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

The question is the same as the title states. In my text book,the following is written

$\ce{Cr^{2+}}$ gets converted to $\ce{Cr^{3+}}$ as the +3 oxidation state hase half filled $\mathrm{t_2g}$ orbitals - thus it is a good reducing agent, on the other hand $\ce{Mn^{3+}}$ gets ocnverted to $\ce{Mn^{2+}}$ as +2 oxidation stae has half filled $\mathrm{d}$-orbital.

Now my doubt is -

• 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 $\ce{Cr}$?

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
• from slide 16 here: alpha.chem.umb.edu/chemistry/ch371/documents/… it seems like Cr(I) is s1d4 – DavePhD Mar 2 '15 at 19:46