1. Key idea: hydration enthalpy depends mainly on charge density

For ions with the same charge (here all are $M^{2+}$), the magnitude of hydration enthalpy depends largely on:

• ionic size
• charge density
• crystal field stabilization in water

More negative (more exothermic) hydration enthalpy is associated with smaller ions and/or extra stabilization.

The question asks for the lowest enthalpy of hydration. In this context for transition metal ions, this means the least negative / least exothermic hydration enthalpy.

2. Compare the given ions

The ions are:

• $\mathrm{Cr}^{2+} : 3d^4$
• $\mathrm{Mn}^{2+} : 3d^5$
• $\mathrm{Fe}^{2+} : 3d^6$
• $\mathrm{Co}^{2+} : 3d^7$

Across the period from $\mathrm{Cr}^{2+}$ to $\mathrm{Co}^{2+}$, ionic radius generally decreases, so hydration enthalpy should generally become more negative.

However, there is an important exception due to crystal field stabilization energy (CFSE) in octahedral aqua complexes.

3. Check CFSE for high-spin $M^{2+}$ ions in octahedral water complexes

Since water is a weak-field ligand, these are high-spin octahedral ions.

• $\mathrm{Cr}^{2+} (d^4)$: CFSE $=0.6\Delta_o$
• $\mathrm{Mn}^{2+} (d^5)$: CFSE $=0$
• $\mathrm{Fe}^{2+} (d^6)$: CFSE $=0.4\Delta_o$
• $\mathrm{Co}^{2+} (d^7)$: CFSE $=0.8\Delta_o$

Thus, $\mathrm{Mn}^{2+}$ has zero CFSE, so it gets no extra stabilization on hydration.

Therefore its hydration enthalpy is expected to be least negative, i.e. the lowest among these options.

4. Option-wise conclusion

• A: $\mathrm{Cr}^{2+}$ — has some CFSE, so not the lowest.
• B: $\mathrm{Mn}^{2+}$ — zero CFSE, so hydration enthalpy is lowest. ✅
• C: $\mathrm{Fe}^{2+}$ — has CFSE, so more negative than $\mathrm{Mn}^{2+}$.
• D: $\mathrm{Co}^{2+}$ — has even greater CFSE, so not the lowest.

5. Final answer

The ion with the lowest enthalpy of hydration is: $\boxed{\mathrm{Mn}^{2+}}$

So the correct option is B.