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

Options

A. a)`OF_(2)ltKO_(2)ltBaO_(2)ltO_(3)`
B. 2)`BaO_(2)ltKO_(2)ltO_(3)ltOF_(2)`
C. 3)`BaO_(2)ltO_(3)ltOF_(2)ltKO_(2)`
D. 4)`KO_(2)ltOF_(2)ltO_(3)ltBaO_(2)`

Answer

B

Explanation

Let the oxidation number of oxygen in following compounds is x .
In `OF_(2)`
`x+(-1)2=0`
`x=+2`
In `KO_(2)`
`+1+(x xx2)=0`
`2x=-1`
`x=-(1)/(2)`
In `BaO_(2)`
`+2+(x xx2)=0`
`2x=-2`
`x=-1`
In `O_(3)` , oxidation number of oxygen is zero because oxidation number of an element in free state or in any of its allotropic form is always zero .
Thus , the increasig order of oxidation number is
`{:(BaO_(2)lt,KO_(2)lt,O_(3)lt,OF_(2)),(-1,-(1)/(2),0,+2):}`