Slides34

Lecture #36 December 5, 2005

CURMUDGEON GENERAL'S WARNING. These "slides" represent highlights from lecture and are neither complete nor meant to replace lecture. It is advised not to use these as a reliable means to replace missed lecture material. Do so at risk to healthy academic performance in 09-105.

Lecture outline

Transition Metal Complexes

Ligand Bonding to central ion's hybrid orbitals

Crystal field theory told us nothing about the actual bonds. To begin our discussion of bonding in transition metal complexes, we return to consideration of hybrid atomic orbitals. Here is a review of hybrid atomic orbitals for atoms which can have expanded octets as would be the case for our transition metal ions.

The highest occupied atomic orbital in Cr³⁺ and the lowest unoccupied orbitals are shown. The d²sp³ hybrids can be constructed from appropriate choices here.

A review of d²sp³ hybrid atomic orbitals

The ammonia ligands bind to the transition metal ion. In theory, the bond forms in the region of overlap between the transition metal ion's hybrid atomic d²sp³ orbital and the ligand's hybrid sp³ atomic orbital. Two electrons go here, both from the ligand, to form a sigma bond.

Looking in just the xy plane where the sigma bonds to four NH₃'s are located as is the d_xy non-bonding orbital. The ligand atomic orbital which would overlap with the transition metal hybrid orbital (a non-bonding interaction) is shown on the x- and y-axes.

Fe(H₂O)₆²⁺ is our next example.

If we allow iron's six valence electrons to occupy the 3d orbital as indicated, each water will contribute a lone pair of electrons to a bond between its sp³ orbital and iron's d²sp³ orbital formed as indicated

Fe(CO)₆²⁺ is our next illustration. Since CO is a strong field ligand, we should not be surprised to see a low-spin complex ion electron configuration show up by necessity.

Bonding of the CO ligands to the d²sp³ hybrid orbitals chosen as illustrated is consistent with the complex ion being diamagnetic.

Looking in just the xy plane, we see that the electron in the d_xy nonbonding orbital is repelled by chlorine's lone pair in its p-orbital, weaking Cl's bond to the metal.

Can we now explain why CO (and CN-) are strong field ligands?

Let's remind ourselves of the underlying valence electron structure associated with CO, including the empty anti-bonding molecular orbitals.

With CO sigma bonded to the metal just as NH₃ and Cl^- were, the CO has its empty pi-anti-bonding orbital placed so that a delocalized molecular orbital in conjunction with the metal's d_xy becomes possible. Note the constructive interference possibility.

Here, then is the delocalized orbital into which the previously though-to-be-non-bonding electron is placed, giving the CO bond to the metal some partial pi-bond character...making it a stronger bond!

ZnCl₄^2- is the next example.

Ten valence electrons fill the zinc 3d orbital. The four ligands, each contributing a lone pair to bond formation, utilize the hybrids formed from Zn's next four orbitals, the 4s and the three 4p's corresponding to sp³ hybrids and the observed tetrahedral geometry for complexes of this type.

Ni(CO)₄²⁺ is our last illustration.

The arrangement which is consistent with the observed properties of the complex (geometry and magnetic properties) and other complexes similar to it arises from use of four Ni orbitals that hybridize to give dsp² hybrids. Although we have not discussed this combination before, it does correspond to four equal orbitals arranged in a square planar distribution.