When chlorine gas is passed through a hot solution of \(NaOH\), a disproportionation reaction occurs. The zero oxidation state of chlorine changes to:
A. 0 to +5
B. 0 to -1
C. 0 to +3
D. 0 to +1
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A, B only
In a disproportionation reaction the same element is both oxidised and reduced. With hot and concentrated alkali, chlorine (oxidation state 0) gives chloride and chlorate:
\( \displaystyle \ce{3Cl2 + 6NaOH} \xrightarrow{\text{hot, conc.}} \ce{5NaCl + NaClO3 + 3H2O} \)
A. 0 to +5 — True. Cl in \( \ce{NaClO3} \): +1 + x + 3(−2) = 0, so x = +5 (oxidation).
B. 0 to −1 — True. Cl in NaCl is −1 (reduction).
C. 0 to +3 — False. No chlorite (\( \ce{ClO2-} \), Cl +3) is formed in this reaction.
D. 0 to +1 — False. Hypochlorite (Cl +1) forms only with cold and dilute alkali:
\( \displaystyle \ce{Cl2 + 2NaOH} \xrightarrow{\text{cold, dil.}} \ce{NaCl + NaOCl + H2O} \)
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Chlorine – “… With cold and dilute alkalies chlorine produces a mixture of chloride and hypochlorite but with hot and concentrated alkalies it gives chloride and chlorate.
2NaOH + Cl2 → NaCl + NaOCl + H2O (cold and dilute)
6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O (hot and conc.)”