a) Complete the following chemical equations for reactions :
(i) ${\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+{\mathrm{S}}_2{\mathrm{O}}_3^{2-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)⟶$
(ii) ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)⟶$

(B)Give an explanation for each of the following observations : 
(i) The gradual decrease ‘n’ 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-orbital configuration d⁴, Cr²⁺ ion is a reducing agent but Mn³⁺ ion is an oxidising agent.

(A)


(B)(i) The actinoid contraction is greater than lanthanoid contraction due to poorer shielding of 5f electrons as they are extended in space beyond 6s and 6p orbitals whereas 4f orbitals are buried deep inside the atom. 

(ii) Due to presence of more unpaired electrons and use of all 4s and 3d electrons in the middle of series. 

(iii) Cr²⁺ has the configuration 3d⁴ which easily changes to d³ due to stable half-filled t₂g orbitals. Therefore Cr²⁺ is reducing agent. While Mn²⁺ has stable half-filled d⁵ configuration. Hence Mn³⁺ easily changes to Mn²⁺ and acts as oxidising agent.