Electrolysis of molten NaCl gives:

The correct answer is Option (4) → \(Na\) at cathode and \(Cl_2\) at anode

In molten NaCl the only species present are \( \ce{Na^{+}} \) and \( \ce{Cl^{-}} \) ions; there is no water. Reduction takes place at the cathode and oxidation at the anode:

    Cathode: \( \ce{Na+(melt) + e- -> Na(l)} \)

    Anode: \( \ce{Cl-(melt) -> 1/2Cl2(g) + e-} \)

Why the other options are incorrect

• Option (1) \(H_2\) at cathode, \(Cl_2\) at anode and \(NaOH\) solution — Incorrect. These are the products of electrolysis of aqueous NaCl, where water is reduced in preference to \( \ce{Na^{+}} \): \( \ce{H2O(l) + e- -> 1/2H2(g) + OH-(aq)} \); the \( \ce{OH^{-}} \) left with \( \ce{Na^{+}} \) forms NaOH.

• Option (2) \(Na\) at cathode, \(Cl_2\) at anode and \(NaOH\) solution — Incorrect. NaOH needs water; there is no water in the melt.

• Option (3) \(H_2\) at anode, \(Cl_2\) at cathode — Incorrect. It reverses the electrodes; \( \ce{Cl^{-}} \) loses electrons, which can happen only at the anode, and there is no hydrogen in molten NaCl.

NCERT (2026-27 edition):
Products of Electrolysis – “… The products of electrolysis depend on the different oxidising and reducing species present in the electrolytic cell and their standard electrode potentials. … For example, if we use molten NaCl, the products of electrolysis are sodium metal and Cl₂ gas. Here we have only one cation (Na⁺) which is reduced at the cathode (Na⁺ + e^– → Na) and one anion (Cl^–) which is oxidised at the anode (Cl^– → ½Cl₂ + e^–).”