1. For the complex [CrF6]3– the peak maxima in the UV-Vis spectra are found at 275, 427, and 638 nm with absorption coefficients between 10 and 100 M–1cm–1 while for [VF6]4– the peak maxima are found at 464, 512, and 953 nm with similar absorption coefficients. Find B and 10Dq for each complex. Predict the spin-only magnetic moment for the two complexes. Which of these metals is lower on the Irving-Williams series and which is higher? The g-value for [CrF6]3– was found to be 1.98. Estimate the spin-orbit coupling constant. Would you expect splitting , shoulders, or observable spin-forbidden transitions in the [CrF6]3– optical spectrum? Why or why not? Hint: The spin-orbit coupling constant in Cr3+(g) is 275 cm–1.
For [CrF6]3–, d3 in Oh
ν1 = 638 nm = 15670 cm–1
ν2 = 427 nm = 23420 cm–1
ν3 = 275 nm = 36360 cm–1
Therefore 10Dq = ν1 = 15600 cm–1
15B = ν3 + ν2 – 3ν1 = 36360 + 23420 – 3(15670) so B = 851 cm–1
S = 3/2 or n = 3 so μ = [n(n+2)]½ = [3(3+2)]½ = 3.87 μB
For [VF6]4–, d3 in Oh
ν1 = 953 nm = 10490 cm–1
ν2 = 512 nm = 19530 cm–1
ν3 = 464 nm = 21550 cm–1
Therefore 10Dq = ν1 = 10490 cm–1
15B = ν3 + ν2 – 3ν1 = 21550 + 19530 – 3(10490) so B = 641 cm–1
S = 3/2 or n = 3 so μ = [n(n+2)]½ = [3(3+2)]½ = 3.87 μB
Since 10Dq(V2+) < 10Dq(Cr3+), V2+ is lower on the Irving-Williams series than Cr3+. This is consistent with the pattern of the other metals.
For [CrF6]3–, g = 1.98 = 2.0023 – 8λ/10Dq
So λ = (2.0023 – 1.98)(10Dq)/8 = (2.0023 – 1.98)(15600)/8 = 43 cm–1
The spin-orbit coupling constant for the complex is small, ~16% of the free-ion value, so there should be no observable splitting or shoulders from spin-orbit coupling. Further, d3 in Oh is not Jahn-Teller active, so again no splitting, broadening, or shoulders are expected. Finally, the small value of λ for the complex suggests that the intensity of spin-forbidden transitions should be very low, probably below being observable.
2. For the reaction [Co(NH3)5(H2O)]3+ + X– → [Co(NH3)5X]2+ the rate constants for X– = Cl–, Br–, N3–, and SCN– are all similar, differing by less than a factor of 2. What does this suggest about the mechanism of the substitution? Would you expect the equilibrium constants for these reactions to be less than 1 or greater than 1. Briefly explain. SCN– is ambidentate: which atom do you think is bonding to the metal in this reaction? Briefly explain. Based on this data, is N3– a weak ligand or a strong ligand. Briefly explain.
Since the nature of the substituting ligand has little effect on the rate of the substitution reaction, the rate law must not include [X–], which implies a dissociative mechanism (D).
All of the substituting ligands shown are lower on the spectrochemical series than H2O, with the possible exception of azide ion. Since the LFSE for the product will lower for the product than the reactants, the equilibrium constants are expected to be less than 1.
To be consistent with the rate constants it seems likely that the S end of the thiocyanate ligand, the weaker bonding site, is likely where the binding occurs to the cobalt ion.
Likewise, for the rate constants to be consistent azide must be a weaker ligand than water, thus a weak ligand.
3. Using [PtCl4]2– (D4h) as a starting material, design syntheses for cis-[PtCl2(NO2)(NH3)]– and trans-[PtCl2(NO2)(NH3)]–.
cis-[PtCl2(NO2)(NH3)]–
[PtCl4]2– + NH3 → [PtCl3(NH3)]– + Cl–
[PtCl3(NH3)]– + NO2– → cis-[PtCl2(NO2)(NH3)]– + Cl–
Cl– is a stronger trans director than NH3 so substitution will occur trans to the chloride ligand. Since NH3 is the stronger ligand, the nitrite ion will substitute for a weaker chloride ligand, giving the cis product.
trans-[PtCl2(NO2)(NH3)]–
[PtCl4]2– + NO2– → [PtCl3(NO2)]2– + NH3
[PtCl3(NO2)]2– + NH3 → trans-[PtCl2(NO2)(NH3)]– + Cl–
NO2– is a stronger trans director than Cl– so substitution will occur trans to the nitrite ligand, giving the trans product.
4. Are the following complexes stable by the EAN rule? Why or why not? [V(CO)6], [Cr(η6-C6H6)2], trans-[Ir(CO)Cl(P(C6H5)3)2] (Vaska’s complex), [Ni(η4-COD)2] (COD = 1,5-cyclooctadiene). Which, if any, of these complexes might be considered as a catalyst for alkene hydrogenation or hydroformulation? Briefly explain.
[V(CO)6]
V0 is d5 and each CO contributes 2 e– so the total is 5 + 6(2) = 17 e–, so is not expected to be stable. (Experimentally, [V(CO)6] is stable: there is an internal disproportionation to give [V(CO)6]+ and [V(CO)6]–, 16 and 18 e– complexes, respectively.)
[Cr(η6-C6H6)2]
Cr0 is d6 and each η6-benzene contributes 6 e– so the total is 6 + 2(6) = 18, stable.
trans-[Ir(CO)Cl(P(C6H5)3)2]
Ir+ is d8, CO contributes 2 e–, Cl– contributes 2 e– and each triphenylphosphine contributes 2 e– so the total is 8 + 1(2) + 1(2) + 2(2) = 16, likely stable. (It is stable.)
[Ni(η4-COD)2]
Ni0 is d10 and each η4-COD contributes 4 e– so the total is 10 + 2(4) = 18, stable.
Only trans-[Ir(CO)Cl(P(C6H5)3)2] is expected to be a useful catalyst: it is coordinately unsaturated and can support oxidative addition (to Ir3+) and reductive elimination
[V(CO)6] and [Cr(η6-C6H6)2] are coordinately saturated so it would be difficult to add a substrate to the catalyst to initiate catalysis.
[Ni(η4-COD)2] might work since it is not coordinately saturated and could be oxidized, but the COD ligand is bulky and might prevent addition of a substrate. COD is a good leaving group, though.
5. Briefly explain why the nanoparticles of a metal are usually superior to the bulk metal when used as a catalyst in reduction reactions.
The surface area of nanoparticles are much greater on a per gram or per mole basis so there are more sites for potential catalytic activity. Further, there are significantly more edges and corners in nanoparticles, which tend to be more active sites for many reactions.