Equivalent weight of `MnO_(4)^(ɵ)` in acidic neutral and basic media are in ratio of:

Options

A. 3: 5: 15
B. 5: 3 : 1
C. 5: 1 : 3
D. 3 : 5 : 5

Answer

D

Explanation

(d) `MnO_(4)^(-) +8H^(+) 5e^(-) rarr Mn^(2+) + 4H_2O` (Acidic medium)
`MnO_(4)^(-) + 2H_2O + 3e^(-) rarr MnO_2, + 4OH` (Basic medium)
`MnO_(4)^(-) + 2H_2O +3e^(-) rarr MnO_2 + 4OH` (Neutral medium) If M is mol. wt. of `KMnO_(4)`, then its Eq. wt. in acidic, basic and neutral media respectively are (M/5) : (M/3) : (M/3) or 3 : 5 : 5