JEE Main 2022 · 26 July · morning · Single correct · +4 / −1
Reaction of BeCl₂ with LiAlH₄ gives :
(A) AlCl₃
(B) BeH₂
(C) LiH
(D) LiCl
(E) BeAlH₄
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)
Answer and solution
Answer (b) (A), (B) and (D)
Solution
Chemistry ma’am · 3 steps
LiAlH₄ hands its hydrides to beryllium and takes the chlorides, so balance the swap.
- What LiAlH4 does. LiAlH₄ is a hydride donor. With a metal chloride it swaps ions: hydride (H⁻) goes to the metal, and chloride goes to Li and Al. This is the standard way to make BeH₂, which cannot be made easily from Be and H₂ directly.
2BeCl₂ + LiAlH₄ → 2BeH₂ + LiCl + AlCl₃
- Check the balance. Four H atoms come from LiAlH₄ and end up in two BeH₂. Four Cl atoms come from two BeCl₂: one goes to LiCl and three go to AlCl₃.
- So. The products are AlCl₃ (A), BeH₂ (B) and LiCl (D), which is option (b).
The idea to keep. When LiAlH₄ meets a metal chloride, the metal becomes a hydride and the chlorides end up as LiCl and AlCl₃. Count H and Cl to confirm.
Answer from the NTA final answer key. Solution by PrepWiser.