Why is Cr²⁺ is a good reducing agent but Mn³⁺ is a good oxidising agent?

Question

The question is the same as the title states. In my textbook,^{[NCERT Chemistry I for Class 12 , pg no 217.]} the following is written

$\ce{Cr^{2+}}$ gets converted to $\ce{Cr^{3+}}$ as the +3 oxidation state has half-filled $\mathrm{t_{2g}}$ orbitals; thus it is a good reducing agent.

On the other hand $\ce{Mn^{3+}}$ gets converted to $\ce{Mn^{2+}}$ as its +2 oxidation state has a half-filled d-subshell.

Now my questions are:

Why can't the $\mathrm{t_{2g}}$ argument be used for $\ce{Mn}$?
Why can't the half-filled $\mathrm{d}$-orbital argument be used for $\ce{Cr}$?

$\ce{Mn^3+}$ could be oxidized to $\ce{Mn^4+}$, which would have a half-filled $\mathrm{t_{2g}}$ set. As oxidation and reduction both can take place in aqueous medium, why wouldn't $\mathrm{t_{2g}^3}$ or $\mathrm{d^3}$ configuration be more stable than $\mathrm{d^5}$ configuration in aqueous medium?

A similar observation is made in case of $\ce{Fe^2+}$ and $\ce{Cr^2+}$, where it observed that $\ce{Cr^2+}$ is more powerful reducing agent than $\ce{Fe^2+}$ (due to reasons explained above).

If possible, I would like to see the crystal field splitting calculations that show this to be the case.

since you have talked about aqueous solution there is term enthalpy which dominates overall change. the difference is hydration enthalpy! — आर्यभट्ट, May 29 '20 at 11:10
Duplicates don't necessarily have to be done in chronological order, see e.g. the answers at https://chemistry.meta.stackexchange.com/q/4611/16683, but anyway I don't plan to pronounce on which one should be closed as the other (not right now, at least; that can be a decision for later). My main point was that the questions are very similar. — orthocresol, May 29 '20 at 13:19
http://ncert.nic.in/textbook/pdf/lech108.pdf link for the textbook — Chemist, May 29 '20 at 19:50

score -1 · Answer 1 · edited Aug 22 '19 at 07:43

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Why can't the $\mathrm{t_{2g}}$ argument be used for $\ce{Mn}$ ?

Because the number of $\mathrm{t_{2g}}$ electrons is not changing upon going from $\ce{Mn^3+}$ to $\ce{Mn^2+}$. In both $\ce{Mn^2+}$ and $\ce{Mn^3+}$ there are 3 $\mathrm{t_{2g}}$ electrons. I think the point they are trying to make with $\ce{Cr^3+}$ being half filled is that there are 3 $\mathrm{t_{2g}}$ electrons and no $\mathrm{e_g}$ electrons. $\ce{Mn^3+}$ and $\ce{Mn^2+}$ both have $\mathrm{e_g}$ electrons.

You could make a similar argument for $\ce{Mn^3+}$ being oxidized to $\ce{Mn^4+}$, which would have half filled $\mathrm{t_{2g}}$ and empty $\mathrm{e_g}$.

Why can't the half-filled d-orbital argument be used for Cr?

Because neither $\ce{Cr^3+}$ nor $\ce{Cr^2+}$ has a half filled d subshell. $\ce{Cr^3+}$ has 3 electrons while $\ce{Cr^2+}$ has 4 electrons.

edited Aug 22 '19 at 07:43

S R Maiti

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answered Mar 02 '15 at 19:07

DavePhD

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I was asking why not $\ce{Cr^+}$. – Soham Mar 02 '15 at 19:11
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And actually, $\ce {Cr^2+} $ also has 3 $\ce {t_2g}$ electrons. – Soham Mar 02 '15 at 19:13
would Cr+ be d5 or s1d4? – DavePhD Mar 02 '15 at 19:15
i don't know why it would be s1d4 as d5 would be completely half filled and s is in the outer shell – Soham Mar 02 '15 at 19:21
from slide 16 here: http://alpha.chem.umb.edu/chemistry/ch371/documents/2.Chromiumacetate_001.pdf it seems like Cr(I) is s1d4 – DavePhD Mar 02 '15 at 19:46
Won't Cr5+ be at d5 as well? How do we use that argument only for Mn3+->Mn2+ ? – shre_sudh_97 Feb 15 '16 at 10:47
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Why not Mn3+->Mn4+? Then this would become a half filled t2g , right? Why can't we apply this argument? – shre_sudh_97 Feb 15 '16 at 10:47
Answer in incomplete – Chemist Jun 03 '20 at 04:01

Why is Cr²⁺ is a good reducing agent but Mn³⁺ is a good oxidising agent?

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