In which fo the following has the oxidation number of oxygen been arragned in increasing order ?

Options

A. `KO_2 lt OF_2 lt O_3 lt BaO_2`
B. `BaO_2 lt KO_2 lt O_3 lt OF_2`
C. `BaO_2 lt O_3 lt OF_2 lt KO_2`
D. `OF_2 lt KO_2 lt BaO_2 lt O_3`

Answer

B

Explanation

Let x be the oxidation number (O.N.) of O atom in a given compound:
O.N. of O in ` BaO_2 rArr (+2) +2x rArr =0 rArr x=- 1`
O.N of O in `KO_2 rArr (+1) + 2x =0 rArr x=- 1//2`
O.N. of O in `O_3 rArr 0` (Free state or uncombined state )
O.N. of O in ` OF_2 rArr x+ 2(-1) =0 rArr x= +2`
`:. Baoverset(-1)(O_(2)) lt Koverset(-1//2)(O_(2)) lt overset(0)(O_(2)) lt overset(+2)(OF_(2))`