Question

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

Hint:

Elements with unstable electronic configurations in certain oxidation states (like \( Cr^{2+} \) or \( Mn^{3+} \)) are more likely to either gain or lose electrons to reach more stable configurations.

Answer 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 instability makes \( Cr^{2+} \) a strong reducing agent because it readily loses electrons. In contrast, \( Mn^{3+} \) has an electronic configuration of \( [Ar] 3d^4 \), which is also unstable. It tends to gain electrons to achieve the more stable \( Mn^{2+} \) configuration, \( [Ar] 3d^5 \). Hence, \( Mn^{3+} \) acts as a strong oxidizing agent, readily accepting electrons. \bigskip Step 2: 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