In aqueous electrolysis the species discharged at each electrode is decided by competing reduction or oxidation potentials, not merely by the dissolved salt. At the cathode, the two candidates for reduction are Na^+ and water. The reduction potential of water (producing hydrogen gas) is far less negative than that required to deposit sodium metal, so water is reduced preferentially: 2H_2O + 2e^- → H_2 + 2OH^-. Hence hydrogen gas is liberated at the cathode while sodium ions remain in solution. Chlorine is incorrect because it is produced at the anode, not the cathode, owing to overpotential effects favouring chloride oxidation. Oxygen is also an anode product and arises only when chloride is dilute. Sodium vapour is wrong because sodium cannot be deposited from aqueous solution; it would immediately react with water. This selective discharge is central to the chlor-alkali process described in NCERT. Plausibility check: cathodes carry reduction, and the easiest reduction available in water is hydrogen evolution, fully consistent with the chosen answer.