The correct decreasing order of the following ‘Xe’ compounds to act as both fluorinating and oxidising agent is.
(I) XeF₆ (II) XeF₄ (III) XeF₂

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$X e F_{6} > X e F_{4} > X e F_{2}$

$X e F_{4} > X e F_{6} > X e F_{2}$

$X e F_{6} = X e F_{4} > X e F_{2}$

$X e F_{2} > X e F_{4} > X e F_{6}$

answer is B.

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Detailed Solution

The oxidation state of central atom, Xe increase from +2 to +4 to +6 in $X e F_{2}, X e F_{4} a n d X e F_{6}$ respectively.
Thus, its tendency to accept electrons also Increases and thus, its oxidising power increases.
Also XeF₆ can give more fluorine than XeF₄ and XeF₂. So the fluorinating agent order is $X e F_{6} > X e F_{4} > X e F_{2}$ .
Hence the correct decreasing order of the given ‘Xe’ compound at as both fluorinating and oxidising agent is $X e F_{6} > X e F_{4} > X e F_{2}$ .