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JEE Advance - Chemistry (2011 - Paper 1 Offline - No. 20)

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 NH3 dissolves O and gives an intense blue solution.
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 NH3 dissolves O and gives an intense blue solution.
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 NH3 dissolves O and gives an intense blue solution.
The final solution contains :
$${[Pb{(N{H_3})_4}]^{2 + }}$$ and $${[CoC{l_2}]^{2 - }}$$
$${[Al{(N{H_3})_4}]^{3 + }}$$ and $${[Cu{(N{H_3})_4}]^{2 + }}$$
$${[Ag{(N{H_3})_2}]^ + }$$ and $${[Cu{(N{H_3})_4}]^{2 + }}$$
$${[Ag{(N{H_3})_2}]^ + }$$ and $${[Ni{(N{H_3})_6}]^{2 + }}$$

Explanation

M is copper and N is silver nitrate according to the equation

$$\mathop {Cu}\limits_{(M)} + \mathop {2AgN{O_3}}\limits_{(N)} \to \mathop {Cu{{(N{O_3})}_2}}\limits_{Blue} + 2Ag$$

The solution of copper nitrate is blue in colour which is formed by the reaction of copper with silver nitrate.

$$AgN{O_3} + NaCl \to AgCl + NaN{O_3}$$

silver chloride precipitates out (white colour ppt.) when silver nitrate reacts with sodium chloride (double displacement reaction).

$$AgCl + 2N{H_4}OH \to [Ag{(N{H_3})_2}]Cl + 2{H_2}O$$

AgCl is soluble in NH4OH.

$$Cu{(N{O_3})_2} + 4N{H_4}OH \to [Cu{(N{H_3})_4}]{(N{O_3})_2} + 4{H_2}O$$

Copper nitrate on reaction with ammonia or ammonium hydroxide gives intense blue solution which is due to $$[Cu{(N{H_3})_4}]$$.

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