Question 35

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

Solution

Lithium aluminum hydride ($$LiAlH_4$$) acts as a strong reducing agent and hydride donor. When it reacts with beryllium chloride ($$BeCl_2$$), a metathesis (double displacement) reaction takes place, yielding beryllium hydride, lithium chloride, and aluminum chloride.

The balanced chemical equation for the reaction is:

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

$$BeH_2$$ is formed as a polymeric hydride.
$$LiCl$$ and $$AlCl_3$$ are formed as byproducts.

The correct option is C: $$BeH_2 + LiCl + AlCl_3$$

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