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)`.
In compound `(B)` :

Options

A. Boron is `sp^(2)` hybridised
B. `B` is triangular planar molecule
C. It is a Lewis base
D. Dimer

Explanation

To solve the problem step by step, we will identify the compounds formed at each stage of the reactions involving boron and its derivatives.

### Step 1: Reaction of Boron with Oxygen
- **Reaction**: Boron (B) reacts with oxygen (O2) at 700°C.
- **Product (A)**: The product formed is Boron Trioxide (B2O3).

### Step 2: Reaction of Compound A with Carbon and Dry Chloride
- **Reaction**: B2O3 reacts with carbon (C) and dry chloride (Cl).
- **Product (B)**: The product formed is Boron Trichloride (BCl3) along with carbon monoxide (CO).

### Step 3: Reduction of Compound B
- **Reaction**: BCl3 is reduced with Lithium Aluminum Hydride (LiAlH4).
- **Product (C)**: The product formed is Diborane (B2H6), along with Lithium Chloride (LiCl) and Aluminum Chloride (AlCl3).

### Step 4: Reaction of Compound C with Ammonia
- **Reaction**: B2H6 reacts with ammonia (NH3).
- **Product (D)**: The product formed is a complex of boron and nitrogen, specifically Boron Amide (BH2NH3) and Borohydride (BH4^-).

### Step 5: Heating of Compound D
- **Reaction**: Heating of the product D.
- **Product (E)**: The product formed is Borazine (B3N3H6), which is often referred to as inorganic benzene.

### Step 6: Reaction of Compound C with Sodium Hydride
- **Reaction**: B2H6 reacts with Sodium Hydride (NaH).
- **Product (F)**: The product formed is Sodium Borohydride (NaBH4).

### Summary of Compounds:
- **(A)**: B2O3 (Boron Trioxide)
- **(B)**: BCl3 (Boron Trichloride)
- **(C)**: B2H6 (Diborane)
- **(D)**: BH2NH3 (Boron Amide)
- **(E)**: B3N3H6 (Borazine)
- **(F)**: NaBH4 (Sodium Borohydride)

### Final Question: In Compound B
- **Analysis of Compound B (BCl3)**:
- Boron in BCl3 is **sp² hybridized**.
- The molecular geometry is **trigonal planar**.
- BCl3 does not behave as a Lewis base; instead, it acts as a Lewis acid due to the incomplete octet of boron.
- BCl3 does not form dimers under normal conditions.