Reaction of BeCl2\mathrm{BeCl_2}BeCl2 with LiAlH4\mathrm{LiAlH_4}LiAlH4 gives :
(A) AlCl3\mathrm{AlCl_3}AlCl3
(B) BeH2\mathrm{BeH_2}BeH2
(C) LiH
(D) LiCl
(E) BeAlH4\mathrm{BeAlH_4}BeAlH4
Choose the correct answer from options given below :
Official answer: B
2BeCl2+LiAlH4→2BeH2+LiCl+AlCl3\mathrm{2BeCl_2+LiAlH_4\rightarrow2BeH_2+LiCl+AlCl_3}2BeCl2+LiAlH4→2BeH2+LiCl+AlCl3.
(B)