The periodic table of the elements is printed on the insert. You make use the
back for scratchwork but enter all work to be graded in the space provided
with each question.
1) (46 points) Provide the following information for each of the following
compounds: its IUPAC name, the structure of the portion of the compound that
is a complex, and the hybridization of the transition- metal atom.
If geometrical isomerism is possible, provide the structures of all the isomers.
a) Na[FeCl4]
name of the compound: sodium tetrachloroferrate(III)
hybridization of the iron atom: sp3 (tetrahedral complex)
4-Coordinate complexes in which ALL ligands are halide ions typically adopt a tetrahedral
geometry. Iron(III) is a d5 case so square planar is not expected.
b) Ni(CN)2F2
name of the compound: dicyanodifluoronickel(IV)
hybridization of the nickel atom: dsp2 (if square planar), sp3 (if tetrahedral)
Nickel(IV) is d6 so a square-planar complex is not guaranteed.
A tetrahdedral complex might be possible.
On the other hand, a square-planr complex might also occur. In that case,
both cis (Z) and trans (E) isomers are possible.
c) K3[Cr(C2O4)3].3H2O
name of the compound: potassium trioxalatochromate(III) trihydrate
hybridization of the chromium atom: d2sp3 (octahedral)
The oxalate ligand is bidentate so this is a hexacoordinate, octahedral complex.
The complex has neither a plane nor a center of symmetry so there will be a pair of
optical isomers, R and S.
2) (33 points) This problem deals with two binary compounds of vanadium and
fluorine: vanadium(II) fluoride and vanadium(V) fluoride.
a) Vanadium exhibits the oxidation numbers of +2 and +5 in these compounds.
Are these values consistent with the atomic structure of vanadium? Discuss.
From its position in the periodic table, vanadium has the electronic configuration
[Ar](3d)3(4s)2. Upon ionization, the two 4s electrons are easily lost leading to the
formation of V(II). Ionization of the three 3d electrons or their involvement in
covalent bonding leads to the upper bound for the oxidation number, +5.
b) The melting points of vanadium(II) fluoride and vanadium(V) fluoride are
1400 K and 293 K, respectively. Account for these striking differences.
The 4s electrons are easily ionized so ionic bonding is likely in vanadium(II) fluoride.
The very high melting point reflects the high lattice energy. Ionization of all 5
valence electrons has a prohibitively high energy cost so the bonding in vanadium(V) fluoride
is covalent. VF5 molecules with weak intermolelar forces exist in its crystal. A very
low melting point results.
c) Vanadium(II) fluoride is a blue substance and vanadium(V) fluoride is colorless.
Account for these striking differences.
The vanadium cation is present in an octahedral hole in the fluoride lattice.
V(II) has two d electrons and the transition from the t2g MO to the eg* MO involves a
low energy photon, i.e. one in the visible spectrum. V(V) has no d electrons so electrons
must be excited from the low energy bonding MO's to the high energy non-bonding MO's.
A high-energy UV photon is required.
d) Vanadium(II) is very reactive and its aqueous solutions are not stable. Explain why.
V(II) has a low oxidation number and is a strong reducing agent. It can reduce the water.
3) (18 points) This problem addresses the crystal structure of vanadium(III) fluoride
and related compounds. The values of useful ionic radii and transition ratios of ionic
radii are provided below:
ionic species: F- V+3 Fe +3 Al+3
ionic radius(Å): 1.20 0.78 0.69 0.50
transition ratio: 0.732 0.414 0.225 0.155
structure (coord. #): cubic (8)<...>octahedral (6)<...>tetrahedral (4)<...>triangular (3)<...>linear (2)
a) Predict the crystal structure of vanadium(III) fluoride.
Be complete but succinct in your description.
The significantly larger fluoride ion forms a closed packed, probably a cubic closed
packed lattice. The ratio of ionic radii, 0.78/1.20 = 0.65, lies between the cubic
and tetrahdedral cutoff, so the V(III) cation will fit nicely in the octahedral holes.
There is one octahedral hole per fluoride but the fluorides outnumber the V(III)'s by
a ratio of 3:1 so only a third of the octahedral holes are filled. The coordination
number (number of fluoride neighbors) is 6 (octahedral!). The coordination number of
the fluoride ion is 6/3 = 2.
b) A reviewer of vanadium chemistry noted that vanadium(III) fluoride is isostructural
with iron(III) fluoride but not with aluminum(III) fluoride.
(isostructural = similar crystal structure). Provide an explanation for this observation.
From their ionic radii, both V(III) and Fe(III) can fit into the octahedral holes in a
fluoride lattice and both would be expected to have an identical crystal lattice.
The aluminum cation is much small and the ratio of radii, 0.42, is on the boundary.
Al(III) cations presumably fill tetrahedral holes in aluminum(III) fluoride so its
crystal structure is quite different.
4) (18 points) A sample of vanadium(III) nitrate is dissolved in water yielding
a pale blue solution. A large excess of sodium thiocyanate (NaSCN) is added and
the color of the solution changes to pale red.
a) What is the formula of the vanadium complex in each of the aqueous solutions?
In the blue solution, water is the only ligand so the species is V(H2O)6+3. In the
red solution, the thiocyanate ions displace the waters. If all the waters are
displaced, the species would be V(SCN)6-3.
b) Which species, water or thiocyanate, is a stronger ligand?
Present an argument for your conclusion.
Thiocyanate is the stronger ligand. A blue color indicates that a relatively low-energy
photon is absorbed. In the case of the red solution, a relatively high-energy photon
is absorbed. The species that absorbs the more energetic photon exhibits the stronger
metal-ligand binding.
5) (41 points) An inorganic chemist synthesized a compound X from vanadium, chlorine,
and water and obtained the following analytical results:
i) 0.00100 mole of X was shown by spectroscopy to contain 0.00100 mole of vanadium.
ii) A chloride-specific electrode was used to show that the concentration of free chloride in a 0.0100 M solution of X was 0.0100 M.
iii) 0.00100 mole of X was dissolved in water and a large excess of silver nitrate was added. The solution was boiled for one day and 0.00300 mole of silver chloride was produced.
iv) 0.00100 mole of X was gently heated in an oven at 100ºC and lost 0.00200 mole of water.
v) 0.00100 mole of X was exhaustively dehydrated at 1000ºC and lost 0.00600 mole of water.
a) Determine the molecular formula of the compound X.
Per mole of X, 6 total waters, two are loosely held so 4 are ligands.
Per mole of X, 3 total chloride anions, one is loosely held so two are ligands.
[V(H2O)4Cl2]Cl.2H20
b) What is the IUPAC name of the compound?
tetraaquadichlorovanadium(III) chloride dihydrate
c) Draw the molecular structure of the portion of the compound that is a complex. If geometrical isomers are possible, draw the structures of all geometrical isomers consistent with the analytical data.
Six monodentate ligands are bonded to the V(III) so an octahedral complex is expected.
The complex will have cis (Z) and trans (E) isomers.
d) Draw a molecular-orbital energy-level diagram for the complex species whose structure was given in (c). Fill the energy levels with the appropriate number of electrons.
Refer to the handout on MO theory of complexes. The diagram for an octahedral complex applies. The 6 bonding MPO's are filled with 12 electrons (two per ligand). V(III) is d2 so the remaining two electrons go into two of the three t2g non-bonding MO's.
e) How many electrons occupy bonding molecular orbitals? Non-binding orbitals? Antibonding orbitals?
bonding: 12 (2 per ligand)
non-bonding: 2 (V(III) is d2)
anti-bonding: 0
f) Determine the number of unpaired electrons in the compound. n = 2 irrespective
of whether we have strong- or weak-field ligands.
g) The analytical data show that some of the water molecules in X are loosely held.
What is the nature of binding for these molecules?
Hydrogen bonding between the loosely held waters and the waters that are ligands
is a likely mechanism.
26 NOvember 1999