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

To solve the given problem, we need to determine variables X and Y, and then compute X + Y.

Firstly, let's address the compound $\mathrm{KMnO}_{4}$ acting as an oxidizing agent in acidic medium. In this scenario, Mn is reduced from +7 oxidation state to +2:

• Initial oxidation state of Mn = +7
• Final oxidation state of Mn = +2

The difference X is the change in oxidation state:

X = 7 - 2 = 5

Next, we consider the number of 'd' electrons in the brown-red precipitate at the end of the acetate ion test with neutral ferric chloride (typically Fe(OH)₃ which is red-brown).

The oxidation state of iron in Fe(OH)₃ is +3. In this state, the electron configuration of Fe is:

• Atomic number of Fe = 26: Configuration for Fe (neutral) = [Ar] 4s² 3d⁶
• For Fe³⁺: 3d⁵ (since 2 electrons are removed from 4s and 1 from 3d)

Thus, the number of 'd' electrons, Y = 5.

Therefore, the computed value:

X + Y = 5 + 5 = 10

This value is within the provided range 10, 10.

Answer 2

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).