$$\ce{Fe2O3 + 6HI -> 2FeI2 + I2 + 3H2O}$$

Why don't we get $\ce{FeI3}$? After all, iron's oxidation state is $+3$ in the reagent.

Should one just memorize that up to bromine, it's $\ce{FeX3}$, and below it's $\ce{FeX2}$?

  • 9
    $\begingroup$ Because the iodide will reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$. $\endgroup$
    – bon
    Commented May 23, 2016 at 10:08
  • 2
    $\begingroup$ There is no $\ce{I^{3-}}$ ion. $\endgroup$
    – bon
    Commented May 23, 2016 at 10:11
  • 2
    $\begingroup$ Look at the redox potentials for iodide, bromide and chloride here. $\endgroup$
    – bon
    Commented May 23, 2016 at 10:17
  • 2
    $\begingroup$ To quote wikipedia "Ferric iodide, a black solid, is not stable in ordinary conditions, but can be prepared through the reaction of iron pentacarbonyl with iodine and carbon monoxide in the presence of hexane and light at the temperature of −20 °C, with oxygen and water excluded" Housecroft and Sharpe or Greenwood and Earnshaw (don't have them to hand) syas similar. So it does exist, but is fairly unstable, both thermodynamically (as covered above) AND kinetically. $\endgroup$
    – Ian Bush
    Commented Jan 16, 2020 at 12:37
  • 1
    $\begingroup$ @Zenix Varied chemist had noted varied colors of iron(II) iodide. To quote from this paper: "The color of ferrous iodide, therefore, is not by any means settled, the balance of evidence in the chemical literature being in favor of grey or white" $\endgroup$ Commented Jul 24, 2021 at 2:59

3 Answers 3


The standard reduction potentials for the following half reactions can be found here.

$$ \begin{align} \ce{Fe^3+(aq) + e- &-> Fe^2+(aq)} &\quad E^\circ &= \pu{+0.77 V} \\ \ce{I2(s) + 2 e- &-> 2 I-(aq)} &\quad E^\circ &= \pu{+0.54V} \\ \ce{Br2(l) + 2 e- &-> 2 Br-(aq)} &\quad E^\circ &= \pu{+1.07V} \\ \ce{Cl2(g) + 2 e- &-> 2 Cl-(aq)} &\quad E^\circ &= \pu{+1.36V} \end{align} $$

You can see from this that only iodide is a strong enough reducing agent to reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$ at standard conditions. Even with non-standard concentrations it will be very difficult to get bromide to do the reduction because the difference in electrode potential is large.

The trend in electrode potentials for the halogens can be explained in terms of the increasing electronegativity going from iodine to chlorine which increases the first electron affinity. It just so happens that the crossover point with the iron reduction is between iodine and bromine.


As it is pointed out in other answers and in literature, it is indeed thermodynamically unstable and its reaction synthesis has unfavorable pathways. But, is it "non-existent"? Not quite. There was a possibility of its existence indicated by the fact that hydrated ferric oxide dissolves in hydriodic acid, yielding a brown solution. Its thermodynamic limitation was known in aqueous medium, so its synthesis was thought to be carried in non-aqueous medium. It was the year 1989 that the synthesis was somehow achieved. Diiodotetracarbonyl(II) iron was photochemically reacted with iodine in n-hexane medium to yield the product:

$$\ce{2(OC)4FeI2 + I2 ->[h\nu][C6H14] 2FeI3 + 8CO }$$

However, there were some limitation to this reaction:

  1. The photochemical conversion was limited due to photochemically induced decomposition:

$$\ce{2FeI3 ->[h\nu] 2FeI2 + I2}$$

  1. It is metastable and hence attainment of pure iron(III) iodide was quite difficult
  2. The iodine concentration was needed to be precise. More iodide led to formation of complex:

$$\ce{FeI3 + I- -> FeI4-}$$

Ref.: Ferric iodide as a nonexistent compound, K. B. Yoon and J. K. Kochi Inorganic Chemistry 1990 29 (4), 869-874, DOI: 10.1021/ic00329a058

  • 1
    $\begingroup$ One idea: It may not he iron(III) iodide as such that is unstable, but the combination of the separate ions which are prone to react in the usual way. The difference is that when iron is bonded with iodine there is a lot of covalent character, which in my answer shows up as the molecule becoming a soft acid. Were we to prevent the ions from forming and keep away hard bases (like water or THF) that would upset the iron-iodine bonds, we might better sustain $\ce{FeI3}$. $\endgroup$ Commented Jul 24, 2021 at 15:02

Iron(III) iodide as a binary salt is highly unstable/transitory, but stable complexes are known with appropriate ligands.

Pohl et al. [1] first synthesized such a complex, $\ce{FeI3(SC(N(CH3)2)2)}$, actually oxidizing iron(II) iodide with elemental iodine in the presence of a carefully controlled amount of $\ce{((CH3)2N)2CS}$. The iron(III) iodide, despite the high oxidation state of iron, acts as a soft acid, coordinating to sulfur instead of nitrogen (picture from Ref. [1]):

enter image description here

Barnes et al. [2] report that reaction of iron metal with trimethylarsine diiodide affords the complex $\ce{FeI3(As(CH3)3)2}$, a trigonal bipyramidal complex with a structure similar to that of lighter halide complexes (picture from Ref. [2]).

enter image description here

In a variation, Bernstein an Herbstein [3] describe the salt ferrocenium triiodide, $\ce{[Fe(C5H5)2]+I3^-}$, in which iron(II) incorporated into ferrocene reduces elemental iodine and the iodide ion then complexes with additional iodine. Thereby iron(III) is combined with the triiodide ion. The authors even indicate that the triiodide ion is symmetrical, a condition alkali metal triiodides fo not achieve.

These reactions, together with the oxidation of the iron(II) iodide-carbonyl complex reported in Nilay Ghosh's answer, share some common characteristics:

  • The iodine comes from nonionic sources, even where the reactant has it in the -1 oxidation state as in Ref. [2].

  • The environment avoids water and other hard bases, which would presumably bind to the iron (III) and displace iodide ions. The iron(III) iodide, which has significant covalent character and would be a softer acid than ionic $\ce{Fe^{3+}}$, is instead bound to soft bases.

These features suggest that $\ce{FeI3}$, as such, is not all that unstable; rather it is the combination of $\ce{Fe^{3+}}$ and $\ce{I^-}$ ions that doesn't hold up. If the iron(III) iodide were to be formed in a process that avoids the use or generation of the separate ions, it would more likely be sustained.

Compare these results with cerium(IV) chloride.


  1. Siegfried Pohl, Ulrich Bierbach, Wolfgang Saak; "FeI3SC(NMe2)2, a Neutral Thiourea Complex of Iron(III) Iodide", Angewandte Chemie International Edition in English (1989) 28 (6), 776-777. https://doi.org/10.1002/anie.198907761

  2. Nicholas A. Barnes, Stephen M.Godfrey, Nicholas Ho, Charles A.McAuliffe, Robin G.Pritchard; "Facile synthesis of a rare example of an iron(III) iodide complex, [FeI3(AsMe3)2], from the reaction of Me3AsI2 with unactivated iron powder", Polyhedron (2013) 55, 67-72. https://doi.org/10.1016/j.poly.2013.02.066

  3. Tsur Bernstein and F. Herbstein (1968). "The crystal structure of ferricinium triiodide, (C5H5)2FeI3". Acta Crystallographica Section B Structural Crystallography and Crystal Chemistry. 24. 1640-1645. 10.1107/S0567740868004784.

  • 1
    $\begingroup$ A side note: same holds with cobalt(III) and chloride; cobalt(III) chloride does not exist per se, but hexaamminecobalt(III) chloride does (and is stable enough to be sold commercially). $\endgroup$ Commented Oct 17, 2021 at 6:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.