woelen - 7-12-2009 at 00:21
In one of my experiment I noticed that thionylchloride shows a very peculiar and also rather interesting with iodic acid and iodine pentoxide. I did
some more investigations on this topic and found that ANY inorganic compound of iodine which I have shows the same peculiar reaction, regardless of
oxidation state of the iodine.
I took 0.5 ml of SOCl2 and added a pinch of solid potassium iodide. When this is done, then the liquid turns deep red, like red wine. A beautiful
color. Most of the KI dissolves, but not all. When water is added to the mix, then the SOCl2 quickly reacts with the water, but most noteworthy is
that quite a lot of sulphur is formed. A light yellow precipitate is formed. Besides the smell of SO2 there also is another smell, a typical rotten
sulphur/sulfide smell, but not exactly like H2S.
In another experiment I took a pinch of iodine and added 0.5 ml of SOCl2. The ikodine dissolves for the greater part in the SOCl2 giving a deep purple
solution. When water is added to this, then the same is observed as with KI. A light yellow precipitate of sulphur is formed.
In a third experiment I took a pinch of iodine pentoxide and added 0.5 ml of SOCl2. Initially not much seems to happen (but this is common for I2O5,
this material is slow in its reactions at room temperature). When the liquid is heated somewhat, then the I2O5 dissolves, giving a colorless gas (or
is this boiling of SOCl2??) and the liquid turns brown/red. The reaction is somewhat exothermic, you can feel that the liquid remains warm as long as
the reaction is going on. Again, when a large excess of water is added, then the SOCl2 reacts, but also quite a lot of sulphur is precipitated.
In a fourth and final experiment I used KIO3 and this almost gives the same results as I2O5, albeit even more slowly.
I did a counter experiment with a pinch of NaBrO3 in 0.5 ml of SOCl2. When this is done, then the liquid turns very pale yellow. When a lot of water
is added, then a fairly violent reaction sets in but finally the liquid becomes clear and colorless. When more KBrO3 is added, then the liquid becomes
orange and bromine vapor is slowly released from the liquid.
So, what strikes me is that in all used oxidation states of the iodine, there is formation of sulphur when the liquid is mixed with water. Especially
for the iodate and iodine pentoxide the formation of sulphur surprises me. This actually means that SOCl2 is reduced (S goes from oxidation state +4
to oxidation state 0), while a strong oxidizer is added. I do not have an explanation for this, maybe some of you can shed some light on this?
Nicodem - 7-12-2009 at 09:40
Maybe SOCl2 is not getting reduced. This would be nearly impossible given that I2O5 or KIO3 can only be oxidized further to periodates, which is
unimaginable with SOCl2. Instead perhaps iodine in its higher oxidation states catalyses some sort of disproportionation of SOCl2. I do not see any
other viable redox reaction that would yield sulfur with all the examples you gave (provided that the yellow precipitate is sulphur at all).
Try with an only tiny amount of I2O5 (like 1 mol%) in SOCl2, for example, 46 mg of I2O5 in 1 ml of SOCl2. If you still get a comparable amount of a
yellow precipitate as when using larger amount of I2O5 or other iodo compounds, then the reaction is most likely a iodate catalysed
disproportionation. However, to be honest I see no mechanism or reaction for such a thing. The formation of elemental sulphur seems an indication that
SOI2, SO or some other such unstable species might be intermediates. These are very unstable and disproportionate to sulfur and SO2. Other candidates
would be S2Cl2 or SCl2 which disproportionate to sulfur upon hydrolysis. Sulphur is expected to form by the decomposition of SOI2 formed by the
reaction of KI with SOCl2. In fact it is known that the reaction of SOCl2 and KI gives sulphur among other things:
"In fact, treatment of pure thionyl chloride with potassium iodide, yields but little thionyl iodide and results only in the formation of sulphur,
sulphur dioxide and iodine. " (cf. DOI: 10.1007/BF03046547, see also DOI: 10.1007/BF03046548)
But how can this happen with iodine(V) species, I have no idea. I could utmost imagine some sort of iodine oxochlorides to form from the reaction of
I2O5 or KIO3 with SOCl2, but this would only yield SO2 as a side product, and no sulphur.
Edit: You might get some ideas from this report on SOCl2. For one thing it describes SOCl2 can indeed disproportionate on heating into SCl2, S2Cl2, SO2 and Cl2. This reaction (at
least in reverse) is catalysed by charcoal. A related disproportionation between SCl2, SO3, SO2 and SOCl2 is catalysed by SbCl3, a Lewis acid (I2 is a
fairly strong soft Lewis acid!).
[Edited on 7/12/2009 by Nicodem]
Edit: Yes kmno4, you got them right. Links now corrected. That's what I get when blindly trusting Springer gives correct DOI numbers (or maybe doi.org
has it wrong?).
[Edited on 7/12/2009 by Nicodem]
kmno4 - 7-12-2009 at 11:10
Given links do not work but I think they lead to Rao's articles about iodide-sulphur-thionyl iodide interactions:
http://www.springerlink.com/content/g3v4886g432p/?p=133b1448...
Besides, SOCl2 and I(-) reaction is investigated in another article :
( http://dx.doi.org/10.1021/ja01176a501 )
"The reaction may be formulated as
2SOCl2 + 2I(-) —> 2ICl2(-) + SO2 + S"
I also would try KIO4 with SOCl2 or some SOCl2 and NaCl (for example) for a short time and add some water to this mixture.
Currently I do not have SOCl2, I cannot check this myself., but I think that S presence is caused by reactions of products of SOCl2 decomposition
with water( as Nicodem wrote). It is also possible that traces of I2 cause catalytic decomposition of SOCl2.
[Edited on 7-12-2009 by kmno4]
Pyrovus - 8-12-2009 at 01:30
My guess is that like stated by the posters above, it is probably some sort of iodide promoted/catalysed disproportionation reaction. In the case of
the higher oxidation states of iodine, what is probably going on is that some the thionyl chloride is being oxidised and in turn reducing the iodine
to iodide, which in turn triggers the disproportionation by the same mechanism as when iodide is used.
woelen - 8-12-2009 at 14:14
I agree with Pyrovus. I also think that all the higher oxidation state compounds are reduced to iodide (or iodine) and then the sulphur-producing
reaction sets in. It could be a disproportionation, but only in the presence of water. Without water the liquid remains clear.
The provided links are an interesting read (at least the parts, which I have access to) and the mention of the yellow compound is an interesting one.
As i wrote in my first post, with iodide, the solution becomes wine-red, while with iodine the solution becomes purple. The difference is clearly
visible. It might be that the red color is due to a mix of yellow and purple. Next weekend, when I can experiment in daylight, I can make two test
tubes, one with iodide + thionylchloride in the dark, and the other in daylight. Let's see what is the difference after an hour or so...