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0.1 mole of XeF(6) is treated with 1.8 g...

0.1 mole of `XeF_(6)` is treated with 1.8 g of water.The product obtained is

A

`XeO_(2)F_(2)`

B

`XeO_(3)`

C

`Xe+XeO_(3)`

D

`XeOF_(4)`

Text Solution

Verified by Experts

The correct Answer is:
D
`XeF_(6)+H_(2)OtoXeFO_(4)+2HF`
0.1 mole 0,1 mole
`XeF_(6)+2H_(2)OtoXeO_(2)F_(2)+4HF`
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