KMnO4 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 precipitat

Question

KMnO4 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 _______

Prashant Kotian · Accepted Answer

Correct answer: .. **Step 1: Find X (oxidation state change of Mn in KMnO4)** In KMnO4, Mn has oxidation state +7. In acidic medium, KMnO4 is reduced to Mn²⁺ (the product). X = |+7 − (+2)| = 5 **Step 2: Identify the brown-red precipitate in acetate ion test** When neutral FeCl₃ reacts with acetate ions, a brown-red precipitate of ferric acetate complex forms, which hydrolyzes to produce **Fe(OH)₃** (brown-red precipitate). **Step 3: Find Y (d electrons in Fe(OH)₃)** Fe(OH)₃ contains Fe³⁺ ions. Electronic configuration of Fe: [Ar] 3d⁶ 4s² Fe³⁺ loses 3 electrons (two from 4s and one from 3d): [Ar] 3d⁵ Number of d electrons in Fe³⁺ = 5 Y = 5 **Step 4: Calculate X + Y** X + Y = 5 + 5 = 10 Therefore, the answer is **10**

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