images/batch-001/07167a3d725a.webp" data-orsrc="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/07167a3d725a.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2025 (Online) 4th April Morning Shift Chemistry - Redox Reactions Question 9 English Explanation">

X is difference in oxidation state.

$$7-2=5$$

So $X=5$

$$\begin{aligned} & 6 \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O} \\ & \rightarrow\left[\mathrm{Fe}_3\left(\mathrm{OH}_2\right)\left(\mathrm{CH}_3 \mathrm{COO}\right)_6\right]^{\oplus}+2 \mathrm{H}^{\oplus} \\ & {\left[\mathrm{Fe}_3(\mathrm{OH})_2\left(\mathrm{CH}_3 \mathrm{COO}\right)_6\right]^{\oplus}+4 \mathrm{H}_2 \mathrm{O}} \\ & \rightarrow \underset{\text { Brown red ppt }}{\left[\mathrm{Fe}(\mathrm{OH})_2\left(\mathrm{CH}_3 \mathrm{COO}\right]\right.}+\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}^{\oplus} \end{aligned}$$

$\mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0$ contains 5 d electrons

So $Y=5$

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

19:T56b,

JEE Main 2025 (Online) 4th April Morning Shift Chemistry - Redox Reactions Question 9 Hindi Explanation

X ऑक्सीकरण अवस्था में अंतर है।

$$7-2=5$$

अतः $X=5$

Redox Reactions

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