(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+)`.

Answer

(a) (i) `Cr_(2)O_(7)^(2-)(aq)+3H_(2)(S)(g)+8H^(+)(aq) to 2Cr^(3+)+3S+7H_(2)O`
(ii) `2Cu^(2+)(aq)+2I^(-)(aq) to 2Cu^(+)(aq)+I_(2)(S)`
(b) (i) It is because `V` in lower oxidation state is less stable than `Cr` which is less stable than `Mn`. That is why `MrO_(4)^(-)` is best oxidising agent and `VO_(2)^(+)` is least.
(ii) ` Mn(25)` has electronic configuration `[Ar]4S^(2) 3d^(5)`,electronic configuration of `Mn^(2+) ` is `[Ar] 4s^(@) 3d^(5)`. After losing `2` electrons, it has half filled d-orbital, which is more stable that is why `Mn^(2+)` has exceptionally high third ionization energy .i.e., the energy required to remove third electrons is very high.
(iii) it is because in `Cr^(3+), d^(3)` is more stable in aqueous solution than `Fe^(3+)(d^(5))` i.e., `Cr^(3+)` is more stable than `Fe`.
`OR` (a) (i) `8MnO_(4)^(-)(aq)+3S_(2)O_(3)^(2-)(aq)+H_(2)O(I) to 8MnO_(2)(S)+2OH^(-)(aq)+6SO_(4)^(2-)(aq)`
(ii `Cr_(2)O_(7)^(2-)(aq)+6Fe^(2+)(aq)14H^(+)(aq) to 2Cr^(3+)(aq)+6Fe^(3+)(aq)+7H_(2)O(I)`
(b) (i) it is due to presence of unpaired electrons they cannot undergo f-f transition .
(ii) it is because in `Cr^(3+)` gets oxidised to `Cu^(2+)` which is more stable and its hydration energy can overcome second ionsation energy.