Cr2+
Mn2+
Fe2+
Co2+
The enthalpy of hydration (\(\Delta_{\text{hyd}}H\)) of metal ions is influenced by their charge and ionic radius. A higher charge and smaller ionic radius generally lead to a more negative (stronger) \(\Delta_{\text{hyd}}H\), due to stronger interaction with water molecules.
Let's evaluate the 3d-metal ions given:
The strength of hydration depends upon the size and electronic configuration of the ion:
Since Mn2+ with its half-filled d-orbital configuration is more stable, it gives the lowest enthalpy of hydration when compared to other metal ions with full or partially filled d-orbitals.
Conclusion: Among the given options, Mn2+ will give the lowest \(\Delta_{\text{hyd}}H\) when dissolved in water due to its stable half-filled electronic configuration, leading to weaker interactions with water molecules.
| Δhyd H (M+2) | |
|---|---|
| Cr | – 1925 |
| Mn | – 1862 |
| Fe | – 1560 |
| Co | – 1640 |
Mn2+ has the lowest ΔhydH.
So, the correct option is (B): Mn2+.




Enthalpy Change refers to the difference between the heat content of the initial and final state of the reaction. Change in enthalpy can prove to be of great importance to find whether the reaction is exothermic or endothermic.
dH = dU + d(PV)
The above equation can be written in the terms of initial and final states of the system which is defined below:
UF – UI = qP –p(VF – VI)
Or qP = (UF + pVF) – (UI + pVI)
Enthalpy (H) can be written as H= U + PV. Putting the value in the above equation, we obtained:
qP = HF – HI = ∆H
Hence, change in enthalpy ∆H = qP, referred to as the heat consumed at a constant pressure by the system. At constant pressure, we can also write,
∆H = ∆U + p∆V
To specify the standard enthalpy of any reaction, it is calculated when all the components participating in the reaction i.e., the reactants and the products are in their standard form. Therefore the standard enthalpy of reaction is the enthalpy change that occurs in a system when a matter is transformed by a chemical reaction under standard conditions.