(a) Complete the following chemical equations:
(i) ` Cr_(2) O_(7)^(2-) (aq) + H_(2) S(g) + H^(+) (aq) to `
(ii) ` Cu^(2+)(aq) + I^(-) (aq) to `
(b) How would you account for the following :
(i) The oxidising power of oxoanions are in the order `VO_(2)^(+) lt Cr_(2)O_(7)^(2-) lt Mn O_(4)^(-) `
(ii) The third ionization enthalpy of manganess (Z = 25) is exceptionally high.
(iii) `Cr^(2+)` is a stronger reducing agent than `Fe^(2+)`.

Explanation

(a) (i) `Cr_(2) O_(7)^(2-) (aq) + 3H_(2)S(g) + 8H^(+)(aq) rarr 2Cr^(2+) (aq) +7H_(2)O(l) + 3S(s). ` (ii) `2Cu^(2+)(aq) +4I^(-)(aq) rarr Cu_(2)I_(2)(s) +I_(2)(g)`
(b) (i)This is due to increasing stability of the lower species to which they are reduced.
(ii) Because 3rd electron has to be removed from stable half-filled3d-orbitals `(._(25)Mn= 3d^(5) 4s^(2))`.
(iii)`E^(@) ( Cr^(3+) //Cr^(2+))-ve(0.41V) ,E^(@)(Fe^(3+) //Fe^(2+))= +ve( +0.77V)`
Hence, `Cr^(2+)` is easily oxidized to `Cr^(3+)` but `Fe^(2+)` cannot be easily oxidized to `Fe^(3+)`]
or (a) (i) `8 MnO_(4)^(-)(aq) + 3S_(2)O_(3)^(2-) (aq) + H_(2)O(l) rarr 8 MnO_(2)(s) +6 SO_(4)^(2-) + 2OH^(-) (aq) `
(ii) `Cr_(2) O_(7)^(2-) (aq) +6Fe^(2+)(aq)+ 14H^(+)(aq) rarr 2Cr^(3+)(aq) + 6Fe^(3+) (aq) + 7H_(2)O(l)`
(b) (i) Becaue they have completely filled 4f subshell. (II) `Mn^(2+)` has maximum number of unpaired electrons.
(iii) Because `Cu^(+)` in aqueous solution undergoes disproportionation `: 2 Cu^(+)(aq) rarr Cu^(2+)(aq) +Cu(s)`
(as `E^(@)` value is favourable for this reaction ) ]