When a metal rod M is dipped into an aqueous colourless concetrated solution of compound N, the solution turns light blue. Addition of aqueous NaCl to the blue solution gives a white precipitate O. Addition of aqueous `NH_3` dissolves O and gives an intense blue solution.
Q. The metal rod M is
`Fe`
`Cu`
`Ni`
`Co`

Options

A. `Fe`
B. `Cu`
C. `Ni`
D. `Co`

Answer

B

Explanation

`Cu(M)+AgNO_(3)(N)`(queous colourless solution) `to` Resultant solution contains `Cu(NO_(3))_(2)` (blue solution) and `AgNO_(3)` (colourless solution)
Note: Here it is considered that complete `AgNO_(3)` is not utilized in the reaction.
`AgNO_(3)(aq)+NaCl(aq) to AgCl darr`(white) `(O)+NaNO_(3)`
Solution containing white ppt. of `AgCl` also contains `Cu(NO_(3))_(2)` which developed deep blue colouration with aqueous `NH_(3)` solution
`AgCl darr ("white")+2NH_(3)(aq.) to [Ag(NH_(3))_(2)]^(+)`
`Cu(NO_(3))_(2)(aq.)+4NH_(3) (aq.) to [Cu(NH_(3))_(4)](NO_(3))_(2)` (deep blue solution)
So, Metal rod `M` is `Cu`.
The compound `N` is `AgNO_(3)` and the final solution contains `[Ag(NH_(3))_(2)]^(+)` and `[Cu(NH_(3))_(4)]^(2+)`