CHEM231
Problem Set #5
Due April 11, in class.
Level 1 problems
1. (a) What are the two geometries possible for Ni(NH3)2Br2? square planar and tetrahedral
(b) What one measurement would allow you to determine the real geometry of this Ni(II) complex? Explain how you would interpret the result of this experiment. Magnetic susceptibililty would indicate without a doubt which geometry is observed. A square planar geometry is diamagnetic becuase the 8 d-electrons will fill all but the highest d (x2-y2) orbital. A tetrahedral geometry will be paramagnetic with 2 unpaired electrons in the T2 set of orbitals. The observed magnetic moment would be ~2.8 B. M.
Many of you suggested using spectroscopy: the Td complex woudl be expected to have a higher intensity d-d transition because there is no inversion adn so it is allowed. The square planar complex would be expected to have a weaker intensity absorption because it is centrosymmetric. This might work, but if there were charge transfer absorption in addition toe d-d transitions, this method would not be foolproof. The magnetic measurement is foolproof.
2. An intense electronic absorption near 500 nm occurs for [Fe(II)(bpy)3]. What type of transition is this? A MLCT transition from the pi system of the bpy ligand to the Fe(2+). d-d transittions would not be intense. While the complex with D3 symmetry does not have an inversion, the immediate inner coordination environment —what the Fe ion sees— is 6 N atoms in an octahedral. So teh spectroscopy is determined mainly by the symmetry of the attached atoms.
3. How many d-d absorptions do you expect to see in the electronic spectrum of Cr(3+) in an octahedral field? There are three: A1g--> T2g, adn two differentA1g--> T1g transitions.
4. Explain why the color of trans-[Co(III)en2F2]+ is less intense than that from cis-[Co(III)en2F2]+. The trans isomer will have an inversion center causing the LaPorte selectron rule to predict forbidden—weak—transition intensities. The cis isomer has no inversion, and so teh d-d transiton are allowed and more intense.
5. a) Calculate the spin-only value of the magnetic moment of the hexacyanomangaese(4+) complex.
[MnCN6]2- is Mn(4+) and d3. Three unpaired electrons are expected to produce a magnetic moment of 3.8 B. M.
b) How many spin-allowed electronic transitions are expected for the hexaaquo Mn(4+) complex and what are their assignments?
A high spin d3 ion has three transitions as listed in the answer to #3.
6. What is the ground electronic state of Co(II) as a free ion (in a spherical field)?
The ground electronic state electron configuration would be : one electron each in orbitals with m(l) = 2 and =1, two paired electrons in each of the other orbitals with m(l) = 0, -1 and -2. Hence L=2+1 = 3==> a F state. Two unpaired electrons give a triplet multiplicity, so the ground state term is 3F.
Level 2 problems
1. Red crystalline NiCl2(PPh3)2 is diamagnetic. On heating to 387 K a blue-green form of the complex is obtained which has a magnetic moment of 3.18 B. M. at 295 K.
(a) Suggest an explanation. 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.
2. For metal oxyanions of the general formula [MO4]2-, generally the first row compounds are highly colored and those analogous compounds of the 2nd and 3rd row are often white. For example, we saw in class that chromate is bright yellow and permanganzte is intensely magenta, but their heavier relatives [MoO4]2-, and [ReO4]2-, respectively, are both white. Can you account for this observation?
The transition observed in these compoounds is LMCT, from filled pi orbitals of the oxo-ligands to teh metal. The energy of the transition is related to the energy separation of the oxygen p-orbitals and the metal d-orbitals. As one moved fromt the first row to the second adn third row, the energies of the metla d-orbitals increases because the metal is more electropositive. Hence there would be an increase in energy between the O p-orbitals. For the heavier metals, the color of absorption is shifted into the short wavelength UV region, out of the visible region.
3. A sample of Fe(detc)3 has a measured magnetic moment of 2.7 B. M. What does this suggest about the electronic structure of this complex and the crystal field of the detc ligand? Detc = diethyldithiocarbamate, -S2CN(Et)2.
This suggests that the complex undergoes spin-equilbirum, that is, it is a mixture of low spin (d1, 1.79 B.M.) and high spin (d5, 5.9 B.M.) complexes. Fe(III) is one metal ion known to often form complexes that do spin crossover, especially with S-donor liagnds.
4. 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 liagnd.