Question

(a) Complete the following chemical equations for reactions :
`(i) MnO_(4(aq))^(-) + S_(2)O_(3(aq))^(2-) + H_(2)O_((l)) to `
(ii) `Cr_(2)O_(7 (aq))^(2-) + H_(2)S_((g)) + H_((aq))^(+) to `
(b) Give an explanation for each of the following observations :
(i) The gradual decrease in size (actinoid contraction) from element to element is greater among the actinoids than that among the lanthanoids (lanthanoid contraction.)
(ii) The greatest number of oxidation states are exhibited by the members in the middle of a transition series.
(iii) With the same d-orbitals configuration `(d^(4)) Cr^(2+)` ion is a reducing agent but `Mn^(3+)` ion is an oxidising agent .

Answer 1

(a) (i) `8Mn O_(4(aq))^(2-) + 3S_(2)O_(3(aq))^(2-) + 8H_(2)O(l) to 8MnO_(2(aq)) + 6SO_(4(aq))^(2-) + 2OH_((aq))^(-)`
(ii) `Cr_(2)O_(7 (aq))^(2-) + 3H_(2)S_((g)) + 8 H_((aq))^(+) to 2Cr_((aq))^(3+) + 7H_(2)O(l) + 3S(s)`
(b) (i) The actinoid contraction is similar to lanthanoid contraction , this contraction is due to poor shielding by 5 f electrons in the actinoids than that by 4 f electrons in the lanthanoids.
(ii) The highest oxidation in the middle of the transition series is due to the use of all 4s and 3d electrons.
(iii) `Cr^(2+)` has the configuration `d^(4)` and easily changes to `d^(3)` has half filled orbitals and hence stable . Therefore , `Cr^(2+)` is reducing . On the other hand , `Mn^(2+)` is more stable due to half filled `d^(5)` configuration and `Mn^(3+)` easily changes to `Mn^(2+)` and hence is oxidising .