# Does fluoride anion ever act as a strong field ligand?

In the case of $$\ce{[Fe(H2O)6]^3+}$$ there are $$5$$ unpaired electrons as water acts as a weak field ligand. But instead of fluoride acting as a weak field ligand in the $$\ce{[MnF6]^4-}$$ , why does it act as a strong field ligand? In what conditions does it act as a strong field ligand and a weak field ligand?

• Can anybody share their thoughts on this question? Many also consider $\ce{F-}$ to be SFL with $\ce{[NiF6]^2-}$ but not $\ce{[NiF6]^4-}$ – Sir Arthur7 Aug 18 '20 at 7:04
• Even with [PtF4]2- behaves as a strong field ligand I think (low spin) – user600016 Aug 18 '20 at 8:14
• You're asking the wrong question. Whether F- "behaves" as a strong- or weak-field ligand in these complexes has nothing to do with fluoride itself, because fluoride is a constant factor in all of these examples. It's not some magical property of fluoride that's being turned on or off depending on what complex it's in. You should really be asking about the metal and its oxidation state. – orthocresol Aug 18 '20 at 8:16
• @orthocresol Yes that's why I added another example, wherein both the central metal atom and the ligands are same, the only difference being the oxidation state of the metal atom. So my question is rather which "magical property" of the metal/ its oxidation state is being turned "on/off" in such cases? – Sir Arthur7 Aug 18 '20 at 10:03
• May be it related to ligand EN and metal oxidation state?this what CFT tell us. – Jack Rod Aug 18 '20 at 11:50