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$\mathrm{BeH}_2$ can be prepared by the reaction of
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Verified Answer
The correct answer is:
$\mathrm{BeCl}_2$ with $\mathrm{LiAlH}_4$
When $\mathrm{BeCl}_2$ reduced in presence of $\mathrm{LiAlH}_4$ then we obtained $\mathrm{BeH}_2$.
$$
\mathrm{BeCl}_2 \xrightarrow{\stackrel{4[\mathrm{H}]}{\mathrm{LiAlH}_4} \xrightarrow{\longrightarrow}} \mathrm{BeH}_2+2 \mathrm{HCl}
$$
$$
\mathrm{BeCl}_2 \xrightarrow{\stackrel{4[\mathrm{H}]}{\mathrm{LiAlH}_4} \xrightarrow{\longrightarrow}} \mathrm{BeH}_2+2 \mathrm{HCl}
$$
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