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

Question

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?

Answer 1

Related question with same answer but in a different context of the 4f block: Why don't we see these lanthanide species?

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.