Answer:
Option A
Explanation:
Acidified K2Cr2O7, CuSO4, H2O2, Cl2, O3, FeCl3. and HNO3 oxidise aq. iodide to iodine. Alkaline KMnO4 oxidise aq. iodide to IO3-.
Na2S2O3 is a strong reducing agent which on reaction with I2 produces I-
$N_{2}S_{2}O_{3}+I_{2}\rightarrow2I^{-}+N_{2}S_{4}O_{6}$
Therefore, no reaction takes place between Na2S2O3 and iodide ion. Hence, the correct integer is (7)