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Question

A compound of Xe and F is found to have 53.5% Xe. What is the oxidation number of Xe in this compound?

A
4
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B
0
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C
+4
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D
+6
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Solution

The correct option is D +6
Let the total weight of the compound is 100 g.

So the amount of Xe present is 53.5 g and of F is 46.7 g.

Now we find moles of Xe is 53.5133=0.4 and of F is 46.719 is 2.45.

Divide both by 0.4 so that we get a simple whole-number ratio, we get

Moles of Xe=0.40.4=1

Moles of F=2.450.4=6

So the formula is XeF6.

The oxidation state of Xe is +6.

Hence, the correct option is D

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