JEE Main 2020 · 7 January · evening · Single correct · +4 / −1
In the following reaction, products (A) and (B) respectively, are :
NaOH + Cl₂ → (A) + side products
(hot and conc.)
Ca(OH)₂ + Cl₂ → (B) + side products (dry)
- (a) NaOCl and Ca(OCl)₂
- (b) NaClO₃ and Ca(ClO₃)₂
- (c) NaOCl and Ca(ClO₃)₂
- (d) NaClO₃ and Ca(OCl)₂
Answer and solution
Answer (d) NaClO₃ and Ca(OCl)₂
Solution
Chemistry ma’am · 4 steps
Hot alkali pushes chlorine all the way to chlorate; dry slaked lime stops at hypochlorite, bleaching powder.
- Same reaction type. In both reactions chlorine disproportionates: some Cl goes down to -1 as chloride, and some goes up. The conditions decide how far up it goes.
- Hot, conc. NaOH. In hot, concentrated alkali the hypochlorite formed first disproportionates again. Chlorine ends up at +5 as chlorate: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O. So (A) is NaClO₃.
- Dry slaked lime. Passing chlorine over dry Ca(OH)₂ is how bleaching powder is made. Chlorine only reaches +1 as hypochlorite: 2Ca(OH)₂ + 2Cl₂ → Ca(OCl)₂ + CaCl₂ + 2H₂O. So (B) is Ca(OCl)₂.
- So. (A) is NaClO₃ and (B) is Ca(OCl)₂, which is option (d).
The idea to keep. Cold and dilute (or dry) conditions with chlorine give hypochlorite, OCl⁻. Hot and concentrated alkali gives chlorate, ClO₃⁻, because hypochlorite itself disproportionates on heating.
Answer from the NTA final answer key. Solution by PrepWiser.