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Q4.21(i):
How would you account for the following: (i) Of the \(d^{4}\) species, \(Cr^{2+}\) is strongly reducing while manganese(III) is strongly oxidising.

Solution :

Step 1: Electronic Configuration and Stability Analysis

To analyze the redox behavior of $Cr^{2+}$ and $Mn^{3+}$, we must first establish their electronic configurations based on the Aufbau principle and the relative stability of $d$-orbital occupancy.

  • For $Cr^{2+}$ (Atomic Number $Z=24$): The neutral Cr atom is $[Ar] 3d^{5} 4s^{1}$. Removing two electrons yields: $Cr^{2+} = [Ar] 3d^{4}$.
  • For $Mn^{3+}$ (Atomic Number $Z=25$): The neutral Mn atom is $[Ar] 3d^{5} 4s^{2}$. Removing three electrons yields: $Mn^{3+} = [Ar] 3d^{4}$.
[Per Hund’s Rule and the Aufbau Principle, the $d^{4}$ configuration is common to both, yet their chemical potentials differ significantly due to the nuclear charge variation.]

Step 2: Thermodynamic Analysis of $Cr^{2+}$ (Reducing Behavior)

The reducing nature of a species is determined by its tendency to lose an electron to form a more stable state. $Cr^{2+}$ undergoes oxidation to $Cr^{3+}$.

Reaction: $Cr^{2+} (aq) \rightarrow Cr^{3+} (aq) + e^{-}$

The resulting configuration of $Cr^{3+}$ is $[Ar] 3d^{3}$. According to Crystal Field Theory in an octahedral field, the $t_{2g}$ orbitals are lower in energy than the $e_{g}$ orbitals. In the $3d^{3}$ configuration, the three electrons occupy the three $t_{2g}$ orbitals, satisfying the half-filled $t_{2g}$ subshell stability criterion. This configuration is exceptionally stable in aqueous media. Consequently, the high negative value of the standard electrode potential ($E^{\circ} (Cr^{3+}/Cr^{2+}) = -0.41 V$) confirms that $Cr^{2+}$ is a strong reducing agent.

Step 3: Thermodynamic Analysis of $Mn^{3+}$ (Oxidizing Behavior)

The oxidizing nature is defined by the tendency to gain an electron to reach a more stable state. $Mn^{3+}$ undergoes reduction to $Mn^{2+}$.

Reaction: $Mn^{3+} (aq) + e^{-} \rightarrow Mn^{2+} (aq)$

The resulting configuration of $Mn^{2+}$ is $[Ar] 3d^{5}$. This is a spherically symmetric, half-filled $d$-subshell ($d^{5}$), which confers high exchange energy and thermodynamic stability. [Reference: Stability of half-filled subshells]. Because the product ($Mn^{2+}$) is significantly more stable than the reactant ($Mn^{3+}$), the reduction potential ($E^{\circ} (Mn^{3+}/Mn^{2+}) = +1.57 V$) is high, making $Mn^{3+}$ a strong oxidizing agent.

Visual Representation of Orbital Occupancy

Cr2+ (3d4) -> Cr3+ (3d3) ↑ ↑ ↑ ↑ → ↑ ↑ ↑ Mn3+ (3d4) -> Mn2+ (3d5) ↑ ↑ ↑ ↑ → ↑ ↑ ↑ ↑ ↑

Comparative Summary

Species Process Driving Force
$Cr^{2+}$ Oxidation ($d^4 \rightarrow d^3$) Formation of stable half-filled $t_{2g}^3$ configuration.
$Mn^{3+}$ Reduction ($d^4 \rightarrow d^5$) Formation of highly stable half-filled $d^5$ configuration.

Final Solution: $Cr^{2+}$ is strongly reducing because it loses an electron to attain the stable $t_{2g}^3$ electronic configuration, whereas $Mn^{3+}$ is strongly oxidizing because it gains an electron to attain the exceptionally stable $d^5$ half-filled configuration.


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