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 n

1.	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.
Answer» Solution :(i) `Cr^(2+)` is REDUCTION as its configuration CHANGES from `d^(5)` to `d^(3)`, the latter having half filled `t_(2g)` level whereas oxidation of MN from `Mn^(2+)` to `Mn^(3+)` results in half filled do configuration, which is more stable. (ii) In a transition metals series the oxidation state first increases and then decreases, At the middle it is maximum due to GREATER number of unpaired electron in (n-1)d and ns-orbitals.