MCQ
$BeCl _{2}$ reacts with $LiAlH _{4}$ to give ....
- A$Be + Li \left[ AlCl _{4}\right]+ H _{2}$
- B$Be + AlH _{3}+ LiCl + HCl$
- ✓$BeH _{2}+ LiCl + AlCl _{3}$
- D$BeH _{2}+ Li \left[ AlCl _{4}\right]$
✓
Answer
Correct option: C.
$BeH _{2}+ LiCl + AlCl _{3}$
c
$2 BeCl _{2}+ LiAlH _{4} \rightarrow 2 BeH _{2}+ LiCl + AlCl _{3}$
$2 BeCl _{2}+ LiAlH _{4} \rightarrow 2 BeH _{2}+ LiCl + AlCl _{3}$
This is the method to prepare $BeH _{2}$
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