Amphoteric oxide behavior identifies aluminum

Steps:

• Y burns in O₂ to form a white solid oxide.
• The oxide dissolves in excess HCl(aq), forming a chloride solution.
• Adding NaOH(aq) to this solution produces a white precipitate of hydroxide.
• Excess NaOH(aq) dissolves the precipitate, forming a soluble complex.

Why C is correct:

• Aluminum forms Al₂O₃ (white solid), which reacts with HCl to AlCl₃ (soluble); AlCl₃ + NaOH → Al(OH)₃ (white ppt), and excess NaOH → [Al(OH)₄]⁻ (soluble aluminate), per amphoteric hydroxide definition.

Why the others are wrong:

• A. Na: Na₂O dissolves in HCl to NaCl, but adding NaOH forms no precipitate.
• B. Mg: MgO dissolves in HCl to MgCl₂; Mg(OH)₂ precipitates with NaOH but remains insoluble in excess.
• D. P: P₄O₁₀ forms phosphoric acid with water/HCl, no hydroxide precipitate with NaOH.

Final answer: C