Which of the following 3d-metal ion will give the lowest enthalpy of hydration $$\Delta_{hyd}H$$ when dissolved in water?

We are asked which 3d-metal ion among $$Cr^{2+}$$, $$Mn^{2+}$$, $$Fe^{2+}$$, and $$Co^{2+}$$ has the lowest (least negative) enthalpy of hydration.

The enthalpy of hydration of transition metal ions generally increases (becomes more negative) across the 3d series due to increasing nuclear charge and decreasing ionic radius. However, this trend is not smooth — it follows a double-humped curve explained by Crystal Field Stabilisation Energy (CFSE).

In an octahedral field of water ligands, each d-electron in the $$t_{2g}$$ set contributes $$-0.4\Delta_o$$ and each electron in the $$e_g$$ set contributes $$+0.6\Delta_o$$ to the CFSE. The electronic configurations and CFSE values for the given ions are:

Case 1: $$Cr^{2+}$$ has the configuration $$d^4$$ ($$t_{2g}^3 e_g^1$$), giving CFSE = $$3(-0.4) + 1(0.6) = -0.6\Delta_o$$.

Case 2: $$Mn^{2+}$$ has the configuration $$d^5$$ ($$t_{2g}^3 e_g^2$$), giving CFSE = $$3(-0.4) + 2(0.6) = 0$$. Since water is a weak field ligand, the high-spin configuration is adopted, and $$Mn^{2+}$$ has zero CFSE.

Case 3: $$Fe^{2+}$$ has the configuration $$d^6$$ ($$t_{2g}^4 e_g^2$$), giving CFSE = $$4(-0.4) + 2(0.6) = -0.4\Delta_o$$.

Case 4: $$Co^{2+}$$ has the configuration $$d^7$$ ($$t_{2g}^5 e_g^2$$), giving CFSE = $$5(-0.4) + 2(0.6) = -0.8\Delta_o$$.

Since $$Mn^{2+}$$ has zero CFSE, it receives no extra stabilisation from crystal field effects during hydration. This means it falls on the baseline of the double-humped curve and has the lowest (least negative) enthalpy of hydration among the given ions.

Hence, the correct answer is Option B.