An inorganic iodide (A) on heating with a solution of KOH gives a gas (B) and a solution of a compound. The gas (B) on ignition in air gives a compound (C ) and water. Copper sulphate is finally reduced to the metal on passing (B) through its solution.
Select the correct statement from the following for the gas (B).

Options

A. Its solution in water does not decompose in presence of light.
B. It can be prepared by the alkaline hydrolysis of white phosphorus.
C. It is non- inflammable owing to the presence of `P_(2)H_(4)`.
D. It can act as oxidising agent.

Answer

To solve the problem, we need to analyze the reactions and identify the gas (B) produced, as well as the properties of this gas based on the given statements. Let's break down the solution step by step.

### Step 1: Identify the Inorganic Iodide (A)
We start with an inorganic iodide (A) that reacts with KOH. A suitable candidate for this iodide is phosphonium iodide (PH4I).

### Step 2: Reaction of A with KOH
When phosphonium iodide (PH4I) is heated with KOH, the following reaction occurs:
\[ \text{PH}_4\text{I} + \text{KOH} \rightarrow \text{PH}_3 + \text{KI} + \text{H}_2\text{O} \]
From this reaction, we can see that the gas (B) produced is phosphine (PH3).

### Step 3: Reaction of Gas (B) on Ignition
Next, we need to examine what happens when phosphine (B) is ignited in air. The reaction is:
\[ 4 \text{PH}_3 + 4 \text{O}_2 \rightarrow 2 \text{P}_2\text{O}_5 + 6 \text{H}_2\text{O} \]
This shows that the compound (C) formed is diphosphorus pentoxide (P2O5).

### Step 4: Reduction of Copper Sulfate
The problem states that when phosphine (B) is passed through a solution of copper sulfate (CuSO4), it reduces copper sulfate to copper metal. The reaction can be summarized as:
\[ 3 \text{CuSO}_4 + 2 \text{PH}_3 \rightarrow \text{Cu}_3\text{P}_2 + 3 \text{H}_2\text{SO}_4 \]
This confirms that phosphine acts as a reducing agent.

### Step 5: Analyze the Statements
Now we will evaluate the statements provided in the question regarding gas (B):

1. **Statement 1**: "Its solution in water does not decompose in the presence of light."
- **Analysis**: This statement is false. Phosphine decomposes in the presence of light to form red phosphorus and hydrogen gas.

2. **Statement 2**: "It can be prepared by the alkaline hydrolysis of white phosphorus."
- **Analysis**: This statement is true. Phosphine can indeed be prepared from white phosphorus (P4) in the presence of an alkaline solution.

3. **Statement 3**: "It is non-inflammable owing to the presence of P2H4."
- **Analysis**: This statement is false. Pure phosphine is non-inflammable, but if it contains impurities like P2H4, it can become highly inflammable.

4. **Statement 4**: "It can act as an oxidizing agent."
- **Analysis**: This statement is false. Phosphine acts as a reducing agent, as it can be oxidized to phosphine oxides.

### Conclusion
Based on the analysis, the only correct statement regarding gas (B) (phosphine) is Statement 2.

### Final Answer
The correct statement for the gas (B) is: **It can be prepared by the alkaline hydrolysis of white phosphorus.**