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 `(E)` is.

Options

A. Inorganic benzene
B. Borazine
C. Borazon
D. Diborane

Explanation

To solve the given problem step by step, we will analyze the reactions involving boron and its compounds.

### Step 1: Reaction of Boron with Oxygen
When boron (B) reacts with oxygen (O2) at a temperature of 700°C, it forms boron trioxide (B2O3).

**Compound (A)**: B2O3

### Step 2: Reaction of B2O3 with Carbon and Chlorine
Next, boron trioxide (B2O3) reacts with carbon (C) and dry chlorine (Cl2). This reaction produces boron trichloride (BCl3) and carbon monoxide (CO).

**Compound (B)**: BCl3

### Step 3: Reduction of BCl3 with Lithium Aluminum Hydride
The compound BCl3 can be reduced using lithium aluminum hydride (LiAlH4). This reduction yields diborane (B2H6), along with lithium chloride (LiCl) and aluminum chloride (AlCl3).

**Compound (C)**: B2H6

### Step 4: Reaction of B2H6 with Ammonia
When diborane (B2H6) reacts with ammonia (NH3), it forms a compound known as borazane (B2H6·2NH3).

**Compound (D)**: B2H6·2NH3

### Step 5: Heating of Borazane
Upon heating borazane (B2H6·2NH3), it decomposes to form inorganic benzene, which is also known as borazole (B3N3H6).

**Compound (E)**: B3N3H6 (Inorganic Benzene)

### Step 6: Reaction of B2H6 with Sodium Hydride
Finally, when diborane (B2H6) reacts with sodium hydride (NaH), it produces sodium borohydride (NaBH4).

**Compound (F)**: NaBH4

### Summary of Compounds:
- **(A)**: B2O3
- **(B)**: BCl3
- **(C)**: B2H6
- **(D)**: B2H6·2NH3
- **(E)**: B3N3H6 (Inorganic Benzene)
- **(F)**: NaBH4