Q.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.

Al$^{3+}$ + NaOH gives white Al(OH)$_3$ (A), which dissolves in excess NaOH as the soluble aluminate complex $[\text{Al(OH)}_4]^-$ (B). A dissolves in dilute HCl to give AlCl$_3$ (C). Strongly heating A (Al(OH)$_3$) drives off water to give Al$_2$O$_3$ (D, …

1. Identifying the white precipitate (A). A metal salt of X reacting with NaOH to give a white precipitate that is soluble in excess NaOH is the classic signature of an amphoteric hydroxide. Among common metals, this points to aluminium:

2. Dissolution in excess NaOH (B). Al(OH)$_3$ is amphoteric and dissolves in excess alkali to form a soluble tetrahydroxoaluminate complex:

3. Dissolution in dilute HCl (C). Al(OH)$_3$ also dissolves in dilute acid (confirming its amphoteric character) to give aluminium chloride:

4. Strong heating of A gives D, used to extract the metal. Heating Al(OH)$_3$ strongly drives off water, forming alumina: $2\text{Al(OH)}_3 \xrightarrow{\Delta} \text{Al}_2\text{O}_3 (\text{D}) + 3\text{H}_2\text{O}$ …

Work backward from the diagnostic clues: a white precipitate that redissolves in excess NaOH signals an amphoteric hydroxide (Al(OH)$_3$); confirm with its behaviour in dilute HCl and on strong heating (loses water to give the oxide used industrially to extract the metal, i.e. alumina) — this triangulates uniquely …

• Identifying X as zinc (which also gives an amphoteric white hydroxide) instead of aluminium — the clue "(D) is used to extract the metal" (i.e. via electrolysis of alumina, the Hall–Héroult process) is decisive for aluminium. …