Uploaded October 2016 | Updated September 2026, 1 week ago
Much like our experiment with Selenium, Tellurium forms polycations in conc. H2SO4 as well. In this case, the color is red. Back then we had an 8-membered crown-shaped ring.
Tellurium, however, forms a different polycation under the same conditions. This time a square is formed which consists of 4 Tellurium atoms. As the system contains 6 pi-electrons* and it is flat we call this a Hueckel aromatic compound.
Besides Te4(2+), S4(2+) and Se4(2+) can form under different conditions as well and those are aromatic, too. Why does this compound form so fast in comparison to a sulfur polycation which requires Oleum to form ? Sulfur is much more electronegative and the cation possesses a much higher acidity. Thus, Tellurium is easier to react it in H2SO4 and no more reactive reagent is required. There are multiple ways to make the Te4(2+) and depending on the method and counterion it can further polymerize forming Te8(4+) as dimer, or [Te4(2+)]x as polymeric chain where the Te-squares are connected via the vertices.
Note:
* If you are unsure how to determine the amount of electrons here, this is the official calculation to this problem:
- Number of valence electrons = 4 x 6 = 24
from this some electrons have to be subtracted
- Number of sigma-electrons = 8
- Amount of lone pairs 2 x 4 = 8
- Charge = 2
If you subtract these 18 from 24 you get 6 pi-electrons in the system. Next to determine aromaticity you can use the Frost-Musulin-diagram. Take the shape of your ring and place it on a vertex. Each vertex is now an energy level. If you take benzene for example you will see that there are two times two vertices on the same height. Those are degenerate energy levels. Each of these "layers" have to be either completely empty or fully filled to yield a stable compound. Unpaired situations will lead to reactive radicals, which is why there are aromatic, anti-aromatic and non-aromatic compounds and the Hueckel-rules exist.
Ok so we take our square and place it on an edge we get 4 energy levels (where each vertex is). Two of those will be at the same hight and therefore be degenerate. Now we know that we have 6 pi-electrons and each energy level can be filled with two electrons. This will fill the first and the second degenerate "layer" which is a stable situation making it aromatic. It is also flat, which is a requirement for aromatic compounds as well.
Much like our experiment with Selenium, Tellurium forms polycations in conc. H2SO4 as well. In this case, the color is red. Back then we had an 8-membered crown-shaped ring.
Tellurium, however, forms a different polycation under the same conditions. This time a square is formed which consists of 4 Tellurium atoms. As the system contains 6 pi-electrons* and it is flat we call this a Hueckel aromatic compound.
Besides Te4(2+), S4(2+) and Se4(2+) can form under different conditions as well and those are aromatic, too. Why does this compound form so fast in comparison to a sulfur polycation which requires Oleum to form ? Sulfur is much more electronegative and the cation possesses a much higher acidity. Thus, Tellurium is easier to react it in H2SO4 and no more reactive reagent is required. There are multiple ways to make the Te4(2+) and depending on the method and counterion it can further polymerize forming Te8(4+) as dimer, or [Te4(2+)]x as polymeric chain where the Te-squares are connected via the vertices.
Note:
* If you are unsure how to determine the amount of electrons here, this is the official calculation to this problem:
- Number of valence electrons = 4 x 6 = 24
from this some electrons have to be subtracted
- Number of sigma-electrons = 8
- Amount of lone pairs 2 x 4 = 8
- Charge = 2
If you subtract these 18 from 24 you get 6 pi-electrons in the system. Next to determine aromaticity you can use the Frost-Musulin-diagram. Take the shape of your ring and place it on a vertex. Each vertex is now an energy level. If you take benzene for example you will see that there are two times two vertices on the same height. Those are degenerate energy levels. Each of these "layers" have to be either completely empty or fully filled to yield a stable compound. Unpaired situations will lead to reactive radicals, which is why there are aromatic, anti-aromatic and non-aromatic compounds and the Hueckel-rules exist.
Ok so we take our square and place it on an edge we get 4 energy levels (where each vertex is). Two of those will be at the same hight and therefore be degenerate. Now we know that we have 6 pi-electrons and each energy level can be filled with two electrons. This will fill the first and the second degenerate "layer" which is a stable situation making it aromatic. It is also flat, which is a requirement for aromatic compounds as well.
 (Pd in HCl/H2O2)](https://i.ytimg.com/vi/wQA5K90Ijqg/mqdefault.jpg "Palladium Chemistry: [Pd(Cl4](2-) (Pd in HCl/H2O2)
Palladium is an interesting element. Being a Platinum-group metal and standing between Nickel and Platinum one might expect it to be unreactive. Funnily this is not the case and some people even joke it could be more reactive than Nickel. Obviously, this might be a bit too much but the reactivity can be compared to the one of silver with one important difference: Palladium usually does not build up many passive layers. Meaning that it can dissolve in HCl for example.
In this video, however, we cheated a little and this is first to make it shorter and second to make it more visible. Usually, oxygen in HCl is enough to slowly dissolve Palladium but we used H2O2 instead of the oxygen here. Now you might say there is no big difference but there is. The redox reaction now changed to the one of an acidified Peroxide which has a huge redox potential being capable of even dissolving gold and platinum to some extent!
And PdCl2 forms Pd(IV) in H2O2 as well!
Note however that we also tried it with HCl and Oxygen and a lot of boiling and it turned yellow as well!
What forms here is the complex [PdCl4](2-). While Nickel(II) already shows some square planar geometry, especially when working with cyano-Ligands for Palladium almost all complexes are square planar.
Also the coordination number of 4 is really common in Palladium chemistry. So if you have to guess, a square planar geometry is always a good guess when dealing with Pd(II). Even the aqua-complex only has 4 Ligands.")
