Question

$N_2{H}_4$ reduces $I{O}_3^{-}$ to $I^{+}$ in acidic medium as :
$N_2{H}_4+I{O}_3^{-}+H^{+}⟶N_2+I^{+}+H_{2}O$

Intermediate in the above reaction is $I_2$ which can be detected by :

• A. taking solvent $CC{I}_4$ in which $I_2$ is soluble and thus layer becomes violet
• B. starch as indicator which turns blue
• C. both (a) and (b)
• D. None of the above

Answer 1

The correct option is C both (a) and (b)

Both $I_2$ and $CC{l}_4$ are non-polar in nature while water is polar. Thus $I_2$ is soluble in $CC{l}_4$. And Amylose in starch is responsible for the formation of a deep blue color in the presence of iodine. The iodine molecule slips inside of the amylose coil. Iodine Reagent: Iodine is not very soluble in water, therefore the iodine reagent is made by dissolving iodine in water in the presence of potassium iodide. This makes a linear triiodide ion complex with is soluble that slips into the coil of the starch causing an intense blue-black color.