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

1 mole of `Mn^(+7)O_4^(-)` changes of `Mn^(+2)` by gaining of 5 electrons.
i.e., 5 moles of electrons are lost by 1.67 moles of `M^(x+)`
`therefore` 1 mole of `M^(x+)` will lose electrons `=5/1.67=3.01`
Since `M^(x+)` changes to `MO_3^(-)` where oxidation state of M= +5 .