Uploaded October 2016 | Updated September 2026, 1 week ago
In this video we used the Copper powder which we made in the previous video. To a very weak (slightly acidified with HCl) solution of CuSO4 (blue) some Copper powder was added and heated.
It first turned all colorless then a white powder precipitated.
What is happening is that Cu(II) and Cu(0) synproportionate at elevated temperatures to form Cu(I) which is precipitated by the Chloride in the HCl to CuCl.
Something similar can be done to Fe(II) solutions. If those are stored above Iron turnings it will reduce any Fe(III) which could be formed from Oxygen in the air.
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The question is why is Cu(II) so stable in Aqueous conditions. It needs to perform some serious Jahn-Teller distortion to somehow stabilize its d9 system while Cu(I) would have a stable and closed d10 ? I think we mentioned it before but solvation energy, so solvents can influence the redox-chemistry as well.
First of all, Cu(I) is stable in Solid State an in some other solvents than water, namely those which coordinate via softer bases than water as Cu(I) is softer than Cu(II). With a full d-shell Cu(I) can form according to the VSEPR model an sp³-hybride which means Cu(I) likes to appear in for example tetrahedral coordinations while Cu(II) forms due to Jahn-Teller distortion and a not fully filled d-shell an octahedral geometry.
Now we could list different environments around Cu(I) and Cu(II) and in general none is preferred over the other it depends on the situation. Higher coordination can bring more stabilization energy while Cu(I) tends to form stronger bonds to pi-acceptor ligands.
So we have to be more specific. For all those lovely blue colored Cu(II) compounds we are usually referring to [Cu(H2O)6](2+). So why is it stable in aqueous conditions ?
Let's 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, let's 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.
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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 won't 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.
In this video we used the Copper powder which we made in the previous video. To a very weak (slightly acidified with HCl) solution of CuSO4 (blue) some Copper powder was added and heated.
It first turned all colorless then a white powder precipitated.
What is happening is that Cu(II) and Cu(0) synproportionate at elevated temperatures to form Cu(I) which is precipitated by the Chloride in the HCl to CuCl.
Something similar can be done to Fe(II) solutions. If those are stored above Iron turnings it will reduce any Fe(III) which could be formed from Oxygen in the air.
------------------------------------------
The question is why is Cu(II) so stable in Aqueous conditions. It needs to perform some serious Jahn-Teller distortion to somehow stabilize its d9 system while Cu(I) would have a stable and closed d10 ? I think we mentioned it before but solvation energy, so solvents can influence the redox-chemistry as well.
First of all, Cu(I) is stable in Solid State an in some other solvents than water, namely those which coordinate via softer bases than water as Cu(I) is softer than Cu(II). With a full d-shell Cu(I) can form according to the VSEPR model an sp³-hybride which means Cu(I) likes to appear in for example tetrahedral coordinations while Cu(II) forms due to Jahn-Teller distortion and a not fully filled d-shell an octahedral geometry.
Now we could list different environments around Cu(I) and Cu(II) and in general none is preferred over the other it depends on the situation. Higher coordination can bring more stabilization energy while Cu(I) tends to form stronger bonds to pi-acceptor ligands.
So we have to be more specific. For all those lovely blue colored Cu(II) compounds we are usually referring to [Cu(H2O)6](2+). So why is it stable in aqueous conditions ?
Let's 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, let's 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 won't 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.
 (???) [no audio]](https://i.ytimg.com/vi/p8TuAZuQeTU/mqdefault.jpg "Cobalt Chemistry: [Co(CO3)3](3-) (???) [no audio]
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.")
](https://i.ytimg.com/vi/q70jHweiIog/mqdefault.jpg "Iron Chemistry: [Fe(CN)5(NO2)](4-)
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.")
