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Q4.3:
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Solution :

Given Theoretical Foundation

The stability of the $+2$ oxidation state in the first row of transition elements ($3d$-series: $Sc$ to $Zn$) is governed by the electronic configuration of the metal ions ($M^{2+}$) and the corresponding energy required for their formation ($I.E_1 + I.E_2$) relative to the hydration enthalpy or lattice energy.

Step 1: Analysis of Electronic Configurations

As the atomic number increases from $Sc$ ($Z=21$) to $Mn$ ($Z=25$), the $M^{2+}$ ions are formed by the removal of two $4s$ electrons. The resulting configurations are as follows:

Element Atomic Number ($Z$) $M^{2+}$ Configuration
$Sc$ 21 $3d^1$
$Ti$ 22 $3d^2$
$V$ 23 $3d^3$
$Cr$ 24 $3d^4$
$Mn$ 25 $3d^5$

Step 2: Thermodynamic Stability Analysis

The stability of the $+2$ state is determined by the total energy change involved in the transition: $M(g) \rightarrow M^{2+}(aq)$. This involves the sum of the first and second ionization enthalpies ($\Delta_iH_1 + \Delta_iH_2$) and the hydration enthalpy ($\Delta_{hyd}H$).

[Per the Hess's Law and Born-Haber cycle logic]: The formation of $M^{2+}$ ions is endothermic, but the energy investment is offset by the hydration enthalpy in aqueous solutions. As we progress from $Sc^{2+}$ to $Mn^{2+}$, the nuclear charge increases, which decreases the size of the ions.

Step 3: The Role of Exchange Energy and Half-Filled Stability

As we move towards $Mn^{2+}$, the $3d$ subshell becomes progressively filled. Specifically, $Mn^{2+}$ achieves a configuration of $3d^5$.

  • Electronic Stability: The $d^5$ configuration represents a half-filled subshell, which confers extra stability due to maximum exchange energy and symmetrical distribution of electrons [Per Hund’s Rule and Pauli’s Exclusion Principle].
  • Effective Nuclear Charge: The increase in effective nuclear charge from $Sc$ to $Mn$ makes the removal of electrons progressively more difficult; however, the stability of the resulting $M^{2+}$ ion is enhanced significantly upon reaching the $d^5$ state.

Step 4: Visualizing the 3d-Orbital Filling

Sc²⁺ V²⁺ Mn²⁺ Stability

Conclusion

The stability of the $+2$ state increases because the ionic radii decrease and the enthalpy of hydration becomes more exothermic with increasing atomic number. Crucially, the $Mn^{2+}$ ion represents the pinnacle of this trend within the first half of the series due to the exceptional stability of the $3d^5$ configuration, which results from the half-filled $d$-orbital arrangement.

Final Solution: The $+2$ oxidation state becomes increasingly stable due to the increase in effective nuclear charge and the attainment of a stable, half-filled $3d^5$ electronic configuration in the case of Manganese ($Mn^{2+}$), which possesses high exchange energy and symmetrical electronic distribution.


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