7
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • $\begingroup$ since you have talked about aqueous solution there is term enthalpy which dominates overall change. the difference is hydration enthalpy! $\endgroup$
    – आर्यभट्ट
    Commented May 29, 2020 at 11:10
  • $\begingroup$ 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. $\endgroup$
    – orthocresol
    Commented May 29, 2020 at 13:19
  • $\begingroup$ ncert.nic.in/textbook/pdf/lech108.pdf link for the textbook $\endgroup$
    – Chemist
    Commented May 29, 2020 at 19:50

1 Answer 1

Reset to default
-1
$\begingroup$

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.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ I was asking why not $\ce{Cr^+}$. $\endgroup$
    – Soham
    Commented Mar 2, 2015 at 19:11
  • 1
    $\begingroup$ And actually, $\ce {Cr^2+} $ also has 3 $\ce {t_2g}$ electrons. $\endgroup$
    – Soham
    Commented Mar 2, 2015 at 19:13
  • $\begingroup$ would Cr+ be d5 or s1d4? $\endgroup$
    – DavePhD
    Commented Mar 2, 2015 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
    Commented Mar 2, 2015 at 19:21
  • 1
    $\begingroup$ Why not Mn3+->Mn4+? Then this would become a half filled t2g , right? Why can't we apply this argument? $\endgroup$
    – shre_sudh_97
    Commented Feb 15, 2016 at 10:47