Download Problem Set 7 - Bryn Mawr College

Problem Set 7
Due: March 31, 2016
Coordination Metal Isomers, Geometric and Electronic Structures
Answer all of the following questions on a separate piece of paper. Show all work.
Level 1 Questions
1. For each of the coordination complexes below, draw a crystal field diagram including the d-
electrons.
a. Fe(acac)3 Fe(III) is d5
b. [Fe(CN)6]4- Fe(II) is d6, CN- is a strong field ligand so all 6 e- are in the lower energy
T2g set of orbitals in Oh CF diagram above.
c. Ni(CO)4 as a Td complex. Ni(0) is d10, so the Td crystal field splitting with T2 orbitals
higher than E orbitals has all these filled.
d. [Ni(Cl)4]2- as a D4h complex This was a mistake!! Ni(II) is d8, Cl- is a weak field
ligand so its actual structure is Td, not square planar. I believe I intended to write
[Ni(CN)4]2-with cyanide not chloride. 
e. fac-W(Cl)3(NCMe)3 (note: NCMe is acetonitrile) W(III) is d3 and like Cr(III) wants to
be octahedral to place all 3 e- are in the lower energy T2g set of orbitals in Oh.
f. cis-[Ir(Cl)2(phen)2]+ Ir(III) is d6, the two phen liagnds put strong field ligands at 4 of
the 6 sites so all 6 e- are in the lower energy T2g set of orbitals in Oh
g. Zr(Cl)4 Zr (IV) is d0 so that there is no CF stability involved with no d e-, and the
complex adopts simply a Td geometry, with T2 orbitals higher than E orbitals and all
these are empty.
2. Chromium(III) chloride hydrate is completely dissolved in water. What species might be
present in solution? Identify the one(s) you feel are most likely.
CrCl3.6H2O is the solid. In aqueous solution the most likely species are [Cr(3+)(H20)6]3+,
[Cr(3+)(Cl)(H20)5]2+,[Cr(3+)(Cl)2(H20)6] +, but probably very little Cr(3+)(Cl3(H20)3, as this would
be a neutral and therefore insoluble species in polar water. Depending on the concentration of
chromium chloride made, different species predominate. At very low concentrations
[Cr(3+)(H20)6]3+will predominate whereas at higher concentrations [Cr(3+)(Cl)2(H20)6] + is
favored.
Level 2 Questions
3. (a) Draw and label all the isomers possible for the complex Ru(phen)2(Br)2 where phen =
1,10′-phenanthroline.
Cis- dibromo as delta and lamba enantiomers and trans-di-bromo
4.
(b) What is the formal oxidation state of ruthenium in the complex Ru(phen)2(Br)2? Ru(II), or 2+
(c) What are the valence electrons on ruthenium in the complex Ru(phen)2(Br)2 ? d6
(d) What is the spin state of Ru(phen)2(Br)2? Low spin since the are 4 site occupied by strong
field pen ligands
(e) What is the CFSE for Ru(phen)2(Br)2? 6 (-0.4 delta) + 2P = -2.4 delta + 2P. The reason why it
is only 2P (not 3P) is explained in the book: there are 2 additional units
of pairing required as compared to the fivefold degenerate e- configuration of 6 e- (where
one is already paired).
(f) Does the replacement of Br as a ligand by DMSO (dimethyl sulfoxide) create additional
isomerization possibilities? Explain. Yes, there would be many more
isomers possible for two reasons:
1) DMSO could replace either just one Br- or both Br- ligands
2) DMSO can bind through either the sulfur or the oxygen. Though as a softer 2nd
row metal ion, Ru prefers binding through the S- atom.