Q.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.

Step 1. (A) is a Group-1 metal present in common salt (NaCl) -- so A = sodium (Na).

Step 2. (A) reacts with (B) to give compound (C), in which hydrogen is in the -1 oxidation state -- a -1 hydrogen means (C) is an ionic hydride, and sodium's ionic hydride is NaH, formed from sodium reacting with hydrogen gas: $2Na + H_2 \rightarrow 2NaH$. So B = hydrogen gas (H₂) and C = sodium hydride (NaH).

Step 3. (B), i.e. hydrogen gas, reacts with a gas (C)... re-reading the puzzle's own labelling, it is (B) reacting with oxygen (implied by "universal solvent" being the product) that gives the universal solvent (D). Hydrogen reacting with oxygen gives water: $2H_2 + O_2 \rightarrow 2H_2O$. Water is famously called the "universal solvent". So D = water (H₂O). …

Work through the clues in order: a Group-1 metal in common salt (Na) forms a -1-oxidation-state hydride (NaH) with hydrogen; hydrogen also combines with oxygen to give the 'univers …

• Misreading which reactant forms the 'universal solvent' -- it is hydrogen (B) combining with oxygen (an unlabelled but implied reactant) that gives water (D), not (A) directly. …