Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified `KMnO_(4)` for complete oxidation ?

Options

A. `FeSO_(3)`
B. `FeC_(2)O_(4)`
C. `Fe(NO_(2))_(2)`
D. `FeSO_(4)`

Answer

d

Explanation

Reduction half reaction
`MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O`
no of electrons involved in the reqction =5
(a) oxidation of `FeSO_(3)` to `Fe_(2)(SO_(4))_(3)`
`Fe^(2+)rarrFe^(3+)+e^(-)`
`SO_(3)^(2-)+H_(2)OrarrSO_(4)^(2-)+2H^(+)+2e^(-)`
(b) oxidation of `FeC_(2)O_(4)`
`Fe^(2+)rarrFe^(3+)+e^(-)`
`C_(2)O_(4)^(2-)rarr2CO_(2)+2e^(-)`
(c ) oxidation of `Fe(NO_(2))_(2)`
`Fe^(2+)rarrFe^(3+)`
`2NO_(2)^(-)+2H_(2)Orarr2NO_(3)+4H^(+)+4e^(-)`
(d) oxidation of `FeSO_(4)`
`Fe^(2+)rarrFe^(3+)+e^(-)`
no of electronic involved =1 since `FeSO_(4)` gives least number of electons therefore it will required least amount of L `MnO_(4)` for its oxidation