Q15 / 160 · TS EAMCET 2018 (05 May Shift 1) · Chemistry

$\mathrm{BeH}_2$ can be prepared by the reaction of

  • Correct answer
    A
    $\mathrm{BeCl}_2$ with $\mathrm{LiAlH}_4$
  • B
    Be with $\mathrm{H}_2$
  • C
    Be with water
  • D
    Be with liquid ammonia

Solution

When $\mathrm{BeCl}_2$ reduced in presence of $\mathrm{LiAlH}_4$ then we obtained $\mathrm{BeH}_2$. $$ \mathrm{BeCl}_2 \xrightarrow{\stackrel{4[\mathrm{H}]}{\mathrm{LiAlH}_4} \xrightarrow{\longrightarrow}} \mathrm{BeH}_2+2 \mathrm{HCl} $$

Hence, the correct answer is option: 'A'.

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