Talk:Tetrakis(triphenylphosphine)palladium(0)

Square planar?

The article states Pd(PPh₃)₄ has a square planar geometry around palladium. I tried making a model of this molecule - and it was apparent a square planar geometry would be very sterically hindered. So I looked in Greenwood & Earnshaw, and this is what I found in the chapter Nickel, Palladium and Platinum: Compounds of the type [M(PR₃)₄], of which [Pt(PPh₃)₄] has been most thoroughly studied... ...are tetrahedral molecules...

Whoever orignally wrote the square planar bits, can you state your reference(s), because this content would appear to be erroneous.

Cheers Ben 12:24, 11 June 2006 (UTC).

I wasn't the person who wrote the square planar bits, but I think that was correct. Building a model is very inaccurate as Palladium Phosphorous bonds are long, so steric hindrance isn't a problem. Palladium and platinum are extremely different, as the presence of the f electrons for platinum alters things dramatically. In addition, if you look up any Palladium cross coupling mechanism, they go via square co-ordinate intermediates, as there is cis-trans isomerisation as part of the intermediate. I'll try and find some references, but as I'm writing a thesis at the moment, don't actually have much time.

Acidcat 12:15, 10 November 2007 (UTC)AcidCat

Acidcat: nonsense! Tetrahedral complex with four phosphorus (not phosphorous) ligands around a d¹⁰ Pd(0) center. --Smokefoot 13:43, 10 November 2007 (UTC)

I think Pd(II) is usually square planar, but that's not the case here. --Itub 12:38, 12 November 2007 (UTC)

Pd(II) with 4 ligands: d8 + 4*2 electrons = 16, which would suggest a square planar geometry. Pd(0) with 4 ligands: d10 + 4* 2 electrons = 18 e, which would suggest a Td geometry more. --Rifleman 82 13:26, 12 November 2007 (UTC)

You're correct that the 16 e compound is square planar and the 18 e compound is tetrahedral. An 18 e square planar compound would be energetically unfavourable due to the high energy of the d_x²–y² orbital, so the complex would prefer (distorted) tetrahedral geometry. Regarding cis/trans isomerism, if the palladium complex were synthesised as demonstrated in the article, it would be exclusively trans, not cis. This is because the trans effect of the phosphine ligands is greater than the chloride ligands. See Chapter 27 in Greenwood & Earnshaw, Chemistry of the Elements (2^nd edition), where it states that "Zero dipole moments indicate that the [PdX₂(PR₃)₂] compounds are invariably trans, whereas those of platinum may be either cis or trans...". I've changed the equations accordingly. See also my comment below regarding the references, which say nothing about what is stated in the text in the article. Kingky (talk) 13:19, 29 August 2014 (UTC)

References incorrect

References 2 and 3 are used to describe the preparation of [Pd(PPh₃)₄] but they, in fact, describe palladium phosphite compounds (i.e., with OPR₃ ligands). These should be replaced. A good start would be Greenwood & Earnshaw, Chemistry of the Elements (2^nd ed.), chapter 27. Kingky (talk) 13:09, 29 August 2014 (UTC)

Incorrect. Reference 3 describes the title compound. --Smokefoot (talk) 13:35, 29 August 2014 (UTC)