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Compounds Cu(I).

Copper subgroup trends | History of discovery | Preparation | Gold extraction . | Electronic Configurations & Oxidation States | Free elements | History | History | Synthesis | Mercury removal |


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  4. Compounds of Ag
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These compounds can be prepared by Cu(II) reduction. Cu2О is formed at the action of hydrazine, hydroxylamine, aldehydes, glucose on Cu(ОH)2:.

4Cu(ОH)2 + N2H4 = 2Cu2O + N2 + 6H2O

or 4CuO 2Cu2O + О2

2CuI +2NaOH = Cu2O + H2O + 2NaI

 

CuOН does not exist. It decomposes immediately at the moment of liberation on Cu2O and H2O. Cu2O has predominately basic properties. It react with acids::

Cu2O + 4HCl = 2H[CuCl2] + H2O

Cu2O + H2SO4 = Cu + CuSO4 + H2O.

Salts Cu(І) can be obtained by Cu(II) reduction:

2CuSO4 + 2NaCl + SO2 + 2H2O = 2CuCl¯ + Na2SO4 + 2H2SO4

CuCl2 + 2HCl + Cu = 2H[CuCl2] (+H2O) = 2HCl + 2CuCl¯

CuI formation is a result of redox process that proceeds quantitavely:

2CuSO4 + 4KI = 2CuI¯ + I2 + 2K2SO4

Salts Cu(I) are colourless that are usually insoluble in water but at the action of ammonia and some other ligands can form soluble complexes:

CuCl + 2NH3 = [Cu(NH3)2]Cl

CuCl + HCl = H[CuCl2]

Complexes of Cu(I) with monodental ligands with coordination number 2 have maximal stability (for instance, [Cu(NH3)2]+, [CuCl2]-, [Cu(СN)2]-) and have linear configuration (Cl-Cu-Cl)-. Complexes with coordination number 4 are tetrahedral. Some complexes are oxidized by oxygen:

4[Cu(NH3)2]+ + 4NH3 + 4NH4+ + O2 = 4[Cu(NH3)4]2+ + H2O

This reaction is used to absorb oxygen from gaseous mixtures.

In general, equilibrium 2Cu+ Cu2+ + Cu is easily displaced depending on conditions. The presence of I-, CN- that form stable compounds with Cu(I) displaces equilibrium to Cu+ formation. The presence of SO42-, ClO4- shows the contrary action.

 

Compounds Cu(III). These compounds are obtained by oxidation of Cu(II). The deep red Cu2О3 forms at the action of Na2O2 or K2S2O8 on Cu(ОН)2 or Na2[Cu(ОH)4]:

2Na2[Cu(ОH)4] + Na2O2 = Cu2О3 + 6NaOH + H2O

Potassium superoxide being smelted with CuO gives potassium cuprate (III):

2KO2 + 2CuO = 2KCuO2 + O2

This crystalline substance is blue or deep blue. It is completely decomposed by water. Compounds of Cu(III) are unstable and behave as very strong oxidising agents:

2Cu2О3 + 4H2SO4 = 4CuSO4 + O2 + 4H2O.


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