What is the oxidation state of xenon in XeO₂F₂?

Question

Quizplus · Accepted Answer

The oxidation state of oxygen is usually -2, and the oxidation states of sulfur and fluorine are usually -2 and -1, respectively. Therefore, for the compound to be neutral (overall charge of 0), we can solve for the oxidation state of xenon as: 
(+1) + (-2) + 12x(-2) + (-1) + 12x(-2) + (-1) + 12x(-2) + (-1) + (-2) + (-1) + (-2) + (-1) + (-2) + (-1) = 0
Simplifying this expression, we get: 
-2x + (-24) + (-1) + (-24) + (-1) + (-24) + (-1) + (-2) + (-1) + (-2) + (-1) + (-2) + (-1) + (-2) = 0
-2x - 74 = 0 
-2x = 74 
x = -37 
Since xenon is a noble gas and does not usually form bonds with other elements, it has an oxidation state of 0. However, in this compound, it is bonded to oxygen, sulfur, and fluorine, which are more electronegative than xenon. Therefore, xenon is losing electrons and has a positive oxidation state, which is +6.

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What Is the Oxidation State of Xenon in XeO₂F₂