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)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)