Boron reacts with oxygen at `700^@C` to give `(A)`. Compound `(A)` reacts with carbon and dry chloride to give `(B)` and carbon monoxide. `(B)` on reduction with `LiAlH_(4)` gives `( C)` along with `LiCl` and `AlCl_(3). ( C)` on reaction with ammonia gives `(D)`. Which on heating gives `(E).(C)` on reaction with `NaH` gives `(F)`.
Compound `(F)` behaves as.

Options

A. Reducing agent
B. Oxidising agent
C. Complexon
D. Buffer

Explanation

To solve the problem step by step, we will outline the reactions and identify the compounds involved:

### Step 1: Reaction of Boron with Oxygen
Boron reacts with oxygen at `700°C` to form:
\[ \text{B} + \text{O}_2 \rightarrow \text{B}_2\text{O}_3 \]
**Compound (A)** is **Boron Trioxide (B₂O₃)**.

### Step 2: Reaction of Compound (A) with Carbon and Dry Chloride
The next reaction involves boron trioxide reacting with carbon and dry chlorine:
\[ \text{B}_2\text{O}_3 + \text{C} + \text{Cl}_2 \rightarrow \text{BCl}_3 + \text{CO} \]
**Compound (B)** is **Boron Trichloride (BCl₃)**.

### Step 3: Reduction of Compound (B) with Lithium Aluminium Hydride
Now, we reduce boron trichloride with lithium aluminium hydride:
\[ \text{BCl}_3 + \text{LiAlH}_4 \rightarrow \text{B}_2\text{H}_6 + 3\text{LiCl} + \text{AlCl}_3 \]
**Compound (C)** is **Diborane (B₂H₆)**.

### Step 4: Reaction of Compound (C) with Ammonia
Diborane reacts with ammonia to form:
\[ \text{B}_2\text{H}_6 + 3\text{NH}_3 \rightarrow \text{BH}_2\text{NH}_3 \text{ (twice)} + \text{BH}_4^- \]
**Compound (D)** is **Ammonium Borohydride (BH₂NH₃)**.

### Step 5: Heating of Compound (D)
When compound (D) is heated, it decomposes to form:
\[ \text{BH}_2\text{NH}_3 \rightarrow \text{B}_3\text{N}_3\text{H}_6 + \text{H}_2 \]
**Compound (E)** is **Borazine (B₃N₃H₆)**.

### Step 6: Reaction of Compound (C) with Sodium Hydride
Diborane reacts with sodium hydride to produce:
\[ \text{B}_2\text{H}_6 + \text{NaH} \rightarrow \text{NaBH}_4 \]
**Compound (F)** is **Sodium Borohydride (NaBH₄)**.

### Step 7: Behavior of Compound (F)
Sodium borohydride (NaBH₄) behaves as a:
- **Reducing agent** in various organic reactions.

### Final Answer
The correct option is that compound (F) behaves as a **Reducing agent**.

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