Step-by-step solution:

1. Analyze the first observation: A metal rod M is dipped into a colorless concentrated aqueous solution of compound N, and the solution turns light blue.

   • This color change indicates that the metal M is being oxidized and entering the solution as a hydrated metal ion that has a characteristic blue color.
   • The light blue color of an aqueous solution is a classic sign of the hydrated copper(II) ion, $[Cu(H_2O)_6]^{2+}$.
   • This suggests that the metal rod M is Copper (Cu). The oxidation reaction is: $Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-$.
   • For this reaction to occur, the cation in the solution of compound N must have a higher standard reduction potential than copper ($E^0_{Cu^{2+}/Cu} = +0.34 V$). Since the solution of N is colorless, a good candidate for the cation in N is $Ag^+$ ($E^0_{Ag^+/Ag} = +0.80 V$). So, compound N could be silver nitrate, $AgNO_3$.
   • The overall reaction would be: $Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)$. The solution turns light blue due to the formation of $Cu^{2+}(aq)$.

2. Analyze the second observation: Addition of aqueous $NaCl$ to the blue solution gives a white precipitate O.

   • The blue solution now contains $Cu^{2+}$ ions and, if the reaction in step 1 was not complete, some unreacted cations from compound N (e.g., $Ag^+$).
   • When $NaCl$ is added, it provides $Cl^-$ ions.
   • $Cu^{2+}$ ions form $CuCl_2$ which is soluble in water.
   • $Ag^+$ ions react with $Cl^-$ ions to form a characteristic white precipitate of silver chloride, $AgCl$.
   • Reaction: $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$.
   • This observation is consistent with our hypothesis that N is an $Ag^+$ salt and the white precipitate O is $AgCl$.

3. Analyze the third observation: Addition of aqueous $NH_3$ dissolves O and gives an intense blue solution.

   • This step involves two reactions happening in the mixture.
   • Reaction 1: Dissolution of precipitate O. The white precipitate O ($AgCl$) reacts with aqueous ammonia to form a soluble complex, diamminesilver(I) chloride. This is a characteristic test for silver halides (except AgI). $AgCl(s) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq) + Cl^-(aq)$ This explains why the precipitate O dissolves.
   • Reaction 2: Formation of an intense blue solution. The $Cu^{2+}$ ions in the solution react with aqueous ammonia to form the tetraamminecopper(II) complex ion, $[Cu(NH_3)_4]^{2+}$, which has a characteristic deep blue or intense blue color. $Cu^{2+}(aq) + 4NH_3(aq) \rightarrow [Cu(NH_3)_4]^{2+}(aq)$ This explains the formation of the intense blue solution.

4. Evaluate other options:

   • A: Fe (Iron): Iron forms $Fe^{2+}$ (light green) or $Fe^{3+}$ (yellow-brown) ions in solution, not light blue.
   • C: Ni (Nickel): Nickel forms $Ni^{2+}$ ions, which are green in aqueous solution, not light blue.
   • D: Co (Cobalt): Cobalt forms $Co^{2+}$ ions, which are pink in aqueous solution, not light blue.

5. Conclusion: The entire sequence of reactions and color changes is perfectly consistent with the metal M being Copper (Cu). The properties observed (light blue solution with $Cu^{2+}$, formation of $AgCl$ precipitate, dissolution of $AgCl$ with ammonia, and formation of deep blue $[Cu(NH_3)_4]^{2+}$ complex) are all standard qualitative analysis tests for copper and silver ions.