Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110

General Chemistry

Summer 2006

Lecture Notes::Lec 24_5 July

© R. Paselk 2006
 
     
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The Chemistry of the Elements

Ligands and Stereoisomers, cont.

So the hybrid model seems to have done a good job of predicting the geometry of the octahedral complex for iron, but will it do as well for other complex ion geometry? Let's look at tetracoordinate systems:

Tetracyanonickelate(II) = [Ni(CN)4]2-. For the nickel ion the electronic configuration will be:

Ni2+: [Ar]

3d
4s
4p

Hybridizing (note that we only need to hybridize four orbitals and thus chose the four lowest energy empty orbitals, the s, d, and two of the p's) we get:

Ni2+: [Ar]

3d
dsp2
4pz
Giving a complex ion with the configuration below:

Ni2+: [Ar]

3d
dsp2
4pz

Note that this should be square planar, not tetrahedral because of the unhybridized p-orbital (there would be no place for it in a tetrahedron). (You may recall that with Representative Elements the VSEPR square-planar geometry had a d2sp3 hybrid with four bonded atoms and two lone pairs (pp 422-3 in your text). The difference is in that case we had six electron pairs instead of four and expansion into the d-shell with the same energy level as the s- and p-shells.)

Tetracyanozincate(II) = [Zn(CN)4]2-. For the zinc ion the electronic configuration will be:

Zn2+: [Ar]

3d
4s
4p

This gives an ion with a very stable filled d-subshell, so only the s and p orbitals will be involved in hybridization, giving the old familiar sp3 hybrid orbitals and a tetrahedral geometry:

Zn2+: [Ar]

3d
sp3

The valence bond hybrid-orbital approach does do a pretty good job of predicting the geometry of these ions and allowing us to visualize what these complex ions must look like. However, it also has some serious failings.

Fe3+: [Ar]

3d
d2sp3

its easy to see the low spin state, but there is now way to accommodate a high spin state - there are no other orbitals to allow the electrons to unpair into.

Crystal Field Theory

An early alternate, but still simplistic, theory to account for spectral and spin properties in transition metal complexes is the Crystal Field Theory of Hans Bethe. Bethe took a completely different approach to explaining transition metal properties by assuming independent, non-covalently bonded ions in a crystal lattice. In this case the ligands will affect the metal ion via the fields imposed upon the ions by their polarities and/or charges.

As result, crystal field theory just looks at the d-orbitals (remember, in transition metal ions the s-electrons will have been removed, so only need to look at d-electrons and orbitals). The ligands can then be assumed to split the energy levels of the d-orbitals depending on their proximity and orientation. Let's consider the case of an octahedral complex (similar arguments can be made for square planar and tetrahedral complexes as well).

Let's consider two complex ions of iron(III) as examples:

Weak field:

eg
deltao is small
t2g

Strong field:

eg
   

deltao is large
t2g

If we look at the field splitting diagrams (overhead), we can see that there will be no difference between strong and weak field ligands for d1-d3 metals because there is no pairing, so electrons will stay in the t2g energy level.

Weak field:

eg
deltao
t2g

 

Strong field:

eg
    deltao
t2g

However, with d4-d7 metals, there is at least one electron pair for strong field ligands in the t2g energy level, and at least one empty orbital in the eg energy level. This allows for a decrease in pairing with a weak field ligand, and thus differering paramagnetism between weak field and strong field ligands, e.g. for d5 metals:

Weak field:

eg
deltao
t2g

 

Strong field:

eg
    deltao
t2g

For d8& d9 metals on the other hand, there is at least one electron in each higher energy, eg orbital, so boosting an electron from the t2g orbital would result in no advantage (net pairing and thus repulsion remains unchanged), and there is thus no difference between the electron diagrams of the weak and strong ligand complexes.

Finally, for d10 all of the d orbitals are filled, so no transition is possible.

Color and Crystal Field Theory

As I've said before, visible color in complexes results from the availability of a small energy gap between occupied and unoccupied orbitals. This energy gap must be in the visible range (l=400nm, violet = 5.0x10-19J/photon = 300kJ/mol - l=750nm, deep red = 2.6x10-19J/photon = 160kJ/mol; from E = hn = hc/l). The only electronic energy gap generally small enough in the transition metals to give these colors will be a d-d transition as modeled in Crystal Field Theory. Thus color is due to the absorption of a complementary color photon in a transition from t2g to eg.

As expected ions with no d-electrons or completely filled d-subshells (d10) are colorless since d-d transitions are not possible in these cases.

How about intensity? Notice that we not only see differences in the colors of complexes, but also the intensity of the colors (e.g. the colors of the water [light blue] and ammonia [intense, deep blue] complexes of copper). An explanation of the intensity differences seen in transition metal complexes is the occurrence of allowed and forbidden transitions.

In quantum mechanics some transitions are said to be forbidden. This means that they are very improbable (often they are impossible in classical theory). One example of a forbidden transition is one where an electron must change its spin in order to pair up with another electron. Since electrons have a low probability of changing spin, the opportunity for an atom to absorb a photon AND have an electron change it's spin simultaneously is extremely unlikely, so it is forbidden.

For example, if we look at the high-spin (weak-field ligands), d5, situation:

Ground state:

eg
deltao
t2g

Excited state:

eg
deltao
t2g

Note that in the ground state, each of the orbitals has one electron, and all of the electrons share the same spin, as discussed in Chem 109. Thus, in going to the excited state, one of the electrons must flip spins, in order to pair up with the electron already occupying the excited orbital. We would thus expect a complex in this state to be very pale in color.

Note that the paramagnetism also changes, going from 5/2 in the ground state to 3/2 in the excited state (this change in paramagnetism is readily measured by putting the samples between the poles of a magnet and weighing them).

Now if we look at the same system in the low spin configuration (strong field ligands):

Ground state:

eg
    deltao
t2g

Excited state:

eg
    deltao
t2g

Note here that no change in spin is required of the excited electron, so there is a much higher probability of the transition occurring - we expect this complex to have a much more intense coloration. Note in this case the paramagnetism also remains the same, which again can be readily measured.

Would we expect to see the same colors in these two situations for the same central atom? No, since the ligands are setting up different energy levels, as indicated in the diagrams.

As an example, let's look at manganese(II), a d5 ion:

The various ligands can be arranged in order of increasing field strength (and thus color) to give the so-called Spectrochemical Series:

Strong Field
-CN

 
-CO
-NO2
-NH3
-NCS
-OH2
-OH
-F
-Cl, -SCN
-Br

Weak Field
-I

Note that this series is only approximate, it varies with metal ion and metal oxidation state. But one can predict reasonably well for a single metal ion at a single oxidation state, particularly if one looks at the extremes of the series. It also works OK for multiple ligands, if the ligands are all either strong or weak.

The ligands tend to shift toward the strong field as the charge of the central atom increases (e.g. -NH3 behaves as a weak field ligand with Cobalt(II), but is a strong field ligand for Cobalt(III). You can imagine this occurs because as the charge increases, the ligands are drawn in closer to the metal ion, and thus interact more strongly with the metals d-orbitals.

Tetrahedral Complex Ions: Similar arguments can be made for the tetrahedral geometry of complex ions. Note here that the lobes of a tetrahedrally symmetrical complex do not align with the x, y, and z axis, but rather fall between them. Thus the 3dx2-y2 and 3dz2 orbitals will now be at the lower energy while the 3dxy, 3dxz and 3dyz orbitals will be higher energy. Essentially splitting is the inverse of octahedral splitting in that sense. You should look at the diagrams and read the materials in your text on tetrahedral geometry, pp 1015-18.


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