Question

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 1

Correct Answer - (i) It is because `Cr^(3+)` loses electron to become `Cr^(3+)` which is more stable due to half filled `t_(2g)` orbitals. Whereas `Mn^(3+)` will gain electrons to become `Mn^(2+)` which is more stable due to half filled d-orbitals.
(ii) It is due to large of unpaired electrons in d-orbitals in middle of the series.