Question

The products formed in the reaction of \( BeCl_2 \) with \( LiAlH_4 \) are

• A. Be, \( Li[AlCl_4] \), \( H_2 \)
• B. \( BeH_2 \), \( LiCl \), \( AlCl_3 \)
• C. Be, \( AlH_3 \), \( LiCl \), \( HCl \)
• D. \( BeH_2 \), \( Li[AlCl_4] \)

Hint:

- Lithium aluminium hydride (\( LiAlH_4 \)) is a strong reducing agent used to reduce metal halides.
- Beryllium chloride (\( BeCl_2 \)) reacts with \( LiAlH_4 \) to form beryllium hydride (\( BeH_2 \)).
- The side products include lithium chloride (\( LiCl \)) and aluminium chloride (\( AlCl_3 \)).

Answer 1

The Correct Option is B

Step 1: Understanding the Reaction Between \( BeCl_2 \) and \( LiAlH_4 \) 
- The reaction between beryllium chloride (\( BeCl_2 \)) and lithium aluminium hydride (\( LiAlH_4 \)) results in the reduction of \( BeCl_2 \) to \( BeH_2 \). 
- The complete reaction can be written as: \[ BeCl_2 + 2 LiAlH_4 \rightarrow BeH_2 + 2 LiCl + AlCl_3 \]

 Step 2: Identifying the Products 
- The major products are: 
- \( BeH_2 \) (beryllium hydride), 
- \( LiCl \) (lithium chloride), 
- \( AlCl_3 \) (aluminium chloride). 

Step 3: Verifying the Correct Option 
- The correct option should contain \( BeH_2 \), \( LiCl \), and \( AlCl_3 \). 
- This matches option (2). Thus, the correct answer is \( (2) BeH_2, LiCl, AlCl_3 \).