Consider the following compounds : KO2, H2 O2 and H2 SO4 The oxidation state of the underlined elements in them are, respectively,
- +1, -2, and +4
- +4, -4, and +6
- +1, -1, and +6
- +2, -2, and +6
Answer
(3) +1, -1, and +6
Solution
KO2 (superoxide, O2-): K is an alkali metal, oxidation state = +1.
H2O2 (peroxide): 2(+1) + 2x = 0 x = -1
H2SO4: 2(+1) + x + 4(-2) = 0 x = +6
Answer: (3)