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 _______ .

Hint:

Determine the oxidation states and the number of d-electrons to find the values of X and Y.

Answer 1

Correct Answer: 10

1. Oxidation states of Mn: 
- Reactant: $\mathrm{Mn}^{7+}$ 
- Product: $\mathrm{Mn}^{2+}$ 
- Difference in oxidation states: $X = 7 - 2 = 5$ 
2. Brown red precipitate: - The brown red precipitate is $\mathrm{Fe}(\mathrm{OH})_2(\mathrm{CH}_3\mathrm{COO})_n$. 
- $\mathrm{Fe}^{3+}$ has 5 d-electrons. - Therefore, $Y = 5$. 
3. Calculate $\mathrm{X}+\mathrm{Y}$: \[ \mathrm{X} + \mathrm{Y} = 5 + 5 = 10 \] 
Therefore, the correct answer is (10).