The magnetic moment (\( \mu \)) of an ion is a property directly proportional to its number of unpaired electrons, calculated via:
\[
\mu = \sqrt{n(n+2)} \, \mu_B,
\]
where \( n \) is the count of unpaired electrons and \( \mu_B \) is the Bohr magneton.
Step 1: Determine Electronic Configuration and Unpaired Electrons.
For \( \text{Mn}^{2+} \): Atomic number of Mn is 25. Neutral Mn configuration: \( [\text{Ar}] \, 3d^5 4s^2 \). \( \text{Mn}^{2+} \) configuration: \( [\text{Ar}] \, 3d^5 \). All 5 electrons in the \( 3d \) orbital are unpaired, so \( n = 5 \).
For \( \text{Fe}^{3+} \): Atomic number of Fe is 26. Neutral Fe configuration: \( [\text{Ar}] \, 3d^6 4s^2 \). \( \text{Fe}^{3+} \) configuration: \( [\text{Ar}] \, 3d^5 \). Similar to \( \text{Mn}^{2+} \), all 5 electrons in the \( 3d \) orbital are unpaired, resulting in \( n = 5 \).
Step 2: Calculate Magnetic Moment.
Using the formula with \( n = 5 \):
\[
\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \mu_B.
\]
Both \( \text{Mn}^{2+} \) and \( \text{Fe}^{3+} \) exhibit a magnetic moment of approximately \( 5.92 \, \mu_B \).
Step 3: Conclusion.
\( \text{Mn}^{2+} \) and \( \text{Fe}^{3+} \) possess identical magnetic moments due to having 5 unpaired electrons each. Thus, the correct selection is \( \mathbf{(4)} \).