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

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-4

answer is D.

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Detailed Solution

Xe=53.5% .. F= 46.5%
Relative number of atoms Xe
$= \frac{53 .5}{131 .2} = 0 .4 and F = \frac{46 .5}{19} = 2 .4$

Simple ratio Xe = 1 and F=6; Molecular formula is XeF₆O.N.of Xe is +6

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