`A_(1) " and " A_(2)` are the two ores of a metal M. They undergo the following changes. Identify the orea `A_(1) " and " A_(2)`.
`I_(2)+ppt overset("Dil.HCl")underset(KI)larr A_(1) overset("Calcination")to "Black solid"+CO_(2)+H_(2)O`
`A_(2) overset("Roasting")to "Metal"+gasoverset(K_(2)Cr_(2)O_(7))underset(H_(2)SO_(4))to "Green colour"`

Explanation

The data indicates that the ore `A_(1)` is basic copper carbonate and the ore `A_(2)` cuprous sulphide. The equations involved are as follows :
`A_(1): underset(CuCl_(2)+CO_(2)+H_(2)O)underset(darr"Dil. HCl")underset("Basic copper carbonate")(CuCO_(3).Cu(OH)_(2))overset("Calcination")to underset(("Black solid"))(CuO+CO_(2)+H_(2)O`
`2CuCl_(2)4KI to underset(ppt.)(Cu_(2)I_(2))+4KCl+I_(2)`
`A_(2) :2Cu_(2)S+3O_(2) overset("Roasting")to 2Cu_(2)O+2SO_(2)`
`Cu_(2)S+2Cu_(2)O to underset(("Metal"))(6Cu)+underset(("Gas"))(SO_(2))`
`SO_(2)` gas is a strong reducing agent and gives green colouration with acidified `K_(2)Cr_(2)O_(7)` solution.
`3SO_(2)+K_(2)Cr_(2)O_(7)+H_(2)SO_(4) to K_(2)SO_(4)+underset(("Green colour"))(Cr_(2)(SO_(4))_(3))+H_(2)O`