JEE Chemistry 2025 Question with Solution

Given: $\mathrm{KMnO}_{4}$ acts as an oxidising agent in acidic medium, and $Y$ is the number of $d$ electrons in the brown red precipitate formed in the acetate ion test with neutral ferric chloride.

Find: The value of $\mathrm{X}+\mathrm{Y}$.

In acidic medium, manganese in $\mathrm{KMnO}_{4}$ is reduced from oxidation state $+7$ to $+2$.

The brown red precipitate is taken as a ferric compound containing $\mathrm{Fe}^{3+}$. For iron, neutral configuration is $[\mathrm{Ar}]\,4s^2 3d^6$, so for $\mathrm{Fe}^{3+}$ the configuration is $3d^5$.

Therefore,

Therefore, the required numerical value is $10$.

Given: $\mathrm{KMnO}_{4}$ acts as an oxidising agent in acidic medium.

Find: $\mathrm{X}+\mathrm{Y}$, where $X$ is the change in oxidation state of manganese and $Y$ is the number of $d$ electrons in the brown red precipitate species containing ferric ion.

1. In $\mathrm{KMnO}_{4}$, manganese is in oxidation state $+7$.
2. In acidic medium, permanganate is reduced to $\mathrm{Mn}^{2+}$.

3. The brown red precipitate formed with neutral ferric chloride is treated in the solution as a ferric species, so iron is in oxidation state $+3$.
4. Iron has atomic number $26$ and electronic configuration $[\mathrm{Ar}]\,4s^2 3d^6$.
5. Removing three electrons to form $\mathrm{Fe}^{3+}$ gives $3d^5$.

Hence,

So, the answer is $10$.