For the reaction
`M^(x+)+MnO_(4)^(ө)rarrMO_(3)^(ө)+Mn^(2+)+(1//2)O_(2)`
if `1 "mol of" MnO_(4)^(ө)` oxidises `1.67 "mol of" M^(x+) "to" MO_(3)^(ө)`, then the value of `x` in the reaction is

Options

A. 5
B. 3
C. 2
D. 1

Answer

c

Explanation

`MnO_(4)^(-)+5e^(-)rarrMn^(2+)`
since 1 mole of `MnO_(4)^(-)` accept 5 mole of electrons therefore 5 moles of electrons are lost by 1.67 moles of `M^(x+)`
since `M^(x+)` change to `MO_(3)^(-)` by accepting 3 electrons
`therefore` oxidation state of M
x=+5-3=+2