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  • Reaction of  with <img alt="\mathrm{ \mathrm{LiAlH}_{4}}" src="/latex-image/?%5Cmath

Reaction of \mathrm{\mathrm{BeCl}_{2}} with \mathrm{ \mathrm{LiAlH}_{4}} gives:

\mathrm{\text { (A) } \mathrm{AlCl}_{3}}

\mathrm{\text { (B) } \mathrm{BeH}_{2}}

\mathrm{\text { (C) } \mathrm{LiH}}

\mathrm{\text { (D) } \mathrm{LiCl}}

\mathrm{\text { (E) } \mathrm{BeAlH}_{4}}
Choose the correct answer from options given below :

Option: 1

\mathrm{(A),(D) \text { and }(E)}


Option: 2

\mathrm{(\mathrm{A}),(\mathrm{B}) \text { and }(\mathrm{D})}


Option: 3

\mathrm{\text { (D) and (E) }}


Option: 4

\mathrm{(B),(C) \text { and }(D)}


Answers (1)

best_answer

Reaction of \mathrm{BeCl_{2}} with \mathrm{LiAlH_{4}} is a method for formation of \mathrm{BeH_{2}} giving \mathrm{LiCl} and \mathrm{AlCl_{3}} as biproducts.

\mathrm{\begin{aligned} \\ &2 \mathrm{BeCl}_2+ \mathrm{LlAlH}_4 \longrightarrow 2 \mathrm{BeH}_2+ \mathrm{LiCl}+\mathrm{AlCl}_3 \end{aligned}}

Hence, the correct answer is Option (2).

Posted by

Irshad Anwar

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