Reaction of BeO with ammonia and hydrogen fluoride gives A which on thermal decomposition gives $\mathrm{BeF_2}$ and $\mathrm{NH_4F}$. What is 'A'?
Solution
$$
\mathrm{BeO}+\mathrm{NH}_{3}+\mathrm{HF} \longrightarrow\left(\mathrm{NH}_{4}\right)_{2} \underset{(\mathrm{A})}{\mathrm{BeF}_{4}} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}+\mathrm{NH}_{4} \mathrm{~F}
$$
<br/><br/>
Compound $\mathrm{A}$ is $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{BeF}_{4}$
About this question
Subject: Chemistry · Chapter: s-Block Elements · Topic: Alkali Metals
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