1.

The oxidation number of oxygen in KO_(3), Na_(2)O_(2) is

Answer»

`3, 2`
`1,0`
`0, 1`
`-0.33, -1`

Solution :LET OXIDATION number of oxygen `=x`
O.N. of oxygen in `KO_(3)=1+3x=0`
`:. X=-1//3=-0.33`
O.N. of oxygen in `Na_(2)O_(2)=2xx1+2x=0"":. X=-1`.


Discussion

No Comment Found

Related InterviewSolutions