JEE Mains · Chemistry · STD 11 -p-Block elements - I
\(BeCl _{2}\) reacts with \(LiAlH _{4}\) to give ....
- A \(Be + Li \left[ AlCl _{4}\right]+ H _{2}\)
- B \(Be + AlH _{3}+ LiCl + HCl\)
- C \(BeH _{2}+ LiCl + AlCl _{3}\)
- D \(BeH _{2}+ Li \left[ AlCl _{4}\right]\)
Answer & Solution
Correct Answer
(C) \(BeH _{2}+ LiCl + AlCl _{3}\)
Step-by-step Solution
Detailed explanation
\(2 BeCl _{2}+ LiAlH _{4} \rightarrow 2 BeH _{2}+ LiCl + AlCl _{3}\) This is the method to prepare \(BeH _{2}\)
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