On addition of conc. \(H_2SO_4\) to a chloride salt, colourless fumes are evolved but in the case of iodide salt, violet fumes come out. This is because
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → HI gets oxidized to \(I_2\)
Concentrated sulphuric acid first displaces the hydrogen halide from the salt:
\( \displaystyle \ce{NaCl + H2SO4 -> NaHSO4 + HCl}\quad(\text{colourless fumes}) \)
\( \displaystyle \ce{NaI + H2SO4 -> NaHSO4 + HI} \)
Concentrated \( \ce{H2SO4} \) is also an oxidising agent. HCl is a poor reducing agent, so it escapes unchanged. HI is a strong reducing agent (the H–I bond is the weakest of the hydrogen halides), so it is oxidised to iodine, which comes off as violet vapour, while the acid is reduced to \( \ce{SO2} \):
\( \displaystyle \text{Oxidation: }\ce{2I^- -> I2 + 2e^-} \)
\( \displaystyle \text{Reduction: }\ce{SO4^2- + 4H+ + 2e^- -> SO2 + 2H2O} \)
\( \displaystyle \ce{2HI + H2SO4 -> I2 + SO2 + 2H2O} \)
Why the other options are incorrect
Option (1) \(H_2SO_4\) reduces HI to \(I_2\) — Incorrect. Going from HI (I = −1) to \( \ce{I2} \) (0) is an oxidation; \( \ce{H2SO4} \) oxidises HI, it does not reduce it.
Option (2) HI is of violet colour — Incorrect. Hydrogen iodide is a colourless gas; the violet colour is that of \( \ce{I2} \) vapour.
Option (4) HI change to \(HIO_3\) — Incorrect. Sulphuric acid is not strong enough to take iodine to +5; iodic acid is made with concentrated nitric acid, and it is not violet in any case.
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):
Sulphuric Acid – “… Hot concentrated sulphuric acid is a moderately strong oxidising agent. In this respect, it is intermediate between phosphoric and nitric acids. Both metals and non-metals are oxidised by concentrated sulphuric acid, which is reduced to SO2.”