Horza2002 Posted February 23, 2010 Posted February 23, 2010 Hi, Its revision time again! I came across this metal complex [Co(NH3)5 (NCS)]2- This implies that the Co is in the -1 oxidation state and therefore a d10 species. My question is this, how can a metal have a negative oxidation state and still form a complex. Surely the negative charge would repel the incoming, potential ligands; especially the anion thiocyante. Also if its a d10 species, it will have no CFSE and therefore be very labile and will exchange with the solvent it is present in...
UC Posted February 23, 2010 Posted February 23, 2010 That would appear to be what is commonly referred to as a typo. The majority of cobalt complexes you will find have a Co(III) ion at the core, and if that - in your equation was actually a +, the +2 charge would make perfect sense. What reference is this from?
Horza2002 Posted February 23, 2010 Author Posted February 23, 2010 Yer after a little bit of thought and discussion I also came to it must be +2 not -2. It was a question on last years exam paper from our university. The lecture who set it is not here anymore and so we couldn't go ask him.
John Cuthber Posted February 24, 2010 Posted February 24, 2010 (edited) I think that cobalt might be in a negative oxidation state (-1 as it happens). It would be odd with NH3 as a ligand. If the question had been about the rather better known NaCo(CO)4 what would you have said about it? Is there any more information about it in the question? Edited February 24, 2010 by John Cuthber
Horza2002 Posted February 24, 2010 Author Posted February 24, 2010 The questions is: Predict the products of the electron transfer reactiong giving your reasoning. Sketch the structure of the electron transfer transition state that is consistent with your reaction. [Co(NH3)5(NCS)]2- + [Cr(OH2)6]2+ ==> ? Well the thiocyano group is a bridging ligand and so I would expect an inner sphere mechanism with the sulphur displacing a water from the chromium complex. If I make the assumption that it is a typo and it is actually Co(III) d6, then the Cr(II) d4, would become Co(II) d7 and Cr(III) d3. The Cr(III) would then be low spin and reasonably inert.
John Cuthber Posted February 24, 2010 Posted February 24, 2010 With that other information it looks very much like it's a typo.
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