$BeCl_2$ reacts with $LiAlH_4$ to give ....

  • A
    $Be + Li[AlCl_4] + H_2$
  • B
    $Be + AlH_3 + LiCl + HCl$
  • C
    $BeH_2 + LiCl + AlCl_3$
  • D
    $BeH_2 + Li[AlCl_4]$

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