$\mathrm{KMnO}_{4}$ 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 _______ .
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:
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:
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.
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).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,