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Total number of unpaired electrons(s) pr...

Total number of unpaired electrons(s) present in both cationic and anionic part of compound `O_(2)[PtF_(6)]`.

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The correct Answer is:
`2`
`O_(2)^(+)[PtF_(6)]^(-)`
`O_(2)^(+)` has one unpaired `e^(-)`
`[PtF_(6)]^(-)` has one unpaired `e^(-)`, because Pt is in +5 oxidation state so total unpaired electron is 2.
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