JEE Main 2022 · 27 Jun · Shift 1 · Chemistry Q68

ChemistrySingle correct+4 / -1Question 68 of 90

$\mathrm{BeCl_2}$ reacts with $\mathrm{LiAlH_4}$ to give :

\mathrm{Be+Li[AlCl_4]+H_2}

\mathrm{Be+AlH_3+LiCl+HCl}

\mathrm{BeH_2+LiCl+AlCl_3}

\mathrm{BeH_2+Li[AlCl_4]}

Answer and worked solution

Official answer: C

Hydride transfer. $\mathrm{2BeCl_2+LiAlH_4\rightarrow2BeH_2+LiCl+AlCl_3}$ .

(C)

Topics:Important compounds of s-block Hydrides

JEE Main 2022 · 27 Jun · Shift 1, Chemistry question 68