At the end of the reaction the solution had such a color.
DraconicAcid - 17-1-2021 at 11:02
I suspect you will also get some formation of iodine and CuI.vano - 17-1-2021 at 11:28
I thought so too, but I will find out exactly tomorrow when I dry.DraconicAcid - 17-1-2021 at 11:43
There's really no comparison between potassium tetracyanocadmate and copper dibromodiiodocadmate. First, cyanide is a much better ligand to a
transition metal than a halide ion. Secondly, the halide ions are just as likely to coordinate to copper as they are to cadmium (unlike potassium,
which just doesn't do simple coordination chemistry, unless you count crown ethers in organic solvents).
It's well-known that [Hg(SCN)4](2-) forms precipitates with metals such as copper and zinc, but I suspect the structure isn't as simple as the formula
suggests. vano - 17-1-2021 at 12:11
Yes I agree with you. It is true that the halide ion acts likely on cadmium and copper. However, in addition to thiocyanate, copper also produces
tetraiodomercurate, when the solution contains mercury ions, a red compound is formed upon mixing the solution. I also tried to make
dibromodiiodomercurate and it also precipitated instantly, if the compound precipitated instantly after the exchange reaction then it does not matter
that the halide ion acts the same if potassium salt is not in excess. But the case of cadmium is different from that of mercury, but it was worth a
try. woelen - 17-1-2021 at 12:26
I think that the combination of copper(II) and iodide is not a stable one. Copper(II) oxidizes iodide to iodine, at least partially. Probably that is
the cause of the brown color.
What would be interesting is to see whether you can make a copper(I) complex of cadmium, maybe Cu2[CdBr4]?vano - 17-1-2021 at 12:42
@Woelen I can try tomorrow. I will use potassium bromide instead of iodide.
[Edited on 17-1-2021 by vano]Bezaleel - 17-1-2021 at 12:55
I think that the combination of copper(II) and iodide is not a stable one. <snip>
That's true, but it
does not seem that it excreted any iodine. When I added KI to CuSO4 solution, the iodine just floated on top of the solution. In vano's photo, it
stayed in suspension, at least. vano - 17-1-2021 at 13:09
Bezaleel you are right. The reaction between copper and potassium iodide is different.woelen - 18-1-2021 at 00:17
Depends on how much KI you add to CuSO4. If you add excess KI, then no solid iodine will be visible, all of it will remain in solution as brown I3(-)
ion. I do not know the precise conditions of vano's experiment, but I can imagine that some iodine is formed, which remains dissolved and that the
resulting copper(I) forms an insoluble solid with some cadmium-complex. This could explain the combination of brown color and the solution not being
clear.vano - 18-1-2021 at 07:55
@woelen it does not work. But I think a K2[CdBr4] complex formed in the solution, Because when I added the copper sulfate solution, the solution did
not change color, it was just blue. That is, there were no free bromine anions in the solution.Bedlasky - 18-1-2021 at 10:35
Vano: But Woelen spoke about Cu2[CdBr4], not about Cu[CdBr4]. You need firstly reduce copper with metabisulfite, than add this solution in to
[CdBr4]2- solution. Mercury form analogous red precipitate.vano - 18-1-2021 at 10:49
Oh, I thought he was talking about divalent copper. I will try with copper monochloride. Wait a minute.vano - 18-1-2021 at 11:08
I did it twice. First in a test tube, then in a flask, but nothing happened. If you have other ideas I can try it too. I also made Cu2[HgI2Br2] .
