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Diborane on heating to 200^@C with ammon...

Diborane on heating to `200^@C` with ammonia it gives

A

`B_2 H_6. 2NH_3`

B

`[BH_2(NH_3)_2]^+ [BH_4]^-`

C

`BN`

D

`B_3 N_3 H_6`

Text Solution

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The correct Answer is:
To solve the question regarding the reaction of diborane (B2H6) with ammonia (NH3) when heated to 200°C, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Reactants**: The reactants in this reaction are diborane (B2H6) and ammonia (NH3). 2. **Understand the Reaction Conditions**: The reaction occurs at a temperature of 200°C. This temperature is significant as it influences the products formed. 3. **Write the Balanced Reaction**: When diborane reacts with ammonia at elevated temperatures, the reaction can be represented as: \[ 3 \text{B}_2\text{H}_6 + 6 \text{NH}_3 \rightarrow 2 \text{B}_3\text{N}_3\text{H}_6 + 12 \text{H}_2 \] This indicates that three moles of diborane react with six moles of ammonia to produce two moles of borazine (B3N3H6) and hydrogen gas (H2). 4. **Identify the Product**: The main product of this reaction at 200°C is borazine (B3N3H6), which is also known as borazole. 5. **Conclusion**: Therefore, when diborane is heated with ammonia at 200°C, the products formed are borazine (B3N3H6) and hydrogen gas (H2). ### Final Answer: Diborane (B2H6) on heating with ammonia (NH3) at 200°C gives borazine (B3N3H6) and hydrogen gas (H2). ---
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