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The question is this:

On the basis of CFT, predict with justification which of the following two complex ions would be more stable [Mn(CN)6]4- or [Fe(CN)6]4-.

Attempt at solution:

Both the complexes are low spin. Filling in the d electrons we get 5 electrons in t2g for Mn2+ and 6 electrons in t2g for Fe2+. There are no electrons in eg for both.

Mn2+ has one unpaired electron while Fe2+ has completely filled t2g level. From here I was in a dilemma because Fe has fully occupied t2g level as well as a higher charge density because the effective ionic size of Fe2+ in low spin is 61 pm while Mn2+ in low spin has size 67 pm. However, due to Jan Teller effect the Mn compound will gain more stability due to the unsymmetrical filling of the t2g level. I know that Jan Teller distortion for t2g is less pronounced, so I was wondering if it was enough to make the complex more stable than Fe2+ complex?

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    $\begingroup$ A simpler approach, somewhat ridiculous, I agree. But look : $\ce{Fe^{2+}}$ is {Ar} + 6 electrons. If 6 $\ce{CN^-}$ are approaching, each one brings one doublet. The iron outer shell has now : 6 + 6·2 = 18 electrons. This is like the Krypton atom. Manganese ion cannot do this. $\endgroup$
    – Maurice
    Commented Apr 19, 2020 at 10:35
  • $\begingroup$ CFT is outdated and deprecated: chemistry.stackexchange.com/questions/131780/… $\endgroup$
    – Nilay Ghosh
    Commented Apr 19, 2020 at 11:07
  • $\begingroup$ Horrible question as the word "stability" has little meaning unless you state with respect to what change you are considering some is stable or not. But I suspect chem.libretexts.org/Bookshelves/Inorganic_Chemistry/… may help you give them the answer they want. $\endgroup$
    – Ian Bush
    Commented Apr 19, 2020 at 12:05
  • $\begingroup$ @Ian Bush I read that site and drew the above conclusions from there. I just want to know if they are correct. I think stability refers to the complex with the least energy. $\endgroup$
    – Neo Darwin
    Commented Apr 19, 2020 at 12:13

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