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NEET CHEMISTRYThe d-and f-Block ElementsMedium

Question

In the neutral or faintly alkaline medium, KMnO4KMnO_4 oxidises iodide into iodate. The change in oxidation state of manganese in this reaction is from:

A

+6 to +5

B

+7 to +4

C

+6 to +4

D

+7 to +3

Step-by-Step Solution

In a neutral or faintly alkaline medium, permanganate (MnO4MnO_4^-) oxidises iodide (II^-) to iodate (IO3IO_3^-). The reaction is given as: 2MnO4+H2O+I2MnO2+2OH+IO32MnO_4^- + H_2O + I^- \rightarrow 2MnO_2 + 2OH^- + IO_3^- In this reaction, manganese is reduced from the +7 oxidation state in MnO4MnO_4^- to the +4 oxidation state in MnO2MnO_2 .

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 Elementsneutralfaintlyalkalinemediumoxidises

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