3
$\begingroup$

How does $\ce{P(C2H5)3}$ acts as a ligand and forms dπ-dπ bonds with transition elements. Why not pπ-dπ bond?

I understand that transition elements have d subshell available to accept electrons. But what about the electrons from phosphorus? From Chemistry Class 12 (NCERT), Part 1, p. 168.:

Another factor which affects the chemistry of nitrogen is the absence of d orbitals in its valence shell. Besides restricting its covalency to four, nitrogen cannot form dπ-pπ bond as the heavier elements can e.g., $\ce{R3P=O}$ or $\ce{R3P=CH2}$ ($\ce{R}$ = alkyl group). Phosphorous and arsenic can form dπ-dπ bond also with transition metals when their compounds like $\ce{P(C2H5)3}$ and $\ce{As(C6H5)3}$ act as ligands.

$\endgroup$
  • 8
    $\begingroup$ What do you mean by dπ-dπ? PEt3 acts as a $\sigma$ donor towards metal centres via the lone pair on phosphorus. It's also a $\pi$ acceptor but the acceptor orbital is the $\ce{P-C}$ $\sigma^*$ orbital, not the 3d orbitals of P. $\endgroup$ – orthocresol Feb 16 '17 at 15:37
  • 1
    $\begingroup$ @orthocresol My NCERT textbook says 'Phosphorous and arsenic can form dπ-dπ bonds also with the transition metals when their compounds like P(C2H5)3 and As(C6H5)3 act as ligand.' $\endgroup$ – Reeshabh Ranjan Feb 16 '17 at 18:24
  • 1
    $\begingroup$ @Reeshabh -- That is outdated. The outer-$d$-orbital bonding theory was invented to force-fit pairwise covalent bonds with observed valences. Molecular orbital theory which allows delocalized bonds over multiple atoms can explain the valences without need for "outer $d$" irbitals. $\endgroup$ – Oscar Lanzi Sep 26 '17 at 12:58
9
$\begingroup$

As I already mentioned in my comment, phosphines such as $\ce{PEt3}$ can bind to transition metals with partially filled d orbitals as a σ-donor as well as a π-acceptor.

The σ interaction is simply a donation of electrons from the lone pair on phosphorus to empty metal d orbitals.

Phosphine to metal sigma donation

The π interaction is backdonation of electrons from filled metal d orbitals to an empty orbital on the phosphine. Traditionally this was thought to be the 3d orbitals on phosphorus, hence "dπ–dπ". However, as Martin pointed out in the comments this is a rather outdated mode of thinking. The current accepted view is that the acceptor orbitals are $\ce{P-R}$ σ* orbitals (to be precise, a linear combination of them that has appropriate symmetry to overlap with metal d orbitals).

Metal to phosphine pi backbonding

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.