USEFUL DATA USED IN ANSWERING THE QUESTIONS

ion: Li+ Na+ K+ Rb+ Cs+ Cr+3 Cl-

ionic radius, r(Å): 0.80 0.95 1.33 1.48 1.69 0.53 1.81

NAME:___ANSWER KEY__ Chem 51, 4th. Midterm Exam, Fall, 2002

An insert with a periodic table and ionic radii are included with the examination questions. You may use the back for scratch work but enter ALL work to be graded in the space provided with each question.

1) (50 points) For each of the following compounds, give its IUPAC chemical name and the hybridization of the transition-metal atom. Draw the structure of the complex portion of each. If geometrical and optical isomers are possible, draw the structure of all isomers.

The structures are found on Figure 1.

a) Na[Cr(H2O)2(CN)2(C2O4)]

name: sodium diäquadicyanoöxalatochromate(III)

There are seven d electrons and octahedral coordination is expected. Note that oxalate is a bidentate ligand. Structures (i) and (ii) have a plane of symmetry and therefore are not chiral, i.e. they do not have optical isomers. Structures (iii) and (iv) are non-superimposable mirror images and hence constitute a pair of optical isomers.

b) [Ni (NH3)2 F2]

name: diämminedifluoronickel(II)

Nickel(II) has eight d electrons so square-planar coordination is expected with a hybridization scheme of dsp2.

c) Which of the two substances would be more suitable for crystallization from an organic solvent such as ethyl alcohol? Briefly explain.

Compound 1a which has a sodium cation and a complex anion exhibits ionic bonding in addition to covalency. It will not be very soluble in typical organic solvents and crystallization if possible without destroying the compound would have to do conducted in water. Compound 1b on the other hand exhibits solely covalent bonding and would therefore show solubility in ethanol.

2) (55 points) The number of unpaired electrons in the transition-metal complex hexaämminecobalt(II) chloride was found from a measurement of its paramagnetic susceptibility to be three. Answer the following questions about this compound.

a) Draw the molecular orbital (MO) diagram for the complex with the correct number of electrons. Show just the occupied energy levels. Label each energy level as bonding, non-bonding, or antibonding.

Cobalt(II) is a d7 case, i.e. seven d electrons. We have a total of 12+7 = 19 electrons with 12 electrons coming from the 6 ligands. 12 electrons fill the lowermost bonding molecular orbitals. The next 7 are used to fill the eg* and t2g molecular orbitals in a manner consistent with the the experimental datum: 3 unpaired electrons.

_· _ _· _ antibonding eg* orbitals

· · · · · · non-bonding t2g orbitals

 

 

 

 

· · set of 6

· · · · · · bonding

· · · · orbitals

 

b) Compared with other complexes, is the metal-ligand bond in the cobalt-ammonia complex relatively strong or weak? Explain.

We have a high spin case which results from a energy small gap between the eg* and t2g orbitals. This is a consequence of a weak ligand-metal interaction so the bond is relatively weak. The bond is further weakened by placing two electrons in antibonding molecular orbitals.

c) Air must be excluded in the preparation of the complex as the cobalt is easily oxidized from cobalt(II) to cobalt(III). Provide an explanation for the ease of oxidation.

It was stated in class that transition metals to the left are easily oxidized from the +2 to the +3 state. This fact reflects the relatively low effective nuclear charge. This state of affairs is aided by having two electrons in antibonding molecular orbitals. The electron that is removed would come from the eg* orbital.

d) Will the complex be lightly or darkly colored? Explain.

Lightly colored. Octahedral complexes in which all ligands are identical possess a center of symmetry and a greatly reduced extinction coefficient.

e) Suppose that the substance is heated in an oven until all the ammonia is driven off. Would the color of the resulting product be different? Briefly discuss.

With the removal of all the ammonia ligands, the chloride anions become by default the ligand. The halide-metal interaction is expected to be weaker than the ammonia-metal interaction. The absorption spectrum will shift to the red end of the spectrum and the transmitted light to the blue. The cobalt(II) ions are small and would be expected to fill tetrahedral holes in a cp lattice of chloride ions. Hence, each cobalt would be surrounded by 4 chloride anions. Tetrahedra lack a center of symmetry and a great increase in the extinction coefficient would be expected.

3) (20 points) Most alkali halides have the same crystal structure as NaCl but there are some surprises. In the case of the chloride salts, LiCl, NaCl, and KCl exhibit one type of structure and CsCl and RbCl a second. Provide an explanation for this anomaly

If one calculates the ratio of the cation ionic radius to the anion ionic radius, one discovers that the ratio in the cases of LiCl, NaCl, and KCl is less than the 0.73 transition between cubic coordination and octahedral coordination. Cesium and rubidium cations are larger and the ratio in this case exceeds 0.73. Hence, in the case of the three, the halide anions form a cubic closing packing (ccp) lattice whose octahedral holes are filled with the cations. In the case of the later two salts, a cubic coordination (coordination number of 8) is possible. This requires a rearrangement of the chloride lattice to primitive cubic. The holes in the new lattice are cubic holes that are filled by the cations.

One further comment is in order. LiCl, NaCl, and KCl would not be expected to exhibit cubic coordination for the cations. The cations are too small relative to the anions. In the case of CsCl and RbCl, cubic coordination of the cations is possible BUT octahedral coordination is also possible! In fact, two forms of RbCl have been found.

 

4) (10 points) Granite consists primarily of quartz and feldspar. Metamorphic rocks can be produced from the application of high temperature and pressure on granite. They typically contain silicates such as garnet, zircon, and topaz which have discrete SiO4-4 anions. On the level of chemical structure, what chemical transformations are involved in the process of metamorphosis?

 

Quartz and feldspar, the principal constituents of granite, are prime examples of network solids in which there is a continuous network of Si-O covalent bonds throughout the crystal. In the case of garnet and zircon there are discrete silicate anions. The conversion of quartz and feldspar to garnet and related minerals requires conditions that between down the extensive network of covalent bonds.

 

5) (15 points) Aspirin is a pro drug which is converted in the body to salicylic acid, the active ingredient. The crystal structure of salicylic acid has the following features:

The salicylic acid molecules are found in pairs as shown below.

All atoms in the molecule are in the same plane.

In the structure of C6H5OH, the OH group is perpendicular to the plane of the C's.

Provide an explanation for the special features of the structure of salicylic acid.

Hydrogen-bonding is responsible for the special structural features. A pair of intermolecular hydrogen bonds formed between the carboxyl groups on two salicylic acid molecules links the two molecules together. On each salicylic acid molecule, an additional intramolecular hydrogen bond between the OH directly bonded to the benzene ring and the carbonyl (C=O) group on the carboxyl group. This interaction locks the OH group into the plane of the molecule, a geometry quite different from that in phenol (C6H5OH).

The structure of salicylic acid that is displayed here was obtained from the Cambridge Crystallographic Database which was established by Olga Kennard, a distinguished crystallographer. 4 molecules of salicylic acid are found in each unit cell. Pairs of molecules are related by symmetry as shown above. The symmetry is a consequence of hydrogen bonding, not the reverse.

 

ex51_04_02.htm, WES, 1 Dec. 2002