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

The Transition Metals, cont.

Ligands and Stereoisomers, cont.

• 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.

  • Polydentate ligands can have different numbers of bonding sites.
    • Polydentate ligands are often referred to as chelating agents and their complexes are known as chelates (from the Greek for claw).
      • Chelating agents can have 2, 3, 4, 5, or 6 points of attachment with hexacoordinate central atoms.
    • One of the most common and simplest of polydentate ligands is ethylenediamine (often abbreviated en in complex ion formulae): H₂NCH₂CH₂NH₂. Each of the nitrogens has a free electron pair which can act as a Lewis base to interact with the Lewis acid sites of the central metal ion (e.g. tetrachloro(ethylenediamine)cobaltate(III) ion, [CoCl₂(en)]^-.
      • Note that, due to its small size ethylenediamine can only bond cis- corners, its too short to reach around for trans- bonding.
    • Polydentate ligands like ethylenediamine give geometrical isomers, cis- and trans-
    • Polydentate ligands like ethylenediamine can also give rise to enantiomerism (mirror image or chiral isomers) in hexacoordinate complex ions. (models).

 

Transition Metal Ion Colors & Complex Ion Bonding

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

• Why are transition metal ions often colored?
  • Must have small electronic energy differences in ions or complexes to have transitions in the visible region (recall hydrogen atoms spectra - ground level spectral lines are all in UV).
  • Apparently the outermost (valence) electrons in transition metal ions have access to orbitals at only slightly higher energies.
    • Note that metal ions of the Representative element Groups I-IV are not colored.
    • Note also lack of color in Sc, Zn, Cd, and Hg - all explained by a lack of valence electrons in ions.
• How do we account for the small differences in energy levels required to explain these colors?

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.

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 3d_xy, 3d_xz, and 3d_yz 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 3d_x²-y² and 3d_z² 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 3d_xy, 3d_xz, and 3d_yz 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.

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.

• It turns out that species like iron(III) can form complex ions with different spins. That is the degree of paramagnetism can differ significantly. Thus Iron has low spin (spin 1/2) and high spin (spin 5/2) complex ions. Looking at the hybrid orbital prediction for its electronic configuration:

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.

• In addition, the particular spin state taken by the cation in a complex is dependent on the ligands. Obviously a localized electron theory cannot account for the effect of a species not even taken into account in the model.
• Finally, there is no way to explain the colors of these complexes. Looking at our electronic configuration diagram, there is no way to come up with additional orbitals with closely related energies as we would expect for the generation of the colors seen in so many transition metal compounds.

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).

• In the case of an octahedral complex we will consider that the ligands are at the corners of the octahedron, that is placed equidistant from the metal ion's center along the positive and negative axis of an x,y,z coordinate system.
• The ligands will then have a relatively strong interaction with the 3d_x²-y² and 3d_z² orbitals, which are also oriented along these axis (note that in the 3d_z² the "donut" is in the x-y plane). You can think of the ligands being in contact with the orbitals:

• The interaction of the ligands with the 3d_xy, 3d_xz, and 3d_yz orbitals on the other hand will be significantly less since they will lie between the orbital lobes on the diagonals of the axis as shown for the 3d_xy orbitals below (the images for the other two orbitals are identical, just substitute different axis sets):

• As a result of these interactions we see that the five d-orbitals will be split into two sets, with the two along the axis raised to a higher energy, while the three between the axis go to a lower relative energy.
• If we consider that the energy difference is delta_o (the subscript o refers to octahedral), then:
  • The higher energy orbitals, designated e_g (3d_x²-y² and 3d_z²) will have an energy of +3/5delta_o relative to the average d-orbital energy.
  • The lower energy orbitals t_2g (3d_xy, 3d_xz, and 3d_yz) will have an energy of -2/5delta_o relative to the average d-orbital energy.

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

• The fluoride ion, F^-, is a weak-field ligand, so the hexafluoroferrate(III) complex ion will exhibit a weak-field effect, the e_g and t_2g energy levels will be closely spaced, so the repulsive interactions between electrons will dominate over the energy difference and the electrons will spread out, giving a high-spin complex (spin = 5/2):

Weak field: eg deltao is small t2g

• The cyanide, CN^-, is a strong-field ligand, so the hexacyanoferrate(III) complex ion will exhibit a strong-field effect, the e_g and t_2g energy levels will be well spaced, so the energy difference will dominate over the repulsive interactions between electrons and the electrons will clump, giving a low-spin complex (spin = 1/2):

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 t_2g 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 t_2g energy level, and at least one empty orbital in the e_g 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, e_g orbital, so boosting an electron from the t_2g 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 t_2g to e_g.

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 high spin, weak field complex, hexaaquamanganese(II) ion, [Mn(H₂O)₆]²⁺ is a very light pink. The intensity of the absorption of the complementary color is low, as expected since a forbidden transition is needed.
• The low spin, strong field complex, hexacyanomanganate(II) ion, [Mn(CN)₆]^4- is an intense violet. In this case the allowed transition results in a strong absorption by the complex of the complementary photon. Note also the quite different colors absorbed.

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 3d_x²-y² and 3d_z² orbitals will now be at the lower energy while the 3d_xy, 3d_xz and 3d_yz 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.

© R A Paselk

Last modified 5 July 2006