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.
The compound (C ) :

Options

A. has `sp^(3)` hydbridisation of central atom(s)
B. has sixteen sigma bonds.
C. is used as a dehydrating agent.
D. all of these.

Answer

To solve the problem step by step, we will identify the compounds A, B, and C based on the information provided in the question.

### Step 1: Identify Compound A
The question states that an inorganic iodide (A) reacts with KOH to produce a gas (B) and a solution of a compound. A common inorganic iodide that fits this description is phosphonium iodide (PH₄I).

**Reaction:**
\[ \text{PH}_4\text{I} + \text{KOH} \rightarrow \text{PH}_3 + \text{H}_2\text{O} + \text{KI} \]

### Step 2: Identify Compound B
From the reaction above, we see that the gas (B) produced is phosphine (PH₃).

### Step 3: Identify Compound C
The gas (B), which is phosphine (PH₃), is ignited in air. When phosphine burns in the presence of oxygen, it forms phosphorus pentoxide (P₂O₅) and water.

**Reaction:**
\[ 4 \text{PH}_3 + 6 \text{O}_2 \rightarrow 2 \text{P}_2\text{O}_5 + 6 \text{H}_2\text{O} \]

Thus, the compound (C) formed is phosphorus pentoxide (P₂O₅).

### Step 4: Reduction of Copper Sulfate
The question also states that copper sulfate (CuSO₄) is reduced to copper metal when phosphine (B) is passed through its solution. This is consistent with the reducing properties of phosphine.

**Reaction:**
\[ 3 \text{CuSO}_4 + 2 \text{PH}_3 \rightarrow \text{Cu}_3\text{P}_2 + 3 \text{H}_2\text{SO}_4 \]

### Conclusion
Based on the above steps, we conclude that:
- Compound A is phosphonium iodide (PH₄I).
- Compound B is phosphine (PH₃).
- Compound C is phosphorus pentoxide (P₂O₅).

### Final Answer
The compound (C) is **P₂O₅** (phosphorus pentoxide).