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

We need to find the products of the reaction between BeCl₂ and LiAlH₄.

We start by writing the balanced reaction since LiAlH₄ is a strong reducing agent that acts as a source of hydride ions (H⁻). The reaction is:

$$2BeCl_2 + LiAlH_4 \rightarrow 2BeH_2 + LiCl + AlCl_3$$

Next, the products formed are (A) AlCl₃, (B) BeH₂, and (D) LiCl. Aluminium chloride is formed when Al³⁺ combines with Cl⁻ ions; beryllium hydride is formed when Be²⁺ receives hydride ions from LiAlH₄; and lithium chloride is formed as a by-product.

In this reaction, the hydride (H⁻) from LiAlH₄ replaces the chloride in BeCl₂, forming BeH₂. The displaced chloride ions combine with Li⁺ and Al³⁺ to form LiCl and AlCl₃. Note that BeAlH₄ (Option E) is not formed in this reaction, and neither is LiH (Option C).

Therefore, the correct answer is Option B.