The disproportionation of chlorine in cold dilute NaOH (3 Cl₂ + 6 OH⁻ → 5 Cl⁻ + ClO₃⁻ + 3 H₂O is the hot reaction). At cold and dilute conditions, Cl₂ + 2 OH⁻ →:
A2 Cl⁻ + H₂O only
B2 ClO⁻ + H₂ gas
CClO₃⁻ + Cl⁻ ions
DCl⁻ + ClO⁻ + H₂O
Answer & Solution
Correct answer: D. Cl⁻ + ClO⁻ + H₂O
1. In cold dilute NaOH, Cl₂ disproportionates to Cl⁻ (reduced) and ClO⁻ (oxidised).
2. Cl₂ + 2 OH⁻ → Cl⁻ + ClO⁻ + H₂O.
3. Cl in Cl₂ is 0; in Cl⁻ it is −1 (reduction); in ClO⁻ it is +1 (oxidation).
4. The hot and concentrated version produces Cl⁻ and ClO₃⁻ (a +5 state).
_Source: NCERT Class 11 Chem Ch 8 §8.2 Disproportionation_
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