Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110

General Chemistry

Fall 2003

Lecture Notes::Lec 35_1 December

© R. Paselk 2003
 
     
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The Chemistry of the Elements

Stereoisomerism, cont.

Last time we looked at coordination numbers 2 and 4, today we will look at coordination number 6:

 

Polydentate (multidentate) ligands or Chelating Agents: Ligands occur which can bind to the central atom of a complex in more than one place. These ligands can give rise to additional kinds of isomerism.

 

Transition Metal Ion Colors & Complex Ion Bonding

Introductory discussion of color and electronic energy levels in atoms and ions.

To approach these questions in greater depth we need to discuss the bonding in complexes from a quantum point of view. We'll approach this much as we did when we introduced molecular bonding theory, beginning with a hybrid orbital (localized electron orbital) model and then going to molecular approximations to give us energy levels etc.

 

Hybrid Orbital Theory for Transition Metal Ions

As we've mentioned before the interactions between ligands and transition metal ions are best considered to be covalent in nature. So how do we come up with a set of bonds giving octahedral symmetry with a transition metal ion? Let's look at one of the more intensely colored complex ions we've seen in lab, ferricyanide ion (more properly, hexacyanoferrate(III) ion): [Fe(CN)6]3-.

First, let's look at the electronic configuration of the electrons in the ground-state free atom:

Fe (Z = 26): [Ar]
3d
4s
4p

To get the ferric ion we remove three electrons:

Fe3+ (Z = 26): [Ar]
3d
4s
4p

(Note - You can see why iron prefers to lose two electrons [the s-electrons] to give ferrous ion, or three electrons [the s-electrons, plus one d-electron to give a stable half-filled d-shell.]

Now the question is, how do we hybridize the remaining orbitals? Remember that in hybridization the orbitals must overlap to give good addition. Also, we need to come up with hybrid orbitals along the x-, y-, and z-axis to give octahedral symmetry, so we expect to hybridize with the p-orbitals which lie along thee axis, and the s-orbital which is symmetrical over all axis. If we look at the d-orbitals we see that the 3dxy, 3dxz, and 3dyz orbitals lie on the diagonals of these axis, and thus do not overlap with the p-orbitals as required to give good hybridization: (computer animations below)

On the other hand the 3dx2-y2 and 3dz2 orbitals do have their lobes aligned along the axis:

But how do we get the electrons arranged so that these orbitals are available for hybridization?

First, recall that though the hybridization model is for the localized electron picture, that doesn't mean it applies to an isolated atom or ion, Thus we will assume that the ligands will influence the energy levels of our ion in some complex way, and for our purposes we will assume that it results in all of the d-electrons "bunching up" into the 3dxy, 3dxz, and 3dyz orbitals giving:

Fe3+ (Z = 26): [Ar]
3d
4s
4p

If we now "mix" the two empty d-orbitals and the s- and p-orbitals we will get the d2sp3 hybrid orbital set:

Fe3+ (Z = 26): [Ar]
3d
d2sp3

Our hybridized ferric ion can now act as a Lewis acid to accept six electron pairs from cyanide ion acting as a Lewis base. Note that the cyanide ion has been shown experimentally to donate the electron pair on the carbon atom, not the nitrogen pair: [:C:::N:]-

The electronic configuration for iron in the final complex is then:

Fe3+ (Z = 26): [Ar]
3d
d2sp3

where the electrons in the d2sp3 hybrid orbitals are all contributed by the ligands.


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Last modified 1 December 2003