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.

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    $\begingroup$ I was asking why not $\ce{Cr^+}$. $\endgroup$ – Soham Mar 2 '15 at 19:11
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    $\begingroup$ And actually, $\ce {Cr^2+} $ also has 3 $\ce {t_2g}$ electrons. $\endgroup$ – Soham Mar 2 '15 at 19:13
  • $\begingroup$ would Cr+ be d5 or s1d4? $\endgroup$ – DavePhD Mar 2 '15 at 19:15
  • $\begingroup$ i don't know why it would be s1d4 as d5 would be completely half filled and s is in the outer shell $\endgroup$ – Soham Mar 2 '15 at 19:21
  • $\begingroup$ from slide 16 here: alpha.chem.umb.edu/chemistry/ch371/documents/… it seems like Cr(I) is s1d4 $\endgroup$ – DavePhD Mar 2 '15 at 19:46

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