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The average Xe-F bond energy is 34Kcal//...

The average `Xe-F` bond energy is `34Kcal//mol` , first `I.E`. Of `Xe` is `279Kcal//mol` , electron affinity of `F` is `85Kcal//mol` . Then, the enthalpy chnage for the reaction
`XeF_(4)rarrXe^(+)+F^(-)+F_(2)+F` will be

A

`367Kcal//mol`

B

`425Kcal//mol`

C

`292Kcal//mol`

D

`392Kcal//mol`

Text Solution

Verified by Experts

The correct Answer is:
A

`DeltaH=4E_(Xe-F)+DeltaH_("ionisation")[XerarrXe^(+)]`
`-DeltaE_(F-F)-DeltaH_(eg)[FrarrF^(-)]`
`=4xx34+279-85-38=136+279-123`
`=415-213=292Kcal//mol`
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