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Cu^(2+) on reaction with potassium iodid...

`Cu^(2+)` on reaction with potassium iodide gives

A

`CuI_3`

B

`Cu(I_3)_2`

C

`Cu_2I_2`

D

`CuI`

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Consider the following statements: Statement-1: Cu^(2+) ions are reduced to Cu^(+) by potassium iodide and potassium cyanide both, when taken in excess. Statement-2: H_(2)S will precipitate the sulphide of all the metals from the solution from the solutions of chlorides of Cu,Zn and Cd if the solution is aqueous. Statement-3: The presence of magnesium is confirmed in qualitative analysis by the formation of a white crystalline precipitate of MgNH_(4)PO_(4) . Statement-4: Calmol on reaction with potassium iodide gives red precipitate.

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