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1.
The lambda max = 450 nm for [Cr(NH3)6]3+.
The lambda max = 420 nm for [Co(NH3)6]3+.
Assume P= 18,000 cm-1 for both complexes.

(b) Calculate the delta(o) and CFSE in kJ/mol for each complex.
(Recall the relationship between wavelength, nm, and wavenumber, cm-1 and that
1.1962 x 10-2 x E (cm-1)= E (kJ/mol).

Cr: 450 nm corresponds to 22,222 cm-1; using the conversion relationship, delta = 266 kJ/mol.

Co: 420 nm corresponds to 23,800 cm-1; using the conversion, delta = 285 kJ/mol.

Co(3+) is d6 and low spin, therefore CFSE = 2.4 delta - 2P or CFSE = 684 kJ/mol - 2P

Since delta = 285 kJ/mol for the Co(3+) complex, 2P must be less than 285 kJ/mol (or P< 143 kJ/mol) for the low spin state to be energetically favorable.

(a) Which reaction, (i) or (ii), is predicted to have larger value for its Kf and why?

Reaction (i) will have a larger Kf. While both reactions use bidentate ligands and so benefit from the chelate effect, the first reaction uses ligands with N-atom donors. Since Ru(2+) is a soft ion it will prefer teh N atoms of phen ligands to the harder O atoms of oxalate.

3. (answer) Due to its d9 electron count, the Cu2+ ion with six ligands L of any type will undergo a Jahn-Teller distortion because the d9 configuration will create an electronic degeeracy in teh Eg orbitals of an Oh symmetry complex. The point symmetry of the hexa-coordinated cupric ion is D4h.

4. Using CFT applied to coordination complexes, explain why the two very strong field ligands CO and CN- on nickel make Ni(CO)4 and [Ni(CN)4] 2-, 4-coordinate complexes that have two different geometries.

Ni(CO)4 is Ni(O) and has a d10 configuration. With a full d-shell, regardless of geometry, there will be no CFSE (LFSE). The complex therefore adopts the geometry that places the carbonyl ligands as far away from each other as possible.

[Ni(CN)4]2- is Ni(II) and is a d8 ion. For Ni, especiallywith strong ligand fields, a square planar geometry is favored. In this D4h symmetry, all 8 e- are in the lowest energy d orbitals and the one very high energy x2-y2 orbital is empty

5. Nitric oxide, NO, reacts with many metals and binds as the nitrosyl ligand, NO+.
Predict the structure of this ligand binding to metals and the crystal field strength of
this ligand.

NO+ is isoelectronic with carbon monoxide and so can be considered to have an identical N-triple bonded to O. As such, it is expected to bind throguh the more electropositive N atom and to be a strong field, pi-acid ligand.

6. NiCl2(PPh3)2 is red and diamagnetic. On heating to 387 K this complex changes color forming a blue-green complex which has a magnetic moment of 3.18 B. M. at 295 K.
(a) Suggest an explanation for the change in magnetic properties.

The change in magnetic moment immediately implies a change in geometry has occure, starting as s quare planar and rearranging to tetrahedral at high temperature.
(b) Is there any significance to the change in color?

The low temperature, square planar form is red: therefore it absorbes about 500nm. The hghg temperature, tetrahedral form is blue-green: therefore it absorbes about 650nm. This indeed makes sense because the sq. pl. CF diagram has d(x2-y2) at very high energy, and so this would result in a higher energy (shorter wavelength) absorption. Tetrahedral crystal field is typically small adn so is also consistent with absorptio at low energy, longer wavelength.

7. A new protein has been isolated from a squishy pink marine animal. A conspicuous feature of the new protein is an amino acid sequence corresponding to a metal binding site. The residues of this binding site are two cysteines, one histidine and one methionine. It is known that this metalloprotein is involved in one-electron redox reactions and it is speculated that the metal is the catalytic site of the redox reaction. Which first row transition metal ion(s) could be the metal used by this marine protein?

The characteristic residues of amino acids cysteine, methionine and histidine are the -SH, the S-CH3 and the imidazole groups. These are potential S atom and N atom donors to metals. These atoms are all soft and so will prefer to bind a soft metal. Of the first row metals, Cu(+) is the softest. Cu is also able to do 1 electron redox reaction, easily moving between Cu(I) to Cu(II).

8. (answer) Mo(CO)3(pyridine)3 is a Mo(0) complex having a d6 electronic configuration. Having a full T2g set of d-orbitals means it is greatly stabilized by having 3 CO, pi-acid ligands to lower the energy of the T2g orbitals.

Mo(CO)3(O)3 is a Mo(6+) complex having a d0 electronic configuration. While the empty d-orbitals can receive e- from the pi-base oxide ligands, the CO ligands will not have any pi- interactions and make the Mo-CO bonds weak. Hence this complex does not exist.

9. Chromium is a metal of many colors! [chroma, Greek for “color”] One green pigment used in paints as well as in eyeshadows is chrome green, Cr2O3. (Note that chromium in chrome green adopts an octahedral site surrounded by oxide ions in the solid state). Chromium takes on a completely different hue in the yellow pigment chrome yellow, or lead chromate, whose empirical formula is [Pb+2][CrO4-2]. Is the green color of chrome green produced by the same type of electronic transition that causes the yellow color of chrome yellow?

No, the colors do not originate from the same type of electronic transision. The green color of chrome green is due to a d-d transition in Cr(3+) in an octahedral field whereas the yellow of chrome yellow must be due to a LMCT since that pigment has Cr as a d0 C4(6+) ion.

Predict the relative saturation (i.e., absorptivity or extinction coefficient) of chrome green versus chrome yellow. Defend your answers with detailed explanations.

Since chrome green is due to a d-d transition in an octahedral field, it is LaPorte forbidden and is expected to be weak (an extinction coefficient less than 100). The LMCT in chrome yellow however would be completely allowed since it will be from O-p orbitals to d orbitals and the ion is tetrahedral giving it a much larger extinction coefficient ( > 1000)>

10. My mother always told me to take iron vitamins with orange juice because of its high concentration of citric acid (structure below). Analyze her suggestion as a chemist: what aspects of the citric acid would be helpful to taking an iron nutritional supplement?

The MO of Pt(CN)4]2- shown is composed of the d(xy) orbital and the pi* orbitals (in-plane) on the 4 CN- ligands. There is positive bonding (overlap) between the Dxy and the pi* on the C, but a phase change of the pi* between the C and N atoms. A diagram illustrating only that bonding interaction is shown below:

Ethylene added to Co(2+)Cl2 must form a tetrahedral complex to explain the deep blue color of high absorptivity.

This could be either a [Co(en)2]2+ ion or even a mixed Co(en)Cl2 species, which might be expected to be more insoluble. When either of these are added to water, the labile Co(2+) undergoes a fast ligand substitution losing all en and/or Cl ligands to make the heaxaaquoCo(2+) ion. This ion is octahedral and so has LaPorte forbidden absorptions accounting for its pale pink color.