1. Identify the blue solution formed when metal rod $M$ is dipped into colourless concentrated solution of $N$

A light blue solution strongly suggests the presence of $Cu^{2+}$ ions in solution.

A standard reaction is: $Cu + 2AgNO_3 \rightarrow Cu(NO_3)_2 + 2Ag$

• $AgNO_3(aq)$ is colourless and concentrated.
• Copper metal rod dipped in it forms $Cu^{2+}$ in solution, giving a light blue colour.
• So:
  • $M = Cu$
  • $N = AgNO_3$

2. Effect of adding aqueous $NaCl$ to the blue solution

The blue solution now contains mainly:

• $Cu^{2+}$ ions
• $Ag^+$ ions may still be present if $AgNO_3$ was concentrated/excess

On adding $NaCl$, a white precipitate forms: $Ag^+ + Cl^- \rightarrow AgCl(s)$

So white precipitate $O$ is: $O = AgCl$

3. Effect of adding aqueous $NH_3$

Silver chloride dissolves in aqueous ammonia due to complex formation: $AgCl(s) + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-$

Also, $Cu^{2+}$ with excess ammonia gives the intense deep blue tetraammine complex: $Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+}$

Thus the final solution contains:

• $[Ag(NH_3)_2]^+$
• $[Cu(NH_3)_4]^{2+}$

4. Match with options

This corresponds to Option C.

5. Check all options briefly

• A: $[Pb(NH_3)_4]^{2+}$ is not the relevant complex here; also cobalt chloride complex is not indicated.
• B: $Al^{3+}$ does not form such a stable tetraammine complex in aqueous solution; inconsistent.
• C: $[Ag(NH_3)_2]^+$ and $[Cu(NH_3)_4]^{2+}$ — exactly matches observations.
• D: Nickel would give a different initial chemistry; also blue solution and white $AgCl$ precipitate point to silver and copper, not nickel.

Therefore, the correct answer is C.