Chemistry · Redox Reactions

JEE Main 2025 — 4 April, Morning Shift — Question 15

$\mathrm{KMnO}_{4}$ acts as an oxidising agent in acidic medium. " $X$ " is the difference between the oxidation states of $M n$ in reactant and product, " $Y$ " is the number of ' $d$ ' electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of $X+Y$ is $\_\_\_\_\_$

Answer: 10

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{KMnO}_{4}^{+7} \rightarrow \mathrm{Mn}^{2+}$ (when act as oxidising agent)

difference in O.S. $=5, \therefore X=5$

$\mathrm{Fe}^{3+} \Rightarrow d^{5} \Rightarrow 5$ d-electrons present,

$\therefore \mathrm{Y}=5$

$\therefore \quad X+Y=10$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Morning Shift
Subject: Chemistry
Chapter: Redox Reactions
Topic: n-Factor, Redox Titrations, Self Indicator & Miscellaneous Cases