[Morphic flip]

I`m trying to find the oxidation state of a metal in a compound that changes throughout a catalytic cycle.
Of these ligands, do I have the oxidation states correct? Then I balance up the ligands and metal to achieve the oxidation state of the metal?

H⁺ = -1
CH₂OH = 0
PPh₃ =-2
H₂CO = 0
(CO)₂ = 0
CHO = -1

IE:
HRh(H₂CO)(CO)₂PPh₃
Oxidation state of Rh is +3 ??

[Albert]

I think it's:

CH2OH = -1
PPh3 = 0

 

[Dan]

HRh(H₂CO)(CO)₂PPh₃
Oxidation state of Rh is +3 ??

I'd say that was Rh(I)

Because:
The H is formally H^-. H₂CO is formaldehyde, so is probably co-ordinated in a n² fashion, and is uncharged. CO and triphenylphosphene are also uncharged.

[Morphic flip]

So to balance the compound it`s only the ligands with a minus charge added up which gives the oxidation state of the metal?

[Dan]

Yeah, only formally charged ligands, like hydride, halides, alkyls, hydroxide, alkoxy etc will increase the ox state of the metal.
Uncharged ligands, like phospenes, carbonyl, alkenes etc. do not increase the ox state of the metal.

eg.

Cr(CO)₆ is a Cr(0) species.

Pt(NH₃)₂(Cl)₂ is a Pt(II) species

[wereworm73]

Usually the formula for hydrides has the hydrogen written after the metallic atom(s), though.

[Morphic flip]

How many electrons would be around the Rh in HRh(H₂CO)(CO)₂PPh₃


I make it 14 without the bond from the PPh₃ ligand, I can`t decide whether the bond should be single or double.
So if it was a single bond from PPh<sub>3</sub>, I`d make it 16 electrons around the Rh.
Thank`s.

[Dan]

yep, 16 electrons. I've never seen a double bond to a phospine ligand.

[Morphic flip]

Thank`s.

[Dan]

Morphic flip, in reply to the message you sent me, I think these are the complexes you are dealing with.

The first one is a 16 electron Rh(I) complex
The second is an 18 electron Rh(III) complex

[Morphic flip]

Hey, cheers for the scans