inorganic chemistry - Why is WF6 stable whereas CrF6 is unknown?

$\ce{F-}$ 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

The answer simply has to do with 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; the only Cr(VI) compounds that we know of are paired with extremely hard bases like the oxide ion, viz. CrO₃, CrO₄²⁻, Cr₂O₇²⁻.

Apparently, fluoride isn't quite hard enough to stabilise Cr(VI). Another factor is sterics - squeezing six fluorines around Cr is relatively difficult as compared to having three or four oxygens.

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.

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 (more obvious if you construct a Frost diagram, but I'm too lazy to do so):

This trend is hardly limited to the Group 6 elements. 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).