D And F Block Elements – JEE Main ChemistryPractice Questions & PYQs

The number of electrons transferred in a redox reaction is equivalent to the net decrease in the oxidation state of the atom undergoing reduction. In the reactant $\mathrm{KMnO}_4$, the oxidation state of Manganese is determined by the equation $1 + \text{Mn} + 4(-2) = 0$, which yields $\text{Mn} = +7$.

For $[\mathrm{MnO}_4]^{-2}$, the oxidation state of Manganese is determined by $\text{Mn} + 4(-2) = -2$, resulting in $\text{Mn} = +6$, which corresponds to a transfer of $7 - 6 = 1$ electron. In the case of $\mathrm{MnO}_2$, the oxidation state is $\text{Mn} + 2(-2) = 0$, resulting in $\text{Mn} = +4$, representing a transfer of $7 - 4 = 3$ electrons. For $\mathrm{Mn}_2\mathrm{O}_3$, the oxidation state is $2\text{Mn} + 3(-2) = 0$, giving $\text{Mn} = +3$, which corresponds to a transfer of $7 - 3 = 4$ electrons. Finally, for $\mathrm{Mn}^{+2}$, the oxidation state is $2$, signifying a transfer of $7 - 2 = 5$ electrons.

The magnetic moment of a transition metal ion is directly proportional to the number of unpaired electrons present in its orbitals. Comparing the ions, $\mathrm{Ti}^{2+}$ contains 2 unpaired electrons, while both $\mathrm{Fe}^{2+}$ and $\mathrm{Cr}^{2+}$ possess 4 unpaired electrons. Since $\mathrm{Mn}^{2+}$ exhibits 5 unpaired electrons, which is the highest count among these choices, it demonstrates the maximum magnetic moment.