Iron(II) sulphate solution is added to a test-tube. Following this, potassium manganate solution is added gradually to the test-tube as well. The observation seen is that the green solution turns pink at first, then potassium manganate turns colourless as manganese(VII) is oxidised to manganese(II).

But Fe(II) is oxidised to Fe(III) ions as well, which are brown in colour, so shouldn’t the solution turn brown instead of colourless?

I have seen these observations experimentally in the lab but still did not understand. I know I’m probably thinking about it the wrong way so any help would be much appreciated.


3 Answers 3


In solution, the iron(III) will be hexaaqua iron(III), 6 water molecules coordinating the iron ion, with the oxygen atoms being at the vertices of an octahedron.

Hexaaqua iron(III) has a high spin d5 electron configuration.

All the d-d transitions are spin forbidden.

Hexaaqua iron(III) is nearly colorless.


Your expectations are not completely unfounded. Even without the addition of special ligands, aqueous solutions of Fe(III) often appear yellow since they contain the ions $\ce{[Fe(OH)(H2O)5]^2+}$ and $\ce{[Fe(OH)2(H2O)4]+}$.

However, your experiment takes place under acidic conditions. (I guess, sulfuric acid was added.) This is apparent because the purple permanganate is reduced to the colourless $\ce{Mn^2+}$ ion:

$$\ce{MnO4- + 8H+ + 5e- <=> Mn^2+ + 4H2O}$$

Under neutral conditions, permanganate would be reduced to dark brown manganese(IV) oxide:

$$\ce{MnO4- + 4H+ + 3e- <=> MnO2 + 2H2O}$$

Under the given acidic conditions, the colourless complex $\ce{[Fe(H2O)6]^3+}$ is stable:

$$\ce{[Fe(OH)2(H2O)4]+ + 2H+ <=> [Fe(OH)(H2O)5]^2+ + H+ <=> [Fe(H2O)6]^3+}$$

Therefore, the solution appears colourless.


This answer mentions about colours of aqueous iron solutions and how to test them (only) $$\ce{10 FeSO4 + 2 KMnO4 + 8 H2SO4 -> 5 Fe2(SO4)3 + 2 MnSO4 + K2SO4 + 8 H2O}$$

Iron(II) salts are usually very pale green in aqueous solution and iron(III) salts are yellow in aqueous solution. You can test what form of iron is present by adding thiocyanate, which gives red precipitate exclusively with iron(III): Place a drop of test solution on a spot plate place and add one drop of 1 per cent ammonium thiocyanate solution. A deep-red colouration may appear. Nitrites should be avoided in acid solution as they form nitrosyl thiocyanate $\ce{NOSCN}$, which yields a red colour, disappearing on heating similar to that with iron(III) $$\ce{Fe^3+ + 3 SCN- ->Fe(SCN)3}$$

And iron(II) can be detected most reliably with 2,2'-bipyridine reagent where deep-red complex bivalent cation $\ce{[Fe(C10H8N2)3]^2+}$ is formed in mineral acid solution. Test a drop of the faintly acidified test solution with one drop of the reagent on a spot plate; a red colouration may be obtained.

  • $\begingroup$ Do you mean to say say that the solution will not turn colourless when the reaction is complete? Because according to my chemistry book it does (though it is not explained why). And I conclude from your answer that Fe(III) ppt are brown and Fe(III) in solution are yellow,am I right? $\endgroup$
    – Sara
    Dec 14, 2014 at 16:42
  • $\begingroup$ @Sara I meant to say that there must be colour, but as it is not there, i just gave you the method to check for which state of iron is there $\endgroup$
    – RE60K
    Dec 15, 2014 at 8:55
  • $\begingroup$ @ADG Why do you say "there must be colour", considering that for high spin Fe(III) all d-d transitions are spin forbidden and Laporte forbidden? $\endgroup$
    – DavePhD
    Dec 15, 2014 at 16:49
  • $\begingroup$ @DavePhD you're last comment was above my knowledge, sorry $\endgroup$
    – RE60K
    Dec 16, 2014 at 3:43

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