Uploaded September 2016 | Updated September 2026, 1 week ago
Sulfur consists S8-crown-shaped rings. That way a zick-zack-structure is maintained.At room temperature the so called alpha-Structure is present. Van-Der-Waals forces hold the S8-crowns in certain staggered orders to form an orthorhombic sulfur.
Heating this slowly to 100°C the Van-Der-Waals forces aren't strong enough anymore and the monoclinic beta-Sulfur is formed. This is yellow, too since no bigger change has happened.
Heating it to around 120°C will form a melt of S8-crowns. Further heat starts to break the S-S-bonds forming different possible substructures which become soon dark red as different polymer fragments form.
For π-Sulfur small rings are expected, at around 160°C, λ-Sulfur forms (the rings break and long chains are formed). ω-Sulfur becomes darker and darker. If cooled µ-Sulfur can be isolated. Here the polymer structure remains for a short period of time and a dark red rubber can be made.
If heated even more sulfur will go into the gas-phase. This can be seen at different points in the video. For lower temperatures it consists of S7-Rings, for higher ones of S4, S3 or S2 molecules.
At really high temperatures S2 is formed as diamagnetic compound (like Oxygen). At some point the colors change for a short time,
as S8 - S4 are red as gases, S3 (said to form at 700°C and 1 mbar pressure) will be blue and S2 (1800°C) will be purple. At really high temperatures and low pressures atomic Sulfur will form.
Sulfur consists S8-crown-shaped rings. That way a zick-zack-structure is maintained.At room temperature the so called alpha-Structure is present. Van-Der-Waals forces hold the S8-crowns in certain staggered orders to form an orthorhombic sulfur.
Heating this slowly to 100°C the Van-Der-Waals forces aren't strong enough anymore and the monoclinic beta-Sulfur is formed. This is yellow, too since no bigger change has happened.
Heating it to around 120°C will form a melt of S8-crowns. Further heat starts to break the S-S-bonds forming different possible substructures which become soon dark red as different polymer fragments form.
For π-Sulfur small rings are expected, at around 160°C, λ-Sulfur forms (the rings break and long chains are formed). ω-Sulfur becomes darker and darker. If cooled µ-Sulfur can be isolated. Here the polymer structure remains for a short period of time and a dark red rubber can be made.
If heated even more sulfur will go into the gas-phase. This can be seen at different points in the video. For lower temperatures it consists of S7-Rings, for higher ones of S4, S3 or S2 molecules.
At really high temperatures S2 is formed as diamagnetic compound (like Oxygen). At some point the colors change for a short time,
as S8 - S4 are red as gases, S3 (said to form at 700°C and 1 mbar pressure) will be blue and S2 (1800°C) will be purple. At really high temperatures and low pressures atomic Sulfur will form.
![Chromium Chemistry: Burning (NH4)2[Cr2O7] (no volcano)](https://i.ytimg.com/vi/uCwS1bTaX-8/mqdefault.jpg "Chromium Chemistry: Burning (NH4)2[Cr2O7] (no volcano)
You are probably familiar with the famous Ammonium Dichromate volcano. If you light some (NH4)2[Cr2O7] it will form Cr2O3 while creating burning sparks that form a sort of volcano. We tried to do that but our Dichromate is quite wet, unfortunately. So we tried a different idea and heated it directly with the burner. The temperature is high enough to vaporize the water and burn the Dichromate, creating much more sparks than usually.
What is happening is that the system contains two parts, in Ammonia the Nitrogen is reduced. If it is oxidized a stable N-N triple bond forms. On the other hand, the Dichromate contains a highly oxidized Chromium(VI). The stable form is Cr(III) so in exchange Cr(VI) is reduced and N(-III) oxidized.
This is also the reason why we didnt dry our Dichromate for the volcano. Many of these combinations of Ammonium and strong oxidizers (e.g. Ammonium Nitrate) tend to combust when heated. The dry compound starts to decompose around the boiling point of water. Although Ammonium Dichromate is usually not that reactive.")
 an NO (+*-) as Ligand](https://i.ytimg.com/vi/uLh69ukYmrU/mqdefault.jpg "Iron Chemistry: [Fe(H2O)5NO](2+) an NO (+*-) as Ligand
This is a very famous test for Nitrate-Ions:
In this case we added some Iron nails for one day to 38% H2SO4 and then added some of the clear solution to a KNO3 solution. If conc. H2SO4 is very carefully added a brown-purple ring forms.
The ring has the composition [Fe(H2O5)NO](2+).
For many years people believed that Iron has the Oxidation state +I in this compound since NO is actually a Nitrosyl NO(+). Other suggestions say its Fe(II) or Fe(III).
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Just thinking about this is sounds strange for this to be Fe(I). Fe(II) is really hard to store as solution since it easily oxidizes to Fe(III). Thus it should become an Fe(III) really easy and not an Fe(I) (where no real compounds exist, some exceptions of course). On the other hand Nitrate is reduced to NO so something clearly has to be oxidized for this to happen which is Fe(II) becoming Fe(III).
We can also look this up in the redox series:
NO3(-) + 4H3O(+) + 3e(-) to NO + 6H2O (U = 0,96 V)
Fe(II) to Fe(III) + e(-) (U = 0,77 V).
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But can NO act in three different ways as a ligand ?
As you can see in this picture, neutral NO is already a radical
http://daten.didaktikchemie.uni-bayreuth.de/umat/stickstoffoxide/no_moschema.gif
So we could write it as NO* for a second. If another electron is added to form NO(-) we reach something that reminds us of Oxygen and if we take an electron away to make NO(+) we can get something which is isoelectronic to Carbonyl (CO). And Oxygen and Carbonyl are famous Ligands. This is why all three seem possible depending on the electronic situation making the Iron a (I), (II) or (III). It gets even harder as there are multiple ways how NO can coordinate to a metal center, but we wont further discuss this here.
Quite new calculations and research show that it is really an Fe(III) and an NO(-) present.
But there is more! If you followed our video series you might have seen videos where we used Nitroprusside. In Nitroprusside an NO(+) is present. If this compound is reduced even further you might expect NO* to form. Analyzing the products showed that for some percent, some say even 25%, Fe(I) forms !
Still many of these compounds remain a mystery and are worth to be further researched.")
 (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.")
