Lithium aluminium hydride can be prepared from the reaction of:
\(LiCl \ and \ Al_2H_6\)
\(LiH \ and \ Al_2Cl_6\)
\(LiCl, Al \ and \ H_2\)
\(LiH \ and \ Al(OH)_3\)
Step 1: Reaction for Preparation of \(\text{LiAlH}_4\)
Lithium aluminium hydride \(\text{LiAlH}_4\) is prepared by the reaction of lithium hydride (\(\text{LiH}\)) with aluminium chloride \(\text{Al}_2\text{Cl}_6\). The reaction is as follows:
\[8\text{LiH} + \text{Al}_2\text{Cl}_6 \rightarrow 2\text{LiAlH}_4 + 6\text{LiCl}.\]
Conclusion: The correct reactants for preparing \(\text{LiAlH}_4\) are \(\text{LiH}\) and \(\text{Al}_2\text{Cl}_6\). Therefore, the correct answer is (2)
Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32