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Answer: 4. \(d^4\) to \(d^5\) configuration

Key idea: a high \(E^\circ\) for \(M^{3+}/M^{2+}\) means \(M^{3+}\) is easily reduced to a very stable \(M^{2+}\).

\(\mathrm{Mn^{3+}}\) is \(d^4\); on reduction it becomes \(\mathrm{Mn^{2+}}\) with the stable half-filled \(d^5\) configuration. This extra stability makes the couple strongly positive. \(\mathrm{Cr^{3+}}\) (\(d^3\)) goes to \(d^4\) and \(\mathrm{Fe^{3+}}\) (\(d^5\)) goes to \(d^6\), which give no such gain. Option 4.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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