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Topic: Determining oxidation states in metal complexes  (Read 857 times)

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Offline LNTH

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Determining oxidation states in metal complexes
« on: June 16, 2019, 08:37:54 PM »
I've been reading about Cobalt C-H activation reactions and wondering about the oxidation states.

In high valent Co(III) C-H activations Cp*Co complexes are largely used.

In most papers, I see [Cp*Co(CO)I2] and [Cp*CoCl2}2] (dimeric complex) as pre-catalyst turning into a cationic complex with AgSbF6 additives, or in some cases cationic Co as [Cp*Co(C6H6)][B(C6F5)4]2 is used.

I know that Cp* can give 5 or 6 electrons and Iodine sometimes give 2 or 1 electrons but in this case, considering the 18e rule, how can I count electrons and figure out Co oxidation states?
« Last Edit: June 16, 2019, 09:42:47 PM by LNTH »

Offline chenbeier

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Re: Determining oxidation states in metal complexes
« Reply #1 on: June 17, 2019, 07:49:25 AM »
The oxidation state is only driven by the anions.

[Cp*Co(CO)I2]  Co2+ becuse 2 I-
[Cp*CoCl2}2] (dimeric complex) Co2+ because of 2 Cl-
[Cp*Co(C6H6)][B(C6F5)4]2 Co2+ because 2 [B(C6F5)4]-

Cp, C6H6 and CO, are neutral ligands.

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