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Why is Cr^(2+) reducing and Mn^(3+) oxid...

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

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`Cr^(2+)` is reducing as its configuration changes from `d^(4)` to `d^(3)` the having half filled `t_(2)` g level. On the other hand , the change from `Mn^(2+)` to ` Mn^(3+)` result in the half filled `d^(5)` configuration which has extra stability .
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Assertion : Cr^(2+) is reducing and Mn^(3+) is oxidising. Reason : Cr^(2+) and Mn^(3+) have d^4 configuration.

The ion Cr^(2+) is redusing agent while that of Mn^(3+) is an oxidising agent though both have 3d^(4) configuration. This is because

Assertion :- Cr^(+2) is a reducing agent and Mn^(+3) is oxidising agent. Reason :- Mn^(+3) has d^(5) configuration.

How would you account for the following: (i) Cr^(2+) is reducing in nature while with the same d-orbital configuration (d^(4)) Mn^(3+) is an oxidising agent (ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation state occurs in the middle of the series.

How would you account for the following? (i) Cr^(2+) is reducing in nature while with the same d-orbital configuration (d^14) Mn^(3+) is an oxidising agent. (ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series. (iii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than for the 3d series.