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Chem 324 Fall 2016
Problem Set 3 ANSWERS
September 25, 2016
This problem set focuses on content from lecture note set 5. Note some of the questions will require finishing
lecture notes 5B.
Additional questions
1. For the following complexes: provide (a) the metal’s oxidation state, (b) the metals’ d-electron count, (c)
the total electron count at the metal, and (d) sketch the d-orbital energy levels for the complex and put
the d-electrons in these orbitals in their most likely configuration.
(a)
V(IV), d1, total of 13 electrons.
This is an octahedral structure
(b)
Ti(0), d4, total of 14 electrons.
This is a trigonal bipyramidal structure.
Even though the ligands are strong field, still fill the bottom 4 orbitals for this structure in “high spin”
configuration.
(c)
Ir(I), d8, total of 16 electrons.
This is a square planar structure
(d)
Co(II) d7, total of 15 electrons.
This is a tetrahedral structure
(e)
Ni(II) d8, total of 20 electrons.
This is an octahedral structure
(f)
Cu(I) d10, total of 18 electrons.
This is a tetrahedral structure
(g)
Fe(II), d6, total of 18 electrons.
This is an “octahedral” structure (remember, Cp ligand is structurally analogous to three cis- ligands).
Overall the ligands are mostly pi acceptors (CO and Cp-) so the most likely configuration is low-spin.
(h)
Ru(II), d6, total of 18 electrons.
This is an “octahedral” structure (4-benzene is is structurally analogous to two cis- ligands).
The ligands are all pi acceptors so the most likely configuration is low-spin.
2. Explain the electroneutrality principle in relation to ionic vs. covalent bonding in metal complexes.
The ionic model of bonding maximizes the charge separation between ligands and metal and usually leads to
unrealistically high positive charges on the metal ion. In contrast, the purely covalent model puts unrealistically
low charges (or large negative charges) on the metal. The electroneutrality principle basically favours an
intermediate position, somewhere in between ionic and covalent, such that charge distribution is evenly
distributed between metal and ligands such that no atom has an excessive charge as much as possible (ideally
between -1 and 1).
3. In one diagram, sketch side-by-side the d-orbitals for the following three complexes and explain the
relative positioning of the d-orbitals in the three complexes:
(a) [Fe(CN)6]4(b) Fe(NH3)6]2+
(c) [FeCl6]4-
Assume that the eg* orbitals are at more or less the same energy. The t2g orbitals change as a function of the
ligand bonding capabilities. Starting with the middle complex: ammonia is a sigma donor ligand ONLY – it has no
pi bonding capabilities at all. NH3 is somewhere in the middle of the spectrochemical series so its not easy to
predict whether this would be HS or LS. The hexacyano complex has good pi accepting ligands, which has the
effect of converting the (formerly non-bonding)t2g otbitals in to bonding ones- and as such they go down in
energy. This increases Δo and favours the low-spin configuration. In contrast, chloride is a classic pi donor ligand,
which makes t2g orbitals antibonding and as such higher in energy relative to the ammonia complex. As a result
Δo is smaller for this complex and the high spin configuration is preferred.
4. (a) Explain the Dewar-Chatt-Duncanson model for bonding between a metal and ethylene (below left).
(b) Shown below (right) is an alternate formulation of a metal-C2H4 complex. Note in particular the
geometry of the ligand. How would you describe the bonding in this kind of complex (e.g. what’s the
charge on the ligand and how do you think about the orbitals/electrons involved?
(a) There are two components to metal-ethylene bonding:
1. Sigma type donation from ethylene HOMO . ethylene is a sigma donor in this respect.
2. Pi type “back bonding from metal to ethylene LUMO. Ethylene is a pi acceptor here.
(b) The “alternate picture has the CH bonds bending back away from the metal. This suggests a change in
hybridization at the carbons to (sort of) sp3. Also the bonds directly from M to the carbons suggest
convention M-C sigma bonds. The correct interpretation of this is that the C2H4 ligand here is a dianion.
As such this ligand donates two pairs of electrons and now formally occupies two cis coordination sites.
This ligand can be thought of as two methyl (CH3) anions which have been bonded together (or
alternatively, doubly deprotonated ethane (C2H6)
5. We have not discussed the MO diagrams of 3 coordinate complexes. It turns out that, e.g. trigonal planar
complexes also follow a sixteen-electron rule (like square planar) instead of the more common eighteen
electron rule. Why?
With only three ligands, there would only be three M-L sigma bonding molecular orbitals, which would have 6
electrons. Filling ALL of the metal-based (d) molecular orbitals would take 10 more electrons, for a total of 16.
Any more electrons than this would then necessarily go into M-L antibonding orbitals.