Why does Mn (II) behave as "hard"? Why, within HSAB, does $\ce{Mn^{2+}}$ behave as a hard acid?

If we go across the third block, then all metals to the left of manganese form soft divalent cations (e.g. $\ce{Ti^{2+}}$, $\ce{V^{2+}}$, $\ce{Cr^{2+}}$) and all to the right are borderline (e.g. $\ce{Fe^{2+}}$, $\ce{Co^{2+}}$, $\ce{Ni^{2+}}$, $\ce{Cu^{2+}}$, $\ce{Zn^{2+}}$).

An example of the effect of its "hardness" on reactivity is that $\ce{Mn^{2+}}$ forms more stable complexes with oxalate ($\beta=10^{3.8}$) than with ethylenediamine ($\beta=10^{2.2}$).
It has to do with the fact that $\ce{Mn^{2+}}$ is a high-spin $d^5$ ion (in most cases) which is afforded special stability, as all frontier $d$ orbitals are half-filled. This reduces affinity for more covalent (soft) interactions that would involve disrupting the $d^5$ state by addition of electrons. So, the character is more ionic.

Editing after some additional searching: This also explains why ions such as $\ce{Fe^{3+}}$ are also hard. However, it seems to contradict in cases like organocuprates, where it would seem that the formation of $\ce{Cu(I)}$ would result in a similarly stabilized $d^{10}$ ion (same with silver). In these cases, is it simply because polarizability of transition metals in that region outweigh the $d^{10}$ stabilization?