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NEET CHEMISTRYThe d-and f-Block ElementsMedium

Question

When neutral or faintly alkaline KMnO4KMnO_4 is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is:

A

I2I_2

B

IO4IO_4^-

C

IO3IO_3^-

D

IOIO^-

Step-by-Step Solution

Potassium permanganate (KMnO4KMnO_4) acts as a strong oxidising agent. Its reaction product with iodide depends on the pH of the medium:

  1. In acidic medium: Iodide (II^-) is oxidised to iodine (I2I_2).
  2. In neutral or faintly alkaline solution: Iodide (II^-) is oxidised to iodate (IO3IO_3^-). The manganese is reduced from +7 (MnO4MnO_4^-) to +4 (MnO2MnO_2).

The specific reaction equation provided in the NCERT text is: 2MnO4+H2O+I2MnO2+2OH+IO32MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^-

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from The d-and f-Block Elements. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThe d-and f-Block Elementsneutralfaintlyalkalinetreatedpotassium

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