Uploaded September 2016 | Updated September 2026, 1 week ago
Elemental Arsenic reacts with conc. Nitric Acid.
What makes this reaction so special ?
Arsenic might appear to be metallic, but if you look at it's Chloride for example you will notice that it is not a Arsenic(III) and Chloride(-I) but a molecule. Even more interesting is the fact that there is no Arsenic Nitrate. Much like Non metals dissolving it in Nitric Acid will form the correspondent Acid like Arsenous or Arsenic Acid. Selenium does the same thing.
Now depending on conc. and the acid which is used it might form Arsenous Acid or Arsenic Acid. This is a problem that some elements have. Two oxidation states form at once and have to be somehow separated. Mercury for example forms Hg(I) and Hg(II) if dissolved in Nitric Acid. The chemistry is way different for each of them so a clean product is really important for any experiment.
With Arsenic Acid it is often heated dry so the As(III) will sublimate. We thought of a way to maybe oxidize it like in our Iodine video.
The next problem is that in some way an acid can be extracted but the dry version is the Oxide. So there is also an equilibrium between being Arsenic(V)Oxide and Arsenic Acid.
This is the reason why we usually buy clean Arsenic compounds instead of making them. Literature on this topic is quite controversial as some swap the acids for example.
We will show the reaction of Arsenic with other Acids as well leaving out the actual question for the Oxidation state.
Elemental Arsenic reacts with conc. Nitric Acid.
What makes this reaction so special ?
Arsenic might appear to be metallic, but if you look at it's Chloride for example you will notice that it is not a Arsenic(III) and Chloride(-I) but a molecule. Even more interesting is the fact that there is no Arsenic Nitrate. Much like Non metals dissolving it in Nitric Acid will form the correspondent Acid like Arsenous or Arsenic Acid. Selenium does the same thing.
Now depending on conc. and the acid which is used it might form Arsenous Acid or Arsenic Acid. This is a problem that some elements have. Two oxidation states form at once and have to be somehow separated. Mercury for example forms Hg(I) and Hg(II) if dissolved in Nitric Acid. The chemistry is way different for each of them so a clean product is really important for any experiment.
With Arsenic Acid it is often heated dry so the As(III) will sublimate. We thought of a way to maybe oxidize it like in our Iodine video.
The next problem is that in some way an acid can be extracted but the dry version is the Oxide. So there is also an equilibrium between being Arsenic(V)Oxide and Arsenic Acid.
This is the reason why we usually buy clean Arsenic compounds instead of making them. Literature on this topic is quite controversial as some swap the acids for example.
We will show the reaction of Arsenic with other Acids as well leaving out the actual question for the Oxidation state.
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If Co(II) solutions are treated with alkaline H2O2 Co2O3 and CoO2 may form. If however the precipitate from Sodium Bicarbonate and Cobalt Chloride is oxidized with a 10 M H2O2 at cold temperatures a green product froms.
Well we used worm solutions, 3% Hydrogen Peroxide and Sodium Carbonate but still got a green compound. Of course we cant be sure about its composition that way. It could also be the different oxides but the green color at least indicates some of the complex might have actually formed that way.
The Co(II) is oxidized to Co(III) which forms an insoluble complex with Na2CO3, Na3[Co(CO3)3]. Interesting enough the carbonate can be easily substituted for other ligands making this complex a great precursor to other Co(III) compounds.
We already discussed the chemistry of Co(II) and Co(III) in another video and showed that the oxidation potential is highly dependend on the system you use Cobalt in. Cobalt Chemistry: [Co(CO3)3](3-) (???) [no audio]](https://i.ytimg.com/vi/p8TuAZuQeTU/mqdefault.jpg)
![Palladium Chemistry: Palladium(IV) in Aqua Regia ?
At least we wanted to show a Palladium compound which was not square planar. Once you switch from the oxidation state +II up to +IV the octahedral geometry shows again.
In this video, we demonstrate the similarity to Platinum. We dissolve some elemental Palladium in Aqua Regia. According to literature now the octahedral H2[PdCl6] is formed. You dont find this composition very often in books because some Pd(II) forms as well and the Pd(IV) seems to be converted into Pd(II) once you evaporate it. Still, the dark red-brown color is an indication for Pd(IV) at least according to the books we read.
Much like for Platinium you can also precipitate this using K(+) or NH4(+), the problem being here that many of these salts also precipitate with Pd(II) so the test is not as sensitive as with Platinum. Palladium Chemistry: Palladium(IV) in Aqua Regia ?](https://i.ytimg.com/vi/pLi1dVY3q1M/mqdefault.jpg)

In this video, we show some of the interesting properties of a compound called Sodium Nitroprusside. Often used to analyse Sulphides and Sulphites it shows some interesting chemistry, too.
Here we convert the NO-Ligand in [Fe(CN)5(NO)]2- to NO2 in [Fe(CN)5(NO2)](4-) using Potassium Hydroxide. While basic the reaction shifts towards the yellow NO2-compound. As later H2SO4 is added the pale red Nitroprusside Forms again.
Now you might ask yourself why NO(+) and NO2(-) shift upon addition of OH(-). This reaction is quite similar to a video we have already uploaded. Back then, we tried to add a polysulfide to the same compound and said a NOS-Ligand would form. The Sulphide was a substitute for HS(-) which is the heavier form of OH(-) thus they create a similar compound. We notice that nucleophiles readily attack the Nitrosyl-Nitrogen.
As CN is quite inert here and the Iron doesnt react anyways the complex is a stable substitute for NO(+) and can be used to do reactions with it. It would be interesting to also do the same experiment with Selenides and Tellurides. Those however are quite expensive unfortunately. Maybe we will try to make a Polyselenide and Telluride again and use that.
At the moment we also try to change the NO2(-) for a AsO2(-), yes a Fe-As-coordination. We tried this more than once and it is mentioned in literature to be orange but unfortunately so is the reactant itself. Iron Chemistry: [Fe(CN)5(NO2)](4-)](https://i.ytimg.com/vi/q70jHweiIog/mqdefault.jpg)
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Depending on how x varies the colour changes, while a big concentration of Isothiocyanate (4 x N-coordination) is red, 5x an S-coordination seems to be blue. Ruthenium Chemistry: Thiocyanate color change](https://i.ytimg.com/vi/q9CmMW8Te28/mqdefault.jpg)
![Chromium Chemistry: [2xfaster] Chromium in Gas-Phase (Chromylchloride)
Chromyl Chloride is a well known compound in organic chemistry. We tried using its property to become airborne quite easily and tested if the gaseous Chromium would react with some Hydrogen Peroxide in another vial to form the blue Peroxo-complex.
So we prepared two vials, one containing some 3% Hydrogen Peroxide and diluted H2SO4 and the other containing a mixture of KCl and K2Cr2O7. To the second one a few drops of conc. H2SO4 were added. It starts to foam and an orange (hard to see on camera) gas forms. On the video it looks quite pale but it was quite dense in reality. On the glass walls you can see some red drops of liquid Chromyl Chloride. We covered both vials with a beaker and waited. After some time a blue complex forms proofing that the yellow gas is not only Chlorine but also a gaseous Chromium compound.
To understand the structure of Chromylchloride a bit more it is easier to remind yourself that this compound can be prepared if Chromic Acid is reacted with Hydrochloric Acid, too.
Now we had this in the Polychromate video already. Chromic acid is a tetrahedron having two Cr-O bonds and two Cr-OH bonds. Back then we added acid to form a Cr-O-Cr bond. But what if we dont have the H(+) acting here but the Cl(-) ? The Cl(-) could substitute the OH(-)
Cr-OH + HCl to Cr-Cl + H2O. this happens to both of the Cr-OH groups forming the CrO2Cl2. Now the water reacts with this compound so the would would have to be captured here. One way to do this is using conc. H2SO4 which is strongly hygroscopic. And the HCl can be substituted by NaCl for example to reduce the amount of water even more. And we showed that Chromic acid forms if Chromates meet Acids, so the setup can be reduced to having a Dichromate or Chromate, a Chloride salt and some conc. H2SO4. Chromium Chemistry: [2xfaster] Chromium in Gas-Phase (Chromylchloride)](https://i.ytimg.com/vi/qeQLe06sV0E/mqdefault.jpg)

If Thiocyanate is added to a Bismuth(III) solution the color will change to yellow upon the formation of [Bi(SCN)6](3-). Bismuth Chemistry: [Bi(SCN)6](3-)](https://i.ytimg.com/vi/qzjkOAttUHw/mqdefault.jpg)

![Mercury Chemistry: Mercury(II)chromate
Much like the brown Mercury(I)chromate, we prepared before using HgCl2 instead of Hg2Cl2 a corresponding Hg[CrO4] can be prepared as well. After heating, you can quickly see the bright red color of Hg[CrO4]. Mercury Chemistry: Mercury(II)chromate](https://i.ytimg.com/vi/toITTSeY9Og/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. Chromium Chemistry: Burning (NH4)2[Cr2O7] (no volcano)](https://i.ytimg.com/vi/uCwS1bTaX-8/mqdefault.jpg)
 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. Iron Chemistry: [Fe(H2O)5NO](2+) an NO (+*-) as Ligand](https://i.ytimg.com/vi/uLh69ukYmrU/mqdefault.jpg)
![Tellurium Chemistry: Aromatic Polycation Te4(2+)
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. Tellurium Chemistry: Aromatic Polycation Te4(2+)](https://i.ytimg.com/vi/uSOnzhbSh6A/mqdefault.jpg)