$\mathrm{KMnO}_4$ acts as an oxidising agent in acidic medium. ' X ' is the difference between the oxidation…

$\mathrm{KMnO}_4$ 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 $\mathrm{X}+\mathrm{Y}$ is ______.

Solution

$\underset{(\mathrm{O}. \mathrm{A})}{\stackrel{+7}{\mathrm{KMnO}}} \xrightarrow{\text { Acidic medium }} \mathrm{Mn}^{2+}$
X is difference in oxidation state.
$6 \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O} \rightarrow\left[\mathrm{Fe}_3\left(\mathrm{OH}_2\right)\left(\mathrm{CH}_3 \mathrm{COO}\right)_6\right]^{\oplus}+2 \mathrm{H}^{\oplus}$
$\left[\mathrm{Fe}_3(\mathrm{OH})_2\left(\mathrm{CH}_3 \mathrm{COO}\right)_6\right]^{\oplus} \text{(Brown red ppt)} +4 \mathrm{H}_2 \mathrm{O}\rightarrow\left[\mathrm{Fe}(\mathrm{OH})_2\left(\mathrm{CH}_3 \mathrm{COO}\right]+\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}^{\oplus}\right.$
$\mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0$ contains 5 d electrons
So $Y=5$
$X+Y=5+5=10$

Asked in: JEE Main 2025 (04 Apr Shift 1)

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