Uploaded September 2016 | Updated September 2026, 1 week ago
There is an interesting way to test for Iodides. Thallium(I) is oxidized using Bromine to Tl(III) and Tl(III) is reacted with Iodide to form Iodine which will be identified using starch.
Still many books and people believe in a Tl(III)-Iodide which is formed if Tl(I)-Iodide is heated with Iodine in Methanol or Ethanol.
In this experiment we will try this:
Part one of the video is the precipitation of Thallium(I)-Iodide. To a TlNO3 solution some solid NaI is added. The insoluble Thallium(I)-Iodide is yellow. Remember, Thallium and Silver share a quite similar chemisty, AgI is yellow and insoluble as well.
Some Methanol is added an Iodine in dissolved. The mixture is heated till it boils. It becomes darker and transparent again and a black compound settles on the bottom of the test tube.
In Methanol this compound can be expressed a TlI3 with a polyiodide so I3(-), while as solid compound on the bottom of the tube an Ion pair [Tl](+) [I3](-) forms.
There is an interesting way to test for Iodides. Thallium(I) is oxidized using Bromine to Tl(III) and Tl(III) is reacted with Iodide to form Iodine which will be identified using starch.
Still many books and people believe in a Tl(III)-Iodide which is formed if Tl(I)-Iodide is heated with Iodine in Methanol or Ethanol.
In this experiment we will try this:
Part one of the video is the precipitation of Thallium(I)-Iodide. To a TlNO3 solution some solid NaI is added. The insoluble Thallium(I)-Iodide is yellow. Remember, Thallium and Silver share a quite similar chemisty, AgI is yellow and insoluble as well.
Some Methanol is added an Iodine in dissolved. The mixture is heated till it boils. It becomes darker and transparent again and a black compound settles on the bottom of the test tube.
In Methanol this compound can be expressed a TlI3 with a polyiodide so I3(-), while as solid compound on the bottom of the tube an Ion pair [Tl](+) [I3](-) forms.
![Mercury Chemistry: Hg3[AsO3]2 and Hg3[AsO4]2
In this video, we make and compare a Mercury(II)-Arsenite and -Arsenate by mixing HgCl2 with NaAsO2 and NaHAsO4. Now usually we do not make every possible compound just to increase the number of videos but for this, we wanted to test something. The Arsenate is, according to literature yellow, while the Arsenite is colorless, but develops a yellow color which is believed to be due to oxidation. We wanted to test this and made both solutions.
At first, you cannot see any precipitate (right jar = Arsenate) but after a while, and a second addition of more Arsenate you can see a yellow turbid color. For the Arsenite, it happens much fast but the solution turns yellow as well, much as expected. If you compare the two after a while you can see that they have the same type of yellow in their solution. 24h later the Arsenate had precipitated as yellow compound while the Arsenite formed a colorless one.
This proves, that the compounds differ in color and only a small fraction, the parts in solution are yellow, probably due to oxygen in the water and the stirring process and that it probably is some oxidation of the Arsenite.
We also tested both up to 350°C for thermochromism but nothing happened. Mercury Chemistry: Hg3[AsO3]2 and Hg3[AsO4]2](https://i.ytimg.com/vi/kCN6V7e5-Eo/mqdefault.jpg)

 and [CoCl4](2-) you probably thought this could be possible with similar compounds as well.
So what we did here was to mix CuCl2 with H2O and HCl to a ratio where the nearly colorless, blue [Cu(H2O)6](2+) is turned into the green [CuCl4](2-) when it is heated. Copper Chemistry: Copper Thermochromism](https://i.ytimg.com/vi/l6AYp2X_iq4/mqdefault.jpg)

. So why is it stable in aqueous conditions ?
Lets assume we had Copper(I) in a gas phase, free from any solvation- or hydration energy. To make Cu(II) we require a lot of Ionizationenergy. This makes the reaction really endothermic. If we take it and dissolve it into Acetonitrile, lets say CuCl which is soluble in Acetonitrile, it will dissolve and stay as Cu(I). The reason for this is that the solvationenergy in Acetonitrile is too low to overcome this ionization energy. Water however produces a quite large Energy and Cu(II) can be formed under disproportionation.
Now what happens in this experiment ?
2 Cu(I) to Cu(II) and Cu(0) is an equilibrium. In water it is exothermic forming Cu(II). And if you work with Cu(II) solutions it wont turn into Cu(I) (unless you add an reducing agent like Sulfite) as no Cu(0) is present to shift the equilibrium back. As we know from Le Chatelier, if a reaction is exothermic, heat causes the system to go back to the endothermic side. So Cu(I) forms and combines with the Cl(-) which is present to form CuCl. CuCl is insoluble in water as the lattice energy is higher than the hydration energy and thus Cu(I) remains stable discoloring the blue solution over time. Copper Chemistry: Synproportionation Cu(II) and Cu(0)](https://i.ytimg.com/vi/mGx4hOx7YdY/mqdefault.jpg)





