How would you account for the following ? (i) The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series. (ii) The E^(@) value for the Mn^(3+)//Mn^(2+) couple is much more positive than that for Cr^(3+)//Cr^(2+) couple or Fe^(3+)//Fe^(2+) couple. (iii) The highest oxidation state of a metal is exhibited in its oxide or fluoride.

How would you account for the following ? (i) The atomic radii of the

1.	How would you account for the following ? (i) The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series. (ii) The E^(@) value for the Mn^(3+)//Mn^(2+) couple is much more positive than that for Cr^(3+)//Cr^(2+) couple or Fe^(3+)//Fe^(2+) couple. (iii) The highest oxidation state of a metal is exhibited in its oxide or fluoride.
Answer» Solution :(i) This is due to intervention of 4f ORBITALS which must be filled before the 5d series of elements BEGIN. The filling of 4f before 5d results in a regular decrease in atomic radii called lanthanoid contraction, which compensates for the expected increase in atomic size with increasing atomic NUMBER. The net result is that second and third d series exhibit similar radii. (ii) More positive `E^(@)` value for `Mn^(3+)//Mn^(2+)` is due to particularly stable `Mn^(2+)(d^(5))` (half-filled) configuration. Low value for Fe is due to extrastability of `Fe^(3+)(d^(5))` and therefore lower stability of `Fe^(2+)`. Similarly `E^(@)` value for `Cr^(3+)//Cr^(2+)` is lower because of lower stability of `Cr^(2+)`. (iii) This is because oxygen and fluorine are most electronegative elements. They have the maximum capacity to WITHDRAW electrons from the metals to ENABLE them to exhibit the maximum oxidation state.