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XeF(2) reacts with SbF(5) to form...

`XeF_(2)` reacts with `SbF_(5)` to form

A

`[XeF]^(+)[SbF_(6)]^(-)`

B

`[XeF_(3)]^(-)[SbF_(6)]^(-)`

C

`Xe^(-)[PtF_(6)]^(+)`

D

`XeF_(4)`.

Text Solution

Verified by Experts

The correct Answer is:
A
`XeF_(2)` is a fluoride donor whereas `SbF_(5)` is a fluoride acceptor.
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