# Why is Cr²⁺ is a good reducing agent but Mn³⁺ is a good oxidising agent? [duplicate]

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! – Jack Rod May 29 at 11:10
• Duplicates don't necessarily have to be done in chronological order, see e.g. the answers at 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 at 13:19
• ncert.nic.in/textbook/pdf/lech108.pdf link for the textbook – Chemist May 29 at 19:50

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.

• I was asking why not $\ce{Cr^+}$. – Soham Mar 2 '15 at 19:11
• And actually, $\ce {Cr^2+}$ also has 3 $\ce {t_2g}$ electrons. – Soham Mar 2 '15 at 19:13
• would Cr+ be d5 or s1d4? – DavePhD Mar 2 '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 2 '15 at 19:21
• 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

Cr2+ which has d4 configuration gets converted to Cr3+ which has d3 has half-filled t2g orbital

Which makes it good reducing agent

while In case of Mn3+ which had d4 Configuration gets converted to Mn2+ which has d5 configuration giving it half –filled d-configuration has more stability which makes it good oxidising agent.

Cfse consideration

Also Cr2+ gives negative CFSE while Cr+ having Zero CFSE makes it less stable compared to Cr2+.

In case of Mn, it always forms low spin complexes. So, it will have more negative CFSE in case of Mn2+ than Mn3+ .Mn4+ is highly charged which make it unstable and highly oxidising.

Sources quora

Credits-Golu parte