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:

Options

A. `Cu+Dil.H_(2)SO_(4) to CuSO_(4)+H_(2)(g)`
B. `Cu+dil.HNO_(3) to Cu(NO_(3))_(2)+N_(2)O(g)`
C. `CuSO_(4)(aq.)+KCN(excess)toK_(2)[Cu(CN)_(4)]`
D. `CuSO_(4)(aq.)+NH_(4)OH to Cu(OH)_(2)darr`

Answer

D

Explanation

`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)])`