JEE Main 2025 Apr 4 Shift 1, Chemistry Q22: Oxidation and Reduction

JEE Main2025Apr 4, Shift 1Chemistry

KMnO $_{4}$ acts as an oxidising agent in acidic medium. 'X' is the difference between the oxidation states of Mn 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 _____.

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Solution

Finding X: In acidic medium, KMnO $_{4}$ (Mn in $+ 7$ ) is reduced to Mn $^{2 +}$ ( $+ 2$ ). So $X = 7 - 2 = 5$ .

Finding Y: The acetate-ion test with neutral FeCl $_{3}$ first gives a red colouration of $[F e_{3} (O H_{2}) (C H_{3} C O O)_{6}]^{+}$ ; on boiling, a brown-red precipitate of basic ferric acetate $[F e (O H)_{2} (C H_{3} C O O)]$ separates. In this species iron is in the $+ 3$ state.

Fe $^{3 +}$ : $[A r] 3 d^{5} 4 s^{0}$ contains 5 d-electrons, so $Y = 5$ .

$X + Y = 5 + 5 =$ 10.

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JEE Main 2025 Apr 4 Shift 1, Chemistry Q22: Oxidation and Reduction

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