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Q4.2:
Why are Mn^{2+} compounds more stable than Fe^{2+} towards oxidation to their +3 state?

Solution :

Step 1: Electronic Configuration of Mn²⁺ and Fe²⁺

To determine the relative stability of these ions towards oxidation, we must first analyze their ground-state electronic configurations based on the Aufbau principle and Hund’s rule of maximum multiplicity.

  • Manganese ($Z=25$): The neutral atom configuration is $[Ar] 3d^5 4s^2$. Upon forming the $Mn^{2+}$ ion, the two electrons are removed from the $4s$ orbital.
    Configuration of $Mn^{2+}$: $[Ar] 3d^5$
  • Iron ($Z=26$): The neutral atom configuration is $[Ar] 3d^6 4s^2$. Upon forming the $Fe^{2+}$ ion, the two electrons are removed from the $4s$ orbital.
    Configuration of $Fe^{2+}$: $[Ar] 3d^6$

Step 2: Stability Analysis via Exchange Energy

The stability of $d$-orbital configurations is governed by the symmetry and filling of the $d$-subshell. [Per Hund’s Rule and the principles of Exchange Energy]:

  • The $Mn^{2+}$ ion possesses a $3d^5$ configuration. This is a half-filled $d$-subshell. According to symmetry principles, half-filled and fully-filled shells possess extra stability due to the symmetrical distribution of electron density and maximum exchange energy.
  • The $Fe^{2+}$ ion possesses a $3d^6$ configuration. Removal of one electron ($Fe^{2+} \rightarrow Fe^{3+} + e^-$) results in a $3d^5$ configuration. Since $3d^5$ is highly stable, there is a strong thermodynamic driving force for $Fe^{2+}$ to lose an electron and achieve the $3d^5$ state.

Step 3: Evaluation of Oxidation Potentials

The oxidation process involves the transformation of the divalent ion to the trivalent state:

Ion Configuration Result of Oxidation ($+2 \rightarrow +3$)
$Mn^{2+}$ $3d^5$ (Stable) $3d^4$ (Less stable)
$Fe^{2+}$ $3d^6$ (Unstable) $3d^5$ (Highly stable)

The conversion of $Mn^{2+} (3d^5)$ to $Mn^{3+} (3d^4)$ involves the disruption of the stable half-filled configuration, which requires a significant amount of energy. Conversely, the conversion of $Fe^{2+} (3d^6)$ to $Fe^{3+} (3d^5)$ is energetically favorable because it results in the formation of a stable, half-filled $d$-subshell.

Mn²⁺ (3d⁵) : Stable Fe²⁺ (3d⁶) : Unstable → Fe³⁺ (3d⁵)

Step 4: Conclusion

Because $Mn^{2+}$ has a stable half-filled $d$-orbital configuration, it resists oxidation as any change would decrease its stability. $Fe^{2+}$, having an extra electron beyond the half-filled state, readily loses it to attain the stable half-filled configuration. Therefore, $Mn^{2+}$ is more stable towards oxidation than $Fe^{2+}$.

Final Solution: $Mn^{2+}$ compounds are more stable because the $Mn^{2+}$ ion has a $3d^5$ electronic configuration, which is a half-filled subshell providing extra stability. Oxidation to $Mn^{3+}$ would disrupt this configuration, whereas oxidation of $Fe^{2+}$ ($3d^6$) to $Fe^{3+}$ ($3d^5$) allows the ion to achieve this stable half-filled state.


More Questions from Class 12 Chemistry The d- and f-Block Elements Exercises


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