JEE Main 2022 · 27 June · morning · Single correct · +4 / −1
BeCl₂ reacts with LiAlH₄ to give :
- (a) Be + Li[AlCl₄] + H₂
- (b) Be + AlH₃ + LiCl + HCl
- (c) BeH₂ + LiCl + AlCl₃
- (d) BeH₂ + Li[AlCl₄]
Answer and solution
Answer (c) BeH₂ + LiCl + AlCl₃
Solution
Chemistry ma’am · 4 steps
LiAlH₄ swaps hydride for chloride on beryllium; nothing is reduced to the metal.
- What LiAlH4 does. LiAlH₄ is a store of hydride ions, H⁻. With a metal halide it carries out an exchange: the hydride goes to the metal and the chloride goes to Li and Al.
- Apply it to BeCl2. Beryllium stays in the +2 state and trades its two Cl⁻ for two H⁻, giving beryllium hydride, BeH₂. This is the standard laboratory route to BeH₂.
2BeCl₂ + LiAlH₄ → 2BeH₂ + LiCl + AlCl₃
- Check the balance. Be: 2 on each side. Cl: 4 on the left, 1 + 3 = 4 on the right. H: 4 from LiAlH₄, 4 in 2BeH₂. Everything matches.
- So. The products are BeH₂, LiCl and AlCl₃, which is option (c).
The idea to keep. When LiAlH₄ meets a metal chloride, think exchange first: H⁻ moves onto the metal, Cl⁻ moves onto Li and Al. That is how BeH₂ is made.
Answer from the NTA final answer key. Solution by PrepWiser.