Identify \('X'\) along with oxidation state of halogen in the given reaction:
\(3Cl_2 + 6NaOH \longrightarrow 5NaCl + X + 3H_2O\) (hot and conc.)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(NaClO_3\), +5
Chlorine reacts with alkali by disproportionation: part of it is reduced to \( \ce{Cl^-} \) and part is oxidised. With cold, dilute NaOH the oxidised product is hypochlorite; with hot and concentrated NaOH it is chlorate.
With hot and concentrated NaOH, X = NaClO3 (sodium chlorate). The oxidation state of Cl in it:
\( (+1) + x + 3(-2) = 0 \Rightarrow x = +5 \)
Check by electrons: five Cl atoms go from 0 to −1 (gain 5 e−) and one Cl goes from 0 to +5 (loses 5 e−).
Why the other options are incorrect
Option (1) \(NaClO_2\), +3 — Incorrect. Sodium chlorite (Cl +3) is not formed when chlorine reacts with NaOH.
Option (3) \(NaClO_2\), +7 — Incorrect. Wrong compound and an oxidation state that does not fit it; Cl is +3 in \( \ce{NaClO2} \), and +7 belongs to \( \ce{NaClO4} \).
Option (4) \(NaClO_3\), +1 — Incorrect. Right compound, wrong oxidation state; +1 is the state of Cl in NaOCl, the product with cold, dilute NaOH.
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 – …
