How is XeF(6) prepared from the XeF(4) ?...

How is `XeF_(6)` prepared from the `XeF_(4)` ? Write the chemical equation for the reaction.

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To prepare `XeF_(6)` from `XeF_(4)`, we need to oxidize `XeF_(4)` to `XeF_(6)`. The oxidation state of xenon in `XeF_(4)` is +4, and in `XeF_(6)`, it is +6. To achieve this oxidation, we can use an oxidizing agent. ### Step-by-Step Solution: 1. **Identify the Reactants and Products**: - Reactant: `XeF_(4)` - Product: `XeF_(6)` - Oxidizing Agent: `O_2F_2` (dioxygen difluoride) ...

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XeF_(4) are reaction with H_(2) gives

`XeF_(2)`

`XeF_(6)`

XeF_(6) (s) can be prepared by:

Xe:`F_(2)` in 1:20 mole ratio and high pressure (60-70 bar) and temperature 573 K

`Xe:F_(2)` in 1:5 mole ratio and pressure (7 bar) and temperature 873 K

From product of (b), followed by an interaction with `O_(2)F_(2)` at 143 K

Taking Xe in excess at atmospheric pressure (1 bar ) and temperature (673 K)

How is `XeF_(6)` prepared from the `XeF_(4)` ? Write the chemical equation for the reaction.

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XeF_(4) are reaction with H_(2) gives

XeF_(6) (s) can be prepared by:

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CBSE COMPLEMENTARY MATERIAL-P-BLOCK ELEMENTS-LONG ANSWER TYPE QUESTIONS