Problem Set #6 – (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 you answer with appropriate CFT arguments.

The change in ligands will have not effect on the Ni(II) spin state. This is because Ni(II) is d8 and in a Oh crstal field htere is only one way to put 8 electrons into the five d orbitals: all the stabilized T2g orbitals (xz,yz and xy) are filled and there is one electron in each of the destabilized Eg orbitals (z2, x2-y2). The difference will show up in the CFSE where the crystal field splitting parameter is much larger for ammonia ligands.

2. The magnetic moments of the first row metal hexaaquo complexes, which are all high spin, are:

a) Ti(3+): d1, so mu = (1(1+2))^1/2 = 1.73

b) Cr(2+): d4, so mu = (4(4+2))^1/2 = 4.90

c) Fe(3+): d5, so mu = (5(5+2))^1/2 = 5.9

d) Ni(2+): d8 with n=2, so mu = (2(+2))^1/2 = 2.83

e) CR(3+): d3, so mu = (3(3+2))^1/2 = 3.87

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(level 2 problems)

1.   (a) What is the most likely geometry for PdBr2(NH3)2 ?  Defend your answer with appropriate CFT arguments.

Because Pd is a 2nd row metal d8 ion, it has a large CF splitting and is ALWAYS square planar, never tetrahedral.

(b) One isomer of PdBr2(NH3)2 is unstable with respect to a second isomer.  The isomerization process can be followed by IR spectroscopy.  The IR of the first isomer shows absorptions at 480 and 460 cm-1 assigned to Pd-N stretching modes. During the isomerization the band at 460 gradually disappears and the band at 480 shifts to 490 cm-1.  Explain these observations.

The square planar complex can occur as cis and trans isomers. The first complex is the cis isomer with C2v symmetry. Stretches of the two cis- Pd-N bonds have symmetries A1 and B1 and are both IR-active and these can be assigned to the two Pd-N stretches (symmetric 480 cm-1 and antisymmetric 460 cm-1 ). The trans isomer has symmetry D2h and in this point group only one stretching mode (B3u) is IR-active, the antisymmertic Pd-N stretch at 490 cm-1. So the first species formed is the cis isomer that subsequently rearranges to the trans isomer which must be more thermodynamically favored.

2. 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.