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 metal rod M is
`Fe`
`Cu`
`Ni`
`Co`

Options

A. `Fe`
B. `Cu`
C. `Ni`
D. `Co`

Explanation

To determine the identity of the metal rod M, we can analyze the information given in the question step by step.

### Step-by-Step Solution:

1. **Identify the Reaction with the Metal Rod:**
- When the metal rod M is dipped into a colorless concentrated solution of compound N, the solution turns light blue. This indicates that the metal rod is likely reacting with the solution to form a complex or a new compound that imparts a blue color.

2. **Consider Possible Metal Candidates:**
- The options provided are `Fe`, `Cu`, `Ni`, and `Co`. Among these, copper (Cu) is known to form blue solutions, particularly when it forms copper(II) ions in solution.

3. **Formation of Light Blue Solution:**
- The light blue color suggests the formation of a copper(II) complex, likely copper(II) nitrate (Cu(NO₃)₂) when copper reacts with a nitrate solution. This confirms that M could be copper.

4. **Addition of Aqueous NaCl:**
- The question states that adding aqueous NaCl to the blue solution results in a white precipitate O. This is characteristic of silver chloride (AgCl) formation, which occurs when silver ions from the solution react with chloride ions from NaCl.

5. **Dissolving the White Precipitate with Ammonia:**
- The addition of aqueous ammonia (NH₃) dissolves the white precipitate O and gives an intense blue solution. This is indicative of the formation of a complex ion, specifically the [Cu(NH₃)₄]²⁺ complex, which is known for its intense blue color.

6. **Conclusion:**
- Based on the observations, the metal rod M must be copper (Cu), as it is the only metal among the options that fits all the criteria described in the question.

### Final Answer:
The metal rod M is **Cu** (Copper).

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