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When metal $X$ is treated with sodium hydroxide, a white precipitate $(A)$ is obtained, which is soluble in excess of $NaOH$ to give soluble complex $(B)$. Compound $(A)$ is soluble in dilute $HCl$ to form compound $(C)$. The compound $(A)$ when heated strongly gives $(D)$, which is used to extract metal. Identify $(X), (A), (B), (C)$ and $(D)$. Write suitable equations to support their identities.
Solution
The given metal $X$ gives a white precipitate with sodium hydroxide and the precipitate dissolves in excess of sodium hydroxide. Hence, $X$ must be aluminium. The white precipitate (compound $A$ ) obtained is aluminium hydroxide. The compound $B$ formed when an excess of the base is added is sodium tetrahydroxoaluminate $(III)$.
$2 \mathrm{Al}+3 \mathrm{NaOH} \rightarrow \mathrm{Al}(\mathrm{OH})_{3} \downarrow+3 \mathrm{Na}^{+}$
$(\mathrm{X})$
$(\mathrm{ppt})$
$\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{NaOH} \rightarrow \mathrm{Na}^{+}\left[\mathrm{Al}(\mathrm{OH})_{4}\right]^{-}$
$(A)$ Sodium tetrahydroxo aluminate $(III)$
Now, when dilute hydrochloric acid is added to aluminium hydroxide, aluminium chloride (compound $\mathrm{C}$ ) is obtained.
$\mathrm{Al}(\mathrm{OH})_{3}+3 \mathrm{HCl} \rightarrow \mathrm{AlCl}_{3}+3 \mathrm{H}_{2} \mathrm{O}$
$(A)$ $\quad$ $(C)$
Also, when compound $\mathrm{A}$ is heated strongly, it gives compound $\mathrm{D}$. This compound is used to extract metal $X$. Aluminium metal is extracted from alumina. Hence, compound D must be alumina.
$2 \mathrm{Al}(\mathrm{OH})_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \mathrm{O}$
