1.

The unique behaviour of CU, having a positive E^(@) (reduction potential) accounts for its inability to liberate H_(2) from acids,. Only oxidising acids (nitric acid and hot concentrated sulphuric acid) react with Cu. The high energy of transform Cu(s) to Cu^(2+)(aq.) is not balanced by its hydration enthalpy. On the other hand, All Cu(II) halides are known except iodide. in this case, Cu^(2+) oxidises I^(-) to I_(2): 2Cu^(2+)+4I^(-) to 2CuI(s)+I_(2) However, copper (I) compounds are unstable in aqueous solution annd undergo disproportionation. 2Cu^(+)(aq.)toCu^(2+)(aq.)+Cu The stability of Cu^(2+)(aq.) rather than Cu^(+)(aq) is due to the much more negative Delta_(Hyd) of Cu^(2+)(aq.) than Cu^(+)(aq.) Q. Select the correct chemical change:

Answer»

`CU+Dil.H_(2)SO_(4) to CuSO_(4)+H_(2)(g)`
`Cu+dil.HNO_(3) to Cu(NO_(3))_(2)+N_(2)O(g)`
`CuSO_(4)(AQ.)+KCN(EXCESS)toK_(2)[Cu(CN)_(4)]`
`CuSO_(4)(aq.)+NH_(4)OH to Cu(OH)_(2)darr`

Solution :`Cu^(2+)(aq.)+2CN^(-)(aq.)tounderset("Unstable")(Cu(CN)_(2)) OVERSET(I.M.R.)to CuCN darr+(CN)_(2)uarr`
`CuCN darr+underset("Excess")(3KCN) hArr underset("Colourless "soln."")(K_(3)[Cu(CN)_(4)])`


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