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Although I am okay with rationalising bridging halides and $\mu$-2 CO I can't seem to figure out others, including $\mu$-2 hydride, CN- and NO.

When considering the electron count of bridging ligands, this is my rationale so far:

-Neutral bridging ligands contribute to the overall complex the number of electrons they would have when terminally bonded. These electrons are given in any proportion. This means that $\mu$-2 bridging CO gives 1 electron to a metal centre, for example.

-Ionic counting gives even numbers of electrons for ligands; I am unsure if this is still the case for bridging ligands - I suspect not?

Considering NO, I would therefore believe it to be a 2 electron donor to one metal and a 1 electron donor to the second metal.

Considering hydride, more specifically the dimer form of titanocene, I have been taught that each Ti centre is 16 electrons. I have been told, separately, that each hydride contributes half an electron in neutral counting, and 1 electron in ionic. This would lead to a 16 electron complex for "Titanocene" if there is a Ti-Ti bond, however this source: https://link.springer.com/content/pdf/10.1007%2Fs40828-015-0010-4.pdf seems to imply that it is in fact an 18 electron complex, with each individual hydride contributing more than its total electron count? Is this source correct (and my lecturer wrong) in how you would count this complex?

Considering CN-, this is similar to hydride I believe, and should be treated in the same way - is this correct?

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  • $\begingroup$ Are you using L or X? $\endgroup$ – Zhe May 15 '18 at 11:54
  • $\begingroup$ Sorry, I'm not quite sure what you mean? $\endgroup$ – K.P. May 16 '18 at 14:35
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    $\begingroup$ If you're counting electrons, you're using the L method (counting ligands as neutral) or the X method (counting some ligands as anionic). Which are you using? $\endgroup$ – Zhe May 16 '18 at 15:05
  • $\begingroup$ I'm using predominantly the L method, although I am also considering the X method for hydride and CN-. $\endgroup$ – K.P. May 17 '18 at 10:37
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In the covalent (L) method, $\mu$-hydrides should be treated as a normal hydride (1-electron ligand) for one metal. The $\ce{M-H}$ bond itself should be treated as a 2-electron ligand for the other method. This explains the two entries for hydride in the link you provided:

mu-hydride titancoene

Cyano-ligands are isoelectronic to carbonyl ligands, so I'll leave that to you as an exercise to figure out.

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  • $\begingroup$ My issue with cyano ligands is that although they are isoelectronic to carbonyls, they have a negative charge. Therefore my treatment of carbonyls are: 1) Terminal ligand, 2 electron donors, 2) mu-2 bridging, 1 electron donor to each centre in both L/X methods. However, with CN-, I treat terminal, L method as a 1 electron donor and X method as a 2 electron donor, therefore shouldn't the bridging considerations still be different? Charge notwithstanding it would also behave as a 1 electron donor per bridging centre however it is similar to hydride in how you treat the terminal ligand? $\endgroup$ – K.P. May 18 '18 at 15:08
  • $\begingroup$ Therefore wouldn't that suggest that you could treat the cyano ligand reasonably as how you treat the hydride in the bridging case? $\endgroup$ – K.P. May 18 '18 at 15:12
  • $\begingroup$ That would be fine. Alternatively, by analogy to CO is perfectly fine as well. Consider a complex with 2 bridging cyano ligands. Change these to bridging CO ligands and increase the charge of the overall complex by 2. $\endgroup$ – Zhe May 18 '18 at 17:34

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