Mister Junk Pile - 2-5-2012 at 22:26
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.
AndersHoveland - 2-5-2012 at 22:50
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.
Mister Junk Pile - 3-5-2012 at 01:29
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.