December 08, 2019, 09:22:53 AM
Forum Rules: Read This Before Posting


Topic: Hydrogenation with palladium  (Read 313 times)

0 Members and 1 Guest are viewing this topic.

Offline Catheryba

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +0/-0
Hydrogenation with palladium
« on: October 09, 2019, 06:00:08 PM »
Hi, first of all thank you x for accepting me in the forum! I have a doubt to expose you. In the palladium-hydrogen bond (pd-H2) I assumed a sigma donation from hydrogen to the orbital eg empty of palladium and a pigreco retrodonation from one of the t2g orbitals filled with palladium to hydrogen. I'm getting closer: pd (0) has electronic configuration [kr] 4d10 or what is the same [kr] 4d8 5s2. Considering the latter, it would mean having 3 full and two half-full orbitals. Suppose those with higher energy are the dz2 and the d dx2-y2 and promote 1 electron in the dx2-y2 in order to leave the dz2 empty (in truth in the isolated palladium all the orbitals d should be degenerate ...). At this point the empty dz2 interacts with the full sigma of the H2 and a first complex is formed in which the two hydrogen atoms, still bound together, also bind to palladium. At this point palladium, using one of its full orbitals (any one of t2g), can bind to the asterisked sigma of the hydrogen and give an electronic doublet. At this point a new palladium-hydrogen bond is formed, and the H-H bond is cleaved as the ligand interaction is broken by the presence of 2 electrons in the asterisked sigma. At this point a new complex is formed in which palladium is oxidized to 2+ because the donated electrons are attributed to hydrogen. In reality I do not agree since, although generally the metals have less electronegativity than the ligand, in our case both hydrogen and palladium have electronegativity 2.20. Once the electrons have been transferred to the asterisked sigma of H-H which orbitals are involved in the H-Pd-H bond? Should we assume that the splitting of the hydrogen molecule, due to the acceptance of an electron pair, leads to the formation of two hydride ions that bind to the two empty orbitals of palladium (ie to the dz2 and to the ex-orbital d solid)? From the moment in which H-Pd-H reacts with an alken, a sigma p-p donation from the alkenes to the orbital of palladium and a pigreco retoning from one of the t2g orbitals filled with palladium to the algene antilegic oleocyte should occur. The problem is that I no longer have empty orbitals at this point. You might think about promoting the dx2-y2 electrons but I don't know if it's correct. Maybe you could think of hybridizing a 5s2 orbital already full with a 5p to get two half-full sp orbitals or, following the promotion, have a full sp and an empty sp and then do a further promotion to move the electrons from dx2-y2 to sp empty. Clearly it is only a hypothesis, but I would like to understand in detail how the reduction of an alkene at the level of the molecular orbitals involved occurs. Waiting for your reply, I wish you all a good evening.

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 3202
  • Mole Snacks: +277/-57
Re: Hydrogenation with palladium
« Reply #1 on: October 11, 2019, 09:36:40 AM »
Hi Catheryba,

Can you confirm you're speaking about atomic palladium? Up to some 3000K palladium is essentially solid or liquid, that is, polyatomic.

Sponsored Links