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

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 $\ce{FeX2}$?

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    $\begingroup$ Because the iodide will reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$. $\endgroup$
    – bon
    May 23 '16 at 10:08
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    $\begingroup$ There is no $\ce{I^{3-}}$ ion. $\endgroup$
    – bon
    May 23 '16 at 10:11
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    $\begingroup$ Look at the redox potentials for iodide, bromide and chloride here. $\endgroup$
    – bon
    May 23 '16 at 10:17
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    $\begingroup$ 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
    Jan 16 '20 at 12:37
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    $\begingroup$ @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" $\endgroup$ Jul 24 at 2:59
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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.

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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

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    $\begingroup$ 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 $\ce{FeI3}$. $\endgroup$ Jul 24 at 15:02
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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

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.

References

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

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  • $\begingroup$ 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). $\endgroup$ 2 days ago
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Actually, some have voiced an opinion that FeI3 can apparently be created with difficulty, and is reputedly very unstable, decomposing into FeI2 and I2. Similarly, with CuI2, which is likewise unstable. To quote a source reference. $$ FeCl3 + 3KI = FeI3 + 3KCl ; FeI3 = FeI2 + I2 $$

Observation-

Large anions diminish the lattice energy of the higher halide to such an extent that the higher halide may be thermodynamically unstable.

Sample Solution-

Iodide ion is a good reducing agent and hence reduces metal in higher oxidation state into lower Ones.

Even, more interesting is the fact that FeI3 is extremely photosensitive! See this ebook, which describes FeI3 as "very difficult-to-prepare" and as "intensely black" in appearance, which is somewhat definite for a compound arguably that some claim does NOT exist.

So, in light, I would expect:

I- + hv -> .I + e-(aq)

Fe(III) (aq) + e- (aq) -> Fe(II)

.I + .I = I2

which means, in light, an even shorter half-life with potential transient atomic iodine presence and creation of elemental iodine.

So per reference above, one may be able to prepare (albeit briefly) some FeI3 (aq) by adding FeCl3 to a source of iodide (like aqueous KI). Here also is an interesting (but still speculative) supporting source, to quote from the abstract:

This study first reports that ferric chloride (FeCl3) can lead to the formation of iodinated coagulation byproducts (I-CBPs) from iodide-containing resorcinol solution or natural waters. The unwanted I-CBP formation involved the oxidation of iodide by ferric ions to generate various reactive iodine species, which further oxidize organic compounds. Although the oxidation rate of iodide by FeCl3 was several orders of magnitude slower than that by chlorine or chloramine, most of the converted iodide under the ferric/iodide system was transformed into iodine and iodinated organic compounds rather than iodate. Formation of four aliphatic I-CBPs was observed, and four aromatic I-CBPs were identified by gas chromatography mass-spectrometry and theoretical calculation. Coagulation of iodide-containing waters with FeCl3 also produced I-CBPs ranging from 12.5 ± 0.8 to 32.5 ± 0.2 μg/L as I. These findings call for careful consideration of the formation of I-CBPs from coagulation of iodide-containing waters with ferric salts.

The liberation of iodine together with the formation of iodinated coagulation byproducts (supportive of possible atomic iodine formation with any light exposure from the highly photo-sensitive FeI3) possibly suggests a photo-assisted breakdown of FeI3 (aq), in my speculation.

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  • $\begingroup$ To teachers, I apology for voicing an opinion that FeI3 may indeed exist. But, I see interesting experimental evidence that it may indeed behave as expected. Further, the possible existence of a highly photosensitive FeI3 and associated active iodine radical presence is, in my opinion, noteworthy, in itself. For those, like myself, interested in recent scientific thought, with potential patent applications, welcome. $\endgroup$
    – AJKOER
    Jan 16 '20 at 2:18
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    $\begingroup$ I don't really see how any of this is an opinion. You can't have an opinion about factual things, such as whether FeI3 can be made, just like I can't have an opinion that 2+2=4 (or 5 for that matter). What it is is speculative, and this should be (and has been) made clear. || Beyond that: the first source is not exactly what I would consider the most reliable. The second seems more reputable, but deserves to be credited beyond "this ebook", which makes it sound like a random book that a random person self-published online. As a whole, the answer would benefit from better formatting. $\endgroup$
    – orthocresol
    Jul 23 at 22:17

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