Question

A group-1 metal (A) which is present in common salt reacts with (B) to give compound (C) in which hydrogen is present in –1 oxidation state. (B) on reaction with a gas (C) to give universal solvent (D). The compound (D) on reacts with (A) to give (E), a strong base. Identify A, B, C, D and E. Explain the reactions.

Answer 1

The metal that belongs to Group-1 which is present in common salt is Sodium (A).

The metal reacts with (B) to give compound (C) in which hydrogen is present in a -1 oxidation state.

\[\ce{2Na + H2 -> 2NaH}\]

Hence, the (B) is hydrogen, and (C) is sodium hydride. (B) on reaction with gas to give universal solvent (D).

\[\ce{H2 + O2 -> 2NaOH}\]

The universal solvent (D) is water. The compound (D) on reaction with (A) to give (E) which is a strong base.

\[\ce{H2O + 2Na -> 2NaOH}\]

The Compound (E) is Sodium hydroxide.

• A – Sodium (Na)
• B – Hydrogen (H₂)
• C – Sodium Hydride (NaH)
• D – Water (H₂O)
• E – Sodium hydroxide (NaOH)