Bonding in alkene complexes


When the structure of Zeise's salt (K[PtCl3(C2H4)].H2O had been examined, a few features became apparent which exist in all alkene complexes. Through X-Ray diffraction it was revealed that the ethene was bonded side with respect to the platinum chloride plane. Raman spectroscopy showed that the C=C bond length had increased from 1.34 Ao (free ethene C=C length) to 1.37 Ao. NMR showed that the angles of H on the CH2 had been distorted and moved away from the metal.
The bonding in alkene somewhat mirrored the bonding present in metal-cabonyl complexes. The bonding in alkene complex consists of two parts.
1/ A forward donation from the pie bonding orbital of the alkene into the empty orbital of the metal. Also known as than sigma par of the bond.
Interaction between the metal sigma orbital and C=C pie orbital.
2/ A back donation from a filled d-orbital on the metal into the empty pie-orbitals on the alkene also known as the pie part of the bond.
Interaction between the metal pie orbital and C=C antibonding pie orbital.

Thefore the whole bonding can be displayed using the molecular orbital theory.



In alkene complexes the extent to which back bonding occurs depends upon the nature of the alkene, the metal and the other li ligands already present. Metals such as titanium and vanadium posses high energy d-orbitals. Therefore backbonding is favoured in so as to offload excess electron density to the π* orbital of the alkene. In extreme case two distinct σ metal carbon bonds can form resulting in a complex known as metallacyclopropane.


The degree of backbonding also depends on the alkenes ability to accept electrons. Therefore more back-bonding is present in alkenes with electron withdrawing groups.


The ligands attached to the metals can also play a magor role on how the alkanes bond to the metal. With the Zeise salt the alkene bond perpendically to the plane of chlorine. However this is not observed in complexes containing a Zerovalent metal or possessing 2ligands such as phosphine. The reason for this is that the dxy and dyz are of different energies due to the ligands present. The dxz is of higher energy as it lies in the plain therefore is capable of bette backbonding than dxz. Hence the alkene adopts an in plane conformation.


Therefore we can conclude that what effects backbonding are:


1/ The energy of the frontal orbitals in the metal.
2/ Steric hindrance.
3/ Electron withdrawing groups on the alkene.


[History of organometallics] [18 electron rule] [Carbonyl compounds] [Alkene complexes] [Alkyne complexes] [Arene complexes] [Cyclopentadienyl complexes] [Industrial uses of oganometallics] [Glossary] [Links]