A' on heating at `700^(@)C` in air gives a white infusible amorphous powder (B) which is decomposed when heated in the currect of steam to give white powder 'C' and a gas 'D'.'D' turns red litmus blue and in aqueous solution , gives reddish brown ppt with `K_(2)HgI_(4)`. compound 'C on strong heating gives 'E.
'A' is

Options

A. B
B. Si
C. `P_(4)`
D. `N_(2)`

Explanation

To solve the problem step by step, we need to analyze the information provided and deduce the compounds involved in the reactions.

### Step 1: Identify Compound A
The question states that compound A, when heated at 700°C in air, produces a white infusible amorphous powder (B). Given that the process involves heating in air, we can deduce that compound A is likely a metal or metalloid that reacts with oxygen to form an oxide.

**Hint:** Think about elements that can form oxides and are stable at high temperatures.

### Step 2: Determine Compound B
The white infusible amorphous powder (B) formed from A is likely a metal oxide. Based on the context, we can consider that A is boron (B). When boron is heated in air, it forms boron oxide (B2O3), which is a white infusible powder.

**Hint:** Consider the properties of boron and its oxides.

### Step 3: Analyze the Reaction with Steam
When compound B (B2O3) is heated in the current of steam, it decomposes to form a white powder (C) and a gas (D). The reaction can be represented as:
\[ \text{B2O3} + 3 \text{H2O} \rightarrow 2 \text{H3BO3} + \text{NH3} \]
Here, H3BO3 (boric acid) is the white powder (C), and ammonia (NH3) is the gas (D).

**Hint:** Remember that steam (water vapor) can react with certain oxides to form acids.

### Step 4: Identify Gas D
The gas D produced is ammonia (NH3), which is known to turn red litmus blue, indicating its basic nature.

**Hint:** Think about the properties of ammonia and its behavior in aqueous solutions.

### Step 5: Analyze the Reaction of Compound C
When compound C (H3BO3) is heated strongly, it decomposes to form compound E. The reaction can be represented as:
\[ 2 \text{H3BO3} \rightarrow \text{B2O3} + 3 \text{H2O} \]
Compound E is thus B2O3 (boron trioxide).

**Hint:** Consider the thermal stability of boric acid and what it decomposes into upon heating.

### Final Answer
Based on the deductions made, we conclude that:
- A is **Boron (B)**.
- B is **Boron oxide (B2O3)**.
- C is **Boric acid (H3BO3)**.
- D is **Ammonia (NH3)**.
- E is **Boron trioxide (B2O3)**.

### Summary
- **A**: Boron (B)
- **B**: Boron oxide (B2O3)
- **C**: Boric acid (H3BO3)
- **D**: Ammonia (NH3)
- **E**: Boron trioxide (B2O3)