Why are fluorides of transition metals unstable in low oxidation states?

I would think that since fluorine and oxygen are highly electronegative, it's obvious that they be stable at high oxidation states. However, $\ce{Cl}$ is also highly electronegative, but it does exist in low oxidation states with transition metals.

For example $\ce{CuCl}$ exists but not $\ce{CuF}$ and $\ce{TiCl2}$ exists but not $\ce{TiF_{2}}$. Why is it so?

Fluorides of transition metals unstable in low oxidation states

This is the statement given in my high school chemistry text book but no explanation provided. I am interested to know the reason for it.


1 Answer 1


You can take the example of $\ce{CuF}$ which is not known.

Reason: $\ce F$ is such an electronegative element that it will always oxidise $ \ce{Cu}$ to $\ce{Cu^2+}$ and not $\ce{Cu^1+}$. Hence, whenever $\ce{Cu}$ reacts with flourine, copper(II) flouride is formed. The following reaction illustrates this:

$$\ce{Cu + F2 -> CuF2}$$

$\ce{CuF2}$ loses fluorine at temperatures above $\pu{950 °C}$. $\ce{CuF}$ will be formed initially and as your question says, it is highly unstable so it will be transformed to some other stuff and the reactions are:

$$\ce{2CuF2 -> 2CuF (unstable) + F2}$$

$$\ce{2CuF -> CuF2 + Cu}$$

Similarly, you can illustrate this for $\ce{VF2}$.

  • 1
    $\begingroup$ Your answer meanders around in a non-meaningful way. If anything, the final two reactions are relevant. For example, what happens if copper(I) ions and fluoride ions meet in solution? In the case of copper(II) and iodide, a redox reaction will take place due to the respective standard potentials forming copper(I) iodide and iodine. But fluoride ions won’t be reduced any more in solution. $\endgroup$
    – Jan
    Oct 18, 2017 at 13:21
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    $\begingroup$ Actually it could disproportionate, en.m.wikipedia.org/wiki/Copper(I)_fluoride. CuF is better known in complexes where the copper is also bound to a softer base. $\endgroup$ Nov 18, 2017 at 1:03

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