Sodium chlorate(V) from chlorine disproportionation

Steps:

• Identify sodium chlorate(V) as NaClO3, where chlorine is in +5 oxidation state.
• Recall chlorine reacts with NaOH via disproportionation, varying by conditions.
• Cold, dilute NaOH yields NaClO (hypochlorite, Cl+1); hot, concentrated NaOH yields NaClO3 (chlorate, Cl+5).
• Chlorine in water forms HOCl (hypochlorous acid), not NaClO3.

Why D is correct:

• Option D provides the specific reagents for NaClO3 formation, per the 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O reaction under hot, concentrated conditions.

Why the others are wrong:

• A produces NaClO3 but lists incorrect conditions (hot concentrated NaOH is right, but choice mismatches).
• B yields NaClO via Cl2 + 2NaOH → NaCl + NaClO + H2O.
• C forms HCl + HOCl, no sodium salt.

Not enough information for full choice D details.

Final answer: D