3
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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:

  • more electrons are removed from the metal, so you need more ionisation energies
  • you need to generate more $\ce{X-}$ anions, so you need to break more $\ce{X-X}$ bonds (bond dissociation energies) and add more electrons to them (electron affinities).

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.

$\endgroup$
  • $\begingroup$ Why is the M-F bond stronger, though? I mean, even the 3d series metals are quite big, so the lattice will probably not stabilize there...? $\endgroup$ – YUSUF HASAN Dec 28 '18 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.