Sciencemadness Discussion Board

SnI4: A nice covalent compound of tin

woelen - 17-9-2009 at 11:21

Some time ago, another member from this forum (Jor, credits go to him as well) came up with a nice idea of making SnI4, which is an interesting covalent and volatile compound of tin.




It can be made using only mundane chemicals and for many members its preparation may be feasible. It has a nicely colored vapor and in the liquid state it is blood red. I decided to make a webpage about this interesting compound:

http://woelen.homescience.net/science/chem/exps/SnI4/index.h...


[Edited on 17-9-09 by woelen]

JohnWW - 17-9-2009 at 12:12

That would be a possible route to making organotin(IV) compounds. Possible reaction schemes would be to react SnI4 with alkyl- and aryl-lithium or -sodium compounds, in an inert solvent, with LiI or NaI as insoluble (in that solvent) byproduct. It would also be hydrolyzed by alcohols and phenols to alkoxy and phenoxy-compounds.

Picric-A - 17-9-2009 at 12:23

Nice work Woelen!
what possible uses does organotin compounds have? maybe a possible less toxic substitute for tetraethyl lead in petrol?

UnintentionalChaos - 17-9-2009 at 14:44

If I'm not mistaken, you can transmetallate a grignard with SnX4, forming the SnR4. I wonder how one goes about making SnR3X though, which can be converted to SnR3H in-situ with a hydride source.

Jor - 18-9-2009 at 05:42

Thank you very much. I really like these pictures, much more so then the ones I make :) :
http://amateurchemie.nl/viewtopic.php?f=20&t=286&sta...

I'm not sure but if using such a large excess of tin, are you sure your product is not contaminated with SnI2 ? Possibly that's the yellow compound on the test-tube walls?

Using CS2 indeed seems like a better alternative, although I think it may be more feasible to use DCM when doing this on a larger scale, due to the toxicity and extreme flammability of CS2.

Very nice woelen ;)

Does anyone know how to clean my 'dirty' SnI4 from iodine. I decanted the small amount of solvent left (see link above) from from the crystals, and washed once with a small amount of DCM to remove the iodine. The red crystals from wich the liquid was decanted are almost free of iodine, but the dark crystals obtained by evaporating the decanted liquid, contain quite some iodine. What would be the best way to remove the iodine without losing too much SnI4?

woelen - 18-9-2009 at 10:16

Dissolve the material in DCM and add a little bit more tin. The iodine then will react with the tin, while the SnI4 does not react.

Tin(IV) iodide does not react with excess tin. I have my preparation from a book from Vanino, which has a section on the preparation of SnI4. The yellow color of the material is due to the fine division. You can see this effect with e.g. red prussiate of potash, which is red in the crystalline state and yellow when crunched to a very fine powder.

12AX7 - 18-9-2009 at 12:38

Quote: Originally posted by Jor  
Thank you very much. I really like these pictures, much more so then the ones I make :) :
http://amateurchemie.nl/viewtopic.php?f=20&t=286&sta...


Ahh, your pictures are quite good except one thing: lighting! Bring a few halogen lights next time, place them around to spread out the shadows. You might also consider using a flash, but only a little, not so much that it drowns out the scene.

Tim

woelen - 18-9-2009 at 12:51

Jor, indeed is the use of CH2Cl2 a better alternative when the reaction is scaled up. I also tried the experiment with CH2Cl2 and it works almost as well as with CS2. Just use a little bit more of solvent, because iodine dissolves a little bit harder in CH2Cl2 than in CS2. The reaction between the iodine and tin is fast, also when CH2Cl2 is used. Refluxing is not necessary when powdered tin is used, but probably is necessary when globules or molten droplets are used.

UnintentionalChaos - 20-9-2009 at 08:02

I would like to report the success of this synthesis using perchloroethylene as a solvent. Due to it's (relatively) high boiling point, it is not as easy to separate from the SnI4. I achieved this by heating a thin layer of the solution in the bottom of a beaker strongly until the whitish vapor of condensing perc was replaced by the yellow "smoke" of SnI4 and promptly removing from the heat source. I did lose quite some product stuck to the bottom of the beaker though. However, perchloroethylene may be more readily available to some experimenters than DCM or CS2. I think I will make some chloroform for the next time, however.

1.60g I2
Excess (about 1g) mossy tin
About 10ml of Perchloroethylene

Mixture readily reacted. Heating in a hot water bath sped up the reaction threatening violent boiling of the perc due to the exotherm. After several hours at room temperature, the solution was a transparent, dark orange with no tinge of iodine color remaining. Some small crystals of SnI4 had formed, which were dissolved by heating and the solution was filtered through a very small plug of cotton before evaporation.

Isolated compound weighed 1.51g.

blogfast25 - 20-9-2009 at 08:38

Interesting...

Next: SnBr4?

benzylchloride1 - 20-9-2009 at 13:47

Quote: Originally posted by UnintentionalChaos  
If I'm not mistaken, you can transmetallate a grignard with SnX4, forming the SnR4. I wonder how one goes about making SnR3X though, which can be converted to SnR3H in-situ with a hydride source.
. The Trialkyl tin halide can be prepared by a metathesis reaction between a tin(IV) halide and a tetraalkyl tin complex. An example would be the synthesis of diethyltin dichloride. The procedure for this can should be able to be adapteed to the synthesis of the trialkyl tin halide.
Procedure:
10g of tetraethyl tin and 11.1g of tin(IV) chloride, equimolar amounts were placed a a 100ml RBF. A condenser with a calcium chloride tube was attached to the flask. The mixture is then heated at 210 to 220 Celsius in a silicone oil bath for 15 minutes. The mixture was air cooled and then cooled in a cold water bath until the product crystallized. The product was then recrystallized from a boiling hydrocarbon solvent with a boiling point range of 110 to 140 Celsius, xylene probably could be used. The mixture was suction filtered hot and cooled to room temperature. The white needles of diethyltin dichloride were filtered off and dried. Reference; Angelici, Synthesis and Technique in Inorganic Chemistry, 2nd edition.
Can sodium borohydride be used for forming the trialkyltin hydride? I have 250g of tributyltin chloride and at least 30g of sodium borohydride. This reagent has many uses in synthesis. All of the procedures I have seen have used lithium aluminum hydride as the hydride source.
I have prepared tin(IV) iodide and its triphenylphosphine complex which is very dark red in color. Tin (IV) bromide can be prepared by slowly dropping dry bromine on to a slight excess of mossy tin, briefly refluxing and then distilling the tin halide at atmospheric pressure using a propane torch. The tin (IV) bromide soon crystallized in the receiving flask. The tin (IV) bromide was then used for the synthesis of tetraethyl tin.

[Edited on 20-9-2009 by benzylchloride1]

[Edited on 20-9-2009 by benzylchloride1]

Jor - 20-9-2009 at 14:41

Do you handle these alkyl-tin compounds at home? These are very toxic and are of very high environmental toxicity.

JohnWW - 20-9-2009 at 14:50

Yes, I have heard of some organotin(IV) compounds, especially tributyltin derivatives, being used in marine anti-fouling paints on the hulls of boats, to prevent the growth of seaweed, barnacles, teredos, mussels, rock-oysters, etc. on the hulls; but that it has fallen into disfavor as the result of causing deformities in non-target fish species.

itchyfruit - 21-9-2009 at 11:16

Would indium powder be ok?I know their not in the same group, but they are next to each other so must have some similar properties.

woelen - 21-9-2009 at 11:29

I can imagine that it also works for indium, making InI3. InI3, however, probably will not be such an interesting compound. I expect it to be a white, somewhat volatile and easily hydrolized compound. I could give it a try, I have (a small quantity of) 200 mesh indium powder.

itchyfruit - 21-9-2009 at 11:56

I was going to try it as i don't seem to have tin powder. don't waste what you have on my dumb theory i shall report back imminently :D

blogfast25 - 21-9-2009 at 12:24

Woelen:

What surprised me was that the reaction proceeded at room temperature.

You could also try and reduce the SnI4 back to metallic Sn, e.g. with Mg. That could be a very exothermic reaction.

densest - 21-9-2009 at 12:29

@woelen - beautiful pictures! I wonder if there's any way to make I in situ to avoid the US restrictions? Gotta think about that one....

@itchyfruit - the ancient Chem Rubber handbook says InI3 is higher-melting (210) and orange-red. Still would be interesting to see it! I don't know which oxidation/coordination state you'd get: InI, InI2, and InI3 are listed.


[Edited on 21-9-2009 by densest]

itchyfruit - 21-9-2009 at 15:05

I tried with hexane and Xylene and i'm not having much joy.
I'll give it another go tomorrow with the di isopropyl ether(i don't really want to open it indoors)

not_important - 21-9-2009 at 18:05

Preparation of InI3 and InI http://www.orgsyn.org/orgsyn/prep.asp?prep=v79p0059

As they mention indium, as well as gallium, wet glass and stick to it rather well.

While In(I) is well defined, the 'inert pair effect' being reasonably active, In(II) is not stable as such in simple compounds. Many of the In(II) compounds listed in the older literature are really mixed valence compounds In(I)In(III)X4 .

Panache - 21-9-2009 at 18:39

tetra alkyl tin catalysts are commonly available as initiators for silicon rubbers. I know i can go not 3 kilometers from where i live and buy both the tin (FOR GENERAL USE SILICON CASTING RESINS, sry caps) and platinum (for food grade silicon rubber applications applications) readily and easily.
However if you don't live in Australia this is irrelevant except for the fact that one may be able to feel safer in relation to their use because the Australian government surely wouldn't allow their general purchase if they were not safe.

woelen - 22-9-2009 at 09:54

Just out of curiosity I tried a few variations on this reaction, all in CH2Cl2:

- In + I2 does not seem to work. The I2 dissolves, but seems not to react. I now have it in a test tube, well stoppered, for more than a hour, but no reaction. I'll leave it around for a few days and see whether it reacts slowly.

- Sn + Br2 works, but much slower than Sn + I2. This surprises me. But the reaction works. I used excess tin with a small drop of bromine, dissolved in 0.5 ml of CH2Cl2. I first dissolved the Br2, then added the tin. After 15 minutes, the solution is colorless. When the CH2Cl2 is added to warm water, then the CH2Cl2 boils away and a clear and colorless liquid remains behind. When some dilute NH3 is added to this, a white gelatinous precipitate is formed, which must be hydrous SnO2. The solution of SnBr4 in CH2Cl2 fumes somewhat, when it is in contact with air.

- Cu + Br2 does not work in CH2Cl2, but works in water. The Cu becomes covered by a black layer and then the reaction stops, due to insolubility of the reaction product in DCM. When a few drops of water are added, then the reaction continues, giving a white/blue turbid liquid, and the copper granules are covered by a white solid. The white/blue liquid contains a mix of solid copper(I)bromide and dissolved copper(II)bromide.

- A funny thing is observed when tin, a really tiny little pinch of tin(IV) iodide and bromine are added to DCM. First, the liquid is red/brown, due to dissolved bromine. It remains like that for a while, but when the color of bromine becomes weaker, then the liquid becomes purple, due to dissolved iodine and finally the liquid becomes very pale yellow, due to dissolved SnI4. So, apparently the SnI4 is converted to SnBr4 and iodine is set free, which can be seen when the amount of bromine is not large enough anymore to mask the color of the small amount of dissolved iodine.

I'll come back on the In + I2 experiment. I'll see how it looks tomorrow.

Jor - 22-9-2009 at 11:51

Quote: Originally posted by woelen  
Just out of curiosity I tried a few variations on this reaction, all in CH2Cl2:

- In + I2 does not seem to work. The I2 dissolves, but seems not to react. I now have it in a test tube, well stoppered, for more than a hour, but no reaction. I'll leave it around for a few days and see whether it reacts slowly.

- Sn + Br2 works, but much slower than Sn + I2. This surprises me. But the reaction works. I used excess tin with a small drop of bromine, dissolved in 0.5 ml of CH2Cl2. I first dissolved the Br2, then added the tin. After 15 minutes, the solution is colorless. When the CH2Cl2 is added to warm water, then the CH2Cl2 boils away and a clear and colorless liquid remains behind. When some dilute NH3 is added to this, a white gelatinous precipitate is formed, which must be hydrous SnO2. The solution of SnBr4 in CH2Cl2 fumes somewhat, when it is in contact with air.

- Cu + Br2 does not work in CH2Cl2, but works in water. The Cu becomes covered by a black layer and then the reaction stops, due to insolubility of the reaction product in DCM. When a few drops of water are added, then the reaction continues, giving a white/blue turbid liquid, and the copper granules are covered by a white solid. The white/blue liquid contains a mix of solid copper(I)bromide and dissolved copper(II)bromide.

- A funny thing is observed when tin, a really tiny little pinch of tin(IV) iodide and bromine are added to DCM. First, the liquid is red/brown, due to dissolved bromine. It remains like that for a while, but when the color of bromine becomes weaker, then the liquid becomes purple, due to dissolved iodine and finally the liquid becomes very pale yellow, due to dissolved SnI4. So, apparently the SnI4 is converted to SnBr4 and iodine is set free, which can be seen when the amount of bromine is not large enough anymore to mask the color of the small amount of dissolved iodine.

I'll come back on the In + I2 experiment. I'll see how it looks tomorrow.

Cu + Br2 does work! I added some extremely fine copper powder (black, stains everything) to a solution of bromine in DCM. I left to stand for 2 hours, and decanted the liquid. The result is a black powder, wich quickly dissolves in water giving a blue solution. This is thus a great way of making anhydrous CuBr2.

woelen - 22-9-2009 at 12:23

Yes, you have very fine copper, I used copper wire (from electricity flexible wires, isolation stripped away). This is a nice synth for anhydrous copper(II)bromide, but only if you have ultra fine copper powder. With coarser copper, the reaction stops once a black layer is on the copper.

itchyfruit - 24-9-2009 at 15:36

Having left these for a couple of days now the Hexane one has turned a light brown colour and the Xylene one is white/clear.
Woelen you appear to be correct :D

Pomzazed - 24-9-2009 at 19:51

Quote: Originally posted by Jor  

I'm not sure but if using such a large excess of tin, are you sure your product is not contaminated with SnI2 ?


Has the melting point test been considered yet?
According to wiki and my old handbook, SnI2 Is also an orange-red colored compound.

woelen - 24-9-2009 at 22:44

A simpler test is to dissolve some of the SnI4 in concentrated HCl and then add a little amount of an easily reduced compound. There are very sensitive tests for tin(II), I need to lookup which one. I'll try one of these methods.

I know of SnI2 as well (I also have an experiment on my website in which this compound is made), but according to my book from Vanino on preparative chemistry, SnI4 can be made in a very pure state from excess Sn, I2 and CS2 used as solvent. That's why in my original experiment I used CS2, but it appears many other solvents work as well.

UnintentionalChaos - 24-9-2009 at 22:55

Quote: Originally posted by Pomzazed  
Quote: Originally posted by Jor  

I'm not sure but if using such a large excess of tin, are you sure your product is not contaminated with SnI2 ?


Has the melting point test been considered yet?
According to wiki and my old handbook, SnI2 Is also an orange-red colored compound.


Just based on electronics, the formation of SnI2 is highly unfavorable in the presence of something that can oxidize it to the tetrahalide. The tetrahalide is isoelectronic with the zero valent platinum group metals,

Furthermore, it obeys the 18 electron rule, which is the transition-metal complex equivalent of the octet rule. It has 10 d-electrons of it's own and 8 are donated, two each from the 4 iodide ligands.

See brauer page 735. Excess tin is used and the product is reported as "analytically pure"

It is very interesting to note that SnI2 can be prepared from aqueous solution and does not seem to form a hydrate. I believe that chromium (III) iodide behaves much the same way, and can be obtained as black crystals.

[Edited on 9-25-09 by UnintentionalChaos]

Pomzazed - 24-9-2009 at 23:25

Ah, my bad forgetting the 18 e- things, thanks for refreshing my memory UnintentionalChaos :)

barley81 - 27-5-2012 at 18:19

Today I tried to use naphtha as a solvent for making SnI4. I put some tin that had been melted and splashed in water into a test tube, and I put some iodine into the tube. I added Ronsonol lighter fluid (about 2mL) and heated the tube to reflux. Over about a minute, the solution turned orange, and excess tin remained. When the tube cooled down, crystals of SnI4 had formed. It was reheated to dissolve the SnI4, some extra solvent was added, and the liquid was transferred to another tube.

The problem was evaporation. The naphtha had a higher boiling point than either methylene chloride or CS2 (I don't know precisely). Therefore, when it was driven off as vapor, some SnI4 came off as well and formed smoke in the air (not too much). It was difficult to drive all the solvent off. This could be because there is a high boiling component in the naphtha. A red liquid at the bottom of the tube remained, and when this cooled, a pellet of SnI4 formed. It had some liquid adhering to it, and was slightly discolored on the surface from decomposition or hydrolysis. Tomorrow I will crush the pellet and seal the sample in a glass tube. This should work a lot better when I get around to distilling my paint stripper.

(Sorry for bumping the thread)

AJKOER - 3-6-2012 at 15:17

On a related compound that someone may wish to make next, I came across two SnCl4 synthesis in this reference: "Analytical chemistry, Volume 1, by Frederick Pearson Treadwell, page 266, link: http://books.google.com/books?ei=6ozFT5qBLof-8ASXuM25Bg&...

First, Aqua regia is claimed to dissolve tin, forming stannic chloride:

3 Sn + 4 HNO3 + 12 HCl = 4 NO (g) + 8 H20 + 3 SnCl4

I am, however, unclear as whether this reaction actually results in a volatile SnCl4 product or forms a hydrate. Interestingly, replacing above HCl with HI (from say HNO3 and NaI) acting on Sn may be an alternate one step path to SnI4 for those without Iodine on hand (or not, given again the presence of water).

Second, on page 270-271, Treadwell notes:

"Stannic chloride is a colorless liquid, which fumes in the air and boils at 120° C. On adding a little water it solidifies, forming crystals of monoclinic hydrates, SnCl4 + 3 H20, SnCl4 + 5 H20, SnCl4 + 8 H20. The salt with 5 H20 is used commercially as a mordant in dyeing.

On adding more water to these hydrates they dissolve, forming a clear solution, which on boiling (the freshly-prepared, dilute solution) gradually becomes turbid, owing to the precipitation of voluminous stannic hydroxide:

SnCl4 + 4 HOH <=> 4 HCl + Sn(OH)4.

If the solution is very dilute it becomes turbid in the cold. The stannic acid thus formed is not precipitated quantitatively, either in the cold or on boiling, because a considerable amount remains in the hydrosole form. By "salting out" the hot solution (best with ammonium nitrate), the stannic acid may be completely precipitated.

A solution of stannic chloride can be most readily obtained for analytical purposes by chlorinating or brominating a solution of stannous chloride.

On adding chlorine to a solution of stannous chloride, stannic chloride is formed in the cold:

SnCl2 + Cl2 = SnCl4.

As, however, chlorine is colorless in a dilute solution, it is difficult to tell when the oxidation is complete: it is more easily ascertained if bromine is used."

Now, I would guess that one does have the option of treating their SnCl4 with HI (from say heating NaHSO4 and NaI) to make SnI4:

NaHSO4 + NaI --> Na2SO4 + HI (g)

4 HI + SnCl4 --> 4 HCl (g) + SnI4

of course, none of these SnI4 synthesis is as elegant and simple as that presented Woelen.

[Edited on 3-6-2012 by AJKOER]

barley81 - 3-6-2012 at 16:32

Nitric acid and hydriodic acid doesn't sound like it'd be too stable. Nitric acid itself can oxidize iodine to iodic acid. I don't think hydriodic acid would stand much of a chance. HNO3 and NaBr already gives some bromine. I would expect NaI and HNO3 (fuming) even to give iodic acid/iodates after a long reaction.

AJKOER - 3-6-2012 at 18:27

Quote: Originally posted by barley81  
Nitric acid and hydriodic acid doesn't sound like it'd be too stable. Nitric acid itself can oxidize iodine to iodic acid. I don't think hydriodic acid would stand much of a chance. HNO3 and NaBr already gives some bromine. I would expect NaI and HNO3 (fuming) even to give iodic acid/iodates after a long reaction.


Well, the reaction of HI and HNO3:

2 HNO3 + 3 HI ==> I2 + 2 NO + 2 H2O

This should be compared to the reaction of HCl and HNO3:

3HCl + HNO3 ==> Cl2 + ClNO + 2 H2O

Source: http://pubs.acs.org/doi/abs/10.1021/jp992666p

where the nitrosyl chloride is visibly apparent as yellow/orange. With dilute HCl (note, the above reaction with the creation of water may dilute the HCl), a hydrolysis reaction occurs:

ClNO + H2O <--> HNO2 + HCl

So if SnI4 happens to be formed(?), would most likely be some multi-step reaction, for example:
3 Sn + 4 HNO3 + H20 ==> 3 H2SnO3
H2SnO3 + 4 HI ==> SnI4 + 3 H20

and/or:
Sn + I2 ==> SnI2
SnI2 + I2 ==> SnI4

and vaporized as shouldn't exist in an aqueous environment.


[Edited on 4-6-2012 by AJKOER]

barley81 - 3-6-2012 at 19:16

Brauer's prep for SnI4 says that no water must be present. Since water is in conc. HNO3 and is produced by oxidation of HI by nitric acid, SnI4 will not be produced, but perhaps SnO2 will be.