I read that the fluorides of 3d metals in lower oxidation states, e.g. $\ce{VF2}$, $\ce{TiF2}$, and $\ce{CuF}$ are thermodynamically unstable. For example, $\ce{CuF}$ disproportionates to $\ce{Cu + CuF2}$, as described on Wikipedia.
On the other hand, their iodides tend to be more stable (so, for example, $\ce{CuI}$ can be easily obtained from a redox reaction between $\ce{Cu^2+}$ and $\ce{I-}$). Why is this the case?
In general, iodides stabilise lower oxidation states and fluorides stabilise higher oxidation states, e.g. $\ce{CuF2}$ versus $\ce{CuI}$. This can be explained with some thermodynamics. Consider
$$\ce{M(s) + $\frac{n}{2}\,$X2 (g/l) -> MX_n (s)}$$
In order to form a halide salt with a higher oxidation state (i.e. larger $n$), you need to pay some energetic costs to get the metal to the higher oxidation state, namely:
This seems bad, so why would any higher oxidation state ever be formed? The answer is that when you pay these costs, you recoup some of the energy by forming more bonds between $\ce{M^n+}$ and $\ce{X-}$. If this bonding is strong enough, this can outweigh the costs described previously, and that makes the higher oxidation state more stable than the lower one.
For fluorine, the $\ce{M-F}$ bond is strong (regardless of whether you consider it to be ionic or covalent) and hence this energetic gain is large. On the other hand, the formation of fluoride ion is not particularly costly: the $\ce{F-F}$ bond is easy to break and fluorine loves picking up electrons, so the electron affinity works in your favour.
Overall, this means that there is a thermodynamic force for formation of fluorides in high oxidation states. Conversely, if you consider iodides instead, the formation of iodide ions is not as easy as the formation of fluoride. And the $\ce{M-I}$ bonding is weaker, too, so there isn't even much incentive to go to higher oxidation states.
Let's look specifically at copper. Arguably the "outlier" here is not the nonexistence of $\ce{CuF}$, but the existence of the other halides with copper in the +1 oxidation state. Earlier transition metals in tge fourth period generally don't do that, even with iodine.
As explained here, copper has an increased second ionization energy. It's not enough to fully stabilize copper(I), but it does bring copper(I) into the game. Combine that with a soft base that takes advantage of the greater polarizabiluty of $\ce{Cu^+}$ versus $\ce{Cu^{2+}}$, and the former can be stabilized. For instance, copper(I) becomes stable in water solution when the soft base thiourea is added to the solution to provide a soft-base ligand. This does not occur with proposed metal(I) using an earlier transition metal in the period.
Among halide ions, chloride and its heavier congeners are sufficiently soft and polarizable to stabilize copper(I) and thus form $\ce{CuCl,CuBr,CuI}$; but the more compact and tightly bound fluoride ion is not. So we get specifically $\ce{CuCl,CuBr,CuI}$ as simple salts. This is still more than we see in the +1 oxidation state with earlier transition metals whose second ionization energies are lower than copper's.
As with the aqueous environment (and again with copper instead of an earlier fourth-period transition metal), $\ce{CuF}$ can become stabilized by providing a softer base that can exploit the greater polarizability of the less-positive metal ion. Wikipedia cites such complexes with triphenylphosphine and with molecular nitrogen.
