TS EAMCET · Chemistry · Hydrogen
\(\mathrm{BeH}_2\) can be prepared by the reaction of
- A \(\mathrm{BeCl}_2\) with \(\mathrm{LiAlH}_4\)
- B Be with \(\mathrm{H}_2\)
- C Be with water
- D Be with liquid ammonia
Answer & Solution
Correct Answer
(A) \(\mathrm{BeCl}_2\) with \(\mathrm{LiAlH}_4\)
Step-by-step Solution
Detailed explanation
When \(\mathrm{BeCl}_2\) reduced in presence of \(\mathrm{LiAlH}_4\) then we obtained \(\mathrm{BeH}_2\). \[ \mathrm{BeCl}_2 \xrightarrow{\stackrel{4[\mathrm{H}]}{\mathrm{LiAlH}_4} \xrightarrow{\longrightarrow}} \mathrm{BeH}_2+2 \mathrm{HCl} \]
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