Aluminium reacts with both dilute HCl and aqueous NaOH to liberate dihydrogen, showing that Al 2 O 3 and Al are amphoteric. With excess hot NaOH the aluminium-containing product is:

Finely divided aluminium metal dust and hydrogen

🧪 Chemistry · p-Block Elements (Groups 13 & 14) · NEET & JEE

Aluminium reacts with both dilute HCl and aqueous NaOH to liberate dihydrogen, showing that Al2O3 and Al are amphoteric. With excess hot NaOH the aluminium-containing product is:

Answer: Sodium tetrahydroxoaluminate(III), Na[Al(OH) 4 ].

A Finely divided aluminium metal dust and hydrogen
B Aluminium hydride
C Sodium tetrahydroxoaluminate(III), Na[Al(OH)4]
D Sodium aluminide

Correct answer: C. Sodium tetrahydroxoaluminate(III), Na[Al(OH)4]

Explanation: Aluminium dissolves in hot NaOH giving sodium tetrahydroxoaluminate(III), Na[Al(OH)4], with evolution of H2, illustrating its amphoteric character.

Diamond's rigid 3D tetrahedral network makes it the hardest natural material, while graphite's flat hexagonal sheets are held together only by weak forces, letting them slide past each other (used as a lubricant and in pencils).

Concept context

Covers the Boron family (Group 13: B, Al) and the Carbon family (Group 14: C, Si, Ge, Sn, Pb). Key topics include Lewis acid behaviour of BF₃, borax bead test, silicates, silicones, allotropes of carbon, and trends in properties down the group.

Read the full p-Block Elements (Groups 13 & 14) notes →

Aluminium reacts with both dilute HCl and aqueous NaOH to liberate dihydrogen, showing that Al<sub>2</sub>O<sub>3</sub> and Al are amphoteric. With excess hot NaOH the aluminium-containing product is:

Concept context

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