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$BeH _{2}+ Li \left[ AlCl _{4}\right]$
This is the method to prepare $BeH _{2}$
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$\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{SOCl}_{2} \longrightarrow \mathrm{A} \xrightarrow[AlCl_3]{Benzene} \mathrm{B} \xrightarrow[-OH]{KCN} \mathrm{C}$
[Assume no volume change on adding $\mathrm{NH}_{3}$ ]