Sciencemadness Discussion Board - Cationic species of iodine, iodine in oxidation state +1/2

Cationic species of iodine, iodine in oxidation state +1/2

woelen - 26-6-2009 at 12:48

I did a simple experiment, in which I dissolved iodine in oleum and this results in formation of a blue cationic species with iodine in oxidation state +1/2. This is a remarkable compound. Not easy to isolate, but just interesting to see this stuff.

Click the link for the description of the experiment and more pictures:

http://woelen.homescience.net/science/chem/exps/iodine_in_ol...

stateofhack - 26-6-2009 at 13:31

Wonderful pictures and great work as usual! thanks woelen!

JohnWW - 26-6-2009 at 16:54

A fractional oxidation state? More likely to be a complex of two or more iodine atoms in different oxidation states, similarly to I3-.

UnintentionalChaos - 26-6-2009 at 17:06

Quote: Originally posted by JohnWW

A fractional oxidation state? More likely to be a complex of two or more iodine atoms in different oxidation states, similarly to I3-.

Actually more likely to be the result of oxidizing an I2 molecule, giving a delocalized +1 charge.

If you'd read the discussion of results on woelen's page, you'd know this. However, the formal charge on each iodine atom would be +1/2.

woelen - 27-6-2009 at 10:01

Oxidation state actually is just a bookkeeping device and does not really need to reflect real charges. E.g. in MnO4(-), the Mn is said to have oxidation state +7, but this does NOT mean that the atom has a charge +7 on it.

In I2(+) both iodine atoms are equivalent, so the total (real) charge of the ion is divided over both iodine atoms. And because no other atoms are part of this ion, one cannot say anything else than that iodine has oxidation state +1/2.

There are more of such fractional oxidation state ions or molecules. Examples are the super oxide ion O2(-) and the complex species Cu2Cl3 (which actually has structure Cl-Cu-(mu-Cl)-Cu-Cl, in which copper has oxidation state +3/2.

indigofuzzy - 27-6-2009 at 18:35

As always, woelen, very nice work!

not_important - 27-6-2009 at 21:29

Indeed, nicely done.

Now if you want a real challange, you could make the corresponding xenon ion Xe2(+1) :-)

UnintentionalChaos - 27-6-2009 at 21:37

Quote: Originally posted by woelen

Oxidation state actually is just a bookkeeping device and does not really need to reflect real charges. E.g. in MnO4(-), the Mn is said to have oxidation state +7, but this does NOT mean that the atom has a charge +7 on it.

In I2(+) both iodine atoms are equivalent, so the total (real) charge of the ion is divided over both iodine atoms. And because no other atoms are part of this ion, one cannot say anything else than that iodine has oxidation state +1/2.

There are more of such fractional oxidation state ions or molecules. Examples are the super oxide ion O2(-) and the complex species Cu2Cl3 (which actually has structure Cl-Cu-(mu-Cl)-Cu-Cl, in which copper has oxidation state +3/2.

It's a lovely experiment, but if there is any appreciable free content of I2+ in the reaction mix, an electron has been completely stripped off somewhere. This can hardly be called book keeping anymore, IMO.

As SO2 is indeed a product of the reduction of SO3, I2 has been correspondingly oxidized. Granted, it is an equilibrium, as with all things, but if SO2 escapes (under heating?), that equilibrium is pushed to the right.

Perhaps I used the wrong word, as formal charge doesn't necessarily reflect charge either.

IIRC, this species can be used to introduce an iodine onto a benzene ring directly.

[Edited on 6-28-09 by UnintentionalChaos]

Ozone - 27-6-2009 at 22:43

Pondering:

I2+ should be a nice electrophile, so yes, an aromatic system might attack it. But...what about competitive addition of SO3 from the oleum?

I am intrigued, I will check the literature tomorrow.

Cheers,

O3

JohnWW - 28-6-2009 at 00:38

Quote: Originally posted by not_important

Now if you want a real challange, you could make the corresponding xenon ion Xe2(+1)

It would require equipment and fluorine-containing reagents outside the reach of, or the ability to handle of, amateurs. That cation has a positive charge on one Xe atom and an unpaired electron on the other, with interchange between them providing some resonance stabilization. But even so, it is unlikely to be particularly stable except at low temperatures. The Xe-Xe stretching frequency in the (quite far) infrared spectrum would be very close to that of the I2 molecule.

not_important - 28-6-2009 at 04:56

Indeed, which is why I put the :-) in there and said it would be a real challenge.

As the attachment shows, reaction media is liquid HF + SbF5, at -30°C the product forms dark-green crystals of Xe2(+)Sb4F21(-) . See T. Drews & K. Seppelt, Angew. Chem. Int. Edn. Engl. 36, 273-4 (1997). While an amateur might successfully produce some anhydrous HF, making SbF5 would be difficult as would be XeF2; purchasing of those would be problematic as well. But it is interesting in that it contains a Xe-Xe bond, SFAIK the longest chemical bond known.

The lower halides of bismuth are within the reach of amateurs, using small scale sealed tube methods. Bi4I4, Bi14I4, and possibly Bi18I4, can be obtained by heating bismuth and iodine. Heating a mixture of Bi and BiCl3 for many days, followed by extraction with benzene, gives "Bi6Cl7" or more properly [ {Bi9(+5)}2 {BiCl5(-2)}4 {Bi2Cl8(-2)} ] The Bi9(+5) cation is a metal cluster with a fractional formal charge on each Bi, while Bi4I4 contains Bi(0) and Bi(+2).

Ozone - 28-6-2009 at 08:24

Fascinating!

Apparently, I2 in 2% oleum is an effective catalyst for the oxidation of alkanes, e.g. methane to methyl bisulfite. The mechanism included cites I2+ followed by a single electron transfer to yield methyl iodide and I. (which seems unlikely or very short-lived as addition to O2 or persulfate to the system did not seem to have an effect) which is said to be in equilibrium with 1/2 I (a charge of +1/2). The .pdf is attached.

Any reasonable activation of the C-H bond is exciting, and apparently, this might be tried in the garage

.

I+ will yield iodobenzene but H2SO4/SO3 leads to the classical sulfonation reaction. IIRC, I2/CuCl2 will give I+. Following this, it might be interesting to compare the reactivity of I2/CuCl2 with the I2/oleum (more "dilute", 2%).

Cheers,

O3

Attachment: Periana 2002 Methane activation via I2+.pdf (87kB)
This file has been downloaded 2224 times

[Edited on 28-6-2009 by Ozone]

woelen - 28-6-2009 at 22:39

@UnintentionalChaos: My excuse for the unclear formulation. After I read my own post, I understand that it is not clearly formulated. What I wanted to say is that oxidation number is a bookkeeping device and that the numbers, assigned to the elements in a molecule or ion do not necessarily have to reflect the real charge. But of course, the sum of the oxidation numbers does tell the real charge. So, in the case of I2(+), the sum of the oxidation numbers equals +1 and this ion really has charge +1. But formally, because both atoms are similar, they have oxidation state +1/2.