Identification of Compound N in Metal Displacement Reaction

To determine the identity of compound $\mathbf{N}$, let us analyze the step-by-step chemical transformations described in the passage:

1. Reaction of Metal $\mathbf{M}$ with Compound $\mathbf{N}$:
   When the metal rod $\mathbf{M}$ is dipped into a colourless aqueous solution of compound $\mathbf{N}$, the solution turns light blue. The light blue colour of the aqueous solution indicates the formation of copper(II) ions, $\text{Cu}^{2+}(\text{aq})$, which means metal $\mathbf{M}$ is Copper ($\text{Cu}$).

   For copper to displace the metal ion from compound $\mathbf{N}$, the metal ion must have a higher reduction potential than $\text{Cu}^{2+}/\text{Cu}$ ($E^\circ = +0.34\text{ V}$). Among the options, silver ion ($\text{Ag}^+$) has a standard reduction potential of $E^\circ_{\text{Ag}^+/\text{Ag}} = +0.80\text{ V}$. Thus, copper reduces $\text{Ag}^+$ to silver metal while itself oxidising to $\text{Cu}^{2+}$:
   $\text{Cu}(s) + 2\text{AgNO}_3(\text{aq}) \rightarrow \text{Cu(NO}_3)_2(\text{aq}) + 2\text{Ag}(s)$
   Therefore, compound $\mathbf{N}$ is silver nitrate, $\text{AgNO}_3$.

2. Precipitation with $\text{NaCl}$:
   When aqueous $\text{NaCl}$ is added to the resulting blue solution containing unreacted $\text{Ag}^+$ ions, a white precipitate $\mathbf{O}$ of silver chloride ($\text{AgCl}$) is formed:
   $\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(s) \quad (\text{white precipitate } \mathbf{O})$

3. Dissolution and Complex Formation with $\text{NH}_3$:
   Addition of aqueous ammonia ($\text{NH}_3$) dissolves the white precipitate $\mathbf{O}$ ($\text{AgCl}$) by forming the soluble diamminesilver(I) complex ion:
   $\text{AgCl}(s) + 2\text{NH}_3(\text{aq}) \rightarrow [\text{Ag(NH}_3)_2]^+(\text{aq}) + \text{Cl}^-(\text{aq})$

   Simultaneously, the excess ammonia reacts with the $\text{Cu}^{2+}$ ions present in the solution to form the tetraamminecopper(II) complex, which gives an intense deep-blue solution:
   $\text{Cu}^{2+}(\text{aq}) + 4\text{NH}_3(\text{aq}) \rightarrow [\text{Cu(NH}_3)_4]^{2+}(\text{aq}) \quad (\text{intense blue solution})$

Thus, the compound $\mathbf{N}$ is $\text{AgNO}_3$.

Correct Option: A