Oxidation of bromide ions by hot concentrated sulfuric acid

Steps:

• Hot concentrated H2SO4 reacts with solid KX to form HX and potassium hydrogen sulfate.
• For easily oxidized halides, H2SO4 further oxidizes HX to X2, reducing itself to SO2.
• Bromide ions have suitable reduction potential for oxidation to Br2 without further reduction products.
• The reaction yields a colored mixture confirming X2 presence.

Why B is correct:

• KBr undergoes 2KBr + 3H2SO4 → 2KHSO4 + Br2 + SO2 + 2H2O, where Br2 forms due to the reduction potential of Br2/Br⁻ (1.07 V) allowing oxidation by SO₄²⁻/SO₂.

Why the others are wrong:

• A. Potassium iodide reduces H2SO4 further to H2S alongside I2 and SO2, altering the mixture beyond simple X2 inclusion.
• C. Potassium chloride forms only HCl gas, as Cl⁻ resists oxidation (reduction potential 1.36 V > SO₄²⁻/SO₂).
• D. Not enough information (option incomplete; assuming KF, no reaction occurs as F⁻ forms stable HF without oxidation).

Final answer: B