Cr(II) and Mn(III) - their oxidizing and reducing properties?

My textbook states that $\ce{Cr^2+}$ is a reducing agent while $\ce{Mn^3+}$ is an oxidizing agent in spite of both having $\ce{d^4}$ configuration. The explanation states that when $\ce{Cr^2+}$ gets oxidized to $\ce{Cr^3+}$, it attains a stable half-filled $\mathrm{t\_{2g}}$ orbitals but when $\ce{Mn^3+}$ gets reduced to $\ce{Mn^2+}$, it attains a stable half-filled $\mathrm{d}$ orbitals.


First, is it a reasonable explanation? If this is the case, why can't $\ce{Cr^2+}$ get reduced to $\ce{Cr+}$ to attain a stable half-filled $\mathrm{d}$ orbitals and $\ce{Mn^3+}$ get oxidized to $\ce{Mn^4+}$ to attain a stable half-filled $\mathrm{t\_{2g}}$ orbitals, *i.e.* in-short is the *vice-versa* true?




Related question with same answer but in a different context of the 4f block: [Why don't we see these lanthanide species?](https://chemistry.stackexchange.com/q/53712/16683)


You have a misconception regarding the stability of oxidation states. The factors you have listed are honestly not very important in determining the stability of a certain oxidation state. They will tip a delicate balance in favour of one oxidation state or another, but they are hardly the sole determining factors.


If you are going from $\ce{M+}$ to $\ce{M^2+}$, you need to consider two main things.


1. You need to ionise the second electron from $\ce{M+}$, denoted $I\_2$. That represents an input of energy.
2. $\ce{M^2+}$ is smaller and more highly charged than $\ce{M+}$, so you can gain some energy back on the basis that the $\ce{M^2+}$ ion is more strongly solvated or forms stronger ionic/covalent bonds with anions.


Loosely speaking, if the energy you get out of (2) is enough to compensate for the energy you lose in (1), then $\ce{M+}$ will act as a reducing agent. And the converse is true as well. It is for this reason that one does not observe Mg(I) compounds (apart from some esoteric organometallics). The $I\_2$ of Mg is so small that it can be recouped extremely easily. Likewise, the $I\_2$ of K is so large that there's no way any stronger bonding can compensate for that.


For the cases you have described:


* $\ce{Cr^2+}$ simply does not act as an oxidising agent because $I\_2$ of Cr is rather small. The same is true of the whole 3d block. How many 3d compounds do you know of with the metal in a +1 oxidation state? Without resorting to organometallic compounds, all I can suggest is copper(I). The reason why none of the other transition metals exist in the +1 oxidation state is precisely because their $I\_2$ values are fairly small.
* Similarly, $\ce{Mn^3+}$ does not act as a reducing agent because $I\_4$ is way too large. Remember that successive ionisation is extremely difficult.


What your textbook **should** have written is:


* The $I\_3$ of Mn is anomalously *large* because you are "losing the stability of a half-filled 3d subshell". Hence, $\ce{Mn^3+}$ is a good oxidising agent.
* The hydration enthalpy of $\ce{Cr^3+}$ is anomalously *large* (for a 3+ ion) because there is a large ligand-field stabilisation energy associated with the $(t\_{2g})^3$ configuration. (This is factor number 2, not factor number 1). Hence, $\ce{Cr^2+}$ is a reducing agent.


These factors are *small* compared to the raw size of the ionisation energies. Yes, the $d^3$ ion $\ce{Mn^4+}$ also enjoys a large LFSE, and its hydration enthalpy would be larger than expected. However, that 4th ionisation energy is just too big, and that simply *cannot* be compensated for by the comparatively tiny increase in LFSE.


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