InterviewSolution
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InterviewSolution
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Why is `Cr^(3+)` reducing and `Mn^(3+)` oxidisin when both have `d^(4)` configuration ? |
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Answer» `Cr^(2+)` has the configuration `3d^(4)` . It can lose electron to form`Cr^(3+)` which has the stable `3d^(3)` configuration ( as it has half- filled `t_(2g)` level - explained in unit). Hence, it is reducing . On the other hand, `Mn^(3+)` also has `3d^(4)` configuration but it can gain electron to form `Mn^(2+)` which has stable `3d^(5)` configuration ( as it is exactly half- filled ). Hence, it is oxidizing. Alternatively. `E^(@)` value for `Cr^(3+) // Cr^(2+) ` is negative `( - 0.41V)` whereas `E^(@) `value for `Mn^(3+) // Mn^(3+)` is positive `( + 1.57V)` . Hence, `Cr^(2+)` ion can easily undergo oxidation to give `Cr^(3+)` ion and , therefore , acts as strong reducing agent where as `Mn^(3+)` can easily undergo reduction to give`Mn^(2+)` and hence acts as oxidizing agent. |
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