The oxidation state of oxygen in `O_2[PtF_6]` is

Options

A. (a) zero
B. (b) `-1//2`
C. (c) `+1`
D. (d) `+1//2`

Answer

D

Explanation

`O_2[PtF_6]` is considered to be formed from `O_2^+` and `[PtF_6]^(1-)`.
In `O_2^(+1)`, oxidation state of `O` is `2x=+1`.
`x=+1/2`