JEE MAIN - Chemistry (2025 - 4th April Morning Shift - No. 24)
$\mathrm{KMnO}_4$ 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
10
Explanation
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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 $Y=5$
$$X+Y=5+5=10$$
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