Reaction of $BeCl_{2}$ with $LiAlH_{4}$ gives.
$(A)$ $AlCl_{3}$; $(B)$ $BeH_{2}$; $(C)$ $LiH$; $(D)$ $LiCl$; $(E)$ $BeAlH_{4}$
Choose the correct answer from options given below.

• A
  $(A)$,$(D)$ and $(E)$
• B
  $(A)$,$(B)$ and $(D)$
• C
  $(D)$ and $(E)$
• D
  $(B)$,$(C)$ and $(D)$