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I came across this compound: $\ce{[Pd(η^2-C2H4)Cl2(H2O)]}$

What does that '$\eta^2$' sign mean? I am guessing that it means that the ligand is bidentate because if it wasn't, I don't think it is possible (well it would be rare) for a metal to be only co-ordinates by 3 ligands.

If this is the case, how can the carbon atom in ethene bond to Pd as it doesn't have a lone pair? Does the double break and hence the 2 carbons bond to the metal?

Ivan Neretin
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Nanoputian
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  • I took the liberty of changing your title. See, it's not really about nomenclature, it's more about the nature of this compound. – Ivan Neretin Oct 01 '15 at 08:52
  • ‘because if it wasn't, I don't think it is possible (well it would be rare) for a metal to be only co-ordinates by 3 ligands.’ There are two chlorines, water and ethylene. Assuming ethylene be bidentate, that would be five ligands; otherwise four. Perfectly normal for palladium. Also, complexes such as $\ce{[Ag(CN)2]-}$ (two ligands) exist. – Jan Oct 01 '15 at 17:32

2 Answers2

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Ethylene is a monodentate ligand and there are 4 ligands total: ethylene, two chlorides, and water.

Regarding hapticity, you can find a great answer here: What does the eta notation mean in the naming of a transition metal complex?

While a chapter can be dedicated to the nature of organometallic bonding, I think it is practical to say that it is the two electrons in the pi-bond that are involved in bonding to the metal center.

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Aaron
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It is not carbon that is donating a pair - instead, it is a $\pi$ bond as a whole. Its two electrons, which were shared between two carbons, are now shared between them and palladium in a kind of multicentered bond.

There is a huge class of compounds where organic molecules are attached to a metal in the same way. With larger conjugated $\pi$-systems, we may have ligands of $\eta^5$ coordination (look up ferrocene) and more, all the way to $\eta^8$ and maybe beyond that.

Ivan Neretin
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  • I would +1 this, but the OP obviously does not know about hapticity, so I feel you should include a note on what the notation actually means. – Jan Oct 01 '15 at 17:33
  • That's covered by the other answer already, so why bother. – Ivan Neretin Oct 01 '15 at 17:36