Diborane reacts with HCl in the presence of `AICl_3` and liberates

Options

A. `H_2`
B. `Cl_2`
C. `BCl_3`
D. `Cl_2 & BCl_3`

Answer

To solve the question regarding the reaction of diborane (B2H6) with hydrochloric acid (HCl) in the presence of aluminum chloride (AlCl3), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Diborane (B2H6)
- Hydrochloric acid (HCl)
- Aluminum chloride (AlCl3)

### Step 2: Understand the Role of AlCl3
Aluminum chloride (AlCl3) acts as a Lewis acid in this reaction. It helps in activating the HCl and facilitates the reaction with diborane.

### Step 3: Write the Reaction
When diborane reacts with hydrochloric acid in the presence of aluminum chloride, the reaction can be represented as:
\[ \text{B}_2\text{H}_6 + \text{HCl} \xrightarrow{\text{AlCl}_3} \text{B}_2\text{H}_5\text{Cl} + \text{H}_2 \]

### Step 4: Identify the Products
From the reaction, we can see that the products formed are:
- B2H5Cl (Dichlorodiborane)
- H2 (Hydrogen gas)

### Step 5: Conclusion
The question asks what is liberated during the reaction. From the products identified, we see that hydrogen gas (H2) is liberated.

### Final Answer
The correct answer is that diborane reacts with HCl in the presence of AlCl3 and liberates **H2 gas**.

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