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JEE Main 4-April-2025 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 4-April-2025 Paper - Shift 14 Marks
Question
$\mathrm{KMnO}_{4}$ 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 _____________.
✓

Answer

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

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