Why are phosphines stronger-field ligands than water?

1 Answer
Mar 31, 2017

Because phosphines tend to be #sigma# donors and #pi# acceptors (at the same time). Water, on the other hand, is primarily a #sigma# donor, so it is a (MUCH) weaker-field ligand on the spectrochemical series than phosphines are, and cannot displace phosphines that are attached to a transition metal.

(Two other similar ligands in behavior are #"CO"# and #"CN"^(-)#.)


A sigma (#sigma#) donor is a molecule whose highest-occupied molecular orbital (HOMO) is a #sigma#-type orbital, formed from two orbitals totally symmetric about the internuclear axis. Examples are overlaps of #np_z# atomic orbitals, #(n-1)d_(z^2)# atomic orbitals, etc.

#sigma# donors tend to come in and donate electron density into a #sigma^"*"# orbital.

A pi (#pi#) acceptor is a molecule whose lowest-unoccupied molecular orbital (LUMO) is a #pi^"*"# type orbital, formed from two orbitals that overlapped sidelong. Examples are overlaps of #np_y# atomic orbitals along the #x# axis, #(n-1)d_(xz)# atomic orbitals along the #x# axis, etc.

#pi# acceptors tend to help stabilize the compound by accepting electron density into their #pi^"*"# LUMO.

Phosphines, however, are both of those at the same time. For example, triphenylphosphine, #Ph_3P-#, have this capability. Consider #R = Ph# for the following:

#bbsigma# donation:
https://upload.wikimedia.org/

#bbpi# acceptance (i.e. backbonding):
https://upload.wikimedia.org

The phenyl groups would help the #pi# acceptor ability of the phosphorus, since they can redistribute the delocalized electron density better than, say, if #R# was alkyl, thereby stabilizing the bond order of the #P-C# bond (thus counteracting the fact that electron density in the antibonding orbital increased, which would have weakened bond strength).

Water does not have the ability to #pi# backbond, so a phosphine's bond with a transition metal is more stabilized, and it is more resistant to incoming water ligands in a displacement reaction.