Mister Junk Pile
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Stability of Cobalt and Iodonium Complexes in Acid/Base?
So, I have a test tomorrow in Inorganic Chemistry. Yes, this is a bit last minute, but these questions probably won't be a huge part of the test.
Mainly, it's just curiosity.
Why is bis(pyridine)iodonium (specifically the nitrate salt) unstable in acid? We added KI (aq) and HCl to a solution of the salt and an orange color
was produced. NaOH + KI only gave a slight yellow color (I can only assume this becomes simply an iodide/iodate solution with a bit of iodine).
My hypothesis is that pyridine is protonated by HCl and can no longer form a complex with I+ which then reacts with I- to produce I2 (but why the
orange color?). Also, I thought about I+ reacting with Cl- and I read that excess HCl will prevent hydrolysis of ICl. Is this true? It just seems
to me that it would readily hydrolyze to HI/oxoacids even in acidic conditions. Although, I have little experience with or knowledge of the
interhalogens.
Pentamminechlorocobalt(III) chloride:
Why is it unstable in KI/acid? This "hint" that was given gives me a pretty good idea:
2Co3+ + 3I- ----> 2Co2+ + (I3)-
This is just similar to the I- oxidation using starch as an indicator. So, my hypothesis is that NH3 is protonated by the acid preventing it from
being the ligand that it is thus opening up Co3+ to reduction by I-. Is this correct?
I think I had a few other questions but I'm going to get this out here in case someone can answer it before tomorrow afternoon. If I think of the
other ones, I'll post them here. Thanks in advance.
"If the freedom of religion, guaranteed to us by law in theory, can ever rise in practice under the overbearing inquisition of public opinion, then
and only then will truth prevail over fanaticism." -Thomas Jefferson
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AndersHoveland
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Does the bis(pyridine)iodonium nitrate react with plain dilute HCl solution alone?
If not, my guess is that the pyridine is getting protonated, while the I[+] group react with the iodide ions I[-] from the potassium iodide. So
iodine, I2, may be forming.
I know that a solution of copper(II) chloride will react with sodium iodide to form a precipitate of copper(I) iodide and free iodine. So something
similar could be happening with the cobalt. Cobalt(III) generally holds extremely strongly to amine complexes.
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Mister Junk Pile
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Yeah, I hadn't thought about Co(III) (or Co(II)?) iodide forming.
And we didn't test with HCl alone. Apparently the author of the lab manual didn't feel the need to suggest that.
"If the freedom of religion, guaranteed to us by law in theory, can ever rise in practice under the overbearing inquisition of public opinion, then
and only then will truth prevail over fanaticism." -Thomas Jefferson
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