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Q4.4:
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Solution :

1. Theoretical Foundation of Stability in 3d Series

The stability of oxidation states in the first transition series ($Sc$ to $Zn$) is governed by the electronic configuration of the metal ion, specifically the stability associated with $d^0$, $d^5$, and $d^{10}$ configurations. These configurations represent empty, half-filled, and completely filled subshells, respectively, which exhibit enhanced stability due to exchange energy and symmetry [Per Hund’s Rule and Aufbau Principle].

2. Comparative Analysis of Stability

The transition elements exhibit variable oxidation states due to the participation of both $(n-1)d$ and $ns$ electrons in bonding. However, the stability of a specific state is determined by the energy required to reach that configuration versus the hydration or lattice energy released upon formation.

Element Configuration ($M^{2+}$) Reason for Stability/Instability
$Sc$ ($Z=21$) $[Ar] 3d^1$ $Sc^{3+}$ is highly stable ($[Ar] 3d^0$) as it attains inert gas configuration.
$Mn$ ($Z=25$) $[Ar] 3d^5$ $Mn^{2+}$ is exceptionally stable due to half-filled $d$-subshell stability.
$Zn$ ($Z=30$) $[Ar] 3d^{10}$ $Zn^{2+}$ is stable due to the completely filled $3d$ subshell.

3. Illustrative Case Studies

A. The Case of Manganese ($Mn$)

In $Mn$ ($3d^5 4s^2$), the $+2$ oxidation state is remarkably stable. The $Mn^{2+}$ ion possesses a $3d^5$ configuration, which is spherically symmetric and energetically favored. In contrast, the $Mn^{3+}$ ($3d^4$) state is less stable and acts as an oxidizing agent, attempting to gain an electron to revert to the more stable $3d^5$ state [Per Crystal Field Theory stabilization concepts].

B. The Case of Scandium ($Sc$)

$Sc$ ($3d^1 4s^2$) predominantly exhibits the $+3$ oxidation state. By losing three electrons, the ion reaches the stable $[Ar]$ configuration ($d^0$), which is iso-electronic with Argon. Any oxidation state lower than $+3$ would leave a valence electron that is easily removed [Per Ionization Enthalpy trends].

4. Graphical Representation of Energy Stability

Oxidation States (Increasing) Stability Peak Stability at $d^5$ or $d^{10}$

5. Conclusion

The stability of oxidation states is not merely a function of losing valence electrons but is dictated by the quantum mechanical stability of the resulting $d$-electron count. Elements with $d^0, d^5,$ and $d^{10}$ configurations demonstrate significantly higher stability in their respective ions because these arrangements minimize inter-electronic repulsion and maximize exchange energy.

Final Solution: The electronic configuration is the primary determinant of stability; ions achieving $3d^0$, $3d^5$, or $3d^{10}$ configurations exhibit maximum stability due to the symmetry of the electronic distribution, while deviations from these configurations result in ions with higher reactivity and lower stability.


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