Problem Set #6 –
Level 1 questions
1. (a) Draw and label all the isomers possible for the complex
Ru(phen)2(Br)2. (click for picture)
(b) What is the formal oxidation state of ruthenium in the complex Ru(phen)2(Br)2? --- Ru(2+)
(c) What are the valence electrons on ruthenium in the complex Ru(phen)2(Br)2 ? -- 4d6
(d) What is the spin state of Ru(phen)2(Br)2?
--Low spin; all 6 electrons in T2g orbitals, because 2 x phen ligands are strong field in addition to 2nd row Ru causes large crystal friled splitting.
(e) What is the CFSE for Ru(phen)2(Br)2? CFSE = 6 x -0.4 delta + 2 P (it's 2P not 3P because the d6 electrons in all degenerate d-orbitals already has one paired set of e-, so the Oh field causes and additional 2P units of pairing energy to be 'paid'.)
(f) Does the replacement of Br as a ligand by DMSO (dimethyl sulfoxide) create additional
isomerization possibilities? Explain. Yes, it creates more isomers for two reasons:
Level 2 questions
1. [Fe(H2O)6]2+ is paramagnetic but [Fe(phen)3] 2+ is diamagnetic. Why?
Water is a weak field ligand and so the d6 electrons are maximally unpaired. Phenanthroline is a strong field ligand causing all 6 e- to pair up in thes T2g set of orbitals.
2. [Fe(H2O)6]2+ is pale green but [Fe(phen)3] 2+ is orange red. Why?
Water is a weak field ligand and so causes a smaller crystal field splitting energy, thus the complex absorbs at a longer wavelength (~700nm). Phenanthroline is a strong field ligand causing a large CF splitting energy and this complex absorbs at a shorter wavelength, ~400 nm.
3. [Cr(H2O)6]3+ is Oh but [Cr(H2O)6]2+ is not. Why?
Cr(3+) is d3, nicely occupying half of the T2g vacancies. Cr(2+) is d4, and with the weak field water ligands will be high spin. However, this causes an electronic degeracy: one electron in the degenerate Eg orbitals. This isn't allowed, so the complex distorts tetragonally to a D4h point symmetry to relieve the problem.