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In $\ce{[Pt3(C2H4)]-}$, how many electrons are there around platinum?

While solving this problem I thought three $\ce{Pt}$ formed a triangle with ethylene above. But I was confused about how the $\ce{C2H4}$ gave electron pairs to $\ce{Pt}$. In addition, the answer gives $16$, but I only see the initial $10$ electrons around a $\ce{Pt}$ and $2$ given by another two $\ce{Pt}$.
What is the exact structure of this coordination compound, and how can it be deduced?

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    $\begingroup$ It is very probably, that they missed three ligands, possibly chloride $\endgroup$ – permeakra Apr 28 '16 at 17:33
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Looking up your proposed sum formula, I arrived at a quantum mechanical paper that used said complex (triangular $\ce{Pt3}$ moiety, whereonto an ethylene molecule coordinates onto the longer edge of the triangle) as a modelling study for the interaction of ethylene with platinum metal surfaces.[1]

As far as I can tell,[2] the authors in now way account for a surface of platinum; their three platinum atoms in vacuum are supposed to model an entire platinum surface. I cannot say whether that type of approach is typically used or whether it is considered good by quantum chemists.

Searching further, I also found a 1981 paper describing $\ce{[Pt3(C2H4)(cod)2(cot)2]}$, a neutral complex that includes two additional cycloocta-1,5-diene ($\ce{cod}$) and two additional cycloocta-1,3,5,7-tetrene ($\ce{cot}$) ligands.[3] In this complex, the bonding situation and electron count can be well answered, but that wasn’t your question.[4]

Furthermore, I find a complex $\ce{[Pt3(C2H4)]-}$ to be highly questionable: Platinum is in the 10th group, ethylene is a singlet molecule so adding a single additional electron will create a radical anion. Another reason why your question and answer cannot be serious, since in no way can a radical end up at $16$ electrons. I would even go a step further and say that it is likely an unstable complex that would decompose depending on which species are close to it. Note that the authors of the first paper calculate an uncharged complex $\ce{[Pt3(C2H4)]}$.


Notes and references

  1. A. Cruz, V. Bertin, M. Castro, Int. J. Quantum Chem. 2000, 80, 298. DOI: 10.1002/1097-461X(2000)80:3<298::AID-QUA3>3.0.CO;2-2.

  2. I am not a quantum chemist, and my last calculation efforts date back to a research project during my master’s studies. Those acquainted with methods will have to comment on how good the paper is.

  3. N. M. Boag, J. A. K. Howard, J. L. Spencer, F. G. A. Stone, J. Chem. Soc., Dalton Trans. 1981, 1051. DOI: 10.1039/DT9810001051.

  4. Two platinum atoms with 18 electrons in a square-planar environment, one with 16 electrons in a trigonal environment. The former coordinated by $\ce{cod}$ and $\ce{cot}$ (each contributing two double bonds), the latter coordinated by two $\ce{cot}$ molecules (each contributing one double bond) and an ethylene. The three platinum centres are too far apart to be considered bonded, according to the authors.
    The bonding of the two 18-electron platinums to their respective $\ce{cot}$ unit is described as $\ce{cot^2-}$ and $\ce{Pt^2+}$, resulting in oxidation states of $\mathrm{+II}$ and $\mathrm{\pm 0}$ for platinum.

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    $\begingroup$ There should simply be PtCl3(C2H4) as permeakra said. $\endgroup$ – Mithoron Apr 28 '16 at 19:57
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    $\begingroup$ @Mithoron That may be possible; however, I prefer not to engange ni such speculation ;) $\endgroup$ – Jan Apr 29 '16 at 7:01

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