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Pomzazed
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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 ?
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Has the melting point test been considered yet?
According to wiki and my old handbook, SnI2 Is also an orange-red colored compound.
Don't stare at me making fumes... I'm just experimenting with some gas...
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woelen
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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.
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UnintentionalChaos
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| 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 ?
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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]
Department of Redundancy Department - Now with paperwork!
'In organic synthesis, we call decomposition products "crap", however this is not a IUPAC approved nomenclature.' -Nicodem
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Pomzazed
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Ah, my bad forgetting the 18 e- things, thanks for refreshing my memory UnintentionalChaos 
Don't stare at me making fumes... I'm just experimenting with some gas...
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barley81
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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)
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AJKOER
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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]
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barley81
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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.
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AJKOER
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| 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]
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barley81
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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.
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