Why is `Cr^(2+)` reducing and `Mn^(3+)` oxidising when both have `d^(4 – InterviewSolution

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1.	Why is `Cr^(2+)` reducing and `Mn^(3+)` oxidising when both have `d^(4)` configuration ?
Answer» `Cr^(2+)` is reducing as its configuration changes from `d^(4)" to "d^(3)` , the latter having a half-filled `t_(2g)` level. On the other hand, the change from `Mn^(2+)" to "Mn^(3+)` results in the half-filled `(d^(5))` configuration which has extra stability.

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