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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: Analyzing Electronic Configurations of Mn and Fe

To understand the relative stability of the divalent cations ($Mn^{2+}$ and $Fe^{2+}$) toward oxidation to their trivalent states ($Mn^{3+}$ and $Fe^{3+}$), we must first determine their ground-state valence shell electronic configurations [Per the Aufbau Principle and Hund's Rule of Maximum Multiplicity].

  • Manganese (Mn): Atomic number $Z = 25$. Ground state: $[Ar] 3d^5 4s^2$.
    • $Mn^{2+}$ ion: $[Ar] 3d^5$.
    • $Mn^{3+}$ ion: $[Ar] 3d^4$.
  • Iron (Fe): Atomic number $Z = 26$. Ground state: $[Ar] 3d^6 4s^2$.
    • $Fe^{2+}$ ion: $[Ar] 3d^6$.
    • $Fe^{3+}$ ion: $[Ar] 3d^5$.

Step 2: Assessing Stability via Exchange Energy and Symmetry

The stability of d-block ions is significantly influenced by the symmetry of the d-orbital subshell and the exchange energy associated with electronic arrangements [Based on Crystal Field Theory and Hund's empirical rules].

Ion Electronic Configuration Stability Context
$Mn^{2+}$ $d^5$ Half-filled subshell ($t_{2g}^3 e_g^2$); high exchange energy.
$Mn^{3+}$ $d^4$ Less stable; lacks the half-filled configuration.
$Fe^{2+}$ $d^6$ Configuration involves one paired electron, decreasing total exchange energy.
$Fe^{3+}$ $d^5$ Half-filled subshell; highly stable.

Step 3: Evaluating Oxidation Potentials

Oxidation involves the loss of an electron ($M^{2+} \rightarrow M^{3+} + e^-$). This process is favorable if the resulting configuration is more stable than the precursor.

For Manganese: The transition from $Mn^{2+}$ ($3d^5$) to $Mn^{3+}$ ($3d^4$) involves the destruction of a highly stable half-filled $d$-subshell. This requires a very high ionization enthalpy and is energetically unfavorable. Thus, $Mn^{2+}$ resists oxidation to $Mn^{3+}$. [Per the principle of Extra Stability of Half-Filled Orbitals].

For Iron: The transition from $Fe^{2+}$ ($3d^6$) to $Fe^{3+}$ ($3d^5$) involves the loss of one electron to achieve the highly stable half-filled $d^5$ configuration. This process is energetically favorable, explaining why $Fe^{2+}$ compounds are readily oxidized to $Fe^{3+}$ in the presence of air.

Visual Representation of d-Orbital Occupancy

Mn2+ (3d5) -> Mn3+ (3d4) ↑ ↑ ↑ ↑ ↑ -> ↑ ↑ ↑ ↑ Fe2+ (3d6) -> Fe3+ (3d5) ↑↓ ↑ ↑ ↑ ↑ -> ↑ ↑ ↑ ↑ ↑

Final Solution: $Mn^{2+}$ is more stable than $Fe^{2+}$ towards oxidation because the $Mn^{2+}$ ion possesses a stable half-filled $3d^5$ configuration, whereas $Fe^{2+}$ has a $3d^6$ configuration, which readily loses an electron to attain the stable half-filled $3d^5$ state in $Fe^{3+}$.


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


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