Temperature-dependent disproportionation of chlorine in NaOH

Steps:

• Chlorine undergoes disproportionation with aqueous NaOH, where Cl(0) forms products with different oxidation states.
• In cold NaOH(aq), reaction is Cl₂ + 2NaOH → NaCl + NaClO + H₂O; Cl oxidation states: -1 in Cl⁻, +1 in ClO⁻.
• In hot NaOH(aq), reaction is 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O; Cl oxidation states: -1 in Cl⁻, +5 in ClO₃⁻.
• Match options to conditions: only D correctly identifies -1 and +5 for hot NaOH(aq).

Why D is correct:

• Hot concentrated NaOH produces chlorate (ClO₃⁻, Cl = +5) and chloride (Cl⁻, -1), per the balanced equation 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O.

Why the others are wrong:

• A: Cold NaOH yields +1 (not -3) in hypochlorite.
• B: Hot NaOH yields +5 (not -5) in chlorate.
• C: Cold NaOH yields +1 (not +5) in hypochlorite.

Final answer: D