(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+).

(a) Complete the following chemical equations : (i) Cr_(2)O_(7)^(2-)(

1.	(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+).
Answer» SOLUTION :(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+)`.