0.225)trigonal
a) Na[Co(ox)2(NH3)2] (ox = oxalate = C2O4-2)
sodium diamminedioxalatocobaltate(III)
The oxalate anion is bidentate and ammonia is monodentate so we have a
coordinartion number of 2(2) + 2(1) = 6. Octahedral complexation is expected.
Two geometrical isomers are possible: trans and cis. The trans isomer possesses a
plane of symmetry so it is not chiral. The cis isomer lacks a center and planes of
symmetry so it is chiral. That is, there is a pair of enantiomers (optical isomers)
that are non-superimposible mirror images.
O
O
O
O
O
O
H3N
Co
NH3
O
O
O
O
O
O
H3N
Co
O
O
O
NH3
O
achiral trans isomer
chiral cis isomer
The mirror image is a third isomer.
b) K[CrF4]
potassium tetrafluorochromate(III)
The fluoride anion is a monodentate ligand so a coordination number of 4(1) = 4
is observed here. Chromium(III) is a d3 case so loading electrons into antibonding
MO’s is not the reason why octahedral coordination is not observed here. Note
however, that only weak field ligands are used here and in these cases, tetrahedral
coordination is observed.
F
Cr
F
F
F
c) [Pd(PH3)2(CN)2] (PH3 = :PH3 = phosphine)
dicyanodiphosphinopalladium(II)
Palladium(II) is a d8 case so a square-planar coordination is preferred. Two
geometrical isomers-cis and trans-are possible. Both have a plane of symmetry
and therefore are not chiral.
PH3
N
C
Pd
PH3
C
PH3
N H3P
Pd
C
C
N
trans or E isomer
cis or Z isomer
N
2) (25 points) Consider the compound K[CrF4] which you examined in problem 1b.
a) Predict the number of unpaired electrons. Present an argument for your
answer. A MO diagram will be instructive.
The fluoride anion is a weak field ligand so the band gap between the 2 nonbondingI(e) and the antibonding (t2) MO’s is small. Cr(III) is a d3 case so the
three 3d electrons are distributed without pairing between the two non-bonding
MO’s and one of the antibonding MO’s. There are 3 unpaired electrons.
b) Predict the color of the solid.
Because of the small band gap, the substance will absorb light at the red end of
the spectrum and transmit light at the blue end. Tetrahedral complexes lack a
center of symmetry and a large molar extinction coefficient is expected. The
substance will be dark blue in color.
c) What would happen chemically if the compound were dissolved in water?
What observations would support your prediction?
Water is a stronger ligand than fluoride and its concentration is 55 M. Upon
dissolution, the chromium will be complexed by 6 water molecules. The complex
will be light colored (small extinction coefficient) as the resulting octahedral
complex has a center of symmetry. The NB-AB band gap will increase so the
wavelength of absorbed light will shift to the blue and therefore the wavelength of
transmitted light will shift to the red.
3) (10 points) Polyacrylamide is a polymer of acrylamide. This polymer can absorb large
amounts of water without dissolving. Provide the chemical basis for this interesting fact
which is utilized by bikers as a means of cooling the neck during a long ride in the desert.
O
H
H
N
H
H
H
acrylamide
Polymerization of acrylamide will occur via the radical addition mechanism. Except for
the ends, the polymer will have no carbon, carbon double bonds. Given its high
molecular weight it will be insoluble in most solvents. However, per monomer the
molecule contains one C=O with two lone pairs and one amino group with one lone pair
and two protons which can participate in hydrogen bonding. That is, per monomer the
molecule contains 3 hydrogen bond acceptors and two hydrogen bond donors. The
molecule is well designed to form extensive hydrogen bonding with water.
4) (30 points) The 3D structure of C60, buckminsterfullerene, is shown below.
a) Predict the crystal structure of solid C60.
The covalently bonded molecule is a perfect dodecahedron and has a spherical
shape. This is an obvious case for closest packing. Cubic closest packing was
reported by H.-B. Burg, E. Blanc, D. Schwarzenbach, S. Lu, M. M. Kappes, and
J. A. Ibers, Angew. Chem., Intl. Ed., (1992), 31, 640. Simply stating hexagonal
structure is not enough. All materials with hcp packing will belong to a
hexagonal space group. Not materials with hexagonal symmetry, e.g. beryl,
illustrate hcp packing. The closest packing is illustrated well by this STM
photograph of a monolayer of C60 deposited on a layer of bismuth. Source: Amin
Bannani, Christian Bobisch, Rolf Möller, Science, (2007) 315, 1824.
b) Would tert-butanol, t-C4H9OH, dissolve in solid C60? Explain. [The molecular
radii of C60 and tert-butanol are 5.16 Å and 2.86 Å, respectively.]
In the cp structure, the lattice of C60 molecules create octahedral and tetrahedral
holes. The ratio of molecular radii (2.86/5.16 = 0.554 > 0.414 so packing of the
tert-butanol in octahedral holes. The alcohol has a polar OH but it also has 3
hydrophobic methyl groups so there is no energetic reason to prevent its
dissolving in the C60.
c) The volume of one molecule of C60 is 576.7 Å3. Calculate the density of solid
C60 in g/cm3. NA = 6.022 x 1023 molecule/mole.
V(per mole) = V(per molecule) x NA =
= (576.7 Å3/molecule)(10-8 cm/Å)3(6.022 x 1023 molecule/mole)
= 347.3 cm3/mole
MW = (12.0 g/mole)(60) = 720 g/mole
D = M/V = (729 g/mole)/(347.3 cm3/mole) = 2.07 g/cm3
d) Which method could be used to validate the model made in part (a)? What
would be required in the way of sample preparation if this method were used?
X-ray diffraction is the method of choice, A good single crystal is needed.
e) What elements of symmetry are present in a molecule of C60?
For the cogniscenti, the molecule belongs to the Ih point group. The molecule
possesses a center of symmetry; 15 equivalent planes of symmetry; and two-,
three-, and five-fold rotational axes of symmetry.
f) Predict the physical properties, e.g. solubility, of C60.
In solid C60, one has discrete molecules with strong intramolecular bonds but
weak van der Waals forces between the molecules. It is a molecular solid. It is
expected to insoluble in water but soluble in hexane and have a relatively low mp
and bp. It will be easily scratched, e.g. a low hardness on the Moh scale.
Diamond, a polymorph, is the classic example of a network solid with a high
hardness, a high mp, and insolubility in all solvents.
C60, buckminsterfullerene