Why is CrO3 an oxidising agent, but WO3 is not?

Question

So apparently CrO₃ can be used as an oxidising agent, but WO₃ can not. I saw this on an assignment question which I will include below:

I think the reason why W⁶⁺ is 6-coordinate and Cr⁶⁺ is 4-coordinate is because W⁶⁺ has a larger ionic radius than Cr⁶⁺ based on being lower in the periodic table and therefore can accommodate more anions around itself, based on basic geometry.

I am not quite sure why WO₃ is not an oxidising agent though because since they are both in the same group and have the same oxidation state, they should have the same number of d electrons in the ion (namely 0 because they both have a noble gas configuration - Cr⁶⁺ is [Ar] and W⁶⁺ is [Xe]). So in theory, their ability to act as oxidising agents (i.e. to accept electrons) should be the same - i.e. not very potent because they have a stable full shell noble gas configuration?

Does anyone have any insights into this?

Answer 1

We can look at both ionization energy and coordination aspects. Both factors differentiate the relative stability of both $\ce{MoO3}$ and $\ce{WO3}$ from the strong oxidizing tendency of $\ce{CrO3}$.

Ionization energy

Wikipedia's table of ionization energies reveals that the fourth through sixth ionization energies of chromium are significantly greater than those of its heavier congener molybdenum. In kJ/mol, the relevant values are:

$\ce{Cr}: IE_4=4743, IE_5=6702,IE_6=8745$

$\ce{Mo}: IE_4=4480, IE_5=5257,IE_6=6641$

Values for tungsten are not given, but they are expected to be similar to molybdenum based on typical periodic trends. Thus although bonds with these elements in high oxidation states (+4 or higher) are expected to be polar covalent instead of predominantly ionic, the electron transfer required to achieve the higher oxidation states is more easily accomplished with molybdenum, and thus probably tungsten, than with chromium; and so higher oxides of the heavier elements would expect to be more energetically favored.

Why is there the difference in ionization energy? In this discussion Nicolau Saker Neto notes that certain valence orbitals, such as $4f$ in the lanthanides and $3d$ in the lightest transition metals like chromium, are more difficult to ionize because they lack radial nodes. The interested reader can review the arguments in that answer.

Coordination

When we have a binary compound with a stoichiometric ratio far from $1:1$, lattice stability may be difficult to achieve because of the forced differences in coordination requirements. If the element with fewer atoms cannot achieve a sufficiently high coordination number, the bonding in a lattice will be limited and the material becomes both less stable and more reactive. Nitrides of alkali metals ($\ce{M3N}$) are a good example; only with lithium, whose relatively small ions enable the nitrogen to assume a coordination number of six or more (there are different allotropes of lithium nitride) without severe steric strain, is the nitride reasonably stable.

Similarly with the Group 6 $\ce{MO3}$ oxides, molybdenum as well as tungsten can form trioxides with six-coordinate metal, allowing more bonding than that in chromium oxide where (presumably) steric interference limits the metal coordination number to four. (Chromium does achieve a coordination number of six in lower oxidation states, where the metal center has more valence-electron density and is larger in size.)