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Author: Subject: About to anhydrate Copper Nitrate, any suggestions ?
AndersHoveland
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[*] posted on 23-2-2013 at 14:17


Quote: Originally posted by woelen  
Making anhydrous Cu(NO3)2 is not easy at all. Before appr. 1960 people believed that this compound did not exist at all. But in the 1960's the anhydrous salt was prepared by reaction of metallic copper with a solution of thoroughly dried NO2 in ethyl acetate at low temperature.

Woelen, I am probably completely wrong here, but I question whether they actually prepared pure copper nitrate by this method. After all, such a reaction would have to result in nitric oxide as a by-product. And is not anhydrous copper(II) nitrate somewhat of a reactive oxidizer? How then can nitric oxide be formed? I wonder if perhaps what these researchers actually prepared was the nitrosyl complex of copper nitrate, NO·Cu(NO3)2. The formation of such an adduct would make the copper nitrate less oxidizing. Such nitrosyl complexes are quite common. I wonder whether the researchers bothered to quantitatively measure the nitrogen oxides resulting from decomposition when their product was heated. They may have just confirmed that no water was emitted, and jumped to an erroneous conclusion that they had prepared the pure copper nitrate.

If you have a link to the paper, we can check to see whether they quantitatively measured the nitrogen oxides given off per gram of their copper nitrate when heated.

[Edited on 23-2-2013 by AndersHoveland]
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woelen
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[*] posted on 24-2-2013 at 04:53


The information I have is not from a paper, but from a book. IIRC it is the book "General Chemistry" from Pauling. The material is described as deep blue and somewhat volatile.

I think that NO.Cu(NO3)2 would have copper in oxidation state +1, similar to the [Fe(NO)](+) complex, which has iron in oxidation state +1. But copper in oxidation state +1, combined with NO2 in the reaction vessel and with the NO3(-) ions in the final compound? Is that possible? So, I'm inclined to believe that they really made Cu(NO3)2 and not some nitroso-complex.




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AndersHoveland
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[*] posted on 24-2-2013 at 23:04


It would not really be in the +1 oxidation state. In fact, if you want to think of it this way, it is probably somewhere between +1 and +2. Do not forget that anhydrous copper(II) nitrate is an oxidizer.

The nitrosyl complex is really more of an adduct. And the common hydrate of copper(II) can be seen as an adduct also. Yet you would not say that the copper in copper(II) nitrate hydrate is in the +1 oxidation state.

Think of it this way: Cu(NO3)2 is like a lewis acid and NO behaves as a lewis base in this adduct. Because nitric oxide exists as a radical, the distinction between it acting as a base and acting as a reducing agent is a blurry one.

The definition of a lewis base is essentially the donation of an electron, with the formation of a covalent bond. The definition of a reducing agent in this case would be if the nitric oxide singularly formed a polar bond.

It is no coincidence that oxidizers are more likely to be acids, and reducing agents are more likely to be bases.

If the nitric oxide merely formed a polar bond, that would require the copper atom to be trivalent, since it is also bonded to two nitrate groups. Not that trivalent copper does not exist, but that is not the case here. The nitric oxide not only forms a bond with the copper atom, it also transfers charge into the nitrate groups. This explains how the copper can bond to 3 different groups - the bond order with each is less than 1. Alternatively this can be understood through resonance structures. Because the nitric oxide both bonds and transfers charge, it is acting as a base, not merely just a reducer.

There is nothing in NO·Cu(NO3)2 for NO2 to oxidize.

NO·CuCl2 is a more well known nitrosyl complex of copper. It is not stable in the presence of water, releasing its nitric oxide. Consider this: If the nitric oxide really had reduced the Cu+2, then a molecule of H2O would not be able to so easily displace it. Thus it is clear the NO is acting as a base in the complex.

[Edited on 26-2-2013 by AndersHoveland]
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