Identification of Metal Rod M in Coordination Reaction

To identify the metal rod $\mathbf{M}$, we analyze the chemical reactions described in the passage step-by-step:

1. Reaction of Metal $\mathbf{M}$ with Solution $\mathbf{N}$:
   When the metal rod $\mathbf{M}$ is dipped into an aqueous colourless solution of compound $\mathbf{N}$, the solution turns light blue. Light blue aqueous solutions are characteristic of hydrated copper(II) ions, $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$.

   This indicates that the metal rod $\mathbf{M}$ is copper ($\text{Cu}$), and compound $\mathbf{N}$ is a salt of a less reactive metal that undergoes a redox/displacement reaction, specifically silver nitrate ($\text{AgNO}_3$):
   $\text{Cu (s)} + 2\text{AgNO}_3 \text{ (aq)} \rightarrow \text{Cu(NO}_3)_2 \text{ (aq)} + 2\text{Ag (s)}$

2. Formation of White Precipitate $\mathbf{O}$:
   The light blue solution contains $\text{Cu}^{2+}$ ions and unreacted $\text{Ag}^+$ ions. Adding aqueous $\text{NaCl}$ to this mixture precipitates the silver ions as a white solid, silver chloride ($\text{AgCl}$):
   $\text{Ag}^+ \text{ (aq)} + \text{Cl}^- \text{ (aq)} \rightarrow \text{AgCl (s)} \quad (\text{White precipitate } \mathbf{O})$

3. Dissolution and Deep Blue Coloration with Aqueous $\text{NH}_3$:
   Upon adding excess aqueous ammonia ($\text{NH}_3$):

   • The white precipitate $\text{AgCl}$ dissolves to form the soluble diamminesilver(I) complex:
     $\text{AgCl (s)} + 2\text{NH}_3 \text{ (aq)} \rightarrow [\text{Ag(NH}_3)_2]^+ \text{ (aq)} + \text{Cl}^- \text{ (aq)}$
   • The $\text{Cu}^{2+}$ ions react with ammonia to form the tetraamminecopper(II) complex, which gives an intense deep blue solution:
     $\text{Cu}^{2+} \text{ (aq)} + 4\text{NH}_3 \text{ (aq)} \rightarrow [\text{Cu(NH}_3)_4]^{2+} \text{ (aq)}$

Thus, the metal rod $\mathbf{M}$ is copper ($\text{Cu}$).

Correct Option: (B)