CHEM231 Problem Set #7

Level 1 problems

1. Will substituting all the water ligands in [Ni(H2O)6]2+. by ammonia ligands to form [Ni(NH3)6]2+ cause a change in spin state for nickel?  Explain your answer with appropriate CFT arguments.  How will the change from aquo ligands to ammine ligands change the CFSE?

No spin state change occurs since for d8 Ni(2+) there is only one way to put the electrons in the T2g ane Eg orbitals!! However, the CSFE of -1.2 delta would be much more negative for the complex with NH3 ligands that are stronger field because the delta for NH3 is larger.

2. Calculate the spin-only value of the magnetic moment for the octahdral, hexaaquo complexes of Ti(3+), Cr(2+), Fe(3+), Ni(2+), V(2+).

For all these complexes, you can calculate the theoretical, spin only magnetic moment = (n(n+2))^1/2.^1/2

[Ti(H2O)6]3+ is Ti(3+) and d1. One unpaired electrons are expected to produce a magnetic moment of 1.79 B. M.

[Cr(H2O)6]2+ is Cr(2+) and d4.Four unpaired electrons are expected to produce a magnetic moment of 4.9 B. M.

[Fe(H2O)6]3+ is Fe(3+) and d5. Five unpaired electrons are expected to produce a magnetic moment of 5.9 B. M.

[Ni(H2O)6]2+ is Ni(2+) and d8. Two unpaired electrons are expected to produce a magnetic moment of 2.8 B. M.

[V(H2O)6]2+ is V(2+) and d3. Three unpaired electrons are expected to produce a magnetic moment of 3.8 B. M.

3. (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.

Level 2 problems

1. Cr(+3) and Co(+2) both make homoleptic chloride complexes (homoleptic means only Cl- ligands), but the chromium forms only an octahedral complex whereas Co(+2) favors formation of a Td.  Use CFT to explain why Cr favors strongly Oh complexes whereas Co(+2) favors Td.

Cr(3+) is most stable in an Oh field because its 3 electrons nicely half-fill the low ebergy T2g orbitals giving a large -1.2 delta CFSE. In a Td geometry, Cr(3+) would have much less CFSE and worse, it would experience Jahn Teller distortion from its configuration of E(2e-) and T2(1e-).

Co(2+) on the other hand, with 7 electrons will always have some e- in the higher energy orbitals. In a Oh field it would have the configuration T2g(4e-) Eg(2e-) and have a spin degeneracy problem in the non-bonding T2g orbitals, which woudlnt be too be energywise. In A Td field it would have the configuration E(4 e-) T2(3 e-) which i favored due to the nicely symmetric and evenly filled d-orbitals. The Oh CFSE of -0.8 delta however would not be much better than the CFSE for a Td field of -0.4 delta. Finally the lower overall charge on a [CoCl4]2- complex compared to the [CoCl6]4- complex is the fainl piece that makes the Td geometry more favored.

2. 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 occured, starting as a diamagnetic square planar complex and rearranging to paramagnetic tetrahedral complex with n=2 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.

3. 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.