15
$\begingroup$

By palladium and platinum carbonyls, I mean the mononuclear, homoleptic, charge-neutral, binary carbonyl complexes of the form $\ce{M(CO)_n}$ and not any heteroleptic complexes like $\ce{Pd(CO)(PPh3)3}$ and similar compounds of platinum which are abundant in chemical industry and used as catalysts.


So, I was reviewing carbonyls of transition metals when I encountered this statement:

Group 10 elements with 10 valence electrons form metal carbonyls $\ce{Ni(CO)4}$ ($10 + 4 \times 2 = 18$ electrons). Curiously $\mathbf{Pd(CO)_4}$ and $\mathbf{Pt(CO)_4}$ are not stable.

I am proposing that heavy metal carbonyls tend to be unstable(?) and decarbonylate to form polynuclear carbonyls. For example: $\ce{Ru(CO)5}$ and $\ce{Os(CO)5}$ being unstable form $\ce{Ru3(CO)12}$, and $\ce{Os3(CO)12}$. But in the case of palladium and platinum, no carbonyl complex seems to exist in stable condition at room temperature. I reviewed two papers which described the Group 10 carbonyl complexes (links below):

  1. Complexes where the carbonyl group is the only ligand present have been reported for all the platinum group metals except palladium.[...] No substantiated reports have been published on platinum or palladium compounds similar to the nickel carbonyl $\ce{Ni(CO)4}$. Thus the polymeric platinum dicarbonyl is the only known carbonyl of these two metals and even this compound is unstable in air.[...]

    The unstable polymeric platinum dicarbonyl $\ce{Pt_n(CO)_{2n}}$ has been obtained as a cherry red precipitate by the action of carbon monoxide on aqueous or ethanolic solutions of Pt(II) compounds (e.g. $\ce{K2PtCl4}$) at temperatures up to 80 °C; or as a purple colloidal precipitate by the action of water on a benzene solution of $\ce{Pt(CO)2Cl2}$ under an atmosphere of carbon monoxide.

  2. It was not until 1972 that a definite existence of the congeners $\ce{Pd(CO)4}$ and $\ce{Pt(CO)4}$ was reported. Using matrix isolation and Raman techniques, data for the products of Pd atoms CO and CO/Ar cocondensation reaction were able to characterize the series of binary complexes $\ce{Pd(CO)_n}$ analogous to $\ce{Ni(CO)4}$, the most stable being $\ce{Pd(CO)4}$ but still unstable than $\ce{Ni(CO)4}$ which explains the failed attempts in its synthesis. Similar technique were used in making $\ce{Pt(CO)4}$ at 4.2-10 K which is stable up to 30 K. (slightly abridged)

The question is: why they are unstable/don't exist at room temperature?


  1. Platinum Metals Rev., 1972, 16, (2), 50 "The Carbonyls of the Platinum Group Metals" By C. W. Bradford, Research Laboratories, Johnson Matthey & Co Limited (link)
  2. J. Am. Chem. Soc., 1973, 95 (22), pp 7234–7241 "Binary carbonyls of platinum, Pt(CO)n (where n = 1-4). Comparative study of the chemical and physical properties of M(CO)n (where M = nickel, palladium, or platinum; n = 1-4) DOI: 10.1021/ja00803a009
$\endgroup$
0
1
$\begingroup$

Tl, dr: It's not just a matter of size. A combination of coordination geometry, node distribution in the $d$-orbital wavefunctions, and atomic size all conspire to weaken $\pi$ overlap in palladium and platinum carbonyls, not only relative to the nickel compound but also relative to earlier transition metal carbonyls in their respective periods.

Ziegler et al. [1] studied the relative stabilities of metal-carbonyl bonds for $\text{G6(CO)}_6$, $\text{G8(CO)}_5$ and $\text{G10(CO)}_4$ carbonyl complexes, where $\text{G6}$ in the first formula is a Group 6 metal and similarly for the second and third formulae. The following general trends were observed:

  1. Within a period, the G6 hexacarbonyl is most stable.

  2. Within each group, the 4th period carbonyl is most stable, the 5th period carbonyl is least stable and the 6th period carbonyl is slightly more stable than the 5th.

From these trends it is seen that palladium and platinum carbonyls are literally in a bad place.

The authors attribute the stability order primarily to $\pi$ backbonding. We may expect such backbonding to be strongest for the G6 hexacarbonyls where the octahedral geometry about the metal atom causes the metal-atom donor orbitals to have pure $\pi$ symmetry. The five- and four-coordinate geometries (trigonal bipyramidal and tetrahedral, respectively) are progressively less favorable with regard to this symmetry. This accounts for effect (1).

Effect (2) is related to relative atomic sizes within each respective group, but also to a more subtle effect. In the $3d$ subshell, which is used as the valence subshell in the 4th period, the lobes have no radial nodes, a condition that allows maximum overlap with the molecular orbitals coming from the carbon monoxide. The $4d$ and $5d$ orbitals used by the heavier metals have radial nodes so poorer overlap. With the 6th period metals, relativistic effects intervene, but this effect (which I cannot tell from the abstract, and the full paper is behind a paywall) is less potent than having more compact orbitals with no radial nodes at all in the $3d$ series.

Reference:

1. Tom Ziegler, Vincenzo Tschinke, and Charles Ursenbach, "Thermal stability and kinetic lability of the metal carbonyl bond. A theoretical study on M(CO)6 (M = chromium, molybdenum, tungsten), M(CO)5 (M = iron, ruthenium, osmium), and M(CO)4 (M = nickel, palladium, platinum)", J. Am. Chem. Soc. 1987, 109, 16, 4825–4837. Link

$\endgroup$
0
$\begingroup$

I don't know definitively but we know that carbon monoxide acts as a strong sigma donor and pi acceptor of electrons when acting as a ligand. Palladium being huge, has dispersed valence electrons and perhaps due to this, sigma donation to the metal is repelled by the orbitals and/or the dispersed valence electrons do not back bond as strongly as lighter metals such as nickel (whose carbonyl complex is quite stable). Metals such as nickel may be able to donate more e density to those empty P orbitals on carbons monoxides carbon due to more compact and dense valence orbitals and thus form stronger bonds.

$\endgroup$
2
  • 2
    $\begingroup$ You'll get more upvotes with a reference to back up this argument. $\endgroup$ – Oscar Lanzi Dec 15 '19 at 22:04
  • 2
    $\begingroup$ Palladium is smaller than Molybdenum, and the same size as Chromium, Iron and Manganese (en.wikipedia.org/wiki/Atomic_radius) all of which form stable carbonyls. Platinum is smaller than palladium $\endgroup$ – Ian Bush Apr 14 '20 at 7:56

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.