I want to ask a question about determining the oxidation state of a Platinum complex.

I was presented today with the following complex and asked to comment on the oxidation state and hence the d-orbital configuration of the central metal ion:

enter image description here

My working was as follows:

The $\ce{Me}$ groups do not contribute to any charge.

The $\ce{SMe2}$ ligands are neutral similar to $\ce{H2O}$ and don't contribute to any charge either.

However, if I take this into account, this would therefore mean that each Platinum central metal ion has the same magnitude of charge, but in opposite magnitudes i.e. $-2$ or $+2$ and this is not valid chemistry.

My other thought was whether the electronegativity of the $\ce{SMe2}$ ligand central S atom compared to the $\ce{Pt}$ meant that the bonding pair of electrons is electronegatively attracted to the S atom, meaning that for each $\ce{Pt}$ atom, the $\ce{Pt}$ atom would lose two of its own electrons, one in each opposite direction giving a +2 oxidation state.

The answer given is +2 for both metal centres but I was unsure using the approaches above whether this was valid.

What other steps am I missing to reach the final oxidation states of the $\ce{Pt}$ centres?

  • $\begingroup$ Is it a neutral molecule or an ion ? $\endgroup$
    – Maurice
    Dec 24, 2019 at 13:37
  • $\begingroup$ Neutral - there is no charge present on the molecule. @Maurice $\endgroup$
    – vik1245
    Dec 24, 2019 at 14:32
  • 4
    $\begingroup$ You can pretty much assume methyl ligands anionic, e.g. there are four $\ce{Me-}$ and two $\ce{Pt^2+},$ and call it a day. This compound is well-known in inorganic synthesis as bis((μ2-dimethylsulfido)-dimethyl-platinum(II)). Also, please note that we prefer not to use MathJax in titles. $\endgroup$
    – andselisk
    Dec 24, 2019 at 14:38
  • $\begingroup$ Is it just by bad memory, or I remember reading that coordination number of pt is ziesse's salt is not defined. Maybe that's the case here too, not sure could someone help with this $\endgroup$ Dec 31, 2019 at 9:09
  • 1
    $\begingroup$ @Zenix be careful there; Pt in Zeise's salt is undoubtedly +2. en.wikipedia.org/wiki/Zeise%27s_salt $\endgroup$ Jan 1, 2020 at 2:27

3 Answers 3


For organometallic complexes it's often easier to throw the "more electronegative element" idea out entirely and focus on the characteristics of the ligands. This is what @andselisk's comment is referring to. Generally ligands are partitioned into two types, $\ce{L}$ and $\ce{X}$:

  • $\ce{L}$ refers to neutral ligands: for example $\ce{CO}$, phosphines, $\ce{H2O}$, alkenes (via the pi bond), ... and $\ce{SMe2}$ falls comfortably in this category.

  • $\ce{X}$ refers to anionic ligands: for example halides, hydroxide, azide, alkyl anions such as $\ce{Me}$, ...

Other ligands can be constructed as sums of these: so for example η4-butadiene would be $\ce{L2}$ (two pi bonds coordinating to the metal), an allyl group would be $\ce{LX}$ (one pi bond plus one anionic centre - ignore resonance), and so on.

The oxidation state of the metal, then, is simply the charge on the metal plus the number of X ligands. In this case each Pt centre is neutral, and has two L and two X ligands. That makes for a total oxidation state of +2.

As another example, the platinum in Zeise's salt has one L ligand (ethylene), three X ligands (chloride), and a negative charge: so its oxidation state is $-1 + 3 = +2$.

Of course, this method will fall apart very quickly when there are non-innocent ligands: $\ce{NO}$ and $\ce{O2}$ are the examples that come to mind, although there are many more. But for relatively simple complexes they suffice.


There is a very similar molecule in the literature this has aryls instead of methyls. (MEYTEP)

M.A.C.Lacabra, A.J.Canty, M.Lutz, J.Patel, A.L.Spek, Huailin Sun, G.van Koten, Inorg.Chim.Acta, 2002, volume 327, page 15. In the Cambridge database this compound is considered as a Pt(II) compound in the paper.

We also have the dimethyl compound with bridging dimethyl sulfide ligands. (XOFNUB)

Datong Song, Suning Wang, J.Organomet.Chem., 2002, volume 648, page 302 which is the compound which the poster was asking about. In the Cambridge database this is considered as Pt(II). The authors regard it as being a Pt(II) compound.

I would vote for it being a Pt(II) compound as the methyl is likely to be an anionic ligand. But there is a big "but".

As it is an organometallic compound we need to be careful about oxidation state. In organometallic chemistry it is often possible to draw the same compound in two different ways with different oxidation states. For example ferrocene can be regarded as an Iron(II) compound with two anionic Cp ligands. As an alternative we can regard the iron as being in the zero oxidation state and the Cp ligands are neutral ligands. Sometimes in organometallic chemistry oxidation state becomes a bit of a joke at times.

Back to our Pt compound, if we consider the methyl ligands to be neutral then the Pt can be in the zero oxidation state the dimethyl sulfide is a ligand which can offer up a lone pair to a metal.


If you go by the "ionic" counting method, we can start by looking at the charge on the ligands:

-The 4 terminal methyl groups are anionic when using the ionic counting method: 4 anions = 4- charge.

-For SMe2, the covalent bond to the methyl groups completes its full shell-no need to assign it a charge as it is happy being bonded with the methyls (doesn't need electrons from the metal).

Total 4- charge divided onto 2 metal centres, means we need 2+ charge at each Pt centre to balance the charge (assuming neutral complex).

Going further, we can assume sulfur is a 2-electron donor with each bond it makes to a metal centre as it still has 2 lone pairs free for donation.

Now, to find geometry let's count the total number of electrons at the metal. 2 sulfurs are 2 electron donors with each bond they make, each metal has 2 bonds from a sulfur = 4 electrons from sulfur. Platinum is group 10 so with a 2+ charge it has d8 configuration - 8 electrons from the metal. As we treated the alkyl groups as being ionic, they are 2 electron donors 2x2=4.

Add them all up 4+8+4 = 16 electron total at the metal.

Although this breaks the 18 electron rule, this is fine for 5d square planar and we know the complex will take square planar geometry at each metal centre from crystal field theory as it is a 5d metal with d8 configuration.

  • 2
    $\begingroup$ The question is asking about oxidation state, not electron count. $\endgroup$ Jan 2, 2020 at 22:18
  • $\begingroup$ See: "and hence the d-orbital configuration of the central metal ion." I wasn't sure what this meant, but I assume it means geometry/crystal field splitting $\endgroup$
    – Ishy
    Jan 3, 2020 at 0:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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