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 states occurs in the middle of the series.

Explanation

(i) `Cr^(2+)` has d-orbital configuration `3d^(4)`. It attains a more stable configuration `3d^(3)` by donating an electron. Thus, it is reducing in nature. On the other hand, `Mn^(3+)` has d-orbital configuration as `3d^(4)`. It attains a more stable `3d^(5)` configuration by gaining one electron. Thus, it acts as 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. This can be understood from the outer electronic configuration of the metals of the transition series. For example, in the first row transition metals, from Sc to Zn, the number of electrons in the d-orbital increases from 1 to 10. Thus from the first element to the middle element, there will be 1 to 5 unpaired electrons all of which are capable of bonding. From the middle to the last element, pairing of electrons will start and the number of unpaired electrons capable of bonding will decrease. Thus, it is in the middle of series that the metal exhibits greatest number of oxidation states.