Lithium aluminium hydride can be prepared from the reaction of

Question

Consider the following statements:
Statement-I: H$_2$Se is more acidic than H$_2$Te.
Statement-II: H$_2$Se has higher bond dissociation enthalpy than H$_2$Te.
In light of the above statements, choose the correct option:

Answer 1

Lithium aluminium hydride ($LiAlH_4$) is an important compound used as a reducing agent in organic chemistry. It is prepared through a chemical reaction between lithium hydride and aluminium chloride. The correct reaction is as follows:

The reaction between lithium hydride ($LiH$) and aluminium chloride ($Al_2Cl_6$) can be written as:

$6LiH + Al_2Cl_6 \rightarrow 2LiAlH_4 + 4LiCl$

This reaction shows lithium hydride reacting with aluminium chloride to form lithium aluminium hydride and lithium chloride as a byproduct.

Explanation of Options:

$LiH$ and $Al(OH)_3$: This combination does not result in the formation of lithium aluminium hydride as $Al(OH)_3$ is not reactive enough to interact with $LiH$ to form $LiAlH_4$.
$LiCl$ and $Al_2H_6$: This pair is incorrect because $Al_2H_6$ (dimer of dihydroaluminium chloride) is not used in such reactions to produce $LiAlH_4$.
$LiCl$, $Al$, and $H_2$: This combination does not lead directly to the formation of $LiAlH_4$ and is not a recognized method of preparation for the compound.
$LiH$ and $Al_2Cl_6$: Correct option. The reaction between lithium hydride and aluminium chloride forms lithium aluminium hydride, as shown above.

Therefore, the correct answer is the combination of $LiH$ and $Al_2Cl_6$ which forms lithium aluminium hydride.

Lithium aluminium hydride can be prepared from the reaction of

The Correct Option is D

Solution and Explanation

Explanation of Options:

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