The ion that cannot be precipitated by both HCl and `H_(2)S` is

Options

A. `Pb^(2+)`
B. `Cu^(+)`
C. `Ag^(+)`
D. `Sn^(2+)`

Explanation

`Cu^(+)` disproportionates in the aqueous solution to a very large extent.
`2Cu^(+)hArrCu^(2+)+Cu`
`K=([Cu^(2+)])/([Cu^(+)]^(2))=1.6xx10^(6)`
Thus, the concentration of `Cu^(+)` ion is very small and hence it cannot be precipitated.