1. Find the point group for the following species: XeO3, hexacarbonylvanadium(0), and tris(2,2'-bipyridine)ruthenium(II) ion.
XeO3
Using VSEPR rules shows that XeO3 pyramidal,
, so the point group is C3v.
hexacarbonylvanadium(0)
V(CO)6 is octahedral,
, so the point group is Oh.
tris(2,2'-bipyridine)ruthenium(II) ion
The structure is an octahedron with three planar chelating ligands,
, so the point group is D3.
2. Find the LFSE in terms of Dq and P and the magnetic moment in terms of Bohr-magnetons for: fac-tribromotrichloroferrate(II) ion, hexacyanoferrate(III) ion, bis(η4-1,5-cyclooctadiene)nickel(0).
fac-tribromotrichloroferrate(II) ion
Fe2+ is d6 and chloride and bromide are weak ligands so this will be a high spin complex: t2g4eg2 (ignoring the symmetry reduction) gives LFSE = 4Dq with 4 unpaired spins so μ = [4(6)]½ = 4.9 μB.
hexacyanoferrate(III) ion
Fe3+ is d5 and cyanide is a strong ligand so this will be a low spin complex: t2g5 gives LFSE = 20Dq – 2P with 1 unpaired spin so μ = [1(3)]½ = 1.7 μB.
bis(η4-1,5-cyclooctadiene)nickel(0)
Ni(0) is d10 so all of the d orbitals are filled: t2g6eg4 (ignoring the symmetry reduction) gives LFSE = 0 with 0 unpaired spins so μ = [0(2)]½ = 0 μB.
3. Consider the reactions:
[MF6]4–(aq) + 6 OH–(aq) → [M(OH)6]4–(aq) + 6 F–(aq) β is large
[M(H2O)6]2+(aq) + 6 OH–(aq) → [M(OH)6]4–(aq) + 6 H2O(l) β is small
Place hydroxide in the spectrochemical series and explain your reasoning.
Since hydroxide will displace fluoride readily (large β, OH– must be a stronger ligand than F–. In contrast, hydroxide does not readily substitute for water (small β), so OH– must be a weaker ligand than H2O. Thus, hydroxide lies between fluoride and water in the spectrochemical series.
4. Use the EAN rule to predict the stability of the following: trans-dichlorotetrakis(triphenylphosphine)nickel(II), chlorotris(triphenylphosphine)rhodium(I), [PtCl3(CH2=CH2)]–.
trans-dichlorotetrakis(triphenylphosphine)nickel(II)
Ni2+ is d8
Cl– donates 2 electrons, 2(2) = 4
:P(Ph)3 donates 2 electrons, 4(2) = 8
8 + 4 + 8 = 20, not predicted to be stable by EAN.
chlorotris(triphenylphosphine)rhodium(I)
Rh+ is d8
Cl– donates 2 electrons, 1(2) = 2
:P(Ph)3 donates 2 electrons, 3(2) = 6
8 + 2 + 6 = 16, likely to be stable by EAN.
[PtCl3(CH2=CH2)]–
Pt2+ is d8
Cl– donates 2 electrons, 3(2) = 6
CH2=CH2 donates 2 electrons, 1(2) = 2
8 + 6 + 2 = 16, likely to be stable by EAN.
5. 10Dq is found to be 17400 cm–1 for [Cr(H2O)6]3+. a) Predict λmax in units of nm for the lowest energy d-d transition. b) Will this wavelength be longer, shorter, or the same if all of the water ligands are substituted with fluoride? Explain.
a) λmax = 107/E(cm–1) = 107/17400 = 575 nm.
b) Since fluoride is a weaker ligand than water, 10Dq will be smaller, which shifts λmax to longer wavelength, maybe 580 to 600 nm.