When \(Br_2\) is treated with aqueous solution of \(NaF, NaCl, NaI\) separately
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Only \(I_2\) is liberated
The oxidising power of the halogens decreases down the group: \( \ce{F2} \) > \( \ce{Cl2} \) > \( \ce{Br2} \) > \( \ce{I2} \). A halogen can therefore oxidise only the halide ions of the halogens below it, displacing them from their salts.
\( \ce{Br2} \) is above iodine, so it oxidises \( \ce{I^{-}} \) to \( \ce{I2} \):
\( \ce{Br2 + 2NaI -> 2NaBr + I2} \)
\( \ce{F2} \) and \( \ce{Cl2} \) are stronger oxidising agents than \( \ce{Br2} \), so \( \ce{Br2} \) cannot oxidise \( \ce{F^{-}} \) or \( \ce{Cl^{-}} \); there is no reaction with NaF or NaCl. Only I2 is liberated.
Why the other options are incorrect
Option (1) F2, Cl2, I2 are liberated — Incorrect. \( \ce{Br2} \) would have to oxidise \( \ce{F^{-}} \) and \( \ce{Cl^{-}} \), which needs a stronger oxidant than bromine.
Option (2) Only F2 and Cl2 are liberated — Incorrect. This is the reverse of the actual result; \( \ce{F^{-}} \) and \( \ce{Cl^{-}} \) are the ions bromine cannot oxidise.
Option (3) Only Cl2 is liberated — Incorrect. Chlorine displaces bromine (\( \ce{Cl2} \) + \( \ce{2Br^{-}} \) → \( \ce{2Cl^{-}} \) + \( \ce{Br2} \)), not the other way round.
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):
Chemical Properties – …
