 an NO (+*-) as Ligand @colorfulchemistry2569](https://i.ytimg.com/vi/uLh69ukYmrU/maxresdefault.jpg)
Uploaded September 2016 | Updated September 2026, 2 weeks ago
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 it's Fe(II) or Fe(III).
..................................................................................
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).
..............................................................................
But can NO act in three different ways as a ligand ?
As you can see in this picture, neutral NO is already a radical
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 won't 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.
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 it's Fe(II) or Fe(III).
..................................................................................
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).
..............................................................................
But can NO act in three different ways as a ligand ?
As you can see in this picture, neutral NO is already a radical
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 won't 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.
 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).
..................................................................................
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).
..............................................................................
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.")
