Why is WF6 stable whereas CrF6 is unknown?

Question

$\ce{F-}$ is a hard base since it is small and relatively polarizable.

Both $\ce{Cr^6+}$ and $\ce{W^6+}$ are hard acids, but shouldn't $\ce{CrF6}$ be favored since it would be a smaller hard acid?

Answer 1

The answer has to do with two things. Note that HSAB theory is dubious at best and doesn't have very useful predictive power, so I am going to avoid talking about it.

(1) The accessibility of the high +6 oxidation state

In Cr, the 3d electrons drop in energy extremely rapidly as you remove electrons. So, it is much harder to remove multiple electrons one after another. On the other hand

In W, the 5d orbitals have two inner radial nodes. The two pockets of electron density close to the nucleus shield the bulk of the electron density, which is far away from the nucleus. So, when you continuously remove electrons from W, the remaining 5d electrons don't drop in energy so rapidly.

This is partly reflected in the ionisation energies for Cr and W (data from NIST Atomic Spectra Database), all units are in eV).

Element	IE1	IE2	IE3	IE4	IE5	IE6
Chromium	6.76651	16.486305	30.959	49.16	69.46	90.6349
Tungsten	7.86403	16.37	26.0	38.2	51.6	64.77

For an alternative perspective, here are Latimer diagrams for Cr, Mo and W (from Shriver et al., Inorganic Chemistry 6ed.); you can see that the stability of the higher oxidation states increases going from Cr to Mo to W (i.e. the reduction potentials are lower)

This trend is hardly limited to the Group 6 elements. $\ce{OsO4}$ features Os(VIII), whereas the highest oxidation state of iron is Fe(VI), which is already extremely difficult to achieve. Likewise Au(III) is pretty normal but Cu hardly goes above Cu(II).

(2) Sterics

Cr(VI) can be made, but the only Cr(VI) compounds that we know of are paired with the oxide ion, viz. $\ce{CrO3}$, $\ce{CrO4^2-}$, $\ce{Cr2O7^2-}$.

So there is likely also an argument here that chromium, being a smaller metal, cannot quite fit six fluorines around it in the same way that tungsten can. It can fit four oxygens, but that's about it.

Answer 2

Orthocresol's answer sums up the comparison between $3d$ valence transition metals on the one hand, and $4d$ or $5d$ valence transition metals on the other. But there is a second aspect to this problem: the $3d$ metals like chromium can still reach high oxidation states with oxide ligands, not so much with fluoride. The key here is not hard/soft basicity but pi electron interactions: oxide ligands are powerful pi donors and can back-donate to the empty, or at least depleted, $3d$ orbitals that such oxidation states as $\ce{Cr(VI), Mn(VII), Fe(VI)}$ imply. Fluoride ions do not do as well here, and so cannot stabilize high oxidation states as readily in the $3d$ series.