When chlorine is passed over slaked lime the main reaction products are:
A. \(Ca(ClO_2)_2\)
B. \(CaCl_2\)
C. \(Ca(OCl)_2\)
D. \(Ca(OCl_2)\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → B and C only
Chlorine reacts with dry slaked lime to give bleaching powder. Chlorine disproportionates: Cl2 (0) goes partly to chloride in CaCl2 (−1, reduction) and partly to hypochlorite in Ca(OCl)2 (+1, oxidation).
\( \displaystyle \ce{2Ca(OH)2 + 2Cl2} \xrightarrow{\text{dry}} \ce{Ca(OCl)2 + CaCl2 + 2H2O} \)
A. Ca(ClO2)2 — Incorrect. Calcium chlorite has Cl in the +3 state; the reaction gives hypochlorite (Cl +1), not chlorite.
B. CaCl2 — Correct. Calcium chloride is formed from the chlorine that is reduced.
C. Ca(OCl)2 — Correct. Calcium hypochlorite, the bleaching part of bleaching powder.
D. Ca(OCl2) — Incorrect. This formula has no \( \ce{OCl2^{2-}} \) ion to go with \( \ce{Ca^{2+}} \); it is not a real product.
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 dry slaked lime it gives bleaching powder. 2Ca(OH)2 + 2Cl2 → Ca(OCl)2 + CaCl2 + 2H2O”