Water soluble salt (x) when heated decomposes into three products A, B and C. Here B and C are two different paramagnetic gases while A is yellow in hot condition Here the salt (x) is

Options

A. `Hg(NO_3)_2`
B. `FeC_2O_4`
C. `ZnSO_4`
D. `Pb(NO_3)_2`

Answer

To solve the question step by step, we need to identify the water-soluble salt (x) that decomposes into three products: A, B, and C, where A is yellow when hot, and B and C are two different paramagnetic gases.

### Step 1: Identify the Products
The problem states that upon heating the salt (x), we get three products: A, B, and C. We know that:
- A is yellow in hot condition.
- B and C are two different paramagnetic gases.

### Step 2: Consider Possible Salts
We need to consider various water-soluble salts and their decomposition products upon heating.

1. **Lead Nitrate (Pb(NO3)2)**:
- When heated, it decomposes into:
\[ 2PbO + 4NO2 + O2 \]
- Here, PbO is yellow when hot, and NO2 and O2 are the gases produced.

### Step 3: Analyze the Color of Product A
From the decomposition of lead nitrate:
- **Product A (PbO)**: This is yellow when hot, which matches the condition given in the question.

### Step 4: Analyze the Paramagnetism of Products B and C
- **Product B (NO2)**:
- The oxidation state of nitrogen in NO2 is +4.
- The electronic configuration of nitrogen in this state leads to unpaired electrons, making NO2 paramagnetic.

- **Product C (O2)**:
- O2 is known to be paramagnetic due to the presence of unpaired electrons in its molecular orbital configuration.

### Step 5: Conclusion
Since all conditions are satisfied:
- A is yellow (PbO).
- B and C are paramagnetic gases (NO2 and O2).

Thus, the water-soluble salt (x) that decomposes to give these products is **Lead Nitrate (Pb(NO3)2)**.

### Final Answer:
The water-soluble salt (x) is **Lead Nitrate (Pb(NO3)2)**.

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