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?

  • 4
    $\begingroup$ Cu(I) has only one positive charge. That softens things up. $\endgroup$ – Oscar Lanzi Feb 26 '17 at 1:07
  • $\begingroup$ @xahoc Thank you for your answer! But I still don't understand: the orbital sets derived from d-orbitals remain populated by the electrons of the metal only, and the configuration remains (t2g)^3 (eg)^2 in all complexes. Is it something to do with the spin exchange energy decreasing as t2g and eg sets get farther apart in energy? Is it possible to show the spin exchange energy depends on two orbitals being close in energy mathematically? $\endgroup$ – GingerBadger Feb 26 '17 at 10:56
  • $\begingroup$ @xahoc Moreover, even if electrons were added to t2g, the situation would be analogous to adding an electron to Ti2+ d orbitals, i.e. neither increasing nor decreasing the overall pairing energy. $\endgroup$ – GingerBadger Feb 26 '17 at 11:04
  • 1
    $\begingroup$ 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. $\endgroup$ – Xahoc Feb 26 '17 at 12:41
  • 1
    $\begingroup$ 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). $\endgroup$ – Xahoc Feb 26 '17 at 12:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.