(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 MnO_(4)^(-)`.
(ii) The third ionisation enthalpy of manganese (Z = 25) is exceptionally high.
(iii) `Cr^(2+)` is a stronger reducing agent than `Fe^(2+)`.

Explanation

(a) (i)
(ii) `2Cu^(2+)(aq)+4I^(-)(aq)toCu_(2)I_(2)(s)+I_(2)(s)`
(b) (i) Oxidising power of oxoanions is in the order :
`VO_(2)^(+)lt Cr_(2)O_(7)^(2-)lt MnO_(4)^(-)`
This is due to increasing stability of the lower species to which they are reduced.
(ii) This is due to exceptional stability of `Mn^(2+)(3d^(5))` which is due to highly stable half-filled configuration.
(iii) `Cr^(2+)` is a stronger reducing agent than `Fe^(2+)`. This is because after losing one electron `Cr^(2+)(d^(4))` changes to `Cr^(3+)(d^(3))` and these electrons are accommodated in lower `t_(2g)` orbitals which gives `Cr^(3+)` greater stability. On the other hand, `Fe^(2+)(d^(6))` on losing one electron change to `Fe^(3+)(d^(5))`. Although half-filled, the configuration is comparative less stable than `Cr^(3+)`.