Question

How would you account for the following ?
(i) With the same d-orbital configuration `(d^(4)) Cr^(2+)` is reducting agent while `Mn^(3+)` is an oxidizing agent.
(ii) The actionoids exhibits a larger numbe of oxidation states than the corresponding members in the lanthanoid series.
(iii) Most of the transition metal ions exhibit characteristic in colours in aqueous solutions.

Answer 1

`Cr^(2+)` can lose -to form `Cr^(3+)` which has stable `3d^(3)` configuration . Hence reducing where as `Mn^(3+)` can gain electron to form `Mn^(2+)` which has stable 3ds configuration hence oxidizing .
(iii) An increase from the first (3d) to the second (4d) series of the elements but the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series . This phenomenon is associated with the intervention of the 4f orbitals which must filled before the (5d) series are of elements began . The filling of 4f before (5d) orbitals results in regular decreases in atomic radii called lanthanoid contraction which esssentially compensates for the expected increase in atomic size and atomic number . The net result of the Lanthanoid contraction is that the second and the third d series exhibit similar radiii.