(a) (i) Which is stronger reducing agent `Cr^(2+)` or `Fe^(2+)` and why?
(ii) Explain why `Cu^(+)` ion is not stable in aqueous solutions.
(iii) Explain why `Ce^(4+)` is a strong oxidising agent.
(b) Describe the oxidising property of `KMnO_(4)` in neutral or faintly alkaline medium for its reaction with iodide ions and thiosulphate ions.

Explanation

(a) (i) `Cr^(2+)` has outer electronic configuration `3d^(4)` and `Fe^(2+)` has outer electronic configuration `3d^(6)`. `Fe^(2+)` after losing one electron will have stable half-filled configuration `3d.^(5)`. Hence, `Fe^(2+)` has greater tendency to lose electron or to act as reducing agent.
(ii) `Cu^(+)` ion does not have sufficiently negative value of heat of hydration. Therefore, `Cu^(+)` is not stable in aqueous solution. Moreover, it undergoes disproportionation as :
`2Cu^(+)toCu^(2+)+Cu`
(iii) Although, `Ce^(4+)` is stable due to its noble gas configuration it acts as a strong oxidising agent because in doing so it reverts to the common +3 oxidation state. Potential of `Ce^(4+)//Ce^(3+)` is +1.74 V.
(b) Oxidising properties of `KMnO_(4)` in neutral or faintly alkaline medium.
Reaction with iodide ions:
`2MnO_(4)^(-)+H_(2)O+I^(-)to2MnO_(2)+2OH^(-)+IO_(3)^(-)`
Reaction with thiosulphate :
`8MnO_(4)^(-)+3S_(2)O_(3)^(2-)+H_(2)O to8MnO_(2)+6SO_(4)^(2-)+2OH^(-)`