The concept of oxidation number (O.N) is very important in under standing redox reaction s it helps to identify the oxidant / reductant in a redox reaction it also helps to (i) find oiut thepossible molecular formula of any neutral compound and (ii) to balance redox reaction For the reaction M^(x+)+MnO_(4)^(-)rarrMO_(3)^(-)+Mn^(2+)+1//2 O_(2) if one mole of MnO_(4)^(-) oxidises 1.67moles of M^(x+) to MO_(3)^(-) then the value of x in the reaction is

The concept of oxidation number (O.N) is very important in under stand

1.	The concept of oxidation number (O.N) is very important in under standing redox reaction s it helps to identify the oxidant / reductant in a redox reaction it also helps to (i) find oiut thepossible molecular formula of any neutral compound and (ii) to balance redox reaction For the reaction M^(x+)+MnO_(4)^(-)rarrMO_(3)^(-)+Mn^(2+)+1//2 O_(2) if one mole of MnO_(4)^(-) oxidises 1.67moles of M^(x+) to MO_(3)^(-) then the value of x in the reaction is
Answer» 5 3 2 1 Solution :`MnO_(4)^(-)+5e^(-)rarrMn^(2+)` since 1 MOLE of `MnO_(4)^(-)` accept 5 mole of electrons therefore 5 moles of electrons are LOST by 1.67moles of `M^(x+)` since `M^(x+)` change to `MO_(3)^(-)` by ACCEPTING 3 electrons `therefore` oxidation state of M x=+5-3=+2