`BCl_(3)+H_(2)+Coverset(1700-1800^(@)C)to`Product(s)
The product(s) of above reaction is/are

Options

A. `C Cl_(4)+B_(2)H_(6)`
B. `B_(4)C+HCl`
C. `B+Cl_(2)+CH_(4)`
D. `B_(2)H_(6)+CHCl_(3)`

Explanation

To solve the question regarding the reaction of BCl₃ (boron trichloride), H₂ (hydrogen), and carbon at a high temperature (1700-1800 °C), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in the reaction are BCl₃ (boron trichloride), H₂ (hydrogen), and carbon.

2. **Understand the Reaction Conditions**: The reaction occurs at a high temperature of 1700-1800 °C. High temperatures can facilitate the formation of complex compounds.

3. **Recognize the Boron Family Chemistry**: Boron (B) is a member of the p-block elements and has unique properties. When reacting with carbon and hydrogen, boron can form boron carbide (B₄C), which is known for its hardness and is used in bulletproof materials.

4. **Write the Balanced Reaction**:
- The reaction can be represented as:
\[ 4 \text{BCl}_3 + 6 \text{H}_2 + 2 \text{C} \rightarrow \text{B}_4\text{C} + 12 \text{HCl} \]
- Here, B₄C (boron carbide) and HCl (hydrochloric acid) are produced.

5. **Identify the Products**: The products of the reaction are:
- B₄C (boron carbide)
- HCl (hydrochloric acid)

6. **Conclusion**: Therefore, the products formed from the reaction of BCl₃, H₂, and carbon at 1700-1800 °C are boron carbide (B₄C) and hydrochloric acid (HCl).

### Final Answer:
The products of the reaction are B₄C (boron carbide) and HCl (hydrochloric acid).

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