When a metal rod M is dipped into an aqueous colourless concetrated 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 `NH_3` dissolves O and gives an intense blue solution.
Q. The compound N is

Options

A. `AgNO_3`
B. `Zn(NO_3)_2`
C. `Al(NO_3)_3`
D. `Ph(NO_3)_2`

Explanation

To solve the question step by step, we will analyze the information provided and deduce the identity of compound N.

### Step 1: Identify the Metal Rod M
The problem states that when a metal rod M is dipped into the solution of compound N, the solution turns light blue. This color change suggests that the metal rod is likely copper (Cu), as copper ions in solution are known to impart a blue color.

**Hint:** Consider the common colors associated with metal ions in solution.

### Step 2: Determine the Nature of Compound N
Since the solution turns light blue upon the introduction of the metal rod, we can infer that compound N must be a solution that contains a species that can react with copper to form copper ions (Cu²⁺). A common compound that fits this description is silver nitrate (AgNO₃), which does not impart color but can react with copper.

**Hint:** Look for compounds that can release metal ions upon reaction with copper.

### Step 3: Analyze the Reaction with NaCl
The problem states that upon adding aqueous NaCl to the blue solution, a white precipitate O is formed. This suggests the formation of silver chloride (AgCl), which is a well-known white precipitate formed when silver ions (Ag⁺) react with chloride ions (Cl⁻).

**Hint:** Consider the solubility rules for salts to identify precipitates formed in reactions.

### Step 4: Examine the Reaction with Ammonia
When aqueous ammonia (NH₃) is added to the solution containing the precipitate O (AgCl), it dissolves the precipitate and gives an intense blue solution. Ammonia can form a complex with copper ions, leading to the formation of a deep blue solution of the tetraamminecopper(II) complex, [Cu(NH₃)₄]²⁺.

**Hint:** Remember that ammonia can form complexes with certain metal ions, changing the color of the solution.

### Conclusion
Based on the above deductions, we conclude that the compound N is silver nitrate (AgNO₃). This is consistent with all the observations made in the problem.

**Final Answer:** The compound N is AgNO₃ (silver nitrate).