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

Chromium in the \( Cr^{2+} \) state exhibits the electronic configuration \( [Ar] 3d^4 \). This configuration is inherently unstable and readily undergoes oxidation to the more stable \( Cr^{3+} \) state, characterized by the electronic configuration \( [Ar] 3d^5 \). Consequently, \( Cr^{2+} \) functions as a potent reducing agent due to its propensity to lose electrons. In contrast, \( Mn^{3+} \) possesses the electronic configuration \( [Ar] 3d^4 \), which is also unstable. This ion tends to gain electrons to achieve the more stable \( [Ar] 3d^5 \) configuration of \( Mn^{2+} \). Therefore, \( Mn^{3+} \) acts as a powerful oxidizing agent because it readily accepts electrons.

The disparity in electronic configurations between \( Cr^{2+} \) and \( Mn^{3+} \) fundamentally accounts for \( Cr^{2+} \) acting as a strong reducing agent and \( Mn^{3+} \) acting as a strong oxidizing agent.