$\mathrm{KMnO}_{4}$ 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 _______ .
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Determine the oxidation states and the number of d-electrons to find the values of X and Y.
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.
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Approach Solution -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).
