Question

Why is \( Cr^{2+} \) strongly reducing while \( Mn^{3+} \) is strongly oxidizing?

Hint:

Elements with an electronic configuration that leads to an unstable oxidation state (such as \( Cr^{2+} \) or \( Mn^{3+} \)) are more likely to undergo reduction or oxidation to attain more stable states.

Answer 1

To solve the problem, we need to explain why \( Cr^{2+} \) is strongly reducing while \( Mn^{3+} \) is strongly oxidizing.

1. Analyze \( Cr^{2+} \) as a Reducing Agent:
A species is reducing if it easily loses electrons (is oxidized). \( Cr^{2+} \) has the electronic configuration \( [Ar] 3d^4 \). When oxidized to \( Cr^{3+} \), it becomes \( [Ar] 3d^3 \), which is a half-filled \( t_{2g}^3 \) configuration in octahedral complexes, highly stable due to exchange energy and symmetry. The standard reduction potential \( E^0 (Cr^{3+}/Cr^{2+}) = -0.41 \, \text{V} \) is negative, indicating \( Cr^{2+} \rightarrow Cr^{3+} + e^- \) is favorable, making \( Cr^{2+} \) a strong reducing agent.

2. Analyze \( Mn^{3+} \) as an Oxidizing Agent:
A species is oxidizing if it easily gains electrons (is reduced). \( Mn^{3+} \) has the configuration \( [Ar] 3d^4 \). When reduced to \( Mn^{2+} \), it becomes \( [Ar] 3d^5 \), a half-filled configuration, which is very stable. The standard reduction potential \( E^0 (Mn^{3+}/Mn^{2+}) = +1.51 \, \text{V} \) is highly positive, indicating \( Mn^{3+} + e^- \rightarrow Mn^{2+} \) is favorable, making \( Mn^{3+} \) a strong oxidizing agent.

Final Answer:
\( Cr^{2+} \) is strongly reducing because it oxidizes to the stable \( Cr^{3+} \) (\( 3d^3 \), half-filled \( t_{2g} \)), with a negative reduction potential (\( E^0 = -0.41 \, \text{V} \)). \( Mn^{3+} \) is strongly oxidizing because it reduces to the stable \( Mn^{2+} \) (\( 3d^5 \), half-filled), with a high positive reduction potential (\( E^0 = +1.51 \, \text{V} \)).