Ok, so I've conducted a new test.
My reasoning was if the following equation: Cu2+
CuI + 1/2I2
works for my complex, I should then see the blue coloration disappear and the white-ish CuI precipitate. I weighed 5g of copper complex and added 3 (2 for the equivalence and 1 more for excess) molar equivalents of KI, the solution stayed blue.
I then added what I had left of KI, I didn't weigh it but there may have been like 10g of it left. Still blue, though darker with some dark precipitate (probably that same green powder). I then noticed the solution lightly smelled of ammonia, so I thought I should probably neutralize it with some acid ? I took a small sample of it, added like 5 drops of 96% H2SO4, and there it was, white precipitate in a brown solution, the CuI and I2. There's little to no doubt that the ammonia in solution plays a role in the formation of the green precipitate, and neutralizing it with an acid allows the iodide ions to react with the Cu2+
But if the basic solution is still blue after adding an excess of KI, would it mean that maybe another complex is made, thus needing more iodide equivalents for the complex to fall out of solution ?
I think I should have added the copper solution to the KI solution, as opposed to what I did, adding the KI to the copper solution, this way I could have seen better (the copper solution is really dark).
The more I experiment on this, the more questions I have
I dont have the time for a filtration tonight, I'll try one of these days.
@AWK, I really don't see how this reaction could lead to basic copper sulfate, I can't seem to understand the side reactions you're talking about.