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According to the reduction potentials given above, Mn3+ is a much stronger oxidizing agent than Fe3+. However, one would expect that Fe3+ with its larger nuclear charge and slightly smaller size should gain electrons more easily than Mn3+. The Mn3+ ion has a 3d4 electron configuration, and the gain of an electron to form Mn2+ gives the relatively stable 3d5 configuration. However, to convert Fe3+ to Fe2+ requires a change from a stable 3d5 configuration to a less stable 3d6 configuration best accounts for this behavior.
The statement "Mn3+ is more electropositive than Fe3+, so it should attract electrons more strongly" is not correct because an electropositive species should not have a greater affinity for electrons than an electronegative one. And even if the statement "Fe3+ has a higher ionization energy than Mn3+, so it attracts electrons more strongly than Mn3+" were a true statement, it doesn't answer the question ("Why does Mn3+ attract electrons more strongly than Fe3+ does?"). The statement "the Mn3+ ion has a 3d4 electron configuration, and the gain of an electron to form Mn2+ gives the relatively stable 3d5 configuration. However, to convert Fe3+ to Fe2+ requires a change from a stable 3d5 configuration to a less stable 3d6 configuration" is correct, and the statement "the Mn3+ ion has a 3d4 electron configuration, and the gain of an electron to form Mn2+ gives the less stable 3d5 configuration. However, to convert Fe3+ to Fe2+ requires a change from a 3d5 configuration to a 3d6 configuration, which is more stable since it is closer to a filled 3d10 configuration" is false: The 3d subshell can hold 10 electrons (2 electrons in each of the five equivalent d orbitals), and stability is associated with half-filled (and filled) sets of equivalent orbitals. Therefore a 3d5 electron configuration (that is, a half-filled 3d subshell) is more stable than a 3d4 or a 3d6 configuration.