Question

When a metal rod $M$ is dipped into an aqueous colourless concentrated 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 $N{H}_3$ dissolves $O$ and gives an intense blue solution.
The final solution contains

• A. ${\left[Pb{\left(N{H}_3\right)}_4\right]}^{2+}$ and ${\left[CoC{l}_4\right]}^{2-}$
• B. ${\left[Al{\left(N{H}_3\right)}_4\right]}^{3+}$ and ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}$
• C. ${\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}$and ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}$
• D. ${\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}$and ${\left[Ni{\left(N{H}_3\right)}_e\right]}^{2+}$

Answer 1

Answer: (C)

$\underset{(N)}{AgN{O}_3}+NaCl\to \underset{(O)}{AgCl\downarrow }+NaN{O}_3$
$AgCl+2N{H}_3\to {\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}C{l}^{-}$
$Cu{\left(N{O}_3\right)}_2+4N{H}_{4}OH\to {\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}$