Read the passage carefully and answer the questions.
Sodium dichromate \( Na_2Cr_2O_7 \) and Potassium dichromate \( K_2Cr_2O_7 \), are strong oxidizing agents; in acidic solution, its oxidising action can be represented as follows:
\( Cr_2{O_7}^{2-}+ 14H^++ 6e^-→2Cr^{3+}+ 7H_2O \) \( (E° = 1.33V) \)
The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. The chromate ion is tetrahedral whereas the dichromate ion consists of two tetrahedra sharing one corner with Cr-O-Cr bond angle of 126°
Potassium permanganate is prepared by fusion of \( MnO_2 \) with an alkali metal hydroxide and an oxidising agent like \( KNO_3 \). This produces the dark green \( K_2MnO_4 \) which disproportionates in a neutral or acidic solution to give permanganate. It is a strong oxidizing agent. Hydrogen ion concentration of the solution plays an important part in the reduction of permanganate to manganate, manganese dioxide and manganese(II) salt.
Potassium permanganate forms dark purple (almost black) crystals which are isostructural with those of \( KClO_4 \). The salt is not very soluble in water (6.4 g/100 g of water at 293 K), but when heated it decomposes at 513 K.
Alkaline/neutral \( KMnO_4 \) oxidizes iodide (\( I^- \)) ion to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(IO_3^-\)
What permanganate is reduced to, and how far it oxidises a substance, depends on the hydrogen ion concentration. In acidic solution \( \ce{MnO4^-} \) takes 5 electrons and becomes \( \ce{Mn^2+} \). In neutral or faintly alkaline solution it takes only 3 electrons and is reduced to \( \ce{MnO2} \):
\( \displaystyle \ce{MnO4^- + 2H2O + 3e^- -> MnO2 + 4OH^-} \)
Under these conditions iodide is oxidised all the way from I(−1) to I(+5), the iodate ion, giving out 6 electrons:
\( \displaystyle \ce{I^- + 6OH^- -> IO3^- + 3H2O + 6e^-} \)
Taking the first half-reaction twice to balance the 6 electrons and adding:
\( \displaystyle \ce{2MnO4^- + H2O + I^- -> 2MnO2 + 2OH^- + IO3^-} \)
So alkaline or neutral \( \ce{KMnO4} \) oxidises iodide to iodate, \( \ce{IO3^-} \).
Why the other options are incorrect
Option (2) \(IO_2^-\) — Incorrect. Iodide is not stopped at the +3 state; the oxidation goes on to the stable iodate ion.
Option (3) \(I_2\) — Incorrect. Iodine is the product in acidic solution: \( \ce{10I^- + 2MnO4^- + 16H^+ -> 2Mn^2+ + 8H2O + 5I2} \).
Option (4) \(I_2O_4\) — Incorrect. This is an oxide of iodine; it is not formed when iodide is oxidised in aqueous solution.
NCERT (2026-27 edition):
In Neutral or Faintly Alkaline Solutions
