Reaction of $\mathrm{BeCl}_{2}$ with $\mathrm{LiAlH}_{4}$ gives :
(A) $\mathrm{AlCl}_{3}$
(B) $\mathrm{BeH}_{2}$
(C) $\mathrm{LiH}$
(D) $\mathrm{LiCl}$
(E) $\mathrm{BeAlH}_{4}$
Choose the correct answer from options given below :
Solution
$$2 \mathrm{BeCl}_{2}+\mathrm{LiAlH}_{4} \rightarrow 2 \mathrm{BeH}_{2}+\mathrm{LiCl}+\mathrm{AlCl}_{3}$$
About this question
Subject: Chemistry · Chapter: s-Block Elements · Topic: Alkali Metals
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