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

Explanation

`Cr(Z=24)=3d^54s^1`
`Cr^(2+)=3d^4`
`Cr^(2+)(3d^4)tounderset(t_(2g)^(3))(C)r^3+(3d^3)+e^(c-)`
`Cr^(2+)` undergoes oxidation and hence is reducing as its configuration changes from `d^4` to `d^3`, the latter having a half filled `t_(2g)^3` level.
On the other hand, the reduction of `Mn^(3+)` to `Mn^(2+)` results in the half-filled `(3d^5)` configuration which has extra stability,
And thus is oxidising.
`Mn(Z=25)=3d^54s^2,Mn^(2+)=3d^5,Mn^(3+)=3d^4`
`Mn^(3+)(3d^4)+e^(c-)toMn^(2+)(3d^5)`
[Undergo reduction and thus is oxidising]