How would you account for the following:

(i) Cr²⁺ is reducing in nature while with the same d-orbital configuration (d⁴) Mn³⁺ is an oxidizing 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.

(i) We know that the stability of a half-filled orbital (like d⁵) or a full filled orbital (like d¹⁰) is always higher than the other configurations. So, an element or an ion would always prefer the half-filled or full-filled configuration to gain stability, thus they always try to go somehow to that configuration.

Now, the electronic configuration of Cr is [Ar]3d⁵4s¹

When Cr loses 2 electrons to become Cr²⁺, the electronic configuration becomes [Ar]3d⁴. But it always tries to gain the more stable Cr³⁺ configuration state by losing another electron and get oxidised to form d³ due to the stable t_2g configuration.

Whereas the electronic configuration of Mn is [Ar]3d⁵4s²

Now, when Mn loses 3 electrons to form Mn³⁺, d⁴ configuration is achieved. So, it would try to gain an electron and get reduced to the more stable d⁵ configuration (Mn²⁺).

So, Cr²⁺ is reducing in nature while Mn³⁺ is an oxidizing agent.

(ii) It is obvious that in the middle of a transition series, due to the presence of more unpaired electrons and more number of partially filled orbitals the metals exhibit the greatest number of oxidation states.