A chemist opened a cupboard and found four bottles containing water solutions, each of which had lost its label.Bottles 1,2,3 contanied colourless solution, while bottle 4 contained a blue solution.The labels from the bottles were lying scattered on the floor of the cupboard.They were:
copper (II) sulphate, Hydrochloric acid
lead nitrate , Sodium carbonate
By mixing samples of the contents of the bottles, in pairs , the chemist made the following observations :
Bottle 1 +Bottle 2 `to` White precipitate is formed.
Bottle 1 +Bottle 3 `to` White precipitate is formed.
Bottle 1 +Bottle 4 `to` White precipitate is formed.
Bottle 2 +Bottle 3 `to` Colourless and odourless gas is evolved.
Bottle 2 +Bottle 4 `to` No visible reaction is observed.
Bottle 3 +Bottle 4 `to` Blue precipitate is formed.
With the help of the above observations answer the following questions.
Which one of the following bottles develops deep blue with aqueous ammonia ?

Options

A. Bottle 1
B. Bottle 2
C. Bottle 3
D. Bottle 4

Explanation

To solve the problem, we need to analyze the observations made by the chemist when mixing the solutions from the four bottles. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the contents of each bottle
- **Bottle 4** contains a blue solution. The only compound among the labels that is blue in color is **Copper (II) Sulfate (CuSO4)**. Therefore, **Bottle 4 = CuSO4**.
- The remaining bottles (1, 2, and 3) must contain the other colorless solutions: Hydrochloric acid (HCl), Lead Nitrate (Pb(NO3)2), and Sodium Carbonate (Na2CO3).

### Step 2: Analyze the observations
1. **Bottle 1 + Bottle 2 → White precipitate is formed.**
- This indicates that one of these bottles contains a solution that can react with the other to form a precipitate. Possible combinations could be Lead Nitrate (which can react with Sodium Carbonate to form Lead Carbonate, a white precipitate).

2. **Bottle 1 + Bottle 3 → White precipitate is formed.**
- Similar to the above, this suggests that one of these bottles also contains a solution that can react to form a white precipitate.

3. **Bottle 1 + Bottle 4 → White precipitate is formed.**
- Since Bottle 4 is Copper Sulfate, this indicates that Bottle 1 must contain a solution that can react with Copper Sulfate to form a precipitate. This could be Sodium Carbonate or Lead Nitrate.

4. **Bottle 2 + Bottle 3 → Colourless and odourless gas is evolved.**
- This suggests that one of these solutions is an acid (likely HCl) reacting with a carbonate (likely Na2CO3) to produce carbon dioxide gas.

5. **Bottle 2 + Bottle 4 → No visible reaction is observed.**
- Since Bottle 4 is Copper Sulfate, this indicates that Bottle 2 does not contain a reactive solution with Copper Sulfate, suggesting that Bottle 2 is likely Hydrochloric Acid (HCl).

6. **Bottle 3 + Bottle 4 → Blue precipitate is formed.**
- This indicates that Bottle 3 must contain a solution that can react with Copper Sulfate to form a precipitate. Since we already established that Bottle 4 is CuSO4, Bottle 3 must be Lead Nitrate (Pb(NO3)2), which can react with CuSO4 to form a blue precipitate of Copper (II) Hydroxide.

### Step 3: Assign the contents of the bottles
- **Bottle 1**: Sodium Carbonate (Na2CO3)
- **Bottle 2**: Hydrochloric Acid (HCl)
- **Bottle 3**: Lead Nitrate (Pb(NO3)2)
- **Bottle 4**: Copper (II) Sulfate (CuSO4)

### Step 4: Determine which bottle develops deep blue with aqueous ammonia
- When Copper (II) Sulfate (CuSO4) is treated with aqueous ammonia (NH4OH), it forms a deep blue complex known as tetraamine copper(II) sulfate, represented as Cu(NH3)4SO4.

Thus, the answer to the question is:
**Bottle 4 (Copper (II) Sulfate) develops deep blue with aqueous ammonia.**