Question

(a) 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

 Step 1: Chromium in the \( Cr^{2+} \) state has an electronic configuration of \( [Ar] 3d^4 \), which is relatively unstable and easily oxidizes to the more stable \( Cr^{3+} \) configuration, \( [Ar] 3d^5 \). This makes \( Cr^{2+} \) a strong reducing agent because it readily loses electrons. On the other hand, \( Mn^{3+} \) has an electronic configuration of \( [Ar] 3d^4 \), which is also unstable. It is prone to gaining electrons to become \( Mn^{2+} \), which has a more stable \( [Ar] 3d^5 \) configuration. Thus, \( Mn^{3+} \) is a strong oxidizing agent because it readily accepts electrons.

 Step 2: Thus, the differences in the electronic configurations of \( Cr^{2+} \) and \( Mn^{3+} \) explain why \( Cr^{2+} \) is a strong reducing agent and \( Mn^{3+} \) is a strong oxidizing agent. \bigskip