Why is Mo(VI) more stable than Cr (VI)?

Acids of $\ce{Mo}^\mathrm{VI}$ are more stable than those of $\ce{Cr}^\mathrm{VI}$ in terms of redox reactivity.

In p-block elements, as we go down the groups, the stability of lower oxidation state keeps on increasing. I understand how this can be explained with the inert pair effect.

On the other hand, as we go down the group from $\ce{Cr}$ to $\ce{Mo}$, why does the opposite happen and there isn't any inert pair effect? Shouldn't the lower oxidation state be more stable down the groups in d-block elements? Wouldn't there be less shielding in $\ce{Mo}$ making is less electropositive than $\ce{Cr}$?

The inert pair effect plays no particular role in the stability or instability of transition metals’ oxidation states. It is due to the large and rising energy difference between a shell’s s- and p-subshells which make the ionisation of the lower electrons more difficult. However, it is equally wrong to say that lower oxidation states be stabilised in lower periods: the most stable halide with the lowest oxidation state a halogen can have is fluoride ($\mathrm{-I}$).

All 3d-metals before iron can be oxidised up to their highest oxidation state while being strong oxidants in that state. This is not so true for the 4d-metals; an indication that the 3d electrons are better attracted to the nucleus. Indeed, the 3d orbitals do not contain radial nodes and their electrons are thus closer to the nucleus. The radial nodes in 4d and higher d-orbitals ‘add distance’ between the nucleus and the corresponding d-electrons (do not copy that wording: it is bad). It is this proximity to the nucleus and the consequential lesser extension of the orbitals into space that draws electrons into them, destabilising the higher oxidation states and stabilising slightly lower ones.

The same effect is true for f-elements; see Nicolau’s answer to a related question on f-element oxidation states’ stabilities.