# 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?
• You can consider ligands with soft character to bind covalently with metals in ways that informally affect the electrons on the d-orbitals of the metal. Consider $\ce{d^6 Mo(CO)_6}$ versus $\ce{d^6 Mo(H_2O)_6}$. The pi backbonding capability of $\ce{CO}$ reduces electron density in the $\ce{d_{xy},d_{xz},d_{yz}}$ orbitals, lowering their energies. You can imagine the introduction of a pi donor such as $\ce{I^-}$ would increase electron density in those same orbitals. While additional bonding to soft/weak Lewis acid like CO may be OK, covalent bonding to soft Lewis bases is disfavored. – Xahoc Feb 26 '17 at 12:41
• Also, in the case of $\ce{d_1,d_2 (Ti^{2+})}$ ions, there is a vacant $\ce{T_{2g}}$ orbital for electrons to enter, which wouldn't involve a pairing penalty while conversion of high spin $\ce{d_5}$ to $\ce{d_6}$ would have such a penalty. (High spin electron transfer would also require a spin-forbidden $\ce{T_{2g}}$ to $\ce{E_g}$ transition, but this is a kinetic, and doesn't affect the question). – Xahoc Feb 26 '17 at 12:51