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